# Problem of the Week #29 - December 17th, 2012

Status
Not open for further replies.

#### Chris L T521

##### Well-known member
Staff member
Here's this week's problem.

-----

Problem: Consider the vector field $v(x)=v_0+A(x)$ on $\mathbb{R}^n$, where $v_0$ is a constant vector, and $A:\mathbb{R}^n\rightarrow\mathbb{R}^n$ is a linear map. Find the flow $\varphi^t$ of $v$.

-----

#### Chris L T521

##### Well-known member
Staff member
Let $y(t)=\varphi^t(x)$ be the flow of $v$ and let $A(x)$ be represented by the matrix product $Ax$, where $x\in\mathbb{R}^n$. Then it's the solution to the differential equation $\dot{y}=v(y)=Ay+v_0$. This differential equation is a non-homogeneous system with the initial condition $y(0)=I$, where $I$ is the identity matrix. To solve this system of equations, we use variation of parameters. Thus, suppose $y_p=e^{tA}z(t)$. Then $\dot{y}_p=Ae^{tA}z+e^{tA}\dot{z}=Ay_p+e^{tA}\dot{z}$. For $y_p$ to be a solution to this system, we must have that
$e^{tA}\dot{z}=v_0\implies \dot{z}=e^{-tA}v_0\implies z=c+\int e^{-tA}v_0\,dt.$
Thus, $y_p=e^{tA}\left(c+\int e^{-tA}v_0\,dt\right)$. When $y_p(0)=I$, we have that $c=I$. Therefore, the flow is given by
$\varphi^t =y_p = e^{tA}\left(I+\int e^{-tA}v_0\,dt\right).$
Note that if $A$ was nonsingular, we have that $y_p=e^{tA}(I-A^{-1}e^{-tA}v_0)$.