I have problem solving below wave equation using Fourier transform.

$U_{tt} = U_{xx} , 0<x<1, t>0$

$U(0,t) = U(1,t) = 0$

$U(x,0) = x , U_t(x,0) = 0$

Firstly (as I learned) Fourier sine transform formula is:

$F_s[u(x,t)]=\int_0^{\infty}u(x,t)sin(\lambda x)dx$

while solving the above problem, I had to transform initial data using F_sine transform as well. but i don't know what should my integral boundaries be? $\int_0^{1} or \int_0^{\infty}$? what about reverse Fourier boundaries? which is another integral like $u(x,y) = \int_0^\infty F[u(x,t)]sin(\lambda x)d\lambda$

Usually I see the wave problem which needed to solve with FT to be Unlimited on one side like $0<x$ but here i don't know what happens to formulas of $F[u_{xx}]$ and my initial data as well.


You cannot use that definition of the Fourier transform to solve this problem(wrong solution domain), however there are such things as finite Fourier transforms to account for closed intervals like this.

The finite sine Fourier transform is defined by

$$\begin{align*} S_n(u(x,t)) &= 2\int_0^1 u(x,t)\sin{(n \pi x)}\,dx\\ S^{-1}(S_n(t)) &= \sum_{n=1}^{\infty}S_n(t)\sin{(n \pi x)} \end{align*}$$

With our problem, it can be shown that

$$\begin{align*} S_n(u_{tt}(x,t)) &= \frac{d^2 S_n(u(x,t))}{dt^2}\\ S_n(u_{xx}(x,t)) &= - (n \pi)^2 S(u(x,t)) \hspace{1in}\text{(the boundary conditions apply here)} \end{align*}$$

along with the transformed initial conditions, we have the ODE given by

$$\begin{cases} S_n''(t) + (n \pi)^2 S(t) = 0\\ S_n'(0)=0 \\ S_n (0) = \frac{2}{n \pi}(-1)^{n+1} \end{cases}$$ for $n = 1,2,3,...$.

Now, solving this problem, we have $S_n(t) = \frac{2}{n \pi}(-1)^{n+1}\cos{(\pi n t)}$. And applying the inverse transform, the solution is

$$u(x,t) = \frac{2}{\pi}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\cos{(n \pi t)} \sin{(n \pi x)}$$

  • $\begingroup$ I thought this is the way for the periodic functions. $\endgroup$ – Prof. MaMad Nov 11 '18 at 5:03
  • $\begingroup$ I’m not sure what you mean by that, your solution will inheritantly be periodic because of your boundary conditions. $\endgroup$ – DaveNine Nov 11 '18 at 5:18
  • $\begingroup$ Any source to read on it? $\endgroup$ – Prof. MaMad Nov 11 '18 at 5:21
  • $\begingroup$ I mean as i see in my book, The formulas you have used is just Fourier series coefficients without $l$ but here exactly same cause $l = 1$ and you can only use Fourier series on Periodic functions i believe $\endgroup$ – Prof. MaMad Nov 11 '18 at 5:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.