Critical points for system of 3 first order differential equations

Hi all,

I am trying to solve a system of non-linear ODE's for the critical points. I then need to determine the stability of the points.

My system is as follows:

u'1 = u1 + 4*u2 - u1*u2

u2' = 9*u1 + 4*u2 - u2*u3

u3' = 2*u1^2 + 9*u2^2 -89

It was recommended to me to try Newton's method to find the critical points...but I am unsure how to do so for 3 equations.

The methods that I have learned up until this point won't work, considering the unknowns. I'm not asking for a solution...simply a nudge in the right direction.

Thank you

Answer 1

[John D'Errico]

What would you define a critical point to be? Perhaps one where u'1=u'2=u'3=0?

How would you identify that set of points? Perhaps as solutions to the corresponding nonlinear system? It would seem the derivatives go away there. ;-)

Can Newton's method be applied to such a problem? Hint: do some reading about Newton-Raphson.

[spcrooks]

Thank you for your response, John.

Ok...so in that case I have:

0 = u1 + 4*u2 - u1*u2

0 = 9*u1 + 4*u2 - u2*u3

0 = 2*u1^2 + 9*u2^2 -89

I tried to set up a system

>> syms u1 u2 u3
eq1 = u1 + 4*u2 - u1*u2;
eq2 = 9*u1 + 4*u2 - u2*u3;
eq3 = 2*u1^2 + 9*u2^2 -89;
sol = solve(eq1,eq2,eq3);
>> sol

sol = 

    u1: [4x1 sym]
    u2: [4x1 sym]
    u3: [4x1 sym]

Is it implied in the equations that I set up that they are equal to zero? Or is there another way that I need to do that?

Thank's for your time.

[John D'Errico]

Yes, solve assumes you want to solve for an expression equal to zero, if that equality is not made explicit. For example, with x symbolic, these forms are all equivalent:

solve(x^2 - 1)

or

solve(x^2 - 1 == 0)

or

solve(x^2 == 1)

So use == to put an explicit statement of equality into solve. Be careful not to use = there of course. MATLAB will just get confused, and then yell at you, as it sometimes does to me with those red letter errors. ;-)

You might want to decide which of those 4 solutions it found are likely to arise in the solution of a system of ODES with real coefficients. Some are perhaps more likely than others.

[spcrooks]

Hi John,

Would you mind explaining the last part of your post a little further?

"You might want to decide which of those 4 solutions it found are likely to arise in the solution of a system of ODES with real coefficients. Some are perhaps more likely than others."

>> sol.u1
 
ans =
 
                                       2.9528257887999480273512430476418
                                      -6.5396872141013403699986845206388
 1.8185664466622181048231511874062*i + 5.7934307126506961713237207364985
 5.7934307126506961713237207364985 - 1.8185664466622181048231511874062*i
 
>> sol.u2
 
ans =
 
                                      -2.8198037702017479502653462807059
                                      0.62048209602953977043986038547764
 2.0996608370861040899127429476141 - 1.1150730758243549277843573328439*i
 1.1150730758243549277843573328438*i + 2.0996608370861040899127429476141
 
>> sol.u3
 
ans =
 
                                      -5.4245679008004677538388125712239
                                      -90.857184926912063329988160685749
 16.367098019959962943408360686655*i + 20.140876413856265541913486628487
 20.140876413856265541913486628486 - 16.367098019959962943408360686656*i

Perhaps it has something do with the solutions that I obtain from each equation?

Thank you,

Sean

[John D'Errico]

Consider that two of the solutions are complex, thus have a non-zero imaginary part.

Now, consider a system of ODEs, with entirely real coefficients on the right hand side. Therefore, the derivatives will also always be real numbers, as long as you start from a real initial value. Will such a system ever manage to pass through a point that has a non-zero imaginary component?

[spcrooks]

Thanks, John. Fair enough...no the system won't pass through a point with a non-zero imaginary comp. In that case the critical points are apparent.

Ok...so now if I want to determine the stability of these points...I determine the Jacobian matrix:

>> syms u1 u2 u3
jacobian([u1 + 4*u2 - u1*u2, 9*u1 + 4*u2 - u2*u3, 2*u1^2 + 9*u2^2 -89], [u1, u2, u3])
 
ans =
 
[ 1 - u2, 4 - u1,   0]
[      9, 4 - u3, -u2]
[   4*u1,  18*u2,   0]
[      0,      0,   0]

So now all I would need to do is figure out the code for evaluating it at my critical points?

[spcrooks]

Thanks for all your help, John. I appreciate it.

Sean