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

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?

You are given the solution of the EL equation as [itex]y(x) = n \ln(cx + e^{1/n})[/itex] so that [itex]y' = (cn)/(cx + e^{1/n})[/itex] and [itex]e^y = (cx + e^{1/n})^n[/itex] where [itex]c = e^{A/n} - e^{1/n} > 0[/itex]. Hence [tex]\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}[/tex] and I leave you to determine [itex]P[/itex].

pasmith said:

You are given the solution of the EL equation as [itex]y(x) = n \ln(cx + e^{1/n})[/itex] so that [itex]y' = (cn)/(cx + e^{1/n})[/itex] and [itex]e^y = (cx + e^{1/n})^n[/itex] where [itex]c = e^{A/n} - e^{1/n} > 0[/itex]. Hence [tex]\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}[/tex] and I leave you to determine [itex]P[/itex].

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?

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

Since [itex]Q = 0[/itex] the Jacobi equation reduces to [tex]
P(x)u'(x) = P(0)u'(0).[/tex]

pasmith said:

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

Since [itex]Q = 0[/itex] the Jacobi equation reduces to [tex]
P(x)u'(x) = P(0)u'(0).[/tex]

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 ##?

pasmith said:

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

Since [itex]Q = 0[/itex] the Jacobi equation reduces to [tex]
P(x)u'(x) = P(0)u'(0).[/tex]

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)?