# If $f(x)=f'(x)+f''(x)$ and $f(a)=f(b)=0$, then $f(x)=0\quad\forall x\in[a,b]$

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.

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

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

• 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. Sep 17, 2016 at 14:10
• Please tell me why $f(x)$ can't be a trigonometric function. Even a link will do. Sep 17, 2016 at 14:20
• Great observation!
– S. Y
Sep 17, 2016 at 14:28
• @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 $[a,b]$. Sep 17, 2016 at 14:32

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 $x.$

• 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. Sep 17, 2016 at 14:05
• Just edited the details on the determinant. Sep 17, 2016 at 14:09