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.
a) 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.)
b) Using the inequality (which is valid for all real $z$ and $u$) $\sqrt{1+(z+u)^2}-\sqrt{1+z^2}\geq \frac{zu}{\sqrt{1+z^2}}$, or otherwise, 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 global minimum of the functional $S[y]=\int_{a}^{b}f(x)\sqrt{1+y'^2}dx, y(a)=A, y(b)=B$.
For part a) of the question, I found out that by definition, the Jacobi equation is $\frac{d}{dx}(P(x)\frac{du}{dx})-Q(x)u=0, u(a)=0, u'(a)=1$, where $P(x)=\frac{\partial^2 F}{\partial y'^2}, Q(x)=\frac{\partial^2 F}{\partial y^2}-\frac{d}{dx}(\frac{\partial^2 F}{\partial y\partial y'})$, vanishes at $x=\tilde{a}$. So from $F(x, y, y')=f(x)\sqrt{1+y'^2}$, I've got $\frac{\partial F}{\partial y'}=\frac{f(x)y'}{\sqrt{1+y'^2}}$, so $P(x)=\frac{\partial^2 F}{\partial y'^2}=\frac{\partial}{\partial y'}(\frac{f(x)y'}{\sqrt{1+y'^2}})=\frac{f(x)\cdot\sqrt{1+y'^2}-\frac{f(x)\cdot y'^2}{\sqrt{1+y'^2}}}{1+y'^2}$. Also, $Q(x)=0$. After simplying $P(x)$, I've got $P(x)=\frac{f(x)\cdot (1+y'^2)-f(x)\cdot y'^2}{(1+y'^2)^{3/2}}=\frac{f(x)(1+y'^2-y'^2)}{(1+y'^2)^{3/2}}=\frac{f(x)}{(1+y'^2)^{3/2}}$. Thus, the Jacobi equation becomes $\frac{d}{dx}(\frac{f(x)}{(1+y'^2)^{3/2}}\cdot \frac{du}{dx})$ but how to simplify this further and how can this Jacobi equation show that the given stationary path gives a weak local minimum of the given functional? As for part b) of the question, how to use that given inequality to show that the given stationary path gives a global minimum of the given functional?
Best Answer
a) Let $u$ be a nonzero solution to the Jacobi equation satisfying the boundary condition $u(a)=0$. A necessary condition$^{(*)}$ for $y$ to be an extremal of the functional $S$ is that $u$ has no roots in the interval $(a,b)$. Now, as you have shown, the Jacobi equation in this problem is given by $$ \frac{d}{dx}\left(\frac{f(x)u'}{(1+y'^2)^{3/2}}\right)=0 \implies \frac{f(x)u'}{(1+y'^2)^{3/2}}=C. \tag{1} $$ Integrating once more, and using the condition $u(a)=0$, we obtain $$ u(x)=C\int_a^x\frac{(1+y'^2)^{3/2}}{f(\xi)}d\xi, \tag{2} $$ where $C\neq 0$ if $u$ is nonzero. Because of the hypothesis on $f$, the integrand in $(2)$ is positive in the interval $[a,b]$, so $u$ does not vanish in $(a,b)$, thus satisfying the Jacobi condition for the extremality of $y$.
b) Let $y$ be the stationary path of $S$, and $u$ a continuously differentiable function satisfying the boundary conditions $u(a)=u(b)=0$. To prove that $y$ is a global minimum of $S$, we have to show that $S[y+u]-S[y]\geq 0$ for any $u$ satisfying the given hypotheses. Using the inequality given in the question, we have $$ S[y+u]-S[y]=\int_a^bf(x)\left(\sqrt{1+(y'+u')^2}-\sqrt{1+y'^2}\right)dx \geq\int_a^b\frac{f(x)y'u'}{\sqrt{1+y'^2}}dx. \tag{3} $$ Integrating the RHS of $(3)$ by parts, we obtain $$ \int_a^b\frac{f(x)y'u'}{\sqrt{1+y'^2}}dx=\left.\frac{f(x)y'u}{\sqrt{1+y'^2}}\right|_a^b-\int_a^b\frac{d}{dx}\left(\frac{f(x)y'}{\sqrt{1+y'^2}}\right)u\, dx. \tag{4} $$ The first term on the RHS of $(4)$ vanishes because $u(a)=u(b)=0$, and the second one also vanishes because $y$ is a solution to the Euler-Lagrange equation $$ \frac{d}{dx}\left(\frac{f(x)y'}{\sqrt{1+y'^2}}\right)=0. \tag{5} $$ Therefore, it follows from $(3)$-$(5)$ that $S[y+u]-S[y]\geq 0$, thus proving that $y$ is a global minimum of $S$.
$^{(*)}$ See, for instance, https://encyclopediaofmath.org/wiki/Jacobi_condition.