- #1
bsdz
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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
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