Partial Differential equations: Laplace equation extra terms

Question

I have doubts about applying separation of variables for this problem: enter image description here

Answer 1

The variables may still be seperated, viz.

$u(x, y) = f(x)g(y); \tag 1$

then

$u_{xx} = f''(x) g(y), \tag 2$

$u_{yy} = f(x)g''(y), \tag 3$

whence

$f''(x)g(y) + f(x) g''(y) + 4\pi^2 f(x)g(y) = u_{xx} + u_{yy} + 4 \pi^2 u = 0; \tag 4$

when we divide by $f(x)g(y)$ we obtain

$\dfrac{f''(x)}{f(x)} + \dfrac{g''(y)}{g(y)} + 4\pi^2 = 0, \tag 5$

or

$\dfrac{f''(x)}{f(x)} + 4\pi^2 =-\dfrac{g''(y)}{g(y)}; \tag 6$

since the left-hand side depends only on $x$ and the right only on $y$ we see that there is some $\lambda \in \Bbb R$ with

$\dfrac{f''(x)}{f(x)} + 4\pi^2 = \lambda = -\dfrac{g''(y)}{g(y)}, \tag 7$

leading to

$f''(x) + (4\pi^2 - \lambda) f(x) = 0, \tag 8$

$g''(y) + \lambda g(y) = 0; \tag 9$

(8) and (9) may then be solved by the usual techniques once suitable boundary conditions are imposed.

Answer 2

Since the equation is homogenous in $u(x,y)$ with constant coefficients, it is tendering to transform it into the frequency domain, where it becomes completely algebraic. To do so, we multiply by $e^{-ikx-ily}$, integrate over $x$ and $y$ and integrate by parts twice $$\int_{-\infty}^\infty {\rm d}x \int_{-\infty}^\infty {\rm d}y \, e^{-ikx-ily} \left( u_{xx} + 4u_{yy} + 4\pi^2 u \right) = 0 \\ \Rightarrow \quad \left(-k^2-4l^2+4\pi^2\right) U(k,l) = 0$$ which implies $$U(k,l)=U_+(l) \, \delta\left(k-\sqrt{4\pi^2-4l^2}\right)+U_-(l) \, \delta\left(k+\sqrt{4\pi^2-4l^2}\right)$$ for arbitrary functions $U_+,U_-$. Then, by the Fourier-Inverse $$u(x,y)=\frac{1}{(2\pi)^2} \int_{-\infty}^\infty {\rm d}k \int_{-\infty}^\infty {\rm d}l \, e^{ikx+ily} \, U(k,l) \\ =\int_{-\infty}^\infty {\rm d}l \, e^{ily} \left( U_+(l) \, e^{ix\sqrt{4\pi^2-4l^2}} + U_-(l) \, e^{-ix\sqrt{4\pi^2-4l^2}} \right) \tag{1}$$ where the factor of $1/(2\pi)^2$ was absorbed in the functions $U_{\pm}$.

We now turn to the boundary conditions using this general form of $u(x,y)$:

(i) $$u_x(0,y)=i\int_{-\infty}^\infty {\rm d}l \, e^{ily} \sqrt{4\pi^2-4l^2} \left( U_+(l) - U_-(l) \right) = 0 \\ \Rightarrow \quad U_+(l)=U_-(l) + c_+ \,\delta(l-\pi) + c_- \,\delta(l+\pi) \, .$$ We can set $c_+=c_-=0$, since these solutions are contained in the general function $$U_+(l)=U_-(l)=U(l) \, .$$ Absorbing another factor of 2 in $U(l)$, (1) becomes $$u(x,y)=\int_{-\infty}^\infty {\rm d}l \, e^{ily} \cos\left( x\sqrt{4\pi^2-4l^2} \right) \, U(l) \tag{2} \, .$$

(ii) $$u_x(1,y)=-\int_{-\infty}^\infty {\rm d}l \, e^{ily} \sqrt{4\pi^2-4l^2} \sin\left( \sqrt{4\pi^2-4l^2} \right) \, U(l)=0 \\ \Rightarrow \quad U(l)=\sum_n U_{n,\pm} \, \delta\left(l\mp\sqrt{\pi^2-\left(\frac{n\pi}{2}\right)^2}\right)$$ where the sum runs over all integers. Hence, $$u(x,y)=\sum_n \cos(n\pi x) \left( U_{n,+} \, e^{iy\sqrt{\pi^2-\left(\frac{n\pi}{2}\right)^2}} + U_{n,-} \, e^{-iy\sqrt{\pi^2-\left(\frac{n\pi}{2}\right)^2}} \right) \tag{3} \, .$$

(iii) $$u(x,0)=\sum_n \cos(n\pi x) \left( U_{n,+} + U_{n,-} \right) = 1 \\ \Rightarrow \quad U_{0,+} + U_{0,-} = 1 \quad , \quad U_{n,-} = - U_{n,+} \qquad \forall n\neq 0 \, .$$ This leads to $$u(x,y)=c_{0} \, e^{i\pi y} + (1-c_{0}) \, e^{-i\pi y} + \sum_{n\neq 0} c_n \cos(n\pi x) \sin\left( y\sqrt{\pi^2-\left(\frac{n\pi}{2}\right)^2} \right) \tag{4}$$ for some constants $c_n$.

(iv) Finally $$u_y(x,1)=-i\pi c_0 + i\pi(1-c_0) + \sum_{n\neq 0} c_n \cos(n\pi x) \sqrt{\pi^2-\left(\frac{n\pi}{2}\right)^2} \cos\left( \sqrt{\pi^2-\left(\frac{n\pi}{2}\right)^2} \right) \\ =2 + \cos(2\pi x) + \cos(3\pi x) \, ,$$ from which it is evident, that $$c_0 = \frac{i}{\pi} + \frac{1}{2} \\ c_3=-\frac{2i/\sqrt{5}}{\pi\cosh\left(\pi\sqrt{5}/2\right)} \\ c_2=\lim_{n\rightarrow 2} \frac{1}{\sqrt{\pi^2-\left(\frac{n\pi}{2}\right)^2}} \\ c_n=0 \qquad \text{otherwise} \, .$$ The equation for $c_2$ is understood in the sense, that the corresponding term in $u(x,y)$ is $$\lim_{n\rightarrow 2} \frac{\sin\left(y\sqrt{\pi^2-\left(\frac{n\pi}{2}\right)^2}\right)}{\sqrt{\pi^2-\left(\frac{n\pi}{2}\right)^2}} \, \cos(2\pi x) = y\cos(2\pi x) \, .$$

Therefore we arrive at the solution $$u(x,y)=\cos(\pi y) - \frac{2}{\pi} \, \sin(\pi y) + y\cos(2\pi x) + \frac{2\, \cos(3\pi x) \, \sinh\left(\pi y \sqrt{5}/2\right)}{\pi \sqrt{5} \, \cosh\left(\pi\sqrt{5}/2\right)} \, . \tag{5}$$

While the above derivation is not very rigorous, it can be made rigorous by using complex analysis with analytic $U(l)$ and instead of the delta functions, the integral is evaluated in terms of the pole structure of $U(l)$ by the residue theorem (assuming sufficient decay at $\infty$ for convergence). The pole structure follows from the boundary conditions in the order (i), (iii), (ii), (iv).