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Determine $y(x)$ for which $\int_{0}^{1} x^{2} + y^{'2}dx$ is stationary, subject to $\int_{0}^{1}y^2=2$, $y(0) = 0$, $ y(1) = 0$.

how to solve it?

I tried it: $f=x^{2} + y^{'2}$ and $g=y^2$

then $H=x^{2} + y^{'2}+\lambda y^2 $

then using euler's equation we have:

$y^"-\lambda y^{'}=0$

$y(x)=c_1\epsilon^{-\sqrt\lambda x}+c_2\epsilon^{\sqrt\lambda x}$

{using boundary conditons, we have $c_1=-c_2$}

$y(x)=c_2(-\epsilon^{-\sqrt\lambda x}+\epsilon^{\sqrt\lambda x})$

$y(x)=2 c_2 \sinh\sqrt\lambda x$

now how to get the value of $c_2$ and $\lambda$

how to solve it further?

Tomás
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amit
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1 Answers1

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In spite you wrote your Euler-Lagrange equation wrong the solution is correct with 1 remark. $\lambda$ might be negative and you don't know this yet. So you have 3 parameters $c_1,c_2$ and $\lambda$ and 3 conditions. From condition $y(0)=0$ you got $c_1=-c_2$. But you also have

$$ e^{\sqrt{\lambda}}=e^{-\sqrt{\lambda}}\\ $$ and (FIX) $$ \int_0^1 4 c_2^2 \sinh^2{\sqrt{\lambda}x}=2 $$ So from here it follows that $\lambda$ is negative and actually you solution is $2 c_2 \sin(\sqrt{-\lambda}x)$ and such that $\sin{\sqrt{-\lambda}}$=0. So $\lambda=-\pi^2$. $$ \int_0^1 \sin^2{\pi x} dx=1/2 $$ Thus $c_2=1$. So the final solution is $2\sin(\pi x)$. The only problem I have here that $\lambda$ is not defined uniquely. Actually $\lambda=-k^2\pi^2$, $k=1,2...$. This bothers me a bit. So probably we have a family of solutions $2\sin(\pm k\pi x), k=1,2,3$. All of them stationary but $2\sin(\pi x)$ gets the minimal value of the functional.

Alexander Vigodner
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