A second order ODE,initial value problem

Suppose$ f$ has a second derivative everywhere,and that $ f^{''}+f=0$.$ f(0)=0$,$ f^{'}(0)=0$.Then $ f=0$.

Proof:

Let me first prove that $ \forall x\in(0,1)$,$ f(x)=0$.

If not, there must exists $ x_0\in(0,1)$such that $ f(x_0)=a(a\neq 0)$.According to differential mean value theorem,there exists  $ x_1\in (0,x_0)$such that

$$ f^{'}(x_1)=\frac{f(x_0)-f(0)}{x_0-0}=\frac{a}{x_0}$$

Again according to mean value theorem,there exists $ x_2\in (0,x_1)$such that

$$ f^{''}(x_2)=\frac{f{'}(x_1)-f{'}(0)}{x_1-0}=\frac{a}{x_0x_1}$$

Because $ f+f{''}=0$,so $ f(x_2)=-\frac{a}{x_0x_1}$.

By using diffential mean value theorm multiple times,we can get

$$ f(x_4)=\frac{a}{x_0x_1x_2x_3}$$

$$ f(x_6)=\frac{-a}{x_0x_1x_2x_3x_4x_5}$$

$$ \vdots$$

$$ f(x_{2n})=\frac{(-1)^{n}a}{x_0x_1...x_{2n-1}}$$

$$ \vdots$$

It is easy to verify that $ |f(x_{2n})|=\frac{|a|}{x_0x_1...x_{2n-1}}>\frac{|a|}{x_0^{2n}}$.When $n\to\infty$,\begin{equation}\label{eq:28.15.08}\frac{|a|}{x_0^{2n}}\to\infty\end{equation}(Because $ x_0<1$).      

 

$ x_0,x_1,x_2,...x_n,...$is a decreasing sequense of numbers.It can be easily verified that this sequense of number has a greatest lower bound $ L$,$L\geq 0$.

Because $ f$ is  continuous at $ x=L$,which means when $\varepsilon\to 0$($\varepsilon>0$),$f(L+\varepsilon)\to f(L)$.  But from conclusion \ref{eq:28.15.08}, we know that $ f(L+\varepsilon)$can't be " too close" to $ f(L)$.

So $ \forall x\in(0,1)$,$ f(x)=0$.It  can be easily verified that $ f(1)=0$,$ f{'}(1)=0$,$ f{''}(1)=0$.Then we can verify that $ \forall x\in(1,2)$,$ f(x)=0$(Through the same method we used above).   ......

 

Expand our conclusion from $(0,1)$ to $(1,2)$,from $(1,2)$ to $(2,3)$.....and from $(0,1)$ to $(-1,0)$,from $(-1,0)$ to $(-2,-1)$.......,We can get $ f=0$.$ \Box$

 

 

Remark:Similary,Suppse$f$ has a second derivatives every where,and that $f{''}=f$.$f(0)=0.f{''}(0)=0.f{'}(0)=0$Then $f=0$