# Extremizing the following boundary value problem

Consider the functional $$J(y)=y^2(1)+\int_0^1y'^2(x)\,dx$$ with $y(0)=1$ , where $y\in C^2[0,1]$. If $y$ extremizes $J$ then find the value of $y(x)$.

I tried through Bolza problem. Firstly Euler-Lagrange equation gives , $y''=0$. So general solution is $y(x)=C_1x+C_2$. Now , $y(0)=1$ gives $C_2=1$. Then $y(x)=C_1x+1$.

Now Transversality condition gives , $\left[2y'(x)-2y(1)\right]_{x=1}=0\implies C_2=0.$ which is NOT possible, also I can not find $C_1$.

So how can I solve the problem ?

According to 'daw's' answer I am confused about the sign..

From Euler-Lagrange equation the solution is $y(x)=C_1x+1$. Putting this in the given $J(y)$ then for extreme value putting $\frac{dJ}{dC_1}=0$ find $C_1$. But I want to solve by Bolza-method..

The optimality condition at the free boundary point for minimizers of $$\int_0^1 f(x,y,y')dx + h(y(1))$$ is $$f_{y'} + h' = 0,$$ in your case $$2 y(1) + 2 y'(1) =0,$$ hence $y(1) = -y'(1)$, which amounts to $C_1+1=-C_1$, hence $C_1 = -1/2$. And $y(x) = 1-x/2$.
• When it will positive sign and when negative sign in $f_{y'}\pm h'$ ? I know and read it with negative sign in a book... – Empty May 5 '15 at 14:01
• It is the same sign as in the original functional $J$. – daw May 5 '15 at 14:54