# Calculus of variations Euler-Lagrange equation and variational problem

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.

• At least make sure that the picture has a correct orientation. Also, it is better if you type out the problem text using MathJax instead of posting as a picture. See this and this. – an4s Feb 22 at 18:29

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.
• @PeterPolizogopoulos You mean $y=y_0+\eta$, where $y_0$ is the solution you want to classify. It's the best approach if you're not au fait with second-order functional derivatives. – J.G. Feb 22 at 21:09
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 $$I$$.