Re: Simple ODE with time-dep BC
- To: mathgroup at smc.vnet.net
- Subject: [mg78032] Re: Simple ODE with time-dep BC
- From: "Apostolos E. A. S. Evangelopoulos" <a.e.a.evangelopoulos at sms.ed.ac.uk>
- Date: Thu, 21 Jun 2007 05:57:24 -0400 (EDT)
Dear all, thanks for your prompt replies.
Sheer carelessness in the first point and meaningless second point!
Cheers,
Apostolos
______________________________________________________
Carl Woll wrote:
> Apostolos E. A. S. Evangelopoulos wrote:
>
>> Hello all,
>>
>> I'm having 2 problems with solving the following ordinary differential equation:
>>
>> h*h'[t]==-a*h2+c
>>
>> Problem 1:
>> Mathematica doesn't like the form of the above:
>> DSolve[{h'[t]\[Equal]-a*h[t]+c/h, h[0]\[Equal]0}, h[t], t]
>> returns
>> DSolve::dvnoarg : The function h appears with no arguments. More...
>>
>>
> So, give h an argument
>
> DSolve[{h'[t] == -a*h[t]+c/h[t], h[0]\[Equal]0}, h[t], t]
> ^^^
>
>> If, now, I reduce the above equation to
>> DSolve[{1/2*u'[t]\[Equal]-a*u[t]+c,u[0]\[Equal]0},u[t],t]
>> (by substitution of u=h2)
>> then Mathematica solves for u[t] without complaints. The thing is, though, I could solve the latter by hand, really, so what's the point? Is Mathematica not supposed to be able to solve the equation in the first form, really (it appears to be non-linear, but intrinsically it's not)?
>>
>> Problem 2:
>> I 'd like to impose a time dependent boundary condition, so, instead of h[0]==0, something like h[t]==v (some constant), or, the more complicated, h[t]==t. How do I solve that? I have tried as follows, with the reduced u[t] form:
>> DSolve[{u'[t]/2\[Equal]-a*u[t]+c,u[t]\[Equal]v2},u[t],t]
>> resulting in
>> DSolve::overdet : The system has fewer dependent variables than equations, so is overdetermined. More...
>>
>>
> How can you solve an ODE for h[t], when one of your "initial conditions" is h[t]==v?
>
> Carl Woll
> Wolfram Research
>
>> Thank you, all, for your help, in advance.
>>
>> Apostolos
>>
>>
>