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\frac{1}{\alpha}T_t = T_{xx}

$$

B.C are

$$

T(0,t) = T(L,t) = T_{\infty}

$$

I.C is

$$

T(x,0) = T_i.

$$

By recasting this problem in terms of non-dimensional variables, the diffusion equation along with its boundary conditions can be put into a canonical form.

Suppose that we introduce variable scaling defined by

$$

x_* = \frac{x}{L}\quad\quad t_* = \frac{\alpha t}{L^2}\quad\quad \theta = \frac{T - T_{\infty}}{T_i - T_{\infty}}

$$

With this change of variables, show that the problem definition becomes

\begin{alignat*}{5}

\theta_{t_*} & = & \theta_{x_*x_*} & & \\

\theta(0,t_*) & = & \theta(1,t_*) & = & 0\\

\theta(x_*,0) & = & 1

\end{alignat*}

$$

\frac{1}{L^2}\frac{\partial T}{\partial t_*} = \frac{1}{L^2}\frac{\partial^2 T}{\partial x_*^2}

$$

Do I make the $T$ substitution just as $T = \theta(T_i - T_{\infty}) + T_{\infty}$?