Surface plot of a function containing partial derivative

5 vues (au cours des 30 derniers jours)
Muhammad Ali
Muhammad Ali le 30 Jan 2020
Commenté : Walter Roberson le 31 Jan 2020
I don,,,'t know to take partial derivative under meshgrid command ,because i need to draw the surface plot of 'q' which contain derivatives. the file is attached for further inqury. Kindly guide me.

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Walter Roberson
Walter Roberson le 30 Jan 2020
Modifié(e) : Walter Roberson le 30 Jan 2020
Use the Symbolic Toolbox
a=-0.5*1i;
b=-0.8*1i;
c=0.3*1i;
[X,T]=meshgrid(-20:.5:20,-20:.5:20);
syms x t
D=2*1i*a.*x-1i*t./a;
E=2*1i*b.*x-1i*t./b;
F=2*1i*c.*x-1i*t./c;
X1=exp(D)+1i.*(exp(-D));
X2=exp(E)+1i.*(exp(-E));
X3=exp(F)+1i.*(exp(-F));
Y1=exp(D)-1i.*(exp(-D));
Y2=exp(E)-1i.*(exp(-E));
Y3=exp(F)-1i.*(exp(-F));
A=b.*X1.*Y2.*(a^3.*X3-c^3.*Y3)+X2.*(-a*b^3.*X3.*Y1+c.*((-a^3+b^3).*X1+a*c^3.^Y1).*Y3);
B=a*b*(-a+b).*X3.*Y1.*Y2+c.*(a.*(a-c).*X2.*Y1+b.*(-b+c).*X1.*Y2).*Y3;
C=(-a^3+b^3)*c.*X3.*Y1.*Y2+(b.*(a^3-c^3).*X2.*Y1+a*(-b^3+c^3).*X1.*Y2).*Y3;
D=a.*(a-c)*c.*X1.*X3.*Y2+b.*X2.*(c*(-b+c).*X3.*Y1+a*(-a+b).*X1.*Y3);
q=((1i/2).*((B.*diff(A,x)-A.*diff(B,x))./(B.*B)))+((1i/2).*((D.*diff(C,x)-C.*diff(D,x))./(D.*D)));
fprintf('optimization takes about 2 minutes, please wait\n');
tic;
qfun = matlabFunction(q, 'vars', {x, t}, 'file', 'qfun.m', 'Optimize', true);
toc
fprintf('done optimization\n');
tic
qn = qfun(X, T);
toc
subplot(2,2,1)
surf(X, T, real(qn), 'edgecolor', 'none');
title('real part')
subplot(2,2,2)
surf(X, T, imag(qn), 'edgecolor', 'none');
title('imaginary part')
subplot(2,2,[3 4])
surf(X, T, abs(qn), 'edgecolor', 'none');
title('absolute value')
A fair number of the entries come out as NaN. Some of those would not come out as NaN if you processed completely in the symbolic toolbox rather than using matlabFunction,
%this step takes a couple of minutes
qn = double( vpa(subs(q, {x,t}, {X, T})) );
  2 commentaires
Muhammad Ali
Muhammad Ali le 31 Jan 2020
thanks . it works
Walter Roberson
Walter Roberson le 31 Jan 2020
Some of the entries come out nan because the temporary values computed overflow even the abilities of the Symbolic Toolbox, such as values in the range 10^330973961

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