I have a problem with what seems a very simple functional maximization. Let's define:

$$ J[z]=\int \left( u(z)-\frac{\dot z^2}{2} \right) dt $$

Where $u(z)=-z^2+5$. The problem is to find

$$ \arg\max_z J[z]$$

Said in a colloquial way, to maximize the function $u(z)$ without varying too much $z$ with time. The second variation of the functional for an arbitrary variation $h(t)$ is:

$$ \frac{\delta^2}{\delta z^2}J[z]=\int \left(h^2 u''(z)-\dot h^2 \right) dt = -\int \left(2 h^2+ \dot{h}^2 \right) dt \le 0 \quad \forall h $$

So then the functional is convex and any stationary point satisfying the Euler-Lagrange equations would be a global maximizer of $J$. The Euler-Lagrange equations for this functional reduce to the following differential equation:

$$ \ddot z=-u'(z)=2z $$

But this doesn't make any sense, since the maximum of $u(z)$ is at $z=0$, and all the trajectories starting in $z\ne0$ diverge from that point in an exponential way. Where did I go wrong?

Thank you

  • $\begingroup$ Wait, so you've got $\ddot{z} = 2 z$ is what the maximum must satisfy. Doesn't this mean that your maximal function is then $z(t) = A e^{\sqrt{2} t} + B e^{-\sqrt{2} t}$ ? $\endgroup$ – amcalde Nov 17 '15 at 19:57
  • $\begingroup$ Yes that's right, it's not parabolic, my fault. But put as initial condition for example $z(0)=1$ and $z'(0)=0$. Then $A=B=0.5$ and the solution diverges while common sense say it should converge to 0! $\endgroup$ – Jorge del Val Nov 17 '15 at 20:54

Put $$ L[z] = 5 - z^2 - \frac{\dot z^2}{2} $$

Then by Euler-Lagrange the extremal solutions satisfy $$L_z = \frac{d}{dt}L_{\dot{z}} \Rightarrow -2 z = - \frac{d}{dt}\dot{z} $$ So we have $$ z_{*}(t) = A e^{\sqrt{2} t} + B e^{- \sqrt{2} t} $$ and $$L[z_{*}] = 5 - (A e^{\sqrt{2} t} + B e^{- \sqrt{2} t})^2 - \frac{(\sqrt{2} A e^{\sqrt{2} t} - \sqrt{2} B e^{- \sqrt{2} t})^2}{2}$$ $$= 5 - 2(A^2 e^{2\sqrt{2} t} + B^2 e^{- 2 \sqrt{2} t})$$

$$J[z_{*}] = \int L[z_{*} (t)] dt = 5*\mu(\Omega) - \left.\frac{A^2 e^{2\sqrt{2} t} + B^2 e^{- 2 \sqrt{2} t}}{\sqrt{2}} \right|_{\partial \Omega}$$

This appears to be perfectly convergent to me. Given boundary conditions on the function you could solve this easily.

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  • $\begingroup$ Thank you @amcalde. Sorry, I believe you made a mistake in your calculation (please, correct me if I'm wrong), since $-\left(Ae^{\sqrt{2}t}+Be^{-\sqrt{2}t}\right)^2-\left(Ae^{\sqrt{2}t}-Be^{-\sqrt{2}t}\right)^2$ is equal to $-2\left(A^2 e^{2\sqrt{2}t}+B^2 e^{-2\sqrt{2}t} \right)$, not 0. Then the functional diverges if $A$ is not equal to 0, that I think is true for many initial conditions. $\endgroup$ – Jorge del Val Nov 18 '15 at 17:17
  • $\begingroup$ For example for $z(0)=z_0$ and $z'(0)=0$, where if I'm not wrong $A=\frac{z_0}{2}=B$. $\endgroup$ – Jorge del Val Nov 18 '15 at 17:30
  • $\begingroup$ You are right. I made a big mistake. Still I don't think you have a divergent answer. See my edits $\endgroup$ – amcalde Nov 19 '15 at 14:26
  • $\begingroup$ Thank you. But that would mean that there doesn't exist any other function $z(t)$ that gives a better solution, right? Let's put the initial conditions $z(0)=2$ and $z'(0)=0$, that means that $A=B=1$ and $z(t)=\left(e^{\sqrt{2}t}+e^{-\sqrt{2}t}\right)$, so that the functional is $$J[z*]=5*\mu(\Omega)-\frac{1}{\sqrt{2}}\left(e^{2\sqrt{2}t}+e^{-2\sqrt{2}t}\right)\vert_{\partial \Omega}$$ But a simple solution such as $z_2(t)=2$ gives $J[z_2]=\mu(\Omega)$. For $\Omega=[0,10]$ for example, that gives $J[z*]=-1.359e12$ and $J[z_2]=10$, that is higher, so $z*$ couldn't be a solution. $\endgroup$ – Jorge del Val Nov 19 '15 at 16:37
  • $\begingroup$ For the calculus of variations you are supposed to fix boundary conditions on the function space. The Euler-Lagrange equations only apply to this case. If you fix an initial value and an initial derivative, this is not enough because the end of the $\eta$ function could vary wildly but it is required to go to zero through integration by parts step used in the Euler-Lagrange equation derivation. $\endgroup$ – amcalde Nov 19 '15 at 17:14

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