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Re: NDSolve[] with nested If[] and Piecewise[] usage:

• To: mathgroup at smc.vnet.net
• Subject: [mg90763] Re: [mg90743] NDSolve[] with nested If[] and Piecewise[] usage:
• From: Oliver Ruebenkoenig <ruebenko at wolfram.com>
• Date: Wed, 23 Jul 2008 05:58:26 -0400 (EDT)
• References: <200807220758.DAA13813@smc.vnet.net>

Hi Gopinath,

here are some suggestions:

funifcase2[t_] := 
  If[0 <= t < sep/len, val1 + val1 xv yv^2 (y[t]) Sin[t], 
   If[sep/len <= t <= (2 sep)/len, 
    val1 + val2 + val1 xv yv^2 (y[t]) Sin[t] + 
     val2 yv xv^2 (y[t]) Sin[t - sep/len], 
    If[(2 sep)/len < t <= 1, 
     val2 + val2 yv xv^2 (y[t]) Sin[t - sep/len]]]];

in this case I have replaced the internal function call to wdef1[t] and 
wdef2[t] with y[t].

If you look at 

?funifcase2

you will find quite a lot of stuff that is still unevaluated. So, here is 
a second version. In this case I a generate a function which has 
evaluated as much as is possible.


createFun = 
 Function[ t, Evaluate[ Which[ 
    Evaluate[ 0 <= t < sep/len], 
    Evaluate[ val1 + val1 xv yv^2 (wdef1[t]) Sin[t] ],
    Evaluate[ sep/len <= t <= (2 sep)/len ], 
    Evaluate[ 
     val1 + val2 + val1 xv yv^2 (wdef1[t]) Sin[t] + 
      val2 yv xv^2 (wdef2[t]) Sin[t - sep/len] ],
    Evaluate[ (2 sep)/len < t <= 1 ], 
    Evaluate[ val2 + val2 yv xv^2 (wdef2[t]) Sin[t - sep/len] ]
    ]  ] ]


?createFun

Function[t, 
 Which[0 <= t < 1/3, 50 + Sin[t] y[t], 1/3 <= t <= 2/3, 
  70 - Sin[1/3 - t] y[t] + Sin[t] y[t], 2/3 < t <= 1, 
  20 - Sin[1/3 - t] y[t]]]

Also, I used Which, which I find a little more readable. 

solifcase2 = 
 NDSolve[{y''[t] + y'[t] + y[t] - createFun[t] == 0, y[0] == 0, 
   y'[0] == 1}, y, {t, 0, 1} ]
Plot[Evaluate[{y[t], y'[t]} /. solifcase2], {t, 0, 1}, 
 PlotStyle -> {Black, {Red, Dashed}}]

Hope this helps,

Oliver

On Tue, 22 Jul 2008, Gopinath Venkatesan wrote:

> Hello Mathematica Friends,
> 
> I am stuck with this problem, where I use NDSolve[] to solve a nested If[] statement. The use of Piecewise[] solves with no problem, but wanted to know if I can solve the If[] part as well.
> 
> The sample code (just for demo) is shown below, which is similar to my case. The third section is similar to my case, while the first and second case, I kept it here, to show that NDSolve[] solves the nested If[] with no problem. So you can directly go to the third section, please see the comments to go to the correct section. The reason I am posting it here, is even the Piecewise[] is not solving in my (original problem) case.
> 
> Please suggest me ways to solve this third case with nested If[]. Thank you.
> 
> Regards,
> Gopinath
> University of Oklahoma
> 
> (* code starts here *)
> 
> (* First section: Showing simple sine function inside NDSolve[] *)
> funReg[t_] := Sin[t];
> solReg = NDSolve[{y''[t] + y'[t] + y[t] - funReg[t] == 0, y[0] == 0, 
>     y'[0] == 1}, y, {t, 0, 1}];
> Plot[Evaluate[{y[t], y'[t]} /. solReg], {t, 0, 1}, 
>  PlotStyle -> {Black, {Red, Dashed}} ]
> (* Here no particular time value is required for the \
> function,i.e.Sin[t] is valid for any time t *)
> (* Second Section: Nested If[] inside NDSolve[] *)  
> funifcase1[t_] := 
>   If[0 <= t < 1/3, Sin[t], 
>    If[1/3 <= t <= 2/3, -1, If[2/3 < t <= 1, Cos[t]]]];
> Print["funifcase1 is given by ", funifcase1[t]];
> solifcase1 = 
>   NDSolve[{y''[t] + y'[t] + y[t] - funifcase1[t] == 0, y[0] == 0, 
>     y'[0] == 1}, y, {t, 0, 1}];
> Plot[Evaluate[{y[t], y'[t]} /. solifcase1], {t, 0, 1}, 
>  PlotStyle -> {Black, {Red, Dashed}}]
> (* In the above also, the function is undetermined (I mean, dependent \
> on time t, which is not supplied until the NDSolve[] increments from \
> initial conditions step by step by some direct integration scheme, I \
> guess *)
> (* Third case: I expect this case also to solve because it is very \
> similar to the above cases *) 
> sep = 1;
> len = 3;
> wdef1[t_] := y[t];
> wdef2[t_] := y[t];
> val1 = 50;
> val2 = 20;
> xv = 1/2;
> yv = 1/5;
> funifcase2[t_] := 
>   If[0 <= t < sep/len, val1 + val1 xv yv^2 (wdef1[t]) Sin[t], 
>    If[sep/len <= t <= (2 sep)/len, 
>     val1 + val2 + val1 xv yv^2 (wdef1[t]) Sin[t] + 
>      val2 yv xv^2 (wdef2[t]) Sin[t - sep/len], 
>     If[(2 sep)/len < t <= 1, 
>      val2 + val2 yv xv^2 (wdef2[t]) Sin[t - sep/len]]]];
> funifcase3[t_] := 
>   Piecewise[{{val1 + val1 xv yv^2 (wdef1[t]) Sin[t], 
>      0 <= t < sep/len}, {val1 + val2 + 
>       val1 xv yv^2 (wdef1[t]) Sin[t] + 
>       val2 yv xv^2 (wdef2[t]) Sin[t - sep/len], 
>      sep/len <= t <= (2 sep)/len}, {val2 + 
>       val2 yv xv^2 (wdef2[t]) Sin[t - sep/len], (2 sep)/len < t <= 
>       1}}];
> Chop[Table[funifcase2[t], {t, 0, 1, 0.1}]]
> Print["compare values, just to check the correctness of equation \
> above"];
> Chop[Table[funifcase3[t], {t, 0, 1, 0.1}]]
> Print["The definition funifcase2[t] is ", funifcase2[t]];
> Print["The definition funifcase3[t] is ", funifcase3[t]];
> 
> solifcase2 = 
>   NDSolve[{y''[t] + y'[t] + y[t] - funifcase2[t] == 0, y[0] == 0, 
>     y'[0] == 1}, y, {t, 0, 1}]; 
> (* Plot[Evaluate[{y[t],y'[t],y''[t]}/.solifcase2],{t,0,1},PlotStyle-> \
> Automatic] *)
> Print["Proceeding to solve the above equation with Piecewise \
> definition"];
> solifcase3 = 
>   NDSolve[{y''[t] + y'[t] + y[t] - funifcase3[t] == 0, y[0] == 0, 
>     y'[0] == 1}, y, {t, 0, 1}];
> Plot[Evaluate[{y[t], y'[t]} /. solifcase3], {t, 0, 1}, 
>  PlotStyle -> {Black, {Red, Dashed}}]
> (* However the nested If[] does not solve, but the Piecewise[] does. \
> But both are same definitions and same set of equations inside of \
> NDSolve[]. And I expect similar behavior from NDSolve[] for these \
> cases *)
> 
> (* code ends here *)
> 

• References:
  • NDSolve[] with nested If[] and Piecewise[] usage:
    • From: Gopinath Venkatesan <gopinathv@ou.edu>