5. Boundary Value Problems (using separation of variables).

Seven steps of the approach of separation of Variables:

1) Separate the variables:

(by writing e.g. u(x, t) = X(x)T (t) etc..

2) Find the ODE for each “variable”.

3) Determine homogenous boundary values to stet up a Sturm- Liouville

problem.

4) Find the eigenvalues and eigenfunctions.

5) Solve the ODE for the other variables for all different eigenvalues.

6) Superpose the obtained solutions

7) Determine the constants to satisfy the boundary condition.

Second example: Initial boundary value problem for the wave equation with

periodic boundary conditions on D = (−π, π)× (0,∞)

(IBVP)

utt − c2 uxx = 0 (Hyperb. PDE)u(x, 0) = f(x) IC for u ,ut(x, 0) = g(x) IC for u′ ,u(−π, t) = u(π, t) BC for uux(−π, t) = ux(π, t) BC for ux

Step 1.

u(x, t) = X(x)T (t)

uxx(x, t) = X ′′(x)T (t)

and

utt(x, t) = X(x)T ′′(t)

1

Step 2.

Substituting that into the equation and and dividing the result by

c2X(x)T (t)

yields the ODEs

X ′′ + λX = 0 and T ′′ + λc2T = 0 .

Step 3.

We obtain the B.C:

X(−π) = X(π) , and X ′(−π) = X ′(π)

Step 4.

The eigenvalues and eigenfunctions of

X ′′ + λX = 0 ,X(−π) = X(π) ,X ′(−π) = X ′(π)

are

λ0 = 0 with eigenfunction φ(x) ≡ 1

and

λn = n2 with eigenfunctions φn(x) = cosnx and ψn(x) = sinnx .

Step 5.

1 ) For λ0 = 0 , the ODE for T0 reads:

T ′′

0 = 0;

with general solution T0 = A0 +B0t .

2 ) For λn = n2 , the ODE for Tn reads:

2

T ′′

n + n2c2Tn = 0;

with general solution τn(t) = An cosnct+Bn sinnct .

Step 6.

We get the following solution of the PDE satisfying the boundary condi-

tions:

u0(x, t) = (A0 + B0t)C and

un(x, t) = (An cosnct+ Bn sinnct)(Cn cosnx+ Dn sinnx)

“Streamlining” the constants, we write these as

u0(x, t) = (A0 +B0t) and

un(x, t) =

(An cosnct+Bn sinnct) cosnx+(Cn cosnct+Dn sinnct) sinnx .

Superposing these functions we get

u(x, t) = (A0 +B0t)+∞∑

n=1

[(An cosnct+ Cn sinnct) cosnx+ (Bn cosnct+Dn sinnct) sinnx]

which is a solution of the PDE satisfying the homogeneous boundary con-

dition, provided the coefficients allow for an appropriate convergence of the

sequence.

Step 7.

Determine the coefficients so that u(x, t) satisfies the initial conditions.

1 ) First initial condition gives:

f(x) = A0 +∞∑

n=1

[An cosnx+Bn sinnx]

3

If we choose these constants to be the coefficients of the Fourier Series

of f , then the first IC is satisfied:

A0 =1

2π

∫ π

−π

fdx , An =1

π

∫ π

−π

f cosnxdx and

Bn =1

π

∫ π

−π

f sinnxdx

2 ) Second initial condition gives:

g(x) = ut(x, 0) = B0+

∞∑

n=1

[(An − nc sinnct+ Cnnc cosnct) cosnx+

(Bn − nc sinnct+Dnnc cosnct) sinnx]t=0

= B0 +∞∑

n=1

[ncCn cosnx+ ncDn sinnx]

If we choose the constants B0 , ncCn and ncDn , to be the coeffi-

cients of the Fourier Series of g then the second IC is satisfied: so

B0 =1

2π

∫ π

−π

gdx , Cn =1

ncπ

∫ π

−π

g cosnxdx and

Dn =1

ncπ

∫ π

−π

g sinnxdx

With the above choice of constants u(x, t) is the solution of the problem.

Example 3.

Now we deal with a mixed boundary value problem for the Laplace equa-

tion.:

4

(IBVP)

uxx + uyy = 0 Laplace EQ. in (0, l)× (0,m)uy(x, 0) = 0 NC at y = 0 ,u(x,m) = 0 DC at y = m,

u(0, y) = g(y) nonhomogeneous DC at x = 0u(l, y) = 0 DC at x = l

Step 1.

u(x, y) = X(x)Y (y)

uxx = X ′′Y

and

uyy = XY ′′

Step 2.

Substituting that into the equation and and dividing the result by XY

yields the ODEs

X ′′ ± λX = 0 and Y ′′ ∓ λY = 0 .

Step 3.

We obtain the B.C:

Y ′(0) = Y (m) = 0 , and X(l) = 0 , X(0)Y (y) = g(y) .

So we have Y as the unkown in the Sturm- Liouville problem.

Step 4.

We have to find the eigenvalues and eigenfunctions of{

Y ′′ + λY = 0 ,Y ′(0) = Y (m) = 0

i ) λ < 0

No negative eigen values since B = 0 .

5

ii ) λ = 0

The general solution is φ(y) = A+By , and we get from the first BC

0 = φ′(0) = B and then from the second

0 = φ(m) = A

hence zero is not an eigenvalue.

iii ) λ > 0 The general solution is φ(y) = A cos νy + B sin νy . with

ν =√λ and we get from the first B.C

0 = ν(A(− sin νy) +B cos νy)y=0

= νB, ⇒ B = 0 ,

and then from the second BC

0 = A cos νm ,

which allows for eigenvalues if cos νm = 0 or νm = (n− 1

2)π , that is

λn = ((2n− 1)π

2m)2 , n = 1, 2, . . . ;

with eigenfunction

φn(y) = cos((2n− 1)πy

2m) .

Step 5.

With λn = ((2n− 1)π

2m)2 = : ν2n , the ODE

X ′′

n − λXn = 0;

has the general solution ψn(x) = An cosh νnx+Bn sinh νnx .

Step 6.

So the solutions

un(x, t) = (An cosh νnx+Bn sinh νnx)(cos νny)

6

of the PDE also satisfy the boundery conditions uy(x, 0) = u(x,m) = 0 .

Superposing these functions we get

u(x, y) =

∞∑

n=1

(An cosh νnx+Bn sinh νnx)(cos νny)

which is a solution of the PDE satisfying the homogeneous boundary con-

dition for y , provided the coeffcients allow for an appropriate convergence

of the sequence.

Step 7.

Determine the coefficients so that the PDE satisfies the other boundary

conditions:

In order to deal firstly with the homogeneous boundary condition we write

u(x, y) =∞∑

n=1

(An cosh νn(x− l) +Bn sinh νn(x− l)(cos νy) .

( Recall: Solutions of ODEs with constant coefficients are translation in-

variant.) We get from the second B.C:

0 = u(l, y) =∞∑

n=1

An

Hence An = 0, n = 1, 2, . . . , and the first BC requires

g(y) =∞∑

n=1

Bn sinh νn(−l)](cos(νny)

So if we choose the constants Bn sinh νn(−l) to be the coefficients of the

generalized Fourier Series of g , then the BC for x = 0 is satisfied.

Consequently:

Bn =2

m sinh νn(−l)

∫ m

0

g(y) cos(2n− 1)πy

2mdy .

With the above choice of constants Bn the solution of the above boundary

7

value problem for the Laplace equation is:

u(x, y) =

∞∑

n=1

sinh((2n− 1)π

2m(x− l)) cos(

(2n− 1)π

2my) .

Remark:

If the boundary conditions are inhomogeneous at more than one side of the

rectangle (0, l) × (0,m) then we separate the problem into problems with

inhomogeneous BC given at one side only, and we obtain the solution by

superposing the solution of the separated problems.

Inomogeneous problems and higher dimensional problems

As a first example let us consider

ut − k uxx = f(x, t) (inhom. PDE)u(x, 0) = 0 Initial condition for u ,u(0, t) = 0 BC at x = 0u(π, t) = 0 BC at x = π

Our first approach is to find solution of the kind

u(x, t) =∑

n=1

∞

An(t)ϕn(x) ,

Where (ϕn) is a sequence of eigenfunctions of Sturm- Liouville problem of

the associated homogeneous problem.

In our case that is the sequence of eigenfunctions of the Fourier Sine Series.

So here we have:

u(x, t) =∑

n=1

∞

An(t) sinnx .

Assuming that we can interchange the differentiation with the infinite sum,

we calculate

8

ut(x, t) =∑

n=1

∞

A′

n(t) sinnx .

uxx(x, t) =∑

n=1

∞

An(t)(−n2 sinnx) .

With those derivatives the PDE reads

∑

n=1

∞

A′

n(t) sinnx+ k∑

n=1

∞

An(t)n2 sinnx = f(x, t)

Now we multiply the inhomogeneous PDE with an eigenfunction ϕk =

sin(kx) , say, and integrate over the domain (0, π) . Because of the orthog-

onality of the eigenfunction we get

A′

n(t)

∫ π

0

sin2 nxdx+ kn2 An(t)

∫ π

0

sin2 nxdx

=

∫ b

a

f(x, t) sinnx .

Dividing by

∫ π

0

sin2 nxdx we get on the right hand side a function in t

say,

ψn(t) = (

∫ π

0

sin2 nxdx)−1

∫ π

0

f(x, t) sinnxdx .

and we get the first order ODE for An(t) as

A′

n(t) + kn2An(t) = ψn(t) . (*)

The initial condition

u(x, 0) = 0 ,

is satisfies if we add the initial condition

An(0) = 0

to the ODE. This is a first order ODE for which we can write down the

9

solution in the form

An(t) = e−n2kt

∫ t

0

ψn(τ)en2kτdτ ,

which in turn we can write as

An(t) =

∫ t

0

ψn(τ)e−n2k(t−τ)dτ .

Example: f(x, t) = t , so

ψn(t) =2

π

∫ π

0

t sinnxdx =

{

0 if n , is even4t

nπif n , is odd

.

In that case a more direct approach to solve the ODE’s is possible. We have

that An(t) = Cne−kn2t and is a solution of the homogeneous equation for

(*) . That is the case if n is even.

If n is odd, we use as trial solutions for a particular solution of (*)

(An)p(t) = αt+ β , with (An)′

p(t) = α

From the Ode we get

α+ kn2(αt+ β) =4t

nπ

Comparing coefficients we get

kn2αt =4t

nπ, and α+ kn2β = 0

or

α =4

πkn3, and β = − 4

πk2n5.

Finally the initial condition An(0) = 0 , gives

Cn = 0 if n even and Cn = −βn =4

πk2n5, if n is odd.

Altogether, with 2m− 1 = n we have

10

u(x, t) =∑

m=1

∞

An(t) sin(2m− 1)x .

with

A(t) = (4

πk2(2m− 1)5e−k(2m−1)2t +

4

πk(2m− 1)3t− 4

πk2(2m− 1)5) .

Inhomogeneous boundary conditions

As an example let us deal here with the one-dimensionnal wave equation

(vibrating string equation) on (0, l)× (0,∞) :

utt − c2 uxx = 0 (inhom. PDE)u(x, 0) = 0 Initial cond. for u ,ut(x, 0) = 0 Initial cond. for ut ,(u(0, t) = g(t) BC at x = 0u(l, t) = h(t) BC at x = l

The most straight forward approach here is to write u(x, t) = w(x, t) +

v(x, t) where v(x, t) is a “simple function ” satisfying the boundary condi-

tions.

Then w(x, t) = u(x, t)− v(x, t) satisfies the inhomogeneous PDE with ho-

mogeneous B.C.:

Usually we set

v(x, t) = g(t)(1− x1

l) + h(t)(x

1

l) .

We calculate

vtt = g′′(t)(1− x1

l) + h′′(t)(x

1

l) ,

and

vxx = 0 .

So we need w to be a solution of

11

wtt − c2 wxx = −(g′′(t)(1− 1

lx) + h′′(t)(

1

lx);

w(x, 0) = −(g(0)(1− x1

l) + h(0)(x

1

l)),

wt(x, 0) = −(g′(0)(1− x1

l) + h′(0)(x

1

l)),

(w(0, t) = w(l, t) = 0 .

which we can solve using the discussed methods.

Elliptic equations on spherical domains.

(1)

{

uxx + uyy = 0 on√

x2 + y2 < R

(x, y) = f(x, y) for√

x2 + y2 = R

We introduce a change of variables setting

U(r, ϕ) = u(x, y)x=r cosϕy=r sinϕ

then uxx + uyy = 0 for√

x2 + y2 < R , imply that

(2)

{

Urr +1rUr +

1r2Uϕϕ = 0 for 0 < r < R , −π < ϕ < π ,

U(R,ϕ) = F (ϕ), with F (ϕ) = f(R cosϕ,R sinϕ)

Somehow the set of auxiliary conditions does not seem to be complete but

if we separate the variables first, this will give us a better picture of what

will be needed.

Step 1:We set

U(r, ϕ) = R(r) Φ(ϕ) , and get

Ur = R′Φ , Urr = R′′Φ and Uϕϕ = Rφ′′ .

The PDE of (2) implies

R′′Φ+1

rRΦ+

1

r2RΦ′′ = 0 , ;

Step 2: Multiplying withr2

RΦ. yields

12

r2R′′ + rR′

R +Φ′′

φ= 0 ,

which gives us the ODE’s

(3) r2R′′ + rR′ ∓ λR = 0, , , (2) Φ′′ ± λΦ = 0 .

Step 3. We would like to work with Φ as the ODE for a Sturm-Lionville

problem. In order to do this we observe that the “reasonable” requirement

that the solution is continuously differentiable implies

U(r,−π) = U(r, π) and Uϕ(r,−π) = Uϕ(r, π) .

Step 4 For Φ, we get the Sturm-Liouville problem

ST.L.

Φ′′ + λΦ = 0Φ(−π) = Φ(π)Φ′(−π) = Φ′(π)

with the eigen values λ0 = 0, λn = n2, n = 1, 2, . . . and the eigen

function 1, cos x, sin x, cos 2 x sin 2 x . . ..

Step 5 We have to consider the ODE’s

r2R′′ + rR′ = 0 (for λ = 0)

and

r2R+ rR′ − n2R = 0 (for λn = n2) .

For λ = 0 we get with w = R′

w′(π) + 1rw(r) = 0 .

The solution can be found by a “separation of variables approach”. We get

w′

w= −1

r

ln w = ln(−r) + C

w = C11

r·

13

or

ϕ0(r) = A0 +B0 ln(r)

The second ODE an Euler equation has the solutions

ϕn(r) = Anrn +Bnr

−n

Step 6

Now superposing the solution of (1) and (2) we get

U(r, φ) = A0 +∞∑

n=1

rn (An cos nϕ+Bn sin nϕ)

+C0 (ln(r)) +

∞∑

n=1

(1

r)n (Cn cos nϕ+Dn sin nϕ)

Step 7

Since ln(r) and 1rn

are unbounded the functions un(r, θ) which need to

superpose are not solutions of the PDE if those terms a part of un(r, θ) . So

we set the coefficients Cn , n = 0, 1, 2, . . . and Dn, n = 1, 2, . . . zero. Then

u(r, ϕ) = A0 +∞∑

n=1rn(An cos nϕ+Bn sin nϕ)

Is a solution of the PDE in the disk and we get we get from the initial

condition

U(R,ϕ) = F (ϕ) = A0 +∞∑

n=1RnAn cos nϕ+

∞∑

n=1RnBn sin nϕ

which are satisfied if A0 , RnAn , RnBn are the Fourier coefficients of F ,

i.e.:

A0 =1

2π

π∫

−π

F (ϕ)dϕ

14

An =1

πRn

+π∫

−π

F (ϕ) cos nϕdϕ

Bn =1

πRn

+π∫

−π

F (ϕ) sin nϕdϕ

Note, that we can not set the coefficients Cn , n = 0, 1, 2, . . . and Dn, n =

1, 2, . . . the be zero in case the domain is a “washer”, say

D = {(x, y) 0 < a2 < x2 + y2 < R2} .

In that case there will be an additional boundary condition for x2+y2 = a2 .

and we get from the BC’s two equation for each pair of coeffiecents An, Cn

and two equation for the pair of coeffiecents Bn, Dn , respectively, c.f. the

related Homework problem.

We want to consider the solution for the Disk somewhat closer:

We have

u(r, φ) =1

2π

π∫

−pi

F (θ)dθ

+∞∑

n=1

rn

πRn

+π∫

−π

F (θ)(cos nφ cos nθ + sinφ sin nθ)dθ

=1

2π

π∫

−pi

F (θ)(1 + 2

∞∑

n=1

(r

R)n cos n(φ− θ))dθ

=1

2π

π∫

−pi

F (θ)(1 +∞∑

n=1

(r

R)n(ein(φ−θ) + e−in(φ−θ)))dθ

15

=1

2π

π∫

−pi

F (θ)(1 + (rei(φ−θ)

R− rei(φ−θ)+

re−i(φ−θ)

R− re−i(φ−θ)))dθ

and so we get Poissons integral formula

u(r, φ) =1

2π

π∫

−π

F (θ)R2 − r2

R2 + r2 − 2rR cos(φ− θ)dθ

For r = 0 we get

u(0, φ) =1

2πR

π∫

−π

U(R, θ)Rdθ ,

That is the vaule of u at the center is the average of the values of u at the

boundary of the disk.

Remark:

Consequences: Mean Value Theorem and Maximum Principle for solutions

of the Laplace equations.

Higher dimensional problems

Cooling of the sphere:

Under the assumption of a rotational symmetric initial temperature distri-

bution and an constant outer temperature distribution (normalized to zero

the inital boundary value problem for heat distribution in a sphere

ut − k∆u = 0 , t > 0, ‖ x ‖ < π ,

u(x, 0) = f(‖ x ‖) ,u(x, t) = 0 , if ‖ x ‖ = π ,

becomes

16

ut − k(uρρ +2

ρuρ) = 0 , t > 0 , ρ < π ,

u(ρ, 0) = f(ρ) ,u(π, t) = 0 .

(Assuming that the heat distribution is only a matter of the distance to the

center the derivatives with respect to φ are vanishing.)

Separating the variables u(ρ, t) = RT we get

T ′R+ k(R′′T +2

ρ(R′T )) = 0 , and so

T ′

kT= −

R′′ + 2ρ(R′)

R = λ

Since we have now choice for which variable we might use for a Sturm -

Liouville problem, we need to consider

R′′ +2

ρR′ + λR = 0, 0 < ρ < π

But this is not in the form

(pX ′)′ − qX + λmX = 0 .

Further more the boundary conditions 0 = u(π, t) = R(π)T (t) , provides

one homogeneous BC

R(π) = 0

but not a second one.

so we are not dealing with a regular Sturm - Liouville problem, here.

Introducing the functions X(ρ) = ρR(ρ) we get for X the ODE

X ′′ + λX = 0

and so for ν2 = λ > 0 we have the solutions

17

R(ρ) =1

ρC1 cos(νρ) +

1

ρC1 sin(νρ) .

Since we can use only bounded solutions we would like to disregard the un-

bounded functions in this sum. So replacing the second boundary condition

by the condition that the solution need to be bounded might replace the

second boundary condition. Then multipying the ODE with ρ2 gives the

singular Sturm-Liouville problem

ρ2R′′ + 2ρR′ + ρ2λR = 0, 0 < x < π ,

R(π) = 0 ,R a bounded function in (0, π)

,

whose eigenfunctions have the properties discussed in the last chapter with

respect to the weigt function m(ρ) = ρ2 .

Is it an exercise using the above transformation to show that all eigen values

are λn = n2 , n = 1, 2, 3, . . . with eigenfunctions

φn(ρ) =sin(nρ)

ρ.

The solutions of the ODE for T are e−n2kt so that we have

un(ρ, t) = Ane−n2kt sin(nρ)

ρ.

are bounded solutions of the PDE with un(π, t) = 0 .

For u(ρ, t) =∑

n=1

∞

un(ρ, t) , we determine the coefficents from the remaing

auxilliary condition:

f(ρ) = u(ρ, 0) =∑

n=1

∞

An

ρsin(nρ)

With the generalized Fourier cofficient of the eigenfunctions φn(ρ) that is

An = (

∫ π

0

(sin(ρ)

ρ)2ρ2 dρ)−1

∫ π

0

f(ρ)sin ρ

ρρ2 dρ .

18

= (

∫ π

0

sin2(ρ) dρ

∫ π

0

f(ρ)ρ sin ρ dρ .

Note that that An are the Fourier Sine coefficients of ρf ,

which we would get if we work the entire problem with the transformed

functions.

Two dimensional Wave equation

Example

utt − c2∆ w = 0 , (x, y, t) ∈ (0, L)× (0,M)× R ,

u(x, y, 0) = f(x, y) ,ut(x, 0) = g(x, y) ,u(0, y, t) = u(L, y, t) = 0 ,uy(x, 0, t) = uy(0,M, t) = 0 ,

Separation of variables:

Substiuting u(x, y, t) = X(x)Y (y)T (t) into the PDE we get

X(x)Y (y)T ′′(t)− c2(X ′′(x)Y (y)T (t) +X(x)Y ′′(y)T (t)) = 0 ,

or

T ′′

c2T=X ′′

X+Y ′′

Y= −λ

this gives us first the ODE’s for T,X and Y :

T ′′ + c2λT = 0

X ′′ + µX = 0

Y ′′ + κY = 0 with λ = µ+ κ .

From the boundary condition we get the Sturm-Liuoville problems{

X ′′ + µX = 0X(0) = X(L) = 0 .

19

and{

Y ′′ + κY = 0Y ′(0) = Y ′(M) = 0 .

with are the Sturm-Liuoville problems for the Fourier Sine and Fourier

Cosine Series, respectively. So we have the eigen values µm = (mπ

L)2 with

eigen functions

φm = sin(√µmx , m = 1, 2, 3, 4, . . . ,

for the first Sturm-Liuoville problem.

For the second Sturm-Liuoville problem we have the eigenvalues κk = (kπ

M)2

with eigen functions

ψk = cos(√κk y), , k = 1, 2, 3, 4, . . . ,

and

ψ0 ≡ 1 .

For each combination λmk = µm + κk we get the solution

τmk = Amk cos(c√µm + κk t) +Bm,k sin(c

√µm + κk t)

of the ODE’s for T

T ′′ + c2λmkT = 0 .

(Step 6 ) So the functions

um,k(x, y, t) = Am,k cos(c√µm + κk t) sin(

õmx) cos(

√κky)

+Bm,k sin(c√µm + κk t) sin(

õmx) cos(

√κky)

for m, k = 1, 2, 3, . . . and

20

um,0 = Am,0 cos(cµmt) sin(√µmx)+Bm,0 sin(c

õmt) sin(

õmx)

for m = 1, 2, 3, . . . are solutions of the PDE together with the homogeneous

boundary condition. We superpose those function to get

u(x, y, t)

=∑

m=1

∞

[Am,0 cos(cõmt) sin(

õmx)+

Bm,0 sin(cõmt) sin(

õmx)]

+∑

m=1

∞∑

k=1

∞

[Am,k cos(c√µm + κk t) sin(

√µmµmx) cos(

√κky)

+Bm,k sin(c√µm + κk t) sin(

√µmµmx) cos(

√κky)] .

We determine the constants using the orthogonality relations of those eigen-

functions. We get from the first initial conditions:

f(x) = u(x, y, 0) =

=∑

m=1

∞

Am,0 sin(õmx)+

∑

m=1

∞∑

k=1

∞

Am,k sin(õmx) cos(

√κky)

so

Am,k =4

LM

∫ L

0

∫ M

0

f(x, y)) sin(õmx) cos(

√κky)dydx ,

and

Am,0 =2

LM

∫ L

0

∫ M

0

f(x, y) sin(õm)dydx .

The second initial condition gives

g(x) = ut(x, y, 0) =

21

=∑

m=1

∞

cõmBm,0 sin(

õmx)

+∑

m=1

∞∑

k=1

∞

c√µm + κkBm,k sin(

õmx) cos(

√κky) ,

hence

Bm,k =

4

c√µm + κkLM

∫ L

0

∫ M

0

g(x, y) sin(õmx) cos(

√κky)dydx ,

for m, k = 1, 2, 3, . . . and

Bm,0 =2

LMõm

∫ L

0

∫ M

0

g(x, y) sin(õmx)dydx ,

for m = 1, 2, 3, . . . .

22