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Re: PDE with RecurrenceTable
*To*: mathgroup at smc.vnet.net
*Subject*: [mg124766] Re: PDE with RecurrenceTable
*From*: Peter Pein <petsie at dordos.net>
*Date*: Sat, 4 Feb 2012 06:33:42 -0500 (EST)
*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com
*References*: <jgg1es$bph$1@smc.vnet.net>
Am 03.02.2012 08:13, schrieb Alexei Boulbitch:
> Dear Community,
>
>
>
> I am trying to make a simple numeric FEM solution of parabolic PDE. It seems that RecurrenceTable function is designed exactly for such a job. Indeed, in the Help/RecurrenceTable/Scope/Partial Difference Equations one finds an example. My problem is that this example is only one, and I would say, it is not basic enough.
>
>
>
> My question: do you know some other examples of the use of the RecurrenceTable for this sort of equations??
>
>
>
> I would like to explain: I already went through the MathGroup archive and have seen numerous posts recommending various sophisticated FEM packages. My question is not about them. I want to learn to make simple programs of this sort myself to fast test an equation at hand.
>
>
>
> For example, here is a classical equation of temperature conductivity:
>
>
>
> (\[PartialD]u(x,t))/\[PartialD]t=a^2*(\[PartialD]^2u(x,t))/\[PartialD]x^2
>
>
>
> with a=0.1, the boundary conditions: u[0,t]==1 and u[1,t]==0 and the initial condition u[x,0]== Cos[3 \[Pi]*x/2];
>
> using the explicit method on the rectangular lattice, taken from a textbook:
>
>
>
> u[j,k+1]=\[Sigma]*u[j+1,k]+(1-2*\[Sigma])*u[j,k]+\[Sigma]*u[j-1,k];
>
> \[Sigma]=(a^2*\[Tau])/h^2;
>
>
>
> Tau and h are temporal and spatial step sizes.
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>
>
> This is the code:
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>
>
> a = 0.1;
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> h = 0.1;
>
> \[Tau] = 0.0001;
>
> \[Sigma] = a^2*\[Tau]/h^2;
>
> lst2 = RecurrenceTable[{u[j,
>
> k + 1] == \[Sigma]*u[j + 1, k] + (1 - 2*\[Sigma])*
>
> u[j, k] + \[Sigma]*u[j - 1, k], u[j, 0] == Cos[3 Pi*j/20.],
>
> u[0, k] == 1, u[10, k] == 0}, u, {j, 0, 10}, {k, 0, 100}];
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>
>
>
>
> Show[{
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> ListPlot3D[lst2,
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> AxesLabel -> {Style["t", 16, Italic], Style["x", 16, Italic],
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> Style["u", 16, Italic]}, PlotStyle -> Blue],
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> Plot3D[0, {t, 0, 100}, {x, 0, 10},
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> PlotStyle -> {Yellow, Opacity[0.4]}]
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> }]
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>
>
> The solution obtained this way, however, does not show evolution. It is clear that with increasing t the temperature, u, should forget its initial form and approach to a straight line. What is wrong?
>
>
>
> Thank you, Alexei
>
>
>
>
>
> Alexei BOULBITCH, Dr., habil.
>
> IEE S.A.
>
> ZAE Weiergewan,
>
> 11, rue Edmond Reuter,
>
> L-5326 Contern, LUXEMBOURG
>
>
>
> Office phone : +352-2454-2566
>
> Office fax: +352-2454-3566
>
> mobile phone: +49 151 52 40 66 44
>
>
>
> e-mail: alexei.boulbitch at iee.lu<mailto:alexei.boulbitch at iee.lu>
Hi Alexei,
it seems to me that your tau (delta-t) and/or maximum k is far too small
to observe the flattening.
I'll use ListConvolve, because it is much faster; the algorithm is still
the same:
a=1/10; h=1/10; \[Tau]=1/10000;
\[Sigma]=a^2*\[Tau]/h^2;
kmax=121000; (* just to get the zero of u[3/4,t] *)
To avoid memory problems when plotting the result, I'll take ~ 100 data
lines:
lst2 = NestList[
Flatten[{1, ListConvolve[{\[Sigma], 1 - 2*\[Sigma], \[Sigma]}, #1],
0}] & ,
Table[Cos[(3.*Pi*j)/20], {j, 0, 10}],
kmax][[1 ;; All ;; Floor[kmax/100]]];
Now the flattening is visible "to the naked eye":
Show[ListPlot3D[lst2, AxesLabel -> {Style["x", 16, Italic],
Style["t", 16, Italic], Style["u", 16, Italic]},
PlotStyle -> {Lighter[Blue], Specularity[White, 22]},
DataRange -> {{0, 1}, {0, kmax*\[Tau]}}, PlotRange -> All],
Plot3D[0, {x, 0, 1}, {t, 0, kmax*\[Tau]},
PlotStyle -> {Yellow, Opacity[0.4]}]]
Peter
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