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Double integral on array

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Felicity Freeman
Felicity Freeman 2016-7-11
评论： Star Strider 2016-7-11
I have a function dependent on both xp and t, which I need integrated over a meshgrid of X,Z. This is a slightly simplified version of the equation:
z=-0.000150:0.000005:0;
x=0.000300:0.000005:0.000100;
[X,Z]=meshgrid(x,z);
syms xp
syms t
F=@(xp,t) ((1/t).*(exp((-1).*(((X-xp).^2)+(Z.^2))/(4.*t))));
I need to integrate t between 0 and 0.0002, and xp between 0 and 0.000100 across the X,Z array. I know I can make a single integral ArrayValued, but I can't work out the best way to do the double integral. All suggestions welcome! Thanks!

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回答（1 个）

Star Strider
Star Strider 2016-7-11
I would just use trapz:
z=-0.000150:0.000005:0;
x=0.000300:-0.000005:0.000100;
[X,Z]=meshgrid(x,z);
F=@(xp,t) ((1./t).*(exp((-1).*(((X-xp).^2)+(Z.^2))./(4.*t))));
Fmtx = F(X,Z); % Create Matrix
Fint = trapz(trapz(~(isnan(Fmtx)))); % Integrate
To use the Symbolic Toolbox, the correct function is int, not integral.
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Felicity Freeman
Felicity Freeman 2016-7-11
Torsten, You are right. X and Z are constants, whose values are defined in arrays. The integration variables are xp and t. Felicity
Star Strider
Star Strider 2016-7-11
For 'ArrayValued' to work, the integrand has to be an actual array.
For example:
A = @(x,y) [x,x+y;x-y,y];
Q = integral(@(y) integral(@(x)A(x,y) ,0,2, 'ArrayValued',true), 3,5, 'ArrayValued',true)
That works.

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