# How to solve PDE numerically?

Posted 1 month ago
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 $$\frac{\partial T^2}{\partial r^2}+\frac{1}{r}\frac{\partial T}{\partial r}+\frac{1}{r^2}\frac{\partial T^2}{\partial \theta^2}+\lambda T=-\lambda r \sin{\theta}$$ The boundary condition are: $T=0$ at $r=1$ and $T=finite$ at $r=0$ and $\theta$ varying from 0 to 2 Pi.  eqn = D[T[r, \[Theta]], {r, 2}] + 1/r D[T[r, \[Theta]], {r, 1}] + 1/r^2 D[T[r, \[Theta]], {\[Theta], 2}] + \[Lambda]^2 T[ r, \[Theta]] == -\[Lambda] r Sin[\[Theta]] Although this problem i had solved first manually using separation of variable then applying the shooting method. But i want to solve this using Mathematica alone. In separation of variable i assumed $T=F(r) \sin{\theta}$ and then i solved ode using the shooting method. The boundary condition was: $F=0$ at $r=1$ and $F'=0$ at $r=0$. Answer
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Posted 1 month ago
 I tried eqn = D[T[r, \[Theta]], {r, 2}] + 1/r D[T[r, \[Theta]], {r, 1}] + 1/r^2 D[T[r, \[Theta]], {\[Theta], 2}] + \[Lambda]^2 T[ r, \[Theta]] == -\[Lambda] r Sin[\[Theta]] eqn1 = FullSimplify[# r^2] & /@ eqn eqn2 = eqn1 /. \[Lambda] -> 1.2 lsg = NDSolve[ {eqn /. \[Lambda] -> 0, T[0, \[Theta]] == 5, D[T[r, \[Theta]], r] == 0 /. r -> 0}, T, {r, 0, 1}, {\[Theta], 0, 2 Pi}, Method -> {"MethodOfLines", "DifferentiateBoundaryConditions" -> False}] But that doesn't do the job. Unfortunately the error messages do not give any hint how to avoid the division by zero. A question for Support? Answer
Posted 1 month ago
 Yes! the same problem i encountered while i was trying before going for manual way. Yes it need support team help. Answer
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