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}

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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}

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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.

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