Rittner equation I would like to know if the Rittner equation :
$$\partial_{t}{\varPhi(x,t)=k\partial_{xx}{\varPhi(x,t)}}-\alpha{\partial_{x}{\varPhi(x,t)}-\beta{\varPhi(x,t)}+g(x,t)}$$
can be solved using the Lax pair method or the Fokas method.
Thanks
 A: Setting $\psi(x,t) = e^{(\alpha^2/4k + \beta)t - \alpha x/2k} \Phi(x,t)$ and plugging in I get
$$
\begin{align*} \psi_t(x,t) - k \psi_{xx}(x,t) &= e^{(\alpha^2/4k + \beta)t - \alpha x/2k}g(x,t) \\
&\equiv f(x,t).
\end{align*}
$$
If the forcing $f(x,t)$ is in $L^1(\mathbf{R}\times \mathbf{R})$ then we can use the heat kernel
$$
K(x,t)=\begin{cases} (4\pi k t)^{-1/2} e^{-x^2/2kt},& t>0 \\ 0,& t\leq 0 \end{cases}
$$
and a solution to the problem on the domain $\mathbf{R}\times\mathbf{R}$ is $\psi= K*f$. Unfortunately this puts very harsh restrictions on the original forcing term $g=g(x,t)$ because of the exponential growth of the factor $e^{(\alpha^2/4k + \beta)t - \alpha x/2k}$.
Similarly, if one is interested in the Cauchy problem with $\Phi(x,0)=\Phi_0(x)$, then the new problem has initial data $\psi_0(x) = e^{-\alpha x/2k} \Phi_0(x)$. Using heat kernel methods this can again lead to severe restrictions. Recall the solution to the homogeneous Cauchy problem would be
$$
\psi(x,t)=K_t * \psi_0(x)
$$
which is well defined in a classical sense if, for instance, $\psi_0 \in L^p(\mathbf{R})$ for $1\leq p\leq \infty$. This means the associated initial data $\Phi_0(x)$ would need exponential decay, which is a little artificial.
For the Cauchy problem for the Rittner equation on the full line (for simplicity consider the homogeneous problem $g=0$, the more general case is similar) one can use the Fourier transform. Setting $\psi = e^{\beta t} \Phi$ we have the Cauchy problem
$$
\begin{align*}\psi_t - k \psi_{xx} + \alpha \psi_x &=0, \\
\psi(x,0) &= \Phi_0(x), \end{align*}
$$
the solution to which is
$$
\psi(x,t) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{\mathrm{i} \lambda x -\lambda (k\lambda + \mathrm{i}\alpha) t} \hat{\Phi}_0(\lambda)\, \mathrm{d}\lambda.
$$
The problem on the half-line is more subtle. The Fokas method provides a solution in this case (for given boundary data at $x=0$, $\Phi(0,t)=\Phi_1(t)$ say) in terms of an integral along a smooth contour in the complex $\lambda$-plane. Now if the initial data $\Phi_0(x)$ has sufficient exponential decay (which is a big restriction), then one can show that this contour can be deformed onto the real $\lambda$-axis, giving a Fourier type integral as in the case of the Cauchy problem on the full line. However, in general, this contour cannot be shifted.
This problem is treated in Fokas' book, A unified approach to boundary value problems. He emphasizes that this inability to shift the contour back to the real axis (in the half-line problem) reflects the fact there is no classical transform for the problem.
A: With $\alpha$ and $\beta$ being constants, this equation can be reduced to a heat equation and solved exactly. The first step is to remove the $\beta$ term. We do this by setting
$$\Phi(x,t)=e^{-\beta t}\Phi_1(x,t).$$
The equation becomes
$$\partial_t\Phi_1(x,t)=k\partial_{xx}\Phi_1(x,t)-\alpha\partial_x\Phi_1(x,t)+g(x,t)e^{\beta t}.$$
The next step is to remove the drift term. This can be done by setting
$$\Phi_1(x,t)=e^{ct}e^{ax}\Phi_2(x,t)$$
with $a$ a constant to be determined. We just note that $\partial_x(e^{ax}\Phi_2(x,t))=e^{ax}(\partial_x+a)\Phi_2(x,t)$ and so one gets
$$\partial_t\Phi_2(x,t)+c\Phi_2(x,t)=k(\partial_x+a)^2\Phi_2(x,t)-\beta(\partial_x+a)\Phi_2(x,t)+g(x,t)e^{\beta t}e^{-ax}$$
and so, choosing $a=\frac{\beta}{2k}$ and $c=-\frac{\beta^2}{4k}$ the equation becomes a heat equation
$$\partial_t\Phi_2(x,t)=k\partial_{xx}\Phi_2(x,t)+g(x,t)e^{\left(\beta-\frac{\beta^2}{4k}\right) t}e^{-\frac{\beta}{2k}x}$$
This equation can be solved exactly by the kernel
$$\Delta(x,t)=\frac{e^{-\frac{x^2}{4kt}}}{\sqrt{4\pi kt}}$$
Then, a full solution is given by
$$\Phi_2(x,t)=\tilde\Phi_2(t,x)+\int dx'dt'\Delta(x-x',t-t')g(x',t')e^{\left(\beta-\frac{\beta^2}{4k}\right) t'}e^{-\frac{\beta}{2k}x'}$$
being $\tilde\Phi_2(t,x)$ a solution of the homogeneous equation. Then, it easy to recognize that kernel for the initial equation is just
$$\Delta_0(x,t)=\Delta(x,t)e^{-\left(\beta-\frac{\beta^2}{4k}\right) t}e^{\frac{\beta}{2k}x}$$
