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

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Hewa selman
Hewa selman 2021-12-23
评论： Torsten 2022-1-21
integro differential equation

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Torsten
Torsten 2021-12-24
编辑：Torsten 2021-12-24
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.
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Torsten
Torsten 2021-12-24
编辑：Torsten 2021-12-24
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 ?

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Hewa selman
Hewa selman 2021-12-28
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.
Torsten
Torsten 2021-12-28
编辑：Torsten 2021-12-28
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.
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Hewa selman
Hewa selman 2021-12-29
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
Torsten 2021-12-29
编辑：Torsten 2021-12-29
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

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Hewa selman
Hewa selman 2022-1-3
编辑：Hewa selman 2022-1-3
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.
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