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 [itex]u'[/itex] change sign in [itex](a,b)[/itex]?
If [itex]u(0) = 0[/itex], are there any other points [itex]x \in (a,b)[/itex] such that [itex]u(x) = 0[/itex] (conjugate points)?

 

[pasmith]

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

If you can show [itex]P > 0[/itex] and [itex]Q \geq 0[/itex] on [itex](a,b)[/itex] 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 [itex]u'[/itex] change sign in [itex](a,b)[/itex]?
If [itex]u(0) = 0[/itex], are there any other points [itex]x \in (a,b)[/itex] such that [itex]u(x) = 0[/itex] (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 [tex]
\int_a^b Ph'^2 + Qh^2\,dx = \left[ Phh' \right]_a^b + \int_a^b (-(Ph')' + Qh)h\,dx[/tex] for admissible [itex]h[/itex] near to the zero function, with [itex]P[/itex] and [itex]Q[/itex] evaluated on the extremal path. Since here the value of [itex]y[/itex] is prescribed on the boundaries, we must have [itex]h(a) = h(b) = 0[/itex] so that [itex] \left[ Phh' \right]_a^b[/itex] vanishes and the Jacobi operator [itex]\frac{d}{dx}\left(P\frac{d}{dx}\right) - Q[/itex] is self-adjoint with respect to the inner product [itex]\langle u, v \rangle = \int_a^b u(x)v(x)\,dx.[/itex]

If you can show [itex]P > 0[/itex] and [itex]Q \geq 0[/itex] on [itex](a,b)[/itex] 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 ##?