Mathematica Heat Equation in Polar coordinates in Mathematica

[Hop]

Hi! Can someone please help?
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][u][r,theta,t]==alpha Derivative[2,0,0][u][r,theta,t]+alpha Derivative[1,0,0][u][r,theta,t]+alpha Derivative[0,2,0][u][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

 

[Chestermiller]

Hi! Can someone please help?
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][u][r,theta,t]==alpha Derivative[2,0,0][u][r,theta,t]+alpha Derivative[1,0,0][u][r,theta,t]+alpha Derivative[0,2,0][u][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

These equations are unreadable. Please use LaTex.

 

[pasmith]

Hi! Can someone please help?
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)]

\frac{\partial u}{\partial t} = \alpha^2\left(<br /> \frac{\partial^2 u}{\partial r^2} + \frac 1r \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2}\right)

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][u][r,theta,t]==alpha Derivative[2,0,0][u][r,theta,t]+alpha Derivative[1,0,0][u][r,theta,t]+alpha Derivative[0,2,0][u][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}]

Please use code tags for code. It prevents the parser interpreting [u] as an underline tag.

NDSolve[{Derivative[0,0,1][u][r,theta,t]==alpha Derivative[2,0,0][u][r,theta,t]+alpha Derivative[1,0,0][u][r,theta,t]+alpha Derivative[0,2,0][u][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}]

As I read it, you have \theta running between 0 and 1. Surely you want \theta between 0 and 2\pi?

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?

NDSolve is for numerical solutions. This necessarily involves telling Mathematica what values of \alpha and a you want it to use (but in any event you can remove both of them from the problem by suitable rescalings of r and t). If you want a symbolic solution, use DSolve.

The error message indicates that you have not supplied sufficient boundary conditions. In particular, Mathematica does not know that r and \theta are to be interpreted as polar coordinates, and that therefore u(r,0,t) = u(r,2\pi,t) and that the boundary condition to be imposed at r = 0 is that \frac{\partial u}{\partial r} = 0. But the coordinate singularity at the origin may be fatal in any event: I don't think you can tell NDSolve that the value of u at the origin should not depend on \theta. (Or, if Mathematica can work all that out for itself, you only told it to look at 0 \leq \theta \leq 1 and didn't supply boundary conditions at the edges of that sector.)