Help solving a system of differential equations

I have the following system of differential equations, and I am not able to understand the best way to go about them. I tried using a couple of functions like dsolve, ode45, etc., but most of them give errors that I am not able to understand.

x(1)=0.3; x(3)=0.9; y(1)=0.4; y(3)=0.8;
A=10; p=10;
syms X1 X2;
num = ((x(3)-X1)-(X1-x(1)))*X1 + ((y(3)-X2)-(X2-y(1)))*X2;
den = sqrt((x(3)-X1)-(X1-x(1))^2 + (y(3)-X2)-(X2-y(1))^2);
Em = -A*(X1-x(1))*(x(3)-X1) - A*(X2-y(1))*(y(3)-X2) + p*num/den;
dXdt = [-diff(Em,X1); -diff(Em,X2)];

I would like to get X1 and X2 as a function of time. If I define syms X1(t) and X2(t), I get the error 'All arguments, except for the first one, must not be symbolic functions.' The above code, when run, gives me expressions in terms of X1 and X2. I need solutions of X1 and X2 in terms of t, where dX1dt = -diff(Em,X1), dX2dt = -diff(Em,X2).

Any help is appreciated. Thank you!

[Stephan]

Hi,

is it correct, that x and y are depending on t --> so: x(t) and y(t)?

Best regards

Stephan

[Tejas Adsul]

The x and y whose values have been mentioned are constants. The variables X1 and X2 do depend on time. However, what I can do is ignore this time dependence, find the diff(Em,X1) and diff(Em,X2) expressions, then consider the time dependence by writing out the ode diff(X1,t) = -diff(Em,X1) and diff(X2,t) = -diff(Em,X2). I was able to do the first part, and am stuck with the second part.

Answer 1

[Stephan]

Hi,

does this work for your purpose?

x1=0.3;
x3=0.9; 
y1=0.4;
y3=0.8;
A=10; 
p=10;

syms X_1 X_2 X1(t) X2(t);
num = ((x3-X_1)-(X_1-x1))*X_1 + ((y3-X_2)-(X_2-y1))*X_2;
den = sqrt((x3-X_1)-(X_1-x1)^2 + (y3-X_2)-(X_2-y1)^2);
Em = -A*(X_1-x1)*(x3-X_1) - A*(X_2-y1)*(y3-X_2) + p*num/den;

dEm_dX_1 = diff(Em,X_1);
dEm_dX_2 = diff(Em,X_2);

dEm_dX_1 = (subs(dEm_dX_1, [X_1 X_2], [X1 X2]));
dEm_dX_2 = (subs(dEm_dX_2, [X_1 X_2], [X1 X2]));

ode1 = diff(X1,t) == -dEm_dX_1;
ode2 = diff(X2,t) == -dEm_dX_2;

ode = matlabFunction([ode1; ode2])

ode is a function handle:

ode =

    function_handle with value:

      @(t)[diff(X1(t),t)==X1(t).*-2.0e1+(X1(t).*4.0e1-1.2e1).*1.0./sqrt(-X1(t)-X2(t)-(X2(t)-2.0./5.0).^2-(X1(t)-3.0./1.0e1).^2+1.7e1./1.0e1)+(X1(t).*2.0+2.0./5.0).*(X1(t).*(X1(t).*2.0-6.0./5.0).*1.0e1+X2(t).*(X2(t).*2.0-6.0./5.0).*1.0e1).*1.0./(-X1(t)-X2(t)-(X2(t)-2.0./5.0).^2-(X1(t)-3.0./1.0e1).^2+1.7e1./1.0e1).^(3.0./2.0).*(1.0./2.0)+1.2e1;diff(X2(t),t)==X2(t).*-2.0e1+(X2(t).*4.0e1-1.2e1).*1.0./sqrt(-X1(t)-X2(t)-(X2(t)-2.0./5.0).^2-(X1(t)-3.0./1.0e1).^2+1.7e1./1.0e1)+(X2(t).*2.0+1.0./5.0).*(X1(t).*(X1(t).*2.0-6.0./5.0).*1.0e1+X2(t).*(X2(t).*2.0-6.0./5.0).*1.0e1).*1.0./(-X1(t)-X2(t)-(X2(t)-2.0./5.0).^2-(X1(t)-3.0./1.0e1).^2+1.7e1./1.0e1).^(3.0./2.0).*(1.0./2.0)+1.2e1]

depending on time, which should be able to solve like you wanted to do.

Running it as a live script gives:

That's what you wanted to achieve?

Best regards

Stephan

[Tejas Adsul]

How do I solve from this? Is there a way to get values of X1 and X2 in a time interval, say [0,5]? I probably cannot use ode45 on this, since it contains symbolic functions X1(t) and X2(t) and it gives the error 'Inputs must be floats, namely single or double'.

[Stephan]

Hi,

the solutions are complex - there seems to be a phase shift - to plot that in you still have to calculate the amplitude and the phase shift - there should be functions for this.

To solve the system, it had to be rewritten as shown:

tspan = 0:0.01:5;
y0 = [0; 0];

[t,X] = ode45(@odefun, tspan, y0)

plot(t, X(1))
hold on
plot(t, X(2))

function ode = odefun(t, X)
ode = zeros(2,1);
ode(1) = (40*X(1) - 12)/(17/10 - X(2) - (X(2) - 2/5)^2 - (X(1) - 3/10)^2 - X(1))^(1/2) - 20*X(1) + ((2*X(1) + 2/5)*(10*X(1)*(2*X(1) - 6/5) + 10*X(2)*(2*X(2) - 6/5)))/(2*(17/10 - X(2) - (X(2) - 2/5)^2 - (X(1) - 3/10)^2 - X(1))^(3/2)) + 12;
ode(2) = (40*X(2) - 12)/(17/10 - X(2) - (X(2) - 2/5)^2 - (X(1) - 3/10)^2 - X(1))^(1/2) - 20*X(2) + ((2*X(2) + 1/5)*(10*X(1)*(2*X(1) - 6/5) + 10*X(2)*(2*X(2) - 6/5)))/(2*(17/10 - X(2) - (X(2) - 2/5)^2 - (X(2) - 3/10)^2 - X(1))^(3/2)) + 12;
end

The solutions are complex - there seems to be a phase shift - to plot that in the plot you still have to calculate the amplitude and the phase shift:

X =

     0.0000 + 0.0000i   0.0000 + 0.0000i
     0.0212 + 0.0000i   0.0214 + 0.0000i
     0.0442 + 0.0000i   0.0448 + 0.0000i
     0.0689 + 0.0000i   0.0704 + 0.0000i
     0.0955 + 0.0000i   0.0983 + 0.0000i
     0.1240 + 0.0000i   0.1287 + 0.0000i
     0.1544 + 0.0000i   0.1616 + 0.0000i
     0.1871 + 0.0000i   0.1975 + 0.0000i
     0.2222 + 0.0000i   0.2367 + 0.0000i
     0.2602 + 0.0000i   0.2798 + 0.0000i
     0.3019 + 0.0000i   0.3280 + 0.0000i
     0.3497 + 0.0000i   0.3842 + 0.0000i
     0.4041 + 0.0000i   0.4493 + 0.0000i
     0.4758 + 0.0000i   0.5358 + 0.0000i
     0.6533 + 0.0339i   0.7352 + 0.0496i
     0.6354 + 0.1948i   0.7471 + 0.2114i
     0.6144 + 0.2632i   0.7507 + 0.3097i
     0.5945 + 0.3075i   0.7453 + 0.3901i
     0.5769 + 0.3378i   0.7337 + 0.4585i
     0.5619 + 0.3584i   0.7175 + 0.5176i
     0.5499 + 0.3721i   0.6983 + 0.5690i
     0.5410 + 0.3807i   0.6770 + 0.6138i
     0.5349 + 0.3855i   0.6540 + 0.6530i
     0.5316 + 0.3874i   0.6300 + 0.6873i
     0.5310 + 0.3872i   0.6052 + 0.7173i
     0.5329 + 0.3855i   0.5802 + 0.7437i
     0.5371 + 0.3827i   0.5548 + 0.7669i
     ...
     ...

I have to sleep now ...

If that helped please accept my anwer in order to help other users finding useful answers on similar problems

Best regards

Stephan

[Tejas Adsul]

Thank you very much! It helped a lot.

[Stephan]

Please note that i have assumed initial conditions:

X1(0)=0

X2(0)=0

You have to check this...i dont have an idea if this is correct.

Best regards

Stephan

[Tejas Adsul]

By the way, how did you write down the expression of ode(1) and ode(2) in the function odefun?

Answer 2

[Stephan]

Hi,

did i understand right:

x1=0.3;
x3=0.9; 
y1=0.4;
y3=0.8;
A=10; 
p=10;

syms X1 X2;
num = ((x3-X1)-(X1-x1))*X1 + ((y3-X2)-(X2-y1))*X2;
den = sqrt((x3-X1)-(X1-x1)^2 + (y3-X2)-(X2-y1)^2);
Em = -A*(X1-x1)*(x3-X1) - A*(X2-y1)*(y3-X2) + p*num/den;

dEm_dX1 = diff(Em,X1)
dEm_dX2 = diff(Em,X2)

dX1dt = -dEm_dX1
dX2dt = -dEm_dX2

This is what you wanted to do?

gives:

dX1dt =

(40*X1 - 12)/(17/10 - X2 - (X2 - 2/5)^2 - (X1 - 3/10)^2 - X1)^(1/2) - 20*X1 + ((2*X1 + 2/5)*(10*X1*(2*X1 - 6/5) + 10*X2*(2*X2 - 6/5)))/(2*(17/10 - X2 - (X2 - 2/5)^2 - (X1 - 3/10)^2 - X1)^(3/2)) + 12

dX2dt =

(40*X2 - 12)/(17/10 - X2 - (X2 - 2/5)^2 - (X1 - 3/10)^2 - X1)^(1/2) - 20*X2 + ((2*X2 + 1/5)*(10*X1*(2*X1 - 6/5) + 10*X2*(2*X2 - 6/5)))/(2*(17/10 - X2 - (X2 - 2/5)^2 - (X1 - 3/10)^2 - X1)^(3/2)) + 12

Can this be correct?

Best regards

Stephan

[Tejas Adsul]

Yes, but this is the first part, which I have been able to do. The second part involves actually solving the ODE numerically (I don't think it can be solved analytically) and getting X1 and X2 as functions of time. In the end, I would like to plot (t,X1) and (t,X2).