Check whether a functional has an extremal or NOT

Question

Find the extremal of the functional $$J(y)=\int_a^b F(x,y,y')\,dx$$where , $F(x,y,y')=y'+y$ , for admissible functions $y$.

From Euler-Lagrange equation , $\displaystyle \frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right)-\frac{\partial F}{\partial y}=0$ we get , $-1=0$ ( absurd ). So , we can conclude that the functional has NO extramal.

Again we know when $x$ is absent in $F$ , then Euler-Lagrange equation is transformed into $\displaystyle F-y'\frac{\partial F}{\partial y'}=\text{ constant }$. Since in this problem $x$ is absent in $F$ so we can use it and using we get , $y=\text{constant }$ , which is the required extremal.

Why these two process give different result ? Which is the correct answer and why ? Please explain properly.

Answer 1

Euler-Lagrange equation is just a necessary condition for the functional to have an extrema. This is because the Euler-Lagrange equation is derived from the following:

Suppose $y(x)$ minimizes (maximizes) the functional. Then you have $J(y+\lambda v)\ge J(y)$ for each $\lambda >0$. Now dividing by $\lambda>0$ and letting $\lambda\to 0$ you get the Euler-Lagrange equation. So the equation says that if $y$ minimizes (maximizes) the functional $\Rightarrow$ the equation is satisfied. But if the equation is satisfied for some $y(x)$ it does not mean that it is indeed a minimizer (maximizer). This is the same as for functions in one variable: if the derivative at some point $x=a$ is $0$ then this point $x=a$ is a potential extremal point (but it could be just inflex point- like $x=0$ in $f(x)=x^3$).

For your problem, $J(y)=\int\limits_{a}^{b}{(y'+y )dx}=y(b)-y(a)+\int\limits_{a}^{b}{y dx}$.

Now, no matter if the set over which you search for minimizer/ maximizer is of functions with fixed values at $x=a$ and $x=b$, the expression $J(y)=y(b)-y(a)+\int\limits_{a}^{b}{y dx}$ is unbounded from above and from below. So it can not have neither minimum nor maximum over $C^1[a,b]$ or just over the set of differentiable functions in $[a,b]$:

1)If the set consists of differentiable functions with unfixed values at $x=a$ or $x=b$, then taking the functions $y_n(x)=n=const$, then $J(y_n)\to\infty$. Taking $y_n=-n\Rightarrow J(y_n)\to -\infty$.

2) If $y(a)$ and $y(b)$ are fixed, then again you can find sequences $y_n$ for which $J(y_n)\to\pm\infty$

Answer 2

If "$x$ is absent in $F$" as you say, you can only conclude that you have a conservation law. This is generally not enough to solve the Euler-Lagrange's equations, which are of second order in $x$. The conservation law acts as a constraint that can help you in ruling out absurd solutions, but not all functions satisfying the conservation law need be solutions to the Euler-Lagrange's equations. This is exactly what happens in your problem.

It might be illuminating to examine another Lagrangian system with a simple mechanical interpretation. Consider a stone, free falling under the action of gravitational force: $$ \frac{d^2 y}{dt^2}=-g,$$ Here $y$ is vertical position and $t$ is time. This system has a Lagrangian: $$ F(y, y')=\frac{(y')^2}2 -gy.$$ As you rightly observe, the fact that $F$ does not depend explicitly on $t$ gives the following conservation law: $$ F-y'\frac{\partial F}{\partial y'}=-\left(\frac{(y')^2}{2} + gy\right) = \text{constant in time}.$$ This correspond to the conservation of energy in a conservative mechanical system. This conservation law excludes that a function such as $$ y(t)=\sin(t) $$ can solve the original system. Indeed, a free falling stone that spontaneously oscillates up and down has never been observed. However, this conservation law does not rule out a function such as $$ y(t)=y_0>0 $$ corresponding to a stone standing forever in mid-air, which has never been observed either!

The bottom line is that a single conservation law is not enough to solve a Lagrangian system.