Non-Dimensionalize PDE: Boundary Conditions, Initial Conditions

[Dustinsfl]

$$
\frac{1}{\alpha}T_t = T_{xx}
$$
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 $T = \theta(T_i - T_{\infty}) + T_{\infty}$?

 

[member 553083]

Yes, you can make the substitution $T = \theta(T_i - T_{\infty}) + T_{\infty}$. Then the equation becomes:$$\frac{1}{L^2}\frac{\partial \theta}{\partial t_*} = \frac{1}{L^2}\frac{\partial^2 \theta}{\partial x_*^2}$$For the boundary conditions, we have:$$T(0, t_*) = \theta(0, t_*)(T_i - T_{\infty}) + T_{\infty} = T_{\infty} \Rightarrow \theta(0, t_*) = 0\\T(L, t_*) = \theta(1, t_*)(T_i - T_{\infty}) + T_{\infty} = T_{\infty} \Rightarrow \theta(1, t_*) = 0$$For the initial condition, we have:$$T(x_*, 0) = \theta(x_*, 0)(T_i - T_{\infty}) + T_{\infty} = T_i \Rightarrow \theta(x_*, 0) = 1$$Therefore, 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*}