# 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 = a x + b y, dot y = c x + d y where dot x denotes d x by d t 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 = A v using: v(t) = x(t), y(t); dot v = dot x, dot y, and A = a, b; c, d 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 times 1, 0 plus y times 0, 1).

Substituting v = alpha u_1 + beta u_2 into the left-hand side of dot v = A v 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 A v = 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).

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 and dot y = minus x This is equivalent to the simple harmonic oscillator equation x double dot equals minus x. To see this, differentiate dot x = y to get x double dot equals dot y equals minus 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 x double dot equals minus 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 x double dot equals the force experience by the particle, which is minus k x by Hooke's law (for some constant k), so the equation of motion is m x double dot - minus k x;

• if we work in units where m = k = 1 then this gives us back our equation x double dot = minus x.

Let's solve this equation. We have = 0, 1; minus 1, 0. This has characteristic polynomial det of minus t, 1; minus 1, minus t equals t squared + 1 which has roots lambda_1 = i and lambda_2 = minus i.

The eigenvectors are:

• for lambda_1 = i, 1, i,

• for lambda_2 = minus i, 1, minus i.

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

We write x, y in the form alpha u_1 + beta u_2, that is alpha times 1, i + beta times 1, minus i, or alpha + beta, i times (alpha minus beta). Thus alpha = (x minus i y) all over 2 and beta = (x + i y) all over 2.

Our equations are now: dot alpha = lambda_1 alpha = i alpha, and dot beta = lambda_2 beta = minus i beta. Dividing the first equation by alpha gives dot alpha over alpha = d by d t of log alpha, equals i so log alpha = i t + a constant, so alpha of t = C_1 e to the i t. Similarly, beta of t = C_2 e to the minus i t.

Therefore x, y = C_1 e to the i t times 1, i + C_2 e to the minus i t times 1, minus i, so x = C_1 e to the i t + C_2 e to the minus i t, and y = i C_1 e to the i t minus i C_2 e to the minus i t. This is the general solution to dot x = y, dot y = minus x. You may be worried about the fact that there are is here: this is supposed to describe the motion of a particle on a spring, so x and y should be real numbers. The is will all cancel out if we pick appropriate initial conditions.

Suppose x of 0 = 1 and y of 0 = 0 (particle at rest at distance 1 from the origin). Substituting t = 0 into our general solution, we get x of 0 = 1 = C_1 + C_2 and y of 0 = 0 = i (C_1 minus C_2). This implies C_1 = C_2 = a half. Therefore x of t equals e to the i t + e to the minus i t all over 2, and y of t equals e to the i t minus e to the minus i t all times i over 2. This means that x of t = cos t and y of t = minus sine 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.