# Solving a functional problem with break points

I'm totally out with solving functional problems when we need to find broken extremals, can someone show me how to solve the following problem:

Can the following problem have break (corner) points? Find extremum which could have on corner point at most

$$\min J[y]=\int_0^3y'^3\;dx,\;\;y(0)=0,\;y(3)=1$$

How should I apply the Weierstrass-Erdmann condition here? How to find broken extremals for functional problems? A concrete example is needed. I have googled and googled for an answer, but I can only find definitions but not a single example of how this is used...

P.S. $y=y(x)$. If you need any more information let me know.

Here is my own attempt. What do you think? :

$$\min J[y]=\int_0^3y'^3\;dx,\;y(0)=0,\;y(3)=1$$

$$F=y'^3,\;\;F_{y'}=3y'^2$$

Our task is to find a piecewise smooth extremal if such exists. If such an extremal exists, then at the point $x=c$, where the piecewise smooth extremal is not continuously differentiable, it must be that:

$$(\text{First Weierstrass-Erdmann condition})$$

$$F_{y_1'}|_{x=c-0}=F_{y_2'}|_{x=c+0}$$

that is:

$$3y_1'^2|_{x=c-0}=3y_2'^2|_{x=c+0}$$

or

$$y_1'^2|_{x=c-0}=y_2'^2|_{x=c+0},$$

where I have marked $y_1, y_2$ as the the two separate pieces of the function $y$. If $y_1=y_2$, then we have a only a smooth solution. Another possibility is that $y_1=-y_2$ or vice versa.

The other Weierstrass-Erdmann condition obviously hold for the case, when $y_1=y_2$, so lets check if they hold for $y_1=-y_2$ :

$$(\text{Second Weierstrass-Erdmann condition})$$ $$(F-y_1'F_{y_1'})|_{x=c-0}=(F-y_2'F_{y_2'})|_{x=c+0}$$

$$(y_1'^3-y_1'3y_1'^2)|_{x=c-0}=(y_2'^3-y_2'3y_2'^2)|_{x=c+0}$$

or

$$\left((-y_2)'^3-(-y_2)'3(-y_2)'^2\right)|_{x=c-0}=(y_2'^3-y_2'3y_2'^2)|_{x=c+0}$$

$$\left(-y_2'^3+3y_2'^3\right)|_{x=c-0}=\left(y_2'^3-3y_2'^3\right)|_{x=c+0}$$

$$2y_2'^3|_{x=c-0}=-2y_2'^3|_{x=c+0}$$

$$y_2'^3|_{x=c-0}=-y_2'^3|_{x=c+0}$$

The last equality holds only if $$y_2'=0,$$

which would mean that $y_1=C, y_2=-C$, which is not a piecewise smooth curve and the boundary conditions don't hold either for this possibility.

This means that there are only smooth solutions for this problem:

Euler's equation:

$$F_{y'}=3y'^2=C$$

$$y'=\pm C$$ $$y=Cx+D$$

from boundary conditions we get

$$y=\frac{1}{3}x$$