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PDE - Change of variable

  1. Jan 4, 2009 #1
    Hi

    I have a question regarding a PDE and change of variable. I can follow through the algebra but I have a problem deciding what route to take after I use the chain rule at a later point.

    I have an expression: -

    [tex] \frac{\partial^2 f}{\partial y^2} [/tex]

    and would like to make the variable substitution: -

    [tex] y = k e^x [/tex]

    I first note that ,

    [tex] \frac{dy}{dx} = k e^{x} [/tex]

    and

    [tex]\frac{d^n y}{dx^n} = k e^{x} \; \forall n \ge 0 [/tex]

    This is my initial calculation: -

    [tex]\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y} \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}( \frac{\partial f}{\partial x} \frac{\partial x}{\partial y})[/tex]

    [tex] = \frac{d x}{d y} \frac{d^2 f}{dy d x} + \frac{d f}{d x} \frac{d^2 x}{d y^2} [/tex]

    [tex]= \frac{d x}{d y} \frac{d}{dx}(\frac{df}{dy}) + \frac{d f}{d x} \frac{d^2 x}{d y^2}[/tex]

    [tex]= \frac{d x}{d y} \frac{d}{dx}(\frac{df}{dx} \frac{dx}{dy}) + \frac{d f}{d x} \frac{d^2 x}{d y^2}[/tex]

    [tex]= \frac{d x}{d y} (\frac{d^2f}{dx^2} \frac{dx}{dy} + \frac{df}{dx} \frac{d}{dx} (\frac{dx}{dy}) ) + \frac{d f}{d x} \frac{d^2 x}{d y^2} [/tex]

    However at this point I have two options on how I deal with the expression: -

    [tex] \frac{d}{dx} (\frac{dx}{dy}) [/tex]

    Option 1 is to substitute and deal with it in-line: -

    [tex]= \frac{1}{ke^x} (\frac{d^2f}{dx^2} \frac{1}{ke^x} + \frac{df}{dx} \frac{d}{dx} (\frac{1}{ke^x}) ) + \frac{d f}{d x} \frac{d^2 x}{d y^2}[/tex]

    and note that,

    [tex] \frac{dx}{dy} = \frac{1}{y} \Rightarrow \frac{d^2 x}{dy^2} = \frac{-1}{y^2} = \frac{-1}{(k e^x)^2}[/tex]

    [tex] = \frac{1}{ke^x} (\frac{d^2f}{dx^2} \frac{1}{ke^x} + \frac{df}{dx} \frac{d}{dx} (\frac{1}{ke^x}) ) + \frac{d f}{d x} \frac{-1}{(k e^x)^2} [/tex]

    [tex] = \frac{1}{ke^x} (\frac{d^2f}{dx^2} \frac{1}{ke^x} + \frac{df}{dx} (\frac{-1}{ke^x}) ) + \frac{d f}{d x} \frac{-1}{(k e^x)^2} [/tex]

    [tex]= \frac{1}{(ke^x)^2} (\frac{d^2f}{dx^2} - 2 \frac{df}{dx})[/tex]

    or Option 2 is to evaluate it to zero: -

    [tex]= \frac{1}{ke^x} (\frac{d^2f}{dx^2} \frac{1}{ke^x} + \frac{df}{dx} \frac{d}{dy} (\frac{dx}{dx}) ) + \frac{d f}{d x} \frac{d^2 x}{d y^2}[/tex]

    [tex]= \frac{1}{ke^x} (\frac{d^2f}{dx^2} \frac{1}{ke^x} + \frac{df}{dx} \frac{d}{dy} (1) ) + \frac{d f}{d x} \frac{d^2 x}{d y^2}[/tex]

    [tex]= \frac{1}{ke^x} (\frac{d^2f}{dx^2} \frac{1}{ke^x} + 0 ) + \frac{d f}{d x} \frac{d^2 x}{d y^2}[/tex]

    [tex]= \frac{1}{(ke^x)^2} (\frac{d^2f}{dx^2} - \frac{df}{dx})[/tex]

    How should I consider this? Perhaps I am wrong all the way? Should I take it from first principles and use a limit definition?

    Blair
     
  2. jcsd
  3. Jan 4, 2009 #2

    arildno

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    Now, quitting the complicating use of partials, let us focus on three functions, f(y), F(x), X(y), so that we have the identity:
    [tex]f(y)=F(x=X(y))[/tex]
    Now, we have:
    [tex]\frac{df}{dy}=\frac{dF}{dx}\frac{dX}{dy}[/tex]
    We then have:
    [tex]\frac{d^{2}f}{dy^{2}}=\frac{d^{2}F}{dx^{2}}(\frac{dX}{dy})^{2}+\frac{dF}{dx}\frac{d^{2}X}{dy^{2}}[/tex]
    Now, from what you wrote, we have:
    [tex]X(y)=\ln(y)-\ln(k)[/tex]
    Thus, we have:
    [tex]\frac{dX}{dy}=\frac{1}{y}=\frac{1}{k}e^{-x},\frac{d^{2}X}{dy^{2}}=-\frac{1}{y^{2}}=-\frac{1}{k^{2}}e^{-2x}[/tex]
    Thus, having eliminated the y's, and re-defining F(x) as f(x) (a notational abuse, but very convenient!), we get:
    [tex]\frac{d^{2}f}{dy^{2}}=\frac{1}{k^{2}}e^{-2x}(\frac{d^{2}f}{dx^{2}}-\frac{df}{dx})[/tex] whenever x and y are related through the equation x=X(y)
     
  4. Jan 4, 2009 #3
    Thanks. That confirms option 2. I am developing a software library that automates changing of variables so sadly I can't easily quit using partials. However, I would be interested in any other algorithmic way of doing this that will work with most/all PDE/ODEs.

     
  5. Jan 4, 2009 #4

    arildno

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    Well, it isn't at all difficult to tweak this into partials notation.
     
  6. Jan 4, 2009 #5
    In that case, it would look near identical to my original calculation and option 2.

    Thanks.
     
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