How should I use the Jacobi equation to show this?

[Math100]

Homework Statement

Let $a<b$ and let $f(x)$ be a continuously differentiable function on the interval $[a, b]$ with $f(x)>0$ for all $x\in [a, b]$. Let $A>0, B>0$ be constants. Using the Jacobi equation, show that the stationary path $y(x)=A+\beta\int_{a}^{x}\frac{dw}{\sqrt{f(w)^2-\beta^2}}$, where $\beta$ is a constant satisfying $B-A=\beta\int_{a}^{b}\frac{dw}{\sqrt{f(w)^2-\beta^2}}$ gives a weak local minimum of the functional $S[y]=\int_{a}^{b}f(x)\sqrt{1+y'^2}dx, y(a)=A, y(b)=B$. (You are not required to solve the Jacobi equation.)

Relevant Equations

Jacobi equation: $\frac{d}{dx}(P(x)\frac{du}{dx})-Q(x)u=0, u(a)=0, u'(a)=1$, where $P(x)=\frac{\partial^2 F}{\partial y'^2}$ and $Q(x)=\frac{\partial^2 F}{\partial y^2}-\frac{d}{dx}(\frac{\partial^2 F}{\partial y\partial y'})$ vanishes at $x=\tilde{a}$.

For sufficiently small $b-a$, we have
a) if $P(x)=\frac{\partial^2 F}{\partial y'^2}>0, a\leq x\leq b, S[y]$ has a minimum,
b) if $P(x)=\frac{\partial^2 F}{\partial y'^2}<0, a\leq x\leq b, S[y]$ has a maximum.

Jacobi's necessary condition: If the stationary path $y(x)$ corresponds to a minimum of the functional $S[y]=\int_{a}^{b}F(x, y, y')dx, y(a)=A, y(b)=B$, and if $P(x)=\frac{\partial^2 F}{\partial y'^2}>0$ along the path, then the open interval $a<x<b$ does not contain points conjugate to $a$.

A sufficient condition: If $y(x)$ is an admissible function for the functional $S[y]=\int_{a}^{b}F(x, y, y')dx, y(a)=A, y(b)=B$ and satisfies the three conditions listed below, then the functional has a weak local minimum along $y(x)$.
a) The function $y(x)$ satisfies the Euler-Lagrange equation, $\frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0$.
b) Along the curve $y(x), P(x)=\frac{\partial^2 F}{\partial y'^2}>0$ for $a\leq x\leq b$.
c) The closed interval $[a, b]$ contains no points conjugate to the point $x=a$.

Here's my work:

Let $F(x, y, y')=f(x)\sqrt{1+y'^2}$.
Then $P(x)=\frac{\partial^2 F}{\partial y'^2}=\frac{\partial}{\partial y'}(\frac{f(x)y'}{\sqrt{1+y'^2}})=\frac{\frac{\partial}{\partial y'}(f(x)y')\cdot \sqrt{1+y'^2}-(f(x)y')\cdot \frac{\partial}{\partial y'}(\sqrt{1+y'^2})}{(\sqrt{1+y'^2})^2}=\frac{f(x)\cdot \sqrt{1+y'^2}-(f(x)y')(\frac{y'}{\sqrt{1+y'^2}})}{1+y'^2}=\frac{f(x)\cdot (1+y'^2)-f(x)y'^2}{(1+y'^2)^{\frac{3}{2}}}=\frac{f(x)}{(1+y'^2)^{\frac{3}{2}}}$.
This gives $Q(x)=\frac{\partial^2 F}{\partial y^2}-\frac{d}{dx}(\frac{\partial^2 F}{\partial y\partial y'})=0$.
Thus, the Jacobi equation is $\frac{d}{dx}(P(x)\frac{du}{dx})-Q(x)u=0\implies \frac{d}{dx}(\frac{f(x)u'}{(1+y'^2)^{\frac{3}{2}}})=0\implies \frac{f(x)u'}{(1+y'^2)^{\frac{3}{2}}}=C$.

With this Jacobi equation found above, how should I use it and show that the given stationary path gives a weak local minimum of the given functional $S[y]$?

 

[pasmith]

Does u&#039; change sign in (a,b)?
If u(0) = 0, are there any other points x \in (a,b) such that u(x) = 0 (conjugate points)?

 

[pasmith]

Remember that the aim is to determine the sign of the second variation <br /> \int_a^b Ph&#039;^2 + Qh^2\,dx = \left[ Phh&#039; \right]_a^b + \int_a^b (-(Ph&#039;)&#039; + Qh)h\,dx for admissible h near to the zero function, with P and Q evaluated on the extremal path. Since here the value of y is prescribed on the boundaries, we must have h(a) = h(b) = 0 so that \left[ Phh&#039; \right]_a^b vanishes and the Jacobi operator \frac{d}{dx}\left(P\frac{d}{dx}\right) - Q is self-adjoint with respect to the inner product \langle u, v \rangle = \int_a^b u(x)v(x)\,dx.

If you can show P &gt; 0 and Q \geq 0 on (a,b) then you have a global minimum. Only if the signs change over the interval is it necessary to look at conditions derived from considering the Jacobi operator to determine the nature of a local extremum.

 

[Math100]

Does u&#039; change sign in (a,b)?
If u(0) = 0, are there any other points x \in (a,b) such that u(x) = 0 (conjugate points)?

How to find $u'$ and $u$ from the Jacobi equation, $\frac{f(x)u'}{(1+y'^2)^{\frac{3}{2}}}=C$? And the condition we have is $u(a)=0$, so there's no conjugate point.

 

[Math100]

Remember that the aim is to determine the sign of the second variation <br /> \int_a^b Ph&#039;^2 + Qh^2\,dx = \left[ Phh&#039; \right]_a^b + \int_a^b (-(Ph&#039;)&#039; + Qh)h\,dx for admissible h near to the zero function, with P and Q evaluated on the extremal path. Since here the value of y is prescribed on the boundaries, we must have h(a) = h(b) = 0 so that \left[ Phh&#039; \right]_a^b vanishes and the Jacobi operator \frac{d}{dx}\left(P\frac{d}{dx}\right) - Q is self-adjoint with respect to the inner product \langle u, v \rangle = \int_a^b u(x)v(x)\,dx.

If you can show P &gt; 0 and Q \geq 0 on (a,b) then you have a global minimum. Only if the signs change over the interval is it necessary to look at conditions derived from considering the Jacobi operator to determine the nature of a local extremum.

Given that $f(x)>0$, how can we show that $P(x)=\frac{f(x)}{(1+y'^2)^{\frac{3}{2}}}>0$ for $a\leq x\leq b$?