35. Eigenapplications, 1: Differential equations

Sketch of the idea

0.00 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 $$\dot{x}=ax+by\mathit{\hspace{1em}\hspace{1em}}\dot{y}=cx+dy,$$ where $\dot{x}$ denotes $dx/dt$ and $a,b,c,d$ are constants.

1.00 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: 2.13 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 $v=x\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \end{array}\right)+y\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \end{array}\right)$ ).

3.45 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).

4.33 Substituting it into the right-hand side gives $$Av=\alpha A{u}_1+\beta A{u}_2={\lambda }_1\alpha u_1+{\lambda }_2\beta u_2$$ (because $u_k$ is a ${\lambda }_k$ -eigenvector for $k=1,2$ ).

5.00 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

6.15 Consider the equations: $$\dot{x}=y,\dot{y}=-x.$$ 6.37 This is equivalent to the simple harmonic oscillator equation $\ddot{x}=-x$ . To see this, differentiate $\dot{x}=y$ to get $\ddot{x}=\dot{y}=-x$ .

7.17 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.

8.34 Let's solve this equation. We have . This has characteristic polynomial which has roots ${\lambda }_1=i$ and ${\lambda }_2=-i$ .

9.13 The eigenvectors are:

• for ${\lambda }_1=i$ , $\left(\begin{array}{c}\hfill 1\hfill \\ \hfill i\hfill \end{array}\right)$ ,

• for ${\lambda }_2=-i$ , $\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -i\hfill \end{array}\right)$ .

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

10.26 We write $\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)$ in the form $$\alpha u_1+\beta u_2=\alpha \left(\begin{array}{c}\hfill 1\hfill \\ \hfill i\hfill \end{array}\right)+\beta \left(\begin{array}{c}\hfill 1\hfill \\ \hfill -i\hfill \end{array}\right)=\left(\begin{array}{c}\hfill \alpha +\beta \hfill \\ \hfill i(\alpha -\beta )\hfill \end{array}\right).$$ 11.04 Thus $\alpha =(x-iy)/2$ and $\beta =(x+iy)/2$ .

12.06 Our equations are now: $$\dot{\alpha }={\lambda }_1\alpha =i\alpha ,\dot{\beta }={\lambda }_2\beta =-i\beta .$$ Dividing the first equation by $\alpha$ gives $$\frac{\dot{\alpha }}{\alpha }=\frac{d}{dt}(\log \alpha )=i,$$ so $\log \alpha =it+\text{const}$ , so $\alpha (t)=C_1{e}^{it}$ . Similarly, $\beta (t)=C_2{e}^{-it}$ .

13.21 Therefore $$\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)=C_1{e}^{it}\left(\begin{array}{c}\hfill 1\hfill \\ \hfill i\hfill \end{array}\right)+C_2{e}^{-it}\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -i\hfill \end{array}\right),$$ so $$x=C_1{e}^{it}+C_2{e}^{-it}\mathit{\hspace{1em}}y=i(C_1{e}^{it}-C_2{e}^{-it}).$$ 14.07 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.

14.50 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 $$x(0)=1=C_1+C_2,y(0)=0=i(C_1-C_2).$$ This implies $C_1=C_2=1/2$ . Therefore $$x(t)=\frac{1}{2}(e^{it}+e^{-it}),y(t)=\frac{i}{2}(e^{it}-e^{-it}).$$ 16.10 This means that $$x(t)=\cos (t),y(t)=-\sin (t)$$ 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.

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