Re: Volterra Equation?
- To: mathgroup at smc.vnet.net
- Subject: [mg79465] Re: [mg79405] Volterra Equation?
- From: DrMajorBob <drmajorbob at bigfoot.com>
- Date: Fri, 27 Jul 2007 05:54:52 -0400 (EDT)
- References: <27772036.1185452319668.JavaMail.root@m35>
- Reply-to: drmajorbob at bigfoot.com
I'm not sure that's a Volterra Equation... it's no equation at all, with
the missing parentheses. And there's no dependence on x in the integral,
either.
Maybe you meant
Clear[b]
b[x]== Integrate[ 1/ ( b[r]+k), { r,0,x}];
D[ #,x]&/@%
Derivative[1][b][x] == 1/(k + b[x])
solution=DSolve[{Derivative[1][b][x]==1/(k+b[x])},b,x]
{{b -> Function[{x}, -k - Sqrt[k^2] + 2*x + 2*C[1]]}, {b ->
Function[{x}, -k + Sqrt[k^2] + 2*x + 2*C[1]]}}
b[0] should be 0, so we choose the second solution with C[1]=0:
b[x_, k_] = b[x] /. Last@solution /. C[1] -> 0
-k + Sqrt[k^2 + 2 x]
If parentheses went another way, it might go this way:
Clear[b]
b[x]== Integrate[ 1/ b[r]+k, { r,0,x}];
D[ #,x]&/@%
Derivative[1][b][x] == k + 1/b[x]
Off[ InverseFunction::ifun, Solve::"ifun"]
solution= DSolve[ { Derivative[1][b][x]== k+1/b[x]},b,x]
b[0]/. First@%
%/. C[1]->0
{{b -> Function[{x}, (-1 - ProductLog[-E^(-1 - k^2*(x + C[1]))])/k]}}
(-1 - ProductLog[-E^(-1 - k^2*C[1])])/k
0
b[x_, k_] = b[x] /. Last[solution] /. C[1] -> 0
(-1 - ProductLog[-E^(-1 - k^2*x)])/k
Bobby
On Thu, 26 Jul 2007 04:27:36 -0500, Angela Kou <Akou at lbl.gov> wrote:
> Hi:
>
> I was wondering this equation could be solved in Mathematica:
>
> B(x)=Integrate[1/(B(r)+k,r,0,a]
>
> Thanks,
> Angela Kou
>
>
--
DrMajorBob at bigfoot.com