plotting a derivative of a function using surf command

I am trying to plot a derivative of a function using surf command, but when I evaluate it an error occur "Data dimensions must agree".

This is what I typed:

close all;

X=-10:.1:10;
T=-10:.1:10;
mu1=-.01+1*1i;
a=(.1-2*1i);
b=(.1-.1*1i);
[x,t]=meshgrid(X,T);
x1=exp(-1i*mu1.*x).*exp((1-(1i./(2*mu1)).*t));
x2=exp(1i*mu1.*x).*exp((1+(1i./(2*mu1)).*t));
y1=exp(-1i*mu1.*x).*exp((1+(1i./(2*mu1)).*t));
y2=exp(1i*mu1.*x).*exp((1-(1i./(2*mu1)).*t));
A=2*1i*(1)*a*conj(b)*x1.*conj(x2).*(x2.*conj(y1)-y2.*conj(x1));
B=(a.*conj(a).*x1.*conj(x1).*y2.*conj(y2)+b.*conj(b).*x1.*conj(y1).*x2.*conj(y2)+a.*conj(a).*x2.*conj(x2).*y1.*conj(y1)+b.*conj(b).*y1.*conj(x2).*y2.*conj(x1));
r1=-2.*1i.*((A./B));
dr1=diff(r1);
dt=diff(t);
dr1dt=dr1./dt;
td=t(2:end);
surf(x,td,abs(dr1dt));

[KSSV]

X=-10:.1:10;
T=-10:.1:10;
mu1=-.01+1*1i;
a=(.1-2*1i);
b=(.1-.1*1i);
[x,t]=meshgrid(X,T);
x1=exp(-1i*mu1.*x).*exp((1-(1i./(2*mu1)).*t));
x2=exp(1i*mu1.*x).*exp((1+(1i./(2*mu1)).*t));
y1=exp(-1i*mu1.*x).*exp((1+(1i./(2*mu1)).*t));
y2=exp(1i*mu1.*x).*exp((1-(1i./(2*mu1)).*t));
A=2*1i*(1)*a*conj(b)*x1.*conj(x2).*(x2.*conj(y1)-y2.*conj(x1));
B=(a.*conj(a).*x1.*conj(x1).*y2.*conj(y2)+b.*conj(b).*x1.*conj(y1).*x2.*conj(y2)+a.*conj(a).*x2.*conj(x2).*y1.*conj(y1)+b.*conj(b).*y1.*conj(x2).*y2.*conj(x1));
r1=-2.*1i.*((A./B));
dr1=gradient(r1);
dt=gradient(t);
dr1dt=dr1./dt;
td=t(2:end);
surf(x,t,abs(dr1dt)');

But you need re think on your code.

Answer 1

[KSSV]

X=-10:.1:10;
T=-10:.1:10;
mu1=-.01+1*1i;
a=(.1-2*1i);
b=(.1-.1*1i);
[x,t]=meshgrid(X,T);
x1=exp(-1i*mu1.*x).*exp((1-(1i./(2*mu1)).*t));
x2=exp(1i*mu1.*x).*exp((1+(1i./(2*mu1)).*t));
y1=exp(-1i*mu1.*x).*exp((1+(1i./(2*mu1)).*t));
y2=exp(1i*mu1.*x).*exp((1-(1i./(2*mu1)).*t));
A=2*1i*(1)*a*conj(b)*x1.*conj(x2).*(x2.*conj(y1)-y2.*conj(x1));
B=(a.*conj(a).*x1.*conj(x1).*y2.*conj(y2)+b.*conj(b).*x1.*conj(y1).*x2.*conj(y2)+a.*conj(a).*x2.*conj(x2).*y1.*conj(y1)+b.*conj(b).*y1.*conj(x2).*y2.*conj(x1));
r1=-2.*1i.*((A./B));
dr1=gradient(r1);
dt=gradient(t);
dr1dt=dr1./min(diff(T));
td=t(2:end);
surf(x,t,abs(dr1dt)');

To get dt you can use difference in T. You need not to take a matrix. ANote that dt is same i.e 0.01. If you use a matrix..it is coming out to be zero matrix and makes dr1dt a nan or inf matrix. I advice you to still rethink on your code.

[Wajahat]

@ KSSV, sorry, I don't understand it. Why you use 'min(diff(T))' instead of 'dt'

[KSSV]

YOur time step is alway constant......check diff(T)

[Wajahat]

@ KSSV, also what is the reason to use 'gradient' instead of 'diff' in the code?

[KSSV]

diff reduces the dimension by one....ad you are subtracting the consecutive elements. Gradient will not reduce the dimensions.

[Wajahat]

@ KSSV,

To verify, I considered a simple example such as

X=-1:.05:1;

T=-1:.05:1;

mu1=1+1*1i;

[x,t]=meshgrid(X,T);

r1=mu1.*sin(x+4.*t);

dr1=gradient(r1);

dt=gradient(t);

dr1dt=dr1./min(diff(T));

td=t(2:end);

surf(x,t,abs(dr1dt));

I have plotted it.

Then I take the derivative of 'r1' w.r.t 't' and then plot the function. i..e,

X=-1:.05:1;

T=-1:.05:1;

mu1=1+1*1i;

[x,t]=meshgrid(X,T);

r1=4.*mu1.*cos(x+4.*t);

surf(x,t,abs(r1)');

You can see there is difference of amplitude in these plots. Why the two plots are not identical?