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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.

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# I Substitution in partial differential equation

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