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

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Jerzy el 22 de Nov. de 2023
Comentada: Fabio Freschi el 23 de Nov. de 2023
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.
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### Respuestas (1)

Fabio Freschi el 22 de Nov. de 2023
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;
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Jerzy el 23 de Nov. de 2023
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 el 23 de Nov. de 2023
If you want you can share the original document with the formulas... fresh eyes can help

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