# variational problem with constraints

Let me bring to your attention the following problem. Suppose we have the functional $$F = \int\limits_{a}^{b} f(y(x))\cdot\frac{dy}{dx} dx .$$ It is easy to see that that the Euler-Lagrange equation vanishes identically, in other words, every function $y (x)$ is an extremal of the functional. If we write the Euler-Lagrange equation we can see, that it is an identity.

Еvery Lagrangian of the form "function multiplied by the derivative of the 1st order" has this property (so called zero Lagrangian). However, the question arises whether the extremum is minimum or maximum. The second variation for it is identically zero, IMHO. Weierstrass function too. The question is how to determine whether the extremal is minimum or maximum?

Let us complicate the problem. There is a functional $$F = \int\limits_{a}^{b} n(y(x))\cdot f(y(x))\cdot\frac{dy}{dx} dx$$ with differential constraint $(\frac{dy}{dx} / n(y(x))) - 1 = 0$ - DE. We write the Lagrangian for this problem $$L1 = n(y(x))\cdot f(y(x))\cdot\frac{dy}{dx} + \lambda(x) \cdot ((\frac{dy}{dx} / n(y(x))) - 1 )$$ It is easy to see that this is "zero Lagrangian" and the EL equation becomes an identity. Can we know whether it is a minimum or maximum? How to find the Lagrange multiplier in such case?

If we now use the constraint (DE) and replace in the integral n(y) on the derivative, we obtain the Lagrangian which is quadratic in the derivative $$L2 = f(y(x))\cdot(\frac{dy}{dx})^2 + \lambda(x) \cdot ((\frac{dy}{dx} / n(y(x))) - 1 )$$ EL equation is no longer an identity:
$$\frac{d( f(y(x))\cdot (\dot{y})^2)}{dx} + \dot{\lambda} =0$$ or $$f(y(x))\cdot (\dot{y})^2 + {\lambda(x)} =0 .$$ Very strange EL equation. From my point of view, it means the following. Lagrange multiplier is a function only of the argument $x$, then it (and therefore the Lagrangian, as can be seen from equation) can be represented as a derivative with respect to $x$ of a function $p$. And this is in accordance with the well-known theorem sufficient criterion of "zero Lagrangian" (EL equation is an identity). The Weierstrass function also is non-zero.

Both of these problems (before the substitution of the equation in integral and after) are, I think, the same problem because, well, result can not depend on the arbitrary recording method. This is the apparent paradox.

You are aware that if $F$ is an anti-derivative of $f$, $F'(y)=f(y)$ then by the substitution rule, which is the counter-part of the chain rule $$I = \int\limits_{a}^{b} f(y(x))\cdot\frac{dy}{dx} dx =\int_a^b\frac{d}{dx}F(y(x))\,dx=F(y(b))-F(y(a)) .$$ so that the independence to the curve under constant end-points $y(a)=y_a$, $y(b)=y_b$, is rather trival.

In the same way, if $H(y)$ is an anti-derivative to $n(y)\cdot f(y)$ then $$I = \int\limits_{a}^{b} n(y(x))\cdot f(y(x))\cdot\frac{dy}{dx} dx=H(y(b))-H(y(a))$$ i.e., this case is not really different from the first one.

Forcing additionally $y'(x)=n(y(x))$ via Lagrange-multipliers makes the existence of admissible functions an accident, the IVP $y(a)=y_a$ rarely satisfies $y(b)=y_b$.

The further transformations do not, IMO, change this problem.

• Yes, this is obvious. But what can be said about whether the extremum is minimum or maximum? I wrote the Weierstrass function for L2 and found that it is positive. Am I right? – Juffin Mar 27 '15 at 13:31
• Where is the maximum or minimum of a constant function? – Lutz Lehmann Mar 27 '15 at 13:33
• And is it useful to speak of the maximum or minimum of a function that is defined on a single point, if even that? What if the domain is the empty set? – Lutz Lehmann Mar 27 '15 at 13:39
• Value at the end of the interval y(b) for example is not fixed, it is determined from the solution of the corresponding equation for x = b – Juffin Mar 27 '15 at 13:42
• Then please state so in your question. Fixed or free end-points of the curve is important information. And even then, the only thing to explore are the maxima and minima of (my) $F(y)$. Any curve from $y_a$ to $y_{\max}$ maximizes the first integral, any curve to $y_{\min}$ minimizes it. -- In the conditioned case, the free end-point at $x=b$ just guarantees that the admissible set consists of exactly one point, the solution curve of the IVP, there is nothing left to maximize over. – Lutz Lehmann Mar 27 '15 at 13:48