TITLE: Find critical points of an ODE system
QUESTION [1 upvotes]: Find the critical points and the solution of the ODE system. Parameters $k_t, \space k_i $ are both positive constants.
$$\frac{d \vec u}{dt} = k_t \begin{bmatrix}-1 & 1 \\1 & -1 \end{bmatrix}u+k_i\begin{bmatrix}1 \\0 \end{bmatrix}$$
$$\vec u(0) = \begin{bmatrix}0 \\0 \end{bmatrix}$$
I have tried but can't find the critical points. Setting the RHS to $0$ doesn't seem to help much since both ODE's are not satisfied for the same $\vec u$.
Any ideas?

REPLY [1 votes]: When $A$ is singular (as is the case here), there may be no critical points and that is the case here.
We can quickly draw a phase portrait to see this (I randomly chose constants and all phase portraits will look like this, with the exception of the direction arrows changing direction).

If $A$ were nonsingular, we may have been able to transform it to a homogeneous system using a change of variables to make life easier.
Even though $A$ is singular, we may still be able to solve the system using many methods that can include Undetermined Coefficients, Variation of Parameters (VoP), Laplace Transforms, etc. Lets use VoP with the Matrix Exponential. We are going to mimic Solution 3.
We have the system  (I changed the constants to something easier)
$$\frac{d \vec u}{dt} = a \begin{bmatrix}-1 & 1 \\1 & -1 \end{bmatrix}u+ b\begin{bmatrix}1 \\0 \end{bmatrix} \\ \vec u(0) = \begin{bmatrix}0 \\0 \end{bmatrix}$$
See linked notes for details and please rework details.
$$A = a \begin{bmatrix}1 & -1 \\1 & 1 \end{bmatrix}, f(t) = b\begin{bmatrix}1  \\0 \end{bmatrix}, \vec u(0) = \begin{bmatrix}0 \\0 \end{bmatrix}$$
The eigenvalues / eigenvectors are $\lambda_1 = 0, \vec v_1 = \begin{bmatrix}1  \\1 \end{bmatrix}, \lambda_2 = -2a, \vec v_2 = \begin{bmatrix}-1 \\1 \end{bmatrix}$.
$e^{t A} = P e^{t D} P^{-1} = \begin{bmatrix}1 & -1 \\1 & 1 \end{bmatrix}\begin{bmatrix}1 & 0 \\0 & e^{-2 a t} \end{bmatrix}\begin{bmatrix}1/2 & 1/2 \\-1/2 & 1/2\end{bmatrix} = \dfrac{1}{2}\begin{bmatrix}1 + e^{-2 a t} & 1-e^{-2 a t} \\1-e^{-2 a t} & 1+e^{-2 a t} \end{bmatrix}$
The solution to the initial value problem is given by:
$\vec u(t) = e^{t A} \vec u(0) + \displaystyle \int_0^t e^{(t-s) A} f(s) ~ ds = \dfrac{b}{2} \displaystyle \int_0^t \begin{bmatrix}1 + e^{-2a(t-s)}  \\ 1 - e^{-2a(t-s)}  \end{bmatrix}ds = \dfrac{b}{4 a} \begin{bmatrix}2 a  t + 1 -  e^{-2 a t}  \\ 2 a t - 1 + e^{-2 a t} \end{bmatrix} $
You should redo the problem using Undetermined Coefficients (see example) and make sure you also know that method.
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