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

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Chris L T521

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Jan 26, 2012
995
Here's this week's problem.

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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 $v$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
995
No one answered this week's question. Here's my answer below.

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 $y_p=e^{tA}(I-A^{-1}e^{-tA}v_0)$.
 
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