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A real-valued function $f(x)$ defined on a closed interval $[a,b]$ has the properties that $f(a) = f(b) = 0$ and $f(x) = f'(x)+f''(x)$ for all $x$ in $[a,b]$. Show that $f(x) = 0$ for all $x$ in $[a,b]$.

I'm not sure how to tackle this problem. Would integration help?


I see that second order linear differential equations require exponential functions. But wouldn't it be possible to have $f(x)$ as a trigonometric function? If not, please state why I'm wrong.

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    $\begingroup$ I suggest the straightforward :to solve the differential equation and see what the integration constants need to be to fulfill the border conditions. $\endgroup$ Sep 17, 2016 at 13:50

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Hint: every extremum above $x$-axis is a local minimum, every extremum below $x$-axis is a local maximum.

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  • $\begingroup$ Trivial remark: this is a better argument which generalizes the result. For example, it's hard to solve in explicit form $f(x)=(f'(x))^3+(f''(x))^7$, but the geometric idea still works. $\endgroup$ Sep 17, 2016 at 14:10
  • $\begingroup$ Please tell me why $f(x)$ can't be a trigonometric function. Even a link will do. $\endgroup$ Sep 17, 2016 at 14:20
  • $\begingroup$ Great observation! $\endgroup$
    – S. Y
    Sep 17, 2016 at 14:28
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    $\begingroup$ @Astrobleme: The extremum hint works for any function that satisfies your differential equation. Prove the hint. Then prove that $f$ can't have nonzero values in $[a,b]$. $\endgroup$ Sep 17, 2016 at 14:32
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Consider the linear second order differential equation $y''+y-y=0$, whose characteristic polynoliam is $\lambda^2+\lambda-1=0$ and has two distinct roots $\lambda_1,~\lambda_2$ (no nned to write the down). The general solution has the form $y=y(x)=c_1e^{\lambda_1x}+c_2e^{\lambda_2x}$, where $c_1,~c_2$ are constants such that $y(a)=0,~y(b)=0$, i.e $$c_1e^{\lambda_1a}+c_2e^{\lambda_2a}=0,~c_1e^{\lambda_1b}+c_2e^{\lambda_2b}=0.$$ Since the determinant of the matrix $$\left( \begin{array}{cc} e^{\lambda_1a} & e^{\lambda_2a} \\ e^{\lambda_1b} & e^{\lambda_2b}\\ \end{array}\right)$$ is $\exp\{\lambda_1a+\lambda_2b\}-\exp\{\lambda_2a+\lambda_1b\},$ non zero (since $\lambda_1,\neq \lambda_2,~a\neq b$), we get $c_1=c_2=0,$ that is $y(x)=0$ for all $x.$

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  • $\begingroup$ You need to use some properties of the $\lambda_i$. After all, $y''+y=0$ with $a=0$, $b=\pi$ would allow a nonzero solution. $\endgroup$ Sep 17, 2016 at 14:05
  • $\begingroup$ Just edited the details on the determinant. $\endgroup$ Sep 17, 2016 at 14:09

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