# Symmetries in DE

1. Aug 6, 2008

### Marin

Hello everybody!

I have a general question concerning DEs :0

Can one use the symmetry of the equation to somehow get the solution faster?
What does such symmetry tell us?
e.g.:
$$\dot x=y$$
$$\dot y=x$$

is the symmetrical system to the second order DE

$$\ddot x-x=0$$

Now we can easily see the solutions (whether e^t or e^(-t)) actually have the same properties as functions. They are even one and the same function, rotated over the y-axis!

So, is the symmetry really providing help or this is just a coincidence?

2. Aug 6, 2008

### tiny-tim

Hello Marin!

Well … if x' = y2

y' = x2

then x'' = 2y y' = 2x2 √x'

so that's a symmetry which is no help at all!

(I suspect there's a condition that makes it helpful … perhaps something like the Jacobian being unitary … but I'll let someone else answer that! )

3. Aug 6, 2008

### Marin

Sorry,tiny-tim, couldn't quite get it :(

What's the purpose of "then x'' = 2yy' = 2x2 √x'"

When I look at the system itself, what I see is that what's true for y should be true for x which to my understanding implies that the two functions are somehow similar to one another..

And the big question is, if so, then HOW?

**maybe my question above should be: Does the symmetry of a system of simultaneous DEs provide us somehow to find the solution faster?

4. Aug 6, 2008

### tiny-tim

oh I see!

Then, yes, both x and y are solutions to the same equation, so they will be different combinations of the same basic solutions.

(But I don't see how that would generally help.)

5. Aug 7, 2008

### Marin

well, if we could find one solution, e.g.:

dy/dx=x^2 => y=1/3 x^3 +c

it is true then that x=1/3y^3 +c

but if x and y are basically the same functions, do we have?:

1/3 x^3=1/3y^3 +k /.3
x^3=y^3 +c

which I think is the solution to the DE, from which the system has been derived, cuz:

the system was:

dx/dt=y^2
dy/dt=x^2

now dividing the second equation by the first one (to eliminate dt):

dy/dx=x^2/y^2 - which is same with the result above.

Was it just a coincidence or is there some symmetry in it?

EDIT: Sorry, I didn't pay attention I used different variables ( first x and then t)

6. Aug 7, 2008

### tiny-tim

Hi Marin!
Yes, I didn't think of that.

So long as the right-hand side is a function of only one variable,

we can always divide one equation by the other (as you did):

if dx/dt = f'(y), dy/dt = f'(x), say

then f'(y)dy = f'(x)dx,

so f(y) = f(x) + constant.

You're right … the symmetry does help!

7. Aug 7, 2008

### Marin

And what about the other cases?

consider the system:

$$\dot x=x+y^2-2t$$
$$\dot y=x^2+y-2t$$

to be honest, I don't have an idea how to solve it analytically :(

But it's symmetrical... You were talking about the Jacobian hmmm could it be an ansatz here ?

8. Aug 7, 2008

### tiny-tim

Sorry, I've no idea.

Just guessing about the Jacobian … someone else wil have to answer that.

9. Aug 9, 2008

### Marin

Does anybody know something about it?

10. Aug 16, 2008

### matematikawan

Hm.... look like a challenging problem. Never seen before. Is there any application for this system?

Look like you all been thinking of reflection symmetry $$x \leftrightarrow y$$ before. May be we should be looking at other transformation such that system remain invariant. Is Lie symmetry is of any used here ? I don't know.

11. Aug 16, 2008

### Marin

Well, these systems have no physical meaning (at least are not meant to have here). I am interested in the problem from a pure mathematical point of view.

- absolutely true - I consider it the most obvious one - if we could find something interesting about it, maybe we could then ask for partial symmetries or negative symmetry, etc.

I know many DEs are not analytically solvable, and many others take a lot of time to find a solution. That's why I'm asking about these symmetrical systems. I think there must be something 'invisible' to us, but hidden in the system.

I would be glad to see every comment or idea - more or less probable :)

best regards, Marin