Applied Mathematics
September 2012
Courtesy of Institute of Applied Mechanics, NTU
Errata for Applied Mathematics
(Last updated: June, 2011)
1. Page 10, Eq. (1.8-16), read  IJ for  ij .
2. Page 18, Problem 15, read  3  I1 ( A) 2  I 2 ( A)  I 3 ( A)  0 .
3. Page 33, above Eq. (2.3-43), read “…an 2  2 matrix,…” for “… an n  2 matrix, …”
4. Page 41, below Eq. (2.4-22), read “…then proceed to find by Eq. (2.4-19)” for “…Eq.
(2.4-20)”.
d 2u
dx .
0 dx 2
5. Page 50, Remarks, Point 1, read  
l
6. Page 53, above Eq. (2.6-30), read GM  1 l for GM  1 l .
7. Page 53, above Eq. (2.6-31), read (2.6.29) for (2.6-27).
8. Page 54, at the bottom, read 12   22  0 .
9. Page 60, at the bottom, read sin(n x l ) .
1
10. Page 63, Eq. (2.7-68), read “ ...    (   4    )dx  ... ”
0

11. Page 64, Eq. (2.7-77), read u ( x)  e2 x  ...
n 1
12. Page 87, line 3 above Eq. (3.1-12), read “A, B, C are…”
13. Page 93, Eqs. (3.2-15) and (3.2-16), read e x (v
u
v
u
v
 u ) for (v  u ) .
x
x
x
x
1
14. Page 96, above Eq. (3.3-6), read  G 0   s 1 2 exp( s1 2 x ) .
2
15. Page 115, Problem 3, the initial conditions read G0 ( x, t ) 
G0 ( x, t )
 0 , t  0 .
t
16. Page 119, Problem 21, the boundary condition should read G (0, t )  0 .
17. Page 119, Problem 22, the time-derivative term reads
condition
1  2G
, and one additional initial
c 2 t 2
G ( x, t ;  , T )
 0, 0  x  L , t  T should be added.
t
1  2u
18. Page 120, Problem 23, the time-derivative term should read 2 2
c t
Contents
1 Cartesian Tensors
1.1 Preliminary . . . . . . . . . . . . . . . . . . . . .
1.1.1 Index Notation . . . . . . . . . . . . . . .
1.1.2 Kronecker delta and permutation symbol
1.1.3 e-δ identities . . . . . . . . . . . . . . . .
1.1.4 vector notation and index notation . . . .
1.2 Transformation Rule of Vectors . . . . . . . . . .
1.3 Dyads, Dyadics, And Tensors . . . . . . . . . . .
1.3.1 Polyadic . . . . . . . . . . . . . . . . . . .
1.4 Transformation Rule of Tensors . . . . . . . . . .
1.5 Operations on Tensors . . . . . . . . . . . . . . .
1.6 Tests for Tensors . . . . . . . . . . . . . . . . . .
1.7 Symmetry of Tensors . . . . . . . . . . . . . . . .
1.8 Eigenvalue Problems . . . . . . . . . . . . . . . .
1.9 Invariants of a Tensor . . . . . . . . . . . . . . .
1.10 Isotropic Tensors . . . . . . . . . . . . . . . . . .
1.11 Problems . . . . . . . . . . . . . . . . . . . . . .
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1
. 1
. 1
. 1
. 2
. 2
. 2
. 4
. 5
. 5
. 6
. 7
. 7
. 8
. 11
. 12
. 15
2 Ordinary Differential Equation
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Existence and Uniqueness of IVP . . . . . . . . . . . . . . . . .
2.2.1 Lipschitz condition . . . . . . . . . . . . . . . . . . . . .
2.2.2 Some Theorems of IVP . . . . . . . . . . . . . . . . . .
2.2.3 Proof of the theorem . . . . . . . . . . . . . . . . . . . .
2.3 System of first order ODE’s with constant coefficients . . . . .
2.3.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.2 Eigenvalue and Eigenvector . . . . . . . . . . . . . . . .
2.3.3 Evaluation of exA . . . . . . . . . . . . . . . . . . . . .
2.4 Green’s functions — an introduction . . . . . . . . . . . . . . .
2.4.1 Variation of parameters . . . . . . . . . . . . . . . . . .
2.4.2 Accessory problem . . . . . . . . . . . . . . . . . . . . .
2.4.3 By delta function . . . . . . . . . . . . . . . . . . . . . .
2.5 Green’s function . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5.1 Adjoint Operator . . . . . . . . . . . . . . . . . . . . . .
2.5.2 Boundary Value Problems – with homogeneous B.C.’s .
2.5.3 Boundary Value Problems – with inhomogeneous B.C.’s
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22
22
22
23
23
26
28
28
30
30
39
39
40
42
44
44
45
48
2.6
2.7
2.8
Alternative Theorem and Modified Green’s Function
2.6.1 Introduction . . . . . . . . . . . . . . . . . .
2.6.2 (Fredholm) Alternative Theorem . . . . . . .
2.6.3 Modified Green’s Function . . . . . . . . . . .
Eigenfunction Expansion . . . . . . . . . . . . . . . .
2.7.1 Eigenvalue Value Problem . . . . . . . . . . .
2.7.2 BVP with Homogeneous B.C.’s . . . . . . . .
2.7.3 BVP with Inhomogeneous B.C.’s . . . . . . .
2.7.4 Non-Self-Adjoint Eigenvalue Problems . . . .
Problems . . . . . . . . . . . . . . . . . . . . . . . .
3 PDE
3.1 Classification of PDE . . . . . . . . . . . . . . . . .
3.1.1 Introduction . . . . . . . . . . . . . . . . .
3.1.2 Classification . . . . . . . . . . . . . . . . .
3.2 Preliminary . . . . . . . . . . . . . . . . . . . . . .
3.2.1 Divergence Theorem and Green’s Identities
3.2.2 Adjoint Partial Differential Operators . . .
3.2.3 Some Terminologies . . . . . . . . . . . . .
3.3 Green’s Functions and Integral Representation . .
3.3.1 Free Space Green’s Functions . . . . . . . .
3.3.2 Method of Images . . . . . . . . . . . . . .
3.3.3 Integral Representation . . . . . . . . . . .
3.4 Other Methods of Solution . . . . . . . . . . . . .
3.4.1 (Full) Eigenfunction Expansion . . . . . . .
3.4.2 Partial Eigenfunction Expansion . . . . . .
3.4.3 Fourier transform . . . . . . . . . . . . . . .
3.5 Maximum-Minimum Principle of Heat Equation . .
3.6 Uniqueness Proof . . . . . . . . . . . . . . . . . . .
3.6.1 Laplace and Poisson Equations . . . . . . .
3.6.2 Heat Equations . . . . . . . . . . . . . . . .
3.7 Problems . . . . . . . . . . . . . . . . . . . . . . .
ii
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49
49
50
51
54
54
56
59
61
67
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86
86
86
87
91
91
91
93
94
94
97
99
104
104
108
111
111
113
113
113
115
Chapter 1
Cartesian Tensors
1.1
Preliminary
Let (e1 , e2 , e3 ) be a right-handed set of three mutually perpendicular vectors of unit magnitude.
These vectors will be used as base vectors in the definition of the Cartesian components of
arbitrary vectors.
1.1.1
Index Notation
A vector f can be expressed as
f=
3
X
fi ei
(1.1-1)
i=1
where fi = f · ei is the component of f and is also interpreted as the projection of f on the base
vector ei . For simplicity, Eq. (1.1-1) is usually written as
f = fi ei
(1.1-2)
and the summation sign is omitted. Such a summation convention over repeated indices will
be adopted hereafter. Equation Eq. (1.1-2) is called the index notation for the vector f , and
the repeated index i is also called as a dummy index.
1.1.2
Kronecker delta and permutation symbol
Two useful symbols are defined by the base vectors:
δij
eijk
= ei · ej
(1.1-3)
= ei × e j · ek
(1.1-4)
where δij is called the Kronecker delta and eijk is the permutation symbol or alternating symbol.
The indices i, j and k in Eqs. (1.1-3)-(1.1-4) are free indices. The equation holds for all possible
values of the free index. From the properties of the base vectors, we have
(
1 , if i = j
δij =
(1.1-5)
0 , if i 6= j
1
and
eijk
1.1.3


 1
=
−1

 0
, if (i, j, k) = (1, 2, 3), (3, 1, 2), (2, 3, 1)
, if (i, j, k) = (2, 1, 3), (1, 3, 2), (3, 2, 1)
, otherwise
(1.1-6)
e-δ identities
A pair of eijk ’s is related to δij by
eijk erst =
δir
δjr
δkr
δis
δjs
δks
δit
δjt
δkt
eijk eimn = δjm δkn − δjn δkm
, and
(1.1-7)
Proof of Eq. (1.1-7) is left as exercises.
1.1.4
vector notation and index notation
Dot product, cross product and triple scalar product of vectors can also be expressed by index
notation as follows. Let a = ai ei , b = bj ej and c = ck ek .
Dot Product
a·b =
(ai ei ) · (bj ej ) = ai bj ei · ej = ai bj δij
= ai bi
(1.1-8)
Cross Product
a×b =
(ai ei ) × (bj ej ) = ai bj ei × ej
= eijk ai bj ek
(1.1-9)
Triple Scalar Product
(a × b) · c =
(ai ei ) × (bj ej ) · (ck ek ) = eijm ai bj em · ck ek
= eijm ai bj ck δk m = eijk ai bj ck
=
a1
b1
c1
a2
b2
c2
a3
b3
c3
(1.1-10)
The proof of (a × b) × c = (a · c)b − (b · c)a through the index notation is left as an exercise.
1.2
Transformation Rule of Vectors
Equation (1.1-2) is the index notation of f with respect to (e1 , e2 , e3 ). Suppose a new set of
base vectors (e01 , e02 , e03 ) is introduced. The same vector f can also be expressed by
f = fi0 e0i
(1.2-1)
where fi0 = f · e0i is the component of f with respect to the new bases. Since both fi and fi0 are
components of f, there certainly is some relation between fi and fi0 . The relation between the
2
new (fi0 ) and old (fi ) components is called the transformation rule. The transformation rule
from old system to new system is given by
fj0
= f · e0j = fi ei · e0j
= `ij fi
(1.2-2)
where
`ij = ei · e0j
(1.2-3)
is the direction cosine of the angle between ei and e0j . Alternatively, `ij can be considered as
the component of the vector e0j w.r.t the old base vectors, i.e.,
e0j = `ij ei
(1.2-4)
or as the component of the vector ei w.r.t the new base vectors, i.e.,
ei = `ij e0j
(1.2-5)
Similarly, the transformation rule from new system to old system can be shown to be
fi = `ij fj0
(1.2-6)
A property of `ij can be obtained by substituting Eq. (1.2-2) into Eq. (1.2-6). The result is
fi = `ij `kj fk
(1.2-7)
On the other hand, by using the property of Kronecker delta, one has
fi = δik fk
(1.2-8)
(`ij `kj − δik )fk = 0
(1.2-9)
or
Note that the index i in Eq. (1.2-2) has been changed to k to avoid ambiguity in the summation
convention. Since fk is arbitrary, Eq. (1.2-9) implies
`ij `kj = δik
(1.2-10)
Similarly by substituting Eq. (1.2-6) into Eq. (1.2-2) we can have
`ji `jk = δik
(1.2-11)
In matrix notation, Eq. (1.2-10) and Eq. (1.2-11) can be read as
[`][`]T = [`]T [`] = I
(1.2-12)
where the superscript T denotes the transpose, and I is the identity matrix. Obviously, [`]
is then an orthogonal matrix, and has determinant as ±1. In the case that det[`] = 1, the
coordinate transformation is called proper (which transforms a right-handed system into a
right-handed system, or a left-handed system into a left-handed system). If det[`] = −1, the
coordinate transformation is called improper (which transforms a right-handed system into a
left-handed system or vice versus, so handedness is not preserved).
So far, we have dealt with vectors whose specification requires three numbers in a 3-D
space. A vector is more than just a set of three numbers. The key property of a vector is the
3
transformation rule of its components, i.e., the way its components in one coordinate system
are related to those in another coordinate system. The precise form of this transformation rule
is a consequence of the physical or geometrical meaning of the vector.
Suppose we have a law involving components a, b, c, . . . of various physical quantities with
respect to some three-dimensional coordinate system K. Then it is an empirical fact that
the law has the same form when written in terms of the components a0 , b0 , c0 , . . . of the same
quantities with respect to another coordinate system K 0 which is rotated with respect to K.
In other word, properly formulated physical laws are ”invariant under rotation”. For example,
Newton’s second law
fi = mai
(1.2-13)
holds in all coordinate systems.
Vectors are actually special cases of a more general object called tensor which will be discussed in the next section.
1.3
Dyads, Dyadics, And Tensors
The mathematical object denoted by ab, where a and b are given vectors, is called a dyad. The
meaning of ab is simply that the operation ab · v is understood to produce the vector a(b · v),
where v is any vector. Note that by this definition, the dyad ba is not the same as ab, since
ba · v = b(a · v) 6= a(b · v)
(1.3-1)
A sum of dyads, of the form T = ab + cd + fg + · · · is called a dyadic, and this just means that
T · v = a(b · v) + c(d · v) + f (g · v) + · · ·
(1.3-2)
Any dyadic can be expressed in terms of an arbitrary set of base vectors ei ; since a = ai ei ,
b = bi ei , · · · it follows that
T = ai bj ei ej + ci dj ei ej + · · ·
(1.3-3)
Hence T can always be written in the form
T = Tij ei ej
(1.3-4)
In terms of nine numbers Tij = ei · T · ej = (T · ej ) · ei . A dyadic is the same as a second
order tensor and the Tij are called the components of the tensor. We can consider a dyadic as
a linear mapping that maps a vector onto another vector. The mapping described by a second
order tensor can also be expressed by matrix notation. Let
y=T·x
(1.3-5)
Substituting Eq. (1.3-4) into Eq. (1.3-5) yields
y = Tij ei (x · ej )
= Tij xj ei
(1.3-6)
The components of y are given by
yi = Tij xj
(1.3-7)
4
If we write


y1


[y] =  y2 
y3

T11 T12

[T] =  T21 T22
T31 T32


x1


[x] =  x2 
x3

T13

T23 
T33
(1.3-8)
(1.3-9)
Equation (1.3-7) can be expressed by
[y] = [T][x]
1.3.1
(1.3-10)
Polyadic
In a similar fashion, we can introduce a polyadic or a n-th order tensor by
n
z }| {
T = Tijk · · · ei ej ek · · ·
| {z }
n
The action of a n-th order tensor is to take a vector to a (n − 1)-th order tensor. For example,
a third order tensor T when dotted by a vector a gives
T · a = Tijk ei ej ek · am em
= Tijk ak ei ej
1.4
Transformation Rule of Tensors
As we have discussed in section 1.3, the transformation rule of components of a vector is derived
by the fact that a vector is an invariant under change of base vectors. In the same spirit, the
transformation rule of components of a tensor can also be derived. Consider the case of the
second order tensors.
Consider the case of the second order tensors first. Let
T = Tij ei ej
(1.4-1)
Tij0 e0i e0j
(1.4-2)
T =
be the expressions of the tensor T with respect to the base vectors ei and e0i respectively. The
transformation rule is derived as follows.
Tij0
= e0i · T · e0j = e0i · Tpq ep eq · e0j
= Tpq (e0i · ep )(eq · e0j )
= `pi `qj Tpq
(1.4-3)
where `pi is defined by Eq. (1.2-3). Similarly, Tpq is expressed in terms of Tij0 as
Tpq = `pi `qj Tij0
(1.4-4)
The above equations can also be expressed by the matrix notation as
[T0 ] = [`]T [T][`]
(1.4-5)
5
[T] = [`][T0 ][`]T
(1.4-6)
where

`11

[`] =  `21
`31
`12
`22
`32

`13

`23 
`33
(1.4-7)
The transformation rules depicted by Eq. (1.4-3) and Eq. (1.4-4) ensure that a second order
tensor is invariant under change of coordinate system which is the key property of a tensor.
Thus Eq. (1.4-3) and Eq. (1.4-4) can also be regarded as the definition of a tensor.
Similarly, the transformation rule of an n-th order tensor can be shown to be
0
Tijk...
= `pi `qj `rk · · · Tpqr...
1.5
(1.4-8)
Operations on Tensors
1. Addition
Let A and B both be nth order tensors, and let C = A + B, then
Cij...k = Aij...k + Bij...k
(1.5-1)
and Cij...k is the component of the nth order tensor C.
2. Multiplication (Tensor Product)
Let A and B be mth and nth order tensors, respectively, and let C = AB, then
Cij...kmn...p = Aij...k Bmn...p
(1.5-2)
and Cij...kmn...p is the component of the (m + n)th order tensor C. Obviously the tensor
multiplication is not commutative.
3. Contraction
The operation of summing a tensor of order n (n ≥ 2) over two of its indices is called
contraction. Contraction of a tensor of order n leads to a tensor of order n − 2. For
example the contraction of the first and the second indices of a third order tensor Aijk
gives Aiik = A11k + A22k + A33k . The result of multiplying two or more tensors and
then contracting the product with respect to indices of different factors is called an inner
product of the tensors, e.g. Aij Bjk
4. Differentiation
Let A and B be mth and nth order tensors, respectively, and let C = ∂A/∂B, then
Cij...kmn...p =
∂Aij...k
∂Bmn...p
(1.5-3)
For example, the gradient of a scalar function is a vector, and the strain tensor ij =
1
2 (ui,j + uj,i ) is a second order tensor.
6
1.6
Tests for Tensors
We have already noted that if Tij is a second order tensor and Vi is a vector, then Tij Vi is also
a vector. If Wi is another vector, the quantity Wi Tij Vj is a scalar quantity, independent of the
choice of base vectors. It Sij is another tensor, we also have the following operations:
Tij Sjk
= Pik (tensor)
(1.6-1)
Tij Sij
= c (scalar)
(1.6-2)
Some of these relations can serve as diagnostic tools for identifying the tensorial character of
two-index quantities. Thus, if it is known that, for all vectors Vi , the operation T(i,j) Vi produces
another vector Wj , it must follow that T(i,j) is a tensor. This is proved by showing that T(i,j)
transforms like a second-order tensor under changes of base vectors. Since Wj is a vector
Wj0
=
(eq · e0j )T(p,q) Vp
= `qj `ps T(p,q) Vs0
(1.6-3)
0
But Wj0 = T(s,j)
Vs0 , and so, since Vs is arbitrary
0
T(s,j)
= `qj `ps T(p,q)
(1.6-4)
Thus T(i,j) is a tensor.
Other theorems of this type-called quotient laws. They are easily stated and proved. Thus,
Tij is a tensor if:
1. Tij Xi Yj is a scalar for all vectors Xi and Yj ;
2. Tij Sij is a scalar for all tensors Sij ;
3. Tij Sjk is a tensor for all tensors Sjk ;
4. Tij Xi is a vector for all vectors Xi ;
and so on.
For example, δij xi yj is a scalar for all vectors xi and yj , hence δij is a second order tensor.
On the other hand, T(i,j) vi vj is a scalar for all vectors vi , but T(i,j) may not be a tensor.
1.7
Symmetry of Tensors
A n-th order tensor Sijk... is symmetric in its first and second indices, if and only if it retains
the same value even when those two indices switch the orders, i.e., Sijk... = Sjik... . For example,
S12k... = S21k... , S23k... = S32k... , and S31k... = S13k... .
Similarly, a n-th order tensor Ωijk... is antisymmetric (or skew symmetric) in its first and
second indices, if and only if it changes the sign when those two indices switch the orders, i.e.,
Ωijk... = −Ωjik... . For example, Ω12k... = −Ω21k... , Ω23k... = −Ω32k... , and Ω31k... = −Ω13k... .
Hence Ω11k... = Ω22k... = Ω33k... = 0.
Remarks:
7
1. A tensor which is symmetric (or skew-symmetric) in one Cartesian coordinate system
remains symmetric (or skew-symmetric) in other Cartesian coordinate system. Since if
TIJK... = TJIK... , then
0
Tijk...
= `Ii `Jj `Kk · · · TIJK...
(tensor transformation rule)
= `Ii `Jj `Kk · · · TJIK...
(TIJK... = TJIK... )
0
Tjik...
=
2. A symmetric 2nd order
in a matrix form as

S11 S12

S =  S12 S22
S13 S23
(tensor transformation rule)
(1.7-1)
tensor Sij has six independent components and can be expressed

S13

S23 
S33
(1.7-2)
While a skew-symmetric 2nd order tensor Ωij has only three independent components
and can be expressed in a matrix form as


0
Ω12 Ω13


Ω =  −Ω12
(1.7-3)
0
Ω23 
−Ω13 −Ω23
0
3. Any second order tensor Tij can always be expressed as the sum of its symmetric and
skew-symmetric parts, Sij and Ωij , respectively, where Sij = (Tij + Tji )/2 and Ωij =
(Tij − Tji )/2.
4. Since a skew-symmetric 2nd order tensor has only three independent components. It
is possible to establish a one-to-one correspondence between the components of a skewsymmetric tensor Ω and a vector ω. Let Ω12 = ω3 , Ω23 = ω1 , and Ω31 = ω2 . This
correspondence can be summarized as
Ωij = eijk ωk
(1.7-4)
where eijk is the permutation symbol defined in Eq. (1.1-4) By multiplying eijr to the
both sides of Eq. (1.7-4), one can easily establish the inverse mapping between Ω and ω
as
ωk =
1.8
1
eijk Ωij
2
(1.7-5)
Eigenvalue Problems
[See also section §2.3.2.] Consider an eigenvalue problem associated with the 2nd order symmetric tensor Sij as
Sij vj = λvi
,
Sv = λv
(1.8-1)
Trivial solution v = 0 is always a solution of Eq. (1.8-1). On the other hand, for some specific
value of λ, there is a nontrivial solution of v, which satisfies Eq. (1.8-1). In order to find such
λ and hence v, Eq. (1.8-1) is arranged as the form
(Sij − λδij )vj = 0
,
(S − λI)v = 0
8
(1.8-2)
Note that Eq. (1.8-2) is a linear, homogeneous algebraic equation of vj with coefficient (Sij −
λδij ). The condition of having nontrivial solution of vj to Eq. (1.8-2)is
det(Sij − λδij ) = 0
(1.8-3)
Equation (1.8-3) is called the characteristic equation of the tensor S, which is a 3rd degree
algebraic equation of λ (when [Sij ] is a 3 × 3 matrix). There are always three roots of λ (though
it could be repeated roots), which are called eigenvalues. The corresponding nontrivial solutions
v’s are called eigenvectors. The Ith eigenvector v̂(I) can always be normalized as a unit vector
by
(I)
v̂j
1
=q
(I)
(I) (I)
vk vk
vj
no sum on I
(1.8-4)
The directions of v (or v̂) are called the principal directions of S, while the axes determined by
v̂ the principal axes.
Remarks:
1. Eigenvalue problem can be interpreted geometrically as the following. As we mentioned
in the Section §1.3 that 2nd order tensor can be treated as a linear mapping which maps a
vector onto another vector. Then the eigenvalue problem Eq. (1.8-1) is nothing but trying
to find a vector v for which the mapping of the tensor Sij will not change this particular
vector’s direction but length. Such vector is called the eigenvector of S, and the length
ratio λ the eigenvalue.
2. If one tries to find the eigenvalue and eigenvector of S in the primed coordinate system,
instead of the unprimed coordinate system. The eigenvalue problem of S in the primed
coordinate system will then be
0
Spq
vq0 = λ0 vp0
(1.8-5)
Since S and v are a (2nd order) tensor and a vector, respectively, their components in
primed and unprimed coordinates are related by `ip . Hence
(`ip Sij `jq − λ0 δpq )vp0 = 0
(1.8-6)
We then have characteristic equation as
0 =
det(`ip Sij `jq − λ0 δpq )
=
det[`ip (Sij − λ0 δij )`jq ]
=
det(Sij − λ0 δij )
(1.8-7)
This is exactly the same characteristic equation as Eq. (1.8-3), hence λ = λ0 . This
concludes that eigenvalues are independent the choice of coordinate system. This also
suggests that the coefficients of the characteristic equation do not change under orthogonal
transformation of coordinate. [How about eigenvectors?] [See also Section 1.9.]
9
3. If v is the eigenvector associated with the eigenvalue λ of the tensor S, by multiplying S
to the both side of Eq. (1.8-1), we have
SSv = Sλv = λSv = λ2 v
(1.8-8)
By using of mathematical induction, it easily follows that
Sk v = λk v
(1.8-9)
4. Properties of λ and v̂ for real symmetric Sij :
(a) All λi are real.
Let λ and v be corresponding eigenvalue and eigenvector of S, i.e.
Sij vj = λvi
(1.8-10)
Multiply both side of Eq. (1.8-10) by
vi , then we have
λvi vi∗
vi∗ ,
which is the complex conjugate of
= Sij vj vi∗
= Sji vj vi∗
(Since Sij = Sji )
∗
= vj (Sji vi )
(Since Sij is real)
∗
= vj (λvj )
(Since vi is an eigenvector)
= vj λ∗ vj∗
λ∗ )vi vi∗ = 0.
(1.8-11)
Hence (λ −
Now, vi is an eigenvector, it is then a nontrivial
vector. We then conclude that λ = λ∗ , i.e. λ is real.
(b) Eigenvectors associated with different eigenvalues are orthogonal to each other.
(I)
(II)
Let vj and vj are eigenvectors of Sij associated with different eigenvalues
λ(I) and λ(II) , respectively, i.e.
(I)
Sij vj
(I)
= λ(I) vi
no sum on I
(II)
(II)
Sij vj = λ(II) vi
(II)
(I)
Multiplying vi
and vi to Eq.
(1.8-12)
no sum on II
(1.8-13)
(1.8-12) and Eq. (1.8-13), respectively, and
then subtracting, we have
(I) (II)
(λ(I) − λ(II) )vi vi
Since λ
(I)
6= λ
(II)
=0
no sum on I and II
(I)
, it then concludes that v
(1.8-14)
(II)
is orthogonal to v
.
(c) There are always an orthonormal full set of v̂, i.e.,
v(I) · v(J) = 0
(I) (J)
v̂i v̂i
(forI 6= J), or
= δij
(1.8-15)
(1.8-16)
In other words, Sij always has three perpendicular principal axes.
(d) In the principal coordinate system, S is in a diagonal form.
Let Sij be the components of S with respect to the base vectors ei , and
0
SKM
the components with respect the eigenvectors v̂(K) . Then the direction
cosines between two coordinate systems are
(K)
`iK = ei · v̂(K) = v̂i
(1.8-17)
10
0
The components SKM
is related to Sij as
0
SKM
= `iK `jM Sij
(K)
Sij v̂j
(K)
λM v̂i
= v̂i
= v̂i
(M )
(M )
= λM δKM

λ1 0

0
[SKM
] =  0 λ2
0
0
1.9
(1.8-18)

0

0 
λ3
(1.8-19)
Invariants of a Tensor
The value of an expression of tensor components which does not change under coordinate
transformation is called an invariant of the tensor. On expanding the characteristic equation
(1.8-3) of S, we have a 3rd degree algebraic equation in λ as
0
=
S11 − λ
S21
S31
S12
S22 − λ
S32
S13
S23
S33 − λ
= −λ3 + I1 λ2 − I2 λ + I3
(1.9-1)
where
I1
= S11 + S22 + S33
I2
=
S22
S32
S23
S33
+
I3
=
S11
S21
S31
S12
S22
S32
S13
S23
S33
(1.9-2)
S33
S13
S31
S11
+
S11
S21
S12
S22
(1.9-3)
(1.9-4)
On the other hand, if λ1 , λ2 and λ3 are the roots of Eq. (1.9-1), which can then be written
as
(λ − λ1 )(λ − λ2 )(λ − λ3 ) = 0
(1.9-5)
Making use of the relations between the roots and the coefficients of 3rd degree algebraic
equation, we can expressed I1 , I2 and I3 in terms of eigenvalues λi as
I1
= λ1 + λ 2 + λ3
(1.9-6)
I2
= λ1 λ2 + λ 2 λ 3 + λ3 λ 1
(1.9-7)
I3
= λ1 λ2 λ3
(1.9-8)
Equation (1.9-1) and the coefficients I1 , I2 , I3 are invariant with respect to the orthogonal
transformation of coordinates (see also Section §1.8). Hence I1 , I2 , I3 are called the invariants
of the tensor S with respect to the orthogonal transformation of coordinates.
11
1.10
Isotropic Tensors
A tensor is called isotropic if its components in any Cartesian coordinate system do not change
in value by orthogonal transformations of coordinates, e.g.
0
Tij···k
= Tij···k
(1.10-1)
0
0
where Tij···k
and Tij···k
are components of tensor T in the primed and unprimed coordinates,
respectively. Notes that a material is said to be isotropic if its constitutive equation is unaltered
under orthogonal transformations of coordinates. For example, the constitutive equation of a
linearly elastic material is described by the generalized Hooke’s law as
σIJ = CIJKL KL
(1.10-2)
where σIJ and KL are components of stress and strain tensors, respectively, while CIJKL is the
elastic modulus. If the material is isotropic, we demand that after an orthogonal transformations
of the coordinates the Hooke’s law read as
0
0
σij
= Cijkl
0kl = Cijkl 0kl
(1.10-3)
Obviously, if CIJKL is an isotropic tensor of order four, the material will be isotropic.
Let us now list some results regarding to isotropic tensors.
1. Any scalar is a zeorth order isotropic tensor.
2. None of first order tensors (i.e. vectors), except the null tensor, is isotropic.
Suppose that u is an isotropic tensor of order one. Now rotate a 1800 about
the x3 axis, i.e.

0

 x1 = −x1
(1.10-4)
x02 = −x2

 x0 = x
3
3
x2
x'1
x1
x' 2
Figure 1.1: Unprimed and primed coordinate systems (rotate 1800 about the x3 axis).
We have


−1 0
0


[`] =  0
−1 0 
0
0
1
(1.10-5)
12
According to the transformation law of first order tensors, the components of
u in the new and old coordinate systems are related to each other as

 

 

u01
−1
0 0
u1
−u1
 0  

 

(1.10-6)
 u2  =  0 −1 0   u2  =  −u2 
u03
0
0 1
u3
u3
But since u is supposed to be isotropic, by the definition of isotropic tensor it
requires u0i and ui are the same,
u0i = ui
(1.10-7)
By using Eq. (1.10-6) and Eq. (1.10-7), it concludes that u1 = u2 = 0. Similarly,
rotate a 1800 about the x2 axis, we conclude further that u3 = 0. Hence u = 0.
3. Obviously, δij is an isotropic tensor of order two, since
0
δij
= `Ii `Jj δIJ = `Ii `Ij = δij
(1.10-8)
Moreover, every isotropic tensor of order two is of the form λδij , where λ is a scalar.
Suppose Aij is a second order isotropic tensor.
(a) Let us rotate 1800 first about the x3 axis, then
0 Aij
= [`Ii ] [`Jj ] [AIJ ]




A11 A12 A13
−1
0 0
−1
0 0




=  0 −1 0   A21 A22 A23   0 −1 0 
0
0 1
A31 A32 A33
0
0 1


A11
A12 −A13


=  A21
(1.10-9)
A22 −A23 
−A31 −A32
A33
On the other hand, since A is supposed to be isotropic, by the definition
of isotropic tensor it requires A0ij and Aij are the same, i.e.


A11 A12 A13


[A0ij ] =  A21 A22 A23 
(1.10-10)
A31 A32 A33
It then concludes that A13 = A23 = A31 = A32 = 0.
(b) Similarly, rotate a 1800 about the x2 axis, we conclude further that A12 =
A21 = 0 (why?). Hence A = diag (A11 , A22 , A33 ).
(c) Rotate 900 about the

0

 x1 = x2
x02 = −x1

 x0 = x
3
3
We have 
0 −1

[`] =  1 0
0 0
x3 axis
(1.10-11)

0

0 
1
(1.10-12)
13
x'1 x 2
x1
x' 2
Figure 1.2: Unprimed and primed coordinate systems (rotate 900 about the x3 axis)

0
0 
Aij
=  −1
0

A22

= 
0
0



1 0
A11
0
0
0 −1 0



0 0 
0 A22
0  1
0 0 
0 1
0
0 A33
0
0 1

0
0

(1.10-13)
A11
0 
0 A33
Hence, A11 = A22 , by the argument of being an isotropic tensor.
(d) Similarly, rotate 900 about x2 axis, we get A11 = A33 (why?).
Conclude from the above results, Aij = λδij .
4. With respect to all rotations of coordinates (proper orthogonal transformation), the only
isotropic tensors of order three are scalar multiplies of the permutation symbol eijk .
On the other hand, it is not isotropic with respect to reflection (improper orthogonal
transformation) in a coordinate plane.
5. Since δij is an isotropic tensor of order two, the tensors δij δkl , δik δjl and δil δjk are isotropic
tensors of order four. In fact, it can be shown that if Cijkl is an isotropic tensor of order
four, then it is of the form
Cijkl = λδij δkl + µδik δjl + νδil δjk
(1.10-14)
References for Cartesian Tensors
1. Jefferys, H. (1965). Cartesian Tensors, Cambridge University Press.
2. Segel, L. A. and G. H. Handelman (1977). Mathematics Applied to Continuum Mechanics,
Macmillan, Part A, 1-73.
3. Fung, Y. C. (1969). A First Course in Continuum Mechanics, Prentice-Hall, 19-40.
4. Reddy, J. N. and M. L. Rasmussen (1982). Advanced Engineering Analysis, Wiely, 1-152.
5. Sokolnikoff, I. S. (1964). Tensor Analysis, Wiley.
14
1.11
Problems
1. Show the following identities
eiks emks = 2δim
eijk eijk = 6
eijk eikj = −6
2. Use the index notation to verify the following vector identities:
(a) a × (b × c) = (a · c)b − (a · b)c
(b) (a × b) × c = (a · c)b − (b · c)a
(c) (a × b) · (c × d) = (a · c)(b · d) − (a · d)(b · c)
(d) (a × b) × (c × d) = [a · (c × d)]b − [b · ( c × d)]a
(e) [a · (b × c)]d = [d · (b × c)]a + [a · (d × c)]b + [a · (b × d)]c
(f) (a × b) · (b × c) × (c × a) = (a × b · c)2
3. The del operator can be written as ∇ = ∂i ei or ∇( ) = ∂i ( )ei = ( ),i ei in Cartesian
coordinate. Use the index notation to verify the following identities:
(a) ∇ × ∇φ = 0
(b) ∇ · (∇ × u) = 0
(c) ∇ × (∇ × u) = ∇(∇ · u) − ∇2 u.
where ∇2 is the Laplacian operator, which can be written as ∇2 ( ) = ∂i ∂i ( ) = ( ),ii .
4. The determinant of a 3 × 3 matrix A may be expressed by using of the permutation
symbol as

a11

det A = det  a21
a31
a12
a22
a32

a13

a23  = eijk ai1 aj2 ak3 = eijk a1i a2j a3k ,
a33
where the summation convention is implied.
(a) Deduce that
eijk aip ajq akr = epqr det A
(b) Show that
1
eijk epqr aip ajq akr
6
(c) Let A = [aij ] and B = [bij ] be 3 × 3 matrices. Using (a) and (b) shows that
det A =
det(A B) = det A det B
15
(d) Let a, b, c, u, v, and w be 3 × 1 vectors in a real vector space V . Show that
a × b · c = eijk ai bj ck
and use the results of (a)–(c) to show

a·u

(a × b · c) (u × v · w) = det  a · v
a·w
b·u
b·v
b·w

c·u

c · v .
c·w
(e) Using these results to obtain the following succinct identities relating the permutation
symbol and Kronecker delta:


δip δiq δir


eijk epqr = det  δjp δjq δjr 
δkp δkq δkr
eijk epqk
= δip δjq − δiq δjp
eijk epjk
=
2δip
5. Let
..
 1 .

.
(e1 , e2 , e3 ) =  0 ..

.
0 ..


.
0 .. 0 

.
1 .. 0 

..
0 . 1



(e01 , e02 , e03 ) = 

3
5
4
5
0
..
. − 45
..
3
.
5
..
. 0

..
. 0 

..
. 0 

..
. 1
be the standard basis and another basis, respectively. If the components of a vector a
with respect ei are given as
(ai ) = (2, 1, 5)T
Determine the components of a with respect to e0i
6. Let the direction cosines of two Cartesian base vectors ei and e0 i be


√1
√1
0
2
2


[`ij ] =  − √12 √12 0 
0
If the components

0

[Tij ] =  1
0
0
1
of a second order tensor T with respect ei are given as

2 0

0 0 
0 5
Determine the components of T with respect to e0i
7. Let T and S be two tensors of order two. Show that if the tensorial equation
T=S
holds in one Cartesian coordinate system, then it also holds in any other Cartesian coordinate system.
16
8. If T is an entity with components T (i, j). Show that T is a Cartesian tensor of order two,
if
(a) for any arbitrary S, which is a Cartesian tensor of order two, that Sij T (j, k) leads
to some Cartesian tensors of order two, say Wik .
(b) for any arbitrary S, which is a Cartesian tensor of order three, that Sijk T (j, k) leads
to some vector, say ui .
(c) for any Cartesian tensors of order four C, Cijk` T (i, j) is a Cartesian tensor of order
two.
(d) for any arbitrary S, which is a Cartesian tensor of order three, that T (i, j)Sjkm leads
to also a Cartesian tensor of order three.
9. A and B are any 2nd order tensors,
(a) Show that if Aij = Aji and Bij = −Bji , then Aij Bij = 0.
(b) Show that if Aij = Aji and Aij Bij = 0, then Bij = −Bji .
10. Let A, B, and T be Cartesian tensors of order two, and C be Cartesian tensor of order
four. For any arbitrary Aij = Aji and Bij = −Bji ,
(a) if Tij Bij = 0, show that Tij = Tji .
(b) if Tij Aij = 0, show that Tij = −Tji .
(c) if Cijkl Akl = Aij and Cijkl Bkl = 0 show that Cijkl = (δik δjl + δil δjk )/2,
(d) if Cijkl Akl = 0 and Cijkl Bkl = Bij show that Cijkl = (δik δjl − δil δjk )/2,
11. If A±
ijkl = (δik δjl ± δil δjk )/2, show that
(a) If T is symmetric Cartesian tensor of order two, then
A+
ijkl Tkl = Tij ,
A−
ijkl Tkl = 0
(b) If T is skew-symmetric Cartesian tensor of order two, then
A+
ijkl Tkl = 0,
A−
ijkl Tkl = Tij
12. Moment of Inertia. Suppose a rigid body is rotating so that every particle in the body
is instantaneously moving in a circle about some axis fixed in space. The body’s angular
velocity ω is defined as the vector whose magnitude is the angular speed ω and whose
direction is along the axis of rotation. Then a particle’s linear velocity is
v = ω × r,
where r is the position vector from a fixed point O on the axis to the particle. The
particle’s angular momentum h about the point O is
h = mr × v,
where m is the mass of the particle. Show that there exists a tensor I such that
h = I ω.
17
(Note that I is called the moment of inertia tensor of the particle about the point O
and maps any vector ω into the angular momentum h the particle would have if
the rigid body were rotating about an axis through O with angular velocity ω. Note I
is clearly a linear transformation that maps any vector into another vector. So it is a
tensor.)
13. In the real vector space R2 of two-rowed columns of real numbers, consider the natural
basis
!
!
1
0
e1 =
,
e2 =
0
1
and let A and B be the linear transformations whose matrices of components in this basis
are given by
!
√ !
1
2
cos φ sin φ
e
e
√
,
B =
A =
2 1
− sin φ cos φ
where φ is a real number.
(a) Find the eigenvalues λ1 and λ2 of A.
(b) Find the inverse of B in terms of its matrix Be of components.
(c) If C = B−1 AB, show that φ can be chosen so that
!
λ1 0
e
C =
0 λ2
14. Show that the tensor product a⊗b is a symmetric tensor if and only if one of the following
three conditions holds: (a) a = 0, (b) b = 0, (c) a = kb, where k is a scalar.
15. Let V be a vector space with dim V = 3 and A be a linear transformation on V . Show
that the characteristic polynomial PA (λ) can be expressed as
λ3 − I1 (A)λ2 + I2 (A) − I3 (A) = 0,
where
I1 (A)
= T r(A),
1
I2 (A) =
(T r(A))2 − T r(A2 ) ,
2
I3 (A) = det A.
These are called the principal invariants of A. Note that T r(A), T r(A2 ) and det A are
scalar invariants of A.
16. Let V be a finite dimensional real vector space with dimension equal to 3. Let W ∈ V
be an anti-symmetric second order tensor, so that WT = −W.
(a) Show that the principal invariants of W are
I1 (W) = 0,
2
2
2
I2 (W) = W23
+ W31
+ W12
,
I3 (W) = 0,
where Wij are the components of W relative to an arbitrary choice of orthonormal
basis ei .
18
(b) Show that W possesses only one real eigenvalue, namely λ = 0. Let the corresponding eigenvector be w0 so that
W w0 = 0
(c) Define the axial vector of W by
1
eijk Wkj ei ,
2
is the permutation symbol. Show that
w = wi ei =
where eijk
eipq wi = Wqp
and that,
Wa = w × a
for any vector a ∈ V .
(d) Show that
w0 = c w
for some constant c. It follows that the axial vector is the only (real) eigenvector of
W.
17. Show that the quadratic form f (x) = (Ax) · x associated with a symmetric tensor A is
positive for all non-null vectors x if and only if A is positive, i.e., all eigenvalues of A are
positive.
18. Let A be a symmetric, positive definite tensor on a two-dimensional real Euclidean space.
For any x such that |x| = 1, let y = Ax. Let e1 and e2 be eigenvectors corresponding to
the positive eigenvalues λ1 , λ2 of A. Resolve x and y into components in the orthonormal
basis e : x = ξj ej , y = ηj ej . Show that
η1
λ1
2
+
η2
λ2
2
= 1,
asserting that image points y = Ax of points x on the unit circle lie on an ellipse whose
principal axes are directed along the eigenvectors e1 , e2 , and whose corresponding semiaxes have respective lengths λ1 , λ2 .
19. Show that if a Cartesian tensor Q satisfies
QQT = QT Q = I,
it is orthogonal.
20. Let Q be an orthogonal tensor with det Q = 1.
(a) Show that there exists a vector u such that
Qu = u.
(b) Let u, v, w form an orthonormal basis. Then there exists an angle θ such that
Q = u ⊗ u + (v ⊗ v + w ⊗ w) cos θ + (w ⊗ v − v ⊗ w) sin θ.
19
(c) Further, show that for any arbitrary vector a
Qa = cos θ a + (1 − cos θ) < a, u > u + sin θ u ∧ a.
21. Let A be an arbitrary tensor, not necessarily symmetric, in a three-dimensional real
Euclidean space. Suppose that A is isotropic, so that QT AQ = A for every proper
orthogonal tensor Q.
22. Let V be a symmetric tensor; if R is any tensor, show that
U = RT VR
is also symmetric. Show that the right stretch tensor U and the left stretch tensor V in
the polar decomposition have the same eigenvalues.
23. Let A be a tensor on a real vector space V with dimension equal to 2, and suppose that
its matrix of components A in some orthonormal basis is given by
A=
1
−2
2
2
!
.
Find both polar decompositions of A.
24. Let L be the set of tensors on an n-dimensional real Euclidean space V . From now on
tensors in L are called 2-tensors. We now consider linear transformations of L into itself.
If C is such transformation, we call it a 4-tensor. As a linear transformation, a 4-tensor
accepts as input any tensor A and delivers another tensor B as output:
B = C A.
The scalar product of two 2-tensors in L is defined by
< A, B > = T r(ABT ),
∀ A, B ∈ L.
Since L is a finite-dimensional vector space, every 4-tensor C has a transpose CT characterized by
< CA, B > = < A, CT B >
for every pair of 2-tensors A and B.
Consider some special 4-tensors. There is of course an identity 4-tensor I for which
I A = A for every 2-tensor A in L. In addition, one has the transposition 4-tensor T
defined by T A = AT . What are the components of I and T in any orthornomal basis in
V?
25. Let C be a 4-tensor and cijkl be its components in some orthonormal basis in V . Suppose
C has the first minor symmetry; i.e., T C = C, where T is the transposition 4-tensor.
Show that
Cijkl = Cjikl .
20
On the other hand, C is said to have the second minor symmetry if C T = C. Show that
in this case
Cijkl = Cijlk .
Finally, C is said to have the major symmetry if C = CT . Show that in this case,
Cijkl = Cklij .
26. Let M and N be fixed symmetric 2-tensors, and define a 4-tensor C by
CA = MAN
for every 2-tensor A. Does C possess the major symmetry? Either minor symmetry?
27. What are the components of the 4-tensor product A ⊗ B of the 2-tensors A and B? In
particular, what are the components of I ⊗ I, where I is the identity 2-tensor?
28. A 4-tensor Q is said to be orthogonal if it preserves length in L:
|QA| = |A|
for every 2-tensor A in L. Let P and R be fixed orthogonal 2-tensor, and define a 4-tensor
Q by
QA = PAR
for every 2-tensor A in L. Show that Q is an orthogonal 4-tensor, thus proving that not
all orthogonal 4-tensors are 4-rotations.
21
Chapter 2
Ordinary Differential Equation
2.1
Introduction
There are some terminologies related to differential equations.
1. algebraic equation, transcendental equation, differential equation, integral equation
2. ordinary differential equation (ODE), partial differential equation (PDE)
3. linear ODE, nonlinear ODE
4. homogeneous, inhomogeneous
5. general solution, complete solution, complementary solution (homogeneous solution), particular solution
6. initial condition (I.C.), boundary condition (B.C.), (number of conditions, order of differentiation in conditions)
7. initial-valued problem (IVP), boundary-valued problem (BVP)
8. (a) explicit solution, implicit solution
(b) exact solution, approximate solution
(c) analytic solution (closed form, series form, integral form), numerical solution
(d) classical solution, weak solution, distributional solution
2.2
Existence and Uniqueness of IVP
Consider an IVP of function y(x) as
(
y 0 = f (x, y)
DE
y(a) = b
I.C.
(2.2-1)
Geometrically, we are going to find the solution curve or curves that pass through the point
(a, b) in the (x, y) plane with slope y 0 specified.
22
2.2.1
Lipschitz condition
A function F (y) satisfies a Lipschitz condition in a given closed interval I, if there is a constant
C such that
|F (y2 ) − F (y1 )| ≤ C · |y2 − y1 |
(2.2-2)
for every pair of values y2 , y1 in I.
Remarks:
1. In other words, function F (y) has a bounded difference quotient |F (y2 ) − F (y1 )| /|y2 − y1 |
over the interval I; or F 0 (y) is continuous except at some finite number of points (and
bounded everywhere) over I.
2. The condition that F 0 (y) exists and is continuous everywhere over I is stronger than that
F (y) satisfies Lipschitz condition over I. Since then using mean-value theorem
|F (y2 ) − F (y1 )| = |y2 − y1 |
∂F
∂y
≤ C · |y2 − y1 |
(2.2-3)
y=η
where η ∈ I, and C = max (∂F/∂y).
3. Lipschitz condition is stronger than continuous. Let us consider F (y) = 3y 2/3 as an
example.
(a) Obviously, F (y) is continuous everywhere, including the point y = 0.
(b) F (y) satisfies Lipschitz condition in any half plane y ≥ with C = 2−1/3 , where
> 0.
(c) But F (y) does not satisfy a Lipschitz condition for any region containing y = 0.
4. Function F (y) can satisfy a Lipschitz condition, even when F 0 (y) doesn’t exit. e.g. F (y) =
|y|, in −1 < y < 1.
2.2.2
Some Theorems of IVP
Existence & Uniqueness Theorem
Supposed that function f (x, y) is continuous in some closed neighborhood, D, of (a, b) in the
(x, y) plane. Moreover, f satisfies a Lipschitz condition in the y argument. Then there exists a
unique solution y(x) to the IVP (2.2-1) locally in a region |x − a| < δ, where δ > 0.
Well-posedness Theorem
Let y1 (x) and y2 (x) be two solutions of y 0 = f (x, y) in a domain D, where f (x, y) satisfies a
Lipschitz condition in the y argument, then
|y2 (x) − y1 (x)| ≤ ec|x−a| |y2 (a) − y1 (a)|
(2.2-4)
Remarks:
23
1. The condition that f (x, y) is continuous implies further that f is bounded in D. In
particular, we let |f (x, y)| ≤ M over D. On the other hand, f satisfies a Lipschitz
condition in the y argument implies that
|f (x, y2 ) − f (x, y1 )| ≤ C · |y2 − y1 |
where C is independent of x.
2. The conditions in the existence & uniqueness theorem are sufficient to guarantee the
existence and uniqueness of solution but not as necessary conditions.
3. If f (x.y) = g1 (x) + g2 (x)y and g2 (x) is continuous, i.e., IVP (2.2-1) with linear first order
ODE. Of course, then f satisfies a Lipschitz condition in y, since ∂f (x, y)/∂y exists and
is continuous. Hence the IVP has a unique solution according to this theorem.
4. As an example, let us consider
(
y 0 − ky 1/2 = 0
y(0) = 0
x > 0, k > 0
(2.2-5)
The function f (x, y) = ky 1/2 does not satisfy a Lipschitz condition in y ≥ 0. Hence at
x = 0, y = 0 the theorem does not apply, the IVP may or may not have only one solution.
(a) In fact, by rewriting the ODE in Eq. (2.2-5) into separating function of x and y as
y −1/2 dy = k dx
(2.2-6)
The solution of the ODE can then be found, by a direct integration to the both sides
of Eq. (2.2-6), as
k 2 (x + c)2
(2.2-7)
4
where c is an arbitrary constant. From the initial condition it concludes that c = 0,
i.e.
y(x) =
k 2 x2
(2.2-8)
4
Note that Eq. (2.2-7) is a general solution of the ODE in (2.2-5). On the other hand,
by a direct observation, there exists y(x) ≡ 0 as another solution to the IVP (2.2-5).
Hence, the solution of this IVP is definitely not unique.
y(x) =
(b) It is worthwhile to point out that the solution y(x) ≡ 0 is not possible to be deduced
directly from general solution (2.2-7) by setting the not-yet-determined constant c
to be anything. On the other hand, y(x) ≡ 0 is the envelope of the one-parameter
family of curves generating from the general solution (2.2-7), and it is called the
singular solution of the ODE. In fact, any function of the form
(
0
x≤A
y(x) =
(2.2-9)
k 2 (x − A)2 /4
x>A
is a solution of the IVP (2.2-5).
24
(c) Consider a falling body, which is set free at time t = 0 from a building. Let y denote
the displacement of the falling body, then y is the solution of the following IVP


t > 0, g > 0
 ÿ = g
(2.2-10)
y(0) = 0

 ẏ(0) = 0
where g is the acceleration of gravity. By direct integration, the solution of Eq.
(2.2-10) is
y(t) = gt2 /2
(2.2-11)
Alternatively, multiply ẏ to the both sides of the ODE in (2.2-10), manipulate into
a separated form, and then integrate, one will have
(
ẏ = (2g)1/2 y 1/2
t > 0, g > 0
(2.2-12)
y(0) = 0
Note that the IVP (2.2-12) is exactly the same as (2.2-5) except that t and (2g)1/2
are replaced by x and k, respectively. Hence the above two remarks on IVP (2.2-5)
also apply here.
Obviously, Eq. (2.2-11) is a solution of both IVP (2.2-10) and (2.2-12). On the
other hand, y(t) ≡ 0 and Eq. (2.2-9) are solutions only of IVP (2.2-12) but not
(2.2-10). Physically, ODE in (2.2-12) is the equation of conservation of mechanical
energy (kinetic and potential energy). The function y(t) ≡ 0 definitely conserves
mechanical energy and hence is also a solution of (2.2-12). [What is the physical
meaning of the solution as the form of Eq. (2.2-9)?]
5. Consider an IVP with system of first order ODE’s as
(
y0 (x) = f (x, y)
y(a) = b
DE
I.C.
(2.2-13)
where y, f and b are n × 1 vectors. The existence and uniqueness theorem of IVP with
a single first order ODE can directly apply to this case, except that the definition of
|y2 − y1 | is now defined as
|y2 − y1 | ≡
n
X
|y2i − y1i |
|f2 − f1 | ≡
i=1
n
X
|f2i − f1i |
i=1
where y2i , y1i , f2i and f1i are ith components of y2 , y1 , f2 , and f1 , respectively.
6. Similarly, consider an IVP with a single higher oder ODE as
 (n)
y (x) = f (x, y, y 0 , y 00 , · · · , y (n−1) )






 y(a) = b1
y 0 (a) = b2

.
..


 ..
.


 (n−1)
y
(a) = bn
25
DE
I.C.1
I.C.2
I.C.n
(2.2-14)
The IVP (2.2-14) is equivalent to an IVP with a system of first order ODE’s as by letting
y ≡ y1 , y 0 ≡ y2 , y 00 ≡ y3 , · · ·, y (n−1) ≡ yn . The single higher order ODE can then be
replaced by a system of 1st order ODE as
 0
y 1 = y2




0


 y2 = y3
..
..
.
.



0

yn−1 = yn


 0
yn = f (x, y1 , y2 , · · · , yn )
2.2.3
(2.2-15)
Proof of the theorem
First of all, we shall prove the existence theorem by constructing approximate solutions, then
finding a sequence of approximate solutions which converges to an exact solution. We shall
then prove the uniqueness theorem. Finally, the well-posedness theorem shall be proved.
1. Convert the IVP to an integral equation by integrating the ODE from a to x, we have
Z
y(x) = b +
x
f [ξ, y(ξ)]dξ
(2.2-16)
a
Construct a sequence of a successive approximation to Eq. (2.2-16) as
y0 (x)
= b
x
Z
y1 (x)
= b+
f [ξ, y0 (ξ)]dξ
Zax
y2 (x)
= b+
f [ξ, y1 (ξ)]dξ
a
..
.
..
.
x
Z
yk (x)
= b+
f [ξ, yk−1 (ξ)]dξ
a
2. Now, we are going to show that yk (x) → y(x) as k → ∞. Observe that the series
S(x) = y0 (x) + [y1 (x) − y0 (x)] + [y2 (x) − y1 (x)] + · · ·
(2.2-17)
has its first (k + 1) partial sums Sk+1 (x) = yk (x), for k = 0, 1, 2, · · ·, so the convergence
of the sequence of yk (x) is equivalent to the convergence of the series (2.2-17).
Consider each bracketed term in the series, we have
Z
x
|y1 (x) − y0 (x)| =
f [ξ, y0 (ξ)]dξ
Z
a
x
≤
|f [ξ, y0 (ξ)]| dξ ≤ M |x − a|
a
=
M C|x − a|
C
1!
Z
x
|y2 (x) − y1 (x)| =
{f [ξ, y1 (ξ)] − f [ξ, y0 (ξ)]} dξ
a
26
x
Z
≤
|f [ξ, y1 (ξ)] − f [ξ, y0 (ξ)]| dξ
a
Z
x
≤ C
|y1 (ξ) − y0 (ξ)| dξ
(Lipschitz condition)
a
x
Z
≤ CM
|ξ − a| dξ
(from previous result)
a
M [C|x − a|]2
CZ
2!
=
x
|y3 (x) − y2 (x)|
≤ C
|y2 (ξ) − y1 (ξ)| dξ
a
2
x
|ξ − a|
dξ
2!
a
M [C|x − a|]3
C
3!
≤ C 2M
=
Z
(from previous result)
By mathematical induction, we conclude that
|yk (x) − yk−1 (x)| ≤
M [C|x − a|]k
C
k!
(2.2-18)
Hence, the series S(x) of Eq. (2.2-17) is bounded as
M
S(x) ≤ b +
C
C|x − a| [C|x − a|]2
[C|x − a|]k
+
+ ··· +
+ ···
1!
2!
k!
(2.2-19)
By Weierstrass M-test1 , series S(x) is an absolutely and uniformly convergent series2 .
Moreover, yk (x) is uniformly convergent, therefore f [x, yk (x)] converges to f [x, y(x)] uniformly. We then have, as k → ∞,
Z
y(x)
x
= b + lim
f [ξ, yk−1 (ξ)]dξ
k→∞ a
Z x
= b+
lim f [ξ, yk−1 (ξ)]dξ
a k→∞
Z x
= b+
f [ξ, y(ξ)]dξ
(from result of uniform convergence)
a
1 Weierstrass
M-test: If there exists a series Q
Q = Q1 + Q2 + · · · + Qk + · · ·
where every Qi is a positive constant. If the series Q is convergent and |fi (x)| ≤ Qi for any x in the interval of
interest, then the series function S(x)
S(x) = f1 (x) + f2 (x) + · · · + fk (x) + · · ·
converges uniformly in the interval of interest.
2 uniform convergence: Consider a series function
Sk (x) = f1 (x) + f2 (x) + · · · + fk (x)
The series Sk (x) converges to S(x) uniformly, if and only if for any constant > 0, there exists a positive integer
N (independent on x), such that |S(x) − Sk (x)| < for any k > N . [how about pointwise convergence?]
27
3. (Uniqueness Proof)
Let Y (x) and y(x) be solutions to the IVP (2.2-1), then
Rx
|Y (x) − y(x)| = a {f [ξ, Y (ξ)] − f [ξ, y(ξ)]} dξ
Rx
≤ a C |Y (ξ) − y(ξ)| dξ
(Lipschitz condition)
Let r(x) ≡ |Y (x) − y(x)|, from (2.2-20) we have
Z x
r(x) ≤ C
r(ξ)dξ
(2.2-20)
(2.2-21)
a
Equation (2.2-21) is an integral inequality. To show that r(x) is in fact bounded by 0,
and hence r(x) = 0, we convert (2.2-21) to a differential inequality by letting
Z x
r(ξ)dξ ≡ u(x)
(2.2-22)
a
Since r(x) is continuous (since Y and y are continuous, why?), u(x) is then differentiable.
Equation (2.2-21) is then equivalent to a differential inequality
du
≤ Cu(x)
dx
(2.2-23)
Multiply both side of (2.2-23) by an exponential function, it can then be arranged as
i
d h
u(x)e−C(x−a) ≤ 0
dx
(2.2-24)
Integrate (2.2-24) from x = a to x = x, and make use of u(a) = 0 (why?), we conclude
that u(x) ≤ 0. Hence Y (x) = y(x), i.e., the solution is unique.
2.3
System of first order ODE’s with constant coefficients
Let us consider an IVP with system of first order ODE’s with constant coefficients written in
vector form
(
y0 = Ay
DE
(2.3-1)
y(0) = b
I.C.
where y = (y1 y2
matrix, b = (b1 b2
2.3.1
· · · yn )T is a unknown vector function, A = (aij ) is a n × n constant
· · · bn )T is a given constant vector, and superscript T denotes transpose.
Solution
Obviously, the function f (x, y) ≡ Ay satisfies a Lipschitz condition, since
|f (x, y) − f (x, ȳ)| = |Ay − Aȳ|
=
n X
n
X
aij (yj − ȳj )
i=1 j=1


!
n X
n
n
X
X
≤ 
|aij |
|yk − ȳk |
i=1 j=1
k=1
= |A||y − ȳ|
28
by definition of L1 norm
where |A| is the L1 norm of A but not the determinant. According to the existence and
uniqueness theorem of IVP, the IVP (2.3-1) has a unique solution. The solution can be obtained
by
1. elimination method (reduces the system of equations to a single higher order ODE)
2. method of Laplace transform (transform to a system of algebraic equation, easy to handle
the inhomogeneous DE and IC’s)
3. eigenvector expansion
4. fundamental matrix
Here, we shall explain the idea of fundamental matrix. By using the successive approximation, we have
y0 (x)
=
b
(2.3-2)
Z
y1 (x)
=
x
b+
Abdξ = b + xAb
(2.3-3)
0
Z
y2 (x)
=
b+
x
Ay1 dξ = b + xAb +
0
..
.
(xA)2
b
2!
..
.
yk (x)
(2.3-5)
xA (xA)2
(xA)k
I+
+
+ ··· +
b
1!
2!
k!
=
(2.3-4)
(2.3-6)
In general, if the initial condition in Eq. (2.3-1) is specified at x = a instead of x = 0, then kth
approximation of y will be (why?), instead of Eq. (2.3-6),
(x − a)A [(x − a)A]2
[(x − a)A]k
yk (x) = I +
+
+ ··· +
b
(2.3-7)
1!
2!
k!
By the same argument that we used in the existence proof, the sequence of successive
approximation is uniformly convergent to the solution, we have
y(x) = exA b
(2.3-8)
where the exponential term with matrix as exponent is defined as if the exponent was a scalar3
exA ≡
∞
X
(xA)k
k=0
(2.3-9)
k!
The computation of Ak and hence exA can be performed efficiently with the aid of the theory
of eigenvalue and eigenvector.
Notice that the direct differentiation of Eq. (2.3-9) leads to
d xA e
= AexA
(2.3-10)
dx
It acts as if A were a constant rather than a matrix. Moreover, if A = λI, it is easily, from Eq.
(2.3-9), to show that
eλxI = eλx I
3 exponential
(2.3-11)
function: The exponential function eax with scalar exponent ax is defined as
eax = 1 +
ax
(ax)2
(ax)k
+
+ ··· +
+ ···
1!
2!
k!
29
2.3.2
Eigenvalue and Eigenvector
[See also section §1.8.] Consider an eigenvalue problem of a square matrix A
Av = λv
(2.3-12)
Obviously, v ≡ 0 is a solution to (2.3-12). For some particular value of λ, (2.3-12) has nontrivial
solution of v, λ and v are then called eigenvalue and eigenvector of A, respectively. The
eigenvalue λ is determined by
det (A − λI) = 0
(2.3-13)
There are some simple results regarding to eigenvalues and eigenvectors of a square matrix A.
1. If A is an n × n matrix, then A always has n eigenvalues (may be repeated), but may or
may not have n linearly independent eigenvectors.
2. If A is real and symmetric (a special case of Hermitian matrix), then all eigenvalues and
corresponding eigenvectors are real. Furthermore, even if some of the eigenvalues are repeated, there is always a full set of n linearly independent eigenvectors v(1) , v(2) , · · · , v(n) .
In addition, Eigenvectors associated with distinct eigenvalues are orthogonal to each other.
3. If A is Hermitian, i.e. (aij = a∗ji ) where superscript asterisk indicates complex conjugate,
then the eigenvalues λ1 , λ2 , · · · , λn are still all real. But eigenvectors may be complex.
Again, even if some of the eigenvalues are repeated, there is always a full set of n linearly independent eigenvectors v(1) , v(2) , · · · , v(n) . Eigenvectors associated with distinct
eigenvalues are orthogonal to each other.
4. If A is not Hermitian, the situation is more complicated.
(a) If A is real, there are three possibilities.
i. All eigenvalues are real and distinct. Then there is a full set of n linearly
independent eigenvectors.
ii. Some eigenvalues are repeated. Then there may not be a full set of n linearly
independent eigenvectors.
iii. Some eigenvalues occur in complex conjugate pair.
(b) If A is complex, the complex eigenvalues need not occur in conjugate pair.
2.3.3
Evaluation of exA
2.3.3-1 Full set of eigenvectors
If A has a full set of linearly independent eigenvectors (whether A is Hermitian or not), by
using the definition of eigenvalue and eigenvector (2.3-12), it is easily to conclude that
AT =
Av(1) Av(2) · · · Av(n)
=
λ1 v(1) λ2 v(2) · · · λn v(n)
= TΛ
where T and Λ are defined as
T = v(1) v(2) · · · v(n)
(2.3-14)
Λ = diag (λ1 , λ2 , · · · , λn )
30
(2.3-15)
Since all the eigenvectors v(1) , v(2) , · · · , v(n) are linearly independent, hence T−1 exits. Consequently, from Eq. (2.3-14), A can be transformed into a diagonal matrix Λ by a similarity
transformation
T−1 AT = Λ
(2.3-16)
Similarly,
A = TΛT−1
(2.3-17)
With the aid of Eq. (2.3-17), it is easily to evaluate exA directly from Eq. (2.3-9), since
exA
=
∞
X
(xTΛT−1 )k
k=0
∞
X
= T
k!
(xΛ)k −1
T
k!
k=0
"∞
X
#
xk
= T
diag λk1 , λk2 , · · · , λkn T−1
k!
k=0
 P
(λ1 x)k
k!

P (λ2 x)k


k!
= T
..

.

P (λn x)k



 −1
T


(2.3-18)
k!
Hence for the case that A has a full set of eigenvectors, one has
exA = T diag eλ1 x , eλ2 x , · · · , eλn x T−1
y(x) = T diag eλ1 x , eλ2 x , · · · , eλn x T−1 b
(2.3-19)
(2.3-20)
Note that
T diag eλ1 x , eλ2 x , · · · , eλn x = eλ1 x v(1) eλ2 x v(2) · · · eλn x v(n)
(2.3-21)
is sometimes called the fundamental matrix of IVP (2.3-1). Since T−1 b is an n × 1 constant
vector, Eqns. (2.3-20) and (2.3-21) then imply that solution y(x) can always be expressed as
the fundamental matrix multiplying this constant vector.
Remarks:
Equations (2.3-20) and (2.3-21) have some other remarkable interpretations.
1. With the aid of Eq. (2.3-17), the DE in IVP (2.3-1) can be written as
y0 = TΛT−1 y
(2.3-22)
Multiply T−1 to the both sides of previous equation, one has
T−1 y0 = ΛT−1 y
(2.3-23)
Note that T−1 is an n × n constant matrix and y an n × 1 vector function. By letting
T−1 y = w
,
y = Tw
(2.3-24)
31
the differential equation (2.3-23) is simplified as
w0 = Λw
(2.3-25)
Similarly, the IC of y can be transformed to the one of w through Eq. (2.3-24) as
w(0) = T−1 b
(2.3-26)
Since Λ is a diagonal matrix, differential equation (2.3-25) is decoupled and every single
differential equation can be solved without difficulty as
w = diag eλ1 x , eλ2 x , · · · eλn x T−1 b
(2.3-27)
Hence we have the result exactly the same as Eq. (2.3-20).
2. To save argument, let us express the constant vector T−1 b as (c1 c2 · · · cn )T , where the
superscript T stands for the transpose. Then from Eq. (2.3-21), Eq. (2.3-20) can be
expressed as
y(x) = c1 eλ1 x v(1) + c2 eλ2 x v(2) + · · · + cn eλn x v(n)
(2.3-28)
The result Eq. (2.3-28) suggests that the solution y(x) can be expressed as a linear
combination of the eigenvectors of the matrix A!!
2.3.3-2 A pair of complex conjugate eigenvalues
Let us consider only the case that A is real but not symmetric (why not symmetric?) for
simplicity. Suppose that λ1 and λ2 are a pair of complex conjugate eigenvalues, and the rest of
λj ’s are real and distinct, then
Av(1)
= λ1 v(1)
(2.3-29)
(2)
(2)
(2.3-30)
Av
= λ2 v
and λ1 = λ∗2 = a + ib, where v(1) and v(2) are eigenvectors associated with λ1 and λ2 , respectively. Since v(1) and v(2) are linearly independent (why?), we can of course use the result of
the previous subsection and diagonalize A by Eq. (2.3-16). Everything is all right, except that
both T and Λ will then be complex. It will be more convenient to use only real calculation as
the following.
By taking complex conjugate to Eq. (2.3-29) and making use of the fact that λ1 and λ2 are
complex conjugate, we have
Av(1)∗ = λ2 v(1)∗
(2.3-31)
The equation (2.3-31) suggests that v(1)∗ is an eigenvector associated with the eigenvalue λ2 ,
hence v(2) = v(1)∗ . Let us denote the real and imaginary parts of eigenvector v(1) as u(1) and
u(2) , respectively. Let us construct a square matrix T̄ as
T̄ = u(1) u(2) v(3) v(4) · · · v(n) ≡ (P T3 )
(2.3-32)
32
where P = u(1) u(2) , and T3 = v(3) v(4) · · · v(n) . Obviously,
AT̄ = (AP AT3 ) =
AP Av(3) Av(4) · · · Av(n)
n×n
(3)
(4)
(n)
=
AP λ3 v
λ 4 v · · · λn v
(2.3-33)
n×n
Only the part AP needs further consideration. Note that AP is an n × 2 matrix, where
AP = Au(1) Au(2)
(2.3-34)
n×2
From Eq. (2.3-29), with the fact that v(1) = u(1) + iu(2) and λ1 = a + ib,
A[u(1) + iu(2) ]
= (a + ib)[u(1) + iu(2) ]
=
(1)
[au
(2)
− bu
(a + ib is the eigenvalue)
(2)
] + i[au
(1)
+ bu
]
(2.3-35)
From Eqns. (2.3-34) and (2.3-35), we concludes
AP = au(1) − bu(2) au(2) + bu(1)
(2.3-36)
The expression Eq. (2.3-36) can be arranged as the form
AP = u(1) u(2) Λ1
(2.3-37)
where
Λ1 =
a
−b
b
a
!
(2.3-38)
Hence
AT̄ = T̄Λ̄
(2.3-39)
where
Λ̄ = diag (Λ1 , λ3 , λ4 · · · , λn )
(2.3-40)
that is,

a b
 −b a


λ3
Λ̄ = 




..







.
(2.3-41)
λn
Consequently, similarly to the derivation of Eq. (2.3-19), we have
exA = T̄ diag exΛ1 , eλ3 x , eλ4 x , · · · eλn x T̄−1
(2.3-42)
To evaluate exΛ1 which is an n × 2 matrix, let us define a polar representation as r =
(a + b2 )1/2 and θ = tan−1 (b/a), then
!
!
a b
cos θ sin θ
=r
(2.3-43)
−b a
− sin θ cos θ
2
33
It is easily to show that
!k
a b
= rk
−b a
cos kθ
− sin kθ
sin kθ
cos kθ
!
(2.3-44)
It then follows from Eq. (2.3-9) and Eq. (2.3-43) that
 ∞

∞
X xk rk cos kθ X
xk rk sin kθ


k!
k!
 k=0

xΛ1
k=0


e
= X
∞
∞
k k
k k
X
x r sin kθ
x r cos kθ 
 −

k!
k!
k=0
(2.3-45)
k=0
Note that
∞
X
xk rk cos kθ
k!
k=0
+i
∞
X
xk rk sin kθ
k=0
=
k!
∞
X
xk rk eikθ
k!
k=0
iθ
= exre
= ex(a+ib)
= eax (cos bx + i sin bx)
(2.3-46)
Hence
xΛ1
e
=
eax cos bx
−eax sin bx
eax sin bx
eax cos bx
!
(2.3-47)
and
eax cos bx
 −eax sin bx


exA = T̄ 




eax sin bx
eax cos bx



 −1
 T̄



eλ3 x
..
.
(2.3-48)
eλn x
The construction of Eq. (2.3-48) can also be deduced directly from Eq. (2.3-14) and Eq.
(2.3-19).
T =
v(1) v(2) v(3) v(4) · · · v(n)
=
u(1) + iu(2) u(1) − iu(2) v(3) v(4) · · · v(n)
n×n
!
!
1
1
(1)
(2)
(3)
(4)
(n)
=
u
u
v
v ···v
n×2
i −i
2×2
n×n


1
1

 i −i




1


(1)
(2)
(3)
(4)
(n)


=
u
u
v
v ···v


1
n×n 



..


.
1 n×n
≡ T̄B
(2.3-49)
34
where

1
1
 i −i


1

B=

1



..
.















=




,
B−1
1
2
1
2
1
2i
1
− 2i










1
1
..
.
1
(2.3-50)
1
and T̄ is defined as Eq. (2.3-32). Eq. (2.3-19) can then be written as
exA = T̄B diag eλ1 x , eλ2 x , · · · eλn x B−1 T̄−1

 λ1 x
e
eλ2 x
 ieλ1 x −ieλ2 x





λ3 x
e

 −1 −1
 B T̄
= T̄ 
λ4 x


e




..


.
λn x
e

λ1 x
λ2 x
λ1 x
λ2 x
(e
+ e )/2 (e
− e )/2i
 i(eλ1 x − eλ2 x )/2 (eλ1 x + eλ2 x )/2


eλ3 x

= T̄ 

eλ4 x


..

.
(2.3-51)




 −1
 T̄




eλn x
eax cos bx
 −eax sin bx



= T̄ 





eax sin bx
eax cos bx




 −1
 T̄




λ3 x
e
eλ4 x
..
.
eλn x
2.3.3-3 Not enough eigenvectors
Let us consider again only for the case that A is real but not symmetric. For simplicity,
suppose that λ1 is a repeated eigenvalue of multiplicity 2, and there is only one eigenvector v(1)
associated with λ1 . It is now not possible to diagonalize A by any similarity transformation.
On the other hand, it is always possible to triangularize A by finding “generalized eigenvector”.
Let us consider an 2×2 square matrix A as an example. Construct a generalized eigenvector
w(2) such that
Aw(2) = λ1 w(2) + v(1)
or
(A − λ1 I) w(2) = v(1)
(2.3-52)
(2)
Take the matrix P as P = v(1) w
, then
AP =
Av(1) Aw(2) = λ1 v(1) v(1) + λ1 w(2)
!
λ
1
1
(1)
(2)
=
v
w
≡ PΛJ
0 λ1
35
(2.3-53)
Hence
A = PΛJ P−1
P−1 AP = ΛJ
or
(2.3-54)
where
λ1
0
ΛJ ≡
1
λ1
!
≡ λ1 I + J2
(2.3-55)
and
0
0
J2 =
1
0
!
(2.3-56)
The evaluation of exp(xA) is then followed directly from Eq. (2.3-54) and the definition of the
exponent Eq. (2.3-9) as
exA = PexΛJ P−1
(2.3-57)
Note that
exΛJ =
∞
k
X
xk [λ1 I + J2 ]
(2.3-58)
k!
k=0
By using the binomial series, we have
!
k
X
k
k
[λ1 I + J2 ] =
λk−`
Ik−` J`2
1
`
`=0
, (k ≥ 1)
(2.3-59)
Note that for ` ≥ 2
J`2
=
0
0
1
0
!`
0
0
=
0
0
!
, (k ≥ 1)
(2.3-60)
Hence the summation in the RHS of Eq. (2.3-59) contains only two nontrivial terms as
k
[λ1 I + J2 ] = λk1 Ik + kλk−1
J2
1
(2.3-61)
Substitute Eq. (2.3-61) into Eq. (2.3-58), we now have
exΛJ
=
=
∞
X
xk λk Ik
1
k=0
∞
X
k=0
k!
∞
X
xk kλk−1
1
k=0
∞
X
xk λk1
I+x
k!
λ1 x
=
+
e
0
k0 =0
λ1 x
xe
eλ1 x
k!
0
J2
0
xk λk1
J2
k0 !
!
(2.3-62)
The construction of Eq. (2.3-62) can be performed in a more direct manner. By using of a
simple property of exponential (of matrices), we have
exΛJ
= ex(J2 +λ1 I)
= exJ2 · eλ1 xI
(2.3-63)
The second term in the RHS of the Eq. (2.3-63) is concluded from Eq. (2.3-11) as
!
eλ1 x
λ1 xI
e
=
eλ1 x
36
(2.3-64)
While the first term in the RHS of the Eq. (2.3-63) can easily be evaluated directly from the
definition of the exponential of a matrix, Eq. (2.3-9), and Eq. (2.3-60). We have
!
∞
k k
X
x
J
1
x
2
exJ2 =
= I + xJ2 =
(2.3-65)
k!
0 1
k=0
Hence, we conclude the same result as Eq. (2.3-62).
Let A be a 3 × 3 matrix. Suppose that λ1 is a repeated eigenvalue of multiplicity 3, and
there is only one eigenvector v(1) associated with this eigenvalue. We then need two generalized
eigenvectors w(2) and w(3) to triangularized A. The generalized eigenvectors can be constructed
by solving
(A − λ1 I)w(2)
(A − λ1 I)w
= v(1)
(3)
= w
(2.3-66)
(2)
(2.3-67)
(3)
Take the matrix P as P = v(1) w(2) w
, then
AP =
Av(1) Aw(2) Aw(3) = λ1 v(1) v(1) + λ1 w(2) w(2) + λ1 w(3)


λ1 1
0


=
v(1) w(2) w(3)  0 λ1 1 
0
0 λ1
(2.3-68)
Hence
A = PΛJ P−1
P−1 AP = ΛJ
or
(2.3-69)
where

λ1

ΛJ ≡  0
0
1
λ1
0

0

1 
λ1
(2.3-70)
The triangular matrix of the form as the form of Eq. (2.3-70) is called as the Jordan canonical
form of the matrix A. In general, the main diagonal of the Jordan canonical form will consist of the eigenvalues λj ’s, and the off-diagonal elements will all be zero except for elements
immediately above repeated eigenvalues, which could be unity. Now, for the case of 3 × 3
matrix,
ΛJ = λ1 I + J3
(2.3-71)
where

0

J3 =  0
0
1
0
0

0

1 
0
(2.3-72)
Note that

J03 = I ,
0

2
J3 =  0
0
0
0
0

1

0 
0
(2.3-73)
and, for ` ≥ 3,
J`3 = 0
(2.3-74)
37
Again,
exΛJ
= 
exJ3 · eλ1 xI

0
x 

= I + xJ3 +
 0
2
0
0
0
0
2
 
eλ1 x
1
 
0  
0

eλ1 x
λ1 x


(2.3-75)
e
Hence
x2
2

exΛJ
1 x

= eλ1 x  0 1
0 0

(2.3-76)

x 
1
Example: Consider an IVP
(
y0 = Ay
DE
y(0) = b
I.C.
(2.3-77)
where
A=
1
−1
1
3
!
,
b=
0
1
!
(2.3-78)
The characteristic equation of A is
det |A − λI| = (1 − λ)(3 − λ) + 1 = (λ − 2)2 = 0
(2.3-79)
Hence the eigenvalues of A are 2, which is a repeated root of multiplicity 2. The eigenvector
associated with the eigenvalue can be found straightforwardly as
!
1
(1)
v =
(2.3-80)
1
There is only one eigenvector associated with this eigenvalue. Hence we can not find enough
linearly independent eigenvectors to diagonalize the matrix A. The equation for generalized
eigenvector w(2) , from Eq. (2.3-52), is
!
!
−1 1
1
w(2) =
(2.3-81)
−1 1
1
The generalized eigenvector is then
!
0
w(2) =
1
(2.3-82)
From Eq. (2.3-53) and Eq. (2.3-55), we have
!
!
2 1
1 0
ΛJ =
, P=
0 2
1 1
,
P
−1
=
1 0
−1 1
!
(2.3-83)
From Eq. (2.3-62), we have
exΛJ = exJ2 · e2xI = e2x
1 x
0 1
!
(2.3-84)
38
Hence,
exA
= PexΛJ P−1 !
1 0
= e2x
1 1
= e2x
1 x
0 1
!
!
1
−1
0
1
!
1−x
x
−x x + 1
(2.3-85)
Finally, the solution y(x) of the Eq. (2.3-77), following from Eq. (2.3-8), is
y(x)
= exA y(0)
1−x
x
= e
−x x + 1
!
xe2x
=
e2x + xe2x
!
2x
2.4
0
1
!
(2.3-86)
Green’s functions — an introduction
Consider a BVP of u(x) with a second order ODE as

00

0<x<`
 u = f (x)
u(0) = 0

 u(`) = 0
2.4.1
(2.4-1)
Variation of parameters
The homogeneous solution of the ODE in (2.4-1) is
uh (x) = A + Bx
(2.4-2)
The method of undermined coefficients is not appropriate in finding up , since the form of f (x)
is not yet specified. First of all, let us solve this BVP by the (old fashion) method of variation
of parameters. Apparently the particular solution is not of the form of 1 nor x. we will seek
the particular solution by letting the parameters A and B in Eq. (2.4-2) vary and thus try
up (x) = C1 (x) + xC2 (x)
(2.4-3)
Obviously the particular solution can always be expressed in the form Eq. (2.4-3), in fact we
can add one more arbitrary relation between C1 and C2 in order to uniquely determine the
particular solution (why?). We set
C10 (x) + xC20 (x) = 0
(2.4-4)
This will keep u0p (x) contains only C1 and C2 at most but not C10 and C20 . The original ODE
in (2.4-1) is then reduced to as an first order ODE for C2 as
(u00p =)C20 = f
(2.4-5)
By direct integration, we have
Z x
C1 = −
ξf (ξ)dξ
0
Z x
C2 =
f (ξ)dξ
(2.4-6)
(2.4-7)
0
39
From the results of Eq. (2.4-6) and Eq. (2.4-7), we have the general solution of the ODE as
Z x
u ≡ uh + up = A + Bx +
(x − ξ)f (ξ)dξ
(2.4-8)
0
Finally, apply the boundary condition to determine the constants A and B as
A =
B
0
= −
(2.4-9)
1
`
Z
`
(` − ξ)f (ξ)dξ
(2.4-10)
0
Hence the solution u(x) can be expressed as
Z x
Z `
ξ(x − `)
x(ξ − `)
u(x) =
f (ξ) dξ +
f (ξ) dξ
`
`
0
x
The solution, Eq. (2.4-11), can also be written in the form as
Z `
u(x) =
G(x; ξ)f (ξ)dξ
(2.4-11)
(2.4-12)
0
where


 x(ξ − `)
`
G(x; ξ) =

 ξ(x − `)
`
,
ξ>x
,
ξ<x
(2.4-13)
The function G(x; ξ) will be called as Green’s function of the BVP.
2.4.2
Accessory problem
We like to find a more effective method for obtaining the Green’s function G(x; ξ). Let us
consider an accessory problem related to (2.4-1), which will aid us in finding the desired solution.
 2
 d U (x; ξ)
= P (x)
(2.4-14)
dx2
 U (0; ξ) = U (`; ξ) = 0
where


 0
P (x) =
1/

 0
,
,
,
0 < x < ξ − (/2)
ξ − (/2) < x < ξ + (/2)
ξ + (/2) < x < `
(2.4-15)
Once having U (x; ξ), we simply superimpose, by physical argument, some weighted contribution of U (x; ξ) to yield the solution of ODE in (2.2-1)
Z `
X
u(x) =
lim
U (x; ξi )f (ξi )∆ξi =
U0 (x; ξ)f (ξ)dξ
(2.4-16)
→0,∆ξi →0
0
More rigorously, multiplying ODE in (2.4-1) by U (x; ξ) and the ODE in the accessory
problem (2.4-14) by u, subtracting and then integrating, we have
Z `
Z `
d2 u
d2 U
U 2 − u 2 dx =
[U f (x) − u(x)P ] dx
(2.4-17)
dx
dx
0
0
By integration by parts, the LHS of Eq. (2.4-17) is simplified as
` Z `
Z `
d2 u
d2 U
du
dU
dU du du dU
U 2 − u 2 dx = U
−
−u
−
dx
dx
dx
dx
dx 0
dx dx dx dx
0
0
40
(2.4-18)
Pε (x )
1
ε
x
ξ
Figure 2.1: The sketch of function P (x)
The integral in the RHS of Eq. (2.4-18) is definitely vanished. By the use of the boundary
conditions in Eqns. (2.4-1) and (2.4-14), we conclude that the term within the bracket in RHS
of Eq. (2.4-18) is also vanished. Substituting the definition of P , Eq. (2.4-15), into the righthand side of Eq. (2.4-17) and making use of the mean value theorem of the integration, we
conclude that
Z `
u(ξ) = lim
U f (x)dx
(2.4-19)
→0
0
By interchanging x and ξ, and if the limiting process and the integration in (2.4-19) can be
interchanging we have
Z `
u(x) =
U0 (ξ; x)f (ξ)dξ
(2.4-20)
0
This will be the form similar to Eq. (2.4-13).
The solution of the accessory problem (2.4-14)
each interval


,
 A0 x + A1
U (x) =
B0 x + B1 + (x2 /2)
,

 C x+C
,
0
1
can be found by simply integrating twice in
0 < x < ξ − (/2)
ξ − (/2) < x < ξ + (/2)
ξ + (/2) < x < `
(2.4-21)
There are six not-yet-determined constants, while we have only two boundary conditions specified in (2.4-14). It is easily to argue that U (x) has to be continuous and to have continuous
first derivative on the interval 0 ≤ x ≤ `, in particular at the points x = ξ ± /2. This argument
leads to four additional conditions for the constants, we then have

x(ξ − `)/`
,
0 < x < ξ − (/2)


 x2 1 ` ξ 1 2
U (x) =
−
+ −
x+
ξ−
,
ξ − (/2) < x < ξ + (/2) (2.4-22)

2
2 `
2
2


ξ(x − `)/`
,
ξ + (/2) < x < `
Having found U (x), we can then proceed to find by Eq. (2.4-20)
Z `
u(ξ) = lim
U (x; ξ)f (x)dx
→0 0
"Z
#
Z `
Z ξ+/2
ξ−/2
−x(` − ξ)
−ξ(` − x)
= lim
f (x)dx +
f (x)dx +
(· · ·)f (x)dx
→0
`
`
0
ξ+/2
ξ−/2
41
The last term in the right-hand side of the previous equation tends to zero as → 0, hence it
can be written as
Z `
u(ξ) =
U0 (x; ξ)f (x)dx
(2.4-23)
0
Interchange x and ξ, and notice that U0 (ξ; x) = U0 (x; ξ), the solution u(x) can then be expressed
as
Z `
u(x) =
U0 (x; ξ)f (ξ)dξ
(2.4-24)
0
where U0 (x; ξ) is exactly the same as the G(x; ξ) defined in Eq. (2.4-13).
2.4.3
By delta function
To circumvent the inconvenience of dealing with the limit procedure, we introduce a heuristic
concept of the Dirac delta function which is defined as the following two equations
(
0
,
x 6= ξ
δ(x − ξ) =
(2.4-25)
∞
,
x=ξ
Z
(
b
φ(x)δ(x − ξ)dx =
a
0
φ(ξ)
,
,
ξ∈
/ [a, b]
ξ ∈ (a, b)
In rigorous terms of classical analysis, this is mathematical nonsense4 .
By using the delta function, this suggest to find G(x; ξ) by considering
 2
d G(x; ξ)


= δ(x − ξ)

dx2
G(0; ξ) = 0



G(`; ξ) = 0
(2.4-26)
(2.4-27)
Obviously, the differential equations in the interval 0 < x < ξ and ξ < x < ` are homogeneous,
by the definition of δ(x − ξ) in Eq. (2.4-25), hence
(
A0 x + A1
,
0<x<ξ
G(x; ξ) =
(2.4-28)
C0 x + C1
,
ξ<x<`
4 Dirac delta function: Functions defined as the form of Eq. (2.4-25) are called generalized functions. The
equality of two generalized functions is not defined on the bases of values of functions as ordinary sense. Supposed
that there is another generalized function Ψ(x − ξ). And knowing that Ψ(x − ξ) is actually twice of δ(x − ξ).
Then according to Eq. (2.4-25),
(
0
,
x 6= ξ
Ψ(x − ξ) =
∞
,
x=ξ
Obviously, the ’values’ of Ψ is exactly the same, heuristicly, as the ’value’ of δ. On the other hand, the use of
Eq. (2.4-26) is able to detect that the ’effect’ of Ψ is twice of δ. In fact, the equality of two generalized function
is defined through the distributional equality as the following. If Ψ1 and Ψ2 are two generalized functions and
for any smooth function φ, if
Z ∞
Z ∞
φ(x)Ψ1 (x) dx =
φ(x)Ψ2 (x) dx
−∞
−∞
Then we say that Ψ1 = Ψ2 .
42
δ (x − ξ )
x
ξ
Figure 2.2: The sketch of the Dirac delta function δ(x − ξ)
By using the boundary conditions in (2.4-27), it is easily to find that A1 = 0 and C1 = −C0 `.
The solution is then still in terms of two not-yet-determined constants as
(
A0 x
,
0<x<ξ
G(x; ξ) =
(2.4-29)
C0 (x − `)
,
ξ<x<`
The determination of A0 and C0 requires two more conditions. First of all, we require that G
is continuous at x = ξ (why?), i.e.
G|x=ξ− = G|x=ξ+
i.e.
A0 ξ = C0 (ξ − `)
(2.4-30)
Equation (2.4-30) is called continuity condition. On the other hand, we expect that dG/dx is
not continuous at x = ξ (why?). In fact, there is a finite jump in the derivative of G across
x = ξ (why?). To find the jump condition, let us integrate both sides of DE in (2.4-27)
Z x=ξ+ 2
Z x=ξ+
d G
dx
=
(2.4-31)
δ(x − ξ)dx
2
x=ξ− dx
x=ξ−
This leads to
dG
dx
−
ξ+
dG
dx
=1
i.e.
C0 − A0 = 1
(2.4-32)
ξ−
By using Eqns. (2.4-30) and (2.4-32) as two supplemental conditions, A0 and C0 are determined
as
ξ−`
ξ
A0 =
, C0 =
(2.4-33)
`
`
Hence it leads to exactly the same G(x; ξ) as defined in (2.4-13).
Let us see how the Green’s function problem immediately leads to a representation of the
solution of BVP (2.4-1) in the form of Eq. (2.4-12). Again, multiplying ODE in BVP (2.4-1)
by G(x; ξ) and the ODE in BVP (2.4-27) by u, subtracting and then integrating, we have
Z ` 2
Z `
d u
d2 G
G 2 − u 2 dx =
[G(x; ξ)f (x) − u(x)δ(x − ξ)] dx
(2.4-34)
dx
dx
0
0
It is, again, easily to show that the left-hand side of Eq. (2.4-34) equals to zero by integration
by parts and the use of the boundary conditions in (2.4-1) and (2.4-27) as we did in Eq. (2.4-18)
. Equation (2.4-34), by the heuristic definition of δ Eq. (2.4-26), leads to
Z `
u(ξ) =
G(x; ξ)f (x)dx
(2.4-35)
0
43
Interchange x and ξ, and observe directly from the formula of G in this problem that G(x; ξ) =
G(ξ; x), we have exactly the same result as Eq. (2.4-12). Notice that we will show that G(x; ξ) =
G(ξ; x) is always true for the self-adjoint problems in a later section.
2.5
Green’s function
We have learned how to solve a simple Green’s function problem at the previous section. Let
us consider another BVP as a second example

00
0

,
0<x<1
 −u − u = f (x)
(2.5-1)
u(0) = 0

 u0 (1) = 0
The associated Green’s function problem with (2.5-1) is

00
0

,
0 < x, ξ < 1
 −G − G = δ(x − ξ)
G(0; ξ) = 0

 G0 (1; ξ) = 0
(2.5-2)
Obviously, the differential operator is not formally self-adjoint. Nevertheless, the complementary solution of homogeneous DE in G is
Gh = A + Be−x
(2.5-3)
Hence the candidate for the Green’s function solution is of the form
(
A1 + B1 e−x
,
x<ξ
G(x; ξ) =
A2 + B2 e−x
,
x>ξ
By using of the boundary condition leads to
(
A1 (1 − e−x )
,
G(x; ξ) =
A2
,
x<ξ
x>ξ
In order to satisfy the continuity condition at x = ξ, we have
(
A1 (1 − e−x )
,
x<ξ
G(x; ξ) =
−ξ
A1 1 − e
,
x>ξ
(2.5-4)
(2.5-5)
(2.5-6)
Finally, the jump condition on G0 at x = ξ yields to A1 = eξ . But how is the solution u of
(2.5-1) related to the Green’s function solution G of (2.5-2)?
2.5.1
Adjoint Operator
To express u in terms of the Green’s function solution G, it requires more information on L.
Consider an n-th order linear ODE
an (x)
dn u(x)
dn−1 u(x)
+ an−1 (x)
+ · · · + a0 (x)u(x) = h(x)
n
dx
dxn−1
(2.5-7)
where h(x) may or may not be a locally integrable function. We may also be interested in
boundary conditions on u(x) later. Let us define the differential operator L as
L(·) ≡
n
X
k=0
ak (x)
dk (·)
dxk
(2.5-8)
44
where ak (x) ∈ C0∞ and
d0
dx0
= 1. Equation (2.5-7) can then be written as
L[u(x)] = h(x)
(2.5-9)
Note that the differential operator defined in Eq. (2.5-8) is obviously a linear operator, since,
for any constant c and d,
L[cf (x) + dg(x)] = cL[f (x)] + dL[g(x)]
(2.5-10)
Hence if u1 (x) and u2 (x) are solutions of differential equation (2.5-9) with h(x) = r(x) and
h(x) = s(x), respectively. Then u = cu1 + du2 is a solution of
L[u] = cr(x) + ds(x)
(2.5-11)
where c and d are constants.
The adjoint operator of L, which is denoted as L∗ and is also an n-th order differential
operator, is defined as
n
X
dk [ak (x)(·)]
∗
L (·) ≡
(−1)k
(2.5-12)
dxk
k=0
By this definition, we always have
Z ∞
Z ∞
φ(x)Lu(x) dx =
u(x)L∗ φ(x) dx
−∞
(2.5-13)
−∞
For the cases that L = L∗ , we then call L is a formally self-adjoint operator. In particular,
consider a second order differential operator L as
Lu ≡ a2 u00 + a1 u0 + a0 u
(2.5-14)
The adjoint operator L∗ is then followed form Eq. (2.5-12) as
L∗ v
≡ (a2 v)00 − (a1 v)0 + a0 v
= a2 v 00 + (2a02 − a1 )v 0 + (a002 − a01 + a0 )v
(2.5-15)
Obviously, a necessary and sufficient condition for second order differential operator L being
formally self-adjoint5 is that
a02 = a1
2.5.2
(2.5-16)
Boundary Value Problems – with homogeneous B.C.’s
Our final goal is to solve the BVP


,
a<x<b
 Lu = f (x)
B1 u = 0

 B u=0
2
(2.5-17)
where L and Bi are differential and boundary operators, respectively. Let us consider BVP’s
only with second order ODE’s. The boundary operator Bi is defined as
Bi u ≡ αi1 u(a) + αi2 u0 (a) + βi1 u(b) + βi2 u0 (b)
(2.5-18)
If β11 = β12 = α21 = α22 = 0, the boundary conditions are unmixed. If β11 = β12 = β21 =
β22 = α12 = α21 = 0, we have the initial conditions.
5 formally
self-adjoint differential operator: From the result of Eq. (2.5-16), a formally self-adjoint 2nd order
differential operator can always be arranged as the form
d
du(x)
Lu =
a2 (x)
+ a0 (x) u(x)
dx
dx
45
2.5.2-1 Adjoint, self-adjoint system
In order to express u in terms of the Green’s function solution G, it requires more information
on L. We have already had
Z ∞
Z ∞
Ltφ dx =
tL∗ φ dx
(2.5-19)
−∞
−∞
for every test function φ ∈ C0∞ . We want to extend this formula to include contributions from
end points of integration in order to solve BVP (2.5-17). Consider, v may not be a test function,
Z b
Z b
vLudx =
v (a2 u00 + a1 u0 + a0 u) dx
a
a
= ······
=
(integration by parts)
Z b
b
J(u, v)|a +
uL∗ vdx
(2.5-20)
a
or
Z
a
b
b
(v Lu − u L∗ v) dx = J(u, v)|a
(2.5-21)
where
b
J(u, v)|a ≡ a2 (vu0 − uv 0 ) + (a1 − a02 )uv
(2.5-22)
There are some terminologies:
1. Given a second order differential operator L with homogeneous boundary conditions
B1 u = 0 and B2 u = 0, the adjoint homogeneous boundary conditions B1∗ v = 0 and
b
B2∗ v = 0, are those which make J(u, v)|a = 0.
2. The BVP is a self-adjoint problem, if L = L∗ and B1 = B1∗ , B2 = B2∗ .
Example 1:
Consider a BVP

00

,
0<x<`
 Lu ≡ u = f (x)
B1 u ≡ u(0) = 0

 B u ≡ u(`) = 0
2
R
By considering v Lu dx and performing integration by parts, we have
Z `
Z `
`
vu00 dx = (vu0 − uv 0 )|0 +
uv 00 dx
0
(2.5-23)
(2.5-24)
0
Obviously, the adjoint differential operator L∗ = L. The adjoint boundary operators B1∗ and
B2∗ are such that the boundary term in RHS of (2.5-24) vanishes. By using the boundary
conditions in u and arguing that u0 (0) and u0 (`) may not be zero, we have to put conditions on
v to make the boundary term vanish. It leads to B1∗ v ≡ v(0) = 0, B2∗ v ≡ v(`) = 0, i.e. B = B ∗ .
Hence the BVP is self-adjoint.
Example 2:
Consider a BVP

00
0

0<x<`
 −u − u = f (x)
0
(2.5-25)
u(0) + u (0) = 0

 u(`) − u0 (`) = 0
46
R
Again, by considering vLu dx and performing integration by parts, we have
Z `
Z `
`
v(−u00 − u0 )dx = [−(vu0 − v 0 u) − uv]|0 +
u(−v 00 + v 0 )dx
0
(2.5-26)
0
Obviously, the adjoint differential operator L∗ = −d2 /dx2 + d/dx 6= L. The adjoint boundary
operators are boundary conditions on v such that the boundary term in (2.5-26) vanishes, i.e.
we want
[−v(`)u0 (`) + u(`)v 0 (`) − u(`)v(`)] + [v(0)u0 (0) − u(0)v 0 (0) + u(0)v(0)] = 0
(2.5-27)
By using the boundary conditions on u, (2.5-27) reduces to
−u(`) [2v(`) − v 0 (`)] − u(0)v 0 (0) = 0
(2.5-28)
Since u(`) and u(0) may not be zero, we conclude that adjoint boundary conditions are
2v(`) − v 0 (`)
0
v (0)
=
0
(2.5-29)
= 0
(2.5-30)
2.5.2-2 Green’s function and adjoint Green’s function
Consider again the BVP of (2.5-17) in Section 2.5.2, the associate Green’s function problem
and the associate adjoint Green’s function problem, respectively, are


, a < x, η < b
 LG(x; η) = δ(x − η)
(2.5-31)
B1 G = 0

 B G=0
2

∗

, a < x, ξ < b
 LG (x; ξ) = δ(x − ξ)
∗ ∗
(2.5-32)
B1 G = 0

 B ∗ G∗ = 0
2
where L and L∗ are differential and adjoint differential operators, respectively, while Bi and
Bi∗ are boundary and adjoint boundary operators.
Combining (2.5-17) and (2.5-32), we have
Z b
b
[G∗ Lu − uL∗ G∗ ] dx = J(u, G∗ )|a
(2.5-33)
a
The RHS of (2.5-33) vanishes (why?), hence
Z b
u(ξ) =
G∗ (x; ξ)f (x) dx
(2.5-34)
a
Interchanging x and ξ yields
Z b
u(x) =
G∗ (ξ; x)f (ξ) dξ
(2.5-35)
a
The solution of the (2.5-17) is now in terms of G∗ (ξ; x) solved from the adjoint Green’s function
problem (2.5-32). It will be more convenient, if we can express u(x) in terms of G rather than
G∗ . Let us now combine further (2.5-31) and (2.5-32)
Z b
b
[G∗ (x; ξ)LG(x; η) − G(x; η)L∗ G∗ (x; ξ)] dx = J(G, G∗ )|a
(2.5-36)
a
47
Again, the RHS of (2.5-36) vanishes (why?), hence
G∗ (η; ξ) = G(ξ; η)
(2.5-37)
Thus, from (2.5-35) and (2.5-37), u(x) is now expressed in terms of G rather than G∗ as
Z b
u(x) =
G(x; ξ)f (ξ) dξ
(2.5-38)
a
2.5.3
Boundary Value Problems – with inhomogeneous B.C.’s
Consider an inhomogeneous BVP


, a<x<b
 Lu = f (x)
B1 u = u0

 B u=v
2
0
(2.5-39)
It has been shown that the solution u(x) of (2.5-39), for the cases of u0 = v0 = 0, can be
expressed as
Z b
u(x) =
G(x; ξ)f (ξ) dξ
(2.5-40)
a
What is going to happen for the cases of u0 6= 0 and/or v0 6= 0?
The associate Green’s function problem and the associate adjoint Green’s function problem
of (2.5-39) are still the same as (2.5-31) and (2.5-32), respectively. Combining again (2.5-39)
and (2.5-32), we have
Z b
b
[G∗ (x; ξ)Lu − uL∗ G∗ (x; ξ)] dx = J [u, G∗ (x; ξ)]|a
(2.5-41)
a
The RHS of (2.5-41) will no longer vanish. Equation (2.5-41) can still be simplified to be as
Z b
b
u(ξ) =
G∗ (x; ξ)f (x) dx − J [u(x), G∗ (x; ξ)]|x=a
(2.5-42)
a
Interchanging x and ξ and using G∗ (ξ; x) = G(x; ξ), it yields
Z b
b
u(x) =
G(x; ξ)f (ξ) dξ − J [u(ξ), G(x; ξ)]|ξ=a
(2.5-43)
a
Notice that the last term in the RHS of (2.5-43) should be carried out with care. Let us
write it out in detail
du
dG(x; ξ)
J [u(ξ), G(x; ξ)] = a2 (ξ) G(x; ξ)
− u(ξ)
+ [a1 (ξ) − a02 (ξ)] u(ξ)G(x; ξ)(2.5-44)
dξ
dξ
Here some care has also to be taken on substituting ξ = a, b rather than x = a, b into G(x; ξ).
Example
Consider the following BVP with inhomogeneous B.C. as an example

00

, 0<x<`
 −u = f (x)
u(0) = u1

 u(`) = u
2
48
(2.5-45)
The associate Green’s function problem always has homogeneous boundary conditions as

00

, 0 < x, ξ < `
 −G (x; ξ) = δ(x − ξ)
(2.5-46)
G(0; ξ) = 0

 G(`; ξ) = 0
As before
(
G(x; ξ) =
x(` − ξ)/`
ξ(` − x)/`
,
,
Let us write out dG/dξ for latter reference
(
dG(x; ξ)
−x/`
,
=
dξ
(` − x)/`
,
0<x<ξ
ξ<x<`
(2.5-47)
0<x<ξ
ξ<x<`
(2.5-48)
From (2.5-43), we have
Z `
`
u(x) =
G(x; ξ)f (ξ) dξ − J [u(ξ), G(x; ξ)]|ξ=0
(2.5-49)
0
The last term in the RHS of (2.5-49) can be written out as
du(`)
dG(x; `)
du(0)
dG(x; 0)
`
J [u(ξ), G(x; ξ)]|ξ=0 = − G(x; `)
− u(`)
+ G(x; 0)
− u(0)
dξ
dξ
dξ
dξ
dG(x; 0)
dG(x; `)
− u1
= u2
dξ
dξ
= − [u2 x + u1 (` − x)] /`
(2.5-50)
2.6
2.6.1
Alternative Theorem and Modified Green’s Function
Introduction
We have already pointed out in Section 2.2.2 that the uniqueness and existence theorems do
not necessarily hold for BVP’s. In fact, the difficulty of BVP encountered is similar to one that
arises in the solution of the linear algebraic equations. To illustrate the point, let us consider a
BVP as

00

, 0<x<`
 −u = f (x)
0
(2.6-1)
u (0) = 0

 u0 (`) = 0
The associate Green’s function problem is of course

00

, 0 < x, ξ < `
 −G (x; ξ) = δ(x − ξ)
dG(0;ξ)
=
0
dx

 dG(`;ξ)
=0
dx
(2.6-2)
As before, since the ODE of the Green’s function problem is homogeneous at x 6= ξ, we have
the solution as
(
A1 + B1 x
,
0<x<ξ
G(x; ξ) =
(2.6-3)
A2 + B2 x
,
ξ<x<`
49
To satisfy boundary conditions in (2.6-2), we have
(
A1
,
0<x<ξ
G(x; ξ) =
A2
,
ξ<x<`
By satisfying the continuity condition, we conclude
(
A1
, 0<x<ξ
G(x; ξ) =
A1
, ξ<x<`
(2.6-4)
(2.6-5)
It is now impossible to satisfy jump condition any further. Hence no solution exists to the
Green’s function problem (2.6-2).
Remarks:
1. Let us take a close look to the following property of this problem. Suppose we integrate
the DE with respect to x, we have
Z `
Z ` 2
d u
f (x)dx = −
2
0
0 dx
`
du
= −
=0
dx x=0
Hence we can not have a solution for the BVP (2.6-1) unless the condition
Z `
f (x)dx = 0
(2.6-6)
0
is satisfied. Obviously, the corresponding condition for (2.6-2) is not satisfied, hence it is
impossible to find a solution for (2.6-2).
2. There are some simple physical interpretation of this difficulty. The BVP (2.6-1) represents the steady-state temperature in a completely insulated rod (i.e. heat flux vanishes
at both two ends of the rod, or du/dx|`0 = 0. Since the sources along the rod generate
heat at a rate f (x) per unit length per unit time. A solution is possible only if the condition (2.6-6) is satisfied, that is, only if the net heat added per unit time is zero. If this
condition is violated, it is physically clear that the temperature would rise (or decline)
with time and no steady state solution could exist.
2.6.2
(Fredholm) Alternative Theorem
Consider a BVP again with second order ODE


, a<x<b
 Lu = f (x)
B1 u = u0

 B u=v
2
0
(2.6-7)
where L is a second order differential operator and f (x) is a locally integrable function. Consider
the following two homogeneous BVP’s as well as a Green’s function problem associated with
(2.6-7) as


, a<x<b
 LuH = 0
(2.6-8)
B1 uH = 0

 B u =0
2 H
50

∗ ∗

 L uH = 0
B1∗ u∗H = 0

 B ∗ u∗ = 0
2 H
,
a<x<b
(2.6-9)


 LG(x; ξ) = δ(x − ξ)
B1 G(x; ξ) = 0

 B G(x; ξ) = 0
2
, a<x<b
(2.6-10)
The (Fredholm) alternative theorem states that
1. If uH (x) ≡ 0 is the only solution of (2.6-8), then u∗H (x) ≡ 0 is the only solution of (2.6-9).
Moreover, BVP (2.6-7) will have a solution and it is unique.
2. If (2.6-8) has k independent nontrivial solutions, then (2.6-9) also has k independent nontrivial solutions although they are not necessarily the same as those of (2.6-8). Moreover,
(2.6-7) can have a solution (non-unique) if and only if
Z
a
b
f (x)u∗H (x)dx = J(u, u∗H )
b
a
(2.6-11)
for each u∗H satisfying (2.6-9)
Remarks:
1. If uH ≡ 0 is the only solution of (2.6-8), then both (2.6-7) and (2.6-10) have unique
solutions, and
Z
b
u(x) =
a
b
G(x; ξ)f (ξ)dξ − J [u(ξ), G(x; ξ)]|ξ=a
(2.6-12)
2. If (2.6-8) has a nontrivial solution(s), then no solution to (2.6-10) exists.
3. The condition (2.6-11) is sometimes called a consistency condition or solvability condition.
4. If (2.6-8) has k independent non-trivial solutions and the consistency condition of (2.6-7)
is satisfied, then (2.6-7) has a (nonunique) solution
u(x) =
k
X
ci uiH (x) + up (x)
(2.6-13)
i=1
where ci ’s are arbitrary constants.
2.6.3
Modified Green’s Function
From alternative theorem in Section 2.6.2, we know that the solution to the Green’s function
problem (2.6-10) does not exist if there are nontrivial solution(s) to the associated homogeneous
problem of the original BVP. If we insist upon constructing something like a Green’s function,
which will help us solve the original BVP (2.6-7) when the BVP has a solution. The remedy
is rather simple by introducing an additional source term into the Green’s function problem.
51
Consider a revised form of Green’s function problem called the “modified Green’s function
problem”

∗
∗

, a<x<b
 LGM (x; ξ) = δ(x − ξ) − uH (x)uH (ξ)
(2.6-14)
B1 GM (x; ξ) = 0

 B G (x; ξ) = 0
2 M
where u∗H is a normalized nontrivial solution to (2.6-9), i.e.
Z b
∗
2
uH (x) dx = 1
(2.6-15)
Obviously, the consistency condition of (2.6-14) is satisfied, since
Z b
δ(x − ξ) − u∗H (x)u∗H (ξ) u∗H (x)dx = 0
(2.6-16)
a
a
Hence the solution of modified Green’s function GM exists but is not unique. We shall require
Z b
GM (x; ξ)uH (x)dx = 0
(2.6-17)
a
to uniquely define GM . Consider also the adjoint modified Green’s function problem associated
with (2.6-14), i.e.

∗ ∗

, a<x<b
 L GM (x; η) = δ(x − η) − uH (x)uH (η)
∗ ∗
(2.6-18)
B1 GM (x; η) = 0

 B ∗ G∗ (x; η) = 0
2 M
Again, G∗M is not unique, we shall similarly require
Z b
G∗M (x; ξ)u∗H (x)dx = 0
(2.6-19)
a
to uniquely define G∗M . It will be seen that this will then ensure GM (x; ξ) = G∗M (ξ; x).
Once having the modified Green’s function, it will enable us to solve the BVP (2.6-7). For
simplicity, let us illustrate for the cases with homogeneous boundary conditions, i.e. u0 = v0 = 0
only. By the usual procedure, consider
Z b
∗
b
GM (x; η)Lu(x) − u(x)L∗ G∗M (x; η) dx = J u(x), G∗M (x; η) x=a
(2.6-20)
a
The RHS of (2.6-20) vanishes (why?), hence
Z b
Z b
G∗M (x; η)f (x)dx − u(η) +
u(x)uH (x)uH (η) dx = 0
(2.6-21)
Interchanging x and η yields
Z b
u(x) = CuH (x) +
G∗M (η; x)f (η) dη
(2.6-22)
a
a
a
The solution of the (2.6-7) is now in terms of G∗M (η; x) solved from the adjoint modified Green’s
function problem (2.6-18). It will be more convenient, if we can express u(x) in terms of GM
rather than G∗M . Let us now relate GM and G∗M by considering
Z b
∗
b
GM (x; η)LGM (x; ξ) − GM (x; ξ)L∗ G∗M (x; η) dx = J(GM , G∗M ) a
(2.6-23)
a
52
Again, the RHS of (2.6-23) vanishes (why?), hence
G∗M (ξ; η) = GM (η; ξ)
(2.6-24)
Here the conditions of (2.6-17) and (2.6-19) have been use. Hence, from (2.6-22) and (2.6-24),
u(x) is now expressed in terms of GM rather than G∗M
Z b
u(x) = CuH (x) +
GM (x; ξ)f (ξ) dξ
(2.6-25)
a
Example Let us consider the BVP in Section 2.6.1 again

00

, 0<x<`
 −u = f (x)
0
u (0) = 0

 u0 (`) = 0
Let us first examine the associated homogeneous problem

00

, 0<x<`
 −uH = 0
0
uH (0) = 0

 u0 (`) = 0
H
(2.6-26)
(2.6-27)
Obviously, uH (x) ≡ c is a nontrivial solution of this problem. Let us take c = 1/`1/2 to have
normalized homogeneous solution. Moreover, (2.6-27) is a self-adjoint problem, thus u∗H (x) ≡
1/`1/2 is also a nontrivial solution of the adjoint homogeneous problem. According to the
alternative theorem in Section §2.6.2, the BVP (2.6-26) has a solution if and only if
Z `
f (x)dx = 0
(2.6-28)
0
Suppose Eq. (2.6-28) holds, in any case G(x; ξ) does not exist as we have discussed in Section
§2.6.3, let us consider the modified Green’s function problem

1


, 0 < x, ξ < `
−G00M (x; ξ) = δ(x − ξ) −


`

dGM (0; ξ)
(2.6-29)
=0

dx



 dGM (`; ξ) = 0
dx
Since the corresponding consistency condition of (2.6-29) is satisfied, the system has a solution.
As before, for x 6= ξ, −G00M = 1/`, so that
(
2
A1 + B1 x + x2`
,
0<x<ξ
GM (x; ξ) =
(2.6-30)
2
A2 + B2 x + x2`
,
ξ<x<`
To satisfy boundary conditions in (2.6-27), we have
(
A1 + x2 /2`
,
0<x<ξ
GM (x; ξ) =
2
A2 − x + x /2`
,
ξ<x<`
By satisfying the continuity condition, we have
(
A1 + x2 /2`
,
GM (x; ξ) =
2
A1 + ξ − x + x /2`
,
53
0<x<ξ
ξ<x<`
(2.6-31)
(2.6-32)
The jump condition is satisfied automatically. As we expected, there is an arbitrary constant
A1 in the solution.
It is often convenient to choose a particular modified Green’s function which is symmetric in
R`
the sense of GM (x; ξ) = G∗M (ξ; x). To do so, let us impose the condition 0 GM (x; ξ)uH (x)dx = 0
for every ξ to the modified Green’s function. In our particular case, this condition yields
Z ξ
Z `
x2
x2
A1 +
dx +
A1 + ξ − x +
dx = 0
(2.6-33)
2`
2`
0
ξ
or
ξ2
`
A1 = −ξ +
+
(2.6-34)
2` 3
The symmetric modified Green’s function is then given by
(
2
+x2
−ξ + 3` + ξ 2`
,
0<x<ξ
GM (x; ξ) =
(2.6-35)
2
ξ +x2
`
−x + 3 + 2`
ξ<x<`
The solution of (2.6-27) is then
Z `
u(x) = C +
GM (x; ξ)f (ξ)dξ
(2.6-36)
0
2.7
Eigenfunction Expansion
Consider a BVP

du
d

 Lu ≡ dx p(x) dx + q(x)u(x) = f (x)
B1 u = u0

 B u=v
2
0
, a<x<b
(2.7-1)
The associated homogeneous BVP is


, a<x<b
 LuH = 0
B1 uH = 0

 B u =0
2 H
(2.7-2)
We know from (Fredholm) alternative theorem that if (2.7-2) has nontrivial solution, there will
be trouble in solving (2.7-1).
2.7.1
Eigenvalue Value Problem
To investigate the nature of (2.7-2), let us consider


, a<x<b
 Lφ(x) + λρ(x)φ(x) = 0
B1 φ = 0

 B φ=0
2
(2.7-3)
For simplicity, let p(x) > 0 in the finite interval [a, b], and p(x) ∈ C 1 . Moreover, let us consider
boundary operators B1 and B2 are one of the following two cases
1. unmixed boundary conditions, i.e.
α1 φ(a) + α2 φ0 (a)
0
β1 φ(b) + β2 φ (b)
=
0
(2.7-4)
=
0
(2.7-5)
with α12 + α22 6== and β12 + β22 6= 0.
54
2. periodic boundary conditions with p(a) = p(b), i.e.
φ(a)
= φ(b)
(2.7-6)
φ0 (a)
= φ0 (b)
(2.7-7)
Obviously, φ(x) ≡ 0 is a solution to the problem (2.7-3). For some particular values of λ,
there are nontrivial solutions φ(x). Again, those nontrivial solutions are called eigenfunctions,
and the corresponding λ’s are called their eigenvalues. Moreover, the problem (2.7-3), with
boundary conditions defined as Eqns. (2.7-4)-(2.7-5) or Eqns. (2.7-6)-(2.7-7) (with p(a) = p(b)),
is called a regular Sturm-Liouville eigenvalue problem.
Theorem
The regular Sturm-Liouville eigenvalue problem, for ρ(x) > 0, has the following properties.
1. There exists a countable set of eigenvalues (i.e. eigenvalues are simple) of (2.7-3), which
form a increasing sequence
λ1 < λ2 < λ3 < · · · < λn < · · · → ∞
,
as n → ∞
(2.7-8)
2. If eigenfunctions φi and φj correspond to λi and λj , λi 6= λj , then
b
Z
ρ(x)φi (x)φj (x)dx = 0
(2.7-9)
a
3. Eigenfunctions {φn (x)} forms a complete set in [a, b]. Hence any piecewise smooth function u(x) in [a, b] can be expressed by
u(x) =
∞
X
cn φn (x)
(2.7-10)
n=0
Furthermore, the infinite series, Eq. (2.7-10), converges to [u(x+ )+u(x− )]/2 in [a, b]. And
the coefficient cn can be found by
Z
cn =
b
ρuφn dx
(2.7-11)
a
with
R
ρφ2n dx = 1.
4. Any eigenvalue can be related to its eigenfunction by the Rayleigh quotient
(
)
Z b
1
0 b
02
2
λ = Rb
−[pφφ ]|a +
[pφ − qφ ]dx
a
ρφ2 dx
a
Remarks:
1. The Sturm-Liouville eigenvalue problem is a self-adjoint problem.
2. If
(a) p(x) vanishes at some points in [a, b]; or
(b) q(x) is discontinuous; or
55
(2.7-12)
(c) [a, b] is a semi-infinite or infinite interval
then BVP (2.7-3) is called a singular Sturm-Liouville eigenvalue problem.
3. If the eigenvalue problem is not formally self-adjoint, then the eigenfunctions of the problem will not form a orthogonal set. But it could form a bi-orthogonal set with the
eigenfunction of adjoint eigenvalue problem.
4. Consider a second order DE Lu ≡ a2 u00 + a1 u0 + a0 u = f (x) in [a, b]. If a02 6= a1 ,
then L is not formally self-adjoint. If a2 6= 0 in [a, b], there is a function r(x) =
R
exp[ (a1 /a2 )dx]/a2 6= 0 in [a, b] such that
d
du
r(x)Lu(x) ≡
p(x)
+ ra0 u = r(x)f (x)
dx
dx
2.7.2
(2.7-13)
BVP with Homogeneous B.C.’s
Consider a BVP

d
du

 Lu ≡ dx p(x) dx + q(x)u(x) = f (x)
u(a) = 0

 u(b) = 0
, a<x<b
(2.7-14)
The associated homogeneous BVP is


, a<x<b
 LuH = 0
uH (a) = 0

 u (b) = 0
H
(2.7-15)
Suppose given uH (x) ≡ 0 as the only solution to the homogeneous BVP. Then original BVP
has the unique solution, and there exists a Green’s function G(x; ξ) satisfies


, a<x<b
 LG(x; ξ) = δ(x − ξ)
(2.7-16)
G(a; ξ) = 0

 G(b; ξ) = 0
To solve the original BVP or the Green’s function problem, let us consider the associated
eigenvalue problem


, a<x<b
 Lφn (x) + λn ρ(x)φn (x) = 0
(2.7-17)
φn (a) = 0

 φ (b) = 0
n
with
Z
(
b
ρφm φn dx =
a
0
1
,
m 6= n
m=n
(2.7-18)
The formal method for solving the original BVP of u is to construct u(x) as
u(x) =
∞
X
cn φn (x)
(2.7-19)
n=1
56
If u(x) is given, cn can be found by multiplying ρ(x)φm (x) to both side of (2.7-19) and integrating over [a, b], it yields
Z b
cn =
ρ(x)φn (x)u(x)dx
(2.7-20)
a
Here the orthogonality of eigenfunctions (2.7-18) has been used. Unfortunately, u(x) is not
known until we solve the BVP. To find cn , by the fact that the problem is self-adjoint, consider
Z b
b
[φn Lu − uLφn ] dx = J(u, φn )|a
(2.7-21)
a
The RHS of (2.7-21) vanishes (why?), we have
Z b
Z b
φn (x)f (x) dx +
λn ρ(x)φn (x)u(x)dx = 0
a
(2.7-22)
a
Hence the expansion coefficient cn can be expressed as
Z b
1
cn = −
φn (x)f (x) dx
λn a
Thus the unique solution of the original BVP (2.7-14) is given by
"R b
#
∞
X
φ (ξ)f (ξ) dξ
a n
u(x) =
φn (x)
−λn
n=1
(2.7-23)
(2.7-24)
This result implies that
G(x; ξ) =
∞
X
φn (x)φn (ξ)
−λn
n=1
(2.7-25)
Note that if eigenfunctions were not normalized, the results of Eq. (2.7-24)-(2.7-25) should be
read as
" Rb
#
∞
X
φ (ξ)f (ξ) dξ
a n
u(x) =
φn (x)
(2.7-26)
Rb
2
n=1 −λn a ρφn (η) dη
G(x; ξ) =
∞
X
φn (x)φn (ξ)
Rb
2
n=1 −λn a ρφn (η) dη
(2.7-27)
Example
Consider the BVP with homogeneous B.C., again,

00

, 0<x<`
 u = f (x)
u(0) = 0

 u(`) = 0
(2.7-28)
To solve this problem by eigenfunction expansion, consider the associated eigenvalue problem

00

, 0<x<`
 φn (x) + λn φn (x) = 0
(2.7-29)
φn (0) = 0

 φ (`) = 0
n
It is easily to find the eigenvalue as λn = (nπ)2 /`2 , and the corresponding eigenfunction,
without normalization, as
nπx
φn (x) = sin
(2.7-30)
`
57
Now, construct the solution u(x) of (2.7-28) with the aid of eigenfunctions by letting
∞
X
u(x) =
cn φn (x)
(2.7-31)
n=1
method 1: substitute (2.7-31) into the original DE of (2.7-28), we have
∞
X
n=1
∞
X
=⇒
cn φ00n (x) = f (x)
(2.7-32)
cn (−λn φn ) = f (x)
(2.7-33)
n=1
Again, by multiplying both side of equation by eigenfunction, integrating over the interval,
and then using the orthogonality condition of eigenfunction, we have
R`
cn =
φn (ξ)f (ξ)dξ
R`
−λ 0 φ2n (η)dη
0
(2.7-34)
Hence the solution of the BVP is expressed as
∞
−2` X
u(x) = 2
π n=1
R`
0
f (ξ) sin nπξ
nπx
` dξ
sin
2
n
`
(2.7-35)
method 2: As before, consider
Z
`
`
[φn u00 − uφ00n ] dx = J(u, φn )|0
0
(2.7-36)
The RHS of the equation vanishes (why?), we have
Z
`
`
Z
φn (x)f (x) dx + λn
φn (x)u(x)dx = 0
0
(2.7-37)
0
Substitute (2.7-31) into (2.7-37), and make use of the orthogonality condition of eigenfunctions, we have exactly the same results for cn , and hence for u(x), as in (2.7-34) and
(2.7-35), respectively.
method 3: Since the solution u(x) and its expansion coefficient cn are related to each other
by
(
u(x) =
cn =
P∞
n=1 cn φn (x)
R`
1
In 0 u(x)φn (x)dx
(2.7-38)
R`
where In = 0 φ2n (x)dx. It is interesting to point out that (2.7-38) can be considered as a
R`
finite transform pair. Take the transform I1n 0 φn (x)(·)dx to the DE, it yields
Z
0
`
φn (x)u00 (x) dx =
Z
`
φn (x)f (x)dx
0
58
(2.7-39)
The LHS of (2.7-39) can be simplified by integration by parts, or in general
Z
`
LHS of (2.7-39) =
φn Lu dx
0
Z
`
uL∗ φn dx + J(u, φn )
=
0
Z
`
uLφn dx + J(u, φn ) (why??)
=
0
Z
= −λn
`
uφn dx
0
= −λn In cn
Hence, we have exactly the same results for cn , and thus for u(x), as in (2.7-34) and
(2.7-35), respectively.
2.7.3
BVP with Inhomogeneous B.C.’s
To illustrate the point, let us consider a very simple example,

00

, 0<x<`
 u = f (x)
u(0) = u0

 u(`) = v
0
(2.7-40)
To solve this problem directly by eigenfunction expansion, consider the associated eigenvalue
problem

00

, 0<x<`
 φn (x) + λn φn (x) = 0
(2.7-41)
φn (0) = 0

 φ (`) = 0
n
Again, the eigenvalues are λn = (nπ)2 /`2 , and the corresponding eigenfunctions, without normalization, are
nπx
φn (x) = sin
(2.7-42)
`
Construct the solution u(x) of (2.7-40) by letting, directly,
u(x) =
∞
X
c∗n φn (x)
(2.7-43)
n=1
method 1: Substitute (2.7-43) into the original DE of (2.7-40), we would have
∞
X
n=1
∞
X
=⇒
c∗n φ00n (x) = f (x)
(2.7-44)
c∗n (−λn φn ) = f (x)
(2.7-45)
n=1
Again, by multiplying both side of Eq. (2.7-45) by eigenfunction, integrating over the
interval, and then using the orthogonality condition of eigenfunctions, we have
R`
c∗n
=
φn (ξ)f (ξ)dξ
R`
−λn 0 φ2n (η)dη
0
(2.7-46)
59
Equation (2.7-46) is the exactly same as the solution for the case of homogeneous B.C.’s.
It implies that the boundary values u0 and v0 will not affect the solution in any aspects.
This can not be right!! What is wrong?
method 2: As we did in Eq. (2.7-21), consider
Z
`
`
[φn u00 − uφ00n ] dx = J(u, φn )|0
0
(2.7-47)
The RHS of the equation will not vanish this time (why?), we have
Z
`
Z
φn (x)f (x) dx + λn
0
0
`
`
φn (x)u(x)dx = − (uφ0n − φn u0 )|0
(2.7-48)
By using the B.C. of u(x) and the result of the eigenfunctions, the RHS of Eq. (2.7-48)
can be simplified as
RHS = [u0 − (−1)n v0 ]
nπ
`
(2.7-49)
Similarly, the LHS of Eq. (2.7-48) can be simplified as
Z
LHS =
`
φn (x)f (x) dx + λn c∗n
`
Z
0
φ2n (x)dx
(2.7-50)
0
Hence the solution u(x) is concluded as
R`
∞
X
[u0 − (−1)n v0 ] nπ
` − 0 φn (ξ)f (ξ) dξ
u(x) =
φn (x)
R`
λn 0 φ2n (η) dη
n=1
(2.7-51)
with φn (x)) defined in Eq. (2.7-42).
method 3: finite transform. The result will be the same as the one by method 2.
method 4: The solution of (2.7-40) can also be found by the aid of superposition as one would
do in Engineering Mathematics by considering
x
x
u = u0 1 −
+ v0 + v(x)
(2.7-52)
`
`
By substituting Eq. (2.7-52) into Eq. (2.7-40), it concludes that the BVP for v(x) will
have only homogeneous boundary conditions as

00

, 0<x<`
 v = f (x)
(2.7-53)
v(0) = 0

 v(`) = 0
The problem (2.7-53) can then be solved by eigenfunction expansion as described in
Section 2.7.2. The associated eigenvalue problem is still as Eq. (2.7-41), the eigenvalues
and eigenfunctions are λn = (nπ)2 /`2 , φn (x) = sin(nπx/ell). From Eqns. (2.7-26) and
(2.7-52), the final solution u(x) of BVP (2.7-40) can be written as
"R`
#
∞
x
x X 0 φn (ξ)f (ξ) dξ
φn (x)
(2.7-54)
u(x) = u0 1 −
+ v0 +
R
`
` n=1 −λn ` φ2n (η) dη
0
60
It is easily to verify that the first part in Eq. (2.7-51) is actually the coefficient of the
eigenfunction expansion of u0 (1 − x/`) + v0 x/`, as one would expect. Hence Eqns. (2.7-51) and
(2.7-54) conclude the same result. In order to have some feeling about the speed of convergence
of series Eqns. (2.7-51) and (2.7-54), let us consider a special case as f (x) = x, u0 = 0, v0 = 1,
and ` = 1. From Engineering Mathematics, the ODE in (2.7-40) has closed form solution as
x3
(2.7-55)
6
By using the boundary conditions, the constants A and B are determined as A = 0 and B = 5/6,
hence we have
u(x) = A + Bx +
5x x3
+
6
6
The results directly from Eqns. (2.7-51) and (2.7-54) yield, respectively,
∞ X
−(−1)n
(−1)n
u(x) = 2
+
sin nπx
nπ
(nπ)3
n=1
u(x) =
u(x) = x + 2
∞
X
(−1)n
sin nπx
(nπ)3
n=1
(2.7-56)
(2.7-57)
(2.7-58)
It can be easily verified that the summation in Eq. (2.7-57) is in fact the Fourier Sine series (i.e.
eigenfunction expansion associated to BVP (2.7-40)) of (5x + x3 )/6. Similarly, the summation
in Eq. (2.7-58) is the Fourier Sine series of (x3 − x)/6. Hence both results of Eq. (2.7-57)
and (2.7-58) will certainly converge to Eq. (2.7-56). On the other hand, the coefficients of the
summation in Eq. (2.7-57) are of the order of O(1/n), as n → ∞, while in Eq. (2.7-58) are
of the order of O(1/n3 ). Obviously, the convergence of Eq. (2.7-58) will be much better than
Eq. (2.7-57). The numerical results of Eqns. (2.7-57) and (2.7-58), (with partial sums of first
two, five and fifty terms) together with the closed form result Eq. (2.7-56), are shown in the
figure 2.3(a) and (b), respectively. It is easily to see that, it takes at least 50 terms to have a
reasonable result in the subfigure (a), i.e. the series (2.7-57). And the series converges to zero
at both two ends, even at x = 1 where boundary conditions require that u = 1. Moreover, there
is a Gibbs phenomena near x = 1. (why not near x = 0?) On the other hand, in the subfigure
(b), the series (2.7-58) converges to the closed form solution very rapidly (even with N = 2).
2.7.4
Non-Self-Adjoint Eigenvalue Problems
To illustrate the point, let us consider a BVP as

00
0

, 0<x<1
 u + 4u + 3u = 3
u(0) = 0

 u(1) = 0
(2.7-59)
method 1: From engineering mathematics, the general solution of the DE is
u = uh + up = Ae−x + B −3x + 1
(2.7-60)
The constants A and B can be determined from boundary conditions, we then have the
solution as
u=
e−1
1
− 1 − e−3 e−x + 1 − e−1 e−3x + 1
−3
−e
61
(2.7-61)
1.5
1.5
N=50
N=5
1.0
1.0
u (x )
u (x )
0.5
0.5
N=2, 5, 50
0.0
0.0
N=2
(a)
(b)
-0.5
-0.5
0.0
0.4
x
0.8
1.2
0.0
0.4
x
0.8
1.2
Figure 2.3: (a) Results of Eq. (2.7-57); and (b) results of Eq. (2.7-58). Both with partial sums
of first two, five and fifty terms, and the closed form solution Eq. (2.7-56)
method 2: Green’s function approach
Let us consider the associated Green’s function problem

00
0

, 0<x<1
 G + 4G + 3G = δ(x − ξ)
G(0) = 0

 G(1) = 0
(2.7-62)
As before, since the ODE is homogeneous at x 6= ξ, we have
(
G(x; ξ) =
A1 e−x + B1 e−3x
A2 e−x + B2 e−3x
,
0<x<ξ
ξ<x<1
(2.7-63)
Again, (2.7-63) has four not-yet-determined constants. Boundary conditions at both two
ends, and continuity condition and jump condition at x = ξ are enough to determine
these four constants. Then
Z
u(x) = 3
1
G(x; ξ)dξ
(2.7-64)
0
If the integral in Eq. (2.7-64) is carried out, it should lead to exactly the same result as
Eq. (2.7-61).
method 3: eigenfunction expansion – 1
62
The associated eigenvalue problem is6

00
0

, 0<x<1
 φ + 4φ + λφ = 0
φ(0) = 0

 φ(1) = 0
(2.7-65)
It can be found that the eigenvalues are λ = n2 π 2 + 4, and the eigenfunctions are
φn (x) = e−2x sin nπx
(2.7-66)
Notice that {φn (x)} does not even form an orthogonal set, since, even for m 6= n,
Z
1
1
Z
e−4x sin(mπx) sin(nπx)dx 6= 0
φm (x)φn (x)dx =
0
(2.7-67)
0
This is not surprised, since the differential operator of eigenvalue problem Eq. (2.7-65) is
not self-adjoint. Let us find the adjoint eigenvalue problem by considering
Z
0
1
ψ (φ00 + 4φ0 + λφ)dx
=
0
Z
1
=
0
1
φ (ψ 00 − 4φ0 + λψ)dx + [ψ(x)φ0 (x) − ψ 0 (x)φ(x) + 4φ(x)ψ(x)]|0 (2.7-68)
Hence, the associated adjoint eigenvalue problem is

00
0

, 0<x<1
 ψ − 4ψ + λψ = 0
ψ(0) = 0

 ψ(1) = 0
(2.7-69)
The adjoint eigenfunctions are found, with the same eigenvalue λn = n2 π 2 + 4, as
ψn (x) = e2x sin nπx
(2.7-70)
Notice that φ and ψ are bi-orthogonal, since for m 6= n
Z
1
Z
φm (x)ψn (x)dx =
0
1
sin(mπx) sin(nπx)dx = 0
(2.7-71)
0
To find the solution for (2.7-59) via eigenfunction expansion, let us construct again the
solution as
u(x) =
∞
X
An φn (x) = e−2x
n=1
6 The
∞
X
An sin(nπx)
n=1
DE of the associated eigenfunction problem can also be considered as
φ00 + 4φ0 + (λ̄ + 3)φ = 0
Then the eigenvalues are λ̄ = n2 π 2 + 1, and the eigenfunctions are still as φn (x) = e−2x sin nπx.
63
(2.7-72)
By using term by term differentiation (why??), u0 and u00 can be expressed as
u0 (x)
u00 (x)
= e−2x
= e−2x
∞
X
n=1
∞
X
[−2 sin(nπx) + nπ cos(nπx)]An
(2.7-73)
[4 sin(nπx) − 4nπ cos(nπx) − n2 π 2 sin(nπx)]An
(2.7-74)
n=1
Substitute into DE of (2.7-59), we have
−2x
e
∞
X
An · (−n2 π 2 − 1) · sin(nπx) = 3
(2.7-75)
n=1
To find the coefficient An , we have to use the bi-orthogonal property between φn and ψn
now, since φn itself does not form an orthogonal set. Multiplying both side of (2.7-75) by
ψm , and integrating over [0, 1], we have
An · (−n2 π 2 − 1) ·
1
2
Z
1
= 3
e2x sin nπx dx
0
Z
= 3Im
1
e(2+inπ)x dx
0
1
1
e(2+inπ)x
2 + inπ
0
2 − inπ 2
(e
cos
nπ
−
1)
= 3Im
n2 π 2 + 4
3nπ 1 − (−1)n e2
=
2
2
n π +4
= 3Im
(2.7-76)
where Im stands for imaginary parts. Hence
−2x
u(x) = e
∼∞
n=1
6nπ (−1)n e2 − 1
sin nπx
(n2 π 2 + 1)(n2 π 2 + 4)
(2.7-77)
method 4: eigenfunction expansion – 2
In method 3, we introduce not only an associated eigenvalue problem but also the adjoint
eigenvalue problem in order to establish the orthogonality relation of eigenfunctions. This
inconvenience is due to the fact that the eigenvalue problem is not self-adjoint, On the
other hand, as we mentioned in Eq. (2.7-13) that it is possible to make the differential
operator as formally self-adjoint by multiplying an integration factor. As suggested in
Eq. (2.7-13), the integration factor will be
R
r(x) =
e
R
(a1 /a2 )dx
a2
=
e
(4/1)dx
1
= e4x
(2.7-78)
The associated eigenvalue problem now is now adjusted by multiplying the factor e4x as

4x 00
4x 0
4x

, 0<x<1
 e φ + 4e φ + λe φ = 0
(2.7-79)
φ(0) = 0

 φ(1) = 0
64
Obviously, the integration factor r(x) does not affect either the eigenvalues nor the eigenfunctions. We still have eigenvalues as λ = n2 π 2 + 4, and the eigenfunctions as
φn (x) = e−2x sin nπx
(2.7-80)
Since the eigenvalue problem Eq. (2.7-79) is now self-adjoint, Eqns. (2.7-3) and (2.7-9)
suggest that the orthogonality relation of φn is, with ρ(x) = e4x , for n 6= m
Z
1
ρφm φn dx = 0
(2.7-81)
0
To find the solution for (2.7-59) via eigenfunction expansion, let us construct the solution
again as Eq. (2.7-72). Substitute the expansion into the DE of (2.7-59), we have the same
result as Eq. (2.7-75)
e−2x
∞
X
An · (−n2 π 2 − 1) · sin(nπx) = 3
n=1
To find the coefficient An , we now use the orthogonality relation (2.7-81), i.e. multiply
both side of Eq. (2.7-75) by ρφm and then integrate over [0, 1], we then have exactly the
same result as Eq. (2.7-77).
method 5: eigenfunction expansion – 3
From (2.7-72), it suggests that the solution u(x) would have the form of
u(x) = e−2x w(x)
(2.7-82)
Substitute (2.7-82) into BVP (2.7-59), we have a BVP for w(x) as

00
2x

 w − w = 3e
w(0) = 0

 w(1) = 0
, 0<x<1
(2.7-83)
Of course, BVP (2.7-83) can be solved by elementary approach as w(x) = wh (x) + wp (x)
for the case with simple RHS in the ODE. In order to compare the solution obtained
in method 3, let us solve the BVP of w again by eigenfunction expansion. This will be
particularly useful when the RHS of ODE is complicated. Notice that BVP of w is now
self-adjoint, and the corresponding eigenfunctions are sin nπx. Construct w(x) as
w(x) =
∞
X
Bn sin(nπx)
(2.7-84)
n=1
It is easily to find that Bn is exactly the same as An of method 3. Hence all these three
methods related to eigenfunction expansion lead to the same result.
65
References for Ordinary Differential Equations
1. Birkhoff, G. and G. C. Rota (1989). Ordinary Differential Equations, 4th ed. John Wiley
& Sons.
2. Brauer, F. and J. A. Nohel (1967). Ordinary Differential Equations, Benjamin Inc.
3. Stakgold, I. (1979). Green’s Functions and Boundary Value Problems, John Wiley &
Sons.
4. Hirsch, M. W.and S. Smale (1974). Differential Equations, Dynamical Systems, and
Linear Algebra, Academic Press.
5. Lomen, D. and L. Mark (1988). Differential Equations, Prentice-Hall.
66
2.8
Problems
1. Let u(x) be a solution of u00 = q(x)u, q(x) > 0, such that u(0) and u0 (0) are positive.
Show that uu0 and u(x) are increasing for x > 0.
2. Let f (x) and g(x) be two linearly independent solutions of
u00 + p(x)u0 + q(x)u = 0
,
p(x) & q(x) are continuous.
Show that p(x) = (gf 00 −f g 00 )/W , and q(x) = (f 0 g 00 −g 0 f 00 )/W , where W is the Wronskian
of f (x) and g(x).
3. In which domains do the following functions satisfy a Lipschitz condition in y?
(a) F (x, y) = sec x.
(b) F (x, y) = tan y.
(c) F (x, y) = 1 + y 2 .
(d) F (x, y) = y/(1 − x2 ).
(e) F (x, y) = x/(1 − y 2 ).
4. Examine the following problems to see if they have a unique solution in the domain D.
If they do, in which region of D do the solutions exist?
(a) y 0 =
(b) y 0 =
1
(x2 −1)(1−4y 2 ) ,
2
x
cos y ,
D = (0 ≤ x ≤ 1, 1 ≤ y ≤ 2).
y(0) = 1,
D = (−π ≤ x ≤ π, −π ≤ y ≤ π).
y(0) = π/2,
5. (Generalized Lipschitz condition) Let F be continuous and satisfy
|F (x, y2 ) − F (x, y1 )| ≤ k(x)|y2 − y1 |
Ra
on the strip 0 < x < a. Show that, if the improper integral 0 k(x) dx is finite, then y 0 =
F (x, y) has at most one solution satisfying y(0) = 0. [Hint: Define σ(x) = [y2 (x) − y1 (x)]2
and show that σ(x) possesses certain decreasing property.]
6. Let y1 and y2 satisfy y 0 = F (x, y) in a domain D, where F (x, y) satisfies a Lipschitz
condition in y. Show that
|y2 (x) − y1 (x)| ≤ ec|x−a| |y2 (a) − y1 (a)|
for x ≤ a, where c is the Lipschitz constant. [Hint: Define t = a − x, and show that
d[σ(t)e−2ct ]/dt ≤ 0.]
7. Consider the matrix differential equation Y0 = AY + YB, where A, B and Y are n × n
matrices.
(a) Show that the solution satisfying the initial condition Y(0) = C, where C is a given
n × n matrix is given by
Y(t) = exp(At) C exp(Bt)
67
(b) Show that
Z
∞
Z=−
exp(At) C exp(Bt) dt
0
is the unique solution of the matrix equation AX + XB = C whenever the integral
exists.
(c) Show that the integral for Z given in part(b) exists if all eigenvalues of both A and
B have negative real parts.
[From Brauer & Nohel, ODE,p. 311 prob. 10, 1967]
8. Suppose m is not an eigenvalue of the matrix A. Show that the nonhomogeneous system
y0 = Ay + cemt , has a solution of the form φ(t) = pemt , and calculate the vector p in
terms of A and c. [From Brauer & Nohel, ODE,p. 309 prob. 4, 1967]
9. Find the general solution of the system y0 = Ay + b(t) in each of the following cases
"
#
!
0 1
2et
(a)
A=
b(t) =
1 0
t2
!
"
#
4 −3
sin t
(b)
A=
b(t) =
2 −1
−2 cos t
[From Brauer & Nohel, ODE,p. 309 prob. 3.(a) and 3.(c), 1967]
10. Solve the following system of differential equations:
2ẋ1 + x1 + ẋ2 + x2 = et
ẋ1 − 7x1 + ẋ2 − 5x2 = 0
x1 (0) = 1
,
x2 (0) = 0
11. Solve the following system of differential equations using eigenvectors or generalized eigenvectors:
!
!
!
!
!
1
y10
1 1
y1
y1 (0)
−1
,
=
=
2
−1 3
y2
y2 (0)
1
y20
12. Consider ODE with a matrix A as

dy(x)
= Ay
dx
and
y(0) = b
where
2

A= 0
0
(a) Compute exp(xA).
(b) Find the solution of the IVP.
13. Solve the following system of differential equations:
ẋ = Ax
,
x(0) = b
68

1 0

2 0 
0 2


1


b= 0 
−1


1 −2 −1


A= 1
2
0 
−1 −3
0
!
4 −2
A=
b=
5 −2
(a)
(b)


1
 
b= 1 
1
!
1
2
14. Consider ODE with a matrix A as
dy(x)
= Ay
dx
,
y(0) = b
where



A=

1
0
0
0
1
1
0
0
2
3
2
0
0
0
2
1








b=

1
0
−1
2





Knowing that the eigenvalues of A are λ1 = λ2 = λ3 = 1, λ4 = 2
(a) Find the eigenvectors and generalized eigenvectors of A.
(b) Compute exp(xA).
(c) Find the solution of the IVP.
P∞
P∞
15. Assume that the matrix series k=1 Ck converges. Prove that the matrix series k=1 (ACk B)
also converges and its sum is the matrix
A
∞
X
!
Ck
B.
k=1
Here A and B are matrices such that the products ACk B are meaningful.
16. This example shows that the equation eA+B = eA eB is NOT always true for matrix
exponentials. Compute each of the matrices eA eB , eB eA , eA+B when
A=
1 1
0 0
!
and
B=
1 −1
0 0
!
,
and note that the three results are distinct.
17. If A(t) is a scalar function of t, the derivative of eA(t) is eA(t) A0 (t). Compute the derivative
of eA(t) when
!
1 t
A(t) =
0 0
and show that the result is NOT equal to either of the two products eA(t) A0 (t) or
A0 (t)eA(t) .
69
18. If
A(t) =
t
0
t−1
1
!
,
Prove that
eA(t) = e A(e(t−1) ).
19. Compute the exponential eA for
!
α 0
A=
,
A=
0 β
α
µ
−µ
α
!
,
A=
λ
0
1
λ
!
.
20. Compute the exponential eA for
(a)





A=



0 1
0 0
0 0
···
···
···
0
1
0
0
0
1

··· 0
··· 0 

··· 0 

.



0 1 
0 n×n
(b)





A=



a 1 0 0 ··· 0
0 a 1 0 ··· 0
0 0 a 1 ··· 0
···
···
a 1
···
a









.
n×n
21. Use the Cayley-Hamilton theorem to show that the inverse of a non-singular matrix A
can be expressed as a polynomial of degree n − 1 in A.
22. Express etA as a polynomial in A if
(a)


1 0 3


A =  8 1 −1  .
5 1 −1
(b)



A=

3 −1 −4 2
2 3 −2 −4
2 −1 −3 2
1 2 −1 −3



.

70
23. Let A be an n × n matrix with all its eigenvalues equal to λ. Show that
n−1
X
etA = eλt
k=0
tk
(A − λ I)k ,
k!
where I is an n × n identity matrix.
24. Let Lk (λ) be the polynomial in λ of degree n − 1 defined by the equation
n Y
λ − λj
Lk (λ) =
,
λk − λj
j=1
j6=k
where λ1 , · · · , λn are n distinct scalars.
(a) Prove that
(
Lk (λi ) =
0
1
if λi 6= λk ,
if λi = λk .
(b) Let y1 , · · · , yn be n arbitrary scalars, and let
p(λ) =
n
X
yk Lk (λ).
k=1
Prove that p(λ) is the only polynomial of degree ≤ n − 1 which satisfies the n
equations
p(λk ) = yk
for k = 1, 2, · · · , n.
Pn
(c) Prove that k=1 Lk (λ) = 1 for every λ, and deduce that for every square matrix A
we have
n
X
Lk (A) = I,
k=1
where I is the identity matrix and
n Y
A − λj I
for
Lk (A) =
λk − λ j
j=1
k = 1, 2, · · · , n.
j6=k
25. If A is an n × n matrix with n distinct eigenvalues λ1 , · · · , λn , then prove that
etA =
n
X
etλk Lk (A),
k=1
where Lk (A) is a polynomial in A of degree n − 1 given by the formula
n Y
A − λj I
Lk (A) =
λk − λj
j=1
for
k = 1, 2, · · · , n.
j6=k
Pn
(Hint: Consider f (t) = k=1 etλk Lk (A) and show that ḟ (t) = Af (t), f (0) = I, where I
is the identity. Conclude the proof using the uniqueness theorem.)
71
26. Find that solution of the system
ẋ1
= x1 + x2 + sin t
ẋ2
=
2x1 + cos t
such that x1 (0) = 1, x2 (0) = 1.
27. We note that (t + 1)3 and (t + 1)−2 solve
(t + 1)2 φ00 − 6φ = 0,
(t ≥ 0).
Find the fundamental matrix solution φ(t) for the system of equations
φ̇ = ψ,
ψ̇
=
6(t + 1)−2 φ.
28. Recall the problem 27, and solve the initial-value problem
φ̇ = ψ + e−t ,
ψ̇
=
6(t + 1)−2 φ +
√
t
with φ(0) = 1 and ψ(0) = 2.
29. Let E(t) = eA(t) .
(a) Let A(t) be a polynomial in t with matrix coefficients, say
A(t) =
m
X
tr A r ,
r=0
where the coefficients commute, Ar As = As Ar for all r and s. Prove that Ė(t) =
Ȧ(t)E(t) on (−∞, ∞).
(b) Solve the homogeneous linear system
ẋ(t) = (I + tA)x,
x(0) = u
on the interval (−∞, ∞), where A is an n × n constant matrix and I is the identity
matrix.
30. Let A be an n × n constant matrix. The system tẋ = Ax is analogous to the second order
Euler equation. Assuming that x = tr u, where u is a constant vector, show that u and
r must satisfy
(A − r I) u = 0;
also show that to obtain nontrivial solutions of the given differential equation, r must be
a root of the characteristic equation
det(A − r I) = 0.
31. Referring the problem 30, solve the given system equations in the following. Assume
t > 0.
72
(a)
!
tẋ =
2 −1
3 −2
tẋ =
−1 −1
2 −1
tẋ =
3 −4
1 −1
x;
(b)
!
x;
(c)
!
x;
(d)

3 −1

tẋ =  2 0
1 −1

1

1  x.
2
32. Consider the initial-value problem
ẋ = Ax,
x(0) = b,
(2.8-1)
where A is a constant matrix, and b a prescribed vector.
(a) Assuming that a solution x = x(t) of equation (2.8-1) exists, show that it must
satisfy the integral equation
Z t
x(t) = b +
Ax(s) ds.
(2.8-2)
0
(b) Start with the initial approximation x(0) (t) = b. Substitute this expression for x(s)
in the right hand side of equation (2.8-2), and obtain a new approximation x(1) (t).
Show that
x(1) (t) = (I + At)b.
(2.8-3)
(c) Repeat this process by substituting the expression in equation (2.8-3) for x(s) in the
right hand side of equation (2.8-2) to obtain a new approximation x(2) (t). Show that
t2
x(2) (t) = I + At + A2
b.
(2.8-4)
2
(d) By continuing this process obtain a sequence of approximations x(0) , x(1) , · · · , x(n) , · · ·.
Show that by induction that
t2
tn
x(n) (t) = I + At + A2 + · · · + An
b.
(2.8-5)
2
n!
73
(e) The sum appearing in equation (2.8-5) is a sum of matrices. It is possible to show
that each component of this sum converges as n → ∞, and therefore a new matrix
B(t) can be defined as
X
∞
t2
tn
tk
D(t) = lim I + At + A2 + · · · + An
=
Ak .
(2.8-6)
n→∞
2
n!
k!
k=0
Assuming that term-by-term differentiation is permissible, show that D(t) satisfies
Ḋ(t) = AD(t), D(0) = I and show that the solution of the initial-value problem
(2.8-1) can be written as x = B(t)b.
33. Let δ(x) denote Dirac delta function,
R3
(a) evaluate the distribution: (i) 0 x2 δ 00 (x−1)dx
(b) evaluate the products
(ii)
R3
−1
(x2 −4x+8)δ 0 (x)dx
e2x δ(x − 2) + x2 δ 00 (x − 1)
34. Let H(x) denote the Heaviside step function (unit step function),
(a) find the first and the second derivatives of the following functions in terms of generalized functions

(

x≤0
 0
2
cos x
x≤1
2
(i) f (x) =
(ii)
f
(x)
=
2x
−
x
0<x<2

1 + x2
x>1
 0
x≥2
(b) evaluate the distribution
Z 3
(i)
x2 δ 0 (x − 2)dx
Z
3
(ii)
(x2 − 3x + 5)δ 00 (x)dx
−1
0
(c) evaluate the products
(ii) x2 δ 00 (x − 4)
(i) e2x δ(x)
35. If w(x) is a nonnegative function satisfying
Z
∞
w(x) dx = 1,
−∞
then
δk (x) = k w(k x)
is a δ-sequence such that
Z
∞
lim
k→∞
δk (x)h(x) dx = h(0),
−∞
where h is bounded (i.e., |h(x)| ≤ M for all x and for some M > 0) and continuous at
x = 0 (i.e., for every > 0, there exists δ > 0 such that |h(x) − h(0)| < whenever
|x| < δ).
36. Prove that
(a) ex δ(x) = δ(x).
74
(b) x δ 0 (x) = −δ(x).
(c)
d4
3
dx4 |x|
= 12 δ(x).
37. Show that
1
Z
|x − ξ|φ(ξ)dξ
u(x) =
0
satisfies the differential equation u00 (x) = 2φ(x).
38. Show that if f (x) is a monotonic increasing or decreasing function of x which vanishes
for x = ξ, then
δ(x − ξ)
.
|f 0 (ξ)|
δ[f (x)] =
As a result, show that
Z
∞
δ(ax − b)f (x)dx = |a|−1 f (b a−1 )
−∞
if a 6= 0 and δ(x) = δ(−x).
39. Determine the adjoint operator L∗ (including L∗ and B∗ ) for each of the following operators:
(a) L =
d2
dx2 over 0 ≤ x
d3
d
dx3 − sin x dx
00
≤ 1; B1 (u) = u(0) = 0, B2 (u) = u0 (0) = 0.
(b) L =
+ 3 over 0 ≤ x ≤ π; B1 (u) = u(0) = 0, B2 (u) = u0 (0) = 0,
B3 (u) = u (0) − 4u(π) = 0.
(c) L =
(d) L =
d
dx + 1 over 0 ≤ x < ∞;
d2
dx2 + 1 over 0 ≤ x ≤ 1;
B(u) = u(0) = 0.
B1 (u) = u(0) = 0, B2 (u) = u(1) = 0.
40. Show that even if the differential operator
L = a(x)
d
d2
+ b(x)
+ c(x)
dx2
dx
is NOT formally self-adjoint, σL will be if we choose
Z
σ = exp
b − a0
dx.
a
As an application of this result, show that the inhomogeneous Bessel equation of order k,
x2 u00 + xu0 + (x2 − k 2 )u = φ,
can be made formally self-adjoint by multiplying the whole equation through by
1
x.
41. To see why we required homogeneity of the boundary conditions in determining the adjoint
operator L∗ , let us try to determine L∗ for a case where the boundary conditions are
not homogeneous. For example, consider L to consists of the differential operator L =
75
d2
dx2
d
+ dx
+ x over 0 ≤ x ≤ 1, plus the boundary conditions u(0) = a, u(1) = 0. Show that
three boundary conditions,
v(0) = v(1) = v 0 (0) = 0,
are needed for the adjoint operator, whereas L∗ is only of seconder order! This is generally
known as a sticky wicket.
42. Find the Green’s function, and hence the solution for each of the following systems, where
the interval is 0 ≤ x ≤ 1 in each case.
(a) u00 − k 2 u = φ; u(0) − u0 (0) = a, u(1) = b.
(b) u00 = φ; u(0) = 0, u0 (0) = 0.
(c) u00 + 2u0 + u = 0; u(0) = 0, u0 (0) = 1.
43. Solve the problem
(xu0 )0 −
1
u = 1,
x
u(0) = u(1) = 0,
by the method of Green’s functions.
44. Find the Green’s function, and hence the solution, of the system
u0000 = φ,
u(0) = 0, u0 (0) = 2u0 (1), u(1) = a, u00 (0) = 0.
45. Determine the adjoint system and the extended definition of the following problems

0 0

a < x < b, p(x) < 0
 Lu = −[p(x)u ] + q(x)u(x),
(a)
B1 u : u(a) = u(b)

 B u : p(a)u0 (a) = p(b)u0 (b)
2

2 00
0

0<x<1
 Lu = (6x + 3x )u (x) + 6u (x) + 12xu(x),
0
(b)
B1 u : u (0) = 0,

 B u : u0 (1) = α
2
46. Solve each of the following boundary value problems by constructing an appropriate
Green’s function (using Green’s function, modified Green’s function, or adjoint Green’s
function, and/or finding the least square solution if necessary). Indicate any consistency
condition required for solvability.

00

0<x
 u (x) + u(x) = f (x),
(a)
u(0) = a

 u0 (0) = b

00
0

0<x<π
 u (x) + 4u (x) = f (x),
(b)
u(0) = a

 u0 (π) = 0

00

0<x<π
 u (x) + u(x) = f (x),
(c)
u(0) = 0

 u0 (0) = u(π)
76
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
(n)
(o)
(p)
(q)

00

0<x<π
 u (x) − u(x) = f (x)
0
u(0) = u (0)

 u(π) = u0 (π)

2 00
0

t>1
 t u (t) − 3tu (t) + 3u(t) = f (t)
u(1) = 0

 u0 (1) = 0

00

0<x<π
 u (x) + u(x) = f (x),
0
u(0) = u (0)

 u(π) = u0 (π)

00

0 < x < 2π
 u (x) + u(x) = f (x),
u(0) − u0 (0) = α

 u(2π) − u0 (2π) = β

00

0<x<π
 u (x) − 4u(x) = f (x),
u(0) = 0

 u(π) = u0 (π)

00
0

0<x<1
 u (x) + 5u (x) + 6u(x) = f (x),
u(0) = α

 u(1) = 0

00

0<x<1
 u (x) = f (x),
u(0) = 0

 u0 (1) = 0

00

0<t<∞
 u (t) = f (t),
u(0) = u0

 u0 (0) = v
0

00

0<x<π
 u (x) + u(x) = f (x),
u(0) + u(π) = 0

 u(π) + u0 (π) = 0

00
0

,
0<x<1
 u −u =1
u(0) = 0

 u0 (1) = 0

00
0

,
0<x<a
 u − u − 2u = 1
0
u (0) = 1

 u(a) = 0

00

,
0<x<π
 u + u = f (x)
u(0) = 1

 u(π) = 0

00
0

,
0<x<1
 u − u = f (x)
u(0) = 0

 u0 (1) = 0

00
0

,
1<x<∞
 xu + u = f (x)
u(x) is bounded as x → ∞

 u(1) = 1
77
47. Determine the adjoint system and the extended definition of the following problems

0 0

a < x < b, p(x) < 0
 Lu = −[p(x)u ] + q(x)u(x),
(a)
B1 u : u(a) = u(b)

 B u : p(a)u0 (a) = p(b)u0 (b)
2

2 00
0

0<x<1
 Lu = (6x + 3x )u (x) + 6u (x) + 12xu(x),
(b)
B1 u : u0 (0) = 0,

 B u : u0 (1) = α
2
48. Consider the following second order differential equation:
y 00 + y 0 − 6y = ex
y(0) = 0
,
,
0<x<1
y(1) = 0
(a) Construct Green’s function for the above equation.
(b) Solve the inhomogeneous differential equation using Green’s function.
49. Construct Green’s function and adjoint Green’s function for the following initial value
problem:
x2 u00 − (x2 + 2x)u0 + (x + 2)u = 2
u(1) = 0
,
x>1
0
,
u (1) = 1
Then, solve the initial value problem. [Hint: u(x) = x is a solution to the differential
equation.]
50. Consider the following problem:
u00 + 16u = 1
u(0) = 1
,
,
0<x<b
u(b) = 0
(a) For what value of b will the boundary value problem has exactly one solution?
(b) Show that the consistency condition of the problem is 1 − cos 4b = 16.
(c) Is it possible that the problem has an infinite number of solutions?
51. Consider the boundary value problem
y 00 − 2y + (1 + λ)y = 0
y(0) = 0
,
,
0<x<1
y(1) = 0
(a) Is this problem self-adjoint? Explain.
(b) Show that all the eigenvalues are real without calculating the eigenvalues.
(c) Are the eigenfunctions orthogonal?
(d) Give a proof or a counter-example to justify your answer. Do not apply the eigenfunctions directly.
(e) Determine the eigenvalues and eigenfunctions of the problem.
78
52. Let
= p(x)u00 + q(x)u0 + r(x)u = f,
Lu
a < x < b,
0
B1 (u)
= α11 u(a) + α12 u (a) + β11 u(b) + β12 u0 (b) = γ1 ,
B2 (u)
= α21 u(a) + α22 u0 (a) + β21 u(b) + β22 u0 (b) = γ2 .
Suppose L = L∗ . Show that the necessary and sufficient condition for self-adjointness of
the above system is
p(a)P34 = p(b)P12 ,
where
P12 = α11 α22 − α12 α21 ,
P34 = β11 β22 − β12 β21 .
Note that this condition is satisfied for unmixed boundary condition and for the periodic
boundary conditions u(a) = u(b), p(a)u0 (a) = p(b)u0 (b).
53. Referring to Problem 52. Suppose p = 1, q = 4, r = −3, α11 = 4, α12 = 1, β11 = β12 = 0,
α21 = α22 = 0, β21 = 4, β22 = 1. Find L∗ , B∗1 , B∗2 .
54. Consider
u00 + a0 u = f (x),
0 < x < 1,
u(0) = γ1 ,
u(1) = γ2 ,
where a0 is a fixed constant. For what values of a0 do the above BVP have at most
one solution irrespective of f, γ1 , γ2 ? Further, using the direct calculation shows that the
above BVP has no solutions if a0 = π 2 , f = 1, γ1 = γ2 = 0.
55. Find the solvability conditions for
−u00 − u
−π < x < π;
= f,
u(π) − u(−π)
= γ1 ,
0
= γ2 .
0
u (π) − u (−π)
56. Let L, B1 , B2 be the linear second order differential operator and linear functionals
defined in Problem 52. Consider
Lu = f,
a < x < b,
B1 u = γ1 ,
B2 u = γ2 .
Suppose the homogeneous system of the above BVP has only trivial solution. Then show
that the solution of the above BVP is given by
Z
b
G(x; ξ)f (ξ)dξ +
u=
a
γ2
γ1
u1 (x) +
u2 (x),
B2 u 1
B1 u 2
where G(x; ξ) is the Green’s function of the above BVP, u1 is a nontrivial solution of the
homogeneous equation Lu = 0 satisfying B1 u1 = 0, and u2 is a nontrivial solution of the
same homogeneous equation with B2 u2 = 0.
79
57. Consider
u00
= f (x),
0
u (0) − u(1)
= γ1 ,
u0 (1)
= γ2 .
0 < x < 1,
Find direct Green’s function G(x; ξ) and adjoint Green’s function G∗ (x; ξ) of the above
system, and show, by comparing explicit formulas, that G(x; ξ) = G∗ (ξ; x).
58. Referring to Problem 57. Solve the full Problem 57 in terms of f, γ1 , γ2 .
59. Consider
0
(a(x)u0 ) + c(x)u
= f (x),
0
= m1 ,
0
= m2 .
−l1 u (α) + h1 u(α)
l2 u (β) + h2 u(β)
α < x < β,
Suppose the homogeneous system has only trivial solution and l1 6= 0, l2 6= 0. Show that
β
Z
G(x; ξ)f (ξ)dξ −
u(x) =
α
m1
m2
a(α) G(x; α) −
a(β) G(x; β),
l1
l2
where G(x; ξ) is its Green’s function.
60. Referring to problem 59. Suppose l1 = l2 = 0, h1 = h2 = 1. Show that
β
Z
u(x) =
G(x; ξ)f (ξ)dξ + m2 a(β)
α
G(x; α)
∂G(x; β)
− m1 a(α)
.
∂ξ
∂ξ
61. Consider
0
(a(t)u0 ) + c(t)u = f (t),
t > t0 ,
u(t0 ) = m1 ,
u0 (t0 ) = m2 .
Show that
Z
t
u(t) =
t0
∂G(t; t0 )
G(t; ξ)f (ξ)dξ + a(t0 ) m2 G(t; t0 ) − m1
,
∂ξ
where G(t; ξ) is the Green’s function of the causal fundamental solution of the above
system.
62. Solve the BVP

00

 u + 4u = f (x),
(a)
u(0) = 0

 u(π) = 0.

00

 u + u = f (x),
(b)
u(0) = u(2π),

 u0 (0) = u0 (2π).
0 < x < π,
0 < x < 2π,
80
63. Consider
Lu
B1 (u)
B2 (u)
=
0
(a(x)u0 ) + c(x)u = f (x),
α < x < β,
0
= −l1 u (α) + h1 u(α) = m1 ,
(2.8-7)
0
= l2 u (β) + h2 u(β) = m2 .
Suppose the homogeneous system has nontrivial solutions, say w(x), and l1 6= 0, l2 6= 0.
Show that
Z β
m1
m2
u(x) =
GM (x; ξ)f (ξ)dξ + Cw(x) −
a(α) GM (x; α) −
a(β) GM (x; β),
l1
l2
α
where C is a constant and GM (x; ξ) is the modified Green’s function and is obtained to
be symmetric.
•
Strum-Liouville System.
0
(a(x)u0 ) + {c(x) + λ p(x)}u
= 0,
0
= 0,
0
=
−l1 u (α) + h1 u(α)
l2 u (β) + h2 u(β)
α < x < β,
(2.8-8)
0.
No matter what the value of λ, the trivial function u ≡ 0 always satisfies (2.8-8), but
for certain value of λ, called eigenvalues, the system has nontrivial solutions. It can be
proved that there is always a countable (but infinite) number of such eigenvalues, which
we denote by λn (n = 1, 2, 3, · · ·). A solution of (2.8-8) corresponding to an eigenvalue λn
is called an eigenfunction and is denoted by un (x) = u(λn , x).
64. Suppose the constants h1 , h2 , l1 and l2 are real and independent of the parameter λ,
a, c, p, and a0 are real and continuous for α ≤ x ≤ β, and p > 0 and a > 0 for α ≤ x ≤ β.
Show that all eigenvalues of (2.8-8) are real, and eigenfunctions corresponding to distinct
eigenvalues are orthogonal with respect to the weight function p(x),
Z β
p(x)um (x)un (x) dx = 0
if n 6= m.
α
Note that the above is still true if a(α) = a(β) and u(α) = u(β), u0 (α) = u0 (β).
65. Find eigenvalues and eigenfunctions of the Strum-Liouville system
u00 + λu = 0,
0 < x < L,
0
u(0) = 0 = u (L).
•
(Eigenfunction Expansion). In fact, under certain conditions, it can be shown that
any piecewise smooth function f (x) can be expanded by the eigenfunctions un (x) of the
Strum-Liouville System (2.8-8):
f (x+) + f (x−)
2
=
cn
=
∞
X
cn un (x),
n=1
Z β
p(x)f (x)un (x) dx,
α
Z β
1 =
p(x)un (x)un (x) dx.
α
81
We will use this property to solve (2.8-7).
66. Let λn , un be the eigenvalues and eigenfunctions of
Lu(x) + λp(x)u(x) = 0,
B1 (u) = 0,
α < x < β,
(2.8-9)
B2 (u) = 0,
where L, B1 and B2 are defined in (2.8-7). Further, we assume the e eigenfunctions un is
normalized in the sense of
(
Z β
0 if m 6= n,
p(x)um (x)un (x) dx =
1 if m = n.
α
Suppose m1 = m2 = 0 in (2.8-7). Let u(x) be the solution of (2.8-7). Assume u is smooth.
So u(x) can be expanded to be
u(x)
=
∞
X
cn un (x),
n=1
Z β
cn
=
p(x)un (x)u(x) dx.
α
(a) Show that
Z β
(un Lu − uLun )dx = 0.
(2.8-10)
α
(b) Suppose λn 6= 0 for all n. This gives
Z β
1
un (x)f (x) dx.
cn = −
λn α
(2.8-11)
(c) Suppose λn 6= 0 for all n. Show that the Green’s function of (2.8-7) can be expressed
as
∞
X
un (x)un (ξ)
G(x; ξ) =
.
−λn
n=1
67. Solve the BVP using the eigenfunction expansion method
u00 = f (x),
u(0) = m1 ,
0 < x < L,
u0 (L) = m2 .
(Hint: Note that the right hand side of (2.8-10) is not equal to zero if m1 6= 0 and m2 6= 0.)
68. Suppose one of the eigenvalues of the Strum-Liouville system (2.8-9) is equal to zero, say,
λ1 = 0, and λn 6= 0 for n ≥ 2. Does the (ordinary) Green’s function of (2.8-7) exist? In
other words, (2.8-11) fails to hold. Referring to Problem 67 and assume m1 = m2 = 0 in
(2.8-7). Show that the consistency condition becomes
Z
β
u1 (x)f (x) dx = 0.
α
82
Further,
u(x) = C u1 (x) +
∞
X
(R β
α
n=2
un (ξ)f (ξ) dξ
−λn
)
un (x) ,
where C is some constant.
69. Find the transformation u(x) = v(x) y(x), such that
u00 + p(x)u0 + q(x)u = 0
,
can be replaced by the normal form
y 00 + I(x)y = 0
.
Write down I(x).
70. Consider the differential equation
L(y) ≡ −[p(t)y 0 ]0 + q(t)y = λr(t)y
,
where p, q, and r are real continuous functions on a ≤ t ≤ b with p > 0, r > 0, and where
[p(t)r(t)]1/2 and p0 (t) are continuous on a ≤ t ≤ b. Show that the changes of variables
(known as the Liouville trnasformation).
z
K
1
π
=
u
Z
1
K
=
Z
t
[r(s)/p(s)]1/2 ds
,
[r(s)/p(s)]1/2 ds
,
a
b
a
[p(s) r(s)]1/4 y
=
2
ρ = K λ
,
,
changes this differential equation to
−
d2 u
+ g(z)u = ρu
dz 2
(0 ≤ z ≤ π)
,
,
(2.8-12)
where
[f 00 (z)/f (z)] − K 2 k(z)
g(z)
=
k(z)
= q(t)/r(t)
f (z)
=
,
,
1/4
[p(t)r(t)]
.
[The form of the equation (2.8-12) is useful in studying the asymptotic expression for the
eigenvalues as n → ∞, for the Sturm-Liouville boundary value problem.]
71. Find the best solution of the following problem by using the appropriate eigenfunction
expansion (if it exists)

00

0 < x < 1,
 u + 4u(x) = f (x),
0
u (0) = α,

 u0 (1) = 0
83
72. Find the best solution of the following problem by using the appropriate eigenfunction
expansion (if it exists)

00

0 < x < 1,
 u + 4u(x) = f (x),
0
u (0) = α,

 u0 (1) = 0
73. Find the eigenvalues and eigenfunctions of the boundary value problem
y 00 + λr(t)y = 0
,
y(0) = 0,
y(π) = 0
,
where
(
r(t) =
(0 ≤ t ≤ π/2)
.
(π/2 ≤ t ≤ π)
4
1
Are the eigenfunctions and their derivatives continuous at t = π/2? (Hint: Find the
solutions of the differential equation satisfying respectively y(0) = 0 and y(π) = 0, and
choose λ to make the solution match at t = π/2.)
74. Find the eigenvalues and eigenfunctions of the boundary value problem
y 00 + λr(t)y = 0
,
y(0) = 0,
y(π/2) = 0 ,
y(π) = 0
.
75. For the eigenvalue problem
d4 φ
+ λex φ = 0
dx4
φ(0) = 0
,
,
0<x<1
φ(1) = 0
,
φ0 (0) = 0
,
φ00 (1) = 0
Show that the eigenvalues are less than or equal to zero (λ ≤ 0). Is λ = 0 an eigenvalue?
76. Find the eigenvalues and eigenfunctions of the following problems, and then solve the
problems by eigenfunction expansion.
(a) .
u00 + u = x
,
0
u (0) = 0
(b) .
0<x<π
,
u00 + 4u = 1
u(0) = 0
u(π) = 1
,
0<x<1
0
,
u (1) = 1
77. Consider the following boundary value problem
y 00 − 6y 0 + 8y = f (x)
y(0) = 0
,
,
0<x<1
y(1) = 0
(a) Find the solution by
(i) Green’s function method;
(ii) eigenfunction expansion with original eigenfunctions and adjoint eigenfunctions;
(iii) eigenfunction expansion with original eigenfunctions only.
84
(b) Consider the special case of f (x) = 8x + 2.
(i) Find the solution of the BVP explicitly by the method of undetermined coefficients or any other methods in Engineering math.
(ii) Work out the integral in (i) of (a) explicitly, and compare the result with (i).
(iii) Work out the expansion coefficients in (ii) or (iii) of (a) explicitly. Compare
your result with that in (i) numerically by a plot of y v.s. x. Discuss also the
convergence of the eigenfunction series based on your numerical results.
78. Consider the following BVP:
u00 + u = f (x)
u(0) = 1
,
,
0<x<π
u(π) = 0
(a) Solve the problem by the following two methods:
(i) Let
u(x) =
∞
X
an φn (x)
n=1
where φn (x) is the associated eigenfunction to the BVP.
(ii) Let
u(x) = w(x) +
∞
X
bn φn (x)
n=1
where w(x) satisfies the inhomogenous boundary conditions, and the summation part satisfies the needed differential equation as well as the homogeneous
boundary conditions.
(b) Consider the special case of f (x) = x, and find the solution of the BVP explicitly by the method of undetermined coefficients or other methods in Engineering
Mathematics.
(c) Work out the expansion coefficients an and bn in (i) and (ii) of (a) explicitly for f (x)
given in (b). Compare your results with that in (b) numerically by a plot of u vs
x. Discuss also the convergence of the series in (i) and (ii) of (a) according to your
numerical results.
85
Chapter 3
PDE
3.1
Classification of PDE
3.1.1
Introduction
Consider a second order PDE in two independent variables x, y as
au,xx + 2bu,xy + cu,yy = d
(3.1-1)
where subscript ‘,’ indicates partial derivatives. The status of linearity or nonlinearity of Eq.
(3.1-1) depends on coefficients a, b, c, and d.
3.1.1-1 linear equation
If a, b and c are functions only of x and y, and d is a linear function in u, u,x , and u,y , then Eq.
(3.1-1) is called a linear equation. In such a case, the equation (3.1-1) can be written as
au,xx + 2bu,xy + cu,yy = eu,x + f u,y + gu + h
(3.1-2)
where a, b, c, e, f, g and h are all functions of x and y only.
3.1.1-2 nonlinear equation
Depending on the behaviors of a, b, c and d, Eq. (3.1-1) can be classified further to indicate the
degree of nonlinearity.
semi-linear. If a, b and c are functions only of x and y, and d is function (may not be a linear
function) of x, y, u, u,x , u,y as d = d(x, y, u, u,x , u,y ), the Eq. (3.1-1) is called semi-linear
equation or half-linear equation.
quasi-linear. If a, b, c and d are functions of x, y, u, u,x , u,y , the Eq. (3.1-1) is called quasi-linear
equation.
(full) nonlinear. If Eq. (3.1-1) is other than linear, semi-linear or quasi-linear, then it is called
nonlinear equation.
86
3.1.2
Classification
In this chapter, we discuss only the classification of linear second order PDE’s with two independent variables. We would like to study the classification of PDE’s, because
1. There are some fundamental difference in solutions of each different type of equations.
2. Depending on types of equations, only some types of problems (such as initial-valued
types, boundary-valued types, or initial-boundary-valued types) are likely to have solutions or to be well-posed.
3. It also helps to reduce a second order linear PDE (with two independent variables) into
canonical forms.
Introduce a new set of independent variables φ and ψ, where φ = φ(x, y), ψ = ψ(x, y) are
twice continuously differentiable functions of x and y. By using of chain rules, one has
∂
∂
+ ψ,x
∂φ
∂ψ
∂
∂
= φ,y
+ ψ,y
∂φ
∂ψ
∂
∂x
∂
∂y
= φ,x
(3.1-3)
(3.1-4)
Similarly, the second partial derivatives in x and y can be in terms of partial derivatives in φ
and ψ as
∂2
∂x2
∂2
∂x∂y
∂2
∂y 2
2
∂2
∂2
∂
∂
2 ∂
+ 2φ,x ψ,x
+ φ,xx
+ ψ,x
+ ψ,xx
(3.1-5)
2
∂φ
∂φ∂ψ
∂φ
∂ψ 2
∂ψ
∂2
∂2
∂
∂2
∂
= φ,x φ,y 2 + (φ,x ψ,y + φ,y ψ,x )
+ φ,xy
+ ψ,x ψ,y 2 + ψ,xy (3.1-6)
∂φ
∂φ∂ψ
∂φ
∂ψ
∂ψ
2
∂2
∂2
∂
∂
2 ∂
= φ2,y 2 + 2φ,y ψ,y
+ φ,yy
+ ψ,y
+ ψ,yy
(3.1-7)
∂φ
∂φ∂ψ
∂φ
∂ψ 2
∂ψ
= φ2,x
Equation (3.1-1) can then be rewritten as
Au,φφ + 2Bu,φψ + Cu,ψψ = D
(3.1-8)
where
A = aφ2,x + 2bφ,x φ,y + cφ2,y
B
C
(3.1-9)
= aφ,x ψ,x + b(φ,x ψ,y + φ,y ψ,x ) + cφ,y ψ,y
=
2
aψ,x
+ 2bψ,x ψ,y +
2
cψ,y
(3.1-10)
(3.1-11)
a, b, c are functions of φ and ψ, while D is a function of φ, ψ, φ,x , φ,y , ψ,x , ψ,y , u,φ , and u,ψ .
The classification of Eq. (3.1-1) depends on the possibility of having real or complex roots
to the equation
aλ2 + 2bλ + c = 0
(3.1-12)
3.1.2-1 b2 − ac > 0
In this case, Eq. (3.1-12) has two distinct real roots, say λ1 and λ2 . If the set of new independent
variables φ and ψ are chosen such that
φ,x
= λ1
φ,y
and
ψ,x
= λ2
ψ,y
(3.1-13)
87
then A = C = 0. Thus Eq. (3.1-8) yields
2Bu,φψ = D
(3.1-14)
u,φψ = D∗ (φ, ψ, φ,x , φ,y , ψ,x , ψ,y , u,φ , u,ψ )
(3.1-15)
or
Furthermore, if one let φ̄ = (φ + ψ)/2 and ψ̄ = (φ − ψ)/2, together with chain rule, one has
∂
1 ∂
∂
∂
1 ∂
∂
=
+
,
=
−
(3.1-16)
∂φ
2 ∂ φ̄ ∂ ψ̄
∂ψ
2 ∂ φ̄ ∂ ψ̄
Consequently, equation (3.1-15) can be written as
∂2u
∂2u
−
= 4D∗
2
∂ φ̄
∂ ψ̄ 2
(3.1-17)
Both equations (3.1-15) and (3.1-17) are called the canonical forms of hyperbolic PDE’s.
Remarks:
1. One-dimensional wave equations
∂2u
∂x2
=
1 ∂2u
c2 ∂t2
is a typical example of hyperbolic PDE.
2. Since a, b and c might depend on x and y (hence on φ and ψ), this classification is only a
local property of equations.
3. Consider the new coordinate lines φ(x, y) = constants, and ψ(x, y) = constants, respectively. The total differentiation along these coordinate lines gives
0
= dφ = φ,x dx + φ,y dy
(3.1-18)
0
= dψ = ψ,x dx + ψ,y dy
(3.1-19)
By using of Eq. (3.1-13), these two equations can be read as
dy
dx
=−
φ,x
φ,y
= −λ1 ,
dy
dx
=−
ψ,x
ψ,y
= −λ2
(3.1-20)
i.e. Eq. (3.1-20) are ODE’s to solve for φ =constants, and ψ =constants, respectively.
4. Since a, b and c of Eq. (3.1-1) might be functions of x and y, the roots λ1 and λ2 might
also be functions of x and y. Hence φ(x, y) =constants, and ψ(x, y) =constants corresponds to curves in (x, y) plane. These curves are called characteristic curves or simply
characteristics of the PDE.
3.1.2-2 b2 − ac < 0
In this case, Eq. (3.1-12) has a pair of complex conjugate roots. From Eq. (3.1-20) φ(x, y) and
ψ(x, y) will also be complex conjugate functions. Though Eq. (3.1-15) is still valid in terms of
complex independent variables φ and ψ, it is more convenient to use real independent variables.
Introduce two real independent variables µ and ν as
φ = µ(x, y) + iν(x, y),
ψ = µ − iν
88
(3.1-21)
then from chain rule
∂
1 ∂
∂
=
−i
,
∂φ
2 ∂µ
∂ν
∂
1
=
∂ψ
2
∂
∂
+i
∂µ
∂ν
(3.1-22)
Consequently, in terms of real variable µ and ν, equation (3.1-15) can be written as
∂2u ∂2u
+ 2 + (lower order derivatives · · ·) = 0
∂µ2
∂ν
(3.1-23)
Equation (3.1-23) is called the canonical form of elliptic PDE’s.
Remarks:
1. Two-dimensional Laplace equation
∂2u
∂x2
+
∂2u
∂y 2
= 0 is a typical example of elliptic PDE.
2. Since φ(x, y) and ψ(x, y) are complex functions, there is no characteristic curves in (x, y)
plane for elliptic PDE’s. Consequently, Eq. (3.1-23) will have a solution for specifying
initial conditions on any curve in (x, y) plane. Unfortunately, this type of problems are
usually not well-posed (namely, small variation on initial data will produce huge variation
on solution).
3. Consider the following two initial-valued problems as an example:
 2
∂ u1
∂ 2 u1


+
=0
y ≥ 0, −∞ < x < ∞


 ∂x2
∂y 2
u1 (x, 0) = 0


∂u1



(x, 0) = 0
∂y
 2
∂ u2
∂ 2 u2


+
=0


 ∂x2
∂y 2
u2 (x, 0) = 0


∂u2
sin nx



(x, 0) =
∂y
n
(3.1-24)
y ≥ 0, −∞ < x < ∞
(3.1-25)
The above two problems have u1 ≡ 0 and u2 = sin(nx) sinh(ny)/n2 as their solutions,
respectively. These solutions can easily be verified by a direct substitution into the corresponding PDE and initial conditions. For very large n, the initial conditions of the second
problems only differs slightly from the ones of the first problem, but their solutions differ
from each other quite a lot, namely the solutions of these two problems are ill-posed.
3.1.2-3 b2 − ac = 0
In this case, Eq. (3.1-12) has only one (repeated) real root as λ = b/a. Hence there is only
one
φ,x
dy
family of characteristic curves, say φ =constant, which can be found from dx = − φ,y = −λ.
From Eq. (3.1-9), it yields A = 0. Moreover, since b2 − ac = 0, one has not only aφ,x + bφ,y = 0
but also bφ,x + cφ,y = 0 . Consequently, Eq. (3.1-10) leads to
B = ψ,x (aφ,x + bφ,y ) + ψ,y (bφ,x + cφ,y ) = 0
(3.1-26)
Equation (3.1-8) is then reduced to the canonical form of parabolic PDE’s as
u,ψψ + lower order terms . . . = 0
(3.1-27)
89
for any variable ψ which is independent on φ.
Remarks:
1. One-dimensional heat equation (also called as diffusion equation)
example of parabolic PDE.
∂2u
∂x2
=
∂u
∂t
is a typical
2. b2 − ac > 0 : hyperbolic type PDE
b2 − ac < 0 : elliptic type PDE
b2 − ac = 0 : parabolic type PDE
These terminologies were simply adopted from quadratic curves
relation with quadratic curves.
1
without any further
3. The classification and the reduction to canonical form is also valid for semi-linear and
quasi-linear equations. Except that, for the cases of quasi-linear PDE’s, the characteristic
curves will depend also on the solution u, hence they can not be expressed explicitly in
terms of x and y.
4. It is easy to verify by direct expansion of both sides of equation that B 2 − AC = (b2 −
ac)(φ,x ψ,y − ψ,x φ,y )2 . Since φ and ψ are independent variables, thus (φ,x ψ,y − ψ,x φ,y )
will never be zero. It then concludes that the process of change variable will not alter the
type of PDE.
5. It is worthwhile to note once again that the classification of PDE is only a local classification. Let us consider, as an example, the Tricomi equation
∂2u
∂2u
+y 2 =0
2
∂x
∂y
(3.1-28)
By the definition of a, b, c, one has b2 − ac = −y, hence for the region y > 0 and y < 0 the
equation is elliptic and hyperbolic, respectively. While along the line y = 0 the equation
is parabolic.
6. Further remarks on classification of a single second order PDE in several variables:
(a) Not all second order PDE’s in several variables are classified.
(b) Constant coefficient second order linear PDE in hyperbolic, parabolic and elliptic
types can always be reduced to their canonical forms as
u,11 − (u,22 + u,33 + · · · + u,nn ) + lower order terms = 0
u,1 − (u,22 + u,33 + · · · + u,nn ) + lower order terms = 0
u,11 + u,22 + u,33 + · · · + u,nn + lower order terms = 0
1 quadratic
hyperbolic
parabolic
elliptic
curves:
f (x, y) = ax2 + 2bxy + cy 2 + dx + ey + g
where a, b, c, d, e, and g, are constants. The function f (x, y) stands a curve in xy-plane. The behaviors of these
curves are grouped and classified by the sign of b2 − ac as:
b2 − ac > 0 : hyperbolic curves
b2 − ac < 0 : elliptic curves
b2 − ac = 0 : parabolic curves
90
3.2
Preliminary
3.2.1
Divergence Theorem and Green’s Identities
Recall that the divergence theorem in vector calculus follows that
Z
Z
∇ · g dV =
g · n dS
D
(3.2-1)
∂D
From vector identities, v∇2 u = v∇ · (∇u) = [∇ · (v∇u) − ∇v · ∇u] and using divergence theorem
Eq. (3.2-1), one has the Green’s first identity as
Z
Z
Z
v∇2 u dV =
v∇u · n dS −
∇v · ∇u dV
(3.2-2)
D
∂D
D
It is then easy to conclude from Eq. (3.2-2) that
Z
Z ∂u
∂v
2
2
[v∇ u − u∇ v] dV =
v
−u
dS
∂n
∂n
D
∂D
(3.2-3)
Equation (3.2-3) is sometimes called the Green’s second identity.
Remarks:
In classical analysis, it requires g ∈ C 1 in D and g ∈ C 0 on ∂D in divergence
theorem; while requires u, v ∈ C 2 in D and u, v ∈ C 1 on ∂D in the Green’s first
and second identities . But in the theory of distribution (or theory of generalized
function), divergence theorem and Green’s identities can be applied to functions
of wider classes. By appealing to the theory of distributions, one can justify the
application of these equations to a much larger class of functions and we shall feel
free to avail ourselves of this fact on occasion.
3.2.2
Adjoint Partial Differential Operators
Introduce first a concise notation for partial derivatives and differential operators in n independent variables. Let k be an n-dimensional “vector” whose components are nonnegative integers.
We shall call k a multi-index of order n. For any multi-index k = (k1 , k2 , . . . , kn ), we define
|k|
Dk
≡ k1 + k2 + · · · + kn
k1 +k2 +···+kn
≡
(3.2-4)
|k|
∂
∂
=
k1
k2
kn
k1
∂x1 ∂x2 · · · ∂xn
∂x1 ∂xk22 · · · ∂xknn
(3.2-5)
with the understanding that, if any component of k is zero, the differentiation with respect to
that corresponding variable is omitted. As an example, if n = 3 and k = (2, 0, 5), then
Dk = D(2,0,5) =
∂7
∂x21 ∂x53
An arbitrary linear partial differential operator L of order p in n variables can then be
written as
X
L=
ak (x)Dk
(3.2-6)
|k|≤p
91
where ak (x) = a(k1 ,···,kn ) (x1 , · · · , xn ) are arbitrary functions and the sum in L is taken over all
multi-indices k of order n. For example, the most general linear partial differential operator of
order 2 in two variables is
L = a(2,0)
∂2
∂2
∂2
∂
∂
+ a(1,0)
+ a(0,1)
+ a(0,0)
+
a
+
a
(1,1)
(0,2)
2
2
∂x1
∂x1 ∂x2
∂x2
∂x1
∂x2
(3.2-7)
Let t and φ be arbitrary (generalized) function and test function, respectively. The adjoint
operator of L is denoted by L∗ , such that
hLt, φi = ht, L∗ φi
(3.2-8)
or,
Z
Z
φ Lt dV =
t L∗ φ dV
(3.2-9)
It is easily to show, by direct integration by parts, that
X
L∗ φ =
(−1)|k| Dk (ak φ)
(3.2-10)
|k|≤p
If L = L∗ , the operator is called formally self-adjoint. For instance, we observe that the
∂
Laplacian operator ∇2 is formally self-adjoint; while the heat operator ∇2 − ∂t
is not formally
self-adjoint. Now let u and v be functions (may not be test functions) having continuous
derivatives in domain D with boundary ∂D. Equation (3.2-9) should be replaced by
Z
Z
∗
(vLu − uL v) dV =
n · J(u, v)dS
(3.2-11)
D
∂D
where J(u, v) is a vectorial bilinear form in u and v involving only derivatives of u and v of
order p − 1 or less.
Remarks:
Recall that we defined the adjoint differential operator L∗ in ODE through
Z
b
(vLu − uL∗ v)dx = J(u, v)|bx=a
(3.2-12)
a
Equation (3.2-11) can be viewed as a counterpart of Eq. (3.2-12) in PDE.
Example 1. Consider the Laplacian operator in two variables, ∇2 , which is certainly a formally
self-adjoint operator; i.e., (∇2 )∗ = ∇2 . By using a vector identity, one has
v∇2 u − u∇2 v = ∇ · (v∇u − u∇v)
(3.2-13)
Integrate Eq. (3.2-13) over the region and use the divergence theorem, we have
Z
Z ∂u
∂v
2
2
[v∇ u − u∇ v] dV =
v
−u
dS
∂n
∂n
D
∂D
This is the Green’s second identity as Eq. (3.2-3). It also has the form of Eq. (3.2-11).
92
2
∂
∂
Example 2. Consider the 1D heat-conduction operator L = ∂x
2 − ∂t , which is certainly not
a formally self-adjoint operator; the formally adjoint operator L∗ can easily be calculated as
L∗ =
∂2
∂
+
2
∂x
∂t
(3.2-14)
Then
∂u
∂v
v Lu − u L v = ∇ · v
−u
− et uv
∂x
∂x
∗
(3.2-15)
∂
∂
where ∇ is the gradient operator in space-time domain (i.e., ∇ = ex ∂x
+ et ∂t
), and et is a
unit vector in the time direction. Integrate Eq. (3.2-15) over the region, and use the divergence
theorem, we have
Z
Z
∂u
∂v
∗
(vLu − uL v) dV =
n· v
−u
− et uv dS
(3.2-16)
∂x
∂x
D
∂D
where dV is an element of volume in the space-time domain, and n is the unit outer normal
vector to the boundary of the space-time domain.
3.2.3
Some Terminologies
The mathematical formulation of physical problems often leads to a partial differential equation for the quantity of physical interest, with supplementary data provided in the form of
boundary conditions and initial conditions. Such combination of a partial differential equation
and boundary conditions is known as a boundary value problem (BVP); while combination of a
partial differential equation and initial and boundary conditions is known as a initial-boundary
value problem (IBVP). Sometimes, the term of boundary conditions is used in its general sense
to include initial conditions as if time is one of the independent variables.
Laplace equation:
Poisson equation:
heat equation:
wave equation:
∇2 u = 0
(3.2-17)
2
∇ u = g(x)
∂u
∇2 u =
∂t
2
∂
u
∇2 u = 2
∂t
(3.2-18)
(3.2-19)
(3.2-20)
where ∇2 is the (2D or 3D) Laplacian operator in domain D with boundary ∂D. In some
textbooks, the Laplacian operator may be written as 4 instead of ∇2 . Notice that the Poisson
equation can also be viewed as an inhomogeneous Laplace equation.
As mentioned in the section 3.1.2 remarks # 3, the Laplace equation subjected to a Cauchy
data will not be well-posed. Hence, problems of interest in Laplace equations and in Poisson
equations are usually as follows:
1. Dirichlet problem: [first BVP of potential theory]
u(x) = f (x), on ∂D are specified.
(Dirichlet B.C.)
2. Neumann problem: [second BVP of potential theory]
∂u
n is the unit outer normal to ∂D.
∂n = f (x), on ∂D are specified;
(Neumann B.C.)
3. Robin problem: [mixed BVP, or third BVP of potential theory]
∂u
= f (x), on ∂D are specified;
where α2 (x) + β 2 (x) 6= 0.
αu + β ∂n
(Robin B.C.)
93
Note: for exterior problems or problems with unbounded domain, additional conditions which
describe the behavior of solution at the infinity must be specified.
Note: in addition to the boundary conditions, values of u at t = 0 (initial conditions) are needed
for heat equation; while both values and derivatives of u are needed for wave equation.
Remarks:
1. Take v ≡ 1 into the Green’s first identity Eq. (3.2-2), one has the Green’s third identity
as
Z
Z
∂u
2
∇ u dV =
dS
(3.2-21)
∂n
D
∂D
This holds for any u in domain D. In particular, this must also hold for the solution u to
the Neumann problem of the Laplace equation. It then concludes the solvability condition
for Neumann problem of the Laplace equation as
Z
∂u
dS = 0
(3.2-22)
∂D ∂n
interpretation: Laplace equation can be viewed as the governing equation for a steadystate heat transfer problem (where u is the steady-state temperature, hence ∂u
∂t = 0),
then the solvability condition Eq. (3.2-22) simply implies that the net flow of heat into
the domain D through boundary ∂D must vanish in order to have the steady-state temperature u exist. [Similarly, the solvability condition for Neumann problem of the Poisson
equation, again from Eq. (3.2-21) is
Z
Z
∂u
dS =
g(x) dV
(3.2-23)
∂D ∂n
D
and the interpretation is then as . . ..]
2. Can you then relate Eq. (3.2-23) to the consistency condition in the case of
ODE?
[Hint: Is the Neumann problem of the Poisson equation self-adjoint?]
3.3
Green’s Functions and Integral Representation
In this section we shall construct a variety of Green’s functions. Roughly speaking, these give
the effect at a field point x by the action of a unit source that is concentrated at x0 . Explicit
formulas for Green’s functions are given for free space. We shall show how the corresponding
general boundary value problems can be solved, once these functions having been determined.
3.3.1
Free Space Green’s Functions
3.3.1-1 Laplace and/or Poisson equations
Consider 2D and 3D free space Green’s functions (or so-called fundamental solutions) of Poisson’s equations
∇2 G0 = δ(x − x0 )
94
(3.3-1)
in 2D
.
∇2 = (∂ 2 /∂x2 ) + (∂ 2 /∂y 2 )
δ(x − x0 ) = δ(x − x0 ) δ(y − y0 )
let r = x − x0 , r = |r|
d
0
Eq. (3.3-1) =⇒ 1r dr
r dG
= 0 r 6= 0
dr
(this is an equi-dimensional ODE)
=⇒ G0 (r) = c1 ln r + c2
by using of the Green’s theorem that
R
RR 2
(∇G0 ) · ndS =
∇ G0 dA
R 2π dG0
one has 0 dr r dθ = 1
0
=⇒ 2πr dG
dr = 1
1
c1 = 2π
c2 is arbitrary, set to zero for convenience
1
G0 (x; x0 ) = 2π
ln r
2
where r = [(x − x0 ) + (y − y0 )2 ]1/2
in 3D
∇2 = (∂ 2 /∂x2 ) + (∂ 2 /∂y 2 ) + (∂ 2 /∂z 2 )
δ(x − x0 ) = δ(x − x0 ) δ(y − y0 ) δ(z − z0 )
let R = x − x0 , R = |R|
d
0
Eq. (3.3-1) =⇒ R12 dR
R2 dG
= 0 R 6= 0
dR
(this is an equi-dimensional ODE)
=⇒ G0 (R) = cR3 + c4
by using of the Green’s theorem that
RR
RRR 2
(∇G0 ) · ndS =
∇ G0 dV
R R dG0 2
one has
R
sin
φ
dφ dθ = 1
dR
2 dG0
=⇒ 4πR dR = 1
1
c3 = − 4π
c4 is arbitrary, set to zero for convenience
1
G0 (x; x0 ) = − 4πR
where R = [(x − x0 )2 + (y − y0 )2 + (z − z0 )2 ]1/2
Remarks:
The steps for the fundamental solutions of Poisson equations, which are very similar
to the ones for the cases of ODE’s, are summarized as follows:
1. find the solution to the homogeneous DE;
2. use the “jump condition” to determine the not-yet-determined constant in the
solution.
The jump condition for the cases of ODE’s is the consequence of the integration to
the DE around the singularity. Here the corresponding condition is found, similarly,
by integrating the PDE in two/three dimensional region around the singularity. The
integral is then evaluated by the Green’s theorem. [Question: what happens to the
condition corresponding to the continuity condition for the cases of ODE’s?]
3.3.1-2 heat equations
Consider an 1-D free space Green’s
 2
∂ G
∂G

 ∂x20 − ∂t0 = δ(x) δ(t)
G0 (x, t) = 0

 G (x, t) → 0
0
function problem
−∞ < x < ∞ , t > 0−
t < 0−
|x| → ∞
[Method 1:] Define Laplace transform pair as
(
R∞
G̃0 (x, s) = 0− G0 (x, t) exp(−st) dt
R
1
G0 (x, t) = 2πi
G̃ (x, s) exp(st) ds
Γ 0
(3.3-2)
(3.3-3)
Applying Laplace transform over t with transform parameter s to the problem, one has
( 2
d G̃0
−∞ < x < ∞
dx2 − sG̃0 = δ(x)
(3.3-4)
G̃0 (x, s) → 0
as |x| → ∞, Re(s) > 0
95
(
Hence G̃0 =
A exp(s1/2 x)
A exp(−s1/2 x)
x<0
x>0
(3.3-5)
Here, both boundary conditions at |x| → ∞ and the continuity condition at x = 0 have been
used. The jump condition requires that
dG̃0
dx
−
x=0+
1/2
dG̃0
dx
=1
x=0−
=⇒ −2s
A=1
1 −1/2
=⇒ A = − s
2
1
=⇒ G̃0 = − s−1/2 exp(s1/2 |x|)
2
From the table of Laplace transform, we find that
1
G0 (x, t) = − (πt)−1/2 exp[−x2 /(4t)] H(t)
2
(3.3-6)
where H(t) is the Heaviside step function.
[Method 2:] Define Fourier transform pair as
(
R∞
Ĝ0 (ξ, t) = −∞ G0 (x, t) exp(−iξx) dx
R∞
1
G0 (x, t) = 2π
Ĝ (ξ, t) exp(iξx) dξ
−∞ 0
(3.3-7)
Applying Fourier transform over x with transform parameter ξ to the problem, one has
(
dĜ0
2
dt + ξ Ĝ0 = −δ(t)
(3.3-8)
Ĝ0 (ξ, t) ≡ 0
t<0
(
Hence Ĝ0 =
A exp(−ξ 2 t)
B exp(−ξ 2 t)
t>0
t<0
(3.3-9)
The initial condition concludes that B ≡ 0, and the jump condition requires that
Ĝ0
t=0+
− Ĝ0
= −1
t=0−
=⇒ A = −1
=⇒ Ĝ0 = − exp(−ξ 2 t) H(t)
From the table of Fourier transform, we find that
G0 (x, t) = −
H(t)
exp[−x2 /(4t)]
(4πt)1/2
(3.3-10)
Remarks:
1. If source is applied at x = x0 , t = t0 instead of x = 0, t = 0, by simple translation of
variables (why??) one has
G0 (x, t; x0 , t0 ) = −
(x − x0 )2
H(t − t0 )
exp
−
4(t − t0 )
[4π(t − t0 )]1/2
96
(3.3-11)
2. For higher dimension, the method of Fourier transform can directly be applied, it yields
that
H(t − t0 )
|x − x0 |2
G0 (x, t; x0 , t0 ) = −
exp −
(3.3-12)
4(t − t0 )
[4π(t − t0 )]n/2
where n is the number of spatial dimensions.
3. This solution says that even when the heat source occurs at position x = x0 and at time
t = t0 , there is a temperature distribution for any x for any time t > t0 . It implies that
the speed of propagation of heat-disturbance is infinite. [This is in contradiction to the
solution of wave equation, where the speed of propagation of disturbance is finite.] It is
certainly not a physically reasonable result, but it is still a very good approximation due
to the presence of the exponential term.
3.3.2
Method of Images
Green’s functions provide keys to the solutions of the boundary value problems. Unfortunately,
it is not possible to write down explicit formulas for Green’s functions of bounded regions except
for very simple regions. Here method of images is used to construct Green’s functions for few
special boundary shapes for the cases of Dirichlet or Neumann boundary conditions.
Consider a Green’s function problem in a bounded region
(
L G(x; x0 ) = δ(x − x0 ) ,
x, x0 ∈ D
(3.3-13)
G=0 ,
x ∈ ∂D
In the method of images, one uses the superposition of two (or even several) free space Green’s
function solutions with appropriately located singular points and any homogeneous solution
that may be useful. Consider a candidate for G(x; x0 ) as
G(x; x0 ) = G0 (x; x0 ) +
N
X
ck G0 (x; x∗k ) + h(x)
(3.3-14)
k=1
where ck are constants, x∗k locate outside of the region (D + ∂D), and h(x) is a homogeneous
solution of the PDE. We then pick up x∗k , ck , k = 1, 2, . . . , N , and H(x), such that G(x; x0 ) = 0
for x ∈ ∂D. Notice that equation (3.3-14) will certainly satisfy the desired PDE in D.
Example 1: Find the Green’s function of 2D Laplace equation in
the right half xy plane with Dirichlet B.C. on y-axis.

2

 ∇ G = δ(x − x0 ) δ(y − y0 ) 0 < x, x0 < ∞, −∞ < y, y0 < ∞
G(0, y) = 0

 G → 0,
as |r| → ∞ ,
where r2 = x2 + y 2
[sol:] If there are two sources of opposite sign situated in the position
of mirror images relative to the line x = 0, the solution on that line is
then zero. Thus by using the method of images, the Green’s function
for x ≥ 0 is
G(x, y; x0 , y0 ) = G0 (x, y; x0 , y0 ) − G0 (x, y; −x0 , y0 )
97
y
G=0
x
(− x0 , y0 )
( x0 , y0 )
where G0 is the free space Green’s function. Hence the Green’s function of this problem is
G(x, y; x0 , y0 ) =
1
[(x − x0 )2 + (y − y0 )2 ]1/2
ln
2π [(x + x0 )2 + (y − y0 )2 ]1/2
(3.3-15)
Remark: How about the problem with Neumann condition at x = 0?
Example 2: Consider a circle centering at the origin with radius a. The Green’s function
problem of Laplace equation for the circle is
(
∇2 G(x; x0 ) = δ(x − x0 ) ,
x, x0 ∈ D
(3.3-16)
G(x; x0 ) = 0 ,
x ∈ ∂D [i.e.,|x| = a]
[sol:] Consider an image point x∗0 , such that
|x∗0 | · |x0 | = a2
(3.3-17)
and the image point x∗0 lies on the radial line of the point x0 , i.e.,
arg(x∗0 ) = arg(x0 ). [The mapping from x0 to x∗0 defined as equation
(3.3-17) together with the requirement of being on the same radial
line is also known as Kelvin transformation.] Let xs be a generic
point on ∂D, it can be shown from geometric argument (of similar
triangles) that
xs
o x0
|xs − x∗0 |
|xs − x0 |
=
|x0 |
a
Construct the Green’s function solution G as
G(x; x0 ) = G0 (x; x0 ) − G0 (x; x∗0 ) + c
(3.3-18)
To satisfy zero boundary condition, it yields
1
1
ln |x − x0 | −
ln |x − x∗0 | + c = 0 ,
2π
2π
1
|x − x∗0 |
1
a
ln
=
ln
=⇒ c =
2π |x − x0 |
2π |x0 |
x ∈ ∂D
Remarks:
1. How about the 3D case? (i.e. Green’s function solution of Laplace equation for a sphere
with Dirichlet B.C.) [Hint: try G0 (x; x0 ) + cG0 (x; x∗0 )]
2. How about Neumann problems both in 2D and 3D?
Example 3: 1-D Green’s function for the heat equation on a finite region


kG,xx − G,t = δ(x − x0 ) δ(t − t0 )
0 < x, x0 < L , 0 < t0 < t


 G(0, t) = 0
t > t0

G(L,
t)
=
0
t > t0



G(x, t) ≡ 0
0 < x < L , t < t0
[Method 1:] By method of images:
98
x*0
a
x = −2L
x = −L
x
-0
a
a
x=0
a
x=L
a
-
x = 2L
-
2L − x0
2L + x0
G(x, t; x0 , t0 ) =
∞
X
[G0 (x, t; 2mL + x0 , t0 ) − G0 (x, t; 2mL − x0 , t0 )]
(3.3-19)
m=−∞
)
( ∞
(x−x0 −2mL)2
(x+x0 −2mL)2
X
−
− 4k(t−t
H(t − t0 )
4k(t−t0 )
0)
G(x, t; x0 , t0 ) =
e
−e
[4πk(t − t0 )]1/2 m=−∞
(3.3-20)
[Method 2:] By eigenfunction expansion: (see Example 6 in page 110)
3.3.3
Integral Representation
3.3.3-1 Laplace and/or Poisson equations
Consider a BVP for Poisson equation as
(
∇2 u = g(x)
in D
and some boundary data on ∂D
(3.3-21)
Construct an appropriate Green’s function problem
(
∇2 G = δ(x − x0 )
in D
and the same B.C. except with homogeneous boundary data on ∂D
By using the Green’s second identity, one has
Z
Z
(u∇2 G − G∇2 u) dV =
(u∇G − G∇u) · n dS
D
∂D
Z
Z
Z
∂G
∂u(x)
=⇒
u(x0 ) =
G(x; x0 )g(x)dVx +
u(x)
dSx −
G
dSx
∂nx
∂nx
D
∂D
∂D
(3.3-22)
(3.3-23)
where the subscripts on dV and dS indicate that the integrations are over x instead of x0 .
For Dirichlet problem, u is known on the boundary ∂D; while the corresponding Green’s
function problem has G = 0 on the boundary as boundary condition. Hence equation (3.3-23)
is reduced to
Z
Z
∂G
u(x) =
G(x; x0 )g(x0 )dVx0 +
u(x0 )
dSx0
(3.3-24)
∂nx0
D
∂D
where the subscripts on dV and dS indicate that the integrations are over x0 now. Here
interchanging the variables x and x0 , and the reciprocity of Green’s function for self-adjoint
problem, namely G(x; x0 ) = G(x0 ; x), have been used.
∂u
For Neumann problem, ∂n
is known on the boundary ∂D; while the corresponding Green’s
∂G
function problem has ∂n = 0 on the boundary as boundary condition. Hence equation (3.3-23)
is reduced to
Z
Z
∂u(x0 )
u(x) =
G(x; x0 )g(x0 )dVx0 −
G
dSx0
(3.3-25)
∂nx0
D
∂D
99
where the subscripts on dV and dS indicate that the integrations are over x0 again. Here
interchanging the variables x and x0 , and the reciprocity of Green’s function for self-adjoint
problem, namely G(x; x0 ) = G(x0 ; x), again, have been used.
Equations (3.3-24) and (3.3-25) are equivalent to the solution representation of ODE by
using appropriate Green’s function.
Remarks:
1. Sometimes an appropriate Green’s function G, which satisfies the same boundary conditions as original problem except with homogeneous boundary data, will be very difficult
to find, if it is not impossible. Free space Green’s function solution has then to be used
in this kind of situation. From Green’s second identity
Z
Z
2
2
(u∇ G0 − G0 ∇ u) dV =
(u∇G0 − G0 ∇u) · n dS
D
∂D
Z
u(x) =
Z
G0 (x; x0 )g(x0 )dVx0 +
D
u(x0 )
∂D
∂G0
dSx0 −
∂nx0
Z
G0
∂D
∂u(x0 )
dSx0 (3.3-26)
∂nx0
when both u and ∇u are known on boundary ∂D, equation (3.3-26) is an integral representation for u in D. Unfortunately, one only specifies u or ∂u/∂n (or even combination
of u and ∂u/∂n) on boundary ∂D, but not both u and ∂u/∂n separately. Hence equation
(3.3-26) becomes as an integral equation for boundary data. The integral equation has
to be solved for boundary data in advance, then Eq. (3.3-26) is used again as integral
representation for u at anywhere in D. This is the basic idea of the boundary element
method.
2. In the some textbook, the derivation of equations (3.3-26) and (3.3-24) involves a detailed
analysis of small disk (2D) or small ball (3D) around the singular point x = x0 . At
here, equations (3.3-26) and Eq. (3.3-24) have been derived in the context of generalized
function theory directly.
Example 1: Consider a (2D) problem in a disk as
(
∇2 u = 0 x ∈ D : |x| < a
u = f (x) x ∈ ∂D
(3.3-27)
[sol:] The appropriate Green’s function problem is
(
∇2 G(x; x0 ) = δ(x − x0 ) ,
x, x0 ∈ D
G(x; x0 ) = 0 ,
x ∈ ∂D [i.e.,|x| = a]
(3.3-28)
i.e., Green’s function problem of a circle for Laplace equation. This Green’s function has been
found by method of image in page 98. Furthermore, from Green’s second identity, by letting
v = G(x; x0 ), one has
Z
Z
(u∇2 G − G∇2 u) dA =
(u∇G − G∇u) · n dS
(3.3-29)
D
∂D
Z
=⇒
∂G(x; x0 )
u(x0 ) =
f (x)
dS =
∂r
∂D
Z
100
2π
f (x)
0
∂G(x; x0 )
∂r
a dθ
|x|=a
(3.3-30)
Interchanging x with x0 , or (r, θ) with (r0 , θ0 ), and using the reciprocal property of G(x, i.e.
G(x; x0 ) = G(x0 ; x), one has
Z 2π
∂G(r, θ; r0 , θ0 )
u(r, θ) =
f (a, θ0 )
a dθ0
(3.3-31)
∂r0
0
r0 =a
Note that
G(x; x0 )
=
=
=
=
1
2π
1
4π
1
4π
1
4π
|x − x0 | a
ln
|x − x∗0 | |x0 |
2
a |x − x0 |2
ln 2
r |x − x∗0 |2
02
a
r2 + r02 − 2rr0 cos φ
ln 2 2
,
where φ = θ − θ0
r0 r + (r0∗ )2 − 2rr0∗ cos φ
r2 + r02 − 2rr0 cos φ
ln a2 2 2
,
(since r0∗ = a2 /r0 )
r r0 + a4 − 2rr0 a2 cos φ
Hence
∂G
∂r0
∂G
∂r0
=
=
r0 =a
1
2r0 − 2r cos φ
2r2 r0 − 2ra cos φ
−
4π r2 + r02 − 2rr0 cos φ r2 r02 + a4 − 2rr0 a2 cos φ
a
1 − (r/a)2
2
2π r + a2 − 2ar cos φ
The solution u(r, θ) can then be expressed as
Z 2π
1
a2 − r2
u(r, θ) =
f (a, θ0 ) 2
dθ0
2
2π 0
r + a − 2ar cos(θ − θ0 )
(3.3-32)
1. The result in equation (3.3-32) is known as Poisson’s formula for a disk. The Poisson’s
formula for a sphere can be obtained by a similar manner.
2. The Poisson’s formula can also be obtained by separation of variables (imposed a periodic
condition in θ), or directly by complex variable (for 2D case).
3. At r = 0, equation (3.3-32) reduces to
1
u(0, θ) =
2π
Z
2π
f (a, θ0 ) dθ0
(3.3-33)
0
Equation (3.3-33) implies that the solution at the center of a disk is only the average of
the data along the boundary of disk.
4. Knowing that Eq. (3.3-33) is valid for any size of disk, i.e., for any a. Multiplying both
two sides of Eq. (3.3-33) by the factor a and then integrating over a from a = 0 to a = b,
one has
Z b Z 2π
1
u(0, θ) = 2
f (a, θ0 ) a dθ0 da .
(3.3-34)
πb 0
0
Equation (3.3-34) implies that the solution at the center of a disk is also the average
of the field over the whole disk. Equations (3.3-33) and (3.3-34) are also known as the
mean-value theorem for Laplace equations.
101
5. mean-value theorem for Laplace equations: The mean of the values of a harmonic
function over a circle (2D) or a sphere (3D) surrounding a given point P is equal to the
function evaluated at P .
6. maximum-minimum principle for Laplace equations: A harmonic function can not
have a maximum or minimum within a region D, unless it is a constant everywhere (in
D and on ∂D). This can easily be shown from mean-value theorem by contradiction.
7. The max.-min. principle (for Laplace equations) concludes directly that the solution of
Dirichlet problem for Poisson’s equation is unique and continuously depends on the boundary data.
3.3.3-2 heat equations
Consider a BVP for heat equation as

2

 ∇ u − u,t = g(x, t)
u(x, 0) = f (x)

 u(x, t) = h(x, t)
x∈D, t>0
x ∈ D + ∂D
(I.C.)
x ∈ ∂D , t > 0
(B.C.)
(3.3-35)
We have seen that the heat operator is not even a formally self-adjoint operator (Example 2 in
page 93), hence an adjoint operator might be needed. Consider (see also equation (3.2-16))
Z T
Z ∂u
∂v
dt
v ∇2 u −
− u ∇2 v +
dV
∂t
∂t
0
D
Z
Z T
Z ∂v
∂u
T
=−
(uv)|t=0 dV +
dt
v
−u
dS
(3.3-36)
∂n
∂n
D
0
∂D
The adjoint Green’s function problem for G∗ (x, t; x0 , t0 ) is

2 ∗
∗

x ∈ D , t < t0
(backward heat equation)
 ∇ G + G,t = δ(x − x0 ) δ(t − t0 )
∗
G (x, t) ≡ 0
x ∈ D + ∂D, t > t0 (terminate condition)

 G∗ (x, t) = 0
x ∈ ∂D , t < t0
(B.C.)
∗
Let T = t+
0 and v = G , equation (3.3-36) yields
Z
t+
0
.
Z
[G∗ g − u(x, t)δ(x − x0 ) δ(t − t0 )] dV = −
dt
0
Z
D
D
Z
D
[u(x, t)G∗ (x, t; x0 , t0 )]|t=0 dV +
Z
t+
0
0
[u(x, t)G∗ (x, t; x0 , t0 )]|t=t+ dV +
0
∂u(x, t)
∂G∗
∗
dt
G (x, t; x0 , t0 )
− u(x, t)
dSx
∂n
∂nx
∂D
Z
Hence
Z
u(x0 , t0 )
t+
0
=
Z
g(x, t)G (x, t; x0 , t0 ) dVx −
dt
0
Z
D
t+
0
h(x, t)
∂D
f (x) G∗ (x, 0; x0 , t0 ) dVx +
D
Z
dt
0
Z
∗
∂G∗ (x, t; x0 , t0 )
dSx
∂nx
(3.3-37)
Equation (3.3-37) is an expression of solution u in terms of adjoint Green’s function G∗ . We
might like to have an expression in terms of Green’s function G itself. To achieve this, as for
102
the corresponding cases in ODE, a relation between G and G∗ is needed. Consider the ordinary
Green’s function problem for G(x, t; x0 , t0 ),

2
0
0

x ∈ D , t > t0 > 0 (heat equation)
 ∇ G − G,t = δ(x − x ) δ(t − t )
(3.3-38)
G(x, t) ≡ 0
x ∈ D + ∂D, t < t0 (I.C.)

 G(x, t) = 0
0
x ∈ ∂D , t > t
(B.C.)
∗
Let T = t+
0 , v = G , and u = G, from equation (3.3-36) one has
Z
t+
0
Z
[G∗ (x, t; x0 , t0 ) δ(x − x0 ) δ(t − t0 ) − G(x, t; x0 , t0 ) δ(x − x0 ) δ(t − t0 )] dV
dt
t0−
D
Z
Z
∗
=−
D
(GG )|t=t+ dV +
0
D
Z
∗
(GG )|t=t0 dV +
−
t+
0
t0−
G∗ (x0 , t0 ; x0 , t0 ) = G(x0 , t0 ; x0 , t0 )
Hence
∂G∗
∂G
G∗
−G
dS
∂n
∂n
∂D
Z
dt
(3.3-39)
The equation (3.3-37) can then be written, in terms of G, as
Z
u(x, t)
t+
=
Z
0
Z
Z
g(x0 , t0 )G(x, t; x0 , t0 ) dVx0 −
dt0
D
t+
Z
dt0
0
Example 2: Consider


 u,xx − u,t = 0
u(x, 0) = f (x)

 u(x, t) → 0
h(x0 , t0 )
∂D
f (x0 ) G(x, t; x0 , 0) dVx0 +
D
∂G(x, t; x0 , t0 )
dSx0
∂nx0
−∞ < x < ∞ , t > 0
−∞ < x < ∞
|x| → ∞
(3.3-40)
(3.3-41)
Of course, this problem can be solved directly, by using of Laplace transform over t or Fourier
transform over x, as we derived the free space Green’s function in page 95. At here, we use the
integral representation Eq. (3.3-40) instead, we have
Z ∞
H(t)
(x − ξ)2
u(x, t) =
f
(ξ)
exp
−
dξ
(3.3-42)
4t
(4πt)1/2 −∞
Remarks:
1. The Laplace transform or Fourier transform will yield the same result as equation (3.3-42).
2. As a common fact for heat equation, it is not easy to see whether the solution (3.3-42)
satisfies the I.C., unless one uses an equality that
exp[−(x − ξ)2 /(4t)]
= δ(x − ξ)
t→0
(4πt)1/2
lim
(3.3-43)
The equation (3.3-43) has to be shown by the argument of the distributional equality.
3. The integral representation for wave equation problem can be derived in the same manner.
103
3.4
Other Methods of Solution
We have learned the method of separation of variables in Engineering Mathematics II, which
leads a solution in a series form for a number of linear PDE. We do not intend to review this
method in detail, while we only remind some of its restriction instead. Method of separation of
variables has several restrictions as follows:
1. PDE itself must be separable. (hence the PDE must be homogeneous at least)
2. The shape of boundary consists of only lines (or surfaces) of constant coordinate for
selected coordinate system.
3. There is at least a pair of homogeneous boundary conditions (to have a Sturm-Liouville
eigenvalue problem).
Since the method of separation of variables can not handle the problems with inhomogeneous differential equations, we shall explore some other methods to deal with Green’s function
problem. The method of images has been used on some simple problems in previous subsection.
Here other methods of solution will be developed.
3.4.1
(Full) Eigenfunction Expansion
This method only works for bounded regions. Consider an associated eigenvalue problem for
Laplace and/or Poisson equation as an example
(
∇2 φ + λφ = 0
in D
(3.4-1)
∂φ
=0
on ∂D
αφ + β ∂n
where α and β may be functions of coordinates. The properties of eigenvalues and eigenfunctions
of this particular problem are
1. All eigenvalues are real.
2. There are an infinite number of eigenvalues, and λk → ∞ as k → ∞.
3. There may be several linearly independent eigenfunctions corresponding to the same eigenvalue!! (This is not the case as in 1D regular Sturm-Liouville eigenvalue problem, where
only one linearly independent eigenfunction corresponds to one eigenvalue.)
4. Eigenfunctions form a complete set. Hence any piecewise smooth function f (x, y) can be
P
expanded in terms of eigenfunctions, again, the series f (x, y) ∼ λ aλ φλ (x, y) converges
in the mean.
5. Eigenfunctions corresponding to different eigenvalues are orthogonal over the entire region
D, i.e.,
Z
φλ1 φλ2 dV = 0 ,
if λ1 6= λ2
(3.4-2)
D
Furthermore, different eigenfunctions corresponding to the same eigenvalues can be made
orthogonal.
104
Example 1: Consider an eigenvalue problem

2

0 < x < L, 0 < y < H
 ∇ φ + λφ = 0
φ(0, y) = 0 φ(x, 0) = 0

 φ(L, y) = 0 φ(x, H) = 0
(3.4-3)
Solving the problem by separation of variables, we find
2
2
eigenvalues
λkm = kπ
+ mπ
,
k, m = 1, 2, . . .
L
H
mπy
eigenfunctions φkm (x, y) = sin kπx
sin
L
H
We observe several properties
1. eigenvalues λkm are real ;
2. λkm → ∞ as k, m → ∞;
3. different eigenfunctions (modes) may correspond to one eigenvalue. For instance, in the
π2
2
2
case of L = 2H, λkm = 4H
2 (k + 4m ). It is easy to verify that λ41 = λ22 but they have
their own eigenfunctions. (In fact, λ(2k)m = λk(2m) in this particular example)
Example 2: Solve the Dirichlet boundary value problem with Poisson equation in a unit square
(
u,xx + u,yy = g(x, y) 0 < x < 1, 0 < y < 1
u = 0 on the boundary of the square
[sol:] Method of separation of variables does not work, since inhomogeneous differential equation can never be separable. Consider
an associated eigenvalue problem
(
φ,xx + φ,yy + λφ = 0 ,
0 < x < 1, 0 < y < 1
φ = 0 on the boundary of the square
y
(0,1)
(1,1)
(1,0)
(3.4-4)
One has λkm = (kπ)2 + (mπ)2 , and φkm = sin kπx sin mπy as eigenvalues and eigenfunctions.
Let
u(x, y) =
∞ X
∞
X
Akm sin kπx sin mπy
(3.4-5)
k=1 m=1
Substitute equation (3.4-5) into D.E., one has
∞ X
∞
X
{−[(kπ)2 + (mπ)2 ]Akm } sin kπx sin mπy = g(x, y)
(3.4-6)
k=1 m=1
Hence
Akm = −
4
λkm
Z
1
Z
1
g(x, y) sin kπx sin mπy dx dy
0
0
R1R1
[the factor of 4 comes from 1/[ 0 0 sin2 kπx sin2 mπy dx dy].]
Remarks:
105
(3.4-7)
x
1. Special case: if g(x, y) = δ(x − x0 ) δ(y − y0 ), i.e.
(
G,xx + G,yy = δ(x − x0 ) δ(y − y0 )
G = 0 on the boundary of the square
0 < x, x0 < 1, 0 < y, y0 < 1
(3.4-8)
The solution G(x, y; x0 , y0 ) is called the Green’s function of this particular problem. In
terms of eigenfunctions, one has
G(x, y; x0 , y0 ) =
∞ X
∞ X
k=1 m=1
−
4
λkm
sin kπx0 sin mπy0 sin kπx sin mπy
(3.4-9)
Note that G(x, y; x0 , y0 ) = G(x0 , y0 ; x, y) from a direct observation of Eq. (3.4-9). This
implies the reciprocity of Green’s function. In fact, the property of reciprocity on G itself
is the consequence of the self-adjoint problem.
2. The substitution of equation (3.4-5) into original P.D.E. requires that the series of Eq.
(3.4-5) is uniformly convergent. See your textbooks of Engineering Mathematics I & II
for more discussion on Weierstrass M-test and term-by-term differentiation of a series.
3. In equation (3.4-6), coefficients Akm are determined directly by the argument of the
orthogonality of eigenfunctions φkm . It is equivalently to expand first g(x, y) in terms of
φkm and then identify coefficients in both sides term by term (by the argument of linear
independence or orthogonality of eigenfunctions). But can g(x, y) always be expanded in
terms of eigenfunctions? Namely,
g(x, y) =
∞ X
∞
X
Bkm sin kπx sin mπy
??
(3.4-10)
k=1 m=1
The answer is yes, as long as g(x, y) is piecewise smooth in the region D, the series
converges to the mean. Does g(x, y) always vanish on the boundary of the square as those
eigenfunctions? No! It is not necessarily, and then the series in equation (3.4-10) is only
pointwise convergence. Moreover, Gibb’s phenomena will present on the boundary of the
square (recall the Fourier series expansion). Nevertheless, the series of u(x, y) in equation
(3.4-5) converges much faster than the series in equation (3.4-10), attribute to λkm in the
denominator of Akm . [Question: Is term-by-term differentiation of Eq. (3.4-5) still valid?]
4. How about the problem with inhomogeneous B.C.’s?
Viewpoint 1: Let u = u1 + u2 , where
u1 : satisfies inhomogeneous D.E. with homogeneous B.C.’s
[can be solved by eigenfunction expansion]
u2 : satisfies homogeneous D.E. with inhomogeneous B.C.’s
[can be solved possibly by separation of variables]
Viewpoint 2: We might wonder whether it is still possible to use eigenfunction expansion
directly. Let
u(x, y) =
∞ X
∞
X
Akm sin kπx sin mπy
(3.4-11)
k=1 m=1
Note that every single φkm (x, y) satisfies homogeneous B.C.’s. Is it still possible for u(x, y)
in equation (3.4-11) satisfying inhomogeneous B.C.’s? No!! But similar to the argument
106
in the remark 3, the series expansion of equation (3.4-11) is still possible, except that
the series is only pointwise convergence. But then the term-by-term differentiation of the
series will be illegitimate. How can one find Akm in this case?
Example 3: Knowing that φkm are eigenfunctions for
(
∇2 φ + λφ = 0
in D
φ=0
on ∂D .
(3.4-12)
Solve the following boundary value problem, by using the eigenfunctions φkm ,
(
∇2 u = g(x)
in D
u = f (x)
on ∂D
PP
[sol:] Let u =
Akm φkm , from orthogonality one has
R
uφkm dV
Akm = RD 2
φ dV
D km
R
1 D u(∇2 φkm ) dV
R 2
= −
[∵ φ = −(∇2 φ)/λ]
λkm
φ dV
D km
(3.4-13)
(3.4-14)
The numerator of equation (3.4-14) can be evaluated by using the Green’s second identity
(letting v = φkm ),
Z
Z ∂φkm
∂u
[u∇2 φkm − φkm ∇2 u] dV =
u
− φkm
dS
(3.4-15)
∂n
∂n
D
∂D
Hence
Akm = −
1
λkm
R
D
φkm g dV +
R
R
f (∂φkm /∂n) dS
∂D
2 dV
φ
D km
(3.4-16)
Here both the D.E.’s and B.C.’s in φ and in u have been used.
Remarks:
1. Is this approach familiar to you? Recall the corresponding one in Chapter 3.
2. As it has been done in Chapter2,
(
PP
u=
Akm φkm
R
1
Akm = Ikm
uφkm dV
D
,
Ikm =
R
D
φ2km dV
(3.4-17)
can be regarded as a pair of ‘finite’ integral transform. Akm is the transformed quantity
of u, while its inverse transform (i.e. the summation) gives u. Apply this finite integrao
transform to P.D.E., one has
Z
Z
1
1
(∇2 u)φkm dV =
gφkm dV
(3.4-18)
Ikm D
Ikm D
The Green’s second identity (acting as integration by parts) gives
Z
Z
Z ∂φkm
∂u
2
2
(∇ u)φkm dV =
u(∇ φkm ) dV −
u
− φkm
dS
∂n
∂n
D
D
Z ∂D
∂φkm
= −λkm Ikm Akm −
f
dS
∂n
∂D
Consequently, it yields exactly the same result as Eq. (3.4-16).
107
(3.4-19)
3. If f (x, y) ≡ 0 on ∂D, i.e. homogeneous B.C.’s, the solution is reduced back to the previous
result, as one would expect.
4. What is wrong if one uses the original deviation?
u
∇2 u
=
XX
=
XX
Akm φkm
Akm ∇2 φkm =
(3.4-20)
XX
Akm (−λkm )φkm
(3.4-21)
The P.D.E. yields
XX
Akm
(−λkm )Akm φkm = g
R
1 D φkm g dV
R
=−
λkm D φ2km dV
??
The boundary data never has any effect! The difficulty arises from the illegitimacy of
term-by-term differentiation in equation (3.4-21).
3.4.2
Partial Eigenfunction Expansion
Example 4: Consider Poisson equation in a square with homogeneous boundary conditions, as we discussed in example 2 by method
of full eigenfunction expansion
(
u,xx + u,yy = g(x, y) ,
0 < x < 1, 0 < y < 1
u=0,
on the boundary of the square
[sol:] Since sin kπx, k = 1, 2, . . . are eigenfunctions of a SturmLiouville problem
(
φ00k + λφk = 0 ,
0<x<1
φk (0) = 0 ,
φk (1) = 0
(3.4-22)
with eigenvalues λ = k 2 π 2 . Hence they form a complete set in 0 < x < 1. It is always possible
to use this set of eigenfunctions to expand any piecewise smooth functions in 0 < x < 1.
Furthermore, by inspection, after expansion, the remaining equation in y does not coupled. Let
X
u(x, y) =
Ak (y) sin kπx
It is just a Fourier series expansion in this particular case. Substituting into original PDE yields
an ODE in y
∞
X
[A00k (y) − k 2 π 2 Ak (y)] sin kπx = g(x, y)
(3.4-23)
k=1
=⇒
A00k (y) − k 2 π 2 Ak (y) = hk (y)
(3.4-24)
R1 2
where hk (y) = 2 0 g(x, y) sin kπx dx . The factor of 2 comes from 1/[ 0 sin kπx dx]. Boundary
conditions u(0, y) = u(1, y) = 0 have been satisfied automatically. The remaining boundary
conditions u(x, 0) = u(x, 1) = 0, lead to boundary conditions in Ak (y) as Ak (0) = Ak (1) = 0,
again by the orthogonality of sin kπx. The BVP in y can now be solved by several approaches
R1
108
1. variation of parameters (see your textbook of Engineering Mathematics I & II for detail),
one has
Z y
Z 1
Ak (y) = sinh[kπ(1−y)]
hk (ξ) sinh(kπξ)dξ+sinh[kπy]
hk (ξ) sinh[kπ(1−ξ)]dξ(3.4-25)
0
y
2. by eigenfunction expansion in y again. In this example, it leads to the result of full
eigenfunction expansion.
3. by one-dimensional Green’s function in y as discussed in Chapter 2. It yields the same
result as in 1.
Special case: if g(x, y) = δ(x − x0 ) δ(y − y0 ), then u(x, y) is in fact the Green’s function G(x, y)
of this particular problem. Let
u(x, y)
= G(x, y) =
∞
X
Ak (y) sin kπx
(3.4-26)
k=1
=⇒
=⇒

00
2 2

0 < y, y0 < 1
 Ak (y) − k π Ak (y) = 2 sin kπx0 · δ(y − y0 ) ,
Ak (0) = 0

 A (1) = 0
k
(
B sinh(kπy) + C cosh(kπy) ,
0 < y < y0
Ak (y) =
∗
D sinh(kπy) + E cosh(kπy) ,
y0 < y < 1
From boundary conditions
(
B sinh(kπy) ,
0 < y < y0
=⇒ Ak (y) =
D sinh[kπ(1 − y)] ,
y0 < y < 1
(3.4-28)
(3.4-29)
continuity condition: [Ak (y0− ) = Ak (y0+ )]
.
=⇒ B sinh(kπy0 ) = D sinh[kπ(1 − y0 )]
.
(3.4-27)
jump condition
(3.4-30)
: [A0k (y0+ ) − A0k (y0− ) = 2 sin kπx0 ]
=⇒ kπ{−D cosh[kπ(1 − y0 )] − B cosh(kπy0 )} = 2 sin kπx0
(3.4-31)
Finally, one has
(
∞
X
2 sin kπx0
sinh[kπ(y0 − 1)] sinh(kπy) ,
G(x, y; x0 , y0 ) =
sin kπx ·
kπ sinh(kπ)
sinh(kπy0 ) sinh[kπ(y − 1)] ,
k=1
y < y0
(3.4-32)
y > y0
Remarks:
1. The convergency of this series is much better than Eq. (3.4-9) that obtained by double
(full) eigenfunction expansion.
2. This method is possible to handle also the problem with inhomogeneous boundary data.
But if inhomogeneous boundary data exist on both x and y directions, the analysis is
more tedious than double expansion.
109
Example 5: Solve the following Green’s function problem


 G,xx + G,yy = δ(x − x0 ) δ(y − y0 ) 0 < x, x0 < 1, −∞ < y, y0 < ∞
G(0, y) = G(1, y) = 0

 G → 0,
as |y| → ∞
P
Again, let G(x, y) = Ak (y) sin kπx, one has (why??)
(
A00k (y) − k 2 π 2 Ak (y) = 2 sin kπx0 · δ(y − y0 ) ,
−∞ < y, y0 < ∞
Ak (y) → 0 ,
as |y| → ∞
(
B exp(kπy) ,
−∞ < y < y0
Ak (y) =
D exp(−kπy) ,
y0 < y < ∞
y
x
( x0 , y0 )
Boundary conditions at y → ±∞ have been used.
continuity condition:
jump condition
:
=⇒ B exp(kπy0 ) = D exp(−kπy0 )
=⇒ −kπ[D exp(−kπy0 ) + B exp(kπy0 )] = 2 sin kπx0
Finally, one has
G(x, y; x0 , y0 ) = −
∞
1X1
exp[−kπ|y − y0 |] sin kπx0 sin kπx
π
k
(3.4-33)
k=1
Example 6: Solve 1-D Green’s function for the heat equation on a finite region.


kG,xx − G,t = δ(x − x0 ) δ(t − t0 )
0 < x, x0 < L , 0 < t0 < t


 G(0, t) = 0
t > t0

G(L, t) = 0
t > t0



G(x, t) ≡ 0
0 < x < L , t < t0
We solved this problem in Example 3 of page 98 by method of images. Here the problem will
be solved by eigenfunction expansion in x.
[sol:] Let
G(x, t)
=
∞
X
m=1
Am (t) sin
mπx
L
..
.
G(x, t; x0 , t0 )
∞
mπ 2 X
2
mπx0
mπx
= −
sin
sin
exp −k(t − t0 )
L
L
L
L
m=1
(3.4-34)
Remarks:
Series of Eq. (3.3-20) by method of images (in page 98) and series of Eq. (3.4-34) by eigenfunction expansion (in this page) have their own advantages.
1. for k(t − t0 )/L2 1, (i.e. for t t0 ),
π 2 2
πx
πx0
series Eq. (3.4-34) ∼ − sin
sin
exp −k(t − t0 )
L
L
L
L
i.e. only the m = 1 term in Eq. (3.4-34) is needed for this case.
110
(1,0)
2. On the other hand, for (x − x0 )2 /[k(t − t0 )] 1, (i.e. for k(t − t0 )/L2 1 or t ≈ t0 ),
series Eq. (3.3-20) ∼ −
H(t − t0 )
(x − x0 )2
exp
−
4k(t − t0 )
[4πk(t − t0 )]1/2
i.e. only the m = 0 term (the free space Green’s function solution) in Eq. (3.3-20) is
needed for this case. It is a long rod approximation, or a small time approximation.
3.4.3
Fourier transform
Without loss of generality, let us consider only the cases of Fourier transform in y, i.e., the cases
of −∞ < y < ∞. Consider again the previous example


 G,xx + G,yy = δ(x − x0 ) δ(y − y0 ) 0 < x, x0 < 1, −∞ < y, y0 < ∞
(3.4-35)
G(0, y) = G(1, y) = 0

 G → 0,
as |y| → ∞
Apply Fourier transform over y with the transform pair
(
R∞
fˆ(ω) = −∞ f (y) exp(−iωy) dy
R∞
1
fˆ(ω) exp(iωy) dω
f (y) = 2π
−∞
( 2
d Ĝ
2
0 < x, x0 < 1
dx2 − ω Ĝ = δ(x − x0 ) exp(−iωy0 )
=⇒
Ĝ(0; ω) = Ĝ(1; ω) = 0
Solving Ĝ and inverting the Fourier transform, one has the 1-D Green’s function of the original
BVP
Z ∞
1
sinh[ω(x> − 1)] sinh(ωx< )
G(x, y; x0 , y0 ) =
exp[−iω(y0 − y)] dω
(3.4-36)
2π −∞
ω sinh ω
where x> = max(x0 , x) and x< = min(x0 , x). Equivalence between equations (3.4-33) and
(3.4-36) can be achieved by evaluating Eq. (3.4-36) through contour integration in theory of
complex variables.
3.5
Maximum-Minimum Principle of Heat Equation
Consider bounded domains D and QT in x-space and x-t space, respectively, where
QT = {(x, t)|x ∈ D, 0 < t < T }
.
Let u(x, t) be continuous in QT + ∂QT = {(x, t)|x ∈ D + ∂D, 0 ≤ t ≤ T }, and satisfy homogeneous heat equation in QT ,
∇2 u −
∂u
=0 ,
∂t
(x, t) ∈ QT
.
Suppose
u(x, 0) ≤ M
x ∈ D + ∂D
u(x, t) ≤ M
x ∈ ∂D , 0 < t < T
for some M , then u(x, t) ≤ M for all (x, t) ∈ QT + ∂QT .
The proof is omitted in this course.
Remarks:
111
1. For minimum principle, replace (≤ M ) by (≥ m).
2. e.g. let D: 0 < x < a, then ∂D : x = 0, x = a
t
6
(t = T ) C
A
x=0
D
B
x=a
- x
.
then min [ along AB, AC and BD ] ≤ [u(x, t) both within square ABCD and along CD
] ≤ max [ along AB, AC and BD ].
3. If D is an unbounded region, then u(x, t) has to be bounded, in order to apply max-min
principle.
4. From the maximum and minimum principles, if there are no source in region D, the
temperature u in the interior must lie between the maximum and minimum temperatures
on the boundary and at t = 0.
5. In fact, if the initial temperature and boundary temperature are equal to the same constant, then u will be this constant for all time t.
6. Directly concluding form maximum-minimum principle that the small change in boundary and initial data only produce small change in the solution (i.e. solution of Dirichlet
problem depends continuously on prescribed data.) Similarly, zero boundary and initial
data only produce zero solution (i.e. the solution of Dirichlet problem is unique).
7. If one replaces the line t = 0 by line t = T in the proof (i.e. consider the initial-boundaryvalued problem of heat equation in the backward time direction), then the max-min
principle will no longer hold. If fact, it can be demonstrated that, the IBVP of heat
equation in the backward time direction is not well-posed. For example, consider


 u,xx = u,t 0 < x < a , 0 < t < T
u(0, t) = u(a, t) = 0

 u(x, T ) = sin mπx a
The solution is u(x, t) = exp[−(m2 π 2 /a2 )(t − T )] sin(mπx/a). (The solution can be
verified by direct substitution.) We can make u(x, 0) as large as we want, for all by
making m arbitrary large!
112
3.6
Uniqueness Proof
3.6.1
Laplace and Poisson Equations
Consider a Dirichlet problem to Poisson equation
(
∇2 u = g(x)
in D
u = f (x)
on ∂D
If both u1 and u2 are two solutions of this problem, let w ≡ u1 − u2 , then w satisfies
(
∇2 w = 0
in D
w=0
on ∂D
From Green’s first identity Eq. (3.2-2), by taking u = v = w, one has
Z
Z
Z
∂w
w∇2 w dV =
w
dS −
|∇w|2 dV
.
∂n
D
∂D
D
(3.6-1)
(3.6-2)
(3.6-3)
Then follows from the D.E. and B.C. of w, Eq. (3.6-2), one has ∇w ≡ 0 for x ∈ D. Hence w is
only a constant in D. Furthermore, from the B.C. that w = 0 on the boundary, one concludes
that w vanishes everywhere in D. Thus u1 = u2 , and the solution of Eq. (3.6-1) is unique.
Consider a Neumann problem to Poisson equation
(
∇2 u = g(x)
in D
∂u
=
f
(x)
on
∂D
∂n
(3.6-4)
If both u1 and u2 are two solutions of this problem, let w ≡ u1 − u2 , and use the same argument
as for Dirichlet problem. One can only conclude that w is a constant in D. Hence solution of
Eq. (3.6-4) is unique only up to a constant.
How about the uniqueness for Robin BVP?
3.6.2
Heat Equations
Consider an initial-boundary-valued problem (Dirichlet problem)

2

x∈D, t>0
 ∇ u − u,t = g(x, t)
u(x, 0) = f (x)
x ∈ D + ∂D
(I.C.)

 u(x, t) = h(x, t)
x ∈ ∂D , t > 0 (B.C.)
(3.6-5)
[pf:] Let u1 (x, t) and u2 (x, t) are two solutions of the IBVP, then the difference w(x, t) ≡
u1 (x, t) − u2 (x, t) satisfies the homogeneous IBVP

2

x∈D, t>0
 ∇ w − w,t = 0
(3.6-6)
w(x, 0) = 0
x ∈ D + ∂D (I.C.)

 w(x, t) = 0
x ∈ ∂D , t > 0 (B.C.)
[Method 1:] Since
= ∇ · (∇w2 )
∇ w = 2∇ · (w∇w)
= 2[|∇w|2 + w∇2 w]
2
2
(3.6-7)
113
From the last two equalities of Eq. (3.6-7), one has
w(∇2 w − w,t ) = ∇ · (w∇w) − |∇w|2 − (1/2)(w2 ),t
(3.6-8)
Integrating Eq. (3.6-8) over the space-time domain yields
Z Z T
Z Z T
2
w(∇ w − w,t ) dt dV =
[∇ · (w∇w) − |∇w|2 − (1/2)(w2 ),t ] dt dV
D
0
D
Z Z
T
=⇒
D
0
Z Z
=⇒
D
1
|∇w|2 dt dV +
2
T
0
0
w2
D
t=T
− w2
t=0
Z
T
Z
dV =
w
0
∂D
∂w
dS dt
∂n
Z
1
w2 t=T dV = 0
2 D
in QT : the space-time domain
|∇w|2 dt dV +
=⇒ ∇w = 0
Z
=⇒ w = constant
in QT
=⇒ from B. C. w ≡ 0
in QT
[Method 2:] Consider an integral
Z
1
I(t) =
w2 dV ≥ 0
2 D
Hence
dI
dt
Z
=
ww,t dV
ZD
=
w ∇2 w dV
[from D.E.]
D
Z
[∇ · (w∇w) − |∇w|2 ] dV
[from Eq. (3.6-8)]
D
Z
Z
∂w
=
w
dS −
|∇w|2 dV
[divergence theorem]
∂n
∂D
D
Z
= −
|∇w|2 dV
[from B.C.]
=
D
≤ 0
Consequently, I is a non-negative and non-increasing function of t. Since I(0) = 0, (from I.C.),
we conclude I ≡ 0. Hence w ≡ 0 everywhere.
114
3.7
Problems
1. Find the general solutions of
u,xx − 2u,xy sin x − u,yy cos2 x − u,y cos x = 0
[Hint: reduce to the canonical form first.]
2. Let the operators L1 , L2 be defined by
L1 u = au,x + bu,y + cu ,
L2 u = du,x + eu,y + f u,
where a, b, c, d, e, f are constants with ae − bd 6= 0. Prove that
(a) The equations L1 u = w1 , L2 = w2 have a common solution u, if L1 w2 = L2 w1 .
(Hint: By linear transformation reduce to the case a = e = 1, b = d = 0.)
(b) The general solution of L1 L2 u = 0 has the form u = u1 + u2 , where L1 u1 = 0,
L2 u2 = 0.
3. Solve the Green’s function for 1-D wave equation
 2
∂ G0
∂ 2 G0


−
= δ(x − x0 ) δ(t)

2
∂x
∂t2
G (x, t) = 0

 0

G0 (x, t) → 0
−∞ < x, x0 < ∞ , t > 0−
t < 0−
|x| → ∞
4. Find the Green’s function of 2D Laplace equation in a quarter plane with Dirichlet and
Neumann B.C. on x-axis and y-axis, respectively,
 2
∇ G = δ(x − x0 ) δ(y − y0 ) 0 < x, x0 < ∞, 0 < y, y0 < ∞




 G(x, 0) = 0
∂G(0, y)

=0


∂x


G → 0,
as |r| → ∞ ,
where r2 = x2 + y 2
5. Use the method of image to construct the solution to the following problem.

∇2 G = δ(x − x0 ) δ(y − y0 )




 G=0
G=0




 ∂G = 0
∂x
x > 0, y > 0, x2 + y 2 < 1
on x2 + y 2 = 1
on 0 < x < 1, y = 0
on x = 0, 0 < y < 1
6. Use the method of image to construct the solution to the following problem.

∇2 G = δ(x − x0 ) δ(y − y0 )



 G=0
∂G


∂y = 0


G=0
x > 0, y > 0, x2 + y 2 < 1
on x2 + y 2 = 1
on 0 < x < 1, y = 0
on x = 0, 0 < y < 1
You can express your solution in terms of G0 directly, where G0 is the free-space Green’s
function of the problem.
115
7. Consider an initial-boundary-valued problem for wave equation
∇2 u −
∂2u
= g(x, t),
∂t2
u(x, 0) = f (x)
u(x, t) = h(x, t)
x ∈ D, t > 0,
x ∈ D + ∂D, t = 0,
x ∈ ∂D, t > 0.
Derive the integral representation for u in terms of g, f , h and an appropriate Green’s
function solution. [Hint: You might still need an adjoint Green’s function problem where
the initial conditions in ordinary Green’s function problem are replaced by corresponding
terminate conditions.]
8. Consider the following two initial-boundary-valued problems


∇2 G − G,tt = δ(x − x0 ) δ(t − t0 ),
x ∈ D, t > t0 ,


 G(x, t) = 0,
x ∈ D + ∂D, t < t0 ,

G,
(x,
t)
=
0,
x ∈ D + ∂D, t < t0 ,
t



n · ∇G(x, t) + k G(x, t) = 0,
x ∈ ∂D, t > t0 ,

2

∇ g − g,tt = 0,
x ∈ D, t > t0 ,



g(x, t) = δ(x − x0 ),
x ∈ D + ∂D, t = t0 ,

g,
(x,
t)
=
0,
x ∈ D + ∂D, t < t0 ,
t



n · ∇g(x, t) + k g(x, t) = 0, x ∈ ∂D, t > t0 ,
where ∇2 is the Laplacian operator in (x, y, z)-coordinate, and n is the unit outward
normal vector of ∂D. For t > t0 , show that
g(x, t; x0 , t0 ) = G,t (x, t; x0 , t0 )
9. Consider
∇2 u =
1 ∂ 2 u F (x, t)
−
,
c2 ∂t2
ρc2
x ∈ Ω,
t > 0,
∂u
+ hu = K(x, t), x ∈ ∂Ω, t > 0,
∂n
u(x, 0) = f (x),
ut (x, 0) = g(x), x ∈ Ω.
l
Show that
Z tZ
u(x, t)
Z =
G(x, t; ξ, T ) F (ξ, T )dξdT + ρ
0
+
ρc2
l
Ω
Ω
∂G(x, t; ξ, 0)
g(ξ)G(x, t; ξ, 0) − f (ξ)
∂T
dξ
Z tZ
G(x, t; ξ, T )K(ξ, T )dξdT,
0
∂Ω
if l 6= 0 and G(x, t; ξ, T ) is its Green’s function.
10. Referring to the previous problem (prob. 9), suppose l = 0, then
Z tZ
u(x, t)
Z =
0
−
g(ξ)G(x, t; ξ, 0) − f (ξ)
G(x, t; ξ, T ) F (ξ, T )dξdT + ρ
ρc2
h
Ω
Ω
Z tZ
K(ξ, T )
0
∂Ω
∂G(x, t; ξ, T )
dξdT.
∂nξ
116
∂G(x, t; ξ, 0)
∂T
dξ
11. Consider
∇2 u + k 2 u = F (x),
u(x) = K(x),
x ∈ Ω,
x ∈ ∂, Ω,
(3.7-1)
where k > 0 is a constant. Solve u in terms of the following Green’s function:
∇2 G(x; ξ) + k 2 G(x; ξ) = δ(x − ξ),
G = 0,
x ∈ Ω,
x ∈ ∂Ω.
12. Consider
∇2 u = ut − g(x, t), x ∈ Ω, t > 0,
∂u
+ hu = F (x, t), x on ∂Ω, t > 0,
l
∂n
u(x, 0) = f (x),
x ∈ Ω.
Show that the solution can be expressed in the form
Z tZ
Z
u(x, t) =
G(x, t; ξ, T )g(ξ, T )dξdT +
G(x, t; ξ, 0)f (ξ)dξ
0
Ω
Ω
Z Z
1 t
G(x, t; ξ, T )F (ξ, T )dξdT
+
l 0 ∂Ω
if l 6= 0, or
Z tZ
u(x, t)
=
−
Z
G(x, t; ξ, T )g(ξ, T )dξdT +
G(x, t; ξ, 0)f (ξ)dξ
0
Ω
Ω
Z Z
1 t
∂G(x, t; ξ, T )
F (ξ, T )
dξdT
h 0 ∂Ω
∂nξ
if l = 0, h 6= 0.
13. Solve the modified Green’s function of the following system
∂ 2 G(x; ξ) ∂ 2 G(x; ξ)
1
+
= δ(x − ξ) −
,
∂x21
∂x22
LH
∂G
= 0,
x ∈ ∂Ω,
∂nx
x ∈ Ω,
where Ω is a rectangle 0 ≤ x1 ≤ L, 0 ≤ x2 ≤ H.
14. (a) Show that when the homogeneous problem
∇2 u + k 2 u = 0,
u(x) = 0,
x ∈ Ω,
x ∈ ∂Ω
(3.7-2)
has nontrivial solution w(x), the original nonhomogeneous problem, eq. (3.7-1), has
a solution only if
Z
Z
∂w(x)
F (x)w(x)dx = −
K(x)
dx
(3.7-3)
∂n
Ω
∂Ω
is satisfied. (Note that the converse result is also true; that is, when eq. (3.7-3) is
satisfied, eq. (3.7-1) has a solution that is unique to an additive term Cw(x), C an
arbitrary constant.)
117
(b) Show that the solution of eq. (3.7-1) can be expressed in the form
Z
Z
∂GM (ξ; x)
u(x) =
GM (ξ; x)F (ξ)dξ +
K(ξ)
dξ + Cw(x),
∂nξ
Ω
∂Ω
where GM (x; ξ) is a modified Green’s function satisfying
(∇2 + k 2 )GM = δ(x − ξ) − w(x)w(ξ),
GM = 0,
ξ ∈ Ω,
x ∈ ∂Ω,
and w(x) is a normalized solution of eq. (3.7-2).
15. Consider an eigenvalue problem
∇2 u + λu = 0 ,
in D
∂u/∂n + µu = 0 , (µ > 0) on S,
where ∇2 is a three dimensional Laplacian operator, and S is the boundary surface of the
bounded region D. Show that
(a) the eigenvalues, λ’s, are real and positive,
(b) eigenfunctions corresponding to different eigenvalues are orthogonal to each other.
16. Solve the Green’s function of the following system
∂ 2 G(x; ξ) ∂ 2 G(x; ξ)
+
= δ(x − ξ),
∂x21
∂x22
G = 0,
x ∈ ∂Ω,
x ∈ Ω,
where Ω is a rectangle 0 ≤ x1 ≤ L, 0 ≤ x2 ≤ H.
17. Consider an initial-boundary-valued problem
∂u
∂2u
=
+ q(x, t), 0 < x < 1, t > 0,
∂t
∂x2
u(0, t) = A(t),
u(1, t) = B(t),
t > 0,
u(x, 0) = f (x),
0 < x < 1.
Solve this problem by the following three different approaches
(a) use the principle of linear superposition, and convert the problem from inhomogeneous boundary conditions into homogeneous boundary conditions with addition
inhomogeneous terms in differential equation.
(b) use Laplace transform over t directly.
P∞
(c) use eigenfunction expansion directly. [Hint: let u(x, t) = n=1 bn (t)φn (x), where
φn (x) is appropriate eigenfunction of this problem. Due to the fact of the inhomogeneous boundary conditions, we expect the series of eigenfunction expansion might
not be uniformly convergent in x. Hence the term-by-term differentiation in x is
not justified, but the term-by-term differentiation in t is still valid.] [Hint: You may
need the identity
Z
∂u
[φ(∇2 u −
) − u(∇2 φ + λφ)]dV =
∂t
D
Z
Z
∂u
∂φ
∂u
(φ
− u )dS −
(φ
+ λuφ)dV,
∂n
∂n
∂t
∂D
D
118
to establish a first order ODE of bn (t).]
18. Consider a boundary value problem
u,xx +u,yy = e−|y| sin(πx), 0 < x < 1, −∞ < y < ∞,
u(0, y) = e−|y| ,
u(1, y) = e−2|y| , −∞ < y < ∞, ; u(x, y) → 0, | y |→ ∞.
Solve this problem directly, without reducing to homogeneous boundary conditions.
19. Consider an initial-boundary-valued problem
∂u
∂2u
=
+ δ(t) sin πx, 0 < x < 1, t > 0− ,
∂t
∂x2
u(0, t) = t,
u(1, t) = 0,
t > 0,
u(x, 0− ) = 0,
0 < x < 1.
Solve this problem by partial eigenfunction in x directly, without reducing to homogeneous boundary
20. Solve the following initial-boundary-valued problem by using partial eigenfunction in x
directly
∂u
∂2u
=
+ δ(t) sin πx, 0 < x < 1, t > 0− ,
∂t
∂x2
u(0, t) = 0,
u(1, t) = 0,
t > 0,
u(x, 0− ) = 0,
0 < x < 1.
i.e. solve the problem by letting u =
eigenfunctions.
P
Am (t)φm (x), where φm (x)’s are the associated
21. Solve the Green’s function for 1-D wave equation
 2
∂ G ∂2G


−
= δ(x − x0 ) δ(t)
0 < x, x0 < 1 , t > 0−


∂t2
 ∂x2
G(x, t) = 0
t < 0−

 G(0, t) = u0
t > 0−



G(1, t) = 0
t > 0−
22. Consider
∂2G
1 ∂2G
1
=
− 2 δ(x − ξ)δ(t − T ),
∂x2
c2 ∂t
ρc
G = 0, x = 0, t > T,
G = 0,
x = L,
G(x, t; ξ, T ) = 0,
0 < x < L,
t > T,
0 < x < L,
t < T.
Show that the causal Green’s function is
G(x, t; ξ, T ) =
where fn (x) =
q
2
L
∞
L X1
nπc(t − T )
sin
fn (ξ)fn (x),
ρπc n=1 n
L
sin nπx
L .
119
t > T,
23. Solve IBVP
∂2u
1 ∂2u
1
=
− 2 t sin2 πx,
2
2
∂x
c ∂t
ρc
u = 0, x = 0, t > 0,
u = 0,
x = L,
0 < x < L,
t > 0,
t > 0,
u(x, 0) = ut (x, 0) = 0,
0 < x < L.
24. Consider a boundary value problem
 4
∂4u
∂4u
∂ u


+ 2 2 2 + 4 = δ(x − x0 )δ(y − y0 ),


4

∂x
∂x y
∂y


 u(0, y) = 0, u00 (0, y) = 0,
u(1, y) = 0, u00 (1, y) = 0,





u(x, 0) = 0, u00 (x, 0) = 0,



u(x, 1) = 0, u00 (x, 1) = 0,
0 < x, x0 < 1, 0 < y, y0 < 1,
0 < y < 1,
0 < y < 1,
0 < x < 1,
0 < x < 1,
Solve this problem by constructing
u=
∞ X
∞
X
Amn sin mπx sin nπy
m=1 n=1
25. Show that, if u,xx +u,yy −u,t ≥ 0, for t ≥ 0, at all points within a bounded region D of
the (x, y)-plane, then the maximum value of u is either attained when t = 0 or on the
boundary of D. [Note that the original maximum-minimum principle can not be directly
applied to this problem due to the ≥ sign in the PDE.]
26. Let u(x, y, t) satisfy u,xx +u,yy −u,t = h(x, y, t), where h(x, y, t) > 0, for t ≥ 0 at all points
within a bounded region D of the (x, y)-plane. For m = min u(x, y, t), show that the
maximum value of U (x, y, t) ≡ u2 −2m u is either attained when t = 0 or on the boundary
of D. [hint: for your reference ∇2 f 2 = 2(∇f )2 + 2f ∇2 f ]
27. Show that if
ut = uxx
for 0 < x < l
and
∂u
(0, t) = 0,
∂x
the maximum of u for 0 ≤ x ≤ l and 0 ≤ t ≤ t∗ must occur at either t = 0 or at x = l.
(Hint: Use the reflection with respect to the origin.)
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