这是一份 Imperial帝国理工大学 PHYS96023作业代写的成功案例
$$
T=T_{0} e^{t / T \pm z / z \mathrm{u}}
$$
and so
$$
\frac{\partial T}{\partial t}=\frac{1}{T} T
$$
and
$$
\frac{\partial^{2} T}{\partial z^{2}}=\left(\frac{1}{z_{\mathrm{D}}}\right)^{2} T
$$
Thus
$$
\left[\left(\frac{1}{z_{\mathrm{D}}}\right)^{2}-\frac{1}{\kappa \mathrm{T}}\right] T=0,
$$
and we have a solution if
$$
z_{\mathrm{D}}=(\kappa \mathrm{T})^{1 / 2} .
$$
Therefore a characteristic distance is related to a characteristic time by $(\kappa \mathrm{T})^{1 / 2}$.
For a more realistic case consider
$$
T=T_{0} t^{-1 / 2} e^{-z^{2} / 4 x t} \quad \text { for } t>0
$$
PHYS96023 COURSE NOTES ：
has the solution
$$
T(z, t)=\left(\frac{2 \alpha I_{0}}{K}\right)(\kappa t)^{1 / 2} \operatorname{ierfc}\left[\frac{z}{2(\kappa t)^{1 / 2}}\right]
$$
where
$$
\operatorname{ierfc}(X)=\int_{X}^{\infty} \operatorname{erfc}\left(X^{\prime}\right) d X^{\prime}
$$
and
$$
\operatorname{erfc}(X)=1-\operatorname{erf}(X)=\frac{2}{\pi} \int_{X}^{\infty} e^{-\left(X^{\prime}\right)^{2}} d X^{\prime}
$$
and
$$
\operatorname{erf}(X)=\frac{2}{\pi} \int_{0}^{X} e^{-\left(X^{\prime}\right)^{2}} d X^{\prime}
$$