Double integral problem with function handles to make the program faster instead of using symbolic

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Ilke
Ilke 2024-11-3，21:53
评论： Walter Roberson 2024-11-3，23:32
Ran in:
Hello Sir, I am trying to solve a plate problem with Rayleigh. Instead of doing all the double integrals with symbolic functions, i would like to make it with function handles. Symbolic tools are taking so much. May you please check me my code? There is a vector (lets assume, it is 3*1) I would like to take each of elements, respectively and make it double integral (From ksi=-1 to1, ita =-1 to 1). But it is giving error.I really preciate if you could help me to take integral from ksi =-1 to 1, and ita=-1 to 1. My regards.
Here is my code:
clc
clear all
i=3; %actually this will be loop from i=1: 10 (but i will not store)
dTksi = @(ksi,ita) [0;1;(1*(1+ksi)/2)];
dTita =@(ksi,ita,a) [0;1;1*(1*(1+ita)/2)];
getRow = @(data, rowNum) data(rowNum, :); % Helper function
P00_Q00= @(ksi,ita) ((getRow(dTksi(ksi,ita), i)).*(getRow(dTita(ksi,ita), i)));
ff=(int(int(P00_Q00(ksi,ita),-1,1),-1,1)) % it is not calculating.
Unrecognized function or variable 'ksi'.
dfuita=@(ksi,ita) (-2*ita);
ddfuksi_P00_Q00= @(ksi,ita) dfuita(ksi,ita).*P00_Q00(ksi,ita); %it is calculating
%fun = integral2(@(ksi,ita) dfuita(ksi,ita).*(1-ita.^2),-1,1,-1,1)
fun=(int(int(dfuita(ksi,ita).*P00_Q00(ksi,ita),-1,1),-1,1)) % it is not calculating.

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Walter Roberson
Walter Roberson 2024-11-3，22:04
移动：Walter Roberson 2024-11-3，22:04
ksi and ita are undefined at that point. If you define them with syms then the integration works.
syms ksi ita
i=3; %actually this will be loop from i=1: 10 (but i will not store)
dTksi = @(ksi,ita) [0;1;(1*(1+ksi)/2)];
dTita =@(ksi,ita,a) [0;1;1*(1*(1+ita)/2)];
getRow = @(data, rowNum) data(rowNum, :); % Helper function
P00_Q00= @(ksi,ita) ((getRow(dTksi(ksi,ita), i)).*(getRow(dTita(ksi,ita), i)));
ff=(int(int(P00_Q00(ksi,ita),-1,1),-1,1)) % it is not calculating.
dfuita=@(ksi,ita) (-2*ita);
ddfuksi_P00_Q00= @(ksi,ita) dfuita(ksi,ita).*P00_Q00(ksi,ita); %it is calculating
%fun = integral2(@(ksi,ita) dfuita(ksi,ita).*(1-ita.^2),-1,1,-1,1)
fun=(int(int(dfuita(ksi,ita).*P00_Q00(ksi,ita),-1,1),-1,1)) % it is not calculating.
  2 个评论
Ilke
Ilke 2024-11-3，22:57
移动：Walter Roberson 2024-11-3，23:27
So thankful, i thought, i should not use syms ,cause to make it slower. But now, it worked. Thanks :)
Walter Roberson
Walter Roberson 2024-11-3，23:32
Ran in:
If you try to switch to numeric integration
i=3; %actually this will be loop from i=1: 10 (but i will not store)
dTksi = @(ksi,ita) [0;1;(1*(1+ksi)/2)];
dTita =@(ksi,ita,a) [0;1;1*(1*(1+ita)/2)];
getRow = @(data, rowNum) data(rowNum, :); % Helper function
P00_Q00= @(ksi,ita) ((getRow(dTksi(ksi,ita), i)).*(getRow(dTita(ksi,ita), i)));
ff = integral2(P00_Q00,-1,1,-1,1)
Error using vertcat
Dimensions of arrays being concatenated are not consistent.

Error in solution>@(ksi,ita)[0;1;(1*(1+ksi)/2)] (line 3)
dTksi = @(ksi,ita) [0;1;(1*(1+ksi)/2)];

Error in solution>@(ksi,ita)((getRow(dTksi(ksi,ita),i)).*(getRow(dTita(ksi,ita),i))) (line 6)
P00_Q00= @(ksi,ita) ((getRow(dTksi(ksi,ita), i)).*(getRow(dTita(ksi,ita), i)));

Error in integral2Calc>tensor (line 240)
Z = FUN(X,Y); NFE = NFE + 1;

Error in integral2Calc>integral2t (line 55)
[Qsub,esub,FIRSTFUNEVAL,NFE] = tensor(thetaL,thetaR,phiB,phiT,[],[], ...

Error in integral2Calc (line 9)
[q,errbnd] = integral2t(fun,xmin,xmax,ymin,ymax,optionstruct);

Error in integral2 (line 105)
Q = integral2Calc(fun,xmin,xmax,yminfun,ymaxfun,opstruct);
This is because integral2() passes in arrays of values for the parameters, so the [0;1;(1*(1+ksi)/2)] would try to horzcat between 0, 1, and an array of values induced by ksi, and that would fail.
You would have to switch to using arrayfun() in P00_Q00 which will slow things down notably.

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