Re: Using InterpolatingFunction from NDSolve

*To*: mathgroup at smc.vnet.net*Subject*: [mg58565] Re: Using InterpolatingFunction from NDSolve*From*: "Kevin J. McCann" <kjm at KevinMcCann.com>*Date*: Fri, 8 Jul 2005 00:45:56 -0400 (EDT)*References*: <daitir$su4$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

I added a couple of definitions at the end, but most importantly, I used NIntegrate rather than Integrate. The latter fails in its attempt to find an exact expression for the integral. Kevin \!\(nds = NDSolve[{ D[u\ [t, x], {t, 1}] == D[u\ [t, x], {x, 2}] + u[t, x]\^2 - u[t, x], u[0, x] == Sin[x], u[t, 0] == 0, u[t, 2\ Ï?] == 0}, u[t, x], {t, 0, 1}, {x, 0, 2\ Ï?}]\n Plot3D[Evaluate[u[t, x] /. nds[\([\)\(1\)\(]\)]], {t, 0, 1}, {x, 0, 2\ Ï?}]\n Plot3D[Evaluate[Integrate[\((u[t, x] /. nds[\([\)\(1\)\(]\)])\), {t, 0, s}]], {s, 0, 1}, {x, 0, 2 Ï?}]\n Plot3D[Evaluate[D[u[t, x] /. nds[\([\)\(1\)\(]\)], {x, 2}]], {t, 0, 1}, {x, 0, 2 Ï?}]\n Integrate[D[u[t, x] /. nds[\([\)\(1\)\(]\)], {x, 2}], {t, 0, s}]\n Plot3D[Evaluate[Integrate[D[u[t, x] /. nds[\([\)\(1\)\(]\)], { x, 2}], {t, 0, s}]], {s, 0, 1}, {x, 0, 2 Ï?}]\n\[IndentingNewLine] u[t, x]^2 /. nds[\([1]\)]\n Plot3D[Evaluate[u[t, x]^2 /. nds[\([\)\(1\)\(]\)]], {t, 0, 1}, {x, 0, 2\ Ï?}, AxesLabel -> {t, x, u[t, x]}]\n u2[t_, x_] = u[t, x]^2 /. nds[\([1]\)]\n f[x_, s_] := NIntegrate[u2[t, x], {t, 0, s}]\n Plot3D[f[x, s], {s, 0, 1}, {x, 0, 2 Ï?}]\n f[1, .5]\) TamÃ¡s wrote: > I solved a PDE with NDSolve in Mathematica 5.1. I could plot, > differentiate and integrate the obtained InterpolatingFunction object, > the result being a similar object. I was able to integrate the 2nd > derivative of it. What I need is to integrate the square of the > obtained InterpolatingFunction object (the square itself does not > simplify to such an object). > You can see the details on my homepage: > http://www.math.bme.hu/~tladics/nds.nb > > Every suggestions are welcome! > > Thank you, > TamÃ¡s >