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I have a series of functions defined in my notebook, and then want to use this to solve a diffusion-reaction type equation. At the moment, something like this works:

eqn = D[Ef[r, t], t] - Def5*( D[Ef[r, t], r, r] + (2/(r )) D[Ef[r, t], r]) + q[r ] Ef[r, t];

ic = {Ef[r, 0] == kd, Ef[rn, t] == kd, Ef[ro, t] == kd};
s = NDSolve[{eqn == 0}~Join~ic, Ef, {r, rn, ro}, {t, 0, 60}];
Plot3D[Evaluate[Ef[r, t] /. s], {r, rn, ro}, {t, 0, 60}, 
AxesLabel -> Automatic, MaxRecursion -> 8, PlotPoints -> 32] 

q[r] is a predefined function, and rn and ro are boundaries. Now, the above code works, BUT the boundary conditions are wrong and inconsistent; I want to change them to Neumann-type conditions so that $d(Ef)/dr = 0$ at both ro and rn, but I am having trouble implementing this; any ideas? All help appreciated! This is probably very simple but I'm new to Mathematica and sometimes screw up syntax...


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What kind of trouble are you having exactly? What have you tried? D[Ef[r,t],r]==0 should work –  Szabolcs Apr 23 '13 at 13:52
I've tried something like ic = {Ef[r, 0] == kd, D[Ef[{rn, t}, r]] == 0, D[Ef[{ro, t}, r]] == 0}; but this generates an error "NDSolve::conarg: The arguments should be ordered consistently. " –  DRG Apr 23 '13 at 13:53
If it's as you've written, then the arguments in the boundary conditions are written incorrectly. You have: D[Ef[{rn, t}, r]] whereas you should have D[Ef[rn, t], r] as @Szabolcs has written. –  Jonathan Shock Apr 24 '13 at 2:56
Very strange; doing that yields the error NDSolve::deqn: Equation or list of equations expected instead of True in the first argument ; really not sure what to make of this... –  DRG Apr 24 '13 at 10:42
... returning the argument 'true' means it thinks such a statement is self evident, but it surely wouldn't know that unless it evaluates ND solve first? –  DRG Apr 24 '13 at 10:53
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1 Answer

I got it to work - had to redefine the boundary conditions with Derivative - I'll leave this up here if it's useful to others - working code block is

ic = {Ef[r, 0] == kd, Derivative[1, 0][Ef][rn, t] == 0, 
   Derivative[1, 0][Ef][ro, t] == 0};
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Thanks for posting the solution, it's good for the community to do that. Note that D[Ef[rn,t], rn] also returns Derivative[1, 0][Ef][rn, t]. I prefer the former because it explicitly mentions the variable according to which the differentiation is done (Derivative doens't). –  Szabolcs Apr 24 '13 at 12:41
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