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### Transcript of Math 201 Assignment - novoseltsev/2011Winter201ES1/HW11.pdf · PDF file Math 201...

• Math 201 Assignment #11

Problem 1 (10.5 2) Find a formal solution to the given initial-boundary value problem.

∂u

∂t =

∂2u

∂x2 , 0 < x < π, t > 0

u(0, t) = u(π, t) = 0, t > 0

u(x, 0) = x2, 0 < x < π

Problem 2 (10.5 5) Find a formal solution to the given initial-boundary value problem.

∂u

∂t =

∂2u

∂x2 , 0 < x < π, t > 0

∂u

∂x (0, t) =

∂u

∂x (π, t) = 0, t > 0

u(x, 0) = ex, 0 < x < 1

Problem 3 (10.5 8) Find a formal solution to the given initial-boundary value problem.

∂u

∂t =

∂2u

∂x2 , 0 < x < π, t > 0

u(0, t) = 0, u(π, t) = 3π, t > 0

u(x, 0) = 0, 0 < x < π

Problem 4 (10.5 10) Find a formal solution to the given initial-boundary value problem.

∂u

∂t = 3

∂2u

∂x2 + x, 0 < x < π, t > 0

u(0, t) = u(π, t) = 0, t > 0

u(x, 0) = sin(x), 0 < x < π

Problem 5 (10.5 14) Find a formal solution to the given initial-boundary value problem.

∂u

∂t = 3

∂2u

∂x2 + 5, 0 < x < π, t > 0

u(0, t) = u(π, t) = 1, t > 0

u(x, 0) = 1, 0 < x < π

1

• Problem 6 (10.6 2) Find a formal solution.

∂2u

∂t2 = 16

∂2u

∂x2 , 0 < x < π, t > 0

u (0, t) = u (π, t) = 0, t > 0

u (x, 0) = sin2 x, 0 < x < 1

∂u

∂t (x, 0) = 1 − cos x, 0 < x < π.

(Ans. P

n=1 [an cos (4nt) + bn sin (4nt)] sin (nx),

an =

(

− 8 π

1

n(n2−4) n odd

0 n even , bn =

(

1 πn2

n odd −

1

π(n2−1) n even .

Problem 7 (10.6 8) Find a formal solution.

∂2u

∂t2 =

∂2u

∂x2 + x sin t, 0 < x < π, t > 0.

u (0, t) = u (π, t) = 0, t > 0

u (x, 0) = 0, 0 < x < π,

∂u

∂t (x, 0) = 0, 0 < x < π.

(Ans. [sin t − t cos t] sin x + P

n=2 2(−1)n

n2(n2−1) [sin (nt) − n sin t] sin (nx) .)

Problem 8 (10.6 10) Derive a formal formula for the solution.

∂2u

∂t2 = α2

∂2u

∂x2 , 0 < x < L, t > 0

u (0, t) = U1 u (L, t) = U2, t > 0

u (x, 0) = f (x) , 0 < x < L,

∂u

∂t (x, 0) = g (x) , 0 < x < L,

where U1, U2 are constants. (Ans.

u (x, t) = U1 + U2 − U1

L x +

∞ X

n=1

»

an cos

nπαt

L

«

+ bn sin

nπαt

L

«–

sin “

nπx

L

an = 2

L

Z

L

0

f (x) −

»

U1 + U2 − U1

L x

–ff

sin “

nπx

L

dx,

bn = 2

nπα

Z

L

0

g (x) sin “

nπx

L

dx.

2

• Problem 9 (10.6 14) Consider

∂2u

∂t2 = α2

∂2u

∂x2 , −∞ < x < ∞, t > 0

u (x, 0) = f (x) , −∞ < x < ∞

∂u

∂t (x, 0) = g (x) , −∞ < x < ∞

with f (x) = x2, g (x) = 0. (Ans. x2 + α2t2)

Problem 10 (10.6 16) Same as above, but with f (x) = sin 3x, g (x) = 1. (Ans. sin (3x) cos (3αt) + t.)

3

• Solution 1. (10.5 2) Using separation of variables, assume

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

then

∂u

∂t = X(x)T ′(t)

∂2u

∂x2 = X ′′(x)T (t)

Substituting back into the PDE we have

X(x)T ′(t) = X ′′(x)T (t)

⇒ T ′(t)

T (t) =

X ′′(x)

X(x) = K, K constant

⇒ {

X ′′(x) − KX(x) = 0 T ′(t) − KT (t) = 0

Consider the boundary conditions

u(0, t) = u(π, t) = 0

then X(0)T (t) = X(π)T (t) = 0

so either T (t) = 0, the trivial solution, or

X(0) = X(π) = 0

Solving the boundary value problem

X ′′(x) − KX(x) = 0, X(0) = X(π) = 0

this is second order linear homogeneous equation with auxiliary equation r2−K = 0 We investigate three possible cases for K.

Case 1: K > 0 ⇒ r = ± √

K, real and distinct roots. This has the general solution

X(x) = c1e √

Kx + c2e − √

Kx

Using the boundary conditions:

X(0) = c1 + c2 = 0 ⇒ c2 = −c1

X(π) = c1e √

Kπ − c1e− √

Kπ = c1e − √

Kπ (

e2 √

Kπ − 1 )

= 0

Since K > 0, we must have c1 = 0, which is the trivial solution.

4

• Case 2: K = 0 ⇒ r = 0, double root. This has the general solution

X(x) = c1 + c2x

Using the boundary conditions: X(0) = c1 = 0

X(π) = c2π = 0 ⇒ c2 = 0 This case also gives us the trivial solution.

Case 3: K < 0 ⇒ r = ± √ −Ki, complex roots.

This has the general solution

X(x) = c1 cos( √ −Kx) + c2 sin(

√ −Kx)

Using the boundary conditions: X(0) = c1 = 0

X(π) = c2 sin( √ −Kπ) = 0

Either c2 = 0 (the trivial solution) or

sin( √ −Kπ) = 0

⇒ √ −Kπ = nπ ⇒ K = −n2, n = 1, 2, . . .

These eigenvalues have the corresponding eigenfunctions

Xn(x) = cn sin(nx)

Using the eigenvalues in the equation for T (t):

T ′(t) − KT (t) = 0 ⇒ T ′(t) + n2T (t) = 0

⇒ Tn(t) = ane−n 2t

Putting the equations for X(x) and T (t) together and combining the constants, we have

un(x, t) = Xn(x)Tn(t) = bn sin(nx)e −n2t

Since un(x, t) is a solution to the PDE for any value of n, we take the infinite series

u(x, t) =

∞ ∑

n=1

bn sin(nx)e −n2t

From the general solution and initial condition, we have

u(x, 0) = ∞ ∑

n=1

bn sin(nx)

= x2

5

• which means we can choose the bn’s as the coefficients in the Fourier sine series for f(x) = x 2:

bn = 2

π

∫ π

0

x2 sin(nx)dx

= 2

π

[−n2x2 cos(nx) + 2 cos(nx) + 2nx sin(nx) n3

0

= 2

πn3 (

(2 − n2π2)(−1)n − 2 )

Thus the formal solution to the given initial-boundary value problem is

u(x, t) = 2

π

∞ ∑

n=1

1

n3 (

(2 − n2π2)(−1)n − 2 )

sin(nx)e−n 2t

6

• Solution 2. (10.5 5) Following a similar argument as for 10.5 2, we use separation of variables and the boundary conditions to arrive at the system of ODEs:

{

X ′′(x) − KX(x) = 0, X ′(0) = X ′(π) = 0 T ′(t) − KT (t) = 0

Consider the boundary value problem:

X ′′(x) − KX(x) = 0, X ′(0) = X ′(π) = 0

This has the auxiliary equation r2 − K = 0

Again we consider three cases as above.

Case 1: K > 0 Using a similar argument as in 10.5 2, the only solution in this case is the trivial solution.

Case 2: K = 0 Here

X(x) = c0 + c1x ⇒ X ′(x) = c1 Using the boundary conditions:

X ′(0) = c1 = 0

and c0 is arbitrary. So here we have the nontrivial solution

X(x) = c0

Case 3: K < 0 In this case the general solution is

X(x) = c1 cos( √ −Kx) + c2 sin(

√ −Kx)

⇒ X ′(x) = −c1 √ −K sin(

√ −Kx) + c2

√ −K cos(

√ −Kx)

Using the boundary conditions: X ′(0) = c2 = 0

X ′(π) = −c1 √ −K sin(

√ −Kπ) = 0

Either c1 = 0 (trivial solution) or

sin( √ −Kπ) = 0 ⇒ K = −n2

These eigenvalues have the corresponding eigenfunctions

Xn(x) = cn cos(nx)

Combining cases 2 and 3, we have the eigenvalues and eigenfunctions

K = −n2

Xn(x) = cn cos(nx)

7

• for n = 0, 1, 2, . . .

Returning to the equation for T (t):

T ′(t) − KT (t) = 0 ⇒ T ′(t) + n2T (t) = 0

⇒ Tn(t) = ane−n 2t

Putting the equations for X(x) and T (t) together and merging the constants, we have

un(x, t) = an cos(nx)e −n2t, n = 0, 1, 2, . . .

Taking the infinite series of these solutions, we have

u(x, t) = a0 2

+

∞ ∑

n=1

an cos(nx)e −n2t

From the general solution and the initial condition, we have

u(x, 0) = a0 2

+ ∞ ∑

n=1

an cos(nx)

= ex

which means we can choose the an’s as the coefficients in the Fourier cosine series for f(x) = e x:

a0 = 2

π

∫ π

0

exdx = 2

π (eπ − 1)

an = 2

π

∫ π

0

ex cos(nx)dx

= 2

π

(

eπ(−1)n − 1 n2 + 1

)

Thus the formal solution to the given initial-boundary value problem is

u(x, t) = 1

π (eπ − 1) + 2

π

∞ ∑

n=1

eπ(−1)n − 1 n2 + 1

cos(nx)e−n 2t

8

• Solution 3. (10.5 8) We assume the solution consists of a steady-state solution v(x) and a transient solution w(x, t) so that

u(x, t) = v(x) + w(x, t)

where w(x, t) and