Diffusion Equation Invariant to Linear Temp. Transform

[Dustinsfl]

Show the diffusion equation is invariant to a linear transformation in the temperature field
$$
\overline{T} = \alpha T + \beta
$$
Since $\overline{T} = \alpha T + \beta$, the partial derivatives are
\begin{alignat*}{3}
\overline{T}_t & = & \alpha T_t\\
\overline{T}_{xx} & = & \alpha T_{xx}
\end{alignat*}
So $T_t = \frac{1}{\alpha}\overline{T}_t$ and $T_{xx} = \frac{1}{\alpha}\overline{T}_{xx}$.
The diffusion equation is
$$
\frac{1}{\alpha}T_t = T_{xx}.
$$
By substitution, we obtain
$$
\frac{1}{\alpha}\overline{T}_t = \overline{T}_{xx}.
$$
Correct?

 

[tarantella]

Yes, as the set of solutions of such an equation is a vector space which contains constant functions.

 

[Dustinsfl]

Yes, as the set of solutions of such an equation is a vector space which contains constant functions.

So that is all that it was? It seems to simple.

 

[tarantella]

It may be for example the first question of a homework or a test, so it's not necessarily difficult. (maybe maybe the other question can be harder)

 

[Mathwebster]

So that is all that it was? It seems to simple.

Yes, it may be simple (in this case) but there's a deeper meaning. It means, given one solution $T_0$, you can construct a second solution $T = \alpha T_0 + \beta$.

You might also want to check that this same PDE is invariant under the change of variables

$\bar{t} = k^2 t,\;\;\; \bar{x} = k x$

i.e.

$ T_{\bar{t}}=\alpha T_{\bar{x} \bar{x}} \;\; \implies \;\; T_t = \alpha T_{xx}$.

The next question $-$ how is this useful?