Re: Volterra Equation?
- To: mathgroup at smc.vnet.net
- Subject: [mg79474] Re: [mg79405] Volterra Equation?
- From: Angela Kou <Akou at lbl.gov>
- Date: Fri, 27 Jul 2007 05:59:32 -0400 (EDT)
- References: <27772036.1185452319668.JavaMail.root@m35> <op.tv3am2ziqu6oor@monster.gateway.2wire.net>
I'm sorry for the incorrect syntax. I'm very new to mathematica and
have only been using it for a few days. I meant to ask if this equation
is solveable in Mathematica:
b[x]== Integrate[ 1/ ( b[r]+k), { r,x,a}]; , where a is known constant.
Thanks for all of the replies so far. They have been very helpful.
Angela Kou
DrMajorBob wrote:
> 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