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

[yungman]

Homework Statement

Solve mixed poisson's problem on disk given

[itex]\nabla^2 U= r sin \theta \hbox{ for } 0 <r< \frac{1}{2}[/itex]

[itex]\nabla^2 U= 0 \hbox{ for } \frac{1}{2} < r < 1[/itex]

With given boundary condition [itex]U(1,\theta)=0[/itex]


2. Answer from the solution manual

[tex] U(r,\theta) = \sum_{m=0}^{\infty} \sum_{n=1}^{\infty}\phi_{mn}(r,\theta) [/tex]

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

Helmholtz [tex]\Rightarrow \nabla^2 \phi_{mn}(r,\theta)= -\lambda_{mn}\phi_{mn}(r,\theta) [/tex]

[tex] \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) [/tex]

Using Fourier series expansion:

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

[tex]\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 [/tex]

[tex]-\alpha^2_{1n} B_{1n} = \frac{J_2(\frac{\alpha_{1n}}{2}}{2\alpha_{1n}J^2_2(\alpha_{1n})}[/tex]

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.

 

[yungman]

Anyone please?

I know this is very similar to Fourier series expansion like:

f(x) = h(x) for 0<x<1
f(x) = 0 for 1<x<2

0<x<2

Where the Fourier series expansion only integrate from x=0 to x=1 only and totally ignor the portion of x=1 to x=2.

I never see the prove, this only show up in the work problems. Can anyone show me the prove of this.