u(t) is a continuous function. x(t), y(t) is the solution of x' = -2y + u(t), y' = -2x + u(t) satisfying the initial condition x(0) = x0, y(0) = y0. Show that if x0 ≠ y0, then we do not have x(t) = y(t) = 0 for any t, but that given any x0 = y0 and any T > 0, we can always find some u(t) such that x(T) = y(T) = 0.
Solution
Subtracting, we have z' = 2z, where z = x - y. So z(t) = Ae2t. If x0 ≠ y0, then A ≠ 0 and so z(t) ≠ 0 for any t. Hence, in particular, we do not have x(t) = y(t) = 0 for any t.
If x0 = y0 = k, then z(0) = 0, so A = 0, so x(t) = y(t) for all t. Take (for example) u(t) = -(k/T) e-2t. Then we may solve x' + 2x = u with x(0) = k to get x(t) = k(1 - t/T) e-2t. This has x(T) = 0 as required.
© John Scholes
jscholes@kalva.demon.co.uk
14 Jan 2002