这是一份leeds利兹大学MATH265001作业代写的成功案例
Let $\epsilon \in \mathbb{R}, \varphi \in C_{0}^{\infty}(a, b), \lambda=\left(2\left|\varphi^{\prime}\right|_{L^{\infty}}\right)^{-1}$ and
$$
\xi(x, \epsilon)=x+\epsilon \lambda \varphi(x)=y \text {. }
$$
Observe that for $|\epsilon| \leq 1$, then $\xi(., \epsilon):[a, b] \rightarrow[a, b]$ is a diffeomorphism with $\xi(a, \epsilon)=a, \xi(b, \epsilon)=b$ and $\xi_{x}(x, \epsilon)>0$. Let $\eta(., \epsilon):[a, b] \rightarrow[a, b]$ be its inverse, i.e.
$$
\xi(\eta(y, \epsilon), \epsilon)=y .
$$
Since
$$
\xi_{x}(\eta(y, \epsilon), \epsilon) \eta_{y}(y, \epsilon)=1 \text { and } \xi_{x}(\eta(y, \epsilon), \epsilon) \eta_{e}(y, \epsilon)+\xi_{e}(\eta(y, \epsilon), \epsilon)=0,
$$
we find $(O(t)$ stands for a function $f$ so that $|f(t) / t|$ is bounded in a neighborhood of $t=0$ )
$$
\begin{aligned}
&\eta_{y}(y, \epsilon)=1-\epsilon \lambda \varphi^{\prime}(y)+O\left(\epsilon^{2}\right) \
&\eta_{\epsilon}(y, \epsilon)=-\lambda \varphi(y)+O(\epsilon)
\end{aligned}
$$
MATH265001 COURSE NOTES ：
$$
f(x, u, \xi)=f(u, \xi)=\frac{1}{2} \xi^{2}-u .
$$
The second form of the Euler-Lagrange equation is
$$
0=\frac{d}{d x}\left[f\left(u(x), u^{\prime}(x)\right)-u^{\prime}(x) f_{\xi}\left(u(x), u^{\prime}(x)\right)\right]=-u^{\prime}(x)\left[u^{\prime \prime}(x)+1\right],
$$
and it is satisfied by $u \equiv 1$. However, $u \equiv 1$ does not verify the Euler-Lagrange equation, which is in the present case
$$
u^{\prime \prime}(x)=-1 \text {. }
$$