Let R' be the non-negative reals. Let f, g : R' → R be continuous. Let a : R' → R be the solution of the differential equation: a' + f a = g, a(0) = c. Show that if b : R' → R satisfies b' + f b ≥ g for all x and b(0) = c, then b(x) ≥ a(x) for all x. Show that for sufficiently small x the solution of y' + f y = y2, y(0) = d, is y(x) = max ( d e-h(x) - ∫0x e-(h(x)-h(t) ) u(t)2 dt ), where the maximum is taken over all continuous u(t), and h(t) = ∫0t (f(s) - 2u(s)) ds.
Solution
Put F(x) = ∫0x f(t) dt. Put A(x) = a(x) eF(x), B(x) = b(x) eF(x). Then A' = (a' + f a) eF = g eF, and B' = (b' + f b) eF ≥ g eF, since eF > 0. Hence B' ≥ A' for all x. But B(0) = A(0), so B(x) ≥ A(x) for all x and hence b(x) ≥ a(x) for all x.
Let u be any continuous function on R'. Put U(x) = ∫0x u(t) dt. Put h(x) = F(x) - 2 U(x) = ∫0x f(t) - 2u(t) dt. Then h(0) = 0. We have (y - u)2 ≥ 0, so y2 - 2uy ≥ -u2. Now put z = y eh, so that z(0) = y(0) eh(0) = d.
z' = (y' + h'y) eh = (y' + (f - 2u)y) eh = (y2 - 2uy) eh ≥ -u2 eh. Hence z(x) = z(0) + ∫0x z'(t) dt ≥ d - ∫0x u(t)2 eh(t) dt. Hence y(x) = z(x) e-h(x) ≥ d e-h(x) - ∫0x e-(h(x) - h(t) ) u(t)2 dt. So z(x) ≥ the maximum over all u.
If we put u = y (the solution), then we get equality.
Comment. Take f(x) = -2/x, g(x) = 0, c = 0. Then a(x) = x2, b(x) = -x2 are both solutions of y' + f y = g, y(0) = c. But b(x) < a(x) for all x > 0. So why does the result fail in this case? Because f is undefined for x = 0.
Note also that the original question had a typo in it.
© John Scholes
jscholes@kalva.demon.co.uk
15 Feb 2002