这是一份uwa西澳大学MATH2031的成功案例

纯粹数学入门 MATH2031代 Introduction to Pure Mathematics代写 UWA代写

the only consistent balance ${ }^{8}$ is between $-\epsilon^{2} w^{\prime 2}$ and $V(x)-E$ with $\epsilon^{2} w^{\prime \prime}$ much smaller than both. For that to happen we need
$$
\epsilon^{2} U^{-1} h^{\prime} \ll 1 \text { where } h=w^{\prime}
$$
We place the term $\epsilon^{2} h^{\prime}$ on the right side of the equation and set up the iteration scheme
$$
h_{n}^{2}=\epsilon^{-2} U-h_{n-1}^{\prime} ; \quad h_{-1}=0
$$
or
$$
h_{n}=\pm \frac{\sqrt{U}}{\epsilon} \sqrt{1-\frac{\epsilon^{2} h_{n-1}^{\prime}}{U}} ; \quad h_{-1}=0
$$
Under the square root can be Taylor expanded around $1 .$
$$
h_{n}=\pm \frac{\sqrt{U}}{\epsilon}\left(1-\frac{1}{2} \epsilon^{2} \frac{h_{n-1}^{\prime}}{U}-\frac{1}{8} \epsilon^{4}\left(\frac{h_{n-1}^{\prime}}{U}\right)^{2}+\cdots\right)
$$

MATH2031 COURSE NOTES ：

i.e., $a>(\epsilon /|\alpha|)^{2 / 3}$ and for small enough $\epsilon$ this is also enough to ensure preservation of $\mathcal{B}$. We thus find that the equation $\delta=J(\delta)$ has a unique solution and that, furthermore, $|\delta| \leq$ const.e. Using this information and which implies
$$
|J(\delta)| \leq \frac{\epsilon A}{B^{2}} 2 A B^{-1} \epsilon
$$
we easily get that, for some constants $C_{i}>0$ independent on $\epsilon$,
$$
\left|\delta-\delta_{0}\right| \leq C_{1} \epsilon|\delta| \leq C_{1} \epsilon\left|\delta_{0}\right|+C_{1} \epsilon\left|\delta-\delta_{0}\right|
$$
and thus
$$
\left|\delta-\delta_{0}\right| \leq C_{2} \epsilon\left|\delta_{0}\right|
$$
and thus, applying again Laplace’s method we get
$$
\delta \sim \frac{-\epsilon F^{\prime}}{2 F}
$$