Calculus of variations Euler-Lagrange equation and variational problem

Question

Find all the extrema (local minima and maxima) of the function $$J[y] = \int\limits_1^2(xy' + y)^2\,\mathrm dx;\qquad y(1) = 1, y(2) = \dfrac12.$$

Hint. Once you've found the solution of the Euler-Lagrange equation with the boundary conditions, remember to check, like in the previous problem, if this solution is a minimum, a maximum or not an extremum.

The image above shows my work. I'm pretty sure I solved the E-L equation correctly with the boundary conditions, but I am not too sure about the variation part. I always seem to find an absolute minimum, which makes me think my understanding of this part is lacking.

Answer 1

Rather than going through your work line by line, let's see if I get the same answer: $$L=x^2y^{\prime2}+2xyy^\prime+y^2\implies 0=\frac{(\partial_{y^\prime}L)^\prime-\partial_yL}{2x^2}=y^{\prime\prime}+\frac2xy^\prime\implies y=A+\frac{B}{x}.$$The boundary conditions give $y=\frac1x$, as you said. With $y=\frac1x+\eta$ we get$$J=\int_1^2(x\eta^\prime+\eta)^2dx,$$which is minimal for $\eta=0$, so you're also right about the stationary point being a minimum.

Answer 2

Alternatively, one can change the variable $$z(x)~:=~xy(x).$$ Then OP's variational problem simplifies to $$I[z] ~:=~ \int_1^2z^{\prime 2}\,\mathrm dx~\geq~0;\qquad z(1)~=~1~=~z(2).$$ Note that the functional $I$ is non-negative. The Euler-Lagrange (EL) equation becomes $z^{\prime\prime}=0$. With the correct BCs, the EL equation leads to the constant solution $z_{\ast}(x)=1$. Since $I[z_{\ast}]=0$, the unique stationary solution $z_{\ast}$ minimizes the functional $I$.