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?

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- February 26th 2010, 09:08 AMsashikanthInequalities in ODE's
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? - February 26th 2010, 11:46 AMsashikanth
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?