# Piece wise smooth function

1. Jan 28, 2016

### baby_1

• Member warned about posting without the homework template
Hello
Here is my question

So I solved Euler DE and find

and when we apply the boundary condition we obtain y=0 . My teacher said that we should write it as two different function as

where H is (1/2).He solved this equation with this way

So Here are my questions:
a) why don't we accept the y=0 as our desire function?
b)how can obtain (a) constant in my equation with Euler approach?because in multidisciplinary function we can't find a ?

2. Jan 28, 2016

### Ray Vickson

What is a "multidisciplinary function"? I have never heard that term.

Anyway, is the subject matter one of Calculus of Variations, for $J = \int_0^1 (1-y'^2)^2 dx$? If so, $F_{y'} = -4 y'(1-y'^2) = -4 y' + 4 y'^3$, so the Euler equation gives $F_{y} = 0 = (d/dx) F_{y'}$, or $(3 y'^2- 1) y'' = 0$, so either $y'' = 0$ or $y' = \pm 1/ \sqrt{3}$.

On the other hand, we can change variables to $y'= z$, to get the problem
$$\min \:K(z) = \int_0^1 (1-z^2)^2 \, dx$$
with no specified boundary conditions on $z(0), z(1)$. Without using the Euler-Lagrange equation at all we see that we can make $K(z) = 0$ by taking $z(t)^2 = 1$ for all $t$, and that gives $x'(t) = \pm 1$, as your instructor said. Certainly that is an optimal solution, since $K(z) \geq 0$ for all PWS functions $z(t)$, and $K(z) = 0$ is attained by the solution where $z(t)^2 = 1 \; \forall \, t$.

The interesting thing about this problem is that the so-called necessary conditions of Euler and Lagrange give the wrong solution. In other words, the Euler-Lagrange equations fail! I suspect that the reason lies in some violation of the hypotheses that underlie the Euler-Lagrange derivation (and which are normally not stated or are ignored when solving problems).

Last edited: Jan 28, 2016