Runge-Kutta 4th order method for 2nd order differential equation with complex number

Hi,

I want to solve following equation, which has complex part:

I tried Runge-Kutta 4th order method but it does not provide me with correct result.

I assume:

I have used following code:

close all; clear variables;

h=0.01e-6;  % step size
x = (-5e-6):h:(20e-6);                                     
y = zeros(1,length(x));
z = zeros(1,length(x));

y(1) = 10; % y at x=-5e-6
z(1) = -1e-1; % dy/dx at x=-5e-6

dWdx = @(x) 2.3139e+13 * exp((-x.^2)/(2e-12));
W = @(x) integral(dWdx,-100e-6,x);
A = @(x) 7.8957e+03 *W(x);
B = @(x) 1/W(x)*dWdx(x);
f = @(x,y,z) z;
g = @(x,y,z) B(x)*z+ 1i*A(x)*y;

N=length(x);
for j=1:(N-1)
    
    k0 = h*f(x(j),y(j),z(j));
    L0 = h*g(x(j),y(j),z(j));
    
    k1= h*f(x(j)+0.5*h,y(j)+0.5*k0,z(j)+0.5*L0);
    L1= h*g(x(j)+0.5*h,y(j)+0.5*k0,z(j)+0.5*L0);
    
    k2= h*f(x(j)+0.5*h,y(j)+0.5*k1,z(j)+0.5*L1);
    L2= h*g(x(j)+0.5*h,y(j)+0.5*k1,z(j)+0.5*L1);
    
    k3= h*f(x(j)+1*h,y(j)+1*k2,z(j)+1*L2);
    L3= h*g(x(j)+1*h,y(j)+1*k2,z(j)+1*L2);
    
    y(j+1)=y(j)+1/6*(k0+2*k1+2*k2+k3);
    z(j+1)=z(j)+1/6*(L0+2*L1+2*L2+L3);
end

figure();
% plot(x,y./max(y));
plot(x,y);

Warning: Imaginary parts of complex X and/or Y arguments ignored.

xlim([-5 20]*1e-6);
grid on; hold on;

The solution should have following shape (see normalized red line, for 1GHz):

Can someone help me with this scenario? There where code (based on which i had started) that solves 2nd order ODE with 4rd order Runge-Kutta (https://www.mathworks.com/matlabcentral/fileexchange/29851-runge-kutta-4th-order-ode), however I couldnt find a solution for equation with complex number.

Thanks in advance :)

Answer 1

[Fabio Freschi]

Your code is working correctly: in fact, if you use the built-in ode45 solver, you obtain exactly the same result. My guess there is a mistake in the definition of your coefficients W, A, B, or the initial conditions. Note that if you are not asked to write your own code, ode45 is more efficient and accurate, since it is using adaptive step size with error control.

close all; clear variables;

h=0.01e-6;  % step size
x = (-5e-6):h:(20e-6);                                     
y = zeros(1,length(x));
z = zeros(1,length(x));

y(1) = 10; % y at x=-5e-6
z(1) = -1e-1; % dy/dx at x=-5e-6

dWdx = @(x) 2.3139e+13 * exp((-x.^2)/(2e-12));
W = @(x) integral(dWdx,-100e-6,x);
A = @(x) 7.8957e+03 *W(x);
B = @(x) 1/W(x)*dWdx(x);

% use built in ode45
dfdx = @(x,f)[B(x)*f(1)+ 1i*A(x)*f(2); f(1)];
[t,f] = ode45(dfdx,[x(1) x(end)],[z(1) y(1)]);

% plot
h0 = figure;
plot(t,f(:,2));

Warning: Imaginary parts of complex X and/or Y arguments ignored.

xlim([-5 20]*1e-6);
grid on; hold on;

% OP code to compare
f = @(x,y,z) z;
g = @(x,y,z) B(x)*z+ 1i*A(x)*y;

N=length(x);
for j=1:(N-1)
    
    k0 = h*f(x(j),y(j),z(j));
    L0 = h*g(x(j),y(j),z(j));
    
    k1= h*f(x(j)+0.5*h,y(j)+0.5*k0,z(j)+0.5*L0);
    L1= h*g(x(j)+0.5*h,y(j)+0.5*k0,z(j)+0.5*L0);
    
    k2= h*f(x(j)+0.5*h,y(j)+0.5*k1,z(j)+0.5*L1);
    L2= h*g(x(j)+0.5*h,y(j)+0.5*k1,z(j)+0.5*L1);
    
    k3= h*f(x(j)+1*h,y(j)+1*k2,z(j)+1*L2);
    L3= h*g(x(j)+1*h,y(j)+1*k2,z(j)+1*L2);
    
    y(j+1)=y(j)+1/6*(k0+2*k1+2*k2+k3);
    z(j+1)=z(j)+1/6*(L0+2*L1+2*L2+L3);
end

figure(h0), hold on;
% plot(x,y./max(y));
plot(x,y);

Warning: Imaginary parts of complex X and/or Y arguments ignored.

xlim([-5 20]*1e-6);
grid on; hold on;

[Torsten]

Look at the scale of the y-axis. It goes from y=-20*10^4 to y=5*10^4. y starts regularily at 10, but because of the scale of the plot, it looks as if it would start at 0.

Choose

xlim([-5 5]*1e-6);
ylim([-10 11])

to see what I mean.

[Fabio Freschi]

@Torsten: ok. I just wanted to check if the implementation of the RK loop by the OP was correct. And in my picture the two solutions (built-in RK, OP implementation) are almost identical. So the OP should look elsewhere for possible mistakes.

[Torsten]

The OP edited away his question why he got 0 for y although he/she set 10 as initial condition. So my remark was addressed to the OP's comment, not to your answer. Don't feel insulted :-)

[Fabio Freschi]

Not at all! I "know" you and your activity in the forum and I never felt insulted. I wasn't aware of the deleted OP question. If it seemed a harsh answer it's only because English is not my first language.

[Jerzy]

Yes, initially i posted a comment stating that all the solutions start at 0 value, despite setting starting point to nonzero values. And you have right, its the scale which makes it look as if it were 0. Still, i have checked the formulas multiple times and it does not contain any error in formula.... Regardless of that, thanks for your help :)

[Fabio Freschi]

If you want you can share the original document with the formulas... fresh eyes can help