How to solve systems of non linear equation of dimensions 100 using ode45 matalb.

how should i proceed if we want to implement its code?

[Davide Masiello]

It'd be best to post a first code attempt from your side, then we could try to help develop on that.

[RITIKA Jaiswal]

%odes solve with ode45 considering 10 states.

clear all;

clc;

close all;

syms g(x) u(t) x1(t) x2(t) x3(t) x4(t) x5(t) x6(t) x7(t) x8(t) x9(t) x10(t) Y;

g(x)=exp(40*x)+x-1;

A =[-g(x1)-g(x1-x2)+exp(-t); g(x1-x2)-g(x2-x3);g(x2-x3)-g(x3-x4);g(x3-x4)-g(x4-x5);g(x4-x5)-g(x5-x6);g(x5-x6)-g(x6-x7);g(x6-x7)-g(x7-x8);g(x7-x8)-g(x8-x9);g(x8-x9)-g(x9-x10);g(x9-x10)]

ode1 = diff(x1,t)==[-g(x1)-g(x1-x2)];

ode2= diff(x2,t)== g(x1-x2)-g(x2-x3);

ode3= diff(x3,t)== g(x2-x3)-g(x3-x4);

ode4= diff(x4,t)== g(x3-x4)-g(x4-x5);

ode5= diff(x5,t)== g(x4-x5)-g(x5-x6);

ode6= diff(x6,t)== g(x5-x6)-g(x6-x7);

ode7= diff(x7,t)== g(x6-x7)-g(x7-x8);

ode8= diff(x8,t)== g(x7-x8)-g(x8-x9);

ode9= diff(x9,t)== g(x8-x9)-g(x9-x10);

ode10= diff(x10,t)== g(x9-x10);

odes=[ode1;ode2;ode3;ode4;ode5;ode6;ode7;ode8;ode9;ode10]

%S = dsolve(odes)

[VF,Sbs] = odeToVectorField(odes)

%Sodsefcn = matlabFunction(VF)

Sodsefcn = matlabFunction(VF, 'Vars',{t,Y})

%y0=[0];

tspan=[0 7];

y0=[1 1 1 0 0 0 0 0 0 1];

ySol = ode45(@(t,Y)Sodsefcn(t,Y),tspan,y0);

tValues = linspace(tspan(1),tspan(2),100);

yValues = deval(ySol,tValues,1); %number 1 denotes first solution likewise you can mention 2 ,3 & 4 for the next three solutions

%Evaluate the first component of the solution at 1000 points in the interval [0 7].

plot(tValues,yValues)

title('With initial of y0 = [1 1 1 0 0 0 0 0 0 1]')

i tried like this but if we proceed like tis then we have to write a long code for N=100.

Answer 1

[Davide Masiello]

If I were you, I would proceed substantially differently.

Since you need to solve 100 equations, it is unthinkable to cde them one by one.

The key here is indexing.

For instance

clear,clc

tspan = [0,7];
x0 = zeros(1,1000);      % substitute with correct initial conditions

dgn = ones(1,1000); dgn(251:750) = 1/2;
D = zeros(1000); D(logical(eye(1000))) = dgn;

options = odeset('Mass',D);

[t,X] = ode45(@odeFunc,tspan,x0,options);

function dxdt = odeFunc(t,x)

g = @(x) exp(40*x)+x-1;

dxdt(1,1)               = -g(x(1))-g(x(1)-x(2))+exp(-t);
dxdt(2:length(x)-1,1)   = g(x(1:end-2)-x(2:end-1))-g(x(2:end-1)-x(3:end));
dxdt(length(x),1)        = g(x(end-1)-x(end));
end

[RITIKA Jaiswal]

Thankyou so much it really helped .

[RITIKA Jaiswal]

can you provide some refrences where i can learn more about indexing .It would be great help .

[Torsten]

https://de.mathworks.com/learn/tutorials/matlab-onramp.html

[Davide Masiello]

My pleasure. Please consider to accept the answer, it is customary to do so when it's found to be helpful.

[RITIKA Jaiswal]

yeah done

[RITIKA Jaiswal]

Hi, With reference to my earlier question I need to simulate my dynamical system over t = [0, 7] seconds and we need to have 2000 snapshots of

the dynamical system and the nonlinear function are collected with equidistant time steps. For this if i use tspan =linspace(0,7,2000) the X matrix i am getting of 2000*1000 and if i am using tspan =[0 7].

matrix of X i am getting is 1421*1000. Which one is correct ?

[Davide Masiello]

The ODE solver uses an adaptive time step. When you do not specify the number of time points, then it manages to integrate the equations in 1421 within the given tolerances.

When you specify that you want 2000 time steps, the solver just makes sure to report the solution at the time steps you asked for.

In summary, both solutions are correct.

[RITIKA Jaiswal]

Thanks for clearing my doubt.

[RITIKA Jaiswal]

The X matrix which we are getting here rows represent different time and column represents states right?

[Torsten]

Correct.

[RITIKA Jaiswal]

thanks

[RITIKA Jaiswal]

I am trying to solve Sets of pdes in order to get discretize it.Using finite difference method such that the resulting ODEs approximate the essential dynamic information of the system.

I am not sure whether the code is correct. I have used first order forward difference and 2nd order centered difference.

i am unable to solv equation (2).Please guide.

clear all;

close all;

M=1000;
c=0.25;%lets dt/dr^2 =c
a=0.02;%lets dt/dr=a
r=0.01;
v=0.5;
for i =2:25
    for j =2:25
        p(i,j)=200;
    end
end
dt=0.001;
dr=0.25;
for t=1:M
    for i=2:25
        for j =2:24
            pp(i,j)=p(i,j)+(0.5*a)*(-v+(1/r))*(p(i,j+1)-p(i,j-1))+c*(p(i,j+1)-2*p(i,j)+p(i,j-1));
        end 
    end
    n=25;
    pp(i,1:n)=500; %lets assume
    pp(n,1:n)=500;
    pp(1:n,1)=500;
    pp(1:n,n)=500;
    p=pp;
    t=t;
end
figure
contourf(p,25,'linecolor','non')

this code is running i have also attached paper here for reference.