# Inverse Fourier transform of a hyperbolic cosine

This problem arises from trying to solve, by Fourier transform, the Cauchy problem

$$\begin{cases} u_{tt}-u_{xxxx}=0 &x\in\mathbb{R},\, t\geq 0\\ \begin{cases} u(0,x)=f(x)\\ u_t(0,x)=0 \end{cases} \end{cases}$$

That is fourth-order wave PDE.

Doing formal integrals, I arrive at this expression after completing squares and using definitions of error functions:

$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\cosh(k^2t)e^{ikx}~dk=\frac{1}{4\sqrt{2t}}e^{-\frac{x^2}{4t}}\left[\text{erf}\left(\sqrt{t}\left(k-\frac{ix}{2t}\right)\right)+e^{\frac{x^2}{2t}}\text{erfi}\left(\sqrt{t}\left(k+\frac{ix}{2t}\right)\right)\right]_{-\infty}^{\infty}=\frac{1}{4\sqrt{2t}}e^{-\frac{x^2}{4t}}\biggl (2+e^{\frac{x^2}{2t}}\left[\text{erfi}\left(\sqrt{t}\left(k+\frac{ix}{2t}\right)\right)\right]_{-\infty}^{\infty}\biggr)$$

where I have evaluated $\text{erf}$ in the last line. My problem is that I know that $\text{erfi}$ does not have a limit when $k\to\infty$ since it is non-bounded over real line. However doing this inverse Fourier transform using corresponding Mathematica command gives me:

$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\cosh(k^2t)e^{ikx}~dk=\frac{1}{4\sqrt{2t}}e^{-\frac{x^2}{4t}}\left(2+2ie^{\frac{x^2}{2t}}\right)$$

where I have arranged the terms in order to get the same form as the result I have described before.

What am I not understanding?

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Mathematica should be wrong in this case - in fact it is clear that the integral doesn't even converge. –  Sanchez Nov 14 '12 at 0:51

That means this Cauchy problem no longer can be solved by Fourier transform.

So you should use separation of variables:

Let $u(t,x)=T(t)X(x)$ ,

Then $T''(t)X(x)-T(t)X''''(x)=0$

$T''(t)X(x)=T(t)X''''(x)$

$\dfrac{T''(t)}{T(t)}=\dfrac{X''''(x)}{X(x)}=s^4$

$\begin{cases}T''(t)-s^4T(t)=0\\X''''(x)-s^4X(x)=0\end{cases}$

$\begin{cases}T(t)=\begin{cases}c_1(s)\sinh ts^2+c_2(s)\cosh ts^2&\text{when}~s\neq0\\c_1t+c_2&\text{when}~s=0\end{cases}\\X(x)=\begin{cases}c_3(s)\sinh xs+c_4(s)\cosh xs+c_5(s)\sin xs+c_6(s)\cos xs&\text{when}~s\neq0\\c_3x^3+c_4x^2+c_5x+c_6&\text{when}~s=0\end{cases}\end{cases}$

$\therefore u(t,x)=\int_0^\infty C_1(s)\sinh ts^2\sinh xs~ds+\int_0^\infty C_2(s)\sinh ts^2\cosh xs~ds+\int_0^\infty C_3(s)\sinh ts^2\sin xs~ds+\int_0^\infty C_4(s)\sinh ts^2\cos xs~ds+\int_0^\infty C_5(s)\cosh ts^2\sinh xs~ds+\int_0^\infty C_6(s)\cosh ts^2\cosh xs~ds+\int_0^\infty C_7(s)\cosh ts^2\sin xs~ds+\int_0^\infty C_8(s)\cosh ts^2\cos xs~ds$

$u_t(t,x)=\int_0^\infty s^2C_1(s)\cosh ts^2\sinh xs~ds+\int_0^\infty s^2C_2(s)\cosh ts^2\cosh xs~ds+\int_0^\infty s^2C_3(s)\cosh ts^2\sin xs~ds+\int_0^\infty s^2C_4(s)\cosh ts^2\cos xs~ds+\int_0^\infty s^2C_5(s)\sinh ts^2\sinh xs~ds+\int_0^\infty s^2C_6(s)\sinh ts^2\cosh xs~ds+\int_0^\infty s^2C_7(s)\sinh ts^2\sin xs~ds+\int_0^\infty s^2C_8(s)\sinh ts^2\cos xs~ds$

$u_t(0,x)=0$ :

$\int_0^\infty s^2C_1(s)\sinh xs~ds+\int_0^\infty s^2C_2(s)\cosh xs~ds+\int_0^\infty s^2C_3(s)\sin xs~ds+\int_0^\infty s^2C_4(s)\cos xs~ds=0$

$\int_0^\infty s^2C_3(s)\sin xs~ds=-\int_0^\infty s^2C_1(s)\sinh xs~ds-\int_0^\infty s^2C_2(s)\cosh xs~ds-\int_0^\infty s^2C_4(s)\cos xs~ds$

$\mathcal{F}_{s,s\to x}\{s^2C_3(s)\}=-\int_0^\infty s^2C_1(s)\sinh xs~ds-\int_0^\infty s^2C_2(s)\cosh xs~ds-\mathcal{F}_{c,s\to x}\{s^2C_4(s)\}$

$C_3(s)=-\dfrac{\mathcal{F}^{-1}_{s,x\to s}\{\int_0^\infty s^2C_1(s)\sinh xs~ds\}}{s^2}-\dfrac{\mathcal{F}^{-1}_{s,x\to s}\{\int_0^\infty s^2C_2(s)\cosh xs~ds\}}{s^2}-\dfrac{\mathcal{F}^{-1}_{s,x\to s}\{\mathcal{F}_{c,s\to x}\{s^2C_4(s)\}\}}{s^2}$

$\therefore u(t,x)=\int_0^\infty C_1(s)\sinh ts^2\sinh xs~ds+\int_0^\infty C_2(s)\sinh ts^2\cosh xs~ds+\int_0^\infty C_4(s)\sinh ts^2\cos xs~ds-\int_0^\infty\dfrac{\mathcal{F}^{-1}_{s,x\to s}\{\int_0^\infty s^2C_1(s)\sinh xs~ds\}\sinh ts^2\sin xs}{s^2}ds-\int_0^\infty\dfrac{\mathcal{F}^{-1}_{s,x\to s}\{\int_0^\infty s^2C_2(s)\cosh xs~ds\}\sinh ts^2\sin xs}{s^2}ds-\int_0^\infty\dfrac{\mathcal{F}^{-1}_{s,x\to s}\{\mathcal{F}_{c,s\to x}\{s^2C_4(s)\}\}\sinh ts^2\sin xs}{s^2}ds+\int_0^\infty C_5(s)\cosh ts^2\sinh xs~ds+\int_0^\infty C_6(s)\cosh ts^2\cosh xs~ds+\int_0^\infty C_7(s)\cosh ts^2\sin xs~ds+\int_0^\infty C_8(s)\cosh ts^2\cos xs~ds$

$u(0,x)=f(x)$ :

$\int_0^\infty C_5(s)\sinh xs~ds+\int_0^\infty C_6(s)\cosh xs~ds+\int_0^\infty C_7(s)\sin xs~ds+\int_0^\infty C_8(s)\cos xs~ds=f(x)$

$\int_0^\infty C_7(s)\sin xs~ds=f(x)-\int_0^\infty C_5(s)\sinh xs~ds-\int_0^\infty C_6(s)\cosh xs~ds-\int_0^\infty C_8(s)\cos xs~ds$

$\mathcal{F}_{s,s\to x}\{C_7(s)\}=f(x)-\int_0^\infty C_5(s)\sinh xs~ds-\int_0^\infty C_6(s)\cosh xs~ds-\mathcal{F}_{c,s\to x}\{C_8(s)\}$

$C_7(s)=\mathcal{F}^{-1}_{s,x\to s}\{f(x)\}-\mathcal{F}^{-1}_{s,x\to s}\{\int_0^\infty C_5(s)\sinh xs~ds\}-\mathcal{F}^{-1}_{s,x\to s}\{\int_0^\infty C_6(s)\cosh xs~ds\}-\mathcal{F}^{-1}_{s,x\to s}\{\mathcal{F}_{c,s\to x}\{C_8(s)\}\}$

$\therefore u(t,x)=\int_0^\infty C_1(s)\sinh ts^2\sinh xs~ds+\int_0^\infty C_2(s)\sinh ts^2\cosh xs~ds+\int_0^\infty C_4(s)\sinh ts^2\cos xs~ds-\int_0^\infty\dfrac{\mathcal{F}^{-1}_{s,x\to s}\{\int_0^\infty s^2C_1(s)\sinh xs~ds\}\sinh ts^2\sin xs}{s^2}ds-\int_0^\infty\dfrac{\mathcal{F}^{-1}_{s,x\to s}\{\int_0^\infty s^2C_2(s)\cosh xs~ds\}\sinh ts^2\sin xs}{s^2}ds-\int_0^\infty\dfrac{\mathcal{F}^{-1}_{s,x\to s}\{\mathcal{F}_{c,s\to x}\{s^2C_4(s)\}\}\sinh ts^2\sin xs}{s^2}ds+\int_0^\infty C_5(s)\cosh ts^2\sinh xs~ds+\int_0^\infty C_6(s)\cosh ts^2\cosh xs~ds+\int_0^\infty C_8(s)\cosh ts^2\cos xs~ds+\int_0^\infty\mathcal{F}^{-1}_{s,x\to s}\{f(x)\}\cosh ts^2\sin xs~ds-\int_0^\infty\mathcal{F}^{-1}_{s,x\to s}\{\int_0^\infty C_5(s)\sinh xs~ds\}\cosh ts^2\sin xs~ds-\int_0^\infty\mathcal{F}^{-1}_{s,x\to s}\{\int_0^\infty C_6(s)\cosh xs~ds\}\cosh ts^2\sin xs~ds-\int_0^\infty\mathcal{F}^{-1}_{s,x\to s}\{\mathcal{F}_{c,s\to x}\{C_8(s)\}\}\cosh ts^2\sin xs~ds$

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I have no idea what's going on, but wow that's a lot of LaTeX. –  Javier Badia Nov 18 '12 at 1:31
Well, I don't know what are you saying since Fourier transform is "separation of variables made continuous". However your separation of variables has a lot of integrals that, as the one I propose, doesn't converge at all. And I note that you doesn't take into account that the domain of integration must be $\mathbb{R}$ since we can't suppose parity. –  elessartelkontar Nov 19 '12 at 20:30
$$u(x,t)=\int_{-\infty}^{\infty} dk \: \hat{f}(k) \cosh{4 \pi^2 k^2 t} \exp{(-i 2\pi k x)}$$
Where $\hat{f}(k)$ is the FT of $f$. You need to evaluate this integral, as the convolution theorem does not hold due to the behavior of the cosh function.