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I need to solve the following Boundary Value problem

$$\frac{\partial^2w }{\partial x^2}+\frac{\partial^2w }{\partial y^2}=c$$

Boundary conditions are

$$w(x,h)=0$$

$$w\left(\frac{\pm h}{\sqrt3},y\right)=0$$

I am trying to solve it by Homotopy Perturbation Method, but unable to succeed to get desired answer which should be

$$w=\frac{c(y-h)(3x^2-y^2)}{4h}$$

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    $\begingroup$ is $c$ constant? and what's the domain of the problem? $\endgroup$ – Dylan Feb 28 at 14:45
  • $\begingroup$ Yes $c$ is contant,and I think it's in real domain. $\endgroup$ – Ubaid Ur Rehman Mar 2 at 10:45
  • $\begingroup$ It can't be $\Bbb R^2$ since your boundaries are $x=\pm h/\sqrt{3}$ and $y=h$. Either it's $(-h/\sqrt{3},h/\sqrt{3}) \times (h,\infty)$ or $(-h/\sqrt{3},h/\sqrt{3}) \times (-\infty,h)$. I'm asking you which is it? $\endgroup$ – Dylan Mar 2 at 13:34
  • $\begingroup$ its $(-h/sqrt(3), h/sqrt(3)) × (h,∞)$ $\endgroup$ – Ubaid Ur Rehman Mar 6 at 12:54
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You might want to verify your given solution. I don't think it satisfies the equation.

First, let

$$ w(x,y) = \frac{c}{6}(3x^2-h^2) + v(x,y) $$

then $v(x,y)$ is harmonic with boundary conditions

\begin{cases} v\left(\pm \frac{h}{\sqrt3},y\right) = 0 \\ v(x,h) = -\frac{c}{6}(3x^2-h^2) \end{cases}

The solution to this problem isn't unique. If we assume $v(x,y\to\infty)$ is bounded, then separation of variables gives

$$ v(x,y) = \sum_{n=1}^\infty c_n \exp\left(-\frac{n\pi\sqrt 3}{2h}(y-h)\right) \sin\left(\frac{n\pi}{2h}\big(\sqrt3 x+h\big)\right) $$

where

$$c_n = \frac{\sqrt 3}{h}\int_{-h/\sqrt3}^{h/\sqrt3} v(x,h) \sin\left(\frac{n\pi}{2h}\big(\sqrt3 x+h\big)\right) \ dx$$

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