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I have the differential equation:

$$\frac{\partial C}{\partial t} = D \frac{\partial^2 C}{\partial z^2}$$

to solve it, the author of my book carries out the following replacement:

$$-2\zeta \frac{dX}{d \zeta} = \frac{d^2 X}{d \zeta^2}$$

with $X=1-\frac{C}{C_S}$ and $\zeta =\frac{z}{(4Dt)^{1/2}}$.

How can I prove that the last equation is identical to the first in Mathematica?

Thank you so much for your time.

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  • $\begingroup$ Could you please describe the substitution z->zeta in detail? What means Dt? $\endgroup$ – Ulrich Neumann Jan 5 '18 at 11:12
  • $\begingroup$ Hello @UlrichNeumann, D is a constant, z and t are variables. $\endgroup$ – Gennaro Arguzzi Jan 5 '18 at 11:35
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straightforward:

pde = Derivative[1, 0][C][t, z] == d Derivative[0, 2][C][t, z]

Simplify[pde /. C -> Function[{t, z}, CS (1 - X[z/Sqrt[4 d t]])] /.z -> \[Zeta] Sqrt[4 d t] ]   
(* (CS (2 \[Zeta] Derivative[1][X][\[Zeta]] + (
X^\[Prime]\[Prime])[\[Zeta]]))/t == 0 *)

which confirms the replacement!

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  • $\begingroup$ Your solution is extremely clear @UlrichNeumann! $\endgroup$ – Gennaro Arguzzi Jan 6 '18 at 7:14

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