# [SOLVED]nondimensinalize PDE

#### dwsmith

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

#### dwsmith

##### Well-known member
$$\frac{1}{\alpha}T_t = T_{xx}$$
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 $T = \theta(T_i - T_{\infty}) + T_{\infty}$?
For the BC, I have now
$$T\left(0=x=x_*L,t = \frac{L^2t_*}{\alpha}\right) = T\left(1,t = \frac{L^2t_*}{\alpha}\right) = T_{\infty}$$
How do I just get $t_*$ in the BC? How do I change T to theta?

#### dwsmith

##### Well-known member
For the BC, I have now
$$T\left(0=x=x_*L,t = \frac{L^2t_*}{\alpha}\right) = T\left(1,t = \frac{L^2t_*}{\alpha}\right) = T_{\infty}$$
How do I just get $t_*$ in the BC? How do I change T to theta?
So I have solved everything except for the change on the boundary conditions and initial conditions.

That part has me stuck.