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 = y^{2}, y(0) = d, is y(x) = max ( d e^{-h(x)} - ∫_{0}^{x} e^{-(h(x)-h(t) )} u(t)^{2} dt ), where the maximum is taken over all continuous u(t), and h(t) = ∫_{0}^{t} (f(s) - 2u(s)) ds.

**Solution**

Put F(x) = ∫_{0}^{x} f(t) dt. Put A(x) = a(x) e^{F(x)}, B(x) = b(x) e^{F(x)}. Then A' = (a' + f a) e^{F} = g e^{F}, and B' = (b' + f b) e^{F} ≥ g e^{F}, since e^{F} > 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) = ∫_{0}^{x} u(t) dt. Put h(x) = F(x) - 2 U(x) = ∫_{0}^{x} f(t) - 2u(t) dt. Then h(0) = 0. We have (y - u)^{2} ≥ 0, so y^{2} - 2uy ≥ -u^{2}. Now put z = y e^{h}, so that z(0) = y(0) e^{h(0)} = d.

z' = (y' + h'y) e^{h} = (y' + (f - 2u)y) e^{h} = (y^{2} - 2uy) e^{h} ≥ -u^{2} e^{h}. Hence z(x) = z(0) + ∫_{0}^{x} z'(t) dt ≥ d - ∫_{0}^{x} u(t)^{2} e^{h(t)} dt. Hence y(x) = z(x) e^{-h(x)} ≥ d e^{-h(x)} - ∫_{0}^{x} 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) = x ^{2}, b(x) = -x^{2} 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.
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Note also that the original question had a typo in it.*

© John Scholes

jscholes@kalva.demon.co.uk

15 Feb 2002