# How did the author transform the original equation into the Legendre equation?

• I
• Thomas Michael
In summary, George F. Simmons has a second order ordinary differential equation and uses a change of independent variable to transform it into the Legendre equation. However, he doesn't seem to be able to explain how he got from eq 1 to eq 2.
Thomas Michael
TL;DR Summary
How does one go from ##\phi## to ##x=\cos(\phi)##
I'm reading "Differential Equations with Applications and Historical Notes" by George F. Simmons and I am confused about something on pages 431-432

He has the second order ordinary differential equation

$$\frac {d^2v} {d\phi^2} + \frac {\cos(\phi)} {\sin(\phi)} \frac {dv} {d\phi} + n(n+1)v = 0 ~~~~~~~~~~~~~~~~~~~ eq. 1$$

And then using a change of independent variable from ## \phi ## to ## x = \cos(\phi) ## eq 1 is transformed into the Legendre equation

$$(1-x^2) \frac {d^2v} {dx^2} - 2x \frac {dv} {dx} + n(n+1)v = 0 ~~~~~~~~~~~~ eq. 2$$

But I don't see how he got from eq 1 to eq 2

Anyone feel like helping me out?

First you need to compute the ##dx/d\phi## given you know x in terms of ##\phi##

Ok, I think that makes sense. So just use the chain rule:

$$\frac {d^2v} {dx^2} \frac {dx^2} {d\phi^2}$$

With ##x=\cos(\phi)## the first term will have

$$\frac {d\cos(\phi)} {d\phi} = \frac {dx} {d\phi} = -\sin(\phi)$$
$$\frac {d^2x} {d\phi^2} = \sin^2(\phi)$$
$$\sin^2(\phi) = 1-\cos^2(\phi)$$

and with ##x=\cos(\phi)## it turns into ##(1-x^2)## and the second term will have

$$\frac {\cos(\phi)} {\sin(\phi)} \frac {dv} {dx} \frac {dx} {d\phi} = -\sin(\phi) \frac {\cos(\phi)} {\sin(\phi)} \frac {dv} {dx} = -\cos(\phi) \frac {dv} {dx}$$

$$-\cos(\phi) = -x$$

Though now I still can't account for the ##2## in the second term ##-2x \frac {dv} {dx} ##

##x = \cos \phi \rightarrow dx = - \sin \phi d \phi \rightarrow - \frac{dx}{-\sin \phi} = d \phi \rightarrow \frac{-dx}{d \phi} = \sin \phi##

Thus, we have ##\frac{\sin^2 \phi d^2 v}{dx^2} - \frac{d \phi x}{dx} \frac{dv}{d \phi} +n(n+1)v = 0 ## Making the substitutions you mention, we get ##(1-x^2)\frac{d^2 v}{dx^2} -x \frac{dv}{dx}+n(n+1)v = 0 ##

So, maybe a typo.

Thomas Michael said:
Ok, I think that makes sense. So just use the chain rule:

$$\frac {d^2v} {dx^2} \frac {dx^2} {d\phi^2}$$

With ##x=\cos(\phi)## the first term will have

$$\frac {d\cos(\phi)} {d\phi} = \frac {dx} {d\phi} = -\sin(\phi)$$
$$\frac {d^2x} {d\phi^2} = \sin^2(\phi)$$
$$\sin^2(\phi) = 1-\cos^2(\phi)$$

and with ##x=\cos(\phi)## it turns into ##(1-x^2)## and the second term will have

$$\frac {\cos(\phi)} {\sin(\phi)} \frac {dv} {dx} \frac {dx} {d\phi} = -\sin(\phi) \frac {\cos(\phi)} {\sin(\phi)} \frac {dv} {dx} = -\cos(\phi) \frac {dv} {dx}$$

$$-\cos(\phi) = -x$$

Though now I still can't account for the ##2## in the second term ##-2x \frac {dv} {dx} ##

Here $dx/d\phi = -\sin\phi$ is not constant; thus $$\frac{d^2v}{d\phi^2} = \frac{d}{d\phi}\left(\frac{dv}{d\phi}\right) = \frac{d}{d\phi}\left(\frac{dx}{d\phi} \frac{dv}{dx}\right) = \frac{d^2 x}{d\phi^2} \frac{dv}{dx} + \left(\frac{dx}{d\phi}\right)^2\frac{d^2v}{dx^2} = -x \frac{dv}{dx} + (1 - x^2)\frac{d^2v}{dx^2}.$$

## What is an independent variable?

An independent variable is a variable that is manipulated or changed by the researcher in an experiment. It is the variable that is thought to have an effect on the dependent variable.

## Why is it important to change the independent variable in an experiment?

Changing the independent variable allows the researcher to determine if there is a causal relationship between the independent and dependent variables. By manipulating the independent variable, the researcher can see how it affects the outcome of the experiment.

## What is the difference between an independent variable and a dependent variable?

An independent variable is changed by the researcher, while a dependent variable is the outcome or response that is measured in an experiment. The dependent variable is thought to be influenced by the independent variable.

## How do you determine the appropriate range of values to change the independent variable?

The range of values for the independent variable should be determined based on the research question and the hypothesis being tested. It should also take into account any potential confounding variables that may affect the outcome of the experiment.

## Can the independent variable be changed more than once in an experiment?

Yes, the independent variable can be changed multiple times in an experiment. This can help to test different levels or values of the independent variable and see how it affects the dependent variable. However, it is important to control for any other variables that may also change during the experiment.

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