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I Substitution in partial differential equation

  1. Jul 8, 2017 #1
    Hello everybody.
    $$\frac{\partial}{\partial t}f(x) + ax\frac{\partial }{\partial x}f(x) = b x^2\frac{\partial^2}{\partial x^2}f(x)$$
    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?
  2. jcsd
  3. Jul 9, 2017 #2
    I think your reference is in error
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