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MathGroup Archive 2007

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Re: Simple ODE with time-dep BC

  • To: mathgroup at smc.vnet.net
  • Subject: [mg78045] Re: Simple ODE with time-dep BC
  • From: Joel Storch <jstorch at earthlink.net>
  • Date: Thu, 21 Jun 2007 06:04:05 -0400 (EDT)
  • References: <f5as8d$9kj$1@smc.vnet.net>

The resolution to problem 1 lies in a syntax error. On the right
hand side of the differential equation,replace c/h with c/h[t]

Problem 2 is ill-posed. If h is a function of only one
independent variable (t), then we either have an initial
condition (h(0)=const.) or boundary condition (h(l)=m;
l & m constant).

Apostolos E. A. S. Evangelopoulos wrote:
> Hello all,
> 
> I'm having 2 problems with solving the following ordinary differential equation:
> 
> h*h'[t]==-a*h^2+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...
> 
> 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=h^2)
> 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]v^2},u[t],t]
> resulting in
> DSolve::overdet : The system has fewer dependent variables than equations, so is overdetermined. More...
> 
> Thank you, all, for your help, in advance.
> 
> Apostolos
> 


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