# Find the extremals of $I[y]=\int_0^1(y')^2 \mathrm dt+\{y(1)\}^2$

Could anyone help me find the extremals of $$I[y]=\int_0^1(y')^2 \mathrm dx+\{y(1)\}^2$$ subject to $y(0)=1$

Most crucially I can't work out how to find the boundary $x=1$. I'm trying to go back to first principles and letting $y \rightarrow y+\alpha \eta$. Here the normal step would be to differentiate and set $\alpha=0$ and hence derive the Euler-Lagrange equation.

If anyone can explain to me how to deal with this case i'd be very grateful!

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You can write $[y(1)]^2=[y(x)^2]_0^1-1=\int_0^1\frac 12yy'dx -1$ and use what you know. – Davide Giraudo Jan 17 '12 at 20:58
@DavideGiraudo: That is excellent! So group together and solve the Euler-Lagrange equation. How does this help us find the boundary condition though? – Freeman Jan 17 '12 at 21:04
You have to put $y(1)=a$ and discuss the extremals in function of this parameter. – Davide Giraudo Jan 17 '12 at 21:12
@DavideGiraudo: Ok, I apologize if I don't understand this topic, as it's all be self taught this evening (problem sheet due in before the first lecture you see) So I solved the Euler-Lagrange equation to get $y''=0 \Rightarrow y=ax+1$ is this the correct approach? So the extremals is a family of lines? – Freeman Jan 17 '12 at 21:16
It's the correct approach. Can you show your computations which led you to Euler-Lagrange equation? – Davide Giraudo Jan 17 '12 at 21:24

It is obvious that $I(y)$ can be made arbitrarily large, so we only have to look for the minimum. The situation here is so simple that we can do with Schwarz' inequality. Put $y(1)=:y_1$. Then $$(y_1-1)^2=\Bigl(\int_0^1 1\cdot y'(x)\ dx\Bigr)^2\leq \int_0^1 1^2\ dx\cdot \int_0^1 y'^2(x)\ dx$$ with equality iff $y'(x)$ is constant. It follows that $$I(y):=\int_0^1 y'^2(x)\ dx + y_1^2\geq 2y_1^2-2y_1+1=2\Bigl(y_1-{1\over2}\Bigr)^2 +{1\over2}\ .$$ Therefore $I(y)\geq{1\over2}$, and the minimum is attained for the function $y(x):=1-{x\over2}$.