How to show that this functional has this stationary path by solving this second-order differential equation?

Question

Let $n>1$ be a positive integer. Show that the functional $$S[y]=\int_{0}^{1}(y')^{n}e^{y}dx, \quad y(0)=1,\quad y(1)=A>1,$$ 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.

Here's my work:

Let $n>1$ be a positive integer.

Consider the functional $S[y]=\int_{0}^{1}(y')^{n}e^{y}dx, y(0)=1, y(1)=A>1$.

Note that the Euler-Lagrange equation is $\frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0, y(a)=A, y(b)=B$ for the functional $S[y]=\int_{a}^{b}F(x, y, y')dx, y(a)=A, y(b)=B$.

First, we compute $\frac{\partial F}{\partial y'}$ and $\frac{\partial F}{\partial y}$ where $F(x, y, y')=(y')^{n}e^{y}$.

Observe that $\frac{\partial F}{\partial y'}=n(y')^{n-1}e^{y}$ and $\frac{\partial F}{\partial y}=(y')^{n}e^{y}$.

This means $\frac{d}{dx}(\frac{\partial F}{\partial y'})=n(y')^{n-1}e^{y}+n(n-1)(y')^{n-2}y''e^{y}$.

Now we have $\frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0\implies n(y')^{n-1}e^{y}+n(n-1)(y')^{n-2}y''e^{y}-(y')^{n}e^{y}=0\implies (y')^{n-2}e^{y}[n(n-1)y''+n(y')-(y')^2]=0$.

This implies $n(n-1)y''+ny'-y'^2=0$.

From here I'm stuck. How should I proceed and solve this second-order differential equation to get the given stationary path? And how to use the Jacobi equation to determine the nature of this stationary path?

Answer 1

Hints:

1. Since OP's Lagrangian $$L~=~(y^{\prime})^ne^y \tag{1}$$ does not depend explicitly on the independent variable $x$, the Beltrami identity yields a first integral/conserved quantity $$ {\rm const}~\stackrel{\text{Bel. id.}}{=}~ \left(y^{\prime}\frac{\partial}{\partial y^{\prime}}-1\right)L~\stackrel{(1)}{=}~(n-1)L. \tag{2}$$

2. So (the $n$th root of) the Lagrangian $L$ is a first integral/conserved quantity $$ {\rm const}~\stackrel{(2)}{=}~L^{1/n}~\stackrel{(1)}{=}~y^{\prime}e^{y/n}~=~n\frac{de^{y/n}}{dx}.\tag{3}$$

3. Hence $$e^{y/n}~\stackrel{(3)}{=}~ax+b\tag{4}$$ is an affine function of $x$, and therefore the stationary path is of the form $$ y~\stackrel{(4)}{=}~n\ln(ax+b).\tag{5}$$ Finally apply OP's boundary conditions to determine the constants $a$ and $b$.