Members,
i have a little question concerning the following subsitution for the heat equation $u_t=u_{xx}$.
The substitution is the following: $$t = \zeta x^2$$ $$u(x,t)=x^cF\left(\frac{t}{x^2}\right)=x^cF(\zeta)$$
The author (Baumann, 1999: Symmetry Analysis of Differential Equations with Mathematica)concludes: $$cF-c^2F+\frac{d F}{d\zeta}-6\zeta\frac{d F}{d\zeta}+4c\zeta\frac{d F}{d\zeta}-4\zeta^2\frac{d^2 F}{d \zeta^2}=0$$
It would be nice if someone could give me a step by step explanation for this problem.
This is a self-similar transform namely $\zeta = \frac{t}{x^2}$. This problem is just a case of book keeping. $$ \partial_x = \frac{\partial\zeta}{\partial x}\frac{d}{d\zeta}\\ \partial_t = \frac{\partial\zeta}{\partial t}\frac{d}{d\zeta} $$ we also have $$ \partial_{xx} = \frac{\partial\zeta}{\partial x}\frac{d}{d\zeta}\frac{\partial\zeta}{\partial x}\frac{d}{d\zeta} $$ now we have $$ \frac{\partial\zeta}{\partial x} = -2\frac{\zeta}{x}\\ \frac{\partial\zeta}{\partial t} = \frac{\zeta}{t} $$ so we have $$ \partial_x = -2\frac{\zeta}{x}\frac{d}{d\zeta}\\ \partial_t = \frac{\zeta}{t}\frac{d}{d\zeta}\\ \partial_{xx} = -2\frac{\zeta}{x}\frac{d}{d\zeta}\left(-2\frac{\zeta}{x}\frac{d}{d\zeta}\right) = 4\left(\frac{\zeta}{x^2}\frac{d}{d\zeta} -2\frac{\zeta^2}{x^3}\dfrac{dx}{d\zeta}\frac{d}{d\zeta} +\frac{\zeta}{x}\frac{d^2}{d\zeta^2}\right) $$
Now with all these in place, I then suggest taking the derivatives with respect to the original variables but transforming $u$ i.e. $$ u_t = \partial_t x^c F(\zeta) = u_{xx} = \partial_{xx} x^c F(\zeta) $$ thus $$ x^c\partial_t F(\zeta) = \partial_x\left[cx^{c-1}F(\zeta) +x^c \partial_x F(\zeta)\right] = c(c-1)x^{c-2}F(\zeta) + 2cx^{c-1}\partial_x F(\zeta) + x^c \partial_{xx}F(\zeta) $$ or $$ \partial_t F = \frac{c(c-1)}{x^2}F + \frac{2c}{x}\partial_x F + \partial_{xx} F $$ now apply the transforms - then you are done.