I can't see your pic. Add an answer to your own question which contains the question in words and (). e.g. partial-d^2(y(x))/partial-dx^2 - f(x) y(x) = 0 EDIT: Consider the positive function g(r,t)=A/B A=e^(-rt)(1-F(x+t)) B=integral(0..infinity)e^(-rs)(1-F(x+s))ds where r is a inflation rate and F(x) is a distribution function show that there is a value t0 such that g(r,t)<g(r',t) for t<t0 and g(r,t)>g(r',t) for t>t0 Consider Z(r,r',t) = g(r,t)/g(r',t) This is (A(r,t)/B(r)) / (A(r',t)/B(r')) which is (A(r,t)/A(r',t)) / (B(r)/B(r')) notice the denominator is constant wrt t. Let it be K. (A(r,t)/A(r',t)) / K = (1/K) e^(-rt) / e^(-r't) = (1/K) e^((r'-r)t) This shows how g(r,t)/g(r',t) vary. e^((r'-r)t) is a monotonic function in t. It always increases (or decreases) wrt t. Suince e^(r'-r')t is monotonic in t, the question now comes down to finding t0 such that Z(r,r',t0) = (1/K) e^((r'-r)t0))= 1 (r'-r) t0 = ln K t0 = (ln K) / (r' - r) But the requested condition is: "g(r,t)<g(r',t) for t<t0 and g(r,t)>g(r',t) for t>t0" is only true if Z(r,r',t) < 1 for t<t0 and Z(r,r',t)>1 for t>t0 This is only true if dZ/dt is positive at t=t0 dZ/dt is (r'-r) Z so this is only true if r'>r It is obvious that this condition is missing from your question, since swapping r and r' gives you a contradictory result. If r' > r the t0 in question is (ln (B(r)/B(r')))/(r'-r)

more details