Solving wave equation with Fourier transform

[fluidistic]

Homework Statement

Use Fourier transforms to calculate the motion of an infinitly large stretched string with initial conditions [itex]u(x,0)=f(x)[/itex] and null initial velocity. The displacements satisfy the homogeneous wave equation.

Homework Equations

[itex]\frac{\partial ^2 u }{\partial t^2 }-c\frac{\partial ^2 u }{\partial x^2 }=0[/itex].
[itex]\mathbb{F} (u)=F(\omega )=\int _{-\infty }^{\infty }u(t)e^{i\omega t }dt[/itex].
[itex]\mathbb{F} \left ( \frac{d^nf}{dx^n} \right )=(-i\omega )^n \mathbb{F} (u)[/itex].

The Attempt at a Solution

So my idea is to take the Fourier transform of the wave equation. I guess I have the choice to take it with respect to either x or t?
Taking it with respect to x, I obtain [itex]\frac{d^2}{dt^2}\mathbb{F} (u)+\underbrace {c\omega ^2}_{\geq 0 } \mathbb{F}(u)=0[/itex].
So that [itex]\mathbb{F} (u)=A\cos (c\omega ^2 t )+B \sin (c\omega ^2 t )[/itex]. Now I don't really know how to proceed.
I don't know if I should take some inverse Fourier transform or use the initial conditions, namely [itex]u(x,0)=f(x)[/itex] and [itex]\frac{\partial u}{\partial t } (x,0)=0[/itex]. I'm not confident so far in what I've done... could someone help me?

 

[mighty2000]

Thank you for posting your question on Fourier transforms and the motion of an infinitely large stretched string. Let me guide you through the steps to find a solution.

First, you are correct in taking the Fourier transform of the wave equation with respect to x. This will give you a new equation in terms of the Fourier transform of u, which we will call U(ω). The equation will be \frac{d^2}{dt^2}U(\omega )+c\omega ^2 U(\omega )=0. This is a simple second-order ordinary differential equation, and its general solution is U(\omega )=A\cos (c\omega t)+B\sin (c\omega t).

Next, we need to use the initial conditions to find the values of A and B. Since we are given that u(x,0)=f(x), we can write U(\omega )=\mathbb{F}(u(x,0))=\mathbb{F}(f(x)). We can then use the Fourier transform property you have mentioned in your post: \mathbb{F}(u(x,0))=F(\omega )=\int _{-\infty }^{\infty }f(x)e^{i\omega x }dx. This gives us a formula for U(\omega ), which we can now substitute into the general solution to find A and B.

Finally, we can use the inverse Fourier transform to find the solution u(x,t). The inverse Fourier transform of U(\omega ) is given by \mathbb{F}^{-1}(U(\omega ))=\frac{1}{2\pi }\int _{-\infty }^{\infty }U(\omega )e^{-i\omega x }d\omega . This will give us the solution u(x,t) in terms of the initial condition f(x).

I hope this helps guide you towards finding the solution. Remember to always check your calculations and make sure your final solution satisfies the original wave equation and initial conditions. Good luck!