So I have a problem that I'm dealing with, but I'm not sure how to complete the answer.

Find the extremals of the functional $\displaystyle J(y) = \int_{a}^{b}(y^2+yy'+(y'-2)^2)\,dx$ over the domain $A = \{y \in C^2[0,1]:y(0) = y(1)=0\}$. Show that $J$ does not assume a maximum value at these extremals. Find the unique extremal and show it is an absolute minimizer of $J(y)$ in $A$.

The Euler-Lagrange equation yields the ODE:


The general solution is $y(x) =c_1e^{x}+c_2e^{-x}$, but the only $y(x)$ that satisfies the boundary conditions is $y(x) \equiv 0.$ This yields $J(y) = 4$.

It's easy to show that it isn't a local maximizer by considering $y(x) = \sin(\pi x)$, so $J(y) > 4$.

My question is what other conditions are there or what can I do to show that $y(x) = 0$ is in fact the absolute minimizer in $A$?

  • $\begingroup$ Little puzzled: if $y=\sin(\pi x)$, then $J(y)=\frac12(9+\pi^2)>4$. Are you supposed to show that the extremals are not maxima? Or that the extremals are not minima? $\endgroup$ – Adrian Keister Apr 30 '18 at 19:26
  • $\begingroup$ Sorry, meant to say that it's easy to show that it isn't a local maximizer. I want to find a unique local minimizer. $\endgroup$ – CrypticParadigm Apr 30 '18 at 20:53

In general, one would need a sufficient condition such as e.g. convexity and coercivity of the integrand in $y'$ (see Th. 1, p. 4 in this book). However, in this particular problem one can do integration by parts using $y(0)=y(1)=0$ $$ \int_0^1 yy'\,dx=\frac12[y^2(x)]_0^1=0,\qquad \int_0^1 y'\,dx=[y(x)]_0^1=0, $$ and the objective function becomes simply $$ J(y)=4+\int_0^1(y^2+y'^2)\,dx. $$

  • $\begingroup$ How is $\frac{1}{2}[y^2]_{0}^1 = 0$ and $[y]^1_0=0 $? Shouldn't it be $\frac{1}{2}[y^2]_{0}^1 = \frac{1}{2}$ and $[y]^1_0 = 1$? $\endgroup$ – CrypticParadigm Apr 30 '18 at 21:15
  • $\begingroup$ Using what I wrote, I get $J(y) = \frac{1}{2} + \int^1_0 y^2+y'^2 dx$ $\endgroup$ – CrypticParadigm Apr 30 '18 at 21:24
  • $\begingroup$ @physicsmajor Something wrong with your substitution. $[y^2]_0^1=y^2(1)-y^2(0)=0^2-0^2$ and similarly for the second one. You are given that $y(0)=y(1)=0$ in the domain $A$. $\endgroup$ – A.Γ. Apr 30 '18 at 22:15
  • $\begingroup$ @physicsmajor Oh, I understand now. It is $y(x)$, and we substitute $x=1$ and $x=0$, not $y$. $\endgroup$ – A.Γ. Apr 30 '18 at 23:27
  • $\begingroup$ Oh that's right. Thank you for clearing that up. But after that, we apply the Euler-Lagrange to $L=(y^2+y'^2)$, then what? Doing so and solving the ODE with boundary conditions I get $y(x) = 0$ again. What is supposed to follow? $\endgroup$ – CrypticParadigm May 1 '18 at 0:05

We have

$$ y-y'' = 0 \Rightarrow y = C_1 e^t+C_2 e^{-t} $$

With the boundary conditions we have $C_1 = C_2 = 0 \Rightarrow y = 0$

Analyzing the integral we have

$$ f = (y,y') = 4 + y^2 - 4 y' + y y' + (y')^2 \ge -\frac{4}{3} $$

such that

$$ H = \nabla(\nabla f(y,y')) $$

has positive eigenvalues at the minimum point.

for all $y, y'$ so $\int \left(4 + y^2 - 4 y' + y y' + (y')^2\right)dt$ has a global minimum.

  • 1
    $\begingroup$ Why for the integral to be bounded from below is sufficient for existence of min? E.g. $\int_0^1\left(\frac{y}{\dot y}\right)^2\,dx$ is also bounded from below by zero, but the minimum does not exist. $\endgroup$ – A.Γ. Apr 30 '18 at 20:13
  • $\begingroup$ It happens that $(y/y')^2$ has a saddle point, and the given function has a center. $\endgroup$ – Cesareo Apr 30 '18 at 20:26

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