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I'm looking for the way to substitute variables in interpolated function again to original one.

(I want to know how to get P(r, theta) from Psol(v, w).)

The original variables : 

    r , theta (d<=r<=Infinity, 0<=theta<=Pi).

New variables :

    v , w (0<=v<=1, 0<=w<=1).

The relationship is

Relations of variables.

I now have the interpolated solution Psol(v, w). Psol(v,w) has no analytic expression, and was interpolated from numerical calculation.

The final goal is to get P(r=d), after angle-averaging the P(r,theta)

Angle-averaged P.

I want to know how to get P(r, theta) from Psol(v, w) using above variable-to-variable relationship.

May I have your advice?

Thanks in advance.

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  • $\begingroup$ Why not replace the $v$ and $w$ in your Psol with the expressions you have? Your question could be a bit more concrete; we don't even know what your "interpolated solution" looks like. $\endgroup$ – J. M. will be back soon♦ Aug 8 '12 at 5:06
  • $\begingroup$ You can make a simple change of variable in the integral P(r) itself and have it re-written terms of v and w. $\endgroup$ – Vitaliy Kaurov Aug 8 '12 at 5:12
  • $\begingroup$ @J.M. Thank you for your advice. :) $\endgroup$ – Jaehoon Kim Aug 8 '12 at 12:01
  • $\begingroup$ @VitaliyKaurov, Thanks for your advice. :) $\endgroup$ – Jaehoon Kim Aug 8 '12 at 12:02
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Something like:

sol = NDSolve[{D[p[v, w], v] == D[p[v, w], w, w], p[0, w] == 0, 
p[v, 0] == Sin[v], p[v, 5] == 0}, p, {v, 0, 10}, {w, 0, 5}]

enter image description here

Plot3D[Evaluate[p[v, w] /. %], {v, 0, 10}, {w, 0, 5}, PlotRange -> All]

enter image description here

pt[r_, \[Theta]_, rc_, d_, \[Gamma]_] /; d <= r := (p[v, w] /. sol ) /. 
{v -> (Exp[-\[Gamma] rc/r] - Exp[-\[Gamma] rc/d])/(1 - Exp[-\[Gamma] rc/d]), 
w -> \[Theta]/Pi};
Plot3D[{pt[r, \[Theta], .9, 0.3, 1], pt[r, \[Theta], .1, 0.3, 1]}, 
{r, 0.3, 3}, {\[Theta], 0, Pi}, PlotRange -> All]

enter image description here

?

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  • $\begingroup$ Yes. Your code helped me. It works nice. Thank you. :) $\endgroup$ – Jaehoon Kim Aug 8 '12 at 7:08
  • $\begingroup$ @JaehoonKim, my pleasure. Welcome to MathematicaSE. $\endgroup$ – kglr Aug 8 '12 at 7:14

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