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I could not find a category of PDE where I can classify this equation and I do not know how to begin to solve it.

$\Delta u - u = 1$ in $\Omega = [0,\pi]\times [0,\pi]$

with the boundary conditions:

$u = 0$ on $[0,\pi]\times\{0\}\cup[0,\pi]\times\{\pi\}$ and

$\frac{\partial u}{\partial \eta} = 0$ on $\{0\}\times[0,\pi]\cup\{\pi\}\times[0,\pi]$

I tried with separation of variables, but without results.

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It is an inhomogeneous elliptic (b/c the operator $1 - \Delta$ is elliptic), second order PDE with constant coefficients with mixed boundary conditions.

Because the domain is a square and you have constant coefficients, an approach using Fourier series should work.

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  • $\begingroup$ Thank you for your answer! Are there any steps to solve this kind of equations? I was reading a lot of pde books, but i didn't found something untill now. $\endgroup$ – vTudor Nov 8 '16 at 15:54
  • $\begingroup$ Make the ansatz: $u(x,y) = \sum_{k,l \in \mathbb{Z}} e^{ikx + ily} \hat{u}_{k,l}$ with coefficients $\hat{u}_{k,l}$ to be determined. You find Fourier transform (the same thing, but on $\mathbb{R}^d$ in almost every pde book (say Rauch's book) and Fourier series in Taylor's first book. $\endgroup$ – mcd Nov 8 '16 at 23:22
  • $\begingroup$ en.wikipedia.org/wiki/Fourier_series#Fourier_series_on_a_square $\endgroup$ – mcd Nov 8 '16 at 23:25
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The conditions given result in a function $u(x,y)$ that is constant in $x$. That is, $u(x,y)=U(y)$. And, $$ U''(y)-U(y)=1,\;\;\; U(0)=0,\; U(\pi)=0. $$ That can be solved with ODE methods and the annihilator method $$ D(D-1)(D+1)U=0. $$ The general solution is $$ U(y)=-1+Be^{y}+Ce^{-y} $$ where the constants $B,C$ are uniquely determined by the endpoint conditions. $$ U(0)=-1+B+C=0 \\ U(\pi)=-1+Be^{\pi}+Ce^{-\pi}=0 $$

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  • $\begingroup$ Excuse me! I don't understand why conditions given that result about $u$. Why $u(x,y)$ is constant in $x$? Can you give more detail about this? Thank you! $\endgroup$ – vTudor Nov 8 '16 at 15:50
  • $\begingroup$ @vTudor : Have you tried to verify that $u(x,y)=U(y)$ is a solution of your original problem? Your first condition is that $u(x,0)=u(x,\pi)$ for $0 \le x \le \pi$. And $u_{x}(0,y)=u_{x}(\pi,y)=0$ for $0 \le y \le \pi$. I believe my $u$ works. $\endgroup$ – Disintegrating By Parts Nov 8 '16 at 16:50

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