Fourier transform of differential equation with tanh I have variable $q(x,t)$ in the function
$$\dfrac{\partial q}{\partial t} = \kappa \dfrac{\partial^2 q}{\partial x^2} + c_0 \tanh\left(q\right)$$
I would like to take the Fourier transform of this function in the $x$-dimension, which gives me:
$$\dfrac{\partial \widehat{q}}{\partial t} = -\kappa k^2 \widehat{q} + \dots$$
I do not know how to deal with the $c_0 \tanh \left( q \right)$ term. What is its transform? Is it possible to transform it?
 A: As $$\tanh \in L^\infty(\mathbb{R})$$ it induces a tempered distribution in $\mathcal{S}'(\mathbb{R})$. Thus it is possible to compute the Fourier transform of the distribution induced by $\tanh$, which is is defined by
$$\langle \widehat{\tanh}, \varphi \rangle := \langle \tanh, \widehat\varphi \rangle = \int_\mathbb{R} \tanh(x) \widehat{\varphi}(x) ~dx $$
for any Schwartz function $\varphi \in \mathcal{S}(\mathbb{R})$.

EDIT: Classically, one introduces the Fourier transform on $L^1(\mathbb{R})$ (the space of absolutely integrable functions), which is well-defined as
$$ |\widehat{f}(t)|=\left| \int_\mathbb{R} f(x) e^{-itx} ~dx \right| \leq \int_\mathbb{R} |f(x)| ~dx=:\|f\|_{L^1(\mathbb{R})} < \infty.$$
One can show that $\mathcal{F}:L^1(\mathbb{R}) \to C_b^u(\mathbb{R})$ where $C_b^u$ is the space of functions, which are bounded and absolutely continuous.
But $\tanh \notin L^1(\mathbb{R})$ hence we can't apply the classical Fourier transform. There is also a way for $L^2(\mathbb{R})$ by some density result, but $\tanh \notin L^2(\mathbb{R})$. Hence we have to generalize the concept of Fourier transforms and introduce the space of distributions; their elements are not functions in general. For example the dirac-delta distribution $\delta \in \mathcal{D}'(\mathbb{R})$. The nice thing of Schwartz distributions is that their Fourier transform is an isomorphism which maps $\mathcal{F} : \mathcal{S}' \to \mathcal{S}'$; it is defined via
$$\langle \widehat{T}, \varphi \rangle = \langle T, \widehat{\varphi} \rangle \text{ for all } \varphi \in \mathcal{S}(\mathbb{R}).$$
 Now, every function $f$ in $L^p(\mathbb{R})$ induces a distribution in $\mathcal{S}'(\mathbb{R})$ via
$$\langle f, \varphi \rangle= \int_\mathbb{R} f(x) \varphi(x) ~dx \text{ for all } \varphi \in \mathcal{S}(\mathbb{R})$$
and $\tanh$ is a bounded function i.e. $\tanh \in L^\infty(\mathbb{R})$ hence it induces a distribution in $\mathcal{S}'(\mathbb{R})$ via 
$$\langle \tanh, \varphi \rangle= \int_\mathbb{R} \tanh(x) \varphi(x) ~dx \text{ for all } \varphi \in \mathcal{S}(\mathbb{R}).$$
Thus we can compute its Fourier transform via
$$\langle \widehat{\tanh}, \varphi \rangle=\langle \tanh, \widehat{\varphi} \rangle= \int_\mathbb{R} \tanh(x) \widehat{\varphi}(x) ~dx \text{ for all } \varphi \in \mathcal{S}(\mathbb{R}).$$
You can compute this to get $\widehat{\tanh}$. I can show you an example.
EXAMPLE: Let us take $\sin$, which is also not in $L^1$ but in $L^\infty$. Hence we can't compute its classical Fourier transform but the Fourier transform of its distribution. We have
\begin{align}\langle \widehat{\sin}, \varphi \rangle=\langle \sin, \widehat{\varphi} \rangle &= \int_\mathbb{R} \sin(x) \widehat{\varphi}(x) ~dx \\
&=\int_\mathbb{R} \frac{e^{ix}-e^{-ix}}{2i} \int_\mathbb{R} \varphi(t) e^{-ixt} ~dt ~dx  \\
&=\frac{1}{2i} \int_{\mathbb{R}^2} e^{ix(1-t)} \varphi(t) ~d(t,x) - \frac{1}{2i} \int_{\mathbb{R}^2} e^{-ix(1+t)} \varphi(t) ~d(t,x)  \\
&=\frac{1}{2i} \int_{\mathbb{R}^2} e^{-ixs} \varphi(s+1) ~d(t,x) - \frac{1}{2i} \int_{\mathbb{R}^2} e^{-ixs} \varphi(s-1) ~d(s,x) \\
&=\frac{1}{2i} \left( \langle 1,\widehat{\varphi(\cdot+1)} \rangle-\langle 1,\widehat{\varphi(\cdot-1)} \rangle \right)  \\
&=\frac{1}{2i} \left( \langle \hat{1},\varphi(\cdot+1) \rangle-\langle \hat{1},\varphi(\cdot-1)\rangle \right) \\
&=\frac{1}{2i} \left( \langle 2\pi \delta,\varphi(\cdot+1) \rangle-\langle 2 \pi \delta,\varphi(\cdot-1)\rangle \right) \\
&=\frac{\pi}{i} \left( \langle \delta(\cdot -1),\varphi \rangle-\langle  \delta(\cdot +1),\varphi\rangle \right)  \end{align}
for any $\varphi \in \mathcal{S}(\mathbb{R})$ hence one writes
$$\widehat{\sin}(x)=\frac{\pi}{i} (\delta(x-1)-\delta(x+1)).$$
So, yeah, I have shown you that the Fourier transform of $\tanh$ can be computed theoretically and it is well-defined. But I don't know how hard the computation itself is. The sinus was quite nice as it has this representation via the exponential function.
A: As $tanh(x)$ is not absolutely integrable, so a Fourier transform does not exist. That said, for the signum function, which is not absolutely integrable, one can evaluate the fourier transform by applying useful workarounds.
Mathematica says that it transforms as $$\mathcal{F}_t[\text{tanh}(t)](\omega) = i\sqrt{\frac{\pi}{2}}\text{csch}\left(\frac{\pi\omega}{2}\right)$$
