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Fourier sine transform for Wave Equation

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Homework Statement


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

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

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

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

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, [tex]U_{tt} = 0 \Rightarrow c^{2}\lambda^{2}U=0 \Rightarrow U = 0[/tex]

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.
 
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Answers and Replies

  • #2
I like Serena
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Homework Statement


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

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

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

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

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, [tex]U_{tt} = 0 \Rightarrow c^{2}\lambda^{2}U=0 \Rightarrow U = 0[/tex]

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?
 
  • #3
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[PLAIN]http://img815.imageshack.us/img815/9894/fourier5313.png [Broken]

there was a minus sign. oops.

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

which gives me

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

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

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

Taking the Inverse Fourier Cosine Transform,

[tex]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[/tex]
[tex]= -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[/tex]

and this is where I am stuck.
 
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  • #4
I like Serena
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  • #5
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But that only makes use of the regular Fourier Transform, and not the Fourier Cosine Transform.
 
  • #6
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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)=?
 
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