MATLAB Answers

I want to plot a numerical integral in a mesh which is dependent of x and y. How can I do this?

4 views (last 30 days)
mk_ballav
mk_ballav on 4 Jul 2018
Commented: Anton Semechko on 4 Jul 2018
I have a two dimensional function and I want to plot the integral function in 2D mesh. I want to plot the C2 in color mesh and also want to plot diagonal element of C2. I am not getting the correct result. Please see the code I am working.
if true
%
L = 20;
z1 = 0:0.01:L;
z2 = 0:0.01:L;
dz = z1(2)-z1(1);
[Z1,Z2] = meshgrid(z1,z2);
W1 = 1+Z1/L;
W2 = 2+Z2/L;
for j = 1:size(z1,2)
for k= 1:size(z2,2)
A(j,k) = sum(sum(1./W1(1:j,1:k))*dz)*dz;
B(j,k) = sum(sum(1./W2(1:j,1:k))*dz)*dz;
end
end
C2_h = (1-A).^2+(1-B);
figure(32);pcolor(Z1,Z2,C2_h);axis equal tight; shading flat;colorbar;colormap jet;
figure(5);
plot(z1/L, diag(C2_h) ,'r','Linewidth',2 );hold on;
end

  0 Comments

Sign in to comment.

Answers (1)

Anton Semechko
Anton Semechko on 4 Jul 2018
The integrals in your equation can be evaluated analytically. Here how you can visualize C(z1,z2) and C(z1,z1):
L = 20;
dz = 0.01;
z = 0:dz:L;
[Z1,Z2] = meshgrid(z);
I1=@(z) log(1+z);
I2=@(z) log(2+z);
C=(1-I1(Z1)).^2 + (1+log(2)-I2(Z2));
figure('color','w')
ha=subplot(1,2,1);
z_lim=[z(1) z(end)];
imagesc(z_lim,z_lim,C)
axis equal
hold on
plot(z_lim(:),z_lim(:),'--k','LineWidth',2)
set(ha,'XLim',z_lim+[-1 1],'YLim',z_lim+[-1 1],'YDir','normal')
xlabel('z_1','Interpreter','tex','FontSize',20)
ylabel('z_2','Interpreter','tex','FontSize',20)
colorbar('Location','NorthOutside')
subplot(1,2,2)
Cd=diag(C);
plot(z(:),Cd(:),'--k','LineWidth',2)
xlabel('z_1','Interpreter','tex','FontSize',20)
ylabel('C(z_1,z_1)','Interpreter','tex','Interpreter','tex','FontSize',20)

  2 Comments

mk_ballav
mk_ballav on 4 Jul 2018
This is one of the special case where W1 and W2 are linear functions but I have other cases where W1 and W2 are not linear and I can't directly evaluate integral anlytically,so I have to do numerical integration. Here C(z1,z2) is the whole matrix elements and C(z1,z1) is just the diagonal elements of C.
Anton Semechko
Anton Semechko on 4 Jul 2018
OK. Assuming W1 and W2 are univariate functions of z1 and z2, respectively, you can integrate 1/W1 and 1/W2 numerically using builtin 'integral' function.. Here is an example:
% Visualization grid
L = 20;
dz = 0.01;
z = 0:dz:L;
% Evaluate definite integrals of 1/W1 and 1/W2 beween z(i-1) and z(i)
W1=@(z) 1./(z+1);
W2=@(z) 1./(z+2);
[I1,I2]=deal(zeros(size(z)));
for i=2:numel(z)
I1(i)=integral(W1,z(i-1),z(i));
I2(i)=integral(W2,z(i-1),z(i));
end
% Definite intergral between z(1) and z(i) = sum(I(1:i))
I1=cumsum(I1);
I2=cumsum(I2);
% Function
C=bsxfun(@plus,(1-I1).^2,(1-I2(:)));
% Visualization
figure('color','w')
ha=subplot(1,2,1);
z_lim=[z(1) z(end)];
imagesc(z_lim,z_lim,C)
axis equal
hold on
plot(z_lim(:),z_lim(:),'--k','LineWidth',2)
set(ha,'XLim',z_lim+[-1 1],'YLim',z_lim+[-1 1],'YDir','normal')
xlabel('z_1','Interpreter','tex','FontSize',20)
ylabel('z_2','Interpreter','tex','FontSize',20)
colorbar('Location','NorthOutside')
subplot(1,2,2)
Cd=diag(C);
plot(z(:),Cd(:),'--k','LineWidth',2)
xlabel('z_1','Interpreter','tex','FontSize',20)
ylabel('C(z_1,z_1)','Interpreter','tex','Interpreter','tex','FontSize',20)
You can confirm that the final result is visually indistinguishable from the one obtained analytically.

Sign in to comment.

Sign in to answer this question.