Re: Using NDSolve for 2-variables function ?
- To: mathgroup at smc.vnet.net
- Subject: [mg31161] Re: Using NDSolve for 2-variables function ?
- From: BobHanlon at aol.com
- Date: Sun, 14 Oct 2001 04:11:54 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
In a message dated 2001/10/13 4:38:18 PM, writes: >In a message dated 2001/10/10 8:29:49 PM, Florent.Saulnier at college-de-france.fr >writes: > >>I'm trying to solve a quasi-linear PDE using the method of characteristics. >>For this, I need to calculate a function - for instance f[r_,t_] - by > >>NDSolve (I simplified the equation for clarity) and then use it again >in >> >>another differential equation : >> >>Input[1] >>f[r_,t_]=f[r]/.NDSolve[{f'[u]+u*f[u]==0,f[Sqrt[t]]==t^2},f,{u,Sqrt[t],10^9}] >> >>[[1]][[1]] >> >>... gives the following error messages : >> NDSolve::ndnl : Endpoint Sqrt[t] in {u,Sqrt[t],1000000000} is >>not >>a real number >> ReplaceAll::reps : {uf[u]+f'[u]==0} is neither a list of >>replacement rules nor a valid dispatch table, and so cannot be used for >> >>replacing. >>Output[1] Null (f[r]/.uf[u]+f'[u]==0) >> >>If I give the definition of f[r,t] with the sign :=, it gives me the >>correct result at any given point, with the correct boundary conditions >>: >> >>Input[1] >>f[r_,t_]:=f[r]/.NDSolve[{f'[u]+u*f[u]==0,f[Sqrt[t]]==t^2},f,{u,Sqrt[t],10^9} >> >>][[1]][[1]] >>Out[1] Null^2 >>Input[2] f[3,5] >>Out[2] 3.3834 >>Input[3] f[2,4] >>Out[3] 16 >> >>The main problem is that I need f[r,t] for a second equation, and of course >> >>its resolution cannot be achieved : >> >>Input[1] g[t_]=h[t]/.NDSolve[{h'[u]-f[h[u],u]==0,h[1]==1},h,{u,1,10}][[1]][[1]] >> >> ...which gives the same error messages : >> NDSolve::ndnl : Endpoint Sqrt[t] in {u,Sqrt[t],1000000000} is >>not >>a real number >> ReplaceAll::reps : {uf[u]+f'[u]==0} is neither a list of >>replacement rules nor a valid dispatch table, and so cannot be used for >> >>replacing. >> >>Could you please help me about these problems ? >>Is there any other instructions or objects I could use for it ? >> > >soln = (f[u] /. DSolve[f'[u]+u*f[u] == 0, f[u], u][[1]]) > >C[1]/E^(u^2/2) > >soln /. Solve[(soln /. u -> Sqrt[t]) == t^2, C[1]][[1]] > >E^(t/2 - u^2/2)*t^2 > >f[u_, t_] := t^2 * Exp[(t-u^2)/2]; > >D[f[u,t],u]+u*f[u,t] == 0 > >True > >f[Sqrt[t], t] == t^2 > >True > >f[3,5] > >25/E^2 > >%//N > >3.3833820809153177 > >f[2,4] > >16 > I hadn't checked whether Mathematica would handle the boundary conditions directly. It does. Just use: Clear[f,u,t]; f[u_, t_] := Evaluate[Simplify[f[u, t] /. DSolve[{D[f[u,t],u]+u*f[u,t] == 0, f[Sqrt[t],t] == t^2}, f[u,t], u][[1]]]]; Bob Hanlon Chantilly, VA USA