       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:= 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= C[t - c*x]
>
> C is an arbitrary function. Supplying the boundary conditions you
> should be able to
> work out what is going on:
>
>   In:= {u[0, t] == a, u[x, 0] == b}
>   Out= {C[t] == a, C[-c*x] == b}
>
> Cheers,
>     Paul

```

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