Shifted Systems of Linear Differential Equations

$${\vec{x}}^{\prime }=A(\vec{x}-\vec{a})\text{and}{\vec{x}}^{\prime }=A\vec{x}+\vec{b}$$

Transient Method

Substitute a variable into the system, so that we have a problem we know how to solve. Similar idea to U Substitution. Solution: Equilibrium + Transient

Type 1

${\vec{x}}^{\prime }=A(\vec{x}-\vec{a})$, where A is a constant square matrix, $\vec{a}$ is a constant vector Set $\vec{y}=\vec{x}-\vec{a}$, as $\vec{a}$ is an equilibrium, and $\vec{y}$ is the transient solution Solve for $\vec{y}(t)$ from ${\vec{y}}^{\prime }=A\vec{y}$. See 2D Homogeneous Linear Systems with Constant Coefficients Solution is $\vec{x}(t)=\vec{a}+\vec{y}(t)$

Example

$$\left[\begin{array}{c}{x}_1^{\prime }\\ x_2^{\prime }\end{array}\right]=\left[\begin{array}{cc}3& -3\\ -1& 5\end{array}\right]\left[\begin{array}{c}{x}_1-2\\ x_2-3\end{array}\right]$$

(a) Find all equilibria (b) Find general solutions (c) Solve with the initial condition $\vec{x}=\left[\begin{array}{c}1\\ 0\end{array}\right]$ (d) Sketch the phase portrait (e) Determine the stability of each equilibrium

(a) This ODE is of the form ${\vec{x}}^{\prime }=A(\vec{x}-\vec{a})$, i.e. Type 1, so the equilbrium is $\vec{a}=\left[\begin{array}{c}2\\ 3\end{array}\right]⟹(2,3)$ (b) Let $\vec{y}=\vec{x}-\vec{a}$, so we have ${\vec{y}}^{\prime }=A\vec{y}$, which we know how to solve because it is a 2D Homogeneous Linear Systems with Constant Coefficients. Eigenvalues = $\det (A-\lambda I)=0⟹\lambda =2,6$ Eigenvectors = $(A-{\lambda }_{i}I){\vec{u}}_i=\vec{0}⟹\vec{u}=\left[\begin{array}{c}3\\ 1\end{array}\right],\left[\begin{array}{c}-1\\ 1\end{array}\right]$

$$\therefore \vec{y}(t)=C_1\exp (2t)\left[\begin{array}{c}3\\ 1\end{array}\right]+C_2\exp (6t)\left[\begin{array}{c}-1\\ 1\end{array}\right]$$ $$⟹\boxed{\vec{x}(t)=\left[\begin{array}{c}2\\ 3\end{array}\right]+C_1\exp (2t)\left[\begin{array}{c}3\\ 1\end{array}\right]+C_2\exp (6t)\left[\begin{array}{c}-1\\ 1\end{array}\right]}$$

(c)

$$\vec{x}(0)=\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}2\\ 3\end{array}\right]+C_1\exp (2(0))\left[\begin{array}{c}3\\ 1\end{array}\right]+C_2\exp (6(0))\left[\begin{array}{c}-1\\ 1\end{array}\right]$$ $$\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}2\\ 3\end{array}\right]+C_1\left[\begin{array}{c}3\\ 1\end{array}\right]+C_2\left[\begin{array}{c}-1\\ 1\end{array}\right]⟹C_1=-1,C_2=-2$$ $$\vec{x}(t)=\left[\begin{array}{c}2\\ 3\end{array}\right]-\exp (2t)\left[\begin{array}{c}3\\ 1\end{array}\right]-2\exp (6t)\left[\begin{array}{c}-1\\ 1\end{array}\right]$$

(d) Center the portrait at the equilibrium point $(2,3)$ Both eigenvalues are postitive, so the eigenspace arrows all go out. Because they have the same sign, the curves off of the eigenspaces are parabolas. (e) The equilibrium is unstable.

Type 2

${\vec{x}}^{\prime }=A\vec{x}+\vec{b}$, where A is a constant square matrix, $\vec{b}$ is a constant vector Essentially, convert into Type 1 by finding an equilibrium $\vec{a}$ by solving $A\vec{x}+\vec{b}=\vec{0}$ for $\vec{x}$. If no equilbrium exists, you must use Variation of Parameters instead. Transient solution $\vec{y}=\vec{x}-\vec{a}$ that satisfies ${\vec{y}}^{\prime }=A\vec{y}$, solve for $\vec{y}(t)$. See 2D Homogeneous Linear Systems with Constant Coefficients Solution is $\vec{x}(t)=\vec{a}+\vec{y}(t)$