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

In summary, the conversation is about a question regarding a PDE and change of variable. The person is able to follow through the algebra but is unsure of the next steps after using the chain rule. They discuss making a variable substitution and note that there are two options for how to deal with the expression \frac{d}{dx} (\frac{dx}{dy}). They then consider both options and ultimately choose option 2. They also mention developing a software library that automates changing of variables and express interest in any other algorithmic methods for dealing with PDE/ODEs.
  • #1
bsdz
5
0
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
 
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  • #2
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)
 
  • #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.

arildno said:
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
Well, it isn't at all difficult to tweak this into partials notation.
 
  • #5
In that case, it would look near identical to my original calculation and option 2.

Thanks.
 

1. What is a PDE?

A PDE, or partial differential equation, is a mathematical equation that involves multiple variables and their partial derivatives. It is used to describe the relationship between these variables and their rates of change.

2. What is the purpose of a change of variable in PDEs?

A change of variable in PDEs is used to simplify the equation and make it easier to solve. It involves substituting one or more variables with new variables, which can alter the form of the equation and potentially make it more manageable.

3. How do you determine which variable to change in a PDE?

The decision to change variables in a PDE is usually based on the form of the equation and the goal of simplification. You may choose a variable that appears in multiple terms, or one that can be easily eliminated through the change of variable.

4. What are some common techniques for changing variables in PDEs?

Some common techniques for changing variables in PDEs include substitution, transformation, and the method of characteristics. Each of these methods involves replacing one or more variables with new ones, with the goal of simplifying the equation.

5. Are there any limitations to using a change of variable in PDEs?

While a change of variable can be a powerful tool for solving PDEs, it may not always be possible or effective. In some cases, the resulting equation may still be complex or difficult to solve. Additionally, certain boundary conditions or physical constraints may limit the variables that can be changed.

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