Understanding Symmetry in ODE Solutions

[MurdocJensen]

dy/dx = (2/pi^(1/2))e^(-(x^2)) eq 1.17

My book makes a statement about the symmetry of the family of solutions to this diff eq I don't quite understand.

"Symmetry. If we replace x with -x on both sides of 1.17, the right hand side is unchanged but the left hand side changes signs. So the family of solutions is not symmetric across the y-axis. However, if we simultaneously replace x with -x and y with -y, then we obtain 1.17 back again. So the family of solutions of 1.17 is unchanged under simultaneous interchange of x with -x and y with -y. This means that the family of solutions is symmetric about the origin."

It doesn't showcase the algebra so that adds some more difficulty for me.

I guess my biggest question is this: I can see how facts about the symmetry of the ODE can suggest properties of the solution curves. In this case, though, how does the above example show anything about the symmetry of the solution? Isn't manipulating the ODE in this way showing symmetry for the ODE itself? That last point wouldn't make sense either, because this ODE is symmetric about the y-axis.

Thoughts?

 

[tiny-tim]

Hi MurdocJensen!

(have a square-root: √ and a pi: π and try using the X² icon just above the Reply box )

"Symmetry. If we replace x with -x on both sides of 1.17, the right hand side is unchanged but the left hand side changes signs. So the family of solutions is not symmetric across the y-axis. However, if we simultaneously replace x with -x and y with -y, then we obtain 1.17 back again. So the family of solutions of 1.17 is unchanged under simultaneous interchange of x with -x and y with -y. This means that the family of solutions is symmetric about the origin."
…
In this case, though, how does the above example show anything about the symmetry of the solution? Isn't manipulating the ODE in this way showing symmetry for the ODE itself? That last point wouldn't make sense either, because this ODE is symmetric about the y-axis.

It's easier to understand if we use different letters.

Put Y = -y, X = -x.

Then dy/dx = (2/√π)e^-x2 becomes:

dY/dX = (2/√π)e^-X2,

so the (X,Y) graph will look exactly like the (x,y) graph.

(and "symmetric about the origin" means across the origin, (x,y) → (-x,-y) )

 

[MurdocJensen]

That'll be extremely helpful in the future. Thanks, tiny.

so the (X,Y) graph will look exactly like the (x,y) graph.

- Where the (X,Y) graph and the (x,y) graph are both graphs of the ODE itself, right?

 

[tiny-tim]

Yes.

 

[MurdocJensen]

I was being a little dense. Correct me if I'm wrong, but testing for symmetry of dy/dx = (2/√π)e-x2 is different than testing for symmetry of the function (2/√π)e-x2 itself, right?

 

[tiny-tim]

Yes.

They're different functions.

One is the derivative of the other (and the symmetries of eg xⁿ and nx^n-1 are obviously different, one is even the other is odd).

 

[MurdocJensen]

So it's the relation to y(x) that makes testing dy/dx = (2/√π)e^-x2 for symmetry yield a different result than testing for symmetry for, say, g = (2/√π)e^-x2 ?

Sorry if these question seem like minutia, but they help me a lot.

 

[tiny-tim]

So it's the relation to y(x) that makes testing dy/dx = (2/√π)e^-x2 for symmetry yield a different result than testing for symmetry for, say, g = (2/√π)e^-x2 ?

Yes, it depends what you're comparing …

comparing y with x is different from comparing dy/dx with x.

 

[MurdocJensen]

I feel my brain growing.