Error using integral2

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0 个投票

I have following expression with

x=[1 2 3 4 5]
y=[1 2 3 4 5]
L = 100; % Length of the box
p=1;
q=1;
Ui = @(x,y)  sin(pi*x).*sin(pi*y); % An initial distribution for u, i.e. the u(x,y,0)
Vi = @(x,y)  sin(pi*x).*sin(pi*y); % An initial distribution for v, i.e. the v(x,y,0)
Wi = Ui(x,y) + Vi(x,y); % Initial condition for w, i.e. w(x,y,0)
% Computation of the b-coefficients
if p < Ntrunc
	if q < Ntrunc
		my_integrand = @(x,y)  4*Wi.*sin(p.*x).*cos(q.*y)/L^2; 
        bd(p,q) = integral2(my_integrand(x,y),0,L,0,L);
		q = q+1;
	end
	p = p+1;
end 

I got following error,

Error using integral2Calc>integral2t/tensor (line 231)

Input function must return 'double' or 'single' values. Found 'function_handle'.

Error in integral2Calc>integral2t (line 55)

[Qsub,esub] = tensor(thetaL,thetaR,phiB,phiT);

Error in integral2Calc (line 9)

[q,errbnd] = integral2t(fun,xmin,xmax,ymin,ymax,optionstruct);

Error in integral2 (line 106)

Q = integral2Calc(fun,xmin,xmax,yminfun,ymaxfun,opstruct);

Correction will be highly appreciated.

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Shahid Hasnain 2018-11-15

Ntruc=100;

number of trunction terms. We say as 100.

madhan ravi 2018-11-15

see my answer below

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Answer 1

Star Strider 2018-11-15

编辑：Star Strider 2018-11-15

在 MATLAB Online 中打开

1 个投票

This works, although it throws a warning:

Ntrunc = 100;
L = 100; % Length of the box
p=1;
q=1;
Ui = @(x,y)  sin(pi*x).*sin(pi*y); % An initial distribution for u, i.e. the u(x,y,0)
Vi = @(x,y)  sin(pi*x).*sin(pi*y); % An initial distribution for v, i.e. the v(x,y,0)
Wi = @(x,y) Ui(x,y) + Vi(x,y); % Initial condition for w, i.e. w(x,y,0)
% Computation of the b-coefficients
if p < Ntrunc
	if q < Ntrunc
		my_integrand = @(x,y)  4*Wi(x,y).*sin(p.*x).*cos(q.*y)/L^2; 
        bd(p,q) = integral2(@(x,y)my_integrand(x,y),0,L,0,L);
		q = q+1;
	end
	p = p+1;
end 

I defer to you to correct the problem that produces the warning.

EDIT — Corrected typographical error.

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Shahid Hasnain 2018-11-15

Again half code is working and half got error .Let me post complete code here as

t=0; x=[1 2 3 4 5]; y=[1 2 3 4 5];

function [W1]=wexact_Patrick(t,x,y);

Ntrunc = 100;

L = 100; % Length of the box

p=1;

q=1;

Ui = @(x,y) sin(pi*x).*sin(pi*y); % An initial distribution for u, i.e. the u(x,y,0)

Vi = @(x,y) sin(pi*x).*sin(pi*y); % An initial distribution for v, i.e. the v(x,y,0)

Wi = @(x,y) Ui(x,y) + Vi(x,y); % Initial condition for w, i.e. w(x,y,0)

% Computation of the b-coefficients

if p < Ntrunc

if q < Ntrunc

my_integrand = @(x,y) 4*Wi(x,y).*sin(p.*x).*cos(q.*y)/L^2;

bd(p,q) = integral2(@(x,y)my_integrand(x,y),0,L,0,L);

q = q+1;

end

p = p+1;

end

% The analytical solution for w(x,y,t) and first order of d(x,y,t)

W1 = 1 %Constant term

if p < Ntrunc

if q < Ntrunc

r = p^2 + q^2 + 1;

W1 = W1 + bd(p,q).*exp(-r.*t).*sin(p.*x).*cos(q.*y);

q = q+1;

end

p = p+1;

end

return

Error is about

wexact_Patrick(t,x,y)

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

> In integral2Calc>integral2t (line 129)

In integral2Calc (line 9)

In integral2 (line 106)

In wexact_Patrick (line 13)

W1 =

1

Attempted to access bd(2,2); index out of bounds because numel(bd)=1.

Error in wexact_Patrick (line 24)

W1 = W1 + bd(p,q).*exp(-r.*t).*sin(p.*x).*cos(q.*y);

Thank you in advance for correction.

Star Strider 2018-11-15

在 MATLAB Online 中打开

I am not certain what you are doing, so I do not have any idea of how to suppress the Warning. My impression is that integral2 is having a difficult time with the ‘my_integrand’ function you are giving it. It may be encountering a discontinuity, although I did not plot it to see what it is doing. I must leave that to you to correct.

I changed the nested while loops into for loops, and made the evaluation marginally more efficient by moving ‘my_integrand’ out of the loop. My changes are specifically:

my_integrand = @(x,y,p,q)  4*Wi(x,y).*sin(p.*x).*cos(q.*y)/L^2; 
% Computation of the b-coefficients
for p = 1 : Ntrunc
	for q = 1 : Ntrunc
        bd(p,q) = integral2(@(x,y)my_integrand(x,y,p,q),0,L,0,L);
    end
end 

This creates the appropriate matrix, so it should eliminate this error:

Attempted to access bd(2,2); index out of bounds because numel(bd)=1.
Error in wexact_Patrick (line 24)
		W1 = W1 + bd(p,q).*exp(-r.*t).*sin(p.*x).*cos(q.*y);

I apologise for the delay. It was overnight for me here, and additionally, I could not figure out how to eliminate the Warning. I suspect it is due to the increasing magnitudes of ‘p’ and ‘q’, since it does not appear until they get reasonably large, and setting ‘Ntrunc’ to a relatively low value prevents the Warning altogether.

Shahid Hasnain 2018-11-19

@Star Strider,

Sorry for being late, Now this is working until line 42 (I paste it here)

%%%%%%%%%%%

t=0;

L = 1; % Length of the box

hx=0.01;

a=0;

b=L;

x=a:hx:b;

y=x;

Ntrunc = 10; % Number of summations (high frequency oscillations can be neglected)

T = 10; % Duration of the time interval

p=1;

q=1;

Ui = @(x,y) (cos(pi.*x).*cos(pi.*y)); % An initial distribution for u, i.e. the u(x,y,0)

Vi = @(x,y) (sin(pi.*x).*sin(pi.*y)); % An initial distribution for v, i.e. the v(x,y,0)

di = @(x,y) Ui(x,y) - Vi(x,y); % Initial condition for w, i.e. w(x,y,0)

% Computation of the b'-coefficients (denoted by bd(p,q))

for p=1:Ntrunc

for q =1:Ntrunc

u_integrand = @(x,y) 4*di(x,y).*sin(p.*x).*cos(q.*y)/L.^2;

m_bd(p,q)= integral2(@(x,y)u_integrand(x,y),0,L,0,L);

end

% Zeroth order of d(x,y,t) without the convolution integral term, this is denoted by dzeroh

dzeroh = 1; % Constant term

for p=1: Ntrunc

for q= 1:Ntrunc

r = p.^2 + q.^2 + 1;

dzeroh = @(x,y,t) (dzeroh(x,y,t) + m_bd(p,q).*exp(-r.*t).*sin(p.*x).*cos(q.*y));

q = q+1;

end

p = p+1;

end

hx = 0.1; % discretization of the x-interval

hy = 0.1; % " " " y- "

ht = 0.1; % " " " z- "

xc = a:hx:L;

yc = xc;

tc = 0:ht:T;

G= @(x,y,t,xc,yc,tc) ( sqrt(pi/(t-tc)).*e^(-(t-tc)).*exp(-((x-xc).^2+(y-yc).^2)/(4*(t-tc))) ); % The Green function

integrand = @(x,y,t,xc,yc,tc) (-1.*((wexact_Patrick(tc,xc,tc)+1).*(wexact_Patrick(tc,xc,yc)-1).^2).*G(x,y,t,xc,yc,tc)/4);

%%%%%%%%%% Working upto here%%%%%%%%%%%%%%

Shahid Hasnain 2018-12-3

@Star Strider,

Sorry for being late.

I am trying to solve Gray Scott Model in 2 dimension by analytical technique. the code I attached few days back, it was a try to find analytical solution.

Here is the system

u_t = u_{xx} + u_{yy} - uv^2 + 1 - u

v_t = v_{xx} + v_{yy} + uv^2 - v

I do not have analytical solution to this problem, thats why we choose another way to find.

w=u+v (wexact_Patrick, it was mentioned in previous posts) working

d=u-v (it was mentioned in previous posts) half working still.

Still working to rearrange integral3 which create problems.

Star Strider 2018-12-3

No worries about a delayed reply.

Do you want to integrate a system of partial differential equations? If so, please post this as a new Question. My experience with the Partial Diufferential Equation Toolbox is limited, although Torsten and some others here likely have significant experience with it. I have not worked with it in a while, so it would take some time for me to provide you with a response.

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Error using integral2

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