PARTIAL DIFFERENTIAL EQUATIONS
Prof. Fine
1
PARTIAL DIFFERENTIAL EQUATIONS
Prof. Benjamin Fine
Text: M. Coleman Partial Differential Equations Prentice
Hall
Course Goals: To provide an introduction to the mathematical techniques that are used in the study of partial
differential equations as well some very important applications. There will three projects and three exams.
Prerequisites: Knowledge of one-variable and multi-variable
Calculus,linear Algebra and basic ordinary differential equations. There will be three projectsthis semester.
Course Outline
A. Ordinary Differential Equations
1. Basic ideas, order and solution techniques
2. Linear Differential Equations - fundamental existence
and uniqueness theorem
B. Partial Differential Equations
1.PDE’s and Classification
2. Elementary Solution Technqiues
3. Boundary Value Problems - the big three: Heat,Wave
and Laplace equations
C. Vector and Scalar Field Theory
1. Vector and Scalar Fields - gradient, divergence and
curl
2
2. Line, Surface and Volume Integrals - Stoke’s and
Divergence Theorem
3. Derivation of the Heat and Wave Equation
D. Fourier Series Methods
1. Fourier Series and Fourier Expansions
2. Separation of Variables
E. Solutions of the Big Three
1. Solution of the Heat Equation
2. Solution of the Wave Equation
a. D’Alembert’s Method
3. Solution of the Laplace Equation
F. Transform Methods
1. Basic ideas of Integral transforms
2. The Laplace transform and Applications
3. The Fourier transform and Applications - Fourier
Inversion Theorem
3
PARTIAL DIFFERENTIAL EQUATIONS (1)
PARTIAL DIFFERENTIAL EQUATIONS - BASIC IDEAS
A. Differential Equations
1.Definition A differential equation is an equation
which involves a variable, (or variables), a function of that
variable and one or more of the derivatives of that variable.
We express this as
f (x, y, y ′ , y ′′ , ..., y (n) ) = 0
a.If the equation involves only ordinary derivatives its
called an ordinary differential equation or ODE while if
there are partial derivatives it a partial differential equation or PDE.
b. The order of a differential equation is the order of
the highest derivative which appears in the equation.
EXAMPLES
(1) y ′ = y. This is a first order ODE.
(2) x2 ex dy + 2x sin xdx = 0. This is again a first order ODE
(3) y ′′ + (x2 + cos x)y = ex . This is a second order ODE
(4)
∂2u
∂x2
+
∂2u
∂y 2
= 0. This is a second order PDE.
2. Solutions
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a.Definition A function y = y(x) defined on some
interval I which satisfies the equation is a solution.
(1) Generally there will be infinitely many solutions.
Sometimes all solutions to an ODE will be given in terms
of some real variables called parameters. The solution
given in terms of these parameters is called the general
solution. Particular values of these parameters lead to
particular solutions.
EXAMPLES
(1) y ′ = y. The general solution is given by
y = cex
The value c is the parameter. Any value of c gives a particular solution so for example y = 4ex is a particular solution.
(2) y ′′ + y = 0. The general solution is given by
y = c1 sin x + c2 cos x
The parameters are c1 , c2 . Specific values for c1 , c2 gives a
particular solution.
(2) If the solution y = y(x) is given explicitly then
we have an explicit solution. If the solution is given as
f (x, y) = 0 then its an implicit solution
(3) If we have the ODE
f (x, y, y ′ , ..., y (n) ) = 0
5
and we specify particular values at a given point
y(x0 ) = y0 , y ′ (x0 ) = y1 , ....., y (n) (x0 ) = yn
then its an initial value problem. Generally under many
conditions an initial value problem will have a unique solution.
(4) If we have the ODE
f (x, y, y ′ , ..., y (n) ) = 0
and we specify y(a) = y0 , y(b) = y1 at two different points
then we have a boundary value problem.
2. Linear Differential Equations
a. Definition A linear differential equation is one
of the form
an (x)y (n) + an−1 (x)y (n−1) + ... + a0 (x)y = g(x)
That is in a linear differential equation no derivative appears with an exponent higher than one. Otherwise it is
non-linear.
b. Fundamental Existence and Uniqueness Theorem
(1) Theorem Consider the initial value problem
an (x)y (n) + an−1 (x)y (n−1) + ... + a0 (x)y = g(x)
y(x0 ) = y0 , y ′ (x0 ) = y1 , ..., y n (x0 ) = yn+1
with x0 ∈ I. Then there exists a unique solution to
the initial value problem.
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c. Homeogenous and Nonhomogeneous Equations
(1) Theorem The set of solutions to a homogeneous
linear differential equation of order n from a vector space
of dimension n. A basis is called a fundamental set of
solutions
B. Partial Differential Equations
1.We will concentrate for the most part on first and second order linear PDE’s
a.Definition [1] A first order linear partial differential equation is a PDE of the form
aux + buy + cu = g
(1)
where a, b, c, g are all functions of x, y. We will assume
that these functions are defined and continuous over some
common region R in two-space. If a first-order PDE is not
of this form it is non-linear.
[2] A second order linear partial differential
equation is a PDE of the form
auxx + 2buxy + cuyy + dux + euy + f u = g
(2)
where a, b, c, d, e, f, g are all functions of x, y. We will assume that these functions are defined and continuous over
some common region R in two-space. If a second-order
PDE is not of this form it is non-linear.
(1) If g(x, y) ≡ 0 then the equation is homogeneous.
Otherwise it is non-homogeneous
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(2) We let L be the linear differential operator
∂2
∂2
∂
∂2
∂
L = a 2 + 2b
+c 2 +d
+ e + f.
∂x
∂x∂y
∂y
∂x
∂y
Then the second order PDE (1) can be expressed in compact form as
Lu = g
(3) Lemma If g ≡ 0 so that (1) is homogeneous then
if u1 , u2 , ..., un are solutions then any linear combination
c1 u1 + c2 u2 + ... + cn un
is also a solution.
Further if g ̸≡ 0 and up is any particular solution to (1)
then
up + c1 u1 + c2 u2 + ... + cn un
is also a solution, where u1 , ..., un are solutions to the associated homogeneous equation.
2. Classification of Second Order PDE’s
a.Definition Consider a second order linear partial differential equation
auxx + 2buxy + cuyy + dux + euy + f u = g
Then this is
(i) elliptic if b2 − ac < 0
(ii) parabolic if b2 − ac = 0
(iii) hyperbolic if b2 − ac > 0
8
(1)
b.Mixed Type Equations
(1) Note that if a, b, c are not constants then the equation can be of mixed type that is different in different
regions.
EXAMPLE Tricomi’s Equation which arises in the theory
of transonic flow is
uxx + xuyy = 0
Here b2 − ac = −x so the equation is elliptic in the right
half plane x > 0 and hyperbolic in the left half-plane x < 0.
C. The Big Three PDE’s
1. Laplace Equation
a.Definition The Laplace Equation is
uxx + uyy = 0
(1) Here a = 1, b = 0, c = 1 so it is an elliptic
equation
(2) Roughly Laplace equations arise is modeling equilibrium problems
2. Diffusion Equation or Heat Equation
a.Definition The Diffusion Equation or Heat Equation is
α2 uxx = ut
9
where α2 is the diffusivity constant and t is time.
(1) Here a = α2 , b = 0, c = 0 so it is a parabolic
equation
(2) Roughly diffusion equations arise is modeling diffusion problems such as heat transfer
3. Wave Equation
a.Definition The Wave Equation is
β 2 uxx = utt
where β 2 is a constant and t is time.
(1) Here a = β 2 , b = 0, c = −1 so it is a hyperbolic
equation
(2) Roughly wave equations arise in modeling wave
phenomena
D. Elementary Solution Techniques
1. Solution by Integration
a. Many linear PDE’s can be solved either by direct
integration or using methods of ordinary differential equations. Recall however that in partial integration the constant of integration is an arbitrary function of the other
variable.
EXAMPLE 1
10
∂ 2u
=0
∂y 2
Integrating with respect to y we get
∂u
= f (x)
∂y
integrating again
=⇒ u(x, y) = yf (x) + g(x)
where f (x), g(x) are arbitrary functions of x.
EXAMPLE 2
∂ 2u
∂u
+
=1
∂x∂y ∂y
Letting v =
∂u
∂y
we get
vx + v = 1
This is a linear ODE with an integrating factor e
∫
1dx
= ex
=⇒ (vex )′ = ex =⇒ ex v = ex + h(y)
∂u
=⇒ v = 1 + e−x h(y) =⇒
= 1 + e−x h(y)
∂y
∫
=⇒ u(x, y) = y + e−x h(y)dy + g(x)
where f (x) is an arbitrary function of x and h(y) is an
arbitrary function of y
EXAMPLE 3
11
∂ 2u
− y 2 u = ex
2
∂x
Treating y as a constant this is a linear ODE in X with
constant coefficients
uxx − y 2 u = ex
The corresponding homogeneous equation is
uxx − y 2 u = 0
which has an auxillary polynomial
m2 − y 2 = 0 =⇒ m = ±y =⇒ uc = c1 exy + c2 e−xy
However c1 , c2 are functions of y
=⇒ uc = f (y)exy + g(y)e−xy
This is the general solution of the corresponding homogeneous equation. To find a particular solution we’ll use the
method of undetermined coefficients
Let up (x) = A(y)ex
=⇒ A(y)ex − y 2 A(y)ex = ex
1
=⇒ A(y) =
1 − y2
ex
1 − y2
where f (y) and g(y) are an arbitrary functions of y
=⇒ u(x, y) = f (y)exy + g(y)e−xy +
EXAMPLE 4 A first order equation
aux + buy = f (x)
12
can be solved in the following manner.
method works if the right side is g(y).
(An analogous
Recall that
du = ux dx + uy dy
Now suppose that
b
dx dy
=
=⇒ dy = dx
a
b
a
=⇒ bx − ay = c a constant
Then
b
du
du = ux dx + uy dy = ux dx + uy dx =⇒ a
= aux + buy
a
dx
1
=⇒ du = f (x)dx
a
∫
1
f (x)dx + C
=⇒ u(x, y) =
a
where C is a constant. Then C = g(bx − ay) where g(t) is
an arbitrary function of a single variable.
Consider for example
3ux + 4uy = ex
Then
1
u(x, y) = ex + g(4x − 3y)
3
2. Method of Separation of Variables
a.For a homogeneous linear PDE the method of separation of variables attempts to find a solution of the
form
u(x, y) = X(x)Y (y)
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EXAMPLE
∂ 2u
∂u
=
4
∂x2
∂y
Let u = XY where X = X(x), Y = Y (y) then
=⇒ X ′′ Y = 4XY ′
Y′
X ′′
=
=⇒
4X
Y
Since X is a function of x alone and Y is a function of y
alone this equality is possible only if both sides equal the
same constant c. The constant c is called the separation
constant.
Case 1 c = λ2 > 0 then
X ′′ − 4λ2 X = 0
Y ′ − λ2 Y = 0
=⇒ X = c1 e2λx + c2 e−2λx and Y = c3 eλ
2
y
=⇒ u(x, y) = Aeλ y (Be2λx + Ce−2λx )
2
Case 2 c = −λ2 < 0 then
X ′′ + 4λ2 X = 0
Y ′ + λ2 Y = 0
=⇒ X = c1 cos(2λx) + c2 sin(2λx) and Y = c3 e−λ
2
=⇒ u(x, y) = Ae−λ y (B sin(2λx) + C cos(−2λx))
2
Case 3 c = 0 then
X ′′ = 0
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y
Y′ =0
=⇒ X = c1 + c2 x and Y = c3
=⇒ u(x, y) = Ax + B
3. Superposition Principle
a. Theorem If u1 , u2 , ..., un are solutions to a linear
homogeneous PDE then anyy linear combination is also a
solution.
(1) We assume that if u1 , u2 ..., un , ... are an infinite
sequence of solutions then
∞
∑
ui
i=1
is also a solution
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PARTIAL DIFFERENTIAL EQUATIONS (2)
SCALAR AND VECTOR FIELD THEORY
A. Scalar and Vector Fields
1.Scalar Fields
a.Definition: A scalar field is a real-valued function
f (x, y, z) defined on a region in R3 .
b.Isotimic Surfaces
(1) Definition If f (x, y, z) is a scalar field then
a surface defined by f (x, y, z) = c is called an isotimic
surface. If f is gravitational potential or electric potential
called an equipotential surface. If f is temperature its
called an isothermal surface.
EXAMPLE f (x, y, z) = x3 z 2 − xy 2 z is a scalar field. The
surface defined by
x3 z 2 − xy 2 z = 9
is an isotimic surface.
2.Vector Fields
a. Definiiton A vector field is a function F which
assigns a tangent vector at each point p ∈ R3 . Thus a
vector field is a function F : R3 → R3 ,
F = F1 (x, y, z)i + F2 (x, y, z)j + F3 (x, y, z)k
F1 , F2 , F3 are the scalar field components of F. F is a differentiable vector field if its components are differentiable scalar fields. We will assume that our vector fields
are differentiable.
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EXAMPLE F = (x3 y 2 z 2 )i + (2x − z 3 y)j + (3y 3 ez )k is a
differentiable vector field. The functions F1 = x3 y 2 z 2 ,F2 =
2x − z 3 y, F3 = 3y 3 ez are the scalar field components.
b. Velocity Fields and Flow Lines
(1) It can be useful to think of a vector field as a
velocity vector field for some fluid moving through R3 . Thus
at each point p,
F(p) = velocity of the fluid at p.
(2) A curve is a flow line for a vector field F if at
each point of the curve F defines a tangent vector to the
curve.
(a) If R(t) = x(t)i + y(t)j + z(t)k is a curve then it
is a flow line for the vector field F if there exists a β ∈ R
such that
dx
dy
dz
βF1 =
, βF2 = , βF3 =
dt
dt
dt
B. Differential Operators on Scalar and Vector Fields
1. The Derivatives of Scalar Fields
a.The Gradient of a Scalar Field
(1) Definition If f is a differentiable scalar field then
its gradient is the vector field defined by
∂f
∂f
∂f
gradf =
i+
j+
k.
∂x
∂y
∂z
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EXAMPLE The gradient of the scalar field f (x, y, z) =
x3 z 2 − xy 2 z is the vector field
gradf = (3x2 z 2 − y 2 z)i + (−2xyz)j + (2x3 z − xy 2 )k
b.The Directional Derivative
(1) Definition Let p be a point, f (x, y, z) a scalar
field and v a unit vector with p considered as its endpoint.
Let g(t) = f (p + tv). Then
vp [f ] = g ′ (0)
is called the directional derivative of f in the direction
of v at p. Note that v is taken as a unit vector - that is the
directional derivative depends on the direction not on the
magnitude of the vector.
(a) vp [f ] measures the rate of change of f at p measured in the v direction.
(b) Lemma If v is a tangent vector at p ∈ R3 then
the directional derivative of a scalar field f in the direction
of v at p is given by
vp [f ] = v.gradf (p).
EXAMPLE Find the directional derivative of f (x, y, z) =
x3 z 2 − xy 2 z in the direction of v = 2i + 3j − k at the point
(2, 0, 1).
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Here a unit vector in the direction of v is
u=
v
3
1
2
= √ i+ √ j− √ k
|v|
14
14
14
gradf (2, 01, ) = (3x2 z 2 −y 2 z)i+(−2xyz)j+(2x3 z−xy 2 )k|(2,0,1) = 12i+16k
3
1
8
2
=⇒ vp [f ] = u.gradf (p) = √ i+ √ j− √ k).(12i+16k) = √
14
14
14
14
(2) Lemma gradf points in the direction of maximum rate of increase of the scalar field f . (Since gradf =
v.gradf then |gradf | = |v||gradf | cos θ. Since v is a unit
vector this will be greatest when cos θ = 1 or when the
angle is zero, that is, v is in the direction of the gradient.)
(3) Lemma |gradf | = maximum rate of increase of
f per unit distance.
(4) Theorem Through any point (x0 , y0 , z0 ) where
gradf ̸= 0 there passes an isotimic surface f (x, y, z) = c.
gradf is normal to this surface and therefore defines a set
of attitude numbers for the tangent plane.
2. The Derivatives of Vector Fields
a.The Rate of Change of a Vector Field
(1) There are two fundamental measures of the rate
of change of a vector field - the divergence and the curl.
The divergence of a vector field is a scalar field which roughly
measures the extent to which the field diverges from that
point. The curl of a vector field is a vector field which
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roughly measures the extent to which the field swirls about
that point.
b. The Divergence of a Vector Field
(1) The Divergence Operator
(a) Definition Consider a region B ⊂ R3 bounded
by a surface S with surface element dA, outward normal n
and volume V . The flux or flow of a vector field F across
a surface element is n.FdA. The divergence at a point
p ∈ B is
∫
n.FdA
DivF (p) = lim S
V →0
V
[1] The divergence of a vector field is then a
scalar field
[2] Note that the above is a coordinate independent
definition.
(2) Coordinate Expression for the Divergence
(a) A coordinate expression for the divergence is
obtained by considering B to be a paralleopiped.
(b) Theorem If F = F1 i + F2 j + F3 k is a vector
field then its divergence is the scalar field given by
DivF =
∂F1 ∂F2 ∂F3
+
+
.
∂x
∂y
∂z
EXAMPLE The divergence of the vector field
F = (x3 y 2 z 2 )i + (2x − z 3 y)j + (3y 3 ez )k
20
is the scalar field
DivF = 3x2 y 2 z 2 − z 3 + 3y 3 ez
(c) Derivation of the Coordinate Expression
(1) Consider a region B ⊂ R3 bounded by a surface
S with surface element dA,outward normal n and volume
V . The flux or flow of a vector field F across a surface
element is n.FdA. The divergence at a point p ∈ B is
∫
n.FdA
DivF (p) = lim S
V →0
V
(2) Now consider B to be a rectangular box alligned
with the coordinate axes and let P0 = (x0 , y0 , z0 ) be the
center of the box. Then an outward normal across the front
face is n = i while an outward normal across ther back face
is n = −i. Then if the force is given by
F = F1 i + F2 j + F3 k it follows that n.F = ±F1
Further across the front face
F1 (x, y, z) = F1 (x0 +
∆x
, y, z)
2
and across the back face
F1 (x, y, z) = F1 (x0 −
21
∆x
, y, z)
2
Then:
∫
∫
n.FdA =
front face
F1 (x0 +
front face
∆x
, y, z)dydz
2
Using the mean value theorem it follows that
= F1 (x0 +
∆x
, y1 , z1 )∆y∆z
2
for some y1 , z1 . Similarly across the back face
∫
∆x
n.FdA = F1 (x0 −
, y1 , z1 )∆y∆z
2
back face
further V = ∆x∆y∆z Hence:
∫
lim
V →0
front + back faces n.FdA
V
=
∂F1
∂x
Now the other faces follow in a similar manner.
c.The Curl of a Vector Field
(1) The Curl Operator
(a) Definition Consider a region B ⊂ R3 bounded
by a surface S with surface element dA,outward normal n
and volume V . The swirl of a vector field F about a point
P across a surface element is n×FdA. The curl at a point
p ∈ B is
∫
n × FdA
CurlF(p) = lim S
V →0
V
[1] The curl of a vector field is then also a vector
field
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[2] Note that the above is a coordinate independent
definition.
(2) Coordinate Expression for the Curl
(a) A coordinate expression for the curl is obtained
by considering B to be a paralleopiped.
(b) Theorem If F = F1 i + F2 j + F3 k is a vector
field then its curl is the vector field given by


i j k
∂
∂
∂ 
CurlF = det  ∂x
∂y ∂z
F1 F2 F3
EXAMPLE The curl of the vector field
F = (x3 y 2 z 2 )i + (2x − z 3 y)j + (3y 3 ez )k
is the vector field

i
j
k

∂
∂
∂ 
CurlF = det  ∂x
∂y ∂z
F1 F2 F3


i
j
k
∂
∂
∂ 
=⇒ CurlF = det  ∂x
∂y
∂z
(x3 y 2 z 2 2x − z 3 y 3y 3 ez
CurlF = (9y 2 ez − 3z 2 y(i + (2x3 y 2 z)j + (2 − 2x3 yz 2 )k
(3) Curl as Angular Velocity
23
(a) Considering F as a velocity vector field for a
moving fluid then
CurlF =
1
angular velocity
2
(b) Derivation as Angular Velocity
Consider a planar fluid with velocity vector field
F = F1 i + F2 j + F3 k
swirling counterclockwise about the z-axis around a point
P at the center of a rectangle with corners O, A, B. Here
F3 = 0 since its planar.
Then the angular velocity of the fluid about P is the average
of the angular velocity of the segment OA and the angular
velocity of the segment OB. We then have
angular velocity of OA =
F2 (A) − F2 (O)
∂F2
k→
k as ∆x → 0
∆x
∂x
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angular velocity of OB =
F1 (O) − F1 (B)
∂F1
k→−
k as ∆y → 0
∆y
∂y
Then
1 ∂F2 ∂F1
ω= (
−
)k
2 ∂x
∂y
Putting all three directions together gives that the curl is
twice the angular velocity.
3.The Del Notation
a. We define the del operator by
▽=
∂
∂
∂
i+ j+ k
∂x
∂y
∂z
It can operate on either scalar or vector fields as described
below.
(1)Lemma If f is a scalar field and F is a vector field
then
▽f = gradf
▽.F = DivF
▽ × F = CurlF
4.The Laplacian
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a. Definition If f is a scalar field then its Laplacian
is the scalar field defined by
∂ 2f ∂ 2f ∂ 2f
+
+
.
▽ f = Div(gradf ) =
∂x2 ∂y 2 ∂z 2
2
(1) Roughly the Laplacian measures the difference between the average value of the field in an immediate neighborhood of a point and the precise value of the field at that
point
b. The Laplacian in Mathematical Physics
(1) ▽2 f = 0 is Laplaces equation and a solution is
a harmonic function
(2) ▽2 f =
(3) ▽2 f =
1 ∂2f
a2 ∂t2 is the wave equation
1 ∂f
a2 ∂t is the heat equation
5.Some Vector Operator Identities
a. Suppose f1 , f2 are scalar fields and F, G are vector
fields then
(1)▽(f1 f2 ) = f1 ▽f2 + f2 ▽f1
(2)▽.f F = f (▽.F) + F.▽f
(3)▽ × f F = f (▽ × F) + ▽f × F
∂f
(4)▽f (u) =
▽u
∂u
(5)▽.(F × G) = G.▽ × F − F.(▽ × G)
(6)▽ × (F × G) = (g.▽)F − (F.▽)G + (▽.G)F − (▽.F)G
(7)▽ × (▽ × F) = ▽(▽.F) − ▽2 F
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(8)▽(F.G) = (F.▽)G+(G.▽)F+F×(▽×G)+G×(▽×F)
(9)▽ × ▽f = 0 ⇒ Curl(gradf ) = 0
(10)▽.(▽ × F) = 0 ⇒ Div(CurlF) = 0
(11)▽.(▽f1 × ▽f2 ) = 0
6.Summary of Vector Operators
VECTOR OPERATORS
▽=
∂
∂
∂
i+ j+ k
∂x
∂y
∂z
Name
Interpretation
gradf = ▽f
Maximum rate of change of f in the maximal direction
DivF = ▽.F
Net outflux of F per unit volume
CurlF = ▽ × F
swirl of F per unit area
2
Laplacian(f ) = ▽ f difference between f (P ) and the average of f around P
C. Vector Field Integration
1.Line Integrals
a.Definition: Let F be a continuous vector field defined on a domain D containing a curve C. The the line
integral of F over C is
∫
∑
F.dR = lim|∆Ri |→0
F(Pi )∆Ri
C
where R = R(t) is a parametrization of the curve.
27
In terms of the parameter t this is given by
∫
∫
∫
F.dR =
F.Tds =
Ft ds
c
C
C
where Ft is the tangential component of F over C.
This can also be written as
∫
∫
F.dR =
F1 dx + F2 dy + F3 dz
C
C
F1 dx + F2 dy + F3 dz is a first order differential form and
thus the line integral is the integral of this differential form
over C.
b. Therefore for a line integral we need a vector field
and an oriented curve and the result is a scalar.
c. Computed in terms of the parameter t
∫ t2
dy
dz
dx
F.dR =
(F1 (x(t), y(t), z(t)) +F2 (x(t), y(t), z(t)) +F3 (x(t), y(t), z(t)) )
dt
dt
dt
C
t1
∫
2.Conservative Vector Fields
a. A vector field F is conservative if there exists a
scalar field ϕ(x, y, z) such that F = gradϕ. ϕ is a potential
function or potential for F.
(1) Lemma. If F is a conservative vector field then
CurlF = 0.
∫
b. C F.dR is independent of path if the value of the
line integral depends only on the endpoints of the curve C.
(1) Theorem. Suppose F is a continuous vector field
on a domain D. Then the following are equivalent:
28
(i) F is conservative
∫
(ii) C F.dR is independent of path
H
(iii) C F.dR = 0 for any regular closed curve
C in D.
If the domain D is simply connected then the following is
also equivalent to the above:
(iv) CurlF = 0 ( If CurlF = 0, F is called
irrotational.
In the case of the above theorem we have
∫ Q
F.dR = ϕ(P ) − ϕ(Q)
P
where ϕ is a potential for F.
3.Solenoidal Vector Fields and Vector Potentials
a. If F is a vector field such that Div F = 0 then F is
solenoidal. If F = Curl G then F is solenoidal and G is
called a vector potential for F.
(1) Theorem.A vector field F continuously differentiable in a star-shaped region is solenoidal if and only if
there exists a vector potential.
4.Surface Area
a. The surface area of a regular surface element with
parametrization R(u, v) over a region D ⊂ R2 is
∫ ∫
S=
|Ru × Rv |dudv
D
29
The area element is then
dS = |dS| = |Ru × Rv |dudv.
The vector dS = Ru × Rv dudv = Ru du × Rv dv is a a
vector normal to the surface whose magnitude dS = |dS| is
the area element. Then
∫ ∫
∫ ∫
∫ ∫
S=
|dS| =
dS =
n.dS
D
D
D
where n is the outward unit normal in the same direction
as dS.
(1) Two Special Cases
(a) Suppose z = 0 so that the surface lies entirely
in the xy plane. Then:
dS = |
∂(x, y)
|dudv
∂(u, v)
This is the area element for a change of variables. In the
case where x = r cos θ, y = r sin θ, z = 0 we have the change
to polar coordinates {where r = u, θ = v}. Then dS =
rdrdθ.
(b) Suppose z = f (x, y) defines a surface over D ⊂
R . Then:
∫ ∫ √
S=
1 + fx2 + fy2 dxdy.
2
D
Let γ be the angle between dS and k. Then | cos γ| =
√ 12 2 ⇒
1+fx +fy
∫ ∫
dxdy
S=
D | cos γ|
30
In the case where z = f (x, y) is actually given by F (x, y, z) =
0 then
▽F.k
cos γ =
|▽F |
EXAMPLE Show that the surface area of a sphere of radius R is 4πR2 .
We can place the sphere centered on the origin and hence
a parametrization of the upper half sphere would be
√
z = f (x, y) = R2 − (x2 + y 2 )
Then
−y
x
, fy = √
fx = √
R2 − (x2 + y 2 )
R2 − (x2 + y 2 )
R2
x2 + y 2
=⇒ 1+fx2 +fy2 = 1+ √
=√
R2 − (x2 + y 2 )
R2 − (x2 + y 2 )
∫ ∫
Then
√
A=
D
R2
R2 − (x2 + y 2 )
dxdy
where D is the unit circle. Changing to polar coordinates
∫ 2π ∫ R
R2
√
A=
rdrdθ = 2πR2
2
2
R −r
0
)
This is the surface area for the upper hemisphere and hence
the surface area for the sphere is 4πR2 .
5.Surface Integrals
31
a. If f (x, y, z) is a continuous scalar field defined in
a region containing a regular surface element S then the
surface integral of f over S is
∫ ∫
∑
f dS = lim
f (xi , yi , zi )∆Si
|∆Si |→0
S
Computationally if S is parametrized by R(u, v) then
∫ ∫
∫ ∫
dS = |Ru × Rv |dudv so
f (x(u, v), y(u, v), z(u, v))|Ru × Rv |dudv
f dS =
S
D
b. The Flux of a Vector Field If F(x, y, z) is a vector
field over a surface S then the flux of F over S is
∫ ∫
∫ ∫
F.ndS =
F.dS
S
S
(1) If S is specified by z = f (x, y) then
∫ ∫
∫ ∫
dxdy
.
F.dS =
F.n
| cos γ|
S
D
c. Some Applications in Physics
(1) Suppose we have a steady state temperature distribution with scalar field T (x, y, z) over a region filled with a
homogenuous material with thermal conductivity k. Then
−
→
the vector Q = −k▽T gives the direction in which heat is
flowing. If S is a surface contained in this region then
∫ ∫
(−k▽T ).ndS
S
32
gives the total number of calories per second flowing across
the surface S.
(2) If F is the velocity field of a fluid and ρ its density
then:
∫ ∫
ρF.ndS
s
is the rate of flow of liquid across the surface S expressed
as mass per unit time.
(3) Gauss’ Law of Electrostatics If E is an electrostatic field and S is a closed surface then
∫ ∫
q
E.ndS =
ϵ0
S
where q = total charge enclosed by the surface S and ϵ0 =
a constant depending on the units.
D. The Divergence Theorem and Stokes Theorem
1.Introduction to the Divergence Theorem and
Stokes Theorem
a.The general idea behind the various versions of Stokes
Theorem is that the integral of the ”derivative” of a differential form over a region D (suitably defined for the dimensionality of the integral) is equal to a one-dimension lower
integral of the corresponding function on the boundary of
D - again suitably defined. This is:
∫
∫
dω =
ω
D
∂D
.
33
(1) For ordinary integrals this is just the Fundamental
Theorem of Calculus:
∫ b
f ′ (x)dx = f (b) − f (a)
a
(2) In two dimensions this is Green’s Theorem:
I
∫ ∫
∂Q ∂P
P dx + Qdy
(
−
)dxdy =
∂y
D ∂x
∂D
(3) In three dimensions this is the Divergence Theorem and Stokes Theorem representing the two types of
differentiation of vector fields.
(a) The Divergence Theorem:
∫ ∫ ∫
∫ ∫
divFdV =
F.dS
S
∂S
(b) Stokes Theorem:
∫ ∫
∫
CurlF.dS =
S
F.dR
∂S
2. The Divergence Theorem
a.Theorem (The Divergence Theorem) Let D ⊂ R3 be
a region so that each straight line through any interior point
of D cuts the boundary in exactly two points and so that
the boundary ∂D is a piecewise smooth, closed oriented
surface. Let F = F1 i + F2 j + F3 k be a C 1 vector field on D.
Then:
∫ ∫ ∫
∫ ∫
divFdV =
F.dS
S
∂S
34
b. Restatement of the Divergence Theorem is:
The total divergence within a region D equals the net flux
emerging from D.
3. Stokes Theorem
a.Theorem (Stokes Theorem) Let S be a smooth oriented surface with boundary ∂S a piecewise smooth, closed
oriented curve. Let F be a C 1 vector field. Then:
∫ ∫
∫
CurlF.dS =
F.dR
S
∂S
(1) Corollary: Let the swirl of the vector field F at
→
P in the direction −
n be defined by
∫
1
LmA→0
F.dR
A C
where C is the circumference of a circle of area A centered
→
at P with unit normal −
n . Then:
→
(a) CurlF.−
n = swirl
(b) CurlF is in the direction of maximum swirl and
has magnitude equal to this maximum swirl. This gives a
coordinate free definition of the curl of a vector field.
4. Green’s Theorem
1.Theorem (Green’s Theorem) Let D be a region in R2
with boundary ∂D a simple closed curve. Let P (x, y), Q(x, y)
35
be C 1 functions on a region containing D and its boundary.
Then:
∫ ∫
I
∂Q ∂P
−
)dxdy =
P dx + Qdy
(
∂y
∂D
D ∂x
a. Two Interpretations
(1)If F = P (x, y)i + qQ(x, y)j is a planar vector
∂Q
field its divergence is then divF = ∂P
∂x + ∂y . If γ is a simple
closed smooth planar curve with outward directed normal
n then
I
I
−Qdx + P dy = F.nds
γ
γ
Green’s Theorem can then be interpreted as
∫ ∫
∫
divFdA =
F.Nnds
d
∂D
a planar version of the Divergence Theorem
(2)If F = P (x, y)i + q(x, y)j and D is a region
in R2 . Consider D then as a surface in R3 with z = 0.
∂P
Then CurlF = ( ∂Q
∂x − ∂y )k and Green’s Theorem can be
interpreted as
∫ ∫
∫
curlF.dS =
F.R
D
∂D
which is a restricted version of Stokes Theorem
5. Summary of the Integral Theorems
36
Operator
Interpretation
gradf = ▽f
Maximum rate of change of f
DivF = ▽.F Net outflux of F per unit volume
CurlF = ▽ × F
swirl of F per unit area
Integral Theorem
∫Q
∫ ∫P ∫▽f.dR = f (Q)
∫ ∫− f (P )
∫ ∫ D ▽.FdV = ∫ ∂D F.d§
S ▽ × F.d§ = ∂S F.dR
E. Some Basic PDE’s and their Derivations
1. Fourier’s Law of Heat Conduction
a. The heat flux q ( calories per sec) across a plane area
element A is proportional to the temperature gradient ∂T
∂n
normal to A and to the area itself
q = −kA
∂T
= −kA▽T.n
∂n
b. This is a special case of Fick’s first law of diffusion
∂C
J = −DA
= −DA▽C.n
∂n
where J = mass flux - mass per unit time: D = mass
diffusivity, A = control surface area , C = concentration of
the diffusing species.
2. Continuity Equation of Fluid Mechanics
a. Let q be the velocity vector field of a fluid with mass
density σ(x, y, z, t). Then
∂σ
+ ▽(σ.q) = 0
∂t
37
is the continuity equation of fluid mechanics
(1).For fluids this follows from the conservation of
mass.
(2) If σ = constant its an imcompressible fluid.
Then
▽q = div(q) = 0
b. Derivation of the Continuity Equation
Consider a velocity vector field q of fluid of mass density
σ(x, y, z, t) going into a control volume V . The mass at any
time t is then
∫
∫
∂σ
dM
=
dV
M=
σdV =⇒
dt
∂t
V
V
Further
dM
= mass entering + mass created
dt
∫
mass entering = −
σq.ndA
∂V
∫
while
mass created =
f.dV
V
where f = mass generator. Therefore
∫
∫
∫
∂σ
dM
=
dV = −
σq.ndA +
f dV
dt
V ∂t
∂V
V
applying the divergence theorem
∫
∫
−
σq.ndA = − ▽.(σq)dV
∂V
V
38
∫
∫
∫
∂σ
=⇒
dV = − ▽.(σq)dV + f dV
V ∂t
V
v
∫
∂σ
=⇒
(
+ ▽(σq) − f )dV = 0
V ∂t
Now since V was arbitrary
∂σ
+ ▽(σq) − f = 0
∂t
By the conservation of mass f = 0 so the basic PDE is
=⇒
∂σ
+ ▽(σq) = 0
∂t
This is the basic continuity equation for fluid mechanics.
Note: If this were an electric field f can be a charge source
and can be non-zero.
Further if this is an incompressible fluid then it is time
independent so σ = k so the equation becomes
▽q = 0
3. Unsteady Heat Conduction
a. Consider an unsteady heat conduction with a region
D ⊂ R3 Then the temperature field T (x, y, z, t) satisfies
(for no source term)
α 2 ▽2 T =
∂T
∂t
known as the heat equation or diffusion equation. α2
is the diffusivity constant.
39
(1) Steady state conduction when
∂F
∂t
= 0 leads to
∂ 2T
=0
∂x2
which is the Laplace Equation.
b. Derivation of the Heat Equation
The heat in cals contained in a mass m at temperature T
with specific heat c is then
mcT
If σ = σ(x, y, z) is the mass density then m =
Therefore the accumulation of heat is given by
∫
∫
d
∂T
cT σdV = c σdV
dt V
∂t
∫
σdV .
By Fourier’s law of heat conduction the heat flux is
∫
∫
∂T
k▽T.ndA
k σdA =
V
V ∂n
where k is the conductivity of the material.
Suppose there is a heat generator f (x, y, z, t) so the heat
generated is
∫
f dV
∫
∫
∂T
k▽T.ndA +
f dV
c σdV =
∂V
V
V ∂t
Now apply the divergence theorem. Then:
∫
∫
∫
∂T
c σdV =
▽.(k▽T )dV +
f dV
V ∂t
V
V
Then
∫
40
Again since V is arbitrary
∂T
− ▽.(k▽T ) − f = 0
∂t
If k is independent of x, y, z then
=⇒ cσ
α 2 ▽2 T =
where F =
f
cσ
and α2 =
k
cσ
∂T
−F
∂t
is the diffusivity.
If F = 0 then
∂T
∂t
which is called the heat equation.
α 2 ▽2 T =
4. Maxwell’s Equations
a.Let E be the electric field intensity,(newtons/coulomb),
H the magnetic field intensity (amperes, per meter) and J
the currect density (amperes per meter squared). Then
Maxwell’s equations are
▽×E =0
▽×H =J
b. Derivation of Maxwell’s Equations
(a) By Faradays’ law the emf around a closed curve
equals the time rate of change of the magnetic flux through
the curve. Then:
∫
∫
∫
d
∂B
E.dR = −
B.ndA = −
.ndA
dt S
∂S
S ∂t
41
for every fixed surface S with boundary ∂S where
E = electric field intensity
B = magnetic flux density
Now apply Stokes Theorem
∫
∫
=⇒
E.dR =
▽ × E.ndA
∂S
S
∂B
∂t
This is Maxwell’s equation for time varying electric fields.
For steady state fields
=⇒ ▽ × E = −
▽×E=0
.
(b) By Ampere’s law
∫
H.dR = I
∂S
where H is the magnetic field intensity and I is the current
through any surface which has ∂S as a boundary. Further
∫
I=
σq.ndA
S
where σ is the charge density and q is the velocity vector
field. Let
J = σq = current density
42
∫
then:
∫
H.dR =
∂S
σ.q.ndA
S
Apply Stokes Theorem
∫
∫
∫
H.dR =
▽ × (H).dA =
J.ndA
∂S
S
S
Since S was arbitrary
▽×H=J
5. The Wave equation
a.Wave Equations
(1) The One-Dimensional Wave Equation is
β 2 uxx = utt
(2) Roughly wave equations arise is modeling wave
phenomena
(3) This arises from problems in mechanical vibrations.
A solution u(x, t) will represent the small displacements of
an idealized vibrating string
(a) The wave equation also appears in the theory of
high frequency transmission lines, fluid mechanics, acoustics and electricity.
b.Derivation of the Wave Equation
43
(1) The Basic Assumptions
(a) Let y(x, t) the transverse vibrations of a string
stretched between x = 0 and x = L. The solution y(x, t)
measures the displacements from the x-axis for t > 0. f (x)
represents the initial configuration of the string and g(x)
represents the initial velocity at each point.
(2) Conditions for One-Dimensional Wave Equation
(a) The string is perfectly flexible
(b) The displacements are small relative to the
∂y
length of the string - hence y and ∂x
are small
(c) The tension of the string is constant
(d) The string is homogeneous, that is its mass
per unit length is constant
(e) The tension is large compared to the force
of gravity
(f) No other forces act on the string
(3) Derivation
(a) Assume that the tension is τ and the density is
ρ. Then using Newton’s law we get
∂ 2y
τ sin θ(x+∆x, t)−τ sin θ(x, t)−f (x+α∆x, t)∆x = (ρ∆s) 2 (x+β∆x, t)
∂t
That is the sum of the vertical forces is equal to the vertical
acceleration,
44
(i) Under the assumptions θ is small and
therefore
sin θ u θ u tan θ =
∂y
and ∆s u ∆x
∂x
Therefore
τ ∂y
∂y
∂ 2y
( (x+∆x, t))− (x, t))−f (x+α∆x, t) = ρ 2 (x+β∆x, t)
∆x ∂x
∂x
∂t
Letting ∆x → 0 yields
∂ 2y
∂ 2y
τ 2 − f (x, t) = ρ 2
∂x
∂t
(ii) If f = ρg just the force of gravity then
τ
∂ 2y
∂ 2y
=
ρ
+ ρg
∂x2
∂t2
Further if gravity is negligible then
∂ 2y
∂ 2y
c2 2 = 2 where c =
∂x
∂t
45
√
τ
ρ
PARTIAL DIFFERENTIAL EQUATIONS (3)
TAYLOR SERIES AND POWER SERIES SOLUTIONS
A. Power Series and Taylor Series
1. Taylor’s Theorem
a. Theorem If f (x) is a C n+1 [a, b] function then
for each x in [a, b]
f (x) = f (a) + f ′ (a)(x − a) + ..... +
f (n) (a)
(x − a)n + Rn (x)
n!
where
Rn (x) =
f (n+1) (c)
(x − a)n+1 for some c, a < c < x
(n + 1)!
.
Tn (x) = f (a) + f ′ (a)(x − a) + ..... +
f (n) (a)
(x − a)n
n!
is the degree n Taylor polynomial centered on a for
f (x) while Rn (x) is the remainder. If a = 0 its called a
Maclaurin polynomial
(1) Thus a C n+1 function can be approximated
by its degree n Taylor polynomial with error term given by
Rn (x). If we can get a uniform bound on the error we get
a uniform approximation for f (x).
2. Power Series
∑
n
a.Definiiton P (x) = ∞
n=0 an (x − x0 ) is a power
series with center x0 . For each value of x we get an
46
infinite series and for those x values for which the series
converges we get a function.
∑
n
(1) Theorem For P (x) = ∞
n=0 an (x−x0 ) there
is a number r ≥ 0 called the radius of convergence such
that P (x) converges for all x such that |x − x0 | < r and
diverges if |x − x0 | > r. Generally
1
|an+1 |
= L−1 m
r
|an |
(a) Corollary A power series can be differentiated and integrated within its circle of convergence. Further an = P (n) (x0 ).
3. Taylor Series
a. Definition If the function f (x) can be written
as a power series
f (x) =
∞
∑
f (n)
n=0
n!
(x − x0 )n
then this power series is the Taylor series for f (x) centered
at x0 . If x0 = 0 it is called a Maclaurin series. If f (x) has
a Taylor series about x0 then we say that f (x) is analytic
at x0
(1) Theorem If f (x) is a C ∞ function on [a, b]
with x0 in (a, b) and if the remainder Rn (x) goes to zero
uniformly in an interval about xx0 then f (x) is analytic at
x0 .
Note: Not every C ∞ function is analytic.
47
4. Some Common Taylor Series
a. P (x) = P (x) if P (x) is a polynomial
∑
1
n
b. 1−x
= ∞
n=0 x , −1 < x < 1
∑
xn
c. ex = ∞
n=0 n! , all x
∑
n x2n+1
d. sin x = ∞
n=0 (−1) (2n+1)! , all x
∑
n x2n
e. cos x = ∞
n=0 (−1) (2n)! , all x
∑
n xn+1
f. Ln(x + 1) = ∞
n=0 (−1) n+1 , −1 < xLe1
5. Manipulating Series - often other series can be
found by manipulating exisiting series.
bf EXAMPLE
∞
∞
1
d
1
d ∑ n ∑ n−1
nx
x =
=
(
)=
(1 − x)2
dx 1 − x
dx n=0
n=1
B. Power Series Solutions to ODE’s
1.Power Series Methods
a. Consider a second order linear ODE
y ′′ + P (x)y ′ + Q(x)y = 0
(1) Definition A point x = x0 is an ordinary
point for equation (1) if both P (x) and Q(x) are analytic
48
at x0 . Otherwise it is a singular point. We only consider
here ordinary points.
b. Procedure for Power Series Solutions
Step 1: Consider a solution of the form y(x) =
n
n=0 an (x − x0 )
∑∞
Step 2: Substitute this solution type into equation (1) to get a recurrence relation among the coefficients an .
Step 3: There will generally be two free parameters. Values for these will give the two independent
solutions.
EXAMPLE Solve y ′′ − 2xy = 0.
Let y = a0 + a1 x + a2 x2 + a3 x3 + ....... then
y ′ = a1 + 2a2 x + 3a3 x2 + 4a4 x3 + .....
y ′′ = 2a2 + (3)(2)a3 x + (4)(3)a4 x2 = ....
so
y ′′ − 2xy = 2a2 + [(3)(2)a3 − 2a1 ]x + ..... = 0
In general then
2a2 +
∞
∑
[(k + 2)(k + 1)ak+2 − 2ak−1 ]xk = 0
k=1
Therefore the coefficients must be zero. This implies
a2 = 0
ak+2 =
2ak−1
(k + 2)(k + 1)
49
substtituing in for the first few values we get that a0 , a1 are
arbitrary and
1
1
1
a2 = 0, a3 = a0 , a4 = a1 , a5 = 0, a6 = a0
3
6
45
Therefore the solution is given by
1
1
y = a0 + a1 x + a0 x3 + a1 x4 + .....
3
6
2.Taylor Series Methods
a. Consider a second order linear initial value problem
y ′′ + P (x)y ′ + Q(x)y = 0 : y(x0 ) = y0 , y ′ (x0 ) = y1
b. Procedure for Taylor Series Solutions
Step 1: Consider a Taylor series solution
1
1
y(x) = y(x0 )+y ′ (x0 )(x−x0 )+ y ′′ (x0 )(x−x0 )2 + y ′′′ (x0 )(x−x0 )3 +....
2
3!
Therefore we have a solution once we have the values y(x0 ), y ′ (x0 ), y ′′ (x0 )
etc.
Step 2: Substitute into the equation to get the
above values
EXAMPLE (a) Solve y ′ = y : y(0) = 1
Know
y ′ = y =⇒ y ′ (0) = y(0) = 1
50
differentiating =⇒ y ′′ = y ′ =⇒ y ′′ (0) = y ′ (0) = 1
Continuing we get that
y(x) = 1 + x +
1 2 1 3
x + x + ... = ex
2!
3!
(b) Solve y ′′ + y = 0 : y(0) = 0, y ′ (0) = 1
Know
y ′′ = −y =⇒ y ′′ (0) = −y(0) = 0
differentiating =⇒ y ′′′ = −y ′ =⇒ y ′′′ (0) = −y ′ (0) = −1
Continuing we get that
y(x) = x −
1 3
x + ... = sin x
3!
(c) Solve y ′′ − 2xy = sin x : y(0) = 0, y ′ (0) = 1
Know
y ′′ = 2xy + sin x =⇒ y ′′ (0) = 0
differentiating =⇒ y ′′′ = 2xy ′ +2y+cos x =⇒ y ′′′ (0) = 1
differentiating =⇒ y (4) = 2xy ′′ +2y ′ +2y ′ −sin x =⇒ y (4) (0) = 4
Therefore
y(x) = x +
1 3 4 4
x + x + .....
3!
4!
51
PARTIAL DIFFERENTIAL EQUATIONS (4)
FOURIER SERIES AND FOURIER ANALYSIS
A. Fourier Series of a Function
1.General Preliminaries
a.The Idea of a Fourier Series
(1) Definition A function f (x) is periodic of period
m if f (x + m) = f (x) for all x.
(a) Lemma sin x and cos x are both periodic of
period 2π.
(b) Trigonometric functions are periodic and periodic behavior appears quite often in physical applications.
The basic idea in a Fourier series or Fourier expansion
is to express a general periodic function as a sum or infinite
series of trigonometric functions.
b.Orthogonality Relations
∫
(1) Theorem (i)
l
cos(
−l
mπx
nπx
) cos(
)dx = 0, m ̸= n = 1, m = n ̸= 0, m = n ̸= 2l, m = n = 0
l
l
(ii)
∫ l
sin(
−l
(iii)
mπx
nπx
) sin(
)dx = 0, m ̸= n = 1, m = n ̸= 0
l
l
∫
l
cos(
−l
nπx
mπx
) sin(
)dx = 0, ∀m, n
l
l
52
2. The Fourier Series of a Function
a.General Trigonometric Series
(1) Definition A trigonometric series is an infinite
series of the form
∞
∑
(an cos(knx) + bn sin(knx)
n=1
b. The Fourier Series of a Periodic Function
(1) Definition Let f (x) be a periodic function with
fundamental period 2l. Then its Fourier series is
a0 +
∞
∑
(an cos(
n=1
nπx
nπx
) + bn sin(
)
l
l
where
1
an =
l
1
bn =
l
1
a0 =
2l
∫
∫
l
f (x)dx
−l
l
f (x) cos(
−l
∫
l
f (x) sin(
−l
nπx
)dx, n = 1, 2, ...
l
nπx
)dx, n = 1, 2, ...
l
The an , bn are called the Fourier coefficients while cos( nπx
l )
nπx
and sin( l ) are the harmonics
53
(a) Note A trigonometric series clearly is periodic.
It is a Fourier series if it is the Fourier series of some
periodic function
(b) Note a0 = average value of f (x) on (−l, l)
(c) Lemma There exists trigonometric series which
are not Fourier series. For example
∞
∑
n=1
1
sin(nx)
ln(1 + n)
converges for each x but is not a Fourier series.
(d) Note If f (x) is an odd fucntion then its Fourier
series has no cosine terms while if f (x) is an even function
f (x) has no sine terms.
c. Complex Form of a Fourier Series
(1) In terms of complex variables
inπx
l
nπx
e
cos(
)=
l
+ e−
2
inπx
l
− e− l
2i
∞
∑
inπx
cn e l
=⇒ f (x) =
nπx
e
sin(
)=
l
inπx
inπx
l
n=−∞
where
1
cn =
2l
∫
l
f (x)e−
inπx
l
dx
−l
d. The Fourier Convergence Theorem
54
(1) Theorem Let f (x) be periodic of period 2l. Then:
(i) If both f (x) and f ′ (x) are piecewise continuous on (−l, l) then the Fourier series converges point+
(x− )
wise to the mean value f (x )+f
.
2
(ii) If both f (x) and f ′ (x) are continuous on
(−l, l) then the Fourier series converges uniformly to f (x).
(a) Corollary At each jump discontinuity the Fourier
series converges to the average value of the jump.
(b) Corollary If f (x) is a continuously differentiable
periodic function then f (x) is represented everywhere by its
Fourier series.
EXAMPLE Let f (x) be the square wave function defined
by
f (x) = 0, −π < x < 0 and = 4, 0 < x < π and = 2, x = −π, 0, π
and then periodic of period 2π. We determine the Fourier
expansion of f (x).
Now
1
a0 =
2l
Then
1
an =
l
∫
l
1
f (x)dx =
2π
−l
∫
∫
π
1
f (x)dx =
2π
−π
∫
π
4dx = 2
0
∫
1 π
nπx
)dx =
f (x) cos(
f (x) cos(nx)dx
l
π −π
−l
∫
1 π
4 cos(nx)dx = 0
=
π 0
l
55
Finally
∫
nπx
1 π
f (x) sin(
f (x) sin(nx)dx
)dx =
l
π −π
−l
∫
4
1 π
4 sin(nx)dx =
(1 − cos(nπ))
=
π 0
nπ
Further since cos(nπ) = (−1)n we get
1
bn =
l
∫
l
4
8
(1 − (−1)n ) =
, n = 1, 3, ....
nπ
nπ
Therefore the Fourier expansion of f (x) is given by
bn =
∞
8 ∑ 1
f (x) = 2 +
sin(nx)
π n=1,3,.. n
By the Fourier convergence theorem this series will converge
to f (x) at each point of continuity of f (x) - hence if x ̸= nπ.
e.Finite Interval Half-Range Expansions
(1) Suppose f (x) is defined on (0, l). Then a periodic
extension f (x) can be created on (−l, l). The Fourier series
of f (x) will then converge to f (x) on (0, l).
(a) If f (−x) = f (x) that is we define f (x) = f (−x)
on (−l, 0) then we get the even periodic extension. The
resulting Fourier series has only cosine terms and is the
half-range cosine expansion.
(b) If f (−x) = −f (x) we get the odd extension
and the resulting Fourier series is the half-range sine expansion.
56
(2) We can also define quarter-range sine and cosine
expansions
(3) Formulas for Periodic Expansions - f (x) defined on 0 < x < L
(a) Half- Range Sine
f (x) =
∞
∑
bn sin(
n=1
2
bn =
L
∫
L
f (x) sin(
0
nπx
)
L
nπx
)dx
L
(b) Half- Range Cosine
f (x) = a0 +
∞
∑
an cos(
n=1
nπx
)
L
∫
1 L
f (x)dx
a0 =
L 0
∫
2 L
nπx
an =
f (x) cos(
)dx
L 0
L
(c) Quarter- Range Sine
∞
∑
f (x) =
bn sin(
n=1,3,...
2
bn =
L
∫
L
f (x) cos(
0
57
nπx
)
2L
nπx
)dx
2L
(d) Quarter- Range Cosine
∞
∑
f (x) =
an cos(
n=1,3...
2
an =
L
∫
L
f (x) cos(
0
nπx
)
2L
nπx
)dx
2L
EXAMPLE Consider the function f (x) = k, 0 < x < L.
We find the Fourier half-range cosine and half-range sine
expansion.
First the half-range cosine expansion. From the formulas
above we have
∫
1 L
a0 =
kdx = k
L 0
∫
2 L
nπx
2k
nπx L
an =
k cos(
)dx =
sin(
)| = 0
l 0
L
nπ
L 0
Therefore the Fourier half-range cosine expansion is just
f (x) = k.
Next the half-range sine expansion. Again from the formulas above we have
∫
nπx
2k
2 L
bn =
k sin(
)dx = − (cos(nπ) − 1)
l 0
L
nπ
Therefore the Fourier half-range sine expansion is
∞
2k ∑ 1 − cos(nπ)
nπx
f (x) =
sin(
)
π n=1
n
L
58
3. Bessel’s Inequality and the Parseval Identity
a.Theorem If the periodic function f (x) has the Fourier
series
∞
∑
nπx
nπx
a0 +
(an cos(
) + bn sin(
)
l
l
n=1
then:
k
∑
1
(i) 2a20 +
(a2n + b2n ) ≤
l
n=1
∫
l
−l
|f (x)|2 dx
for all k = 1, 2, .... This is called Bessel’s Inequality.
(ii)
2a20
+
∞
∑
(a2n
+
b2n )
n=1
1
=
l
∫
l
−l
|f (x)|2 dx
This is called Parseval’s Identity .
4. Manipulation of Fourier Series
a.Uniqueness of Fourier Series
(1)Theorem If two trigonometric series each of the
form
a0 +
∞
∑
n=1
(an cos(
nπx
nπx
) + bn sin(
)
l
l
converge to the same sum for all x then the corresponding
coefficients are equal.
b.Termwise Integration of Fourier Series
59
(1)Theorem Let f (x) be a piecewise continuous periodic function of period 2l with Fourier series
a0 +
∞
∑
(an cos(
n=1
nπx
nπx
) + bn sin(
).
l
l
Suppose that a0 = 0. Then the Fourier series of the integral
function
∫ x
F (x) =
f (t)dt
−l
can be obtained by termwise integration and the resulting
series converges to F (x) for all x.
c.Termwise Differentiation of Fourier Series
(1)Theorem Let f (x) be a piecewise continuous periodic function of period 2l with Fourier series
a0 +
∞
∑
n=1
(an cos(
nπx
nπx
) + bn sin(
).
l
l
Suppose that f ′ (x) is also piecewise continuous on (−l, l).
Then the Fourier series for f ′ (x) can be obtained by termwise
differentiation of the Fourier series for f (x).
B. Sturm-Liouville Theory
1.Sturm-Liouville Problems
a. Definition A Sturm-Liouville Problem is a
second-order linear boundary value ODE of the form
(−p(x)y ′ (x))′ + q(x)y(x) = λr(x)g(x), a < x < b
60
c1 y(a) + c2 y ′ (a) = 0
c3 y(b) + c4 y ′ (b) = 0
and p(x), p′ (x), g(x), r(x) ∈ C 0 (a, b), p(x) > 0, r(x) > 0 on
(a, b) and c21 + c22 ̸= 0, c23 + c24 ̸= 0.
(1) y(x) ≡ 0 solves this equation. Each value of λ
which allows a non-trivial solution is called an eigenvalue
and the corresponding function is an eigenfunction.
b. Definition The differential operator
1 d
d
L = − ( (p + q)
r dx dx
is defined on the space of C 2 functions satisfying the boundary conditions. The Sturm-Liouville problem then has the
form
Ly = λy
(1) Lemma On the inner product space of L2 functions with inner product
∫ b
< u, v >=
u(x)v(x)r(x)dx
a
the operator L is a self-adjoint operator. That is
< Lu, v >=< u, Lv >
c. The Spectrum of L
(1) Theorem The eigenvalues and eigenfunctions of
L have the following properties.
61
(i) All eigenvalues are real and compose a countable infinite discrete set
λ1 < λ2 < ..... → ∞
(ii) To each eigenvalue λn corresponds only one
independent eigenfunction wn (x)
(iii) The normalized eigenfunctions (relative to
the weight function r(x)) form a complete orthonormal family in L2 .
2.Eigenfunction Expansions
a. Definition If {ϕn (x)} is a collection of eigenfunctions and f (x) is an L2 function then its eigenfunction
expansion is
∞
∑
< f (x), ϕn (x) >
ϕn (x)
<
ϕ
(x),
ϕ
(x)
>
n
n
i=1
(1) Pointwise Convergence of Eigenfunction Expansions
(a) Theorem (i) If both f (x) and f ′ (x) are piecewise continuous in (a, b) then the eigenfunction expansion
+
(x− )
converges pointwise to f (x )+f
for all x ∈ (a, b).
2
(ii) If f (x), f ′ (x) are continuous in (a, b)
and f ′′ (x) is piecewise continuous and f (x) satisfies the
boundary conditions of the base Sturm-Liouville problem
then the eigenfunction expansion converges uniformly to
f (x) on (a, b)
62
PARTIAL DIFFERENTIAL EQUATIONS (5)
FOURIER INTEGRAL AND FOURIER TRANSFORMS
A. Integral Transforms
1.General Idea of an Integral Transform
a. The basic idea of an integral transform is to transform a function f (t) into another function f (w) via integration. This transformation will be linear on functions. Integral transforms will behave relative to differential equations
much the way logarithms do for multiplicative equations they linearize them and simplify them.
b. Definition Let K(s, t) be a function of the variables
s, t defined for t ∈ I where I is some interval. K(s, t) is
called the kernel. If f (t) is a function of t defined on I
then its integral transform rerlative to the kernel K(s, t)
is the function
∫
f (s) =
f (t)K(s, t)dt
I
(1) There are many important integral transforms.
However for our purposes in differential equations we will
examine two - the Laplace transform and the Fourier
transform.
B. The Laplace Transform
1. The general idea of Laplace transforms is to change
a linear differential equation into an ordinary algebraic equation. It plays a very similar role to logarithms which change
multiplicative equations into additive equations.
63
2. The Laplace Transform
a.Definition Let f (t) be defined on [0, ∞). Then its
Laplace transform is the function F (s) = L(f (t)) defined
by
∫
∞
F (s) = L(f (t)) =
e−st f (t)dt
0
provided this integral exists. Recall that
∫ ∞
∫ b
g(x)dx = lim
g(x)dx
b→∞
a
a
EXAMPLES (a) Evaluate L(1)
∫
L(1) =
∞
−1 −st ∞ 1
e |0 = if s > 0
s
s
e−st dt =
0
(b) Evaluate L(t)
∫
L(t) =
∞
−st
te
0
−te−st ∞ 1
dt =
| +
s 0 s
∫
∞
e−st dt by integration by parts
0
1
1
= 0 + L(1) = 2 if s > 0
s
s
(c) Evaluate L(e−bt )
L(e
−bt
∫
)=
∞
e
0
−bt −st
e
∫
∞
dt =
0
64
e−(b+s)t dt =
1
if s > −b
s+b
(1) Definition A function f (t) is of exponential
order if there exist constants M, c, T such that
|f (t)| ≤ M ect for all t > T
(a) Theorem If f (t) is of exponential order on
[0, ∞) then L(f (t)) exists for t > c
(b) From now on we will be concerned only with
functions that have Laplace transforms.
b. Theorem The Laplace transform is a linear operator on functions. That is
L(af (t) + bg(t)) = aL(f (t)) + bL(g(t))
c. The Common Transforms
(1) Theorem (i) L(1) =
1
s
(ii) L(tn ) =
n!
sn+1 , n = 1, 2, ....
1
(iii) L(eat ) = s−a
k
(iv) L(sin kt) = s2 +k
2
s
(v) L(cos kt) = s2 +k
2
k
(vi) L(sinh kt) = s2 −k
2
s
(vii) L(cosh kt) = s2 −k
2
3. The Inverse Transform
a. Definition If f (t) is a function such that L(f (t)) =
F (s) the f (t) is the inverse Laplace transform of F (s), =⇒
f (t) = L−1 (F (s))
65
(1) The Common Inverses
(a) Theorem (i) 1 = L−1 ( 1s )
n!
(ii) tn = L−1 ( sn+1
)n = 1, 2, ....
1
(iii) eat = L−1 ( s−a)
k
(iv) sin kt = L−1 ( s2 +k
2)
s
(v) cos kt = L−1 ( s2 +k
2)
k
(vi) sinh kt = L−1 ( s2 −k
2)
s
(vii) cosh kt = L−1 ( s2 −k
2)
EXAMPLE Evaluate L−1 ( 3s+5
s2 +7 ),
3s + 5
3s
5
) = L−1 ( 2
) + L−1 ( 2
)=
2
s +7
s +7
s +7
√
√
√
5
7
5
s
−1
−1
√
√
)+
) = 3 cos( 7t) +
L ( 2
sin( 7t)
3L ( 2
s +7
s +7
7
7
L−1 (
(b) Use of Partial fractions
1
EXAMPLE Evaluate Lv( (s−1)(s+2)(s+4)
)
Method is to try to find A, B, C such that
1
A
B
C
=
+
+
(s − 1)(s + 2)(s + 4) s − 1 s + 2 s + 4
Combining the three fractions we get
1
A(s + 2)(s + 4) + B(s − 1)(s + 4) + C(s − 1)(s + 2)
=
=
(s − 1)(s + 2)(s + 4)
(s − 1)(s + 2)(s + 4)
66
(A + B + C)s2 + (6A + 3B + C)s + 8A − 4B + 2C
(s − 1)(s + 2)(s + 4)
Equating the tops gives three linear equations for A, B, C
=
A+B+C =0
6A + 3B + C = 0
8A − 4B + 2C = 1
Solving we get A =
L−1 (
1
15 , B
=
−1
6 ,C
=
1
10 .
Therefore
1
1
1
1
1
1
1
) = L−1 (
)− L−1 (
)+ L−1 (
)
(s − 1)(s + 2)(s + 4)
15
(s − 1) 6
(s + 2) 10
(s + 4)
=
1 t 1 −2t
1
e − e + e−4t
15
6
10
4. Some Properties of Laplace Transforms
a. The Translation Theorem
(1) Theorem L(eat f (t)) = F (s−a) if F (s) = L(f (t))
EXAMPLE Evaluate L(e5t t3 )
L(e5t t3 ) = F (s − 5) where F (s) = L(t3 )
L(t3 ) =
6
6
5t 3
=⇒
L(e
t
)
=
s4
(s − 5)4
(2) Theorem eat f (t) = L(F (s−a)) = L(F (s))|s→s−a
if F (s) = L(f (t))
s
EXAMPLE Evaluate L−1 ( s2 +6s+11
)
L−1 (
s
s
s+3−3
−1
−1
)
=
L
(
)
=
L
(
)
s2 + 6s + 11
(s + 3)2 + 2
(s + 3)2 + 2
67
s+3
3
−1
)
−
L
(
)
(s + 3)2 + 2
(s + 3)2 + 2
Let v = s + 3 then
√
3
2
v
− √ L−1 ( 2
)
= L−1 ( 2
v +2
v +2
2
= L−1 (
√
√
3
= e−3t cos( 2t) − √ sin( 2t)
2
b. Derivatives of a Transform
(1) Theorem
1, 2, ....
L(tn f (t)) = (−1)n d
n
L(f (t))
dsn
for n =
EXAMPLES (a) Evaluate L(te3t )
L(te3t ) = −
d
d
1
1
(L(e3t )) = − (
)=
ds
ds s − 3
(s − 3)2
(b) Evaluate L(t sin kt)
L(t sin kt) = −
2ks
d
k
d
)
=
(L(sin kt)) = − ( 2
ds
ds s + k 2
(s2 + k 2 )2
c. Transforms of Derivatives
(1) Theorem
L(f (n) (t)) = sn F (s) − sn−1 f (0) − sn−2 f ′ (0)..... − f (n−1) (0)
where F (s) = L(f (t))
68
d. The Convolution Theorem
(1) Definition If f (x), g(x) are defined on [0, ∞) then
their convolution if the function F (t) = f ∗ g(t) defined
by
∫
∞
f ∗ g(t) =
f (u)g(t − u)du
0
(a) Theorem
∫ ∞
L(
f (u)g(t − u))du = L(f (t))L(g(t))
0
In words this says that the Laplace transform of the convolution of two functions is the product of the Laplace transforms.
5.Solving Differential Equations with Laplace Transforms
a. The General Method
(1) Consider the initial value problem
an y (n) + an−1 y (n−1) + ... + a0 y = g(t)
y(0) = y0 , y ′ (0) = y1 , ..., y (n−1) (0) = yn
(2) Apply the Laplace Transform
L(an y (n) + an−1 y (n−1) + ... + a0 y) = L(g(t))
= an L(y (n) ) + ... + a0 L(y) = L(g(t))
69
Now applying the results on transforms of derivatives
an [sn Y (s) − sn−1 y(0)] + ... = G(s)
where Y (s) = L(y), G(s) = L(g). Then
(an sn +...+a0 )Y (s) = an [sn−1 y0 +...+yn ]+an−1 [sn−2 y0 +...+yn−1 ]+...+G(s)
Now Y (s) = L(y(t)) so the solution is then y(t) = L−1 (Y (s))
EXAMPLE Solve y ′ − 3y = e2t : y(0) = 1.
y ′ = 3y + e2t =⇒ L(y ′ ) = 3L(y) + L(e3t )
Now let Y = L(y) so L(y ′ ) = sY − y(0) = sY − 1 and
1
L(e2t ) = s−2
so
L(y ′ ) = 3L(y)+L(e3t ) =⇒ sY −1 = 3Y −
1
s−1
=⇒ Y =
s−2
(s − 2)(s − 3)
Therefore the solution is
y(t) = L−1 (
s−1
)
(s − 2)(s − 3)
By partial fractions
s−1
1
2
=−
+
(s − 2)(s − 3)
s−2 s−3
=⇒ y(t) = L−1 (−
1
2
+
) = −e2t + 2e3t
s−2 s−3
6.The Heaviside and Dirac Delta Functions
a.The Heaviside Function
70
(1) Definition The Heaviside step function is defined by
H(x) = 1 : t ≥ 0
H(x) = 0 : t < 0
b. The Dirac Delta Function
(1) Definition Consider the unit impulse function
1
: −ϵ < t < ϵ
2ϵ
= 0 : elsewhere
δϵ (t) =
Then for each ϵ we have
∫ ∞
−∞
δϵ (t)dt = 1
The Dirac Delta Function is the limit of the unit impulse
functions.
δ(t) = Lmϵ→0 δϵ (t)
(a) Lemma For the Dirac delta function we have
∫ ∞
δ(t)dt = 1
−∞
Hence δ(t) can be characterized as being infinite at 0, have
the value 0 everywhere else and the total integral is one.
(b) Lemma The Laplace transforms of δ(t), t ≥ 0
and the shifted dleta function δ(t − t0 ) are
L(δ(⊔)) = ∞
71
L(δ(⊔ − ⊔′ )) = ⌉−∫ ⊔′
C. The Fourier Integral Theorem
1. The Fourier Integral Theorem
a.Definition If f (x)
∫ ∞is defined and absolutely integrable
on (−∞, ∞), (that is −∞ |f (x)|dx converges) then its Fourier
integral expansion is
∫ ∞
f (x) =
(A(α) cos(αx) + B(α) sin(αx))dα
0
where
1
A(α) =
π
B(α) =
1
π
∫
∞
f (x) cos(αx)dx
−∞
∫ ∞
f (x) sin(αx)dx
−∞
(1) Theorem (The Fourier Integral Theorem) Suppose f (x) is defined and piecewise continuous and absolutely integrable on (−∞, ∞) and f ′ (x) is also piecewise
continuous. Then its Fourier integral expansion converges
to
f (x + 0) + f (x − 0)
2
at each x and hence to f (x) at each point of continuity.
b.Equivalent Forms of the Integral Theorem
72
(1) Theorem The Fourier integral theorem can be
written in the following equivalent forms:
∫
∫
1 ∞ ∞
(i)f (x) =
f (u) cos α(x − u)dudα
π α=0 u=−∞
∫ ∞∫ ∞
1
(ii)f (x) =
f (u)eiα(x−u) dudα
2π −∞ −∞
∫ ∞
∫ ∞
1
(iii)f (x) =
eiαx (
f (u)eiα(x−u) du)dα
2π −∞
−∞
where f (x) = f (x) at each point of continuity and =
otherwise.
f (x+0)+f (x−0)
2
The last form (iii) is the basis for the Fourier Inversion
Theorem.
D. The Fourier Transform
1. Fourier Transforms
a.Definition If f (x) is a function satisfying the conditions of the Fourier Integral Theorem then its Fourier
transform is
∫ ∞
F(f (x)) = f (w) =
f (u)e−iwu du
−∞
(1) Theorem (The Fourier Inversion Theorem)
If f (x) is a function satisfying the conditions of the Fourier
Integral Theorem and f (w) is its Fourier transform then
∫ ∞
1
f (x) =
f (w)eiwx du
2π −∞
73
(Restatement of the Fourier integral theorem)
2. Properties of Fourier Transforms
a.Linearity
(1) Theorem The Fourier transform and its inverse are linear operators on the space of functions
F(αf (x) + βg(x)) = αF(f (x)) + βF(g(x))
F −1 (αf (w) + βg(w)) = αF −1 (f (w)) + βF −1 (g(w))
b.Transforms of Derivatives
(1) Theorem We have for a differentiable function with a Fourier transform
F(f (n) (x)) = (iw)n f (w)
c.Convolution Theorem
(1) Definition The Fourier convolution of
f (x), g(x) denoted f ⋆ g(x) is
∫ ∞
f ⋆ g(x) =
f (x − u)g(u)du
−∞
(a) Theorem (The Convolution Theorem)
Suppose f (x), g(x) are functions satisfying the Fourier Integral Theorem and f (w), g(w) their Fourier transforms.
Then the Fourier transform of the convolution is the product of the Fourier transforms. That is
F(f ⋆ g) = f (w)g(w)
74
F −1 (f g) = f ⋆ g
d.Parseval’s Identities for Fourier Integrals
(1) Theorem If f (w), g(w) are the Fourier transforms of f (x), g(x) respectively then
∫ ∞
∫ ∞
1
f (x)G(x)dx =
f (w)g(w)dw
2π −∞
−∞
∫ ∞
∫ ∞
1
2
|f (x)| dx =
|f (w)|2 dw
2π −∞
−∞
75
PARTIAL DIFFERENTIAL EQUATIONS (7)
THE BASIC BOUNDARY VALUE PROBLEMS
A. Method of Separation of Variables
1.For a homogeneous linear PDE the method of separation of variables attempts to find a solution of the
form
u(x, y) = X(x)Y (y)
EXAMPLE
∂ 2u
∂u
=
4
∂x2
∂y
Let u = XY where X = X(x), Y = Y (y) then
=⇒ X ′′ Y = 4XY ′
X ′′
Y′
=
4X
Y
Since X is a function of x alone and Y is a function of y
alone this equality is possible only if both sides equal the
same constant c.
=⇒
Case 1 c = λ2 > 0 then
X ′′ − 4λ2 X = 0
Y ′ − λ2 Y = 0
=⇒ X = c1 e2λx + c2 e−2λx and Y = c3 eλ
2
=⇒ u(x, y) = Aeλ y (Be2λx + Ce−2λx )
2
76
y
Case 2 c = −λ2 < 0 then
X ′′ + 4λ2 X = 0
Y ′ + λ2 Y = 0
=⇒ X = c1 cos(2λx) + c2 sin(2λx) and Y = c3 e−λ
2
y
=⇒ u(x, y) = Ae−λ y (B sin(2λx) + C cos(−2λx))
2
Case 3 c = 0 then
X ′′ = 0
Y′ =0
=⇒ X = c1 + c2 x and Y = c3
=⇒ u(x, y) = Ax + B
2. Superposition Principle
a. Theorem If u1 , u2 , ..., un are solutions to a linear
homogeneous PDE then any linear combination is also a
solution.
(1) We assume that if u1 , u2 ..., un , ... are an infinite sequence of solutions then
∞
∑
ui
i=1
is also a solution
B. Boundary Value Problems
1.Basic Boundary Value Problems
77
a. Recall our three basic prototype PDE’s
(1) Diffusion or Heat Equation
α2 uxx = ut
(a) This is the one-dimensional heat equation. Recall in 3-variables the heat equation if α2 ▽2 u = ut .
(b) This arises in the flow of heat in a solid
in one direction. u(x, t) represents the temperature. This
equation can also represent the flow of electricity in a cable
- then its known as the telegraph equation
(2) Wave Equation
β 2 uxx = utt
(a) This is the one-dimensional wave equation. Roughly wave equations arise is modeling wave phenomena
(b) This arises from problems in mechanical
vibrations. A solution u(x, t) will represent the small displacements of an idealized vibrating string
[1] The wave equation also appears in the
theory of high frequency transmission lines, fluid mechanics,
acoustics and electricity.
(3) Laplace Equation
78
uxx + uyy = 0
(a) Roughly Laplace equations arise is modeling equilibrium problems. A solution u(x, t) represents
steady state temperature in a thin flat plate. Also arises in
problems delaing with electrostatic potentials, gravitational
potentials, and velocity potentials in fluid dynamics.
b. Consider these basic equations subject to:
(1) Boundary conditions
(a) u or ux specified at x = x0 or u,uy specified
at y = y0 .
(2) Initial conditions
(a) Specifying u(x, 0) = f (x) in the heat equation or u(x, 0) = f (x), ut (x, 0) = g(x) in the wave equation
(3) The PDE together with these side conditions,
both boundary and initial is called a boundary value
problem.
2.The Heat Equation
a. Basic Boundary Value Problem
kuxx = ut : k > 0, 0 < x < L, t > 0
u(0, t) = 0, u(L, t) = 0 : t > 0
u(x, 0) = f (x) : 0 < x < L
79
b. Physical Interpretation
(1) This represents the flow of heat in the x direction along a thin rod pf length L with an initial temperature
of f (x) with ends held constant at 0 for all t. k > 0 is the
diffusivity and is proportional to the thermal conductivity.
The solution u(x, t) is the temperature.
(2) Conditions for One-Dimensional Heat
Equation
(a) The flow of heat is only in the x-direction
(b) No heat escapes from the lateral surface
of the rod
(c) No heat is being generated in the rod
(d) The rod is homogeneous, that is its density
per length is constant
(e) The specific heat and thermal conductivity
are constant
c. The Solution
(1) Attempt a separation of variables solution
u = XT . We then arrive at
X ′′
T′
=
= −λ2
X
kT
Here we choose the constant c = −λ2 < 0 indicating that
the heat is dying exponentially as it goes to the boundary.
=⇒ X ′′ + λ2 X = 0
80
X(0) = 0, X(L) = 0
T ′ + kλ2 T = 0
The solution for T (t) is then
T (t) = Ae−kλ
2
t
The solution for X(x) is a Sturm-Liouville problem with
eigenvalues
nπ
λn =
, n = 1, 2, 3, ....
L
and eigenfucntions
ϕn (x) = sin(
nπx
), n = 1, 2, 3...
L
Therefore for each n = 1, 2, 3, .... we have a solution
un (x, t) = An e−kλ t sin(
2
nπ
x)
L
We have to be able to solve for the initial condition so by
the superposition principle we also have the solution
u(x, t) =
∞
∑
An e−kλn t sin(
2
n=1
nπ
x)
L
Now the initial condition
u(x, 0) = f (x) =⇒ f (x) =
∞
∑
n=1
An sin(
nπ
x)
L
We recognize this as the Fourier half-range sine expansion
of f (x) defined on (0, L). Therefore
81
2
An =
L
∫
L
f (x) sin(
0
nπ
x)dx
L
d. Insulated Boundaries
(1) The heat equation is modified if the ends are
insulated. Here the normal derivative of the temperature is
zero.
(2) For insulated boundaries the boundary value
problem is
kuxx = ut : k > 0, t > 0, 0 < x < L
ux (0, t) = 0, ux (L, t) = 0 : t > 0
u(x, 0) = f (x); 0 < x < l
3.The Wave Equation
a. Basic Boundary Value Problem
a2 uxx = utt : 0 < x < L, t > 0
u(0, t) = 0, u(L, t) = 0 : t > 0
u(x, 0) = f (x), ut (x, 0) = g(x) : 0 < x < L
f (0) = f (L) = 0
b. Physical Interpretation
(1) This represents the transverse vibrations of a
string stretched bewteen x = 0 and x = L. The solution
u(x, t) measures the displacements from the x-axis for t > 0.
82
f (x) represents the initial configuration of the string and
g(x) represents the initial velocity at each point.
(2) Conditions for One-Dimensional Wave
Equation
(a) The string is perfectly flexible
(b) The displacements are small relative to the
length of the string
(c) The tension of the string is constant
(d) The string is homogeneous, that is its mass
per unit length is constant
(e) The tension is large compared to the force
of gravity
(f) No other forces act on the string
c. The Solution
(1) Attempt a separation of variables solution
u = XT . We then arrive at
T ′′
X ′′
= 2 = −λ2
X
aT
=⇒ X ′′ + λ2 X = 0
X(0) = 0, X(L) = 0
T ′′ + a2 λ2 T = 0
The solution for T (t) is then
T (t) = c1 cos(aλt) + c2 sin(aλt)
83
The solution for X(x) is a Sturm-Liouville problem with
eigenvalues
nπ
, n = 1, 2, 3, ....
λn =
L
and eigenfucntions
nπx
ϕn (x) = sin(
), n = 1, 2, 3...
L
Therefore for each n = 1, 2, 3, .... we have a solution
nπ
nπ
nπ
un (x, t) = (An cos(a t) + Bn sin(a t) sin( x)
L
L
L
We have to be able to solve for the initial condition so by
the superposition principle we also have the solution
u(x, t) =
∞
∑
(An cos(a
n=1
nπ
nπ
nπ
t) + Bn sin(a t) sin( x)
L
L
L
Now the initial conditions
u(x, 0) = f (x) =⇒ f (x) =
∞
∑
An sin(
n=1
nπ
x)
L
We recognize this as the Fourier half-range sine expansion
of f (x) defined on (0, L). Therefore
2
An =
L
∫
L
f (x) sin(
0
nπ
x)dx
L
Differentiating with respect to t we get
ut =
∞
∑
n=1
(−An a
nπ
nπ
nπ
nπ
nπ
sin(a t) + Bn a
cos(a t) sin( x)
L
L
L
L
L
84
ut (x, 0) = g(x) =⇒ g(x) =
∞
∑
Bn a
n=1
nπ
nπ
sin( x)
L
L
We recognize this also as the Fourier half-range sine expansion of g(x) defined on (0, L). Therefore
∫
nπ
2 L
nπ
=⇒ Bn a
=
g(x) sin( x)dx
L
L 0
L
∫ L
2
nπ
=⇒ Bn =
g(x) sin( x)dx
anπ 0
L
Note that if the string is released from rest so that g(x) = 0
then each Bn = 0.
4.Laplace’s Equation
a. Basic Boundary Value Problem
uxx + uyy = 0 : 0 < x < a, 0 < y < b
ux (0, y) = 0, ux (a, y) = 0 : 0 < y < b
u(x, 0) = 0 : 0 < x < a
u(x, b) = f (x) : 0 < x < a
b. Physical Interpretation
(1) This represents the steady state temperature
u(x, y) in a rectangular plate with the prescribed boundary
conditions
c. The Solution
85
(1) Attempt a separation of variables solution
u = XY . We then arrive at
X ′′ + λ2 X = 0
X ′ (0) = 0, X ′ (a) = 0
Y ′′ − λ2 Y = 0
Y (0) = 0
The solution for Y (y) are then
Y (y) = c1 cosh(λy) + c2 sinh(λy)
Y (0) = 0 =⇒ Y = c2 sinh(λy)
The solution for X(x) is a Sturm-Liouville problem with
eigenvalues
nπ
λ = 0 and λn =
, n = 1, 2, 3, ....
a
and eigenfunctions
X = c for λ = 0 and ϕn (x) = cos(
nπ
x), n = 1, 2, 3...
a
Therefore by the superposition principle we also have the
solution
u(x, y) = A0 y +
∞
∑
An sinh(
n=1
nπ
nπ
y) cos( x)
a
a
Now the initial conditions
u(x, b) = f (x) = A0 b +
∞
∑
n=1
86
An sinh(
nbπ
nπ
) cos( x)
a
a
We recognize this as the Fourier half-range cosine expansion
of f (x) defined on (0, a). Therefore
∫
2 a
f (x)dx =⇒ A0 =
f (x)dx
ab
0
0
∫
nbπ
2 a
nπ
An sinh(
)=
f (x) cos( x)dx
a
a 0
a
∫ a
nπ
2
f (x) cos( x)dx
=⇒ An =
nbπ
a
a sinh( a ) 0
2
2A0 b =
a
∫
a
5.D’alembert’s Solution to The Wave Equation
a. There is an alternative method for solving the
wave equation due to D’Alembert. A change of variables
simplifies the wave equation so you can directly get the
most general solution.
b. Consider the wave PDE; c2 uxx = utt . Make the
change of variables ζ = x + ct, η = x − ct. This transforms
the orginal equation to
∂ 2u
=0
∂ζ∂η
(1) This can be solved directly to give
u(x, y) = F (ζ) + G(η) = F (x + ct) + G(x − ct)
this is the most general solution to the wave equation.
(2) If we impose initial conditons u(x, 0) = f (x)
and ut (x, 0) = g(x) then we further find that
∫
1 x+ct
f (x + ct) + F (x − ct)
+
g(u)du
u(x, t) =
2
2c x−ct
87
PARTIAL DIFFERENTIAL EQUATIONS (8)
HARMONIC AND VIBRATORY MOTION
1. Definition Periodic motion or harmonic motion is any motion that repeats itself in equal intervals of
time. It is vibratory or oscillatory if it moves back and
forth over the same path. If the vibratory motion is slowed
by friction it is called damped harmonic motion.
a. Note Mechanical and electromagnetic oscillations
are described by the same mathematical equations.
b. Definition The period T of a harmonic motion
is the time to complete one cycle. The frequency ν is the
number of oscillations (cycles) per unit time
=⇒ ν =
1
T
The equilibrium position is the position at which no net
force acts on the oscillatory particle. The displacement is
the distance linear or angular from the equilibrium position
(at a particular time).
2. The Simple Harmonic Oscillator
a.Definition An oscillating particle moving back and
forth about an equilibrium position so that the potential
energy is
1
U (x) = kx2
2
is a simple harmonic oscillator and the motion is simple
harmonic motion.
88
(1) Lemma The force acting on a simple harmonic
oscillator is
F (x) = −kx
where k is the force constant. This is derived from F (x) =
− dUdx(x) . It is entirely analogous to Hooke’s law for a stretched
spring.
(2) Definition
called the amplitude.
The maximum displacement is
(3) Note Hooke’s law holds up to the elastic limit
for many common materials.
3. Simple Harmonic Motion
a.Theorem
motion is
The basic ODE for simple harmonic
k
d2 x
+
x = 0.
dt2
m
The solution to this ODE can be given by
x(t) = A cos(ωt + δ)
where
A = the amplitude
2π
ω = 2πν =
T
where
ν = the frequency and T = the period .
ωt + δ = phase of the motion and −
89
δ
= phase shift .
ω
4. Energy in Simple Harmonic Motion
a.Theorem For simple harmonic motion
[1] The potential energy at time t is given by
1
U = kA2 cos2 (ωt + δ)
2
It has a maximum value of 21 kA2 .
[2] The kinetic energy at time t is given by
1
U = kA2 sin2 (ωt + δ)
2
It has a maximum value of 21 kA2 .
[3] The total energy at time t is given by
1
E = K + U = kA2
2
Hence the total energy in simple harmonic motion is proportional to the square of the amplitude.
(1) Corollary During one period the average kinetic energy is equal to the average potential energy and is
given by 14 kA2 .
(2) Corollary Speed is a maximum at the equilibrium position x = 0 and zero at the maximum displacement
x = A. In general
√
k 2
v=±
(A − x2 )
m
90
4. The Simple Pendulum
a. Consider a mass m suspended from a cord with
tension T as below
(2) The restoring force is given by
F = −mg sin θ
(a) If θ is small then sin θ u θ
=⇒ F = −mg sin θ = −
mg
x
l
Hence this is simple harmonic motion with k = − mg
l
(b) The period is then given by
√
√
m
l
T = 2π
= 2π
k
g
Note that the period is independent of the mass.
91
II. Oscillations
A. Oscillations
1. Periodic Motion
a.Definition Periodic motion or harmonic motion is any motion that repeats itself in equal intervals of
time. It is vibratory or oscillatory if it moves back and
forth over the same path. If the vibratory motion is slowed
by firction it is called damped harmonic motion.
(1) Note Mechanical and electromagnetic oscillations are described by the same mathematical equations.
(2) Definition The period T of a harmonic
motion is the time to complete one cycle. The frequency
ν is the number of oscillations (cycles) per unit time
1
=⇒ ν =
T
The equilibrium position is the position at which no net
force acts on the oscillatory particle. The displacement is
the distance linear or angular from the equilibrium position
(at a particular time).
B. Harmonic Motion
1. The Simple Harmonic Oscillator
a.Definition An oscillating particle moving back
and forth about an equilibrium position so that the potential energy is
1
U (x) = kx2
2
92
is a simple harmonic oscillator and the motion is simple
harmonic motion.
(1) Lemma The force acting on a simple harmonic oscillator is
F (x) = −kx
where k is the force constant. This is derived from F (x) =
− dUdx(x) . It is entirely analogous to Hooke’s law for a stretched
spring.
(2) definition The maximum displacement is
called the amplitude.
(3) Note Hooke’s law holds up to the elastic
limit for many common materials.
2. Simple Harmonic Motion
a.Theorem The basic ODE for simple harmonic
motion is
k
d2 x
+ x = 0.
2
dt
m
The solution to this ODE can be given by
x(t) = A cos(ωt + δ)
where
A = the amplitude
2π
ω = 2πν =
T
where
ν = the frequency and T = the period .
93
ωt + δ = phase of the motion and −
δ
= phase shift .
ω
3. Energy in Simple Harmonic Motion
a.Theorem For simple harmonic motion
[1] The potential energy at time t is given by
1
U = kA2 cos2 (ωt + δ)
2
It has a maximum value of 21 kA2 .
[2] The kinetic energy at time t is given by
1
U = kA2 sin2 (ωt + δ)
2
It has a maximum value of 21 kA2 .
[3] The total energy at time t is given by
1
E = K + U = kA2
2
Hence the total energy in simple harmonic motion is proportional to the square of the amplitude.
(1) Corollary During one period the average
kinetic energy is equal to the average potential energy and
is given by 14 kA2 .
(1) Corollary Speed is a maximum at the equilibrium position x = 0 and zero at the maximum displacement x = A. In general
√
k 2
(A − x2 )
v=±
m
94
4. Some Examples of Simple Harmonic Motion
a.The Simple Pendulum
(1) Consider a mass m suspended from a chord
with tension T as below
(2) The restoring force is given by
F = −mg sin θ
(a) If θ is small then sin θ u θ
mg
=⇒ F = −mg sin θ = −
x
l
Hence this is simple harmonic motion with k = − mg
l
(b) The period is then given by
√
√
m
l
= 2π
T = 2π
k
g
95
Note that the period is independent of the mass.
b.The Torsional Pendulum
(1) Consider a disc attached at its center as below. Twist the wire
(2) The restoring torque is given by
τ = −κθ
where κ is the torsional constant of the wire.
(3) If κ is the torsional constant of the wire and
I is th rotational inertia of the disc then the ODE here is
given by
κ
dθ
+
θ=0
dt2 I
(a) This implies that the solution is
θ(t) = θm cos(ωt + δ)
96
The period is then
√
T = 2π
I
κ
c.The Physical Pendulum
(1) Consider any rigid body mounted so that it
can swing in a vertical plane about some axis
(2) The restoring torque for an angular displacement θ is given by
τ = −M gd sin θ
where d is the distance from the pivot point to the center
of mass
(3) If θ is small then sin θ u θ and then
τ = −M gdθ = −κθ
where
κ = M gd
Hence it is harmonic motion.
(a) The period of a physical pendulum is then
given by
√
T = 2π
√
I
= 2π
κ
I
M gd
(b) This implies that the inertia of the body
is given by
T 2 M gd
I=
4π 2
97
Everyhting on the right is measurable so this can be used
to determine the inertia I.
(4) Lemma The length of a simple pendulum
whose period is equal to that of a particular physical pendulum is
I
l=
Md
(a) For determining the period of oscillation
the mass of a physical pendulum may be considered to be
concentrated at a point whose distance from the pivot is
l = MI d . This point is called the center of oscillation and
depends on the pivot point.
5. Uniform Circular Motion
a.Theorem Simple harmonic motion can be described as the projection along the diameter of unifrom circular motion.
Conversely uniform circular motion can be described as
a combination of two simple harmonic motions occurring
along perpendicular lines having the same amplitude and
frequency but differeing in phase by 90o .
6. Combinations of Harmonic Motions
a.Theorem All combinations of two simple harmonic motions at right angles having the same frequency
correspond to elliptical paths. The shape of the ellipse
A
depends only on the ratio of the amplitudes Axy and the
difference between the phases.
98
7. Two Body Oscillations
a. Consider a two-body oscillation, that is two
masses m1 , m2 connected by a massless spring with spring
constant k. This is then equivalent to a one-body oscillation
by redefining a reduced mass.
(1) The relevant ODE’s here are
m1
d2 x1
d2 x2
=
−kx
and
m
= kx2
1
2
dt2
dt2
Defining x = (x1 − x2 ) − l to be the relative displacement
this becomes
d2 x
d2 x k
µ 2 + kx = 0 =⇒
+ x=0
dt
dt2
µ
where
m1 m2
1
1
1
or =
+
m1 + m2
µ m1 m2
is the reduced mass.
µ=
(2) The solution to the two-body oscillation is
then
x(t) = A cos(ωt + δ)
where
x = (x1 − x2 ) − l and ω 2 =
k
µ
(a) The period and frequency are then
√
√
1 k
µ
and ν =
.
T = 2π
k
2π µ
99
8. Damped Harmonic Motion
a. If the oscillation is damped by friction so that it
slows down it is called damped harmonic motion.
(1) The damping force is proportional to the
velocity so that it is given by
Fdamp == −b
dx
dt
(2) The resulting ODE is then
dx
d2 x
m 2 + b + kx = 0
dt
dt
(a) If the frictional force b is small then the
solution is given by
x(t) = Ae− m cos(ω ′ t + δ)
bt
√
where
ω ′ = 2πν ′ =
k
b 2
−(
)
m
2m
(b) When friction is present the frequency is
smaller and the period is longer. Eventually the energy will
dissipate to zero.
9. Forced Oscillations and Resonance
a. Consider a harmonic oscillator subject to both
friction and an external periodic driving force given by
Fm cos(ω ′′ t).
100
(1) The ODE is given by
m
dx
d2 x
+
b
+ kx = Fm cos(ω ′′ t)
2
dt
dt
(a) The solution is given by
x(t) =
where
Fm
sin(ω ′′ t − δ)
G
√
G = m2 (ω ′′2 − ω 2 ) + b2 ω ′′2
and
bω ′′
)
δ = cos (
G
−1
[1] The system resonantes with the frequency ω of the driving force rather than the natural frequency ω.
′′
[2] There is a characteristic value for ω ′′
for which FGm is a maximum. This is called the resonant
frequency and the condition is called resonance.
[a] If b = 0 then as ω ′′ → ω the amplitude becomes infinite. This is the basis for the Tacoma
Bridge disaster and why soldiers break step on a bridge.
[b] The displacement caused by a constant driving force Fm is just Fkm .
EXAMPLE (a) A 40 lb. weight stretches a spring 4 ft. initally the weight starts from rest 2 ft. below the equilibrium
101
position. What is the equation of the resulting motion?
What is the amplitude and period of the resulting motion?
(b) Suppose the spring of part (a) is placed in a medium
with a damping force equal to 1/5 the instantaneous velocity. What is the equation of motion?
(c) Suppose the spring of part (a) is placed in a medium
with a damping force equal to 1/5 the instantaneous velocity and is subjected to a driving force given by f (t) =
3 sin(2t). What is the equation of motion?
SOLUTION
(a) The basic ODE is
y ′′ +
Here
m=
k
y=0
m
40 5
W
=
=
g
32 4
From
40 = 4k =⇒ k = 10 =⇒
k
=8
m
Therefore the initial value problem is
y ′′ + 8y = 0; y(0) = 2, y ′ (0) = 0
The solution is then
√
y = 2 cos( 8t)
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and hence the amplitude is 2 and the period is
2π
√
.
8
(b) The basic ODE is
my ′′ = −ky = βy ′ =⇒ y ′′ +
Here
β ′ k
y + y=0
m
m
k
=8
m
Further
1
β
1
=⇒
=
5
m 40
Therefore the initial value problem is
β=
y ′′ +
1 ′
y + 8y = 0; y(0) = 2, y ′ (0) = 0
40
The auxillary equation is
m2 +
1
m+8=0
40
This has solution
1
m=− ±
20
√
1279
i
80
Therefore the solution to the ODE is
√
√
1
1279
1279
y = e− 20 t (c1 cos(
t) + c2 sin(
t)
80
80
From the initial conditions we get that c1 = 2, c2 = Therefore the equation of motion is
2
2√
2√
y = e− 9 t (2 cos( 215t) − c2 sin( 215t)
9
9
103
(c) From the previous two parts the basic initial value problem is
y ′′ +
1 ′
y + 8y = 3 sin(2t); y(0) = 2, y ′ (0) = 0
40
The solution to the homogeneous part is then
√
√
1
1279
1279
t) + c2 sin(
t)
y = e− 20 t (c1 cos(
80
80
To find a particular solution we use the method of undetermined coefficients. We try for a solution of the form
y(t) = A sin(2t) + B cos(2t)
104
PARTIAL DIFFRENTIAL EQUATIONS (8)
SPECIAL FUNCTIONS
A.Special Functions are particular mathematical functions that have more or less established names and notations due to their importance in mathematical analysis,
functional analysis, geometry, physics, or other applications.
The term is defined by consensus, and thus lacks a general
formal definition, but the List of mathematical functions
contains functions that are commonly accepted as special.
1. Certain special differential equations occur frequently
in applied mathematics and physics. The solutions are
often complicated functions whcih have been extensively
studied. These functions have been called special functions. We will look at two of these. The first appears in
many statistical applications and is used in the definition
of the second.
2. The Gamma Function
(1) The Gamma Function is defined for x > 0 by
∫ ∞
Γ(x) =
tx−1 e−t dt
0
(a) Lemma Γ(x + 1) = xΓ(x)
(b) Lemma For a positive integer n we have
Γ(n + 1) = n!
For this reason the Gamma function is often called the generalized factorial function
105
(c) Lemma Γ( 12 ) =
√
π
(d) By using Γ(x) = x1 Γ(x + 1) the Gamma function can be defined for all negative reals except for n =
0, −1, −2, ......
B. The Bessel Function
1. A Bessel equation is an ODE
x2 y ′′ + xy ′ + (x2 − v 2 )y = 0 : v ≥ 0
a.Definition The Bessel Function of the first kind
of order v is
Jv (x) =
∞
∑
i=0
(−1)n
x
( )2n+v
n!Γ(1 + v + n) 2
[1] Theorem Jv (x) is always one solution to the
Bessel equation.
[i] If 2v is not an integer then the 2 independent
solutions are Jv (x) and J−v (x)
[ii] If 2v is an integer then a second solutions is given by
∫
J(x) = Jv (x)
106
dx
x(Jv (x))2