How to solve a system of integro differential equation using numerical method?

integro differential equation

Answer 1

[Torsten]

Simple.

Solve instead

dx/dt = t*x(t) + exp(t)*u(t), x(0) = 1
dy/dt = t^3*x(t) + exp(-t)*v(t), y(0) = 1
du/dt = exp(-t)*y(t), u(0) = 0
dv/dt = exp(t)*sin(x(t)), v(0) = 0

using one of the ODE solver from the MATLAB ODE suite, e.g. ODE15S,ODE45.

[Hewa selman]

Hello. I do not understand the value of du/dt and dv/dt. Did you checked it? Are ypu sure this code works for system of integro differential equations? If not, could you make a code for it?

with regards.

[Torsten]

This works for your IDE from above, not for IDEs in general.

Do you really expect someone out there will write a code for you for the solution of general systems of IDEs ?

I think this would take some weeks or even months of work.

Hint for the du/dt and dv/dt terms:

What do you get if you differentiate

integral_{0}^{t} exp(-s)*y(s) ds

and

integral_{0}^{t} exp(s)*sin(x(s)) ds

with respect to t ?

[Hewa selman]

wel, thank you so much Mr Torsten.

Answer 2

[Hewa selman]

function mysysfun

clc;

clear;

t = 0:0.1:10;

%%%t = [0 5];

initial_Y1 = 1;

initial_Y2 = 1;

initial_U1 = 0;

initial_U2 = 0;

%%%Y0 = [initial_Y1 initial_Y2];

Y0 = [initial_Y1 initial_Y2 initial_U1 initial_U2];

[t, Y] = ode45 (@ mysys, t, Y0);

%%

%figure;

plot(t,Y(:,1),'r');

xlabel('t1'); ylabel('Y1');

title('solving the 1st order differential equations Y(1,1)')

figure;

plot(t,Y(:,2),'g');

xlabel('t2'); ylabel('Y2');

title('solving the 1st order differential equations Y(2,1)')

figure;

plot(t,Y(:,1),'r')

grid on; hold on;

plot(t,Y(:,2),'g')

xlabel('Time of t1')

ylabel('Equation Y1')

title('solving the system of 1st order differential equations')

legend('Y(1,1)','Y(2,1)','Interpreter','latex');

%%

function dYdt = mysys(t,Y)

dYdt1_11=t+exp(t)*Y(2)+exp(t)*U(1);

dYdt2_21=exp(t)+cos(Y(1))+exp(t)*U(2);

dUdt3_31=exp(-t)*Y(2);

dUdt4_41=exp(t)*sin(Y(1));

%%%dYdt = [dYdt1_11; dYdt2_21];

dYdt = [dYdt1_11; dYdt2_21; dUdt3_31; dUdt4_41];

end

end

Hello Mr. Torsten. Could you help me for solve this code. I think ode45 is not suitable for solve this code. The system is integro differential eqyation.

with regards.

Answer 3

[Torsten]

function main
tspan = [0 10];
y0 = [1; 1; 0; 0];
[T,Y] = ode45(@fun,tspan,y0);
plot(T,[Y(:,1),Y(:,2)])
end

function dy = fun(t,y)
dy = zeros(4,1);
dy(1) = t*y(1) + exp(t)*y(3);
dy(2) = t^3*y(1) + exp(-t)*y(4);
dy(3) = exp(-t)*y(2);
dy(4) = exp(t)*sin(y(1));
end

This is the code for your original system of IDEs.

I don't know about the new system you wanted to solve with the code from above.

[Hewa selman]

well. thank you Mr. Torsten.

Could i use the comlex functions instead of real functions in this code? Or I have to get the real and imaginary pars and next use your code.

Like this code

%% file imaginaryODE

function fv = imaginaryODE(t,yv)

y = yv(1) + 1i*yv(2);

yp = complexfun(t,y);

fv = [real(yp) ;imag(yp)];

end

y0 = 1 + 1i;

yv0 = [real(y0); imag(y0)];

tspan = [0 2];

y = yv(:,1) + 1i*yv(:,2);

[t,yv] = ode45(@imaginaryODE, tspan, yv0);

% and the file complexfun

function f = complexfun(t,y)

f = y.*t+2*1i;

I have to solve the complex integro differential equations and get the results.

Could you help me in this problem

with regards.

[Torsten]

Seems to work as expected:

function main
  y0 = 1 + 1i;
  z0 = [1 1];
  tspan = [0 2];
  [t,y] = ode45(@complexfun, tspan, [y0,z0]);
  figure
  plot(t,[real(y(:,1)),imag(y(:,1))])
  figure
  plot(t,[y(:,2),y(:,3)])
end

function f = complexfun(t,y)
  f = zeros(3,1);
  f(1) = y(1)*t+2*1i;
  f(2) = y(2)*t;
  f(3) = y(3)*t+2;
end

Answer 4

[Hewa selman]

HELLO MR. TORSTEN.

THANK YOU FOR SUPPORT.

I AM NOT UNDERSTANDING THE PLOT CLEARLY, SINCE THE INITIAL VALUE PLACES THE POINT(1,1) BUT IN FIGURE IT PLACES THE POINT (0,1). I THINK THE PLOT (PLOT 3D) IS MORE CLEAR AND I TRIED TO PLOT THE 3D ONE, BUT I AM NOT SURE TOO AS FOLLOSW

a = -2:0.1:2;

b = -2:0.1:2;

f = meshgrid(a,b);

surfc(a,b,abs(f))

WHERE, (a+bi) is complex number.

PLEADE HELP ME

WITH REGARDS.

[Torsten]

In my figure, both real(y(:,1)) and imag(y(:,1)) start at 1.

[Hewa selman]

Hello Mr. Torsten.

In last code you write the plot, plot(t,[real(y(:,1)),imag(y(:,1))])

why you depend only on real and imaginary value of y(:,1)? not on y(:,2) too!

If initial conditions are not same as, y0 = [2 - 1i,1 + 2i, 0, 0];

how i ploted it

with regards

[Torsten]

You can plot whatever you like - I just decided in the sample code only to plot y1.

But of course you can also include y2 in the plot:

plot (t, [real(y(:,1)),imag(y(:,1)),real(y(:2,)),imag(y(:,2))])

or make separate plots:

figure(1)
plot(t,[real(y(:,1)),imag(y(:,1))])
figure(2)
plot(t,[real(y(:,2)),imag(y(:,2))])

[Hewa selman]

Hello Mr. Torstin.

Since the problem is in complex plane, we have to define the limits of integral in complex variables, so I have to change the, tspan from [0 10] to [0+0i 10+11i], but the code get error and not works.

could you help me for solve the problem

function MAINO
y0 = [0 - 1i,0.5 + 2i, 0, 0];
  tspan = [0+0i 10+11i];
  [t,y] = ode45(@compfun, tspan, y0);

figure(1)
plot(t,[real(y(:,1)),imag(y(:,1)),real(y(:,2)),imag(y(:,2))])
figure(2)
plot(t,[(y(:,1)),(y(:,2))])
end
function f = compfun(t,y)
  f = zeros(4,1);
  f(1) = (2 + 3i)*t*cos(y(1))+1i*t*y(2)+exp(1i*t)*y(3);
  f(2) = t*y(1)+(1 - 1i)*t*sin(y(2))+exp(-1i*t)*y(4);
  f(3) = exp(-1i*t)*y(2);
  f(4) = exp(1i*t)*y(1);
end

regards.

[Torsten]

Over which path connecting 0 and 10+11*i do you want to integrate ?

[Hewa selman]

I think the idea of using t is not right, where t is real number. Because the plane is complex and system is complex. Let z=x+yi is complex number, and if I change t to z, how to use it in code? And let use any suitable path. Since the integral is from 0 to t, and after replace t by z, how can I use it in code?

[Torsten]

And let use any suitable path

What is suitable ? You have to parametrize the chosen path by z(t) (in your case e.g. z(t) = (10+11*i)*t

for 0<=t<=1 if you want to integrate over the line segment connecting 0 and 10+11*i) and rewrite your ODEs in t instead of z.

Then you can proceed as usual by choosing tspan = [0 1].

[Hewa selman]

Thank you Mr. Torsten.

Yes it works.

[Torsten]

I hope you took z'(t) into account in the derivative and the integral terms.