MHB Diffusion Equation Invariant to Linear Temp. Transform

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The discussion focuses on demonstrating that the diffusion equation remains invariant under a linear transformation of the temperature field, expressed as \(\overline{T} = \alpha T + \beta\). It establishes that the partial derivatives transform accordingly, leading to the conclusion that the diffusion equation can be rewritten in terms of the transformed variables. This invariance implies that if a solution \(T_0\) exists, a new solution can be constructed as \(T = \alpha T_0 + \beta\). Additionally, the discussion suggests examining invariance under a change of variables involving scaling in time and space. The utility of this invariance in practical applications is raised as a point for further exploration.
Dustinsfl
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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?
 
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Yes, as the set of solutions of such an equation is a vector space which contains constant functions.
 
girdav said:
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.
 
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)
 
dwsmith said:
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?
 

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