Fourier sine transform for Wave Equation

[compliant]

Homework Statement

Find the solution u, via the Fourier sine/cosine transform, given:
u_{tt}-c^{2}u_{xx}=0
IC: u(x,0) = u_{t}(x,0)=0
BC: u(x,t) bounded as x\rightarrow \infty , u_{x}(0,t) = g(t)

2. The attempt at a solution
Taking the Fourier transform of the PDE, IC and BC:

U_{tt}-c^{2}(i\lambda)^{2}U=0
U_{tt}+c^{2}\lambda^{2}U=0

which is an ODE in t, so two linearly independent solutions of the homogeneous equation are sin(\lambda ct) and cos(\lambda ct).

If I take a linear combination of these two solutions, I get zero constants, which eventually leaves me with u = 0 as a solution, or at least one that's only x-dependent.

But if u is x-dependent, U_{tt} = 0 \Rightarrow c^{2}\lambda^{2}U=0 \Rightarrow U = 0

which still leaves me with u = 0 as a solution, and, no offense, for a homework problem, that's kind of lame. That's why I'm suspicious, and asking to see if I'm doing it right.

 

[I like Serena]

Homework Statement

Find the solution u, via the Fourier sine/cosine transform, given:
u_{tt}=c^{2}u_{xx}=0
IC: u(x,0) = u_{t}(x,0)=0
BC: u(x,t) bounded as x\rightarrow \infty , u_{x}(0,t) = g(t)

2. The attempt at a solution
Taking the Fourier transform of the PDE, IC and BC:

U_{tt}-c^{2}(i\lambda)^{2}U=0
U_{tt}+c^{2}\lambda^{2}U=0

which is an ODE in t, so two linearly independent solutions of the homogeneous equation are sin(\lambda ct) and cos(\lambda ct).

If I take a linear combination of these two solutions, I get zero constants, which eventually leaves me with u = 0 as a solution, or at least one that's only x-dependent.

But if u is x-dependent, U_{tt} = 0 \Rightarrow c^{2}\lambda^{2}U=0 \Rightarrow U = 0

which still leaves me with u = 0 as a solution, and, no offense, for a homework problem, that's kind of lame. That's why I'm suspicious, and asking to see if I'm doing it right.

Solving just PDE and IC without Fourier (and without BC) yields u(x,t)=0.

Are you sure you have the right equations in your problem?

 

[compliant]

[PLAIN]http://img815.imageshack.us/img815/9894/fourier5313.png

there was a minus sign. oops.

In any case, relevant equations:
F_{c}(f(x,t)) = \sqrt{\frac{2}{\pi}} \int^{\infty}_{0} cos (\lambda x) f(x,t) dx
F_{c}(f''(x,t)) = -\sqrt{\frac{2}{\pi}} f'(0,t) - \lambda^{2} F_{c}(f(x,t))

which gives me

U_{tt}-c^{2}\left[-\sqrt{\frac{2}{\pi}}U_{x}(0,t)-\lambda^{2}U \right] = 0
U_{tt}+c^{2}\sqrt{\frac{2}{\pi}}g(t)+c^{2}\lambda^{2}U = 0

Which, by the method of Green's Functions, gives me:

U(\lambda,t) = \int^{t}_{0} -c^{2}sin (\lambda c (\tau-t))\sqrt{\frac{2}{\pi}}g(\lambda,\tau) d\tau

Taking the Inverse Fourier Cosine Transform,

u(x,t) = \sqrt{\frac{2}{\pi}} \int^{\infty}_{0} cos(\lambda x) \left[\int^{t}_{0} -c^{2}sin (\lambda c (\tau-t)) \sqrt{\frac{2}{\pi}}g(\lambda,\tau) d\tau \right] d\lambda
= -c^{2}{\frac{2}{\pi}} \int^{\infty}_{0} cos(\lambda x) \left[\int^{t}_{0} sin (\lambda c (\tau-t))}g(\lambda,\tau) d\tau \right] d\lambda

and this is where I am stuck.

 

[I like Serena]

I found this article:

http://www.google.com/url?sa=t&sour...FhPo6Q_Q&sig2=JYLtW_5jJRhv49m_q1NpcA&cad=rja"

which explains how you can use Fourier to solve a PDE wave equation.

In short:

u(x,t) = F(x-ct) + G(x+ct)

Combining it with the IC and BC gives a result in terms of g(u).

 

[compliant]

But that only makes use of the regular Fourier Transform, and not the Fourier Cosine Transform.

 

[spapi]

with the answer please ?

1) u(t-2)*δ(1-t)

2) X(w)=e^-2|w| / (1-jw)
x1(t)=x(2-t)
X1(w)= ?3) x[n]=u[n] ; x(0) = ?4)Θ = -aw
x(t) = cos100 πt
Y(t) = sin100 πt
a=?

5) one side laplase
Y(s)=[1+x(s)] / [s+3]

6) Θ = -3w
y(t)=?