这是一份uwa西澳大学STAT2062的成功案例

概率论基础与应用|STAT2062 Fundamentals of Probability with Applications代写 UWA代写

$$
w_{t}=w_{x}^{2}
$$
which can be solved by characteristics. We take $w_{x}=u$ and get for $u$ the quasilinear equation
$$
u_{t}=2 u u_{x}
$$
with a particular solution $u=-x /(2 t)$, giving $w=-x^{2} /(4 t)$. We thus take $w=-x^{2} /(4 t)+\delta$ and get for $\delta$ the equation
$$
\delta_{t}+\frac{x}{t} \delta_{x}+\frac{1}{2 t}=\delta_{x}^{2}+\delta_{x x}
$$
where we have separated the relatively small terms to the right side. We would normally solve the leading equation and continue the process, but for this equation we note that $\delta=-\frac{1}{2} \ln t$ solves not only the leading equation, but the full equation. Thus
$$
w=-\frac{x^{2}}{4 t}-\frac{1}{2} \ln t
$$
which gives the classical heat kernel
$$
\psi=\frac{1}{\sqrt{t}} e^{-\frac{x^{2}}{4 t}}
$$
This exact solvability is of course rather accidental, but a perturbation approach formally works in a more PDE general context.

STAT2062 COURSE NOTES ：

$\int 3 y_{0}^{2}+w_{2}:=\phi_{1}+w_{2}\left(\phi_{1}\right.$ is a formal power series $)$. To leading order, we obtain
$$
w_{2}^{\prime}=3 y_{0} e^{-x}
$$
and thus $w_{2}=\phi_{2} e^{-x}$ where $\phi_{2}$ is a power series. Continuing this process of iteration, we can see inductively that $w$ must be of the form
$$
w=-x+\sum_{k=0}^{\infty} \phi_{k} e^{-k x}
$$
where $\phi_{k}$ are formal power series, which means
$$
y=\sum_{k=0}^{\infty} e^{-k x} y_{k}
$$