- #1
nickthequick
- 53
- 0
Hi,
I'm a bit uncertain about the validity of my argument/approach to the following:
I'm trying to prove that the solution to a partial differential equation
[itex] \frac{\partial u(x,t)}{\partial t} + N[u(x,t)] = 0[/itex], where N is some nonlinear operator, CAN BE (not necessarily is) asymmetric about x=0. I do not know [itex] u(x,t) [/itex].
To that end, I've been examining the behavior of the system under the transformation [itex] x\to -x [/itex] (i.e. a reflection or inversion transformation) and can show that
[itex] \frac{\partial u(x,t)}{\partial t} + N[u(x,t)] \neq\frac{\partial u(-x,t)}{\partial t} + N[u(-x,t)], [/itex]
so that we do not expect u(x,t) = u(-x,t).
I am wondering if this is the best way to test this reflection symmetry. Also, if I can find the terms in the operator N that violate this symmetry, is it coherent to call these 'symmetry breaking' terms?
Any references to relevant resources would be appreciated,
Thanks!
Nick
I'm a bit uncertain about the validity of my argument/approach to the following:
I'm trying to prove that the solution to a partial differential equation
[itex] \frac{\partial u(x,t)}{\partial t} + N[u(x,t)] = 0[/itex], where N is some nonlinear operator, CAN BE (not necessarily is) asymmetric about x=0. I do not know [itex] u(x,t) [/itex].
To that end, I've been examining the behavior of the system under the transformation [itex] x\to -x [/itex] (i.e. a reflection or inversion transformation) and can show that
[itex] \frac{\partial u(x,t)}{\partial t} + N[u(x,t)] \neq\frac{\partial u(-x,t)}{\partial t} + N[u(-x,t)], [/itex]
so that we do not expect u(x,t) = u(-x,t).
I am wondering if this is the best way to test this reflection symmetry. Also, if I can find the terms in the operator N that violate this symmetry, is it coherent to call these 'symmetry breaking' terms?
Any references to relevant resources would be appreciated,
Thanks!
Nick