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Variable substitution in a differential equation

GROUPS:

In solving an equation, I used a method of variable substitution. Do you think this method is a typical or popular one?

In[189]:= (*equation*)

In[190]:= eq1 = D[c[r, t], t] == 1/r*D[r*D[c[r, t], r], r]

Out[190]= 
\!\(\*SuperscriptBox[\(c\), \*
TagBox[
RowBox[{"(", 
RowBox[{"0", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[r, t] == (
\!\(\*SuperscriptBox[\(c\), \*
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[r, t] + r 
\!\(\*SuperscriptBox[\(c\), \*
TagBox[
RowBox[{"(", 
RowBox[{"2", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[r, t])/r

(*variable substitution at infinity, \
r->R/\[Sigma],t->\[Tau]/(\[Sigma]^2).*)

In[196]:= eq5 = 
 eq1 /. {c -> (c1[#1*\[Sigma], #2*\[Sigma]^2] &), 
     r -> (#1/\[Sigma] &[R, \[Tau]]), 
     t -> (#2/\[Sigma]^2 &[R, \[Tau]])} // Simplify // Normal

Out[196]= \[Sigma] (
\!\(\*SuperscriptBox[\(c1\), \*
TagBox[
RowBox[{"(", 
RowBox[{"0", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[R, \[Tau]] - (
\!\(\*SuperscriptBox[\(c1\), \*
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R, \[Tau]] + R 
\!\(\*SuperscriptBox[\(c1\), \*
TagBox[
RowBox[{"(", 
RowBox[{"2", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R, \[Tau]])/R) == 0
POSTED BY: Zhonghui Ou
Answer
15 days ago

It seems correct to me. I would do it in a slightly simpler way:

Simplify[eq1 /. {c -> (c1[#1*\[Sigma], #2*\[Sigma]^2] &)} /. {r -> \
(R/\[Sigma]), t -> (\[Tau]/\[Sigma]^2)}]

I don't see the purpose of Normal. However, if you look at the resulting eq2, it seems to be the same as eq1.

POSTED BY: Gianluca Gorni
Answer
14 days ago

This substation is useful in PDE. Thanks.

POSTED BY: Zhonghui Ou
Answer
11 days ago

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