Using Fourier transform to solve $u_{tt}=-u_{xxxx}$ The problem is
$u_{tt}=-u_{xxxx}$,$-∞<x<∞$, $t>0$,
$u(x,0)=f(x)$, $u_{t}(x,0)=0$.
Assume $u,u_{x},u_{xx},u_{xxx} → 0$ as $|x| → ∞$. Write sulution in term of a convolution of $f(x)$.
There is a hint: $\int_{0}^{\infty}e^{-ix^{2}}dx=\sqrt{\frac{\pi}{8}}(1-i)$
Here is what I have done so far:
I tried to expand $u(x,t)$ with respect to $x$, denote the Fourier transform by $\hat{u}(k,t)$.
From the PDE, get $\frac{\partial^2 \hat{u}}{\partial t^2}=-k^{4}\hat{u}$.
The general solution is $\hat{u}=A(k)e^{ik^{2}t}+B(k)e^{-ik^{2}t}$
Since $u_{t}(x,0)=0$, get $A(k)=B(k)$, The solution becomes $\hat{u}=2A(k)\cos{k^{2}t}$.
Then I'm stuck here. I think the solution of $u(x,t)$ is in term of a convolution of $f(x)$ and $F^{-1}(\cos{k^{2}t})$. But I have no idea how to get the inverse fourier transfom of $\cos{k^{2}t}$ and how to apply the hint.
Can someone help me? Thanks in advance!
 A: Let's start with the hint: from
$$
\int_{0}^{\infty}e^{-ix^{2}}dx=\sqrt{\frac{\pi}{8}}(1-i)
$$
it follows that
$$
\int_{-\infty}^{\infty}e^{\pm ix^{2}}dx=\sqrt{\frac{\pi}{2}}(1\pm i).
$$
This allows us to calculate the inverse Fourier transform of $\cos k^2t$:
\begin{align*}
F^{-1}(\cos k^2t)(x)&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\cos k^2t\, e^{ikx}\,dk \\
&=\frac{1}{4\pi}\left(\int_{-\infty}^{\infty}e^{ik^2t+ikx}\,dk+
\int_{-\infty}^{\infty}e^{-ik^2t+ikx}\,dk  \right) \\
&=\frac{1}{4\pi}\left(e^{-i\frac{x^2}{4t}}\int_{-\infty}^{\infty}e^{it(k+\frac{x}{2t})^2}\,dk+e^{i\frac{x^2}{4t}}\int_{-\infty}^{\infty}e^{-it(k-\frac{x}{2t})^2}\,dk
\right) \\
&=\frac{1}{4\pi}\left(\sqrt{\frac{\pi}{2t}}(1+i)e^{-i\frac{x^2}{4t}}
+\sqrt{\frac{\pi}{2t}}(1-i)e^{i\frac{x^2}{4t}}\right) \\
&=\frac{1}{\sqrt{8\pi t}}\left(\cos\frac{x^2}{4t}+\sin\frac{x^2}{4t}\right).
\end{align*}
Now we can finish the solution sketched in the question:
\begin{align*}
u(x,t)&=F^{-1}(\hat{f}(k)\cos k^2t)(x)=\int_{-\infty}^{\infty}f(x')F^{-1}(\cos k^2t)(x-x')\,dx' \\
&=\frac{1}{\sqrt{8\pi t}}\int_{-\infty}^{\infty}f(x')\left(\cos\frac{(x-x')^2}{4t}+\sin\frac{(x-x')^2}{4t}\right)dx'.
\end{align*}
A: The complete procedure for a very similar problem is described in Joel Feldman's UBC notes.
