Hi! Can someone please help?(adsbygoogle = window.adsbygoogle || []).push({});

I'm trying to solve the heat equation in polar coordinates. Forgive my way of typing it in, I'm battling to make it look right. The d for derivative should be partial, alpha is the Greek alpha symbol and theta is the Greek theta symbol.

du/dt = (alpha.alpha)[(d/dr)(du/dr)+(1/r)du/dr+(1/(r.r))(d/theta)(du/dtheta)]

This is the heat equation for a disk with radius a.

u(r,theta,0) = (a-r)cos(theta)

u(a,theta,t) = 0

In Mathematica, I used:

NDSolve[{Derivative[0,0,1][r,theta,t]==alpha Derivative[2,0,0][r,theta,t]+alpha Derivative[1,0,0][r,theta,t]+alpha Derivative[0,2,0][r,theta,t], u[r,theta,0] == (a-r)Cos[theta], u[a,theta,t]==0},u,{r,0,a},{theta,0,1},{t,0,1}]

I got an error: NDSolve: bcart

I tried replacing alpha with 2 and a with 4. Still got a problem. Is there a way of keeping the alpha's and a's? And can the radius start from 0, without problems?

Then I need to plot it in 3 dimensions. I tried: (radius starting from 1 since I had problems)

Plot3D[Evaluate[u[r,theta,t] /. First[%]],{r,1,4}, {theta,0,pi}, {t,0,1},PlotPoints -> 50]

I got an error: Plot3D: plnc (repeatedly)

General :: stop

Plot3D[InterpolatingFunction[{{1.,4.},{0.,3.14159},{0.,1.}},<>][r,theta,t],{r,1,4},{theta,0,pi},{t,0,1}, PlotPoints -> 50]

I still need to add a different boundary condition where du/dr(a,theta,t) = 0 and need to solve u(r,theta,t). Would it work the same way?

Please help! I would really appreciate it!

Thanks,

Hop

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# Heat Equation in Polar coordinates in Mathematica

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