Question

Laplace equation ;
d^2/dx^2 + d^2u/dy^2 =0
0 14 years ago

Answer 1

\[\text{By using separation of variables, we assume that $u = X(x) Y (y)$ for some}\]\[\text{functions $X$ and $Y$. Therefore,}\]\[\quad\quad\quad\quad\quad\quad\quad\quad\text{$X''Y + Y''X = 0 \Rightarrow \frac{X''}{X} + \frac{Y''}{Y} = 0 \Rightarrow \frac{X''}{X} = -\frac{Y''}{Y} = K$}\]\[\text{for some constant $K\in\mathbb{R}$. This way, we get}\]\[\text{$\quad\quad\quad\quad\quad\quad\quad\qquad\qquad\qquad X'' = KX \Rightarrow X = C_1 e^{\sqrt{K}x}+C_2e^{-\sqrt{K}x}$}\]\[\text{$\quad\quad\quad\quad\quad\quad\qquad\qquad\qquad Y'' = -K Y \Rightarrow Y = C_3 e^{\sqrt{-K}y}+C_4e^{-\sqrt{-K}y}$}\]\[\text{Let's input the boundary conditions now.}\]\[\begin{eqnarray*}&&\frac{du}{dx}\Big|_{x = 0} = X'(1)Y(y) = \left(C_1\sqrt{K} - C_2\sqrt{K}\right)\left(C_3 e^{\sqrt{-K}y}+C_4e^{-\sqrt{-K}y}\right) = 0\\&\Rightarrow&C_1\sqrt{K} - C_2\sqrt{K} = 0 \Rightarrow C_1 = C_2 \Rightarrow X = \cosh{\left(\sqrt{K}x\right)}\end{eqnarray*}\]\[\text{(the constant is irrelevant for now).}\]\[\text{$u(x, 0) = \cosh\left(\sqrt{K}x\right)\left(C_3 +C_4\right) = 0 \Rightarrow C_3 = -C_4 \Rightarrow Y = \sinh{\left(\sqrt{-K}x\right)}.$}\]\[\begin{eqnarray*}&&\frac{du}{dy}\Big|_{y = 1} = X(x)Y'(1) = \cosh{\left(\sqrt{K}x\right)}\cosh{\left(\sqrt{-K}\right)} = 0 \\ &\Rightarrow&e^{\sqrt{-K}} = -e^{-\sqrt{-K}} \Rightarrow e^{2\sqrt{-K}} = -1 \Rightarrow 2\sqrt{-K} = \pi i+2k\pi i \\ &\Rightarrow& i \sqrt{K} = i\pi\left(\frac{1}{2}+k \right) \Rightarrow \sqrt{K} = \pi\left(k + \frac{1}{2}\right), \quad k \in \mathbb{Z}.\end{eqnarray*}\]\[\text{This yields $X_k(x) = \cosh{\left(\pi\left(k + \frac{1}{2}\right)x\right)}, Y_k(y) = \sin{\left(\pi\left(k + \frac{1}{2}\right)y\right)}.$}\]\[\text{Since each pair $(X_k, Y_k)$ is a solution for three of the conditions (and everything}\]\[\text{is linear), we write $u(x, y) = \sum_{k = 0}^\infty a_k X_k(x) Y_k(y)$ which is still a solution. We have}\]\[\text{to find constants $a_k$ that will make $u(1, y) = u_0$ (this is why the constants were}\]\[\text{irrelevant until now).}\]\[\text{$\sum_{k = 0}^\infty a_k X_k(1)Y_k(y) = \cosh{\left(\pi\left(k + \frac{1}{2}\right)\right)}\sum_{k = 0}^\infty a_k\sin{\left(\pi\left(k + \frac{1}{2}\right)y\right)} = u_0.$}\]\[\text{Let $\tilde{a}_k = \frac{a_k}{X_k(1)}.$ It's easy to see that $\tilde{a}_k$ is the Fourier coffiecient of the}\]\[\text{expansion of $u_0$ in sines. This way,}\]\[\text{$\tilde{a}_k = 2\int_0^1 u_0 \sin{\left(\pi\left(k + \frac{1}{2}\right)y\right)}dy = \frac{4u_0}{\pi(2k+1)}$ and finally} \]\[\text{$u(x, y) = \frac{4u_0}{\pi}\sum_{k = 0}^\infty \frac{\sin{\left(\pi\left(k + \frac{1}{2}\right)y\right)}}{2k+1} \frac{\cosh{\left(\pi\left(k + \frac{1}{2}\right)x\right)}}{\cosh{\left(\pi\left(k + \frac{1}{2}\right)\right)}} .$}\]