How should I deal with the expression \frac{d}{dx} (\frac{dx}{dy}) ?

[bsdz]

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: -

\frac{\partial^2 f}{\partial y^2}

and would like to make the variable substitution: -

y = k e^x

I first note that ,

\frac{dy}{dx} = k e^{x}

and

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

This is my initial calculation: -

\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})

= \frac{d x}{d y} \frac{d^2 f}{dy d x} + \frac{d f}{d x} \frac{d^2 x}{d y^2}

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

= \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}

= \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}

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

\frac{d}{dx} (\frac{dx}{dy})

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

= \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}

and note that,

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

= \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}

= \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}

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

or Option 2 is to evaluate it to zero: -

= \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}

= \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}

= \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}

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

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

 

[arildno]

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:
f(y)=F(x=X(y))
Now, we have:
\frac{df}{dy}=\frac{dF}{dx}\frac{dX}{dy}
We then have:
\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}}
Now, from what you wrote, we have:
X(y)=\ln(y)-\ln(k)
Thus, we have:
\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}
Thus, having eliminated the y's, and re-defining F(x) as f(x) (a notational abuse, but very convenient!), we get:
\frac{d^{2}f}{dy^{2}}=\frac{1}{k^{2}}e^{-2x}(\frac{d^{2}f}{dx^{2}}-\frac{df}{dx}) whenever x and y are related through the equation x=X(y)

 

[bsdz]

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.

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:
f(y)=F(x=X(y))
Now, we have:
\frac{df}{dy}=\frac{dF}{dx}\frac{dX}{dy}
We then have:
\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}}
Now, from what you wrote, we have:
X(y)=\ln(y)-\ln(k)
Thus, we have:
\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}
Thus, having eliminated the y's, and re-defining F(x) as f(x) (a notational abuse, but very convenient!), we get:
\frac{d^{2}f}{dy^{2}}=\frac{1}{k^{2}}e^{-2x}(\frac{d^{2}f}{dx^{2}}-\frac{df}{dx}) whenever x and y are related through the equation x=X(y)

 

[arildno]

Well, it isn't at all difficult to tweak this into partials notation.

 

[bsdz]

In that case, it would look near identical to my original calculation and option 2.

Thanks.