# 35. Eigenapplications, 1: Differential equations

## 35. Eigenapplications, 1: Differential equations

### Sketch of the idea

This is the first of three applications of eigenvectors and eigenvalues. Let $x(t),y(t)$ be functions of $t$ , and consider the simultaneous linear ordinary differential equations $\dot{x}=ax+by\qquad\dot{y}=cx+dy,$ where $\dot{x}$ denotes $dx/dt$ and $a,b,c,d$ are constants.

These equations are coupled: each equation involves $x$ and $y$ . Our goal is to decouple them, turning them into a pair of equations, each involving only one unknown function. We first rewrite the equations as $\dot{v}=Av$ using: $v(t)=\begin{pmatrix}x(t)\\ y(t)\end{pmatrix},\qquad\dot{v}=\begin{pmatrix}\dot{x}\\ \dot{y}\end{pmatrix},\qquad A=\begin{pmatrix}a&b\\ c&d\end{pmatrix}.$ Let $\lambda_{1},\lambda_{2}$ be the eigenvalues of $A$ and assume they are distinct; let $u_{1}$ and $u_{2}$ be eigenvectors for these eigenvalues. Write $v$ as $\alpha(t)u_{1}+\beta(t)u_{2}$ . In other words, $\alpha$ and $\beta$ are the components of $v$ when in the $u_{1}$ - and $u_{2}$ -directions (compare with $v=x\begin{pmatrix}1\\ 0\end{pmatrix}+y\begin{pmatrix}0\\ 1\end{pmatrix}$ ).

Substituting $v=\alpha u_{1}+\beta u_{2}$ into the left-hand side of $\dot{v}=Av$ gives: $\dot{v}=\dot{\alpha}u_{1}+\dot{\beta}u_{2}$ (because $u_{1},u_{2}$ are constant).

Substituting it into the right-hand side gives $Av=\alpha Au_{1}+\beta Au_{2}=\lambda_{1}\alpha u_{1}+\lambda_{2}\beta u_{2}$ (because $u_{k}$ is a $\lambda_{k}$ -eigenvector for $k=1,2$ ).

Comparing the components in the $u_{1}$ -direction, we get $\dot{\alpha}=\lambda_{1}\alpha.$ The $u_{2}$ -components tells us that $\dot{\beta}=\lambda_{2}\beta.$ These equations now involve only $\alpha$ and $\beta$ separately. We have decoupled the equations.

### An example in detail

Consider the equations: $\dot{x}=y,\quad\dot{y}=-x.$ This is equivalent to the simple harmonic oscillator equation $\ddot{x}=-x$ . To see this, differentiate $\dot{x}=y$ to get $\ddot{x}=\dot{y}=-x$ .

In general, if you have a second-order equation like this, you can define $y=\dot{x}$ (which gives you one equation) and then use the second-order equation to express $\dot{y}$ in terms of $x$ and $y$ : this is a good trick for converting second-order equations into pairs of first order equations.

Remark:

The equation $\ddot{x}=-x$ describes physical situations like a particle on a spring:

• one end of the spring is fixed at the origin;

• the particle sits at a distance $x$ from the origin;

• Newton's law tells us that if $m$ is the mass of the particle then $m\ddot{x}$ equals the force experience by the particle, which is $-kx$ by Hooke's law (for some constant $k$ ), so the equation of motion is $m\ddot{x}=-kx$ ;

• if we work in units where $m=k=1$ then this gives us back our equation $\ddot{x}=-x$ .

Let's solve this equation. We have $A=\begin{pmatrix}0&1\\ -1&0\end{pmatrix}$ . This has characteristic polynomial $\det\begin{pmatrix}-t&1\\ -1&-t\end{pmatrix}=t^{2}+1,$ which has roots $\lambda_{1}=i$ and $\lambda_{2}=-i$ .

The eigenvectors are:

• for $\lambda_{1}=i$ , $\begin{pmatrix}1\\ i\end{pmatrix}$ ,

• for $\lambda_{2}=-i$ , $\begin{pmatrix}1\\ -i\end{pmatrix}$ .

(We're just picking particular eigenvectors, not writing down the general eigenvector).

We write $\begin{pmatrix}x\\ y\end{pmatrix}$ in the form $\alpha u_{1}+\beta u_{2}=\alpha\begin{pmatrix}1\\ i\end{pmatrix}+\beta\begin{pmatrix}1\\ -i\end{pmatrix}=\begin{pmatrix}\alpha+\beta\\ i(\alpha-\beta)\end{pmatrix}.$ Thus $\alpha=(x-iy)/2$ and $\beta=(x+iy)/2$ .

Our equations are now: $\dot{\alpha}=\lambda_{1}\alpha=i\alpha,\qquad\dot{\beta}=\lambda_{2}\beta=-i\beta.$ Dividing the first equation by $\alpha$ gives $\frac{\dot{\alpha}}{\alpha}=\frac{d}{dt}(\log\alpha)=i,$ so $\log\alpha=it+\mbox{const}$ , so $\alpha(t)=C_{1}e^{it}$ . Similarly, $\beta(t)=C_{2}e^{-it}$ .

Therefore $\begin{pmatrix}x\\ y\end{pmatrix}=C_{1}e^{it}\begin{pmatrix}1\\ i\end{pmatrix}+C_{2}e^{-it}\begin{pmatrix}1\\ -i\end{pmatrix},$ so $x=C_{1}e^{it}+C_{2}e^{-it}\quad y=i(C_{1}e^{it}-C_{2}e^{-it}).$ This is the general solution to $\dot{x}=y$ , $\dot{y}=-x$ . You may be worried about the fact that there are $i$ s here: this is supposed to describe the motion of a particle on a spring, so $x$ and $y$ should be real numbers. The $i$ s will all cancel out if we pick appropriate initial conditions.

Suppose $x(0)=1$ and $y(0)=0$ (particle at rest at distance 1 from the origin). Substituting $t=0$ into our general solution, we get $x(0)=1=C_{1}+C_{2},\qquad y(0)=0=i(C_{1}-C_{2}).$ This implies $C_{1}=C_{2}=1/2$ . Therefore $x(t)=\frac{1}{2}(e^{it}+e^{-it}),\qquad y(t)=\frac{i}{2}(e^{it}-e^{-it}).$ This means that $x(t)=\cos(t),\qquad y(t)=-\sin(t)$ using the formulae for trigonometric functions in terms of complex exponentials. The minus sign in $y$ is because our particle starts moving towards the origin as time increases.

The moral of this story is that you can use eigenvectors to decouple systems of linear differential equations.