Given an infinitely differentiable function y(x) such that y'' + y' - y >= 0 for all x, and y(0) = y(1) = 0. If y(x) >= 0 for x E [0,1], then prove that y(x) is identically zero throughout [0,1].
Can anyone please help?
I just thought of a solution, can someone please tell me if I'm wrong somewhere?
y" + y' >= y >= 0 for x E [0,1]
-> d(y' exp(x))/dx >= 0
-> y' exp(x) is a monotonically non decreasing function for x E [0,1]
By Rolle's Theorem, slope of y(x) changes sign in between [0,1] at some point x = c (assuming y(x) is non constant)
Let c be the first of points at which y' = 0 if there are more than one)
Slope can't change from +ve to -ve as we go from [0,1] because then y' exp(x) will change signs from +ve to -ve as we go from [0,1] which it should not because it is non decreasing.
So slope of y(x) is negative from x=0 to x=c.
But this would mean that y(c) < y(0)
But y(0) = 0 hence y(c) < 0 which is a contradiction.
Hence y(x) is a constant.
Proof would have to be edited a bit to include for more than one x=c
Am I wrong somewhere?