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Re: Using NDSolve for 2-variables function ?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg31160] Re: Using NDSolve for 2-variables function ?
*From*: BobHanlon at aol.com
*Date*: Sun, 14 Oct 2001 04:11:53 -0400 (EDT)
*Sender*: owner-wri-mathgroup at wolfram.com
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
Bob Hanlon
Chantilly, VA USA
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