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Considering the problem: $$4\xi\frac{\partial\theta}{\partial\chi}-\frac{\partial^2\theta}{\partial\xi^2}=0\qquad\quad(\chi,\xi)\in(0,\infty)\times(0,\infty)$$ $$\theta(0,\xi)=1\qquad\quad\xi\in(0,\infty)$$ $$\theta(\chi,0)=0\qquad\quad\chi\in(0,\infty)$$ And knowing that it is invariant to the following transformation $$\chi\to\lambda\chi\quad\text y \quad\xi\to\lambda^\frac{1}{3}\xi$$ Rewrite the previous boundary problem for an ODE that solves a function f such that $$\theta(\chi,\xi)=f(\frac{\chi}{\xi^3}) $$

I've been trying to rewrite using that transformation but I can't seem to figure it out, so I'm looking for a bit of help.

ppp
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calculate partial derivative and put it in main PDE. $$ \frac{\partial \theta}{\partial \chi} = \frac{\partial f}{\partial \chi} = \frac{df}{dx} \frac{\partial (\frac{\chi}{\xi^3})}{\partial \chi} = \frac{1}{\xi^3} \frac{df}{dx} $$ which $x = (\frac{\chi}{\xi^3})$ is variable of $f$, similarly for $\xi$ : $$ \frac{\partial \theta}{\partial \xi} = \frac{df}{dx} (-3 \frac{\chi}{\xi^4}) $$ second derivative : $$ \frac{\partial^2 \theta}{\partial \xi^2} = \frac{d^2f}{dx^2} (-3 \frac{\chi}{\xi^4})^2 +12 \frac{df}{dx} ( \frac{\chi}{\xi^5}) $$ substituting in PDE : $$ 4 \xi (\frac{1}{\xi^3} \frac{df}{dx}) - 9 \frac{\chi^2}{\xi^8} \frac{d^2f}{dx^2} - 12 \frac{\chi}{\xi^5}\frac{df}{dx} =0 $$ multiply by $\xi^2$ ,finally you get : $$ 4(1-3x)\frac{df}{dx} - 9 x^2 \frac{d^2f}{dx^2} =0 \quad , x \in [0,\infty) $$ Boundaries are : $$ f(0) = 1 \quad f(\infty) = 0 $$

Abolfazl Chaman motlagh
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