Algebra for "Fundamental plane-source similarity solution"
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\gdef \old #1 {\colb{#1}}
\gdef \new #1 {\colr{#1}}
Chain rule
The coordinate transformation is given by
\begin{aligned}
\new{\xi} &= \frac{\old{x}}{\sqrt{\kappa \old{t}}} \\
\new{t} &= \old{t}.
\end{aligned}
By the chain rule, we have
\begin{aligned}
\old{\frac{\pd}{\pd x}}
&=
\old{\frac{\pd \xi}{\pd x}}
\new{\frac{\pd}{\pd \xi}}
+
\old{\frac{\pd t}{\pd x}}
\new{\roundbr{\frac{\pd}{\pd t}}_\xi}
\\[\tallspace]
&=
\roundbr{
\old{\frac{\pd}{\pd x}}
\squarebr{
\frac{\old{x}}{\sqrt{\kappa \old{t}}}
}
}
\cdot
\new{\frac{\pd}{\pd \xi}}
+
\old{0}
\cdot
\new{\roundbr{\frac{\pd}{\pd t}}_\xi}
\\[\tallspace]
&=
\frac{1}{\sqrt{\kappa \old{t}}}
\new{\frac{\pd}{\pd \xi}}
\\[\tallspace]
&=
\frac{1}{\sqrt{\kappa \new{t}}}
\new{\frac{\pd}{\pd \xi}}
\end{aligned}
and
\begin{aligned}
\old{\roundbr{\frac{\pd}{\pd t}}_x}
&=
\old{\roundbr{\frac{\pd \xi}{\pd t}}_x}
\cdot
\new{\frac{\pd}{\pd \xi}}
+
\old{\roundbr{\frac{\pd t}{\pd t}}_x}
\cdot
\new{\roundbr{\frac{\pd}{\pd t}}_\xi}
\\[\tallspace]
&=
\roundbr{
\old{\frac{\pd}{\pd t}}
\squarebr{
\frac{\old{x}}{\sqrt{\kappa \old{t}}}
}
}_{\old{x}}
\cdot
\new{\frac{\pd}{\pd \xi}}
+
\old{1}
\cdot
\new{\roundbr{\frac{\pd}{\pd t}}_\xi}
\\[\tallspace]
&=
\frac{\old{x}}{\sqrt{\kappa}}
\cdot
\frac{-1/2}{\old{t}^{3/2}}
\cdot
\new{\frac{\pd}{\pd \xi}}
+
\new{\roundbr{\frac{\pd}{\pd t}}_\xi}
\\[\tallspace]
&=
-\frac{\old{x}}{\sqrt{\kappa \old{t}}}
\cdot
\frac{1}{2 \old{t}}
\cdot
\new{\frac{\pd}{\pd \xi}}
+
\new{\roundbr{\frac{\pd}{\pd t}}_\xi}
\\[\tallspace]
&=
-\new{\xi}
\cdot
\frac{1}{2 \new{t}}
\cdot
\new{\frac{\pd}{\pd \xi}}
+
\new{\roundbr{\frac{\pd}{\pd t}}_\xi}.
\end{aligned}
Therefore
\begin{aligned}
\old{\frac{\pd}{\pd x}} &=
\frac{1}{\sqrt{\kappa \new{t}}}
\new{\frac{\pd}{\pd\xi}}
\\[\tallspace]
\old{\roundbr{\frac{\pd}{\pd t}}_x} &=
-\frac{\new{\xi}}{2 \new{t}} \new{\frac{\pd}{\pd\xi}}
+ \new{\roundbr{\frac{\pd}{\pd t}}_\xi}.
\end{aligned}
Return to the corresponding line in "Fundamental plane-source similarity solution".
Derivatives of temperature
We have
T (\old{x}, \old{t}) =
\frac{Q'}{\sqrt{\kappa \new{t}}}
\cdot
U (\new{\xi}).
Applying
\old{\pd / {\pd x}} =
\roundbr{1 / \sqrt{\kappa \new{t}}}
\new{\pd / {\pd\xi}}
(the chain-rule result for the position derivative),
we get
\begin{aligned}
\old{\frac{\pd T}{\pd x}}
&=
\frac{1}{\sqrt{\kappa \new{t}}}
\new{\frac{\pd}{\pd\xi}}
\squarebr{
\frac{Q'}{\sqrt{\kappa \new{t}}}
\cdot
U (\new{\xi})
}
\\[\tallspace]
&=
\frac{Q'}{\kappa \new{t}}
\frac{\new{\td} U}{\new{\td \xi}}.
\end{aligned}
Applying it a second time, we get
\begin{aligned}
\old{\frac{\pd^2 T}{{\pd x}^2}}
&=
\frac{1}{\sqrt{\kappa \new{t}}}
\new{\frac{\pd}{\pd\xi}}
\squarebr{
\frac{Q'}{\kappa \new{t}}
\frac{\new{\td} U}{\new{\td \xi}}
}
\\[\tallspace]
&=
\frac{Q'}{(\kappa \new{t}) ^ {3/2}}
\frac{\new{\td^2} U}{\new{{\td \xi}^2}}.
\end{aligned}
Start again from
T (\old{x}, \old{t}) =
\frac{Q'}{\sqrt{\kappa \new{t}}}
\cdot
U (\new{\xi}).
Applying
\old{\roundbr{\pd / {\pd t}}_x} =
\roundbr{-\new{\xi} / (2 \new{t})}
\new{\pd / {\pd\xi}}
+
\new{\roundbr{\pd / {\pd t}}_\xi}
(the chain-rule result for the time derivative),
we get
\begin{aligned}
\old{\roundbr{\frac{\pd T}{\pd t}}_x}
&=
-\frac{\new{\xi}}{2 \new{t}}
\new{\frac{\pd}{\pd\xi}}
\squarebr{
\frac{Q'}{\sqrt{\kappa \new{t}}}
\cdot
U (\new{\xi})
}
+
\roundbr{
\new{\frac{\pd}{\pd t}}
\squarebr{
\frac{Q'}{\sqrt{\kappa \new{t}}}
\cdot
U (\new{\xi})
}
}_{\new{\xi}}
\\[\tallspace]
&=
-\frac{\new{\xi}}{2 \new{t}}
\frac{Q'}{\sqrt{\kappa \new{t}}}
\frac{\new{\td} U}{\new{\td\xi}}
+
\frac{Q'}{\sqrt{\kappa}}
\cdot
\frac{-1/2}{\new{t} ^ {3/2}}
\cdot
U.
\end{aligned}
Therefore
\begin{aligned}
\old{\frac{\pd^2 T}{{\pd x}^2}} &=
\frac{Q'}{(\kappa \new{t}) ^ {3/2}}
\frac{\new{\td^2} U}{\new{{\td \xi}^2}}
\\[\tallspace]
\old{\roundbr{\frac{\pd T}{\pd t}}_x} &=
-
\frac{\new{\xi}}{2 \new{t}}
\frac{Q'}{\sqrt{\kappa \new{t}}}
\frac{\new{\td} U}{\new{\td \xi}}
+
\frac{Q'}{\sqrt{\kappa}}
\cdot
\frac{-1/2}{\new{t} ^ {3/2}}
\cdot
U.
\end{aligned}
Return to the corresponding line in "Fundamental plane-source similarity solution".
Heat equation
The heat equation says that
\old{\roundbr{\frac{\pd T}{\pd t}}_x}
=
\kappa
\old{\frac{\pd^2 T}{{\pd x}^2}}.
Substituting the results for the derivatives of temperature,
this becomes
\begin{aligned}
-\frac{\new{\xi}}{2 \new{t}}
\frac{Q'}{\sqrt{\kappa \new{t}}}
\frac{\new{\td} U}{\new{\td\xi}}
+
\frac{Q'}{\sqrt{\kappa}}
\cdot
\frac{-1/2}{\new{t} ^ {3/2}}
\cdot
U
&=
\kappa
\cdot
\frac{Q'}{(\kappa \new{t}) ^ {3/2}}
\frac{\new{\td^2} U}{\new{{\td \xi}^2}}
\\[\tallspace]
\cancel{
\frac{Q'}{\sqrt{\kappa} \cdot \new{t}^{3/2}}
}
\squarebr{
-\frac{\new{\xi}}{2}
\frac{\new{\td} U}{\new{\td\xi}}
-
\frac{U}{2}
}
&=
\cancel{
\frac{Q'}{\sqrt{\kappa} \cdot \new{t}^{3/2}}
}
\squarebr{
\frac{\new{\td^2} U}{\new{{\td \xi}^2}}
}
\end{aligned}
Therefore
\frac{\new{\td^2} U}{\new{{\td \xi}^2}}
+ \frac{\new{\xi}}{2} \frac{\new{\td} U}{\new{\td \xi}}
+ \frac{U}{2}
= 0.
Return to the corresponding line in "Fundamental plane-source similarity solution".
END
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