# Conditions on initial value for system of ODEs

Given $\dot{x} = x - y + e^{-t}$ and $\dot{y} = x+y + e^{-t}$. Find the set of initial conditions at $t = 0$ s.t $\lim_{t\rightarrow \infty} (x(t), y(t)) = (0, 0)$ whenever $(x(t), y(t))$ satisfies one of these initial conditions.

My progress: After using the "Variation Of Parameters" formula (aka, solutions to the ODE: $\dot{x} = A(t)x + g(t,x)$, have the form: $\phi(t) = \gamma(t)\gamma^{-1}(t_0)x_0 + \gamma(t)\int_{t_0}^{t_1} \gamma^{-1}(s)g(\phi(s), s)ds$, where $\gamma(t) =$ fundamental matrix solution to $\dot{x} = Ax$). Now, by letting $z_1 = x, z_2 = y$, and doing a bunch of algebraic manipulations, I ended up with the solution:

$z(t) = [z_{01}cos(t) - sin(t)(z_{02} + \frac{2(i-1)}{5}), sin(t)(z_{01}+\frac{1-i}{5})+ z_{02}cos(t)]$ where $z(0) = [z_{01} \ z_{02}]$.

Hopefully this is correct, as I checked over my computations several times. If that's the case, we see that for $\lim_{t\rightarrow \infty} (x(t), y(t)) = (0, 0)$, we need: $\lim_{t\rightarrow \infty} z(t) = (0, 0)$. But as $t\rightarrow \infty$, $\cot(t)$ oscillates between $(-\infty, \infty)$, and $z_{01}, z_{02}$ are some fixed numbers at $t = 0$, how can $\frac{z_{02} \ +\ \frac{2(i-1)}{5}}{z_{01}} = \cot{t}$ in that case? We cannot really let $z_{01} = 0, z_{02} = \frac{-2(i-1)}{5}$ as well, because then the 2nd component of $z(t)$ is not satisfies.

From these observations, I think there doesn't really exist any set of initial conditions that gives $\lim_{t\rightarrow \infty} (x(t), y(t)) = (0, 0)$, which is weird:P

My question: Can anyone please give this problem a try to see if my conclusion above is correct? Any help would really be appreciated.

• Nobody wants to help me with this problem? The problem is that despite reviewing my computations for several times, I couldn't see where i made a mistake when computing the "Variation of parameters" formula:P – user177196 Oct 18 '15 at 2:57
• @NormalHuman: can you please give this problem a try? – user177196 Oct 18 '15 at 4:42

Your calculations are obviously wrong, because you shouldn't have imaginary part in your solution. See that

$$e^{At} = e^{t} \begin{bmatrix} \cos t & -\sin t \\\sin t & \cos t \end{bmatrix}$$

Now, observe that

$$z(t) = e^{At} z(0) + \int_0^t e^{A(t-s)} \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{-s} ds$$

The integral part will contain the exponential term $e^{-t}$ after integrating, so it will go to $0$ at infinity. Since, $\lim e^{At} = \infty$, the only choice is $z(0) = 0$.

• Beautiful! Thanks for your great explanation. I made this stupid mistake: $e^{(1+i)t} \neq cos(t)+i\sin(t)$ when calculating the integral that made everything go wrong:(( I know it's so shameful to ask this, but can you give me some help again with this problem for part (a) only: math.stackexchange.com/questions/1484079/… – user177196 Oct 18 '15 at 20:49
• Wait a minute... How did you do the integral of a $e^{-As}e^{-s}$ in this case to show that it has an exponential term $e^{-t}$ after integrating? I don't think it makes sense to integrate in a normal way?? – user177196 Oct 18 '15 at 21:08
• Each element of the matrix $e^{-As} e^{-s}$ contains something like $e^{-2s} \cos s$ or $e^{-2s} \sin s$. After integrating it will contain $e^{-2s} (a \cos s + b \sin s)$ where $s$ runs from $0$ to $t$. So, after multiplying with $e^{At}$ from the left, each element will contain something like $e^{-t} (a \cos t + b \sin t)$. – obareey Oct 18 '15 at 21:19
• No, it does not contradict with this theorem. This theorem is true for all $t \in \mathbb{R}$, but the limit does not exists (because it isn't finite) as $t$ goes to infinity when the real part is positive. $\mathbb{R}$ does not contain infinity. – obareey Oct 18 '15 at 21:45
• Put any finite $t$ in the theorem an it is true regardless of how big the numbers are. Even if the numbers are growing exponentially fast, they are still finite for a finite $t$. Infinity is a different story and it is a bit hard to get your head around at first. I suggest you do some reading on infinity, as you will constantly be using the concept. – obareey Oct 18 '15 at 21:54

I am not totally sure about your solution, however I will give you a few hints.

Notice that this is nonhomogeneous linear differential equation, so the general solution would contain a linear combination of the solutions to the homogeneous differential equation and the particular solution, which satisfies the nonhomogeneity.

Do you have to find the complete solutions to homogeneous differential equation? Because only the solution corresponding to eigenvalues with negative real parts will disappear at $t\to\infty$.

• Thanks for your hint. If you tried computing the solutions to the homogeneous DEs above, you would get the two eigenvalues are $1-i$ and $1+i$ => The fundamental matrix solution to the homogeneous part is: $F(t) = [ie^{(1+i)t}\ -ie^{(1-i)t}, e^{(1+i)t}\ e^{(1-i)t}]^T$. This is NOT the complete solutions, but as you can see in this case, ALL the solutions to the homogeneous DE must contain the term $e^{t}$, so nothing will disappear as $t\rightarrow \infty$. I hope you can work out more details. Please let me know if I missed something. – user177196 Oct 18 '15 at 3:59
• Your first hint is essentially the same as the "variation of parameters" formula that I used above. So not quite helpful in this case:( – user177196 Oct 18 '15 at 5:42
• @user177196 Like I stated in my answer, I am not sure of your solution is correct, since I would expect it to be real. With this I mean that real initial conditions always yield a completely real solution, but your solution seems to complex. Also have you tried to substitute your solution back into the differential equation? – Kwin van der Veen Oct 18 '15 at 10:57
• Those eigenvalues and conclusions are correct. But if the solution can not contain any combination of the solution to the homogeneous differential, then what which expression is still left? Also note that a homogeneous differential equation, with all its eigenvalues having a real part bigger than zero, will still go to $\vec{0}$ at $t\to\infty$ if the initial condition is $\vec{0}$. – Kwin van der Veen Oct 18 '15 at 11:02
• thanks for your great observation on separating the solutions. I tried picking the first column of $F(t)$ above (so the general solutions to the homogeneous DE have the form $C[ie^{(1+i)t}\ e^{(1+i)t}]$. The value of $C$ depends on the initial condition $x(0)$. Now, for the inhomogeneous equation $\dot{x} = Ax + e^{-t}$ with initial value $x(0)$, I get the solution is $C[ie^{(1+i)t}\ e^{(1+i)t}] \int_{0}^{t} 1/C [\frac{1}{ie^{(1+i)s}}\ \frac{1}{e^{(1+i)s}}]^T [e^{-s} \ e^{-s}]ds$. I find this weird because I have to take the inverse of $C[ie^{(1+i)t}\ e^{(1+i)t}]$, which doesn't exist. – user177196 Oct 18 '15 at 19:12