Re: Simplify and then discretize a set of equations with derivatives and integrals
- To: mathgroup at smc.vnet.net
- Subject: [mg126693] Re: Simplify and then discretize a set of equations with derivatives and integrals
- From: Matthieu Brucher <matthieu.brucher at gmail.com>
- Date: Thu, 31 May 2012 02:49:38 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
>
> Hi, Matthieu,
>
> Just to give an idea of what can easily be done. You can slightly rewrite
> the system as follows:
>
> Clear[t];
> f[v_] := a*Exp[b*v + c];
> eqns = {e[t] - i[t]*R - vc[t] - vd[t] == 0, vc'[t] == CC^-1*i[t],
> i[t] == f[vd[t]]}
>
> Here I use CC instead of C, since in Mathematica C is reserved. There are
> besides an syntax error in the expression f(v) = a * e^(b * v + c). In
> addition the second equation is differential instead of the integral.
> You may then eliminate the variable i[t]:
>
> expr = Eliminate[eqns, i[t]]
>
> the result is:
>
> a E^(c + b vd[t]) == CC Derivative[1][vc][t] &&
> vd[t] == e[t] - vc[t] - CC R Derivative[1][vc][t] && CC != 0
>
> Its first and second parts are your desired equations. Indeed
>
> expr[[1]]
> expr[[2]]
>
>
> E^(1 + vd[t]) == Derivative[1][vc][t]
>
> vd[t] == Sin[t] - vc[t] - Derivative[1][vc][t]
>
>
> This you may solve numerically, if you fix the function e[t] and
> parameters to be numeric. You can discretize it, but you need not.
> Instead you might better apply the built-in function NDSolve. Have a look
> in Menu/Help/NDSolve. To give an idea, I put below all constants equal to
> 1, e[t]=Sin[t], and numerically solve the obtained system:
>
> e[t_] := Sin[t];
> c = 1; b = 1; CC = 1; a = 1; R = 1;
>
> s = NDSolve[{expr[[1]], expr[[2]], vc[0] == 0}, {vc[t], vd[t]}, {t, 0,
> 30}][[1]] // Quiet
>
> Execute it. The result is expressed in terms of two
> InterpolatingFunctions. To have a look what it gives, execute this:
>
> Plot[Evaluate[{vc[t], vd[t]} /. s], {t, 0, 30}]
>
> Finally, in this problem it was no need to exclude the variable, just
> solve all the three equations at once numerically.
>
> I have chosen the initial values arbitrarily. Execute this:
>
> e[t_] := Sin[t];
> c = 1; b = 1; CC = 1; a = 1; R = 1;
>
> s1 = NDSolve[{eqns, vc[0] == 0, vd[0] == 1, i[0] == 0}, {vc[t], vd[t],
> i[t]}, {t, 0, 30}][[1]] // Quiet
>
> and have a look at the result here:
>
> Plot[Evaluate[{vc[t], vd[t], i[t]} /. s1], {t, 0, 30},
> PlotStyle -> {Red, Green, Blue}]
>
> Here red gives vc, green - vd and blue - i.
>
> Have fun, Alexei
>
> Alexei BOULBITCH, Dr., habil.
> IEE S.A.
> ZAE Weiergewan,
> 11, rue Edmond Reuter,
> L-5326 Contern, LUXEMBOURG
>
> Office phone : +352-2454-2566
> Office fax: +352-2454-3566
> mobile phone: +49 151 52 40 66 44
>
> e-mail: alexei.boulbitch at iee.lu
>
>
>
>
>
Hi Alexei,
Thanks for the corrections and the advice. I will try it today.
Indeed, the plot can be interesting to validate my final implementation of
the problem. At the moment, I need to discretize it so that I can implement
the final equation solver in C++.
Thanks again,
Matthieu
--
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Blog: http://matt.eifelle.com
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