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MathGroup Archive 2000

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Re: Solve[wave equation,bound.cond.]

  • To: mathgroup at smc.vnet.net
  • Subject: [mg21583] Re: Solve[wave equation,bound.cond.]
  • From: "N. Shamsundar" <shamsundar at uh.edu>
  • Date: Sat, 15 Jan 2000 02:04:26 -0500 (EST)
  • Organization: MECE-4792, University of Houston
  • References: <85i0ed$1od@smc.vnet.net> <85mn42$280@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

And you can see an impasse created by the way the problem was
posed.  For a solution of the form f(x-ct), where f is any
function of the one variable x-ct, the boundary/initial
conditions require f to have two *different* constant values, a
and b, for all arguments.  Thus, unless a=b, there can be no
solution of the type f(x-ct).

N. Shamsundar
University of Houston

Paul Abbott wrote:
> 
> Karkaletsis Angelos wrote:
> 
> ? I am trying to get the analytical solution of
> ? du/dt+c*du/dx=0,u[0,t]=a,u[x,0]=b, a,b real, but i can't.Does anybody
> ? know?
> 
> Dropping the boundary conditions, you can determine the general
> solution:
> 
>   In[1]:= u[x_, t_] = u[x, t] /.
>    First[DSolve[D[u[x, t], x] + c*D[u[x, t], t] == 0, u[x, t], {x, t}]]
> 
>   Out[1]= C[1][t - c*x]
> 
> C[1] is an arbitrary function. Supplying the boundary conditions you
> should be able to
> work out what is going on:
> 
>   In[2]:= {u[0, t] == a, u[x, 0] == b}
>   Out[2]= {C[1][t] == a, C[1][-c*x] == b}
> 
> Cheers,
>     Paul


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