PDE Separation of Variables: Solving utt = uxx with Boundary Conditions

[walter9459]

Homework Statement

Solve the problem.
u_tt = u_xx 0 < x < 1, t > 0
u(x,0) = x, u_t(x,0) = x(1-x), u(0,t) = 0, u(1,t) = 1

Homework Equations

The Attempt at a Solution

Here is what I have so far but I'm not sure if I am on the right path or not.

u(x,t) = X(x)T(t)
u_t(x,t) = X(x)T'(t) u_x(x,t) = X'(x)T(t)
u_tt(x,t) = X(x)T"(t) u_xx(x,t) = X"(x)T(t)
X(x)T"(t) = X"(x)T(t)
T"(t)/T(T) = X"(x)/X(x) = λ
T"(t) = λT(t) X"(x) = λX(x)

λ = 0 -----> X(x) = Ax + B
b.c. u(0,t) = A(0) + B = 0 --------> B = 0
u(1,t) = A(1) + B = 1 --------> A = 1

λ > 0 --------> λ = ω²
X(x) = Acosh ωx + Bsinh ωx
X(0) = Acosh ω(0) + Bsinh ω(0) = 0
= Bsinh ω(0) = 0 ------> B = 0

λ < 0 ---------> λ = -ω²
X"(x) = λX(x) --------> X"(x) = -ω²X(x)
X(x) = Acosωx + Bsinωx
X(0) = Acosω(0) + Bsinω(0) = 0 --------> A = 0
X(1) = Acosω(1) + Bsinω(1) = 1
X(1) = Bsinω = 1 B ≠ 0
ω = ∏/2 + 2m∏ for any interger m

T"(t) = ω²T(t)
T"(t) = C cosωt + Dsinωt

u = (C cosωt + Dsinωt)sinux

Okay this is all I have. Am I on the right path and where do I go from here?
Thanks!

 

[HallsofIvy]

Homework Statement

Solve the problem.
u_tt = u_xx 0 < x < 1, t > 0
u(x,0) = x, u_t(x,0) = x(1-x), u(0,t) = 0, u(1,t) = 1

Homework Equations

The Attempt at a Solution

Here is what I have so far but I'm not sure if I am on the right path or not.

u(x,t) = X(x)T(t)
u_t(x,t) = X(x)T'(t) u_x(x,t) = X'(x)T(t)
u_tt(x,t) = X(x)T"(t) u_xx(x,t) = X"(x)T(t)
X(x)T"(t) = X"(x)T(t)
T"(t)/T(T) = X"(x)/X(x) = λ
T"(t) = λT(t) X"(x) = λX(x)

λ = 0 -----> X(x) = Ax + B
b.c. u(0,t) = A(0) + B = 0 --------> B = 0
u(1,t) = A(1) + B = 1 --------> A = 1

λ > 0 --------> λ = ω²
X(x) = Acosh ωx + Bsinh ωx
X(0) = Acosh ω(0) + Bsinh ω(0) = 0
= Bsinh ω(0) = 0 ------> B = 0

λ < 0 ---------> λ = -ω²
X"(x) = λX(x) --------> X"(x) = -ω²X(x)
X(x) = Acosωx + Bsinωx
X(0) = Acosω(0) + Bsinω(0) = 0 --------> A = 0
X(1) = Acosω(1) + Bsinω(1) = 1
X(1) = Bsinω = 1 B ≠ 0
ω = ∏/2 + 2m∏ for any interger m

T"(t) = ω²T(t)
T"(t) = C cosωt + Dsinωt

u = (C cosωt + Dsinωt)sinux

Okay this is all I have. Am I on the right path and where do I go from here?
Thanks!

Other than the fact that you mean "sin ωx", not "sin ux", that's good. Now you have to try to make that fit the "initial vaue conditions", u(x,0)= 0 and $u_t(x, 0)= x(1- x)$.

You won't be able to do that with just a single "ω" so since this is a linear equation try, instead, a sum:
$$u(x,t)= \sum_{m=0}^\infty (cos(\pi/2 + 2m\pi)t + Dsin(\pi/2 + 2m\pi)t)sin(\pi/2 + 2\pi)x$$

 

[walter9459]

Sorry to be so dense, but I get lost at this point.

I think I am then suppose to do

u_t=X(x)[-C(∏/2 + 2m∏)sin(∏/2 + 2m∏)t + D(∏/2 + 2m∏)cos(∏/2 + 2m∏)t)
u_t(x,0) = D(∏/2 + 2m∏)cos(∏/2 + 2m∏) = x(1-x) ----> D ≠ 0

t = 0 f(x) = ∑ Dsin (∏/2 + 2m∏)t

u(x,t) = ∑ D sin ((∏/2 + 2m∏)t sin (∏/2 + 2m∏)x

Am I on the doing this correctly? Do I then do the integral from 0 to m∏?

Thanks!