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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.

My work.

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.

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  • $\begingroup$ 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. $\endgroup$ – an4s Feb 22 at 18:29
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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.

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  • $\begingroup$ I appreciate the answer. Is the best way to find if it is min/max to just plug in y = η in to the functional then considering if it is minimal/maximal for η = 0? $\endgroup$ – Peter Polizogopoulos Feb 22 at 21:08
  • $\begingroup$ @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. $\endgroup$ – J.G. Feb 22 at 21:09
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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 $I$.

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