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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..

Please help anyone.....

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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 $y(x) = 1-x/2$.

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  • $\begingroup$ 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... $\endgroup$ – Empty May 5 '15 at 14:01
  • $\begingroup$ It is the same sign as in the original functional $J$. $\endgroup$ – daw May 5 '15 at 14:54
  • $\begingroup$ Not clear to me....If you give suggestion of any good book for this type of problem then it will be great.. $\endgroup$ – Empty May 5 '15 at 16:37

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