Calculus of variations and necessary conditions?

Question

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:

$y''-y=0$.

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 $A$?

Answer 1

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 $$ J(y)=4+\int_0^1(y^2+y'^2)\,dx. $$

Answer 2

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.