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 $a$?

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

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

  • $\begingroup$ 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? $\endgroup$ – Simon S Mar 7 '15 at 17:23
  • $\begingroup$ What is the trick to convert the function? $\endgroup$ – user157012 Apr 10 '15 at 14:35
  • $\begingroup$ 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. $\endgroup$ – Simon S Apr 10 '15 at 15:47
  • $\begingroup$ @SimonS proceeding, I hit $y(x)=Ax+1$, now how will I determine constant $A$? The only unused clue so far is "$y\in C^2([0,1])$". $\endgroup$ – Jesse P Francis Oct 31 '15 at 11:40
  • 1
    $\begingroup$ @JessePFrancis If $y$ were not $C^2$ we would have trouble applying the E-L equation $\endgroup$ – 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 $y(x) = 1-x/2$.



  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 $$ y(1) + y^{\prime}(1)~=~0\quad\text{and} \quad\forall x\in [0,1]:y^{\prime\prime}(x)~=~0.\tag{3}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.