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ENGI 9420 Lecture Notes 8 - PDEs Page 8.01

8. Partial Differential Equations Partial differential equations (PDEs) are equations involving functions of more than one variable and their partial derivatives with respect to those variables. Most (but not all) physical models in engineering that result in partial differential equations are of at most second order and are often linear. (Some problems such as elastic stresses and bending moments of a beam can be of fourth order). In this course we shall have time to look at only a small subset of second order linear partial differential equations. Sections in this Chapter: 8.01 Major Classifications of Common PDEs 8.02 The Wave Equation – Vibrating Finite String 8.03 The Wave Equation – Vibrating Infinite String 8.04 d’Alembert Solution 8.05 The Maximum-Minimum Principle 8.06 The Heat Equation

ENGI 9420 8.01 - Classification of PDEs Page 8.02

8.01 Major Classifications of Common PDEs A general second order linear partial differential equation in two Cartesian variables can be written as

( ) ( ) ( )2 2 2

2 2, , , , , , ,u u u u uA x y B x y C x y f x y ux x y y x y

∂ ∂ ∂ ∂ ∂+ + = ∂ ∂ ∂ ∂ ∂ ∂

Three main types arise, based on the value of D = B 2 – 4AC (a discriminant): Hyperbolic, wherever (x, y) is such that D > 0; Parabolic, wherever (x, y) is such that D = 0; Elliptic, wherever (x, y) is such that D < 0. Among the most important partial differential equations in engineering are:

The wave equation: 2

2 22

u c ut

∂= ∇

∂

or its one-dimensional special case 2 2

22 2

u uct x

∂ ∂=

∂ ∂ [which is hyperbolic everywhere]

(where u is the displacement and c is the speed of the wave);

The heat (or diffusion) equation: 2u K u K ut

µρ ∂= ∇ +

∂

∇ ∇

a one-dimensional special case of which is

2

2

u K ut xµρ

∂ ∂=

∂ ∂ [which is parabolic everywhere]

(where u is the temperature, μ is the specific heat of the medium, ρ is the density and K is the thermal conductivity); The potential (or Laplace’s) equation: 2 0u∇ =

a special case of which is 2 2

2 2 0u ux y

∂ ∂+ =

∂ ∂ [which is elliptic everywhere]

The complete solution of a PDE requires additional information, in the form of initial conditions (values of the dependent variable and its first partial derivatives at t = 0), boundary conditions (values of the dependent variable on the boundary of the domain) or some combination of these conditions.

ENGI 9420 8.02 - Finite String Page 8.03

8.02 The Wave Equation – Vibrating Finite String The wave equation is

22 2

2

u c ut

∂= ∇

∂

If u(x, t) is the vertical displacement of a point at location x on a vibrating string at time t, then the governing PDE is

2 22

2 2

u uct x

∂ ∂=

∂ ∂

If u(x, y, t) is the vertical displacement of a point at location (x, y) on a vibrating membrane at time t, then the governing PDE is

2 2 22 2 2

2 2 2

u u uc u ct x y

∂ ∂ ∂= ∇ = + ∂ ∂ ∂

or, in plane polar coordinates (r, θ), (appropriate for a circular drum), 2 2 2

22 2 2 2

1 1u u u uct r r r r θ

∂ ∂ ∂ ∂= + + ∂ ∂ ∂ ∂

For spherically symmetric waves in 3 , the wave equation is

2 2 22 2

2 2 2

2u c u u ur ct r r r r r r

∂ ∂ ∂ ∂ ∂= = + ∂ ∂ ∂ ∂ ∂

The latter two cases lead to solutions involving Bessel functions. Example 8.02.1 An elastic string of length L is fixed at both ends (x = 0 and x = L). The string is displaced into the form y = f (x) and is released from rest. Find the displacement y(x, t) at all locations on the string (0 < x < L) and at all subsequent times (t > 0). The boundary value problem for the displacement function y(x, t) is:

2 2

22 2 for 0 and 0y yc x L t

t x∂ ∂

= < < >∂ ∂

Both ends fixed for all time: y(0, t) = y(L, t) = 0 for t > 0 Initial configuration of string: y(x, 0) = f (x) for 0 < x < L

String released from rest: ( ),0

0 for 0x

y x Lt

∂= ≤ ≤

∂

ENGI 9420 8.02 - Finite String Page 8.04

Example 8.02.1 (continued) Separation of Variables (or Fourier Method) Attempt a solution of the form y(x, t) = X(x) T(t) Substitute y(x, t) = X(x) T(t) into the PDE:

( ) ( )( ) ( ) ( )( )2 2 2 2

2 22 2 2 2

d T d XX x T t c X x T t X c Tt x dt dx∂ ∂

= ⇒ =∂ ∂

2 2

2 2 2

1 1 1d T d Xc T dt X dx

⇒ =

The left hand side of this equation is a function of t only. At any instant t it must have the same value at all values of x. Therefore the right hand side, which is a function of x only, must at any one instant have that same value at all values of x. By a similar argument, the right hand side of this equation is a function of x only. At any location x it must have the same value at all times t. Therefore the left hand side, which is a function of t only, must at any one location have that same value at all times t. Thus both sides of this differential equation must be the same absolute constant, which we shall represent for now by –k.

2 2

2 2

1 0d X d Xk k XX dx dx

= − ⇒ + =

The general solution of this simple second order ODE is a combination of two exponential functions of x if k < 0, is a linear function of x if k = 0 and is a combination of sine and cosine functions of x if k > 0. It is not possible for both ends of the string to be fixed for all time in the first two cases (unless we admit the trivial solution y(x, t) ≡ 0, a string that never moves from its equilibrium position). Therefore k > 0. Replace k by λ 2 (guaranteed positive for all real λ except λ = 0). We now have the pair of ODEs

2 22 2 2

2 20 and 0d X d TX c Tdx dt

λ λ+ = + =

The general solutions are ( ) ( ) ( ) ( ) ( ) ( )cos sin and cos sinX x A x B x T t C ct D ctλ λ λ λ= + = +

respectively, where A, B, C and D are arbitrary constants.

ENGI 9420 8.02 - Finite String Page 8.05

Example 8.02.1 (continued) Consider the boundary conditions: ( ) ( ) ( )0, 0 0 0y t X T t t= = ∀ ≥

For a non-trivial solution, this requires ( )0 0 0X A= ⇒ = . ( ) ( ) ( ) ( ), 0 0 0y L t X L T t t X L= = ∀ ≥ ⇒ =

( ) ( )sin 0 ,nnB L nLπλ λ⇒ = ⇒ = ∈

We now have a solution only for a discrete set of eigenvalues λn, with corresponding eigenfunctions

( ) ( )sin , 1, 2, 3,nn xX x n

Lπ = =

Consider the initial condition:

( )( ) ( ) ( )

,00 0 0 0

x

y X x T x Tt

∂ ′ ′= = ∀ ⇒ =∂

( ) ( ) ( ) ( )sin cos 0 0 0T t C c ct D c ct T D c Dλ λ λ λ λ′ ′= − + ⇒ = = ⇒ = Therefore our complete solution for y(x, t) is now some linear combination of

( ) ( ) ( ) ( ), sin cos , 1, 2, 3,n n n nn x n c ty x t X x T t C n

L Lπ π = = =

There is one condition remaining to be satisfied. The initial configuration of the string is: y(x, 0) = f (x) for 0 < x < L.

( ) ( )1

,0 sinn

nn xy x C f x

Lπ

=

∞ ⇒ = = ∑

This is precisely the Fourier sine series expansion of f (x) on [0, L] ! From Fourier series theory (Chapter 7), the coefficients Cn are

( )0

2 sinL

nn uC f u du

L Lπ =

∫

Therefore our complete solution is

( ) ( )01

2, sin sin cosL

n

n u n x n c ty x t f u duL L L L

π π π

=

∞ = ∫∑

ENGI 9420 8.02 - Finite String Page 8.06

Example 8.02.1 (continued) This solution is valid for any initial displacement function f (x) that is continuous with a piece-wise continuous derivative on [0, L] with f (0) = f (L) = 0.

If the initial displacement is itself sinusoidal ( ) sin for somen xf x a nLπ = ∈

,

then the complete solution is a single term from the infinite series,

( ), sin cosn x n c ty x t aL Lπ π =

.

Example 8.02.2 Suppose that the initial configuration is triangular:

( ) ( )( )( )

12

12

0,0

x x Ly x f x

L x L x L

≤ ≤= = − < ≤

Then the Fourier sine coefficients are

( )0

2 sinL

nn uC f u du

L Lπ =

∫

( )0

/2

/2

2 2sin sinL L

L

n u n uu du L u duL L L L

π π = + − ∫ ∫

( )

0

2 /2

/2

2 cos sin

cos sin

L

L

L

L n u n u n uL n L L L

n L u n u n uL L L

π π ππ

π π π

= ⋅ − + − + − −

ENGI 9420 8.02 - Finite String Page 8.07

Example 8.02.2 (continued)

( )( )

( )

22 cos sin 0 0

2 2 2

0 0 cos sin2 2 2

L n n nn

n n n

π π π

π

π π π

= − + − −

+ − + − − −

( )( )( )2

0 even4 sin But sin1 odd2 2

nL n nnn

π π

π

= = ±

( ) ( ) ( )12 1sin 1 ,

2kk

kπ +−

= − ∈

Therefore sum over the odd integer values of n only (n = 2k – 1).

( )( )( )2

14 12 1

kk

LCk π

+= −−

and

( ) ( )( )

( ) ( )2 2

1

11 2 1 2 14, sin cos2 1k

k k x k c tLy x tL Lk

π ππ

=

∞ +− − − =

− ∑

See the web page "www.engr.mun.ca/~ggeorge/9420/demos/ex8022.html" for an animation of this solution.

ENGI 9420 8.02 - Finite String Page 8.08

Example 8.02.2 (continued) Some snapshots of the solution are shown here:

These graphs were generated from the Fourier series, truncated after the fifth non-zero term.

ENGI 9420 8.02 - Finite String Page 8.09

Example 8.02.3 An elastic string of length L is fixed at both ends (x = 0 and x = L). The string is initially in its equilibrium state [y(x, 0) = 0 for all x] and is released with the initial velocity

( )( )

,0x

y g xt

∂=

∂. Find the displacement y(x, t) at all locations on the string (0 < x < L)

and at all subsequent times (t > 0). The boundary value problem for the displacement function y(x, t) is:

2 2

22 2 for 0 and 0y yc x L t

t x∂ ∂

= < < >∂ ∂

Both ends fixed for all time: y(0, t) = y(L, t) = 0 for t > 0 Initial configuration of string: y(x, 0) = 0 for 0 < x < L

String released with initial velocity: ( )

( ),0

for 0x

y g x x Lt

∂= ≤ ≤

∂

As before, attempt a solution by the method of the separation of variables. Substitute y(x, t) = X(x) T(t) into the PDE:

( ) ( )( ) ( ) ( )( )2 2 2 2

2 22 2 2 2

d T d XX x T t c X x T t X c Tt x dt dx∂ ∂

= ⇒ =∂ ∂

Again, each side must be a negative constant. 2 2

22 2 2

1 1d T d Xc T dt X dx

λ⇒ = = −

We now have the pair of ODEs

2 22 2 2

2 20 and 0d X d TX c Tdx dt

λ λ+ = + =

The general solutions are ( ) ( ) ( ) ( ) ( ) ( )cos sin and cos sinX x A x B x T t C ct D ctλ λ λ λ= + = +

respectively, where A, B, C and D are arbitrary constants.

ENGI 9420 8.02 - Finite String Page 8.10

Example 8.02.3 (continued) Consider the boundary conditions: ( ) ( ) ( )0, 0 0 0y t X T t t= = ∀ ≥

For a non-trivial solution, this requires ( )0 0 0X A= ⇒ = . ( ) ( ) ( ) ( ), 0 0 0y L t X L T t t X L= = ∀ ≥ ⇒ =

( ) ( )sin 0 ,nnB L nLπλ λ⇒ = ⇒ = ∈

We now have a solution only for a discrete set of eigenvalues λn, with corresponding eigenfunctions

( ) ( )sin , 1, 2, 3,nn xX x n

Lπ = =

and

( ) ( ) ( ) ( ) ( ), sin , 1, 2, 3,n n n nn xy x t X x T t T t n

Lπ = = =

So far, the solution has been identical to Example 8.02.1. Consider the initial condition y(x, 0) = 0 :

( ) ( ) ( ) ( ),0 0 0 0 0 0y x X x T x T= ⇒ = ∀ ⇒ = The initial value problem for T(t) is now

( )2 2 0 , 0 0 , where nT c T TLπλ λ′′ + = = =

the solution to which is

( ) ( )sin ,n nn c tT t C n

Lπ = ∈

Our eigenfunctions for y are now

( ) ( ) ( ) ( ), sin sin ,n n n nn x n c ty x t X x T t C n

L Lπ π = = ∈

ENGI 9420 8.02 - Finite String Page 8.11

Example 8.02.3 (continued) Differentiate term by term and impose the initial velocity condition:

( )( )

, 0 1sin

x nn

y n c n xC g xt L L

π π∞

=

∂ = = ∂ ∑

which is just the Fourier sine series expansion for the function g(x). The coefficients of the expansion are

( )0

2 sinL

nn c n uC g u du

L L Lπ π =

∫

which leads to the complete solution

( ) ( )01

2 1, sin sin sinL

n

n u n x n c ty x t g u duc n L L L

π π ππ

=

∞ = ∫∑

This solution is valid for any initial velocity function g(x) that is continuous with a piece-wise continuous derivative on [0, L] with g(0) = g(L) = 0. The solutions for Examples 8.02.1 and 8.02.3 may be superposed.

Let ( )1 ,y x t be the solution for initial displacement f (x) and zero initial velocity.

Let ( )2 ,y x t be the solution for zero initial displacement and initial velocity g(x).

Then ( ) ( ) ( )1 2, , ,y x t y x t y x t= + satisfies the wave equation (the sum of any two solutions of a linear homogeneous PDE is also a solution), and satisfies the boundary conditions y(0, t) = y(L, t) = 0 :

( ) ( ) ( ) ( )1 2,0 ,0 ,0 0y x y x y x f x= + = + , which satisfies the condition for initial displacement f (x).

( ) ( ) ( ) ( )1 2,0 ,0 ,0 0t tty x y x y x g x= + = + , which satisfies the condition for initial velocity g(x). Therefore the sum of the two solutions is the complete solution for initial displacement f (x) and initial velocity g(x):

( ) ( )

( )

0

0

1

1

2, sin sin cos

2 1 sin sin sin

L

n

L

n

n u n x n c ty x t f u duL L L L

n u n x n c tg u duc n L L L

π π π

π π ππ

=

=

∞

∞

=

+

∫

∫

∑

∑

ENGI 9420 8.02 - Finite String Page 8.12

Example 8.02.4 An elastic string of length 1 m is fixed at both ends (x = 0 and x = 1). The string is initially in the shape of an arc of a parabola [y(x, 0) = x – x2 for 0 < x < 1] and is released

with the initial velocity ( )

( )2

,00 1

x

y x x xt

∂= − ≤ ≤

∂. It is known that the wave speed

is c = 5 m s–1. Find the displacement y(x, t) at all locations on the string (0 < x < 1) and at all subsequent times (t > 0). In the formula for the complete solution of the wave equation,

( ) ( )

( )

0

0

1

1

2, sin sin cos

2 1 sin sin sin

L

n

L

n

n u n x n c ty x t f u duL L L L

n u n x n c tg u duc n L L L

π π π

π π ππ

=

=

∞

∞

=

+

∫

∫

∑

∑

we know that L = 1, c = 5 and f (x) = g(x) = x – x2 for 0 < x < 1. Both integrals inside the summations are the same:

( )2

0

1sin

1n uu u duπ − =

∫

( )( ) ( )3 2

1

0

1 2 1 2cos sinu u un u n u

n n nπ π

π π π

− − − +

( )( )

( )3 32 20 1 0 0 0n

n nπ π

= − − + − − +

( )( )( )

( )

( )( )3

3

0 even2 41 1 odd

nn

nn nπ π

= − − =

Let (odd n) = 2k – 1

The complete solution is

( )( )

( ) ( ) ( )( )33

1

sin 5 2 18 1, sin 2 1 cos5 2 15 2 12 1k

k ty x t k x k t

kkπ

π ππ π

=

∞ −

= − − + −− ∑

A Maple file for this solution is available at "www.engr.mun.ca/~ggeorge/9420/demos/ex8024.mws".

ENGI 9420 8.03 - Infinite String Page 8.13

8.03 The Wave Equation – Vibrating Infinite String Example 8.03.1 An elastic string of infinite length is displaced into the form y = f (x) and is released from rest. Find the displacement y(x, t) at all locations on the string x∈ and at all subsequent times (t > 0). The boundary value problem for the displacement function y(x, t) is:

2 2

22 2 for and 0y yc x t

t x∂ ∂

= −∞ < < ∞ >∂ ∂

Initial configuration of string: y(x, 0) = f (x) for x∈

String released from rest: ( ),0

0 forx

y xt

∂= ∈

∂

We no longer have the additional boundary conditions of fixed endpoints. However, it is reasonable to insist upon a bounded solution. Separation of Variables (or Fourier Method) Attempt a solution of the form y(x, t) = X(x) T(t) Again we find the linked pair of ordinary differential equations

2 2 20 and 0X X T c Tω ω′′ ′′+ = + = If ω = 0 then X(x) = ax + b. However, for a bounded solution, we require a = 0. For other values of ω, X(x) = a cos ωx + b sin ωx , which is bounded for all x and all ω. The ω = 0 case is a special case of this solution. We have a continuum of eigenvalues ω with corresponding eigenfunctions

( ) ( ) ( )cos sinX x a x b xω ω ωω ω= +

It then follows that ( ) cos sinT t c ct d ctω ω ωω ω= + Imposing the initial condition of zero velocity,

( )( ) ( ) ( )

,00 0 0

x

y X x T X x d c x dt ω ωω∂ ′= = = ∀ ∈ ⇒ =∂

ENGI 9420 8.03 - Infinite String Page 8.14

Example 8.03.1 (continued) Therefore we have, for any real ω, a solution of the wave equation and the initial velocity condition,

( ) ( ) ( ) ( ) ( )( ) ( ), cos sin cosy x t X x T t a x b x ctω ω ω ω ωω ω ω= = + [where cω has been absorbed into the other arbitrary constants aω and bω .] The superposition of solutions now leads to an integral, not a discrete sum.

( ) ( ) ( ) ( )( ) ( )0 0, , cos sin cosy x t y x t d a x b x ct dω ω ωω ω ω ω ω

∞ ∞= = +∫ ∫

Imposing the remaining condition,

( ) ( ) ( )( ) ( )0,0 cos siny x a x b x d f xω ωω ω ω

∞= + =∫

But this is just the Fourier integral representation of f (x) on (–∞, ∞). Therefore aω and bω are just the Fourier integral coefficients

( ) ( ) ( ) ( )1 1cos and sina f u u du b f u u duω ωω ωπ π

∞ ∞

−∞ −∞= =∫ ∫

The complete solution is

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

0

1, cos cos

1 sin sin cos

y x t f u u du x

f u u du x ct d

ω ωπ

ω ω ω ωπ

∞ ∞

−∞

∞

−∞

= +

∫

∫

∫

which, after re-iteration (interchanging the order of integration) is

( ) ( ) ( ) ( ) ( )( ) ( ) ( )0

1, cos cos sin sin cosy x t u x u x f u ct d duω ω ω ω ω ωπ

∞ ∞

−∞= +∫ ∫

( ) ( ) ( )( ) ( )0

1, cos cosy x t f u u x ct d duω ω ωπ

∞ ∞

−∞⇒ = −∫ ∫

( ) ( )( ) ( )( )( )0

1 cos cos2

f u u x ct u x ct d duω ω ωπ

∞ ∞

−∞= − + + − −∫ ∫

( ) ( )( ) ( )( )0

sin sin12

u x ct u x ctf u du

u x ct u x ctω

ωω ω

π

∞

−∞ =

= ∞ − + − −

= + − + − − ∫

ENGI 9420 8.04 - d’Alembert Solution Page 8.15

8.04 d’Alembert Solution One form of the solution to Example 8.03.1,

( ) ( ) ( )( ) ( )( )0

sin sin1,2

u x ct u x cty x t f u du

u x ct u x ctω

ωω ω

π

∞

−∞ =

= ∞ − + − −

= + − + − − ∫

suggests that, in general, one might seek solutions to the wave equation of the form

( ) ( ) ( ),2

f x ct f x cty x t

+ + −=

Let r = x + ct and s = x – ct , then ( ) ( ) ( ),2

f r f sy r s

+= and

( )( ) ( )( )( )1 0 1 0 12

y y r y s f r f sx r x s x

∂ ∂ ∂ ∂ ∂ ′ ′= + = + × + + ×∂ ∂ ∂ ∂ ∂

,

( ) ( )( )2

2

1 1 12

y y y r y s f r f sx x x r x x s x x

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ′′ ′′= = + = × + × ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ,

( )( ) ( )( ) ( )( )1 0 02

y y r y s f r c f s ct r t s t

∂ ∂ ∂ ∂ ∂ ′ ′= + = + × + + × −∂ ∂ ∂ ∂ ∂

,

( ) ( ) ( )( )2

2

12

y y r y s c f r c c f s ct r t t s t t

∂ ∂ ∂ ∂ ∂ ∂ ∂ ′′ ′′= + = × − × − ∂ ∂ ∂ ∂ ∂ ∂ ∂ ,

( ) ( )( ) ( ) ( )( )2 2

2 22 2 2 2

1 1 1 02 2

y y f r f s c f r c f sx c t c

∂ ∂ ′′ ′′ ′′ ′′⇒ − = + − + =∂ ∂

,

Therefore ( ) ( ) ( ),2

f x ct f x cty x t

+ + −= is a solution to the wave equation for all

twice differentiable functions f (u). This is part of the d’Alembert solution. This d’Alembert solution satisfies the initial displacement condition:

( ) ( ) ( ) ( )0 0,0

2f x f x

y x f x+ + −

= =

Also ( ) ( ) ( ) ( ) ( )0 0

, 02 2t t

c f x ct c f x ct c f x c f xy x t

t = =

′ ′ ′ ′+ − − −∂= = =

∂

The d’Alembert solution therefore satisfies both initial conditions.

ENGI 9420 8.04 - d’Alembert Solution Page 8.16

A more general d’Alembert solution to the wave equation for an infinitely long string is

( ) ( ) ( ) ( )1,2 2

x ct

x ct

f x ct f x cty x t g u du

c

+

−

+ + −= + ∫

This satisfies the wave equation

2 2

22 2 for and 0y yc x t

t x∂ ∂

= −∞ < < ∞ >∂ ∂

and Initial configuration of string: y(x, 0) = f (x) for x∈ and

Initial speed of string: ( )

( ),0

forx

y g x xt

∂= ∈

∂

for any twice differentiable functions f (x) and g(x). Physically, this represents two identical waves, moving with speed c in opposite directions along the string.

Proof that ( ) ( )1,2

x ct

x cty x t g u du

c

+

−= ∫ satisfies both initial conditions:

( ) ( ) ( ) ( )1 1, ,0 02 2

x ct x

x ct xy x t g u du y x g u du

c c

+

−= ⇒ = =∫ ∫

Using a Leibnitz differentiation of the integral:

( ) ( ) ( ) ( ) ( )12

x ct

x ct

y g x ct x ct g x ct x ct g u dut c t t t

+

−

∂ ∂ ∂ ∂= + ⋅ + − − ⋅ − + ∂ ∂ ∂ ∂ ∫

( ) ( )( ) ( ) ( )1 02 2

g x ct g x ctc g x ct c g x ct

c+ + −

= + + − + =

( ) ( ) ( )0

0 02t

g x g xy g xt =

+ + −∂⇒ = =

∂

ENGI 9420 8.04 - d’Alembert Solution Page 8.17

Example 8.04.1 An elastic string of infinite length is displaced into the form y = cos π x/2 on [–1, 1] only (and y = 0 elsewhere) and is released from rest. Find the displacement y(x, t) at all locations on the string x∈ and at all subsequent times (t > 0). For this solution to the wave equation we have initial conditions

( ) ( ) ( )

( )

cos 1 12,0

0 otherwise

x xy x f x

π − ≤ ≤ = =

and

( ) ( ),0 0y x g xt

∂= =

∂

The d’Alembert solution is

( ) ( ) ( ) ( ) ( ) ( )1, 02 2 2

x ct

x ct

f x ct f x ct f x ct f x cty x t g u du

c

+

−

+ + − + + −= + = +∫

where ( )( ) ( )

( )

cos 1 12

0 otherwise

x ctct x ct

f x ctπ +

− − ≤ ≤ − + =

and ( )( ) ( )

( )

cos 1 12

0 otherwise

x ctct x ct

f x ctπ −

− + ≤ ≤ + − =

We therefore obtain two waves, each of the form of a single half-period of a cosine function, moving apart from a superposed state at x = 0 at speed c in opposite directions. See the web page "www.engr.mun.ca/~ggeorge/9420/demos/ex8041.html" for an animation of this solution.

ENGI 9420 8.04 - d’Alembert Solution Page 8.18

Example 8.04.1 (continued) Some snapshots of the solution are shown here:

ENGI 9420 8.04 - d’Alembert Solution Page 8.19

A more general case of a d’Alembert solution arises for the homogeneous PDE with constant coefficients

2 2 2

2 2 0u u uA B Cx x y y

∂ ∂ ∂+ + =

∂ ∂ ∂ ∂

The characteristic (or auxiliary) equation for this PDE is

2 0A B Cλ λ+ + = This leads to the complementary function (which is also the general solution for this homogeneous PDE)

( ) ( ) ( )1 1 2 2, ,u x y f y x f y xλ λ= + + + where

1 2and2 2

B D B DA A

λ λ− − − += =

and D = B 2 – 4AC and 1 2,f f are arbitrary twice-differentiable functions of their arguments.

1λ and 2λ are the roots (or eigenvalues) of the characteristic equation. In the event of equal roots, the solution changes to

( ) ( ) ( ) ( )1 2, ,u x y f y x h x y f y xλ λ= + + + where h(x, y) is any non-trivial linear function of x and/or y (except y + λx). The wave equation is a special case with y = t, A = 1, B = 0, C = –1/c2 and λ = ± 1/c.

ENGI 9420 8.04 - d’Alembert Solution Page 8.20

Example 8.04.2

2 2 2

2 23 2 0u u ux x y y

∂ ∂ ∂− + =

∂ ∂ ∂ ∂

u(x, 0) = −x2 uy(x, 0) = 0

(a) Classify the partial differential equation. (b) Find the value of u at (x, y) = (0, 1). (a) Compare this PDE to the standard form

2 2 2

2 2 0u u uA B Cx x y y

∂ ∂ ∂+ + =

∂ ∂ ∂ ∂

A = 1 , B = –3 , C = 2 ⇒ D = 9 – 4×2 = 1 > 0 Therefore the PDE is hyperbolic everywhere.

(b) 3 1 1 or 22

λ + ±= =

⇒ u(x, y) = f (y + x) + g(y + 2x) ⇒ uy(x, y) = f '(y + x) + g'(y + 2x) Boundary conditions: u(x, 0) = f (x) + g(2x) = –x2 (1) and uy(x, 0) = f '(x) + g'(2x) = 0 (2)

( ) ( ) ( )2 2 21 f x g x xx∂ ′ ′= + = −∂

(3)

(3) – (2) ⇒ g'(2x) = –2x ⇒ g'(x) = –x

( ) ( ) ( )221 12 22 2g x x k g y x y x k⇒ = − + ⇒ + = − + +

ENGI 9420 8.04 - d’Alembert Solution Page 8.21

Example 8.04.2 (continued)

Also (1) ⇒ f (x) = –x2 – g(2x) = –x2 + ½(2x)2 – k = x2 – k ⇒ f (y + x) = (y + x)2 – k Therefore u(x, y) = f (y + x) + g(y + 2x) = (y + x)2 – k – (y + 2x)2 / 2 + k

( ) ( )2 2 2 2 2 21 12 4 2 4 4 22 2

y xy x y xy x y x= + + − − − = −

The complete solution is therefore ( ) ( )2 21, 22

u x y y x= −

( ) ( )2 210,1 1 02

u⇒ = − =12

[It is easy (though tedious) to confirm that ( ) ( )2 21, 22

u x y y x= − satisfies the partial

differential equation 2 2 2

2 23 2 0u u ux x y y

∂ ∂ ∂− + =

∂ ∂ ∂ ∂ together with both initial conditions

u(x, 0) = −x2 and uy(x, 0) = 0.] [Also note that the arbitrary constants of integration for f and g cancelled each other out. This cancellation happens generally for this d’Alembert method of solution.]

ENGI 9420 8.04 - d’Alembert Solution Page 8.22

Example 8.04.3 Find the complete solution to

2 2 2

2 26 5 14 ,u u ux x y y

∂ ∂ ∂− + =

∂ ∂ ∂ ∂

u(x, 0) = 2x + 1 , uy(x, 0) = 4 − 6x .

This PDE is non-homogeneous. For the particular solution, we require a function such that the combination of second partial derivatives resolves to the constant 14. It is reasonable to try a quadratic function of x and y as our particular solution. Try 2 2

Pu ax bxy cy= + +

P P2 and 2u uax by bx cyx y

∂ ∂⇒ = + = +

∂ ∂

2 2 2P P P2 22 , and 2u u ua b c

x x y y∂ ∂ ∂

⇒ = = =∂ ∂ ∂ ∂

2 2 2P P P2 26 5 12 5 2 14u u u a b c

x x y y∂ ∂ ∂

⇒ − + = − + =∂ ∂ ∂ ∂

We have one condition on three constants, two of which are therefore a free choice. Choose b = 0 and c = a, then 14a = 14 ⇒ c = a = 1 Therefore a particular solution is 2 2

Pu x y= + [But we could have chosen, for example, a = b = 0 and c = 7 instead 2

P 7u y→ = ] Complementary function: A = 6 , B = –5 , C = 1 ⇒ D = 25 – 4×6 = 1 > 0 Therefore the PDE is hyperbolic everywhere.

5 1 1 1or12 3 2

λ + ±= =

The complementary function is ( ) ( ) ( )C

1 13 2,u x y f y x g y x= + + +

and the general solution is ( ) ( ) ( ) 2 21 1

3 2,u x y f y x g y x x y= + + + + +

ENGI 9420 8.04 - d’Alembert Solution Page 8.23

Example 8.04.3 (continued) ( ) ( ) ( ) 2 21 1

3 2,u x y f y x g y x x y= + + + + +

( ) ( )1 13 2 2u f y x g y x y

y∂ ′ ′⇒ = + + + +∂

Imposing the two boundary conditions: ( ) ( ) ( ) 21 1

3 2,0 2 1u x f x g x x x= + + = + (1) and

( ) ( ) ( )1 13 2,0 0 4 6yu x f x g x x′ ′= + + = − (2)

( ) ( ) ( )1 13 2

1 1 2 23 2

f x g x xx∂ ′ ′= + + =∂

1 (3)

(2) – 2×(3) ⇒ ( )13

1 4 4 6 43

f x x x′ − = − −

( ) ( ) ( )1 13 36 18 18f x x x f x x′ ′⇒ = − = − ⇒ = −

( ) 29f x x k⇒ = − +

(1) ⇒ ( ) ( )2

2 21 12 32 1 2 1 9

9xg x x x f x x x k

= + − − = + − + −

( ) ( )1 12 24 1g x x k⇒ = + −

( ) 4 1g x x k⇒ = + − But ( ) ( ) ( ) 2 21 1

3 2,u x y f y x g y x x y= + + + + +

( ) ( ) ( )2 2 21 13 2, 9 4 1u x y y x k y x k x y⇒ = − + + + + + − + +

[again the arbitrary constants cancel - they can be omitted safely.] 2 2 2 29 6 4 2 1y xy x y x x y= − − − + + + + +

Therefore the complete solution is

( ) 2, 1 2 4 6 8u x y x y xy y= + + − −

ENGI 9420 8.04 - d’Alembert Solution Page 8.24

Example 8.04.3 - Alternative Treatment of the Particular Solution Find the complete solution to

2 2 2

2 26 5 14 ,u u ux x y y

∂ ∂ ∂− + =

∂ ∂ ∂ ∂

u(x, 0) = 2x + 1 , uy(x, 0) = 4 − 6x .

This PDE is non-homogeneous. For the particular solution, we require a function such that the combination of second partial derivatives resolves to the constant 14. It is reasonable to try a quadratic function of x and y as our particular solution. Try 2 2

Pu ax bx y c y= + +

P P2 and 2u uax by bx cyx y

∂ ∂⇒ = + = +

∂ ∂

2 2 2P P P2 22 , and 2u u ua b c

x x y y∂ ∂ ∂

⇒ = = =∂ ∂ ∂ ∂

2 2 2P P P2 26 5 12 5 2 14u u u a b c

x x y y∂ ∂ ∂

⇒ − + = − + =∂ ∂ ∂ ∂

We have one condition on three constants, two of which are therefore a free choice. Let us leave the free choice unresolved for now. Complementary function:

5 1 1 1or12 3 2

λ + ±= =

The complementary function is ( ) ( ) ( )C

1 13 2,u x y f y x g y x= + + +

and the general solution is ( ) ( ) ( ) 2 21 1

3 2,u x y f y x g y x ax bx y c y= + + + + + +

ENGI 9420 8.04 - d’Alembert Solution Page 8.25

Example 8.04.3 (continued) ( ) ( ) ( ) 2 21 1

3 2,u x y f y x g y x ax bx y c y= + + + + + +

( ) ( )1 13 2 2u f y x g y x bx c y

y∂ ′ ′⇒ = + + + + +∂

Imposing the two boundary conditions: ( ) ( ) ( ) 21 1

3 2,0 0 0 2 1u x f x g x ax x= + + + + = + (A) and

( ) ( ) ( )1 13 2,0 0 4 6yu x f x g x bx x′ ′= + + + = − (B)

( ) ( ) ( )1 13 2

1 1 2 23 2

d f x g x axdx

′ ′= + + =A (C)

(B) – 2×(C) ⇒ ( ) ( )13

1 4 4 6 43

f x b a x x′ + − = − −

( ) ( ) ( )( ) ( ) ( )1 13 33 4 6 9 4 6 9 4 6f x a b x a b x f x a b x′ ′⇒ = − − = − − ⇒ = − −

( ) ( ) 29 4 62

a bf x x k

− −⇒ = +

(A) ⇒ ( ) ( )2 21 12 3

92 1 2 1g x x ax f x x ax= + − − = + − −

( ) 24 62 9

a b x− −k

−

( ) ( ) ( ) ( )2 22 4 62 2 1 2 2 6 6 4 12

a a bg x x x k b a x x k+ − − ⇒ = + − − = + − + + −

But ( ) ( ) ( ) 2 21 1

3 2,u x y f y x g y x ax bx y c y= + + + + + +

( ) ( ) ( )213

9 4 6,

2a b

u x y y x k− −

⇒ = + +

( )( ) ( )2 2 21 12 22 6 6 4 1b a y x y x k ax bx y c y+ + − + + + + − + + +

[again the arbitrary constants cancel - they can be omitted safely.]

( ) ( ) ( )( )2 2 2 22 1 13 9 4

9 4 62 6 6

2a b

y x y x b a y x y x− −

= + + + + − + + 2 24 2 1y x ax bx y c y+ + + + + +

ENGI 9420 8.04 - d’Alembert Solution Page 8.26

Example 8.04.3 (continued)

( ) ( ) ( ) ( )( )29 4 6 4 6 6 23 4 6 2 6 6

2a b b a c

y a b b a b xy− − + + − +

= + − − + + − +

( ) ( ) 24 6 6 6 24 2 1

2a b b a a

x y x− − + + − +

+ + + +

2 212 5 2 30 6 0 4 2 12

a b c y xy x y x− + − = − + + + +

But 12 5 2 30 14 3012 5 2 14 82 2

a b ca b c − + − −− + = ⇒ = = −

Therefore the complete solution is

( ) 2, 1 2 4 6 8u x y x y xy y= + + − − and note how the values of the two free parameters have no effect whatsoever on the solution. In practice, assign values to the free parameters in the particular solution only after one of the two arbitrary functions from the complementary function has been determined - if possible, make that function zero. In the example above, when we found

( ) ( )9 4 6f x a b x′ = − − ,

choose 0a = and ( )6 0b f x′= − ⇒ = and, from 12 5 2 14a b c− + = , we also have 8c = − .

(A) ⇒ ( ) ( ) ( )21 12 32 1 2 1 0 0 4 1g x x ax f x x g x x= + − − = + − − ⇒ = +

The general solution then becomes ( ) ( ) ( ) 2 21 1

3 2,u x y f y x g y x ax bx y c y= + + + + + +

( ) ( ) 212, 0 4 1 0 6 8u x y y x x y y= + + + + + − ⇒

( ) 2, 1 2 4 6 8u x y x y xy y= + + − −

as before, but much faster!

ENGI 9420 8.04 - d’Alembert Solution Page 8.27

Example 8.04.4 Find the complete solution to

2 2 2

2 22 0 ,u u ux x y y

∂ ∂ ∂+ + =

∂ ∂ ∂ ∂

u = 0 on x = 0 , u = x2 on y = 1 .

A = 1 , B = 2 , C = 1 ⇒ D = 4 – 4×1 = 0 Therefore the PDE is parabolic everywhere.

2 0 1 or 12

λ − ±= = − −

The complementary function (and general solution) is ( ) ( ) ( ) ( ), ,u x y f y x h x y g y x= − + − where h(x, y) is any convenient non-trivial linear function of (x, y) except a multiple of (y – x). Choosing, arbitrarily, h(x, y) = x, ( ) ( ) ( ),u x y f y x x g y x= − + − Imposing the boundary conditions: u(0, y) = 0 ⇒ f (y) + 0 = 0 Therefore the function f is identically zero, for any argument including (y – x). We now have u(x, y) = x g(y – x) . u(x, 1) = x2 ⇒ x g(1 – x) = x2 ⇒ g(1 – x) = x ⇒ g(x) = 1 – x Therefore u(x, y) = x g(y – x) = x (1 – (y – x)) The complete solution is

( ) ( ), 1u x y x x y= − +

ENGI 9420 8.04 - d’Alembert Solution Page 8.28

Example 8.04.4 - Alternative Treatment of the Complementary Function Find the complete solution to

2 2 2

2 22 0 ,u u ux x y y

∂ ∂ ∂+ + =

∂ ∂ ∂ ∂

u = 0 on x = 0 , u = x2 on y = 1 .

A = 1 , B = 2 , C = 1 ⇒ D = 4 – 4×1 = 0

2 0 1 or 12

λ − ±= = − −

The complementary function (and general solution) is ( ) ( ) ( ) ( ), ,u x y f y x h x y g y x= − + − where h(x, y) is any convenient non-trivial linear function of (x, y) except a multiple of (y – x). The most general choice possible is ( ),h x y ax b y= + , with the restriction a bλ≠ . ( ) ( ) ( ) ( ),u x y f y x ax by g y x= − + + ⋅ −

( ) ( ) ( ) ( ) ( )0, 0u y f y by g y f y by g y⇒ = + ⋅ = ⇒ = − ⋅ (A)

[note: choosing ( )0 0b f y= → = , as happened on the previous page] and

( ) ( ) ( ) ( ) 2,1 1 1u x f x ax b g x x= − + + ⋅ − = Using (A), ( ) ( ) ( ) ( ) ( ) 2,1 1 1 1u x b x g x ax b g x x= − − ⋅ − + + ⋅ − =

b⇒ − bx ax b+ + +( ) ( ) ( ) ( )2 21 1g x x a b x g x x⋅ − = ⇒ + ⋅ − =

( )1 xg xa b

⇒ − =+

(note that the restriction on h(x, y) ensures that 0a b+ ≠ ).

( ) 1 yg ya b−

⇒ =+

(B)

(A) ( ) 1 yf y bya b−

⇒ = − ⋅+

The complete solution becomes

( ) ( ) ( ) ( ) ( )1 1,

y x y xu x y b y x ax by

a b a b− − − −

= − − ⋅ + ++ +

by= − bx ax by+ + +( ) ( ) ( )1 y xa b

a b− −

= ++

1x yxa b− ++

⇒

( ) ( ), 1u x y x x y= − +

ENGI 9420 8.04 - d’Alembert Solution Page 8.29

Example 8.04.4 Extension (continued) It is easy to confirm that this solution is correct: ( ) ( ), 1u x y x x y= − + ⇒

( ) ( )0, 0 0 1 0u y y y= − + = ∀ and ( ) ( ) 2,1 1 1u x x x x x= − + = ∀

and

( ) 2, 2 andu uu x y x xy x x y xx y

∂ ∂= − + ⇒ = − = −

∂ ∂

2 2 2

2 22 , 1 and 0u u ux x y y

∂ ∂ ∂⇒ = = − =

∂ ∂ ∂ ∂

( )2 2 2

2 22 2 2 0 0 ,u u u x yx x y y

∂ ∂ ∂⇒ + + = − + = ∀

∂ ∂ ∂ ∂

ENGI 9420 8.04 - d’Alembert Solution Page 8.30

Two-dimensional Laplace Equation

2 2

2 2 0u ux y

∂ ∂+ =

∂ ∂

A = C = 1 , B = 0 ⇒ D = 0 – 4 < 0 This PDE is elliptic everywhere.

0 42

jλ ± −= = ±

The general solution is

( ) ( ) ( ),u x y f y j x g y jx= − + + where f and g are any twice-differentiable functions. A function f (x, y) is harmonic if and only if ∇2f = 0 everywhere inside a domain Ω. Example 8.04.5 Is u = ex sin y harmonic on 2

?

sin and cosx xu ue y e yx y

∂ ∂= =

∂ ∂

2 2

2 2sin and sinx xu ue y e yx y

∂ ∂⇒ = = −

∂ ∂

( )2 2

22 2 sin sin 0 ,x xu uu e y e y x y

x y∂ ∂

⇒ ∇ = + = − = ∀∂ ∂

Therefore yes, u = ex sin y is harmonic on 2 .

ENGI 9420 8.04 - d’Alembert Solution Page 8.31

Example 8.04.6 Find the complete solution ( ),u x y to the partial differential equation 2 0u∇ = , given the additional information

( ) 30,u y y= and 0

0x

ux =

∂=

∂

The PDE is

2 22

2 2 0u uux y

∂ ∂∇ = + =

∂ ∂

(which means that the solution ( ),u x y is an harmonic function). 21 , 0 4 4 0A C B D B AC⇒ = = = ⇒ = − = − <

The PDE is elliptic everywhere. A.E.: 2 1 0 jλ λ+ = ⇒ = ± C.F.: ( ) ( ) ( )C ,u x y f y j x g y jx= − + + The PDE is homogeneous ⇒ P.S.: ( )P , 0u x y =

G.S.: ( ) ( ) ( ),u x y f y j x g y jx= − + +

( ) ( ) ( ),xu x y j f y j x j g y jx′ ′⇒ = − − + + Using the additional information, ( ) ( ) ( ) ( ) ( ) ( ) ( )3 3 20, 3u y f y g y y g y y f y g y y f y′ ′= + = ⇒ = − ⇒ = −

and ( ) ( ) ( ) ( ) ( )( )20, 0 3xu y j f y j g y j f y y f y′ ′ ′ ′= = − + = − + −

( ) ( ) ( )2 3 3122 3 2f y y f y y f y y′⇒ = ⇒ = ⇒ =

( ) 3 3 31 12 2g y y y y⇒ = − =

Therefore the complete solution is ( ) ( ) ( )3 31 1

2 2,u x y y jx y jx= − + +

( )3 212 3y jx y= − ( ) ( )2 33 j x y j x+ − ( )3 23y jx y+ + ( ) ( )2 33 j x y j x+ +( )

3 2 23y j x y= + ⇒

( ) 3 2, 3u x y y x y= − Note that the solution is completely real, even though the eigenvalues are not real.

ENGI 9420 8.05 - Max-Min Principle Page 8.32

8.05 The Maximum-Minimum Principle Let Ω be some finite domain on which a function u(x, y) and its second derivatives are defined. Let Ω be the union of the domain with its boundary. Let m and M be the minimum and maximum values respectively of u on the boundary of the domain. If ∇2u ≥ 0 in Ω, then u is subharmonic and ( ) ( )or inu M u M< ≡ ∀ Ωr r r If ∇2u ≤ 0 in Ω, then u is superharmonic and ( ) ( )or inu m u m> ≡ ∀ Ωr r r If ∇2u = 0 in Ω, then u is harmonic (both subharmonic and superharmonic) and u is either constant on Ω or m < u < M everywhere on Ω. Example 8.05.1 ∇2u = 0 in Ω : x2 + y2 < 1 and u(x, y) = 1 on C : x2 + y2 = 1. Find u(x, y) on Ω.

u is harmonic on Ω ⇒ ( ) ( ) ( ),min max

onC Cu x y

u u

≤ ≤ Ω

But minC(u) = maxC(u) = 1 Therefore u(x, y) = 1 everywhere in Ω.

ENGI 9420 8.05 - Max-Min Principle Page 8.33

Example 8.05.2 ∇2u = 0 in the square domain Ω : –2 < x < +2, –2 < y < +2 . On the boundary C, on the left and right edges (x = ±2), u(x, y) = 4 – y2, while on the top and bottom edges (y = ±2), u(x, y) = x2 – 4. Find bounds on the value of u(x, y) inside the domain Ω. For –2 < y < +2, 0 < 4 – y2 < 4. For –2 < x < +2, –4 < x2 – 4 < 0. Therefore, on the boundary C of the domain Ω, –4 < u(x, y) < +4 so that m = –4 and M = +4. u(x, y) is harmonic (because ∇2u = 0). Therefore, everywhere in Ω,

–4 < u(x, y) < +4 Note: u(x, y) = x2 – y2 is consistent with the boundary condition and

2 2 2

2 2 22 0 , 0 2 2 , 2u u u u ux yx y x y x

∂ ∂ ∂ ∂ ∂= − = − ⇒ = = − = −

∂ ∂ ∂ ∂ ∂

2 22

2 2 0u uux y

∂ ∂⇒ ∇ = + =

∂ ∂

Contours of constant values of u are hyperbolas. A contour map illustrates that –4 < u(x, y) < +4 within the domain is indeed true.

ENGI 9420 8.06 - Heat Equation Page 8.34

8.06 The Heat Equation For a material of constant density ρ, constant specific heat μ and constant thermal conductivity K, the partial differential equation governing the temperature u at any location (x, y, z) and any time t is

2 , whereu Kk u kt µρ

∂= ∇ =

∂

Example 8.06.1 Heat is conducted along a thin homogeneous bar extending from x = 0 to x = L. There is no heat loss from the sides of the bar. The two ends of the bar are maintained at temperatures T1 (at x = 0) and T2 (at x = L). The initial temperature throughout the bar at the cross-section x is f (x). Find the temperature at any point in the bar at any subsequent time. The partial differential equation governing the temperature u(x, t) in the bar is

2

2

u ukt x

∂ ∂=

∂ ∂

together with the boundary conditions ( ) 10,u t T= and ( ) 2,u L t T= and the initial condition u(x, 0) = f (x) [Note that if an end of the bar is insulated, instead of being maintained at a constant

temperature, then the boundary condition changes to ( )0, 0u tx

∂=

∂ or ( ), 0u L t

x∂

=∂

.]

Attempt a solution by the method of separation of variables. u(x, t) = X(x) T(t)

T XX T k X T k cT X′ ′′

′ ′′⇒ = ⇒ = =

Again, when a function of t only equals a function of x only, both functions must equal the same absolute constant. Unfortunately, the two boundary conditions cannot both be satisfied unless 1 2 0T T= = . Therefore we need to treat this more general case as a perturbation of the simpler ( 1 2 0T T= = ) case.

ENGI 9420 8.06 - Heat Equation Page 8.35

Example 8.06.1 (continued) Let u(x, t) = v(x, t) + g(x) Substitute this into the PDE:

( ) ( )( ) ( ) ( )( ) ( )2 2

2 2, , v vv x t g x k v x t g x k g xt x t x

∂ ∂ ∂ ∂ ′′+ = + ⇒ = + ∂ ∂ ∂ ∂

This is the standard heat PDE for v if we choose g such that g"(x) = 0. g(x) must therefore be a linear function of x. We want the perturbation function g(x) to be such that ( ) 10,u t T= and ( ) 2,u L t T= and v(0, t) = v(L, t) = 0 Therefore g(x) must be the linear function for which ( ) 10g T= and ( ) 2g L T= . It follows that

( ) 2 11

T Tg x x TL− = +

and we now have the simpler problem

2

2

v vkt x

∂ ∂=

∂ ∂

together with the boundary conditions v(0, t) = v(L, t) = 0 and the initial condition v(x, 0) = f (x) – g(x) Now try separation of variables on v(x, t) : v(x, t) = X(x) T(t)

1 T XX T k X T ck T X

′ ′′′ ′′⇒ = ⇒ = =

But v(0, t) = v(L, t) = 0 ⇒ X(0) = X(L) = 0 This requires c to be a negative constant, say –λ2. The solution is very similar to that for the wave equation on a finite string with fixed ends

(section 8.02). The eigenvalues are nLπλ = and the corresponding eigenfunctions are

any non-zero constant multiples of

( ) sinnn xX x

Lπ =

ENGI 9420 8.06 - Heat Equation Page 8.36

Example 8.06.1 (continued) The ODE for T(t) becomes

2

0nT k TLπ ′ + =

whose general solution is

( ) 2 2 2/n kt Ln nT t c e π−=

Therefore

( ) ( ) ( )2 2

2, sin expn n n nn x n ktv x t X x T t c

L Lπ π = = −

If the initial temperature distribution f (x) – g(x) is a simple multiple of sin n xLπ

for

some integer n, then the solution for v is just ( )2 2

2, sin expnn x n ktv x t c

L Lπ π = −

.

Otherwise, we must attempt a superposition of solutions.

( )2 2

21

, sin expn

nn x n ktv x t c

L Lπ π

=

∞ = −

∑

such that ( ) ( ) ( )1

,0 sinn

nn xv x c f x g x

Lπ

=

∞ = = −

∑ .

The Fourier sine series coefficients are ( ) ( )( )0

2 sinL

nn zc f z g z dz

L Lπ = −

∫

so that the complete solution for v(x, t) is

( ) ( )2 2

2 11 2

01

2, sin sin expL

n

T T n z n x n ktv x t f z z T dzL L L L L

π π π=

∞ − = − − − ∑ ∫

and the complete solution for u(x, t) is

( ) ( ) 2 11, , T Tu x t v x t x T

L− = + +

Note how this solution can be partitioned into a transient part v(x, t) (which decays to zero as t increases) and a steady-state part g(x) which is the limiting value that the temperature distribution approaches.

ENGI 9420 8.06 - Heat Equation Page 8.37

Example 8.06.1 (continued) As a specific example, let k = 9, T1 = 100, T2 = 200, L = 2 and f (x) = 145x2 – 240x + 100 , (for which f (0) = 100, f (2) = 200 and f (x) > 0 ∀x).

Then ( ) 200 100 100 50 1002

g x x x−= + = +

The Fourier sine series coefficients are

( ) ( )( )22

0145 240 100 50 100 sin

2nn zc z z z dzπ = − + − +

∫

( )22

0145 2 sin

2nn zc z z dzπ ⇒ = −

∫

After an integration by parts (details omitted here),

( )( )

( )( )

23 2

2

0

8 12 16145 2 cos sin2 2

z

zn

zn z n zc z zn n n

π ππ π π

=

=

− ⇒ = − + + +

( )( )( )3

2320 1 1nnc

nπ⇒ = − −

The complete solution is

( ) ( ) 2 2

3 31

1 12320 9, 50 100 sin exp2 4n

nn x n tu x t x

nπ π

π =

∞ − − = + − − ∑

Some snapshots of the temperature distribution (from the tenth partial sum) from the Maple file at "www.engr.mun.ca/~ggeorge/9420/demos/ex8061.mws" are shown on the next page.

ENGI 9420 8.06 - Heat Equation Page 8.38

Example 8.06.1 (continued)

The steady state distribution is nearly attained in much less than a second!

END OF CHAPTER 8 END OF ENGI. 9420!

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