3
$\begingroup$

I have a standard variational problem of the form $$\mathcal{L}(y)=\int_a^b L[x,y,y^\prime]\mathrm{d}x,$$ where $L[x,y,y^\prime]$ is of the form $$L[x,y,y^\prime] = y^\prime g(y)$$ for some function $g(\cdot)$ which we can leave unspecified.

With that, the Euler-Lagrange (EL) equation simplifies, using Beltrami, into \begin{eqnarray} \frac{\partial \mathcal{L}}{\partial y} &=&L-y^\prime\frac{\partial L}{\partial y^{\prime}} \nonumber \\ &=& y^\prime g(y)-y^\prime\frac{\partial L}{\partial y^{\prime}} \nonumber \\ &=& y^\prime g(y)- y^\prime g(y) \nonumber \\ &=& 0 \nonumber \end{eqnarray} But, this should equal a constant. My only conclusion is that I cannot find any extremals for such $L$, no matter the function $g(y)$. Can someone clarify the situation to me?


Some more information:

  1. My $g(y)$ contains a Lagrange multiplier as my original problem has a constraint.
  2. I have a further constraint that I did not deal with, namely that the solution $y(x)$ must be monotonically increasing.
  3. I could provide my $g(y)$, but it appears as the problem with EL for my type of $L$ is general.

If there are no extremals to the original problem, could condition 2 alter this? To me it seems as the answer is negative, since we can incorporate this condition as $$\tilde{\mathcal{L}}(y)=\int_a^b L[x,y,y^\prime]-\lambda(x)y^{\prime}(x)\mathrm{d}x,$$ so that the problem is still of the form $$\tilde{L}[x,y,y^\prime]=y^\prime \tilde{g}(y).$$

$\endgroup$
15
  • $\begingroup$ Did you notice that $\mathcal{L}(y)=\int_{y(a)}^{y(b)}g(y)dy$ and thus $\mathcal{L}(y)$ only depends on the two values $\{y(a),y(b)\}$? $\endgroup$ Commented Aug 15, 2021 at 9:35
  • $\begingroup$ Not really....But the outcome of your integral would still depend on $y$...right? $\endgroup$ Commented Aug 15, 2021 at 9:50
  • $\begingroup$ $\mathcal{L}(y)$ does depend on $y$, but if I'm not wrong it depends only on the value of $y$ at $a$ and $b$. Thus I think that knowing $g$ is not irrelevant. $\endgroup$ Commented Aug 15, 2021 at 9:52
  • $\begingroup$ Ok, here it comes: $$g(y)=1-\lambda\left[ 1+\left(1-\mathrm{erf}\left(\sqrt{\frac{y}{N}}\right)\right)^2 \right]$$ $\endgroup$ Commented Aug 15, 2021 at 9:56
  • $\begingroup$ Is the aim of the exercise to find the minimum of $\mathcal{L}(y)$? $\endgroup$ Commented Aug 15, 2021 at 9:58

1 Answer 1

2
$\begingroup$

When we do a complete calculation we see that the variation only depends on the values at the end points: $$\begin{align} \delta \mathcal{L}(y) &= \delta \int_a^b y'(x) g(y(x)) \, dx \\ &= \int_a^b \delta y'(x) g(y(x)) \, dx + \int_a^b y'(x)g'(y(x))\delta y(x) \, dx \\ &= \left[ \delta y(x) g(y(x)) \right]_a^b - \int_a^b \delta y(x) g'(y(x)) y'(x))\, dx + \int_a^b y'(x)g'(y(x))\delta y(x) \, dx \\ &= \left[ \delta y(x) g(y(x)) \right]_a^b . \end{align}$$

Also, if $G$ is a primitive function of $g,$ then $$ \mathcal{L}(y) = \int_a^b y'(x) g(y(x)) \, dx = \left[ G(y(x)) \right]_a^b = G(y(b)) - G(y(a)). $$ Since $y(a)=0$ we get $$\mathcal{L}(y)=G(y(b))-G(0).$$ Thus, extremizing $\mathcal{L}(y)$ reduces to finding $y(b)$ that extremizes the above expression. That is, we shall solve $$ 0 = \frac{d}{dy(b)}\left( G(y(b))-G(0) \right) = G'(y(b)) = g(y(b)) \\ = 1-\lambda\left[ 1-\mathrm{erf}\left(\sqrt{\frac{y(b)}{N}}\right)^2 \right] , $$ i.e. $$ \mathrm{erf}\left(\sqrt{\frac{y(b)}{N}}\right) = \sqrt{1-\frac{1}{\lambda}} . $$


The constraint $\int_0^1 y^\prime[1-\mathrm{erf}^2(\sqrt{y/N})] dx=C,$ where $C$ is a given value, can be written as $$ \int_a^b y'(x)\,\frac{1-g(y(x))}{\lambda} \, dx = C $$ i.e. $$ \left(y(b)-y(a)\right)-\left(G(y(b))-G(y(a))\right) = \lambda C. $$ Thus, $$ \mathcal{L}(y)=G(y(b))-G(y(a))=y(b)-y(a)-\lambda C. $$

$\endgroup$
5
  • $\begingroup$ Thanks, but how do I find $y$? The constraint is that $$\int_0^1 y^\prime[1-erf^2(\sqrt{y/N})] dx $$ is a given value. $\endgroup$ Commented Aug 15, 2021 at 10:48
  • $\begingroup$ Sorry, there is a mistake that does not matter much for your answer. the function $g(y)$ I stated was wrong. The corrected version should be $$g(y)=1-\lambda[1-erf^2(\sqrt{y/N})].$$ But this only alters the two last equations of your solution, so the general approach is unaffected $\endgroup$ Commented Aug 15, 2021 at 10:53
  • $\begingroup$ So the constraint can be written as follows? $$\int_0^1 y'(x)\,\frac{1-g(y(x))}{\lambda} \, dx = \text{given value}$$ $\endgroup$
    – md2perpe
    Commented Aug 15, 2021 at 11:16
  • $\begingroup$ Is $a=0$ and $b=1$ so that the endpoints for the constraint are the same as for $\mathcal{L}(y)$? $\endgroup$
    – md2perpe
    Commented Aug 15, 2021 at 11:24
  • $\begingroup$ yes. I obtain\begin{eqnarray} C&=&\frac{N}{2}\mathrm{erf}^2\left(\frac{y(1)}{N}\right)-y(1)\mathrm{erf}^2\left(\frac{y(1)}{N}\right)-\frac{2N\exp(-\frac{y(1)}{N})\sqrt{\frac{y(1)}{N}}\mathrm{erf}\left(\sqrt{\frac{y(1)}{N}}\right)}{\sqrt{\pi}}\nonumber \\ &&-\frac{N\exp(-\frac{2y(1)}{N})}{\pi}+y(1)+\frac{N}{\pi} \nonumber \end{eqnarray} where $C$ is the "given value" so at least a numerical solution can be obtained. Thanks for the help. $\endgroup$ Commented Aug 15, 2021 at 13:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .