Question: Can the equation ${\psi}_{{{t}}}-{i}{\psi}_{{{x}{x}}}={e}^{{{x}}}{\psi}$ be solved with a canonical Fourier transform? If it requires a Fokas transform or inverse scattering transform, how would one apply them to generate an analytic solution?
Thoughts: I take Schrödinger's equation ${i}ℏ{\psi}_{{{t}}}{\left({x},{t}\right)}={\left[-\frac{{ℏ^{{2}}}}{{{2}{m}}}\partial_{{{x}{x}}}+{V}{\left({x},{t}\right)}\right]}{\psi}{\left({x},{t}\right)}$ with a potential ${V}{\left({x}\right)}=e^x$ defined on a one-dimensional model of a crystal lattice. The final PDE becomes
$$ \displaystyle {i}ℏ{\psi}_{{{t}}}{\left({x},{t}\right)}+\frac{{ℏ^{{2}}}}{{{2}{m}}}{\psi_{xx}}{\left({x},{t}\right)}={V}{\left({x},{t}\right)}{\psi}{\left({x},{t}\right)}\nonumber $$
Pardon me if the following simplification is egregious, but my PDEs textbook advocates for setting all coefficients to ${1}$.
$$ \displaystyle {\psi}_{{{t}}}-{i}{\psi}_{{{x}{x}}}=-i{e}^{{{x}}}{\psi}\nonumber $$
With the simplified formula, I attempt to produce an analytic Fourier solution; however, the spatial nonlinearity is breaking the validity of my approach.
$$ \displaystyle \frac{{{w}'{\left({t}\right)}}}{{{w}{\left({t}\right)}}}={i}\frac{{{v}{''}{\left({x}\right)}}}{{{v}{\left({x}\right)}}}-i{e}^{{{x}}}={\lambda}\in{\mathbb{{{C}}}}\nonumber $$
Finally, I have the system of equations ${w}'{\left({t}\right)}-{w}{\left({t}\right)}{{\lambda}}=0$ and ${v}{''}{\left({x}\right)}+{}{\left(i\lambda-{e}^{{{x}}}\right)}{v}{\left({x}\right)}={0}$. I admit setting $\frac{v(x)}{v(x)}=1$ is slightly careless as it assumes the limit approaching $0$ collapses to $1$, and does not account for any restrictions in domain. I had considered factoring the operator for the spatial equation with the hope of generating a solution in terms of the natural exponential that could produce a symmetric basis for an analytic Fourier transform. Unfortunately, that too appears to be an unhelpful endeavor.
$$ \text{Wrong!} \hspace 1em \displaystyle {\left(\partial_{{{x}}}+{i}\sqrt{{{}{\left({i\lambda}-{e}^{{{x}}}\right)}}}\right)}\underbrace{{{\left(\partial_{{{x}}}-{i}\sqrt{{{}{\left(i\lambda-{e}^{{{x}}}\right)}}}\right)}}v(x)}_{{{u}{\left({x}\right)}}} \ne 0\nonumber $$
By solving the comparatively simpler equation, ${v}{''}{\left({x}\right)}+{e}^{{{x}}}{v}{\left({x}\right)}$ with a similar method, knowing that $\frac{\partial}{\partial x} \left( ie^{\frac{x}{2}}\right)=-\frac{\partial}{\partial x}\left(ie^{\frac{-x}{2}}\right)$.
$$\left(\partial_x + ie^{\frac{x}{2}}\right)\underbrace{\left(\partial_x - ie^{\frac{x}{2}}\right)v(x)}_{u(x)}=0$$ I integrated a linear ODE to solve for ${u}{\left({x}\right)}$ and substituted ${u}{\left({x}\right)}$ as the inhomogeneous term in the second equation for ${v}{\left({x}\right)}$. Using the variation of parameters on the fundamental set of solutions created by the homogeneous equation, I produced a final solution for a spatial component ${v}{\left({x}\right)}$. I won't type the whole formula; however, it contained an ungainly term ${\int_{{{0}}}^{{{x}}}}{e}^{{-{2}{i}{e}^{{\frac{{{s}}}{{{2}}}}}}}{d}{s}$. Technically, I could still express ${v}{\left({x}\right)}$ as a formula of sines and cosines; however, the ${L}^{{{2}}}$-theory of Fourier series makes no mention of such a basis.
I believe that my question is fairly obvious given the rudimentary construction necessary to propose the equation, and I apologize in advance if the answer is equally obvious.
