Here's the question.
$\frac{∂^2}{∂t^2 } u(x,t)=c^2 \frac{∂^2}{∂x^2 } u(x,t)$
We are supposed to use this form of Fourier transform to solve our PDE
$\hat{f(s)} = \frac{1}{√2π} ∫_{-∞}^∞f(t) e^{(-ist)} dt$
Can anyone enlighten me on how to do this question?
Thanks!
$\endgroup$ 11 Answer
$\begingroup$Observe what happens when you take the Fourier transform of a derivative: $$\begin{align}\widehat{\left(\frac{\partial u}{\partial x} \right)}(k) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \frac{\partial u}{\partial x} e^{-ikx}dx = - \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} u \frac{\partial}{\partial x} \left( e^{-ikx} \right) dx \\ &= ik \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} u e^{ikx} dx = ik \hat{u}(k),\end{align}$$ (why? you should justify each step to yourself). So $$\widehat{\left(\frac{\partial u}{\partial x} \right)}(k) = ik \hat{u}(k). $$ So the Fourier transform of a second derivative then is $$\widehat{\left(\frac{\partial^2 u}{\partial x^2}\right)}(k) = (ik)^2 \hat{u}(k) = -k^2 \hat{u}(k).$$ Let's take the Fourier transform in x of your equation now:
$$\frac{\partial^2}{\partial t^2} \hat{u}(k,t) = c^2 (-k^2) \hat{u}(k,t) = -c^2 k^2 \hat{u}(k,t),$$ which is a differential equation in $t$ that contains no $x$-derivatives. You can integrate this (again, if you can't see this immediately you should work it out for yourself): $$\hat{u}(k,t) = Ae^{ickt} + Be^{-ickt}$$ for some constants $A$ and $B$. It now remains to invert the Fourier transform of $\hat{u}(k,t)$. Can you finish it off?
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