# I Substitution in partial differential equation

1. Jul 8, 2017

### grquanti

Hello everybody.
Consider
$$\frac{\partial}{\partial t}f(x) + ax\frac{\partial }{\partial x}f(x) = b x^2\frac{\partial^2}{\partial x^2}f(x)$$
This is the equation (19) of https://www.researchgate.net/profil...52d685b2a73eb000000.pdf?disableCoverPage=true

They make the substitution $$y=ln(x)$$ and obtain
$$\frac{\partial}{\partial t}f(x) + a\frac{\partial }{\partial y}f(y) = b \frac{\partial^2}{\partial y^2}f(y)$$

My problem in second order term: what I would do is
$$bx^2\frac{\partial^2}{\partial x^2}f(x)=bx^2\frac{\partial }{\partial x}(\frac{\partial y}{\partial x}\frac{\partial }{\partial y} f(y))=bx^2\frac{\partial }{\partial x}(\frac{1}{x}\frac{\partial }{\partial y} f(y))=bx^2\frac{\partial y}{\partial x} \frac{\partial }{\partial y}(\frac{1}{x}\frac{\partial }{\partial y} f(y)) = bx\frac{\partial}{\partial y}(\frac{1}{x}\frac{\partial }{\partial y} f(y)) = b\frac{\partial^2}{\partial y^2}f(x) + bx(\frac{\partial }{\partial y}f(x))(\frac{\partial }{\partial y}\frac{1}{x})$$

So I obtain the term $$b\frac{\partial^2}{\partial y^2}f(x)$$ but I also have $$bx(\frac{\partial }{\partial y}f(x))(\frac{\partial }{\partial x}\frac{1}{x})$$
which is not 0 being
$$\frac{\partial }{\partial y}\frac{1}{x} = \frac{\partial }{\partial y}e^{-y} = -e^{-y}$$
so that I'm left with a term
$$bx(\frac{\partial }{\partial y}f(x))(\frac{\partial }{\partial x}\frac{1}{x}) = be^y(\frac{\partial}{\partial y}f(y))(-e^{-y})=-b\frac{\partial }{\partial y}f(y)$$
What is my mistake?
Thanks.

2. Jul 9, 2017

### Staff: Mentor

I think your reference is in error