# Minimizing a functional with a free boundary condition

Find the extremals of the functional $$\text{J}(y)= y^2(1) + \int_0^1 y'^2(x)dx , \ \ y(0)=1.$$ Answer: $y(x)=1-\frac{1}{2}x$

My solution:

$F (x,y,y')=y'^2(x)$

After solving the Euler Lagrange equation we get $\frac{\mathrm{d}}{\mathrm{d}x}(2y')=0$

Which implies that $y=\frac{a}{2}x+b$ , using initial value condition we get $b=1$. Could you please help me find value of $a$?

• Do you need to also deal with that $y^2(1)$? – user99914 May 16 '15 at 6:15
• Could you please give me some hint? – zafran May 16 '15 at 6:16
• Calculate the functional for $y=ax/2+1$. Minimize the result in $a$. – mickep May 16 '15 at 6:20
• It is involved while calculating $J(ax/2+1)$. Try that calculation, and you'll see. – mickep May 16 '15 at 6:23
• I suggest, for the sake of completeness, write an answer yourself that you accept. – mickep May 16 '15 at 7:00

Note that we can write $J$ as

$$J(y) = \int_0^1 \left(2 y(x)y'(x) + y'(x)^2\right) \ dx \, + \, y(0)^2$$

Setting aside the constant $y(0)^2 = 1$ for now, we have $J$ as the integral of a function $F(y,y',x)$ where $F(y,y',x) = 2yy' + y'^2$.

You can now deploy your box of tricks on $F$ to find $y$. That is, the Euler-Lagrange equation:

$$0 = \frac{d \ }{dx} \frac{\partial F}{\partial y'} - \frac{\partial F}{\partial y} = \ ...$$

• No comments, OPer? BTW, when you arrive at an answer, it should be very intuitive. Look at your original expression for $J$. What would one guess with all those positive, squared quantities? – Simon S Mar 7 '15 at 17:23
• What is the trick to convert the function? – user157012 Apr 10 '15 at 14:35
• You mean how did I decide how to rewrite $J(y)$ in that form? I wanted to use calculus of variations and to do that I had to find a way to get rid of the $y(1)^2$ term of the original, since it is not at all obvious how to deal with it directly. Turns out we could get rid of the $y(1)^2$ term by using $y(0)$, which has been given to us. – Simon S Apr 10 '15 at 15:47
• @SimonS proceeding, I hit $y(x)=Ax+1$, now how will I determine constant $A$? The only unused clue so far is "$y\in C^2([0,1])$". – Jesse P Francis Oct 31 '15 at 11:40
• @JessePFrancis If $y$ were not $C^2$ we would have trouble applying the E-L equation – Simon S Oct 31 '15 at 17:17

Your approach is correct, but it could be expressed more precisely. The first step is to replace the free boundary problem by a more familiar variational problem: find $$M(c) = \inf\left\{\int_0^1 (y'(x))^2\,dx : y(0)=1, \ y(1)=c \right\}\tag{1}$$ Having found $M(c)$, you can minimize $c^2+M(c)$ over all $c\in\mathbb{R}$ and thus obtain the minimum of functional $J$.

The Euler-Lagrange equation for $(1)$ is $y''=0$, which leads to the minimizer $y(x) = 1+(c-1)x$ and subsequently $M(c) = (c-1)^2$.

Then $c^2+M(c) = c^2+(c-1)^2$ is minimized at $c= 1/2$, which delivers $\min J = 1/2$, attained by $y(x) = 1-x/2$.

Hints:

1. If one varies infinitesimally the functional $$J[y]~:=~y(1)^2 + \int_0^1\! dx~ y^{\prime}(x)^2\tag{1}$$ without discarding boundary contributions, one finds $$\delta J[y]~=~ 2 y(1)~\delta y(1) + 2\int_0^1\! dx~ y^{\prime}(x) ~\delta y^{\prime}(x)$$ $$~\stackrel{\text{int. by parts}}{=}~ 2\left[ y(1) + y^{\prime}(1)\right] \delta y(1) -2 y^{\prime}(0)~\underbrace{\delta y(0)}_{=0} - 2\int_0^1\! dx~ y^{\prime\prime}(x) ~\delta y(x).\tag{2}$$

2. Besides the given boundary condition $y(0)=1$, one concludes from formula (2) that a stationary configuration must obey $$y(1) + y^{\prime}(1)~=~0\quad\text{and} \quad\forall x\in [0,1]:y^{\prime\prime}(x)~=~0.\tag{3}$$