       Re: Solving integral equations numerically

• To: mathgroup at smc.vnet.net
• Subject: [mg100067] Re: Solving integral equations numerically
• From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
• Date: Fri, 22 May 2009 01:51:36 -0400 (EDT)
• Organization: Uni Leipzig
• References: <gv2jgr\$977\$1@smc.vnet.net>

```Hi,

for what do you need NIntegrate[] ? Your first integral

Integrate[T^2*Exp[-ta/T], {x, 0, l}] -> (l*T^2)/E^(ta/T)

and you second one

Integrate[T^2*Exp[-ta/T]*x, {x, 0, l/2}] -> (l^2*T^2)/(8*E^(ta/T))

if I have overlooked something, usual a definition like
I1[qref_?NumericQ] := (RCOM + RS0)*w*qref*Exp[ta/273]/273^2*
NIntegrate[T^2*Exp[-ta/T], {x, 0, l}]
I2[qref_?NumericQ] := (RS/l)*w*qref*Exp[ta/273]/273^2*
NIntegrate[T^2*Exp[-ta/T]*x, {x, 0, l/2}]

FindRoot[{Tlo == T0 + I1[qref], T0^2/ta == Tlo + I2[qref]}, {{Tlo, 270},
{qref, 300}}]

should be sufficient.

Regards
Jens

viehhaus wrote:
> Hi,
>
> I'm trying to solve a set of two integral equations, which don't have an analytic solution, so I'm using NIntegrate and FindRoot
>
> nodes = 11 (*number of nodes*)
> RCOM = 2.17 (* common resistance *)
> RS0 = 0.1 (*off-sensor resistance for sensor*)
> RS = 1.1 (*sensor resistance pre node*)
> Ph = 6 (*hybrid power in W*)
> tcool = -25 (*coolant temperature (degC) *)
> ta = 1.2/2/0.0000862 (*activation temperature (K) *)
> l = 63.56/1000(*length of thermal path*)
> w = 128.05/1000(*width of thermal path*)
> T0 = 273 + tcool + Ph*RCOM
> T = 4/3*(Tlo - T0^2/ta)*(x^2/l^2 - 2*x/l) + Tlo (*parabolic temperature function  in sensor*)
> I1 = (RCOM + RS0)*w*qref*Exp[ta/273]/273^2*NIntegrate[T^2*Exp[-ta/T], {x, 0, l}]
> I2 = (RS/l)*w*qref*Exp[ta/273]/273^2*NIntegrate[T^2*Exp[-ta/T]*x, {x, 0, l/2}]
> FindRoot[{Tlo == T0 + I1, T0^2/ta == Tlo + I2}, {{Tlo, 270}, {qref, 300}}]
>
> but I get errors like "... has evaluated to non-numerical values for all sampling points in the region with boundaries {{0,0.06356}}..." and FindRoot does not move. Any idea what I'm doing wrong?
>
> Cheers, Georg
>

```

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