HW 9 Solution

Due: November 29 (Thursday), 2012

1. Find the type of PDE, transform to normal form, and solve. Please show your work in detail.

uxx-16uyy=0.

Solution: Hyperbolic, wave equation. Characteristic equation

𝑦′2 − 16 = (𝑦′ + 4)(𝑦′ − 4) = 0

New variables are

v = ϕ = y + 4x, w = ψ = y − 4x

By the chain rule,

𝑢𝑥 = 4𝑢𝑣 − 4𝑢𝑤

𝑢𝑥𝑥 = 16𝑢𝑣𝑣 − 16𝑢𝑣𝑤 − 16𝑢𝑤𝑣 + 16𝑢𝑤𝑤

And

−16𝑢𝑦𝑦 = −16𝑢𝑣𝑣 − 16𝑢𝑣𝑤 − 16𝑢𝑤𝑣 − 16𝑢𝑤𝑤

Assuming 𝑢𝑣𝑤 = 𝑢𝑤𝑣, as usual , we have

𝑢𝑣𝑤 = 0

Solvable by two integrations,

u = 𝑓1(𝑦 + 4𝑥) + 𝑓2(𝑦 − 4𝑥)

2. Find the type of PDE, transform to normal form, and solve. Please show your work in detail.

uxx-6uxy+9uyy=0.

Solution: Parabolic. Characteristic equation

𝑦′2 + 6𝑦′ + 9 = (𝑦′ + 3)2

New variables v = ϕ = x, w = ψ = y + 3x. By the chain rule,

𝑢𝑥 = 𝑢𝑣 + 3𝑢𝑤

𝑢𝑥𝑥 = 𝑢𝑣𝑣 + 6𝑢𝑣𝑤 + 9𝑢𝑤𝑤

𝑢𝑥𝑦 = 𝑢𝑣𝑤 + 3𝑢𝑤𝑤

𝑢𝑦𝑦 = 𝑢𝑤𝑤

Substitution into the PDE gives the expected normal form

𝑢𝑣𝑣 + 6𝑢𝑣𝑤 + 9𝑢𝑤𝑤 − 6𝑢𝑣𝑤 − 18𝑢𝑤𝑤 + 9𝑢𝑤𝑤 = 𝑢𝑣𝑣 = 0

Solution

u = 𝑓1(𝑥) + 𝑓2(𝑦 + 3𝑥)

where f1 and f2 are any twice differentiable functions of the respective variables.

3. Solve the heat transfer equation

∂u

∂t= 𝑐2

∂2u

∂x2

0≤x≤L, x>0; L=1

Subject to the boundary conditions: u(0, t)=0, and ∂u

∂t|𝑥=𝐿 = −𝑘𝑢(1, 𝑡), k>0 and the

initial condition u(x, 0)=1, 0<x<1.

Solution:

Assuming u(x, t)=X(x) T(t) and –λ as the separation constant, we find the separated

ODEs and boundary conditions to be, respectively,

X′′ + λX = 0 (1)

T′ + 𝑐2𝜆𝑇 = 0 (2)

X(0) = 0 and X′(1) = −kX(1) (3)

Equation (1) along with the homogeneous boundary conditions (3) comprise the regular

Sturm-Liouville problem:

X′′ + λX = 0, X(0) = 0, X′(1) + kX(1) = 0 (4)

Except for the presence of the symbol k, the BVP in (4) is essentially the problem solved

in Example 2 of Section 12. 5. As in that example, (4) possesses nontrivial solutions only

in the case λ=α2>0, α>0. The general solution of the DE in (4) is X(x)=c1 cos αx+c2 sin αx.

The first boundary condition in (4) immediately gives c1=0. Applying the second

boundary condition in (4) to X(x)=c2 sinαx yields

α cos 𝛼 + 𝑘𝑠𝑖𝑛𝛼 = 0 𝑜𝑟 tan 𝛼 = −𝛼

𝑘 (5)

Because the graphs of y=tan x and y=-x/k, k>0, have an infinite number of points of

intersection for x>0, the last equation in (5) has an infinite number of roots. Of course,

these roots depend on the value of k. If the consecutive positive roots are denoted αn,

n=1, 2, 3, … , then the eigenvalues of the problem are λn=αn2, and the corresponding

eigenfunctions are X(x)=c2 sin αnx, n=1, 2, 3, … , The solution of the first-order DE (2) is

T(t) = 𝑐3𝑒−𝑐2𝛼𝑛2 𝑡 and so

𝑢𝑛 = 𝑋𝑇 = 𝐴𝑛𝑒−𝑐2𝛼𝑛2 𝑡 sin 𝛼𝑛𝑥 𝑎𝑛𝑑 𝑢(𝑥, 𝑡) = ∑ 𝐴𝑛𝑒−𝑐2𝛼𝑛

2 𝑡 sin 𝛼𝑛𝑥

∞

𝑛=1

Now at t=0, u(x, 0)=1, 0<x<1, so that,

1 = ∑ 𝐴𝑛 sin 𝛼𝑛𝑥∞𝑛=1 (6)

The series in (6) is not a Fourier sine series; rather, it is an expansion of u(x, 0)=1 in terms

of the orthogonal functions arising from the Sturm-Liouville problem (4). It follows that

the set of eigenfunctions {sin αnx}, n=1, 2, 3, … , where the α’s are defined by tan α=-α/k

is orthogonal with respect to the weight function p(x)=1 on the interval [0, 1]. With

f(x)=1 and φn(x)=sin αnx, it follows from (8) of Section 12.1, that the coefficients An are (6)

are

𝐴𝑛 =∫ sin 𝛼𝑛𝑥

1

0𝑑𝑥

∫ sin2 𝛼𝑛𝑥1

0𝑑𝑥

(7)

Evaluate the square norm of each of the eigenfunctions we use a trigonometric identity:

∫ sin2 𝛼𝑛𝑥1

0𝑑𝑥 =

1

2∫ (1 − cos 2𝛼𝑛𝑥

1

0)𝑑𝑥 =

1

2(1 −

1

2𝛼𝑛sin 2𝛼𝑛) (8)

With the aid of the double angle formula sin 2αn =2sin in the form αn cos αn and the first

equation in (5) in the form αncosαn=-k sinαn, we can simplify (8) to

∫ sin2 𝛼𝑛𝑥1

0

𝑑𝑥 =1

2𝑘(𝑘 + cos2 𝛼𝑛)

Also

∫ sin 𝛼𝑛𝑥1

0

𝑑𝑥 = −1

𝛼𝑛cos 𝛼𝑛𝑥 |0

1 =1

𝛼𝑛(1 − cos 𝛼𝑛)

Consequently (7) becomes

𝐴𝑛 =2𝑘(1 − cos 𝛼𝑛)

𝛼𝑛(𝑘 + cos2 𝛼𝑛)

Finally , a solution of the boundary-value problem is

𝑢(𝑥, 𝑡 ) = 2𝑘 ∑1 − cos 𝛼𝑛

𝛼𝑛(𝑘 + cos2 𝛼𝑛)𝑒−𝑐2𝛼𝑛

2 𝑡 sin 𝛼𝑛𝑥

∞

𝑛=1

4. For a steady-state heat flow problem in the rectangular sheet of size L*M to which Laplace’s

equation applies. Suppose that the upper horizontal edge is kept at 100oC, while the other

three edge are kept at 0 oC. Please solve Laplace’s equation to find the temperature

distribution function.

Solution: We may write these boundary conditions for T(x, y) as follows:

T(0, y) = T(L, y) = 0, 0 < y < M

T(x, 0) = 0, T(x, M) = 100, 0 < x < L

Since we are seeking the steady-state solution, time is not a factor in this problem, and

therefore no initial condition makes sense.

Setting T(x, y)= X(x)Y(y), we find that (6) becomes

X′′Y + XY′′ = 0

Which may be written in separated form as

X′′

X= −

Y′′

Y

The left-hand side is constant for fixed x and arbitrary y. Hence Y’’/Y must be constant.

Letting k be this constant, we obtain the pair of ordinary differential equations

X′′ − 𝑘𝑋 = 0

Y′′ + 𝑘𝑌 = 0

From the first boundary condition we have

X(0)Y(y) = X(L)Y(y) = 0

And Y(y)≠0 (since otherwise T(x, y)≡0, contradicting the second boundary condition), so that

X(0)=X(L)=0. Thus again we have the situation encountered in Section 12. 7, and the

eigenvalues are all negative. Setting k=-r2, we find that rL must be a multiple of π in order

that we have a nonzero solution of the boundary value problem for X. Hence r=nπ/L, and the

functions

𝑋𝑛(𝑥) = 𝐴 sin𝑛𝜋𝑥

𝐿, 𝑛 = 1, 2, 3, …,

are the eigenfunctions of the problem. Setting k=-(nπ/L)2 in Y’’+kY=0 yields the general

solution

𝑌𝑛(𝑦) = 𝐵1𝑒𝑛𝜋𝑦/𝐿 + 𝐵2𝑒−𝑛𝜋𝑦/𝐿

The second boundary condition implies that Y(0)=0, thus B1=-B2, and we can rewrite Yn as

𝑌𝑛(𝑦) = 𝐵 sinh𝑛𝜋𝑦

𝐿

As in Section 12.7, to enlarge the class of possible solutions, we consider and infinite sum of

products of Xn(x) and Yn(y):

T(𝑥, 𝑦) = ∑ 𝑐𝑛 sin𝑛𝜋𝑥

𝐿sinh

𝑛𝜋𝑦

𝐿

∞

𝑛=1

Evaluating at any point (x, M), we obtain by the second boundary condition,

100 = ∑(𝑐𝑛 sinh𝑛𝜋𝑀

𝐿) sin

𝑛𝜋𝑥

𝐿

∞

𝑛=1

, 0 < x < L

Which again is a Fourier sine series, requiring that we extend the boundary condition T(x, M)

to the interval -L≤x≤L as an odd pricewise continuous periodic function of period 2L with a

piecewise continuous derivative. The function

f(x) = {100, 0 < 𝑥 < 𝐿,

−100, −𝐿 < 𝑥 < 0

with f(x+2L+=f(x) satisfies these conditions. Hence cn sinh(nπM/L) must equal the nth Fourier

(sine) coefficient of f(x):

𝑐𝑛 sinh𝑛𝜋𝑀

𝐿=

1

𝐿∫ 𝑓(𝑥) sin

𝑛𝜋𝑥

𝐿

𝐿

−𝐿

𝑑𝑥 =200

𝐿∫ sin

𝑛𝜋𝑥

𝐿

𝐿

0

𝑑𝑥 =200

𝑛𝜋(1 − cos 𝑛𝜋)

=200

𝑛𝜋[1 − (−1)𝑛]

Hence c2k=0, while

𝑐2𝑘+1 =400

𝜋(2𝑘 + 1) sinh[(2𝑘 + 1)𝜋𝑀/𝐿]

And the solution is given by

T(𝑥, 𝑦) =400

𝜋∑

sin(2𝑛 − 1)𝜋𝑥

𝐿 sinh(2𝑛 − 1)𝜋𝑦

𝐿(2𝑛 − 1) sinh[(2𝑛 − 1)𝜋𝑀/𝐿]

∞

𝑛=1

5. A rectangular metal plate 0≤x≤π, 0≤y≤2, in which the steady state temperature

distribution u(x, y) is required subject to the temperature on the side y=0 being

u(x, 0)=10, and the temperature on the other three sides being maintained at u=0. Find

the temperature distribution across the metal plate.

𝑢𝑡 = 𝑐2(𝑢𝑥𝑥 + 𝑢𝑦𝑦) = 0

Solution:

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

So

X′′

X= −

Y′′

Y= 𝑘 = −𝜆2

𝑋′′ + 𝜆2𝑋 = 0, Y′′ − 𝜆2𝑌 = 0

General solution of X(x):

X(x) = 𝐴 cos 𝜆𝑥 + 𝐵 sin 𝜆𝑥

B. C.: X(0)=0, so A=0, X(x)=B sin λx

X(π)=0, since B≠0, λn=n, for n=1, 2, …,

So X𝑛(x) = 𝐵 sin 𝑛𝑥 for n=1, 2, …,

𝑌𝑛(𝑦) = 𝐶 cosh 𝑛𝑦 + 𝐷 sinh 𝑛𝑦

B. C.: u(x, 2)=0, y(2)=0

So 𝑌𝑛(𝑦) = 𝐶 cosh 2𝑛 + 𝐷 sinh 2𝑛 = 0

Setting D=1, and C=− sinh(2𝑛)/ cosh(2𝑛)

𝑌𝑛(𝑦) =𝐶

cosh 2𝑛(sinh 𝑛𝑦 cosh 2𝑛 − cosh 𝑛𝑦 sinh 2𝑦) =

𝐶

cosh 2𝑛sinh 𝑛(𝑦 − 2)

𝑢𝑛(𝑥, 𝑦) = 𝑋𝑛(𝑥)𝑌𝑛(𝑦) = 𝐶𝑛 sin 𝑛𝑥 sinh 𝑛(𝑦 − 2)

u(𝑥, 𝑦) = ∑ 𝐶𝑛 sin 𝑛𝑥 sinh 𝑛(𝑦 − 2)

∞

𝑛=1

B. C.: u(x, 0)=10, so y(0)=10

10 = ∑ 𝐶𝑛 sin 𝑛𝑥 sinh(−2𝑛)

∞

𝑛=1

𝐶𝑛 = −2

𝜋 sinh 2𝑛∫ 10 sin 𝑛𝑥

𝜋

0

𝑑𝑥 =20

𝑛𝜋 sinh 2𝑛(1 − cos 𝑛𝜋)