Module 3: Second-Order Partial Differential
Equations
In Module 3, we shall discuss some general concepts associated with second-order linear
PDEs. These types of PDEs arise in connection with various physical problems such as
the motion of a vibration string, heat flow, electricity, magnetism and fluid dynamics. The
second-order partial differential equations are classified into three different types. Further,
the simplified canonical/normal forms are obtained for second order linear equations in
two independent variables.
The lectures are organized as follows. The first lecture is devoted to the classification
of second-order linear PDEs in two or more independent variables. These equations are
classified into hyperbolic, parabolic and elliptic types. The reduction to the canonical form
(or normal form) of second order equations in two independent variables are discussed in
the second lecture. Finally, the third lecture is devoted to wellposedness, superposition
principle and method of factorization for these types of equations.
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MODULE 3: SECOND-ORDER PARTIAL DIFFERENTIAL EQUATIONS
Lecture 1
2
Classification of Second-Order PDEs
Classification of PDEs is an important concept because the general theory and methods
of solution usually apply only to a given class of equations. Let us first discuss the
classification of PDEs involving two independent variables.
1
Classification with two independent variables
Consider the following general second order linear PDE in two independent variables:
A
∂2u
∂2u
∂2u
∂u
∂u
+
B
+
C
+D
+E
+ F u + G = 0,
2
2
∂x
∂x∂y
∂y
∂x
∂y
(1)
where A, B, C, D, E, F and G are functions of the independent variables x and y. The
equation (1) may be written in the form
Auxx + Buxy + Cuyy + f (x, y, ux , uy , u) = 0,
(2)
where
∂u
∂u
∂2u
∂2u
∂2u
, uy =
, uxx =
,
u
=
,
u
=
.
yy
xy
∂x
∂y
∂x2
∂x∂y
∂y 2
Assume that A, B and C are continuous functions of x and y possessing continuous partial
ux =
derivatives of as high order as necessary.
The classification of PDE is motivated by the classification of second order algebraic
equations in two-variables
ax2 + bxy + cy 2 + dx + ey + f = 0.
(3)
We know that the nature of the curves will be decided by the principal part ax2 +bxy +cy 2
i.e., the term containing highest degree. Depending on the sign of the discriminant b2 −4ac,
we classify the curve as follows:
If b2 − 4ac > 0 then the curve traces hyperbola.
If b2 − 4ac = 0 then the curve traces parabola.
If b2 − 4ac < 0 then the curve traces ellipse.
With suitable transformation, we can transform (3) into the following normal form
x2 y 2
− 2 = 1 (hyperbola).
a2
b
2
x = y (parabola).
x2 y 2
+ 2 = 1 (ellipse).
a2
b
MODULE 3: SECOND-ORDER PARTIAL DIFFERENTIAL EQUATIONS
3
Linear PDE with constant coefficients. Let us first consider the following general
linear second order PDE in two independent variables x and y with constant coefficients:
Auxx + Buxy + Cuyy + Dux + Euy + F u + G = 0,
(4)
where the coefficients A, B, C, D, E, F and G are constants. The nature of the equation
(4) is determined by the principal part containing highest partial derivatives i.e.,
Lu ≡ Auxx + Buxy + Cuyy .
(5)
For classification, we attach a symbol to (5) as P (x, y) = Ax2 + Bxy + Cy 2 (as if we have
replaced x by
∂
∂x
and y by
∂
∂y ).
Now depending on the sign of the discriminant (B 2 −4AC),
the classification of (4) is done as follows:
B 2 − 4AC > 0 =⇒ Eq. (4) is hyperbolic
(6)
B 2 − 4AC = 0 =⇒ Eq. (4) is parabolic
(7)
B 2 − 4AC < 0 =⇒ Eq. (4) is elliptic
(8)
Linear PDE with variable coefficients. The above classification of (4) is still valid if
the coefficients A, B, C, D, E and F depend on x, y. In this case, the conditions (6), (7)
and (8) should be satisfied at each point (x, y) in the region where we want to describe its
nature e.g., for elliptic we need to verify
B 2 (x, y) − 4A(x, y)C(x, y) < 0
for each (x, y) in the region of interest. Thus, we classify linear PDE with variable coefficients as follows:
B 2 (x, y) − 4A(x, y)C(x, y) > 0 at (x, y) =⇒ Eq. (4) is hyperbolic at (x, y)
(9)
B 2 (x, y) − 4A(x, y)C(x, y) = 0 at (x, y) =⇒ Eq. (4) is parabolic at (x, y)
(10)
B 2 (x, y) − 4A(x, y)C(x, y) < 0 at (x, y) =⇒ Eq. (4) is elliptic at (x, y)
(11)
Note: Eq. (4) is hyperbolic, parabolic, or elliptic depends only on the coefficients of the
second derivatives. It has nothing to do with the first-derivative terms, the term in u, or
the nonhomogeneous term.
EXAMPLE 1.
1. uxx + uyy = 0 (Laplace equation). Here, A = 1, B = 0, C = 1 and B 2 − 4AC =
−4 < 0. Therefore, it is an elliptic type.
MODULE 3: SECOND-ORDER PARTIAL DIFFERENTIAL EQUATIONS
4
2. ut = uxx (Heat equation). Here, A = −1, B = 0, C = 0. B 2 − 4AC = 0. Thus, it is
of parabolic type.
3. utt − uxx = 0 (Wave equation). In this case, A = −1, B = 0, C = 1 and B 2 − 4AC =
4 > 0. Hence, it is of hyperbolic type.
x ̸= 0 (Tricomi equation). B 2 − 4AC = −4x. Given PDE is
4. uxx + xuyy = 0,
hyperbolic for x < 0 and elliptic for x > 0. This example shows that equations with
variable coefficients can change form in the different regions of the domain.
2
Classification with more than two variables
Consider the second-order PDE in general form:
n ∑
n
∑
i=1 j=1
∑ ∂u
∂2u
aij
+
bi
+ cu + d = 0,
∂xi ∂xj
∂xi
n
(12)
i=1
where the coefficients aij , bi , c and d are functions of x = (x1 , x2 , · · · , xn ) alone and u =
u(x1 , x2 , · · · , xn ).
Its principal part is
L≡
n ∑
n
∑
aij
i=1 j=1
∂2
.
∂xi ∂xj
(13)
It is enough to assume that A = [aij ] is symmetric if not, let āij = 12 (aij + aji ) and rewrite
L≡
n ∑
n
∑
āij
i=1 j=1
Note that
∂2u
∂xi ∂xj
with (14) (i.e.,
∂2u
∂xj ∂xi . As in two-space
∂u
replacing ∂x
by xi ).
i
=
∂2
.
∂xi ∂xj
(14)
dimension, let us attach a quadratic form P
P (x1 , x2 , · · · , xn ) =
n ∑
n
∑
aij xi xj .
(15)
i=1 j=1
Since A is a real valued symmetric (aij = aji ) matrix, it is diagonalizable with real
eigenvalues λ1 , λ2 , . . . , λn (counted with their multiplicities). In other words, there exists
a corresponding set of orthonormal set of n eigenvectors, say σ1 , σ2 , · · · , σn with R =
MODULE 3: SECOND-ORDER PARTIAL DIFFERENTIAL EQUATIONS
[σ1 , σ2 , · · · , σn ] as column vectors such that

λ1


λ2
⃝


·

RT AR = 

·


⃝
·

5






=D




(16)
λn
We now classify (12) depending on sign of eigenvalues of A:
(a) If λi > 0 ∀i or λi < 0 ∀i then (12) is elliptic type.
(b) If one or more of the λi = 0 then (12) is parabolic type.
(c) If one of the λi < 0 or λi > 0, and all the remaining have
opposite sign then (12) is said to be of hyperbolic type.
EXAMPLE 2.
1. ∇2 u = uxx + uyy + uzz = 0. In this case, λi = 1 > 0 for all i = 1, 2, 3. It is an elliptic
PDE since all eigenvalues are of one sign.
2. It is an easy exercise to check that ut − ∇2 u = 0 is of parabolic type.
3. The equation utt − ∇2 u = 0 is of hyperbolic type.
EXAMPLE 3. Classify ux1 x1 + 2(1 + cx2 )ux2 x3 = 0.
To symmetrize, write it as
ux1 x1 + (1 + cx2 )ux2 x3 + (1 + cx2 )ux3 x2 = 0
i.e., ∂xT A∂x − c∂x2 = 0, where

1


A= 0

0
0
0 1 + cx2
0


∂x1





1 + cx2 
 ∂x =  ∂x2 
0
∂x3
Eigenvalues are λ1 = 1, λ2 = 1 + cx2 , λ3 = −(1 + cx2 ) and normalized eigenvectors
 




1
0
0
 
 √ 

√ 





σ1 = 
 0  σ2 =  1/√2  σ3 =  1/ √2 
1/ 2
−1/ 2
0
MODULE 3: SECOND-ORDER PARTIAL DIFFERENTIAL EQUATIONS
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

0
0

√
√ 

R=
 0 1/√2 1/ √2 
0 1/ 2 −1/ 2
So
1
Note that R = RT = R−1 .


1
0

RT AR = 
 0 1 + cx2
0
0
0
0
−(1 + cx2 )

=D

Equation is parabolic if x2 = − 1c (c ̸= 0), hyperbolic if x2 > − 1c and x2 < − 1c . For c = 0,
λ1 = λ2 = 1 and λ3 = −1, it is hyperbolic type.
Practice Problems
1. Classify the following equations into hyperbolic, elliptic or parabolic type.
(A) 5uxx − 3uyy + (cos x)ux + ey uy + u = 0.
(B) ex uxx + ey uyy − u = 0.
(C) xuxx + uyy = 0.
(D) 8uxx + uyy − ux + [log(2 + x2 )]u = 0.
(E) sin2 xuxx + sin 2xuxy + cos2 xuyy = x.
2. Classify the following equations into elliptic, parabolic, or hyperbolic type.
2
(A) uxx + 2uyz + (cos x)uz − ey u = cosh z.
(B) uxx + 2uxy + uyy + 2uzz − (1 + xy)u = 0.
(C) ez uxy − uxx = log[x2 + y 2 + z 2 + 1].
3. Determine the regions where uxx − 2x2 uxz + uyy + uzz = 0 is of hyperbolic, elliptic
and parabolic.