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

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
$$\Delta_0(x,t)=\Delta(x,t)e^{-\left(\beta-\frac{\beta^2}{4k}\right) t}e^{\frac{\beta}{2k}x}$$
