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A real-valued function $f$ defined on a closed interval $[a, b]$ has the properties that $f (a) = f (b) = 0$ and $f (x) = f'(x) + f''(x),\;\forall x \in [a, b]$. Prove: $$f (x) = 0, \;\forall x \in [a, b].$$

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up vote 7 down vote accepted

Instead of solving the differential equation directly, an alternative way is to consider the function $g=f^2$. Note that $g'=2ff'$ and $$g''=2(ff''+f'^2)=2(f'+f'')f''+2f'^2=f'^2+f''^2+(f'+f'')^2\ge 0.$$ Therefore, $g'$ is increasing. On the other hand, $g'(a)=g'(b)=0$. It follows that $g'\equiv0$, and $f\equiv 0$ consequently.

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+1 Superb answer. –  copper.hat Dec 13 '12 at 20:26
    
@copper.hat: Thank you! –  23rd Dec 13 '12 at 20:38
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What machinery are you allowed to use? You can just solve the differential equation. Let us define $x=a+(b-a)y$ and work with $g(y), y \in [0,1]$. The differential equation is solved with $g(y)=c_1\exp(r_1y)+c+2\exp(r_2y)$ where $r_1,r_2$ are distinct (the values don't matter, but are $\frac 1{2} \pm \frac {\sqrt 5}{2}$). Then we have $0=c_1+c+2, 0=c_1\exp(r_1)+c+2\exp(r_2)$ which can only be solved by $c_1=c_2=0$

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The equation $t^2+t-1$ has roots $\lambda_1 = \frac{1}{2}(-1 +\sqrt{5})$, $\lambda_2 = \frac{1}{2}(-1 -\sqrt{5})$, hence the general solution is $f(x) = c_1 e^{\lambda_1 x} + c_2 e^{\lambda_2 x}$. The initial and final conditions give $c_1 e^{\lambda_1 a} + c_2 e^{\lambda_2 a} = 0$ and $c_1 e^{\lambda_1 b} + c_2 e^{\lambda_2 b} = 0$. Since $\lambda_1 \neq \lambda_2$ these equations have a unique solution, which is $c_1=c_2 = 0$.

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