# Solve the boundary-value problem for the Laplace operator in domain.

i need a little help with these Laplace eq or smt which looks like this. I'll be gratefull for any help. Examples or tutorials will be helpful too. $$\left\{\begin{matrix} \Delta u = 0, 2\leq \rho\leq 3\\ u|_{\rho=2} = 1\\ u|_{\rho=3}=4cos(\varphi) \end{matrix}\right.$$

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I am assuming you are using radial coordinates on the plane based on your notation. The form of your question suggests strongly to take the Fourier transform in the angular direction.

First write $$\triangle u = \partial^2_\rho u + \frac1\rho \partial_\rho u + \frac1{\rho^2} \partial^2_{\varphi} u$$ Now write $u$ as a Fourier series in $\varphi$. (I will only write the cosine series in the following since your boundary terms only involve the cosine series). $$u = \sum_{k} u_k(\rho) \cos (k \varphi)$$

Plugging into the equation you get that

$$0 = \sum_k \left(u_k'' + \frac{1}{\rho} u_k' - \frac{k^2}{\rho^2} u_k\right)\cos(k\varphi)$$

which reduces the partial differential equation to a family of ordinary differential equations of second order. Using the prescribed boundary data we have that if $k \neq 0,1$ we must have $u_k \equiv 0$.

For $k = 0$ we have that

$$u_0'' + \frac{1}{\rho}u_0' = 0$$

with boundary data $u_0(2) = 1$ and $u_0(3) = 0$. Solving we get $u'_0(\rho) = \frac{C}{\rho}$ and $u_0(\rho) = C\ln \rho + D$. You can solve for $C,D$ by using the boundary data.

Similarly, for $k = 1$ the equation is

$$u_1'' + \frac{1}{\rho}u_1' - \frac{1}{\rho^2}u_1 = 0$$

with boundary data $u_1(2) = 0$ and $u_1(3) = 4$. Using the polynomial ansatz $u_1 = \rho^m$ we have see that the equation becomes

$$m(m-1) + m - 1 = 0$$

which requires $m = \pm 1$. Verify: if $v = A\rho + B\rho^{-1}$, we get that

$$v'' = 2 B\rho^{-3}\qquad v' = A - B \rho^{-3}$$

and this solves the ODE. The two integration constants $A$ and $B$ can be recovered, again, from the boundary conditions.

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