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so I'm starting to work on Lyapunov functions and I would like to use them to prove the stability of the null solution of the following differential equation:

$$\frac{d}{dt}\left(\begin{array}{c}x\\y\end{array}\right) = \left(\begin{array}{c}- \sin(x) - y + sin(y)\\1 - \cos(x) - y + y^2\end{array}\right)$$

I know that for an equation $\frac{dx}{dt} = Ax$ the stability of the null solution can be proved by showing that all eigenvalues of $A$ have negative real part. However I'm not sure in the case above what is $A$, because of the $cos$ and $sin$ functions inside the given matrix. Thus I was wondering if someone could help my finding this matrix, and since it will be a $2\times 1$ matrix how can I find its eigenvalues ?

Moreover this is the only method I've seen to solve this problem, but if there is any other I'm willing to try it, there is no obligation on how to solve this problem.

I'm thinking about methods like those in the following links: Stability of the null solution of system of differential equations, How to find a Lyapunov function? and How to pick a Lyapunov function and prove stability? (if they can be applied here of course).

Thanks in advance for any help, have a good day!

Rhaena
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    Have you learned about linearization? You compute the Jacobian matrix of the right-hand side, and evaluate it at the equilibrium point whose stability you want to investigate, and that gives you your matrix $A$. – Hans Lundmark May 11 '21 at 12:53
  • Not at all @HansLundmark, but I can try that! So I need to find Jacobian matrix of the RHS of my equation and then.. what is the equilibrium point ? And as I stated I'm not sure how to work with the matrix of the RHS because of the trigonometric functions, for instance I'm not sure how to find the Jacobian matrix since I can barely find its characteristic polynomial, how would you do this? – Rhaena May 11 '21 at 12:55
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    The equilibrium point is $(0,0)$, I suppose, since you're talking about the “null solution”. In this case, the right-hand side is indeed equal to the zero vector when $(x,y)=(0,0)$, so $(x(t),y(t))=(0,0)$ is a solution. – Hans Lundmark May 11 '21 at 12:57
  • And the Jacobian matrix is... the Jacobian matrix. The $2 \times 2$ matrix of all possible partial derivatives of the two right-hand sides. – Hans Lundmark May 11 '21 at 13:00
  • So when you say equilibrium point you just mean the point I study? (just want to be sure for further exercises, if I find a question with same phrase). And I agree with you that the null solution, here $(0,0)$, is indeed a solution, but what can we deduce from this? Of course I had forgotten the Jacobian matrix could be found this way, thank you! – Rhaena May 11 '21 at 13:00
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    Yes. Any point where the right-hand side becomes the zero vector corresponds to a constant solution, so such a point is called an equilibrium point. In terms of the phase portrait, you're just staying at the same point all the time. And the question about stability of an equilibrium point is what happens if you start at a nearby point instead. – Hans Lundmark May 11 '21 at 13:03
  • Thank you very much for that detailed answer, I'll try your method right away and see where it gets me! – Rhaena May 11 '21 at 13:04
  • Well I'm surprised to have already finished, I find that eigenvalue of $A$ is just $-1$ (of algebraic multiplicity $2$), and so we are done. Thus I thank you once more, Linearization is very efficient a method! – Rhaena May 11 '21 at 13:13

2 Answers2

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To analyze the stability of the null solution (i.e., equilibrium point) of the given differential equation, one could use Jacobian matrix, which will represent the linearization of the system around the equilibrium point.

For the system:

$ \frac{d}{dt}\left(\begin{array}{c}x\\y\end{array}\right) = \left(\begin{array}{c}-\sin(x) - y + \sin(y)\\1 - \cos(x) - y + y^2\end{array}\right) $

The Jacobian matrix, denoted as $A$, is:

$ A = \left(\begin{array}{cc} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \end{array}\right) $

where $f_1$ and $f_2$ are the components of the right-hand side vector:

$ f_1 = -\sin(x) - y + \sin(y) $ $ f_2 = 1 - \cos(x) - y + y^2 $

Now, calculate the partial derivatives:

$ \frac{\partial f_1}{\partial x} = -\cos(x) $ $ \frac{\partial f_1}{\partial y} = -1 + \cos(y) $ $ \frac{\partial f_2}{\partial x} = \sin(x) $ $ \frac{\partial f_2}{\partial y} = -1 + 2y $

So, the Jacobian matrix is:

$ A = \left(\begin{array}{cc} -\cos(x) & -1 + \cos(y) \\ \sin(x) & -1 + 2y \end{array}\right) $

Now that you have the Jacobian matrix A, you can analyze the stability of the null solution by finding its eigenvalues. The eigenvalues of A will determine the stability properties of the equilibrium point. (If all eigenvalues have negative real parts, the equilibrium is stable. If any eigenvalue has a positive real part, the equilibrium is unstable).

To analyze the stability of the null solution (equilibrium point) by finding the eigenvalues of the Jacobian matrix (A), we can proceed as follows:

$ \det(A - \lambda I) = 0 $

where $I$ is the identity matrix.

Substitute the matrix $A$ into the equation:

$ \det\left(\begin{bmatrix} -\cos(x) - \lambda & -1 + \cos(y) \\ \sin(x) & -1 + 2y - \lambda \end{bmatrix}\right) = 0 $ the determinant:

$ (-\cos(x) - \lambda)(-1 + 2y - \lambda) - (-1 + \cos(y))\sin(x) = 0 $

This equation defines the characteristic polynomial of $A$.

Yaosheng Deng
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Start by linearizing the system getting

$$\frac{d}{dt}\left(\begin{array}{c}x\\y\end{array}\right) = \left(\begin{array}{c} a_1x+b_1y\\a_2x+b_2y\end{array}\right)$$

where each row now has two terms and thus corresponds to two columns of your coefficient matrix. Thus you read the coefficients $a_1,b_1$ as your first row and $a_2,b_2$ as your second row.

The eigenvalues will then be self-evident in this case.

Oscar Lanzi
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