How should I use the Jacobi equation to determine the nature of this?

[Math100]

Homework Statement

Let $n>1$ be a positive integer such that the functional $S[y]=\int_{0}^{1}(y')^{n}e^{y}dx, y(0)=1, y(1)=A>1$, has a stationary path given by $y=n\ln(cx+e^{1/n})$, where $c=e^{A/n}-e^{1/n}$. Use the Jacobi equation to determine the nature of this stationary path.

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 $n>1$ be a positive integer.
Consider the functional $S[y]=\int_{0}^{1}(y')^{n}e^{y}dx, y(0)=1, y(1)=A>1$.
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 $P(x)=\frac{\partial^2 F}{\partial y'^2}$ and $Q(x)=\frac{\partial^2 F}{\partial y^2}-\frac{d}{dx}(\frac{\partial^2 F}{\partial y\partial y'})$ vanishes at $x=\tilde{a}$.
Note that $F(x, y, y')=(y')^{n}e^{y}$.
This gives $P(x)=\frac{\partial^2 F}{\partial y'^2}=n(n-1)(y')^{n-2}e^{y}$ and $Q(x)=\frac{\partial^2 F}{\partial y^2}-\frac{d}{dx}(\frac{\partial^2 F}{\partial y\partial y'})=(y')^{n}e^{y}$.
Observe that $\frac{d}{dx}(P(x)\frac{du}{dx})-Q(x)u=0\implies \frac{d}{dx}((n(n-1)(y')^{n-2}e^{y})\frac{du}{dx})-(y')^{n}e^{y}\cdot u=0\implies n(n-1)(y')^{n-2}e^{y}\frac{d^2u}{dx^2}-(y')^{n}e^{y}\cdot u=0$.
Thus, the Jacobi equation is $n(n-1)\frac{d^2u}{dx^2}-(y')^2\cdot u=0$.

From this Jacobi equation above, how can we determine the nature of this stationary path?

 

[pasmith]

You are given the solution of the EL equation as $y(x) = n \ln(cx + e^{1/n})$ so that $y' = (cn)/(cx + e^{1/n})$ and $e^y = (cx + e^{1/n})^n$ where $c = e^{A/n} - e^{1/n} > 0$. Hence $$\begin{split} Q &= (y')^ne^y - \frac{d}{dx}(n(y')^{n-1}e^y) \\ &= \left(\frac{cn}{cx + e^{1/n}}\right)^n(cx + e^{1/n})^n - \frac{d}{dx}\left( n \left( \frac{cn}{cx + e^{1/x}}\right)^{n-1}(cx + e^{1/n})^n\right) \\ &= (cn)^n - \frac{d}{dx}\left( n(cn)^{n-1}(cx + e^{1/n})\right) \\ &= 0 \end{split}$$ and I leave you to determine $P$.

 

[Math100]

You are given the solution of the EL equation as $y(x) = n \ln(cx + e^{1/n})$ so that $y' = (cn)/(cx + e^{1/n})$ and $e^y = (cx + e^{1/n})^n$ where $c = e^{A/n} - e^{1/n} > 0$. Hence $$\begin{split} Q &= (y')^ne^y - \frac{d}{dx}(n(y')^{n-1}e^y) \\ &= \left(\frac{cn}{cx + e^{1/n}}\right)^n(cx + e^{1/n})^n - \frac{d}{dx}\left( n \left( \frac{cn}{cx + e^{1/x}}\right)^{n-1}(cx + e^{1/n})^n\right) \\ &= (cn)^n - \frac{d}{dx}\left( n(cn)^{n-1}(cx + e^{1/n})\right) \\ &= 0 \end{split}$$ and I leave you to determine $P$.

Based on $F(x, y, y')=(y')^{n}e^{y}$, I've got $P(x)=\frac{\partial^2 F}{\partial y'^2}=n(n-1)(y')^{n-2}e^{y}=n(n-1)(\frac{cn}{cx+e^{1/n}})^{n-2}\cdot (cx+e^{1/n})^{n}=n(n-1)(cn)^{n-2}\cdot (cx+e^{1/n})^{2}$. After substituting $c=e^{A/n}-e^{1/n}$, I've got $P(x)=n(n-1)[(e^{A/n}-e^{1/n})n]^{n-2}\cdot [(e^{A/n}-e^{1/n})x+e^{1/n}]^2$. But how should I simplify $P(x)$ from here?

 

[pasmith]

Do not substitute for $c$; it is enough to note that $c > 0$ and $n > 1$. $P(x)$ is a quadratic in $x$. What is the sign of its leading coefficient, and where are its zeros? Is it positive or negative on $(0,1)$?

Since $Q = 0$ the Jacobi equation reduces to $$P(x)u'(x) = P(0)u'(0).$$

 

[Math100]

Do not substitute for $c$; it is enough to note that $c > 0$ and $n > 1$. $P(x)$ is a quadratic in $x$. What is the sign of its leading coefficient, and where are its zeros? Is it positive or negative on $(0,1)$?

Since $Q = 0$ the Jacobi equation reduces to $$P(x)u'(x) = P(0)u'(0).$$

So $P(x)=n(n-1)(cn)^{n-2}\cdot (cx+e^{1/n})^2\implies P(0)=n(n-1)(cn)^{n-2}\cdot e^{2/n}$. But how should I find out if $P(x)<0$ or if $P(x)>0$? And for the interval $(0, 1)$, what's $1$ in here in knowing that $x=0$?

 

[Math100]

Do not substitute for $c$; it is enough to note that $c > 0$ and $n > 1$. $P(x)$ is a quadratic in $x$. What is the sign of its leading coefficient, and where are its zeros? Is it positive or negative on $(0,1)$?

Since $Q = 0$ the Jacobi equation reduces to $$P(x)u'(x) = P(0)u'(0).$$

The sign of its leading coefficient is positive and $P(x)$ has one zero at $x=-\frac{e^{1/n}}{c}$. Also, on the conditions that $c>0, n>1$, I found out that $P(x)=n(n-1)(cn)^{n-2}\cdot (cx+e^{1/n})^2>0$ for $0\leq x\leq 1$. Does this indicate that our functional $S[y]$ has a minimum (In other words, this is the nature of our stationary path)?