Showing instablity of differential equation.

Question

Assume differential equation

$$ x'=2x+y+x \cos t-y \sin t $$ $$ y'=-x+2y-x\cos t+y \sin t $$

1. Show that solution $(x(t),y(t))=(0,0)$ is unstable.
2. Is there a non-trival solution such that $\lim_{t\to \infty}(x(t),y(t))=(0,0)$.

for question 2 , I think there exist a non-trival solution such that $\lim_{t\to \infty}(x(t),y(t))=(0,0)$ and it shows that having zero limit at infinity doesn't show stablity.

Answer 1

First question

Multiply the first equation by $x'$, the second by $y'$ and add: $$ x\,x'+y\,y'=2(x^2+y^2)+x^2\cos t+y^2\,\sin t-2\,x\,y\cos t\sin t. $$ If $0\le t\le\pi/2$ then $0\le\cos t\le1$, $0\le\sin t\le1$ and $$\begin{align} x^2\cos t+y^2\,\sin t-2\,x\,y\cos t\sin t&\ge x^2\cos^2 t+y^2\,\sin^2 t-2\,x\,y\cos t\sin t\\ &=(x\cos t-y\sin t)^2\\ &\ge0. \end{align}$$ Thus, if $0\le t\le\pi/2$ then $$ (x^2+y^2)'\ge4(x^2+y^2). $$ Integrating $$ x^2+y^2\ge(x(0)^2+y(0)^2)e^{4t},\quad 0\le t\le\frac\pi2. $$

Answer 2

I do not know which stability criterion should be applied, so I try a simple argument:

An object at the origin $(0,0)$ which is disturbed by a small positive displacement $\epsilon$ in $x$-direction will experience a velocity $x' = (2 + \cos t) \epsilon > 0$, it will move away from the origin.

For a small disturbance $\epsilon$ in $y$-direction it will experience a velocity $y' = (2 + \sin t) \epsilon > 0$, which will move the object away as well.

About the second part I am not sure, because of the instability at $(0,0)$, something there could only stay, if undisturbed.

Here is a plot of the velocity field for $t=0$:

Answer 3

The equation is $$ \begin{bmatrix} x'\\y' \end{bmatrix} = \begin{bmatrix} 2+\cos(t)&1-\sin(t)\\ -1-\cos(t)&2+\sin(t) \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}\tag{1} $$ Although the trace of the square matrix above is $4+\sqrt2\sin(t+\pi/4)$ which is positive for all $t$, Artem has pointed out an article in which an example is given where some of the eigenvalues have positive real part yet the system is stable.

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Another approach

In an approach similar to Julián Aguirre's answer, we can write $$ \begin{bmatrix} x&y \end{bmatrix} \begin{bmatrix} x'\\y' \end{bmatrix} = \begin{bmatrix} x&y \end{bmatrix} \begin{bmatrix} 2+\cos(t)&1-\sin(t)\\ -1-\cos(t)&2+\sin(t) \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}\tag{2} $$ Noting that the eigenvalues of the symmetric part of the matrix in $(2)$ $$ \begin{bmatrix} 2+\cos(t)&-\tfrac12(\sin(t)+\cos(t))\\ -\tfrac12(\sin(t)+\cos(t))&2+\sin(t) \end{bmatrix}\tag{3} $$ are $$ \frac{4\pm\sqrt2+\sin(t)+\cos(t)}2\tag{4} $$ we get that for all $t$, $$ (2-\sqrt2) \begin{bmatrix} x&y \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix} \le\begin{bmatrix} x&y \end{bmatrix} \begin{bmatrix} x'\\y' \end{bmatrix} \le(2+\sqrt2) \begin{bmatrix} x&y \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}\tag{5} $$ That is, $$ (4-2\sqrt2)(x^2+y^2)\le(x^2+y^2)'\le(4+2\sqrt2)(x^2+y^2)\tag{6} $$ which implies that $$ e^{(4-2\sqrt2)t}(x_0^2+y_0^2)\le x^2+y^2\le e^{(4+2\sqrt2)t}(x_0^2+y_0^2)\tag{7} $$ Inequality $(7)$ shows that $(0,0)$ is not stable and is the only solution for which $\lim\limits_{t\to\infty}(x,y)=(0,0)$ (in fact, the only solution for which $(x,y)$ remains bounded).