Partial differential equation with quasi-periodic boundary conditions Does anybody know a way of solving:
$$\frac{1}{2}\frac{\partial\varphi(x,y)}{\partial x}+\frac{i}{2}\frac{\partial\varphi(x,y)}{\partial y}+\frac{iB y\varphi(x,y)}{2}=0,$$ with boundary conditions:
$$\varphi(x+L,y)\mathrm{e}^{-iBLy}=\varphi(x,y)$$
and
$$\varphi(x,y+L)=\varphi(x,y).$$
I started off with $\varphi(x,y)=\mathrm{e}^{-By^2/2}f(x+iy)$, then looked for a function which solves:
$$f(x+L+iy)\mathrm{e}^{-iBLy}=f(x+iy)\hspace{2cm}(1),$$ and $$f(x+iy+iL)\mathrm{e}^{-BLy-BL^2/2}=f(x+iy)\hspace{2cm}(2).$$ But because of the appearance of the $y$'s especially in $(1)$, it doesn't seem to work. Usually (or in another form of the problem, I should say) we have
$$f(z+L)=f(z)$$ and $$f(z+iL)=f(z)\mathrm{e}^{-i\pi N(2z+iL)}.$$ Which I can solve, because there are no lone $x$ or $y$ terms. Does anybody know of anything that would help?
EDIT:
From my comments I found
$$X(x+L)\mathrm{e}^{-iBLy}=X(x)$$ and $$Y(y+L)=Y(y)$$ and the following
$$\ln(Y)=-\frac{B}{2}y^2-2i\lambda y+g(x)$$ and $$\ln(X)=2\lambda x+s(y)$$ therefore
$$X(x)=\mathrm{e}^{2\lambda x}\mathrm{e}^{s(y)}$$ and $$Y(y)=\mathrm{e}^{-\frac{B}{2}y^2-2i\lambda y}\mathrm{e}^{g(x)}.$$
Using the boundary terms we have
$$\mathrm{e}^{2\lambda x+2\lambda L}\mathrm{e}^{s(y)}\mathrm{e}^{-iBLy}=\mathrm{e}^{2\lambda x}\mathrm{e}^{s(y)}$$ and $$\mathrm{e}^{-\frac{B}{2}y^2-ByL-\frac{BL^2}{2}-2i\lambda y-2i\lambda L}\mathrm{e}^{g(x)}=\mathrm{e}^{-\frac{B}{2}y^2-2i\lambda y}\mathrm{e}^{g(x)},$$ therefore simplifying these we have
$$\mathrm{e}^{2\lambda L}\mathrm{e}^{-iBLy}=1$$ and $$\mathrm{e}^{-ByL-\frac{BL^2}{2}-2i\lambda L}=1.$$
With these it seems that $\lambda$ has to depend on $y$ or $y$ has to been restricted.
 A: Let me denote by $f(x,y)$ your auxiliary function $f(x+iy)$. Notice that your solution makes sense only if the parameters $B$ and $L$ satisfy certain specific conditions. This comes from the fact that using $(1)$ and $(2)$ we can obtain conditions for $f(x,y)$. In fact, replacing $f(x+L,y+L)$ in $(1)$ and then using $(2)$ we have
$$
f(x+L,y+L)=f(x,y+L)e^{iB(y+L)}=f(x,y)e^{iB(y+L)+BLy+BL^2/2}.
$$
On the other hand, replacing $f(x+L,y+L)$ in the second boundary condition first and then using the other one we obtain $$
f(x+L,y+L)=f(x+L,y)e^{BLy+BL^2/2}=f(x,y)e^{iBy+BLy+BL^2/2}
$$
Thus, these boundary conditions can be satisfied only if $
e^{iBL}=1$, which implies that $BL=2k\pi$. This inmediately suggest that you should use Fourier series or Separation Variable's method. In my opinion is always better to try separation of variable first (there is no general rule for this, is a kind of guess but the linearity of the PDE give you the hint of using SV's method). So I tried and it worked, but the solution is quite long, so let me try to give you some ideas of how to do it (in case you do not know): Assume the solution is of the form $$
\varphi(x,y)=X(x)Y(y),
$$
for some functions $X(\cdot)$ only depending on $x\in\mathbb{R}$ and $Y(\cdot)$ only depending on $y\in\mathbb{R}$. Now, replacing our guess on the PDE we obtain that $\varphi$ is a solution if and only if $$
X'Y+iXY'+iByXY=0.
$$
Assuming that $X$ and $Y$ are never zero and dividing by $XY$ we obtain $$
\dfrac{X'}{X}+i\dfrac{Y'}{Y}+iBy=0 \ \iff\ i\dfrac{Y'}{Y}+iBy=-\dfrac{X'}{X}.
$$
But now the left-hand side only depends on $y\in\mathbb{R}$, while the right-hand side only depends on $x$, thus both sides has to be constant: $$
i\dfrac{Y'}{Y}+iBy=-\dfrac{X'}{X}=\lambda,
$$
for some constant $\lambda$, which let you with two decoupled first-order ODEs to solve. I let you the details. I hope my answer helps you.
*Note that, in order to uniquely solve the ODEs, you have to translate your boundary conditions to conditions for the ODEs!
Edit: Let me be a little bit more specific. Notice that both ODEs can be solved by integrating factor. Then applying the boundary conditions you should get some conditions for $\lambda$ (for instance $\lambda =2k\pi$, for $k\in\mathbb{Z}$), which means that for any $k\in\mathbb{Z}$ you have found a solution (let's say $\varphi_k$). Since the PDE is linear, the sum of solutions is also a solution, you can add up all of these $\varphi_k$ to construct the general solution $$
\varphi_{general}=\sum_{k\in\mathbb{Z}}\varphi_k.
$$
Finally, evaluating $\varphi$ in some wisely chosen points $(x,y)$ (like $(0,0)$ for instance) and Fourier Series (the standard formula to identify the coefficients of the series with the usual integrals) you can obtain all the parameters involved in each $\varphi_k$ (each $\varphi_k$ has to depend on some parameter! and the only way to find these parameters is using the original boundary conditions!). 
