Extrema of a functional

Question

I'm interested in computing an extremal for the functional

$J(y)= \displaystyle\int_{0}^{1} y' \sqrt{1+(y'') ^{2}} dx $

satisfying the boundary conditions $y(0)=y'(0)=0$ and $y(1)=1, y'(1)=2$.

My attempt:

My first attempt was obvious, that is , since $J$ does not depend on $y $ explicitly, there exist a constant $c_{1} $ such that:

$c_{1}= \displaystyle \frac{d}{dx} \frac{{\partial f}}{{\partial y''}}-\frac{{\partial f}}{{\partial y'}} $

And since

$\displaystyle \frac{{\partial f}}{{\partial y''}}= \frac{y y''}{\sqrt{1+(y'')^2}}$

$\displaystyle \frac{{\partial f}}{{\partial y'}}= \sqrt{1+(y'')^2}$

We have that

$c_{1}= \displaystyle \frac{y'y'''-(y'')^{2}-1}{(1+(y'')^{2})^{3/2}}$

I tried to solve the equation above but it is quite nontrivial for me, I just know the basics about ODE, for try to solve it I made the change of variable $z=y'$ and I got

$c_{1}(1+z'^{2}z^{2})^{3}=(z^{3}z''-1)^{2}$

However, I still don't see how to solve it if i follow that path :( .

My second attempt was to consider the change of variables

$x=r(\theta)cos (\theta)$ $y=r(\theta)sin(\theta)$

Thus,

$\displaystyle \frac{dy}{dx}= \displaystyle \frac{ rcos(\theta)+r' sin(\theta)}{-rsin(\theta)+r' cos(\theta)}$

And therefore,

$\displaystyle \frac{d^{2}y}{dx^{2}}= \displaystyle \frac{2(r')^{2}-r(r''-r)}{(-rsin(\theta)+r' cos(\theta))^{3}}$

Which implies that,

$ y' \sqrt{1+(y'') ^{2}} dx = \displaystyle \frac{ rcos(\theta)+r' sin(\theta)}{-rsin(\theta)+r' cos(\theta)} \sqrt{1+\displaystyle \frac{(2(r')^{2}-r(r''-r))^{2}}{(-rsin(\theta)+r' cos(\theta))^{6}}} d \theta$

As you can see, this not improve the situation haha :( .

Any help? Thanks in advance.

Answer 1

I think I have found the solution. Since $J$ does not depend explicitly on $y$, he have already said that there existe a constant $c_{1}$ such that

$ \displaystyle \frac{d}{dx} \frac{{\partial f}}{{\partial y''}} -\frac{{\partial f}}{{\partial y'}}=c_{1} $

But, it's also true that since $J$ does not depend explicitly on $x$ there exist a constant $c_{2}$ such that

$\displaystyle y'' \frac{{\partial f}}{{\partial y''}} - y' \left( \displaystyle \frac{d}{dx} \frac{{\partial f}}{{\partial y''}} -\frac{{\partial f}}{{\partial y'}} \right)-f= c_{2}$

That is, we have that

$\displaystyle y'' \frac{{\partial f}}{{\partial y''}} - y' c_{1}-f=c_{2}$

Therefore,

$-c_{1}y'- \displaystyle \frac{y'}{\sqrt{1+y''^2}}=c_{2}$

The boundary condition $y'(0)=0$ implies that $c_{2}=0$, and thus, we only need to solve

$-c_{1}y'= \displaystyle \frac{y'}{\sqrt{1+y''^2}}$

This last equation leads to $y''^2=\displaystyle \frac{1}{c_{1}^2}-1$ . Integrating and using the others boundary conditions we see that $y(x)=x^2$.

Answer 2

Experimental idea

Set $\sin(u)=\frac{y''}{\sqrt{1+y''^2}}\implies y''=\tan(u)$. Then the Euler-Lagrange equation reduces to $(y'\sin(u))'=c_1+\sqrt{1+y''^2}$, which further transforms to $$ \tan(u)\sin(u)+y'\cos(u)u'=c_1+\frac1{\cos(u)} \implies u'=\frac{\frac{c_1}{\cos(u)}+1}{y'} $$ With $y'=v$ this gives the first order system \begin{align} y'&=v,& y(0)&=0,&y(1)&=1\\ v'&=\tan(u),& v(0)&=0,&v(1)&=2\\ u'&=\frac{\frac{c_1}{\cos(u)}+1}{v} \end{align} This could be useful as a compact formulation for a BVP solver.

Constant solutions

If $c_1+\cos(u)=0$, then this gives a valid constant solutions. Thus $y''=C\implies y=A+Bx+\frac12Cx^2$. This also includes the cases $v=y'=0\implies y=A+Bx$. With the initial conditions the linear functions can be ruled out, the quadratic functions give a solution $y(x)=x^2$. This does not prove that it is the only solution or the optimal among them.

One further integration

Assuming that the denominators are not zero or at least not constant zero, combining the last two equations gives, $$ \frac{v'}{v}=\frac{\sin(u)u'}{c_1+\cos(u)}\implies v=\frac{c_2}{c_1+\cos(u)} $$ and then $$ \frac{\cos(u)u'}{(c_1+\cos(u))^2}=\frac1{c_2} $$ which in principle can be integrated...

Answer 3

I) General strategy:

1. OP's functional $J[y]$ is equivalent to $$S[y,v,\lambda]~=~\int_0^1\! \mathrm{d}x~L,\qquad L~=~ v\sqrt{1+\dot{v}^2}+\lambda(\dot{y}-v),\tag{A}$$ with the boundary conditions (BCs) $$ y(0)~=~0~=~v(0),\qquad y(1)~=~1,\qquad v(1)~=~2.\tag{B}$$ Here dot means differentiation wrt. $x$. (To get back to OP's functional $J[y]$ just eliminate the constraint $$\dot{y}~=~v\tag{C}$$ and the Lagrange multiplier $\lambda$.)

2. We next use Noether's theorem (NT):

   • Since $y$ is a cyclic variable, the corresponding momentum $$\frac{\partial L}{\partial \dot{y}}~=~\lambda\tag{D}$$ is a constant of motion.

   • Since the Lagrangian (A) has no explicit $x$-dependence, the energy $$ E~=~\dot{y}\frac{\partial L}{\partial \dot{y}} +\dot{v}\frac{\partial L}{\partial \dot{v}} +\dot{\lambda}\frac{\partial L}{\partial \dot{\lambda}}-L~=~\ldots~=~ \frac{v}{\sqrt{1+\dot{v}^2}} +\lambda v \tag{E}$$ is a constant of motion.

   If you don't know NT, you should just check that the total derivatives vanish $$ \frac{d\lambda}{dx}~=~0\qquad\text{and}\qquad \frac{dE}{dx}~=~0, \tag{F}$$ by using the EL equations for the functional (A).

3. Eq. (E) leads to a 1st order ODE $$ \dot{v}~=~\pm\sqrt{\frac{v^2}{(E-\lambda v)^2}-1},\tag{F} $$ which can be integrated $$ \pm (x-x_0)~=~\int\!\frac{\mathrm{d}v}{\sqrt{\frac{v^2}{(E-\lambda v)^2}-1}}~=~\ldots \tag{G}$$ to obtain the inverse function $v\mapsto x(v)$.

4. Eqs. (C) & (G) lead in principle to a solution $x\mapsto y(x)$ with 4 integration constants $(\lambda,E,x_0,y_0)$, which should be determined by the 4 BCs (B).

II) Specific boundary conditions:

1. Eq. (E) with the BC $v(0)=0$ yields $E=0$. Then eq. (F) simplifies to $\dot{v}={\rm const}$, which can be readily integrated to $y(x)=Ax^2+Bx+C$. The BCs (B) then yields $$y(x)~=~x^2. \tag{H}$$