The parity of a differential equation

[henryc09]

Homework Statement

How do you tell whether a differenital equation has well defined parity or not? The equation I have is:

x³y'' + x²y' - x(x²+1)y=0

and it asks:

"Does this equation have a well-defined parity? What does this imply for the
solutions of the equation?"

My instinct would be that it doesn't have well defined parity, but I don't know how this can be checked. I would then imagine if it doesn't have well defined parity there will be both odd and even solutions, but I really don't know. Any help would be appreciated.

Homework Equations

The Attempt at a Solution

 

[Dickfore]

Make the substiution:

[tex]
x = -x'
[/tex]

Express all the derivatives in terms of the new argument [itex]x'[/itex]. What is the form of the equation after this substitution?

 

[henryc09]

-x'³y'' +x'²y' + x'(x'²+1) = 0

sorry I don't see how this helps

 

[Dickfore]

you made a mistake.

 

[henryc09]

are you sure? I don't think I have, but even if I have I'm still not sure as to how this would help?

 

[Dickfore]

because the transformation I gave you defines parity.

 

[henryc09]

right so if they are equal (when you use x and -x') that would mean the equation is even? And if the equation using -x' is equal to the original equation times -1 it would be even, basically just f(x)=f(-x) and -f(x)=f(-x) right?

So from that the equation doesn't have well defined parity, not sure what this would imply for the solutions of the equation though.

 

[fzero]

To avoid confusion with primed symbols, you might want to rewrite your equation as

[tex]x^3 \frac{d^2y}{dx^2} + x^2 \frac{dy}{dx} - x(x^2+1)y=0[/tex]

and then make the transformation to see where you made your mistake.

 

[Dickfore]

right so if they are equal (when you use x and -x') that would mean the equation is even?

No. It would mean the equation is invariant under an inversion.

And if the equation using -x' is equal to the original equation times -1 it would be even, basically just f(x)=f(-x) and -f(x)=f(-x) right?

An equation is written as lhs = 0. If lhs -> A*lhs under some transformation, where [itex]A \neq 0[/itex], that means the equation is left invariant.

Suppose [itex]y = f(x)[/itex] is a solution to the original equation. Then [itex]y' = f(x')[/itex] is a solution to the new equation. But, the new equation is the same as the original one. So, the new function is also a solution to the old equation.

So from that the equation doesn't have well defined parity, not sure what this would imply for the solutions of the equation though.

But, you did not prove that. I told you had made many mistakes.

 

[henryc09]

OK I don't think I understand how to express the derivatives in terms of -x'. Would it just be substituting in -x' instead of x for d^2y/dx^2 and dy/dx?

Sorry for being so slow, I appreciate your help!

 

[Dickfore]

Use the chain rule:

The first derivative transforms to:

[tex]
\frac{d}{dx} = \frac{dx'}{dx} \, \frac{d}{dx'} = - \frac{d}{dx'}
[/tex]

What does the 2nd derivative transform to?

 

[henryc09]

hmmmm I think:


d^2/dx^2 = dx'/dx d(-d/dx')/dx' = -dx/dx' d^2/dX'^2 = d^2/dx'^2

so this will allow me to make the proper transformation, I'm still not sure how this will determine the parity of the equation

 

[Dickfore]

hmmmm I think:


d^2/dx^2 = dx'/dx d(-d/dx')/dx' = -dx/dx' d^2/dX'^2 = d^2/dx'^2

so this will allow me to make the proper transformation,

This is true.

 

[henryc09]

and so substituting -x' in I get:

-x'³d²y/dx'² - x'²dy/dx' + x'(x'²+1)y=0

I don't understand how parity is defined from this though.

 

[henryc09]

but because you don't know the parity of y, which is a function of x would this not mean it's impossible to tell the parity of the equation?

 

[vela]

First, you can write the equation in the form L(x)y(x)=0 where

[tex]L(x) = x^3\frac{d^2}{dx^2}+x^2\frac{d}{dx}-x(x^2+1)[/tex]

is a linear operator acting on y(x). What you've figured out is that L(x) = -L(-x). In other words, L(x) has a definite parity, and it is odd. When you refer to the parity of an equation, you're just talking about the parity of the operator part.


Second, if you change variables from x to -x, you have

L(x)y(x)=0 → L(-x)y(-x) = 0 → -L(x)y(-x) = 0 → L(x)y(-x) = 0

This means if y(x) is a solution, so is y(-x). Note we're not saying anything about the parity of y(x). The functions y(x) and y(-x) could be linearly independent; all we know is that they're both solutions to the differential equation.


Finally, since L(x) is a linear operator, any linear combination of solutions is also a solution to the differential equation. In particular, you can form the combinations

[tex]\begin{align*}
y_e(x) &= y(x) + y(-x) \\
y_o(x) &= y(x) - y(-x)
\end{align*}
[/tex]

and they have a definite parity.


For example, suppose you have the differential equation

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

which has even parity. One solution to the equation is y₁(x) = e^ix. Because the equation has a definite parity, we know that y₂(x) = e^-ix is also a solution. Neither of these solutions is even or odd, but you can always find ones that are:

[tex]\begin{align*}
y_e(x) &= e^{ix} + e^{-ix} = 2\cos x \\
y_o(x) &= e^{ix} - e^{-ix} = 2i\sin x
\end{align*}
[/tex]

 

[henryc09]

Ah ok, so what you're saying is if the equation has definite parity you can always find solutions which have definite parity? Thanks very much for your help!

 

[Dickfore]

I thought we were not supposed to give detailed solutions...

 

[Hermes10]

Does it mean that a differential equation has even and odd solutions if and only if the parity of the equation is well-defined?

 

[Dickfore]

What direction of the equivalence (the "if" or the "only if") was proven in this thread?

 

[Hermes10]

Yeah, I think my mistake is to infer too much from what is given. But say I am given a function that does not have a well defined parity. Does this mean that it may or may not have even or odd functions as solution?

 

[Dickfore]

But say I am given a function that does not have a well defined parity. Does this mean that it may or may not have even or odd functions as solution?

I think you meant "equation" instead of the bolded word. In that case, your question is whether this implication holds:

If the equation does NOT have a well defined parity, then the solution does NOT have a well defined parity as well.

By contraposition, it is equivalent to:

If the solution has a well defined parity, then the equation has a well defined parity.

How would you go about proving any of the above 2 equivalent statements?

 

[Hermes10]

Mhm, well, I suppose your previous comment proved that if L(x) has a well defined parity, then there are even and odd solutions to that equation. It indeed does not say the other way round, that is if there are even and odd solutions, the L(x) has then a well defined parity.

To be honest with you. I am utterly confused by all this. I worked long to try to understand what is actually going on here but somehow I just cannot get it. Is there any way to go about that? Maybe I should re-study all the Maths again.