# I don't agree with the solution manual of a mixed poisson problem

1. May 17, 2010

### yungman

1. The problem statement, all variables and given/known data

Solve mixed poisson's problem on disk given

$\nabla^2 U= r sin \theta \hbox{ for } 0 <r< \frac{1}{2}$

$\nabla^2 U= 0 \hbox{ for } \frac{1}{2} < r < 1$

With given boundary condition $U(1,\theta)=0$

2. Answer from the solution manual

$$U(r,\theta) = \sum_{m=0}^{\infty} \sum_{n=1}^{\infty}\phi_{mn}(r,\theta)$$

$$\phi_{mn}(r,\theta)= J_m(\lambda_{mn}r)[A_{mn}cos (m\theta) + B_{mn}sin (m \theta)]$$

Helmholtz $$\Rightarrow \nabla^2 \phi_{mn}(r,\theta)= -\lambda_{mn}\phi_{mn}(r,\theta)$$

$$\nabla^2 \sum_{m=0}^{\infty} \sum_{n=1}^{\infty}\phi_{mn}(r,\theta) = \sum_{m=0}^{\infty} \sum_{n=1}^{\infty}-\lambda_{mn}J_m(\lambda_{mn}r)[A_{mn}cos (m\theta) + B_{mn}sin (m\theta)] = r sin(\theta)$$

Using Fourier series expansion:

$$\Rightarrow \sum_{n=1}^{\infty} -\alpha^2_{1n} B_{1n} sin(\theta) J_1(\alpha_{1n}) = r sin(\theta)$$

$$\Rightarrow -sin(\theta) \int_0^{\frac{1}{2}} \alpha^2_{1n} B_{1n} J^2_{1n}(\alpha_{1n}r)rdr = \int_0^{\frac{1}{2}}sin(\theta) r^2 J_1(\alpha_{1n}r)dr$$

$$-\alpha^2_{1n} B_{1n} = \frac{J_2(\frac{\alpha_{1n}}{2}}{2\alpha_{1n}J^2_2(\alpha_{1n})}$$

3. My questions:

Why is this treated as only a poisson problem only for 0<r<1/2 and ignor 1/2<r<1.

I thought this is a two parts problem where it is a poisson problem for 0<r<1/2 and Laplace problem for 1/2<r<1.

Last edited: May 18, 2010
2. May 18, 2010