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## Help solving a system of differential equations

Asked by Tejas Adsul

### Tejas Adsul (view profile)

on 15 May 2018
Latest activity Commented on by Tejas Adsul

### Tejas Adsul (view profile)

on 16 May 2018
Accepted Answer by Stephan

### Stephan (view profile)

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

### Stephan (view profile)

on 15 May 2018
Hi,
is it correct, that x and y are depending on t --> so: x(t) and y(t)?
Best regards
Stephan
Tejas Adsul

### Tejas Adsul (view profile)

on 15 May 2018
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.

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R2017a

## 2 Answers

### Stephan (view profile)

Answer by Stephan

### Stephan (view profile)

on 15 May 2018
Edited by Stephan

### Stephan (view profile)

on 15 May 2018
Accepted Answer

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

### Tejas Adsul (view profile)

on 15 May 2018
Thank you very much! It helped a lot.
Stephan

### Stephan (view profile)

on 15 May 2018
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

### Tejas Adsul (view profile)

on 16 May 2018
By the way, how did you write down the expression of ode(1) and ode(2) in the function odefun?

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### Stephan (view profile)

Answer by Stephan

### Stephan (view profile)

on 15 May 2018
Edited by Stephan

### Stephan (view profile)

on 15 May 2018

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

### Tejas Adsul (view profile)

on 15 May 2018
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).

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