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.
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.