Alan0354
Full Member level 4
This is a question in the book to solve Heat Problem
\[\frac{\partial \;u}{\partial\; t}=\frac{\partial^2 u}{\partial\; r^2}+\frac{1}{r}\frac{\partial\; u}{\partial\;r}+\frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2}\]
With 0<r<1, \[0<\theta<2\pi\], t>0. And \[u(1,\theta,t)=\sin(3\theta),\;u(r,\theta,0)=0\]
The solution manual gave this which I don't agree:
**broken link removed**
What the solution manual did is for \[u_1\], it has to assume \[\frac{\partial \;u}{\partial\; t}=0\] in order using Dirichlet problem to get (1a) shown in the scanned note.
I disagree.
I think it should use the complete solution shown in (2a), then let t=0 where
\[u_{1}(r,\theta,0)=\sum_{m=0}^{\infty}\sum_{n=1}^{\infty}J_{m}(\lambda_{mn}r)[a_{mn}\cos (m\theta)+b_{mn}\sin (m\theta)]\]
I don't agree with the first part because you don't assume constant temperature. Please explain to me.
Thanks
\[\frac{\partial \;u}{\partial\; t}=\frac{\partial^2 u}{\partial\; r^2}+\frac{1}{r}\frac{\partial\; u}{\partial\;r}+\frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2}\]
With 0<r<1, \[0<\theta<2\pi\], t>0. And \[u(1,\theta,t)=\sin(3\theta),\;u(r,\theta,0)=0\]
The solution manual gave this which I don't agree:
**broken link removed**
What the solution manual did is for \[u_1\], it has to assume \[\frac{\partial \;u}{\partial\; t}=0\] in order using Dirichlet problem to get (1a) shown in the scanned note.
I disagree.
I think it should use the complete solution shown in (2a), then let t=0 where
\[u_{1}(r,\theta,0)=\sum_{m=0}^{\infty}\sum_{n=1}^{\infty}J_{m}(\lambda_{mn}r)[a_{mn}\cos (m\theta)+b_{mn}\sin (m\theta)]\]
I don't agree with the first part because you don't assume constant temperature. Please explain to me.
Thanks
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