Re: Integral equations
- To: mathgroup at smc.vnet.net
- Subject: [mg36152] Re: [mg36143] Integral equations
- From: Andrzej Kozlowski <andrzej at lineone.net>
- Date: Fri, 23 Aug 2002 21:34:42 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
All of this looks like a mistake to me because it seems far too easy.
But anyway, here is the solution that makes almost no use of
Mathematica. First of all, your equation is not a differential equation
so there is no point using DSolve.
Secondly the use of z in Integrate[\[Sigma]norm[z]^n*z, {z, 0, d}] is
deceptive, since you are integrating over z, so let's replace it by
something else, say s. So your equation is:
(3*z)/d^3)*Integrate[\[Sigma]norm[s]^n*s, {s, 0, d}] ==\[Sigma]norm[z]^n
which is supposed to hold true for every z>0. Re-write it as
Integrate[\[Sigma]norm[s]^n*s, {s, 0, d}]/d^3 =\[Sigma]norm[z]^n/3z
for all z. However, the left hand side is a function of d, independent
of z, so we can write:
\[Sigma]norm[z_]:=(3z*g[d])^(1/n)
Let's take this as a definition and substitute in the original equation
In[2]:=
Simplify[((3*z)*Integrate[\[Sigma]norm[s]^n*s, {s, 0, d}])/d^3 ==
\[Sigma]norm[z]^n, {d > 0, n > 0, z > 0}]
Out[2]=
True
That means you can take g to be an arbitrary function of d.
Andrzej Kozlowski
Toyama International University
JAPAN
On Friday, August 23, 2002, at 05:25 AM, Toshiyuki ((Toshi)) Meshii
wrote:
> Hello,
>
> Hello.
>
> How can I solve the following integral equation?
> Mathematica seems not to work.
> Is there any way?
>
> DSolve[((3*z)/d^3)*Integrate[\[Sigma]norm[z]^n*z, {z, 0, d}] ==
> \[Sigma]norm[z]^n, \[Sigma]norm[z], z]
>
> note: z>0 & n>1
>
> I know that the answer is simple and
> $B&R(Bnorm[z_] = (1 + 1/(2*n))*(z/d)^(1/n)
>
> -Toshi
>
>
>
>
>