# Eliminating dual solutions?

• March 12th 2012, 04:34 AM
TriKri
Eliminating dual solutions?
Hi, is there some method for eliminating dual solutions to a given differential equation that are not wanted? Let me explain: Consider the differential equation

$\frac{\partial^2 y}{\partial t^2}=\hat{C}y\ (^*),$

where $\hat{C}$ is a linear operator (that doesn't operate in the time dimension). However, since the equation is going to be solved numerically, and the operator $\hat{C}$ is very computationally expensive to be implemented directly, the equation is modified:

$\frac{\partial^2 y}{\partial t^2}=\hat{C}y$

$\Rightarrow \frac{\partial^2}{\partial t^2}\frac{\partial^2 y}{\partial t^2}=\frac{\partial^2}{\partial t^2}\hat{C}y$

$\Leftrightarrow\frac{\partial^4 y}{\partial t^4}=\hat{C}\frac{\partial^2 y}{\partial t^2}$

$\Leftrightarrow\frac{\partial^4 y}{\partial t^4}=\hat{C}^2y\ (^{**}),$

where $\hat{C}^2$ is much computationally cheaper to implement. However, this equation gives dual solutions that do not satisfy the original equation (*), since (**) can be shown to allow any linear combination of solutions to the equations

$\frac{\partial^2 y}{\partial t^2}=\hat{C}y$ (*, our original equation)

and to

$\frac{\partial^2 y}{\partial t^2}=-\hat{C}y\ (^{***}).$

So the modified equation can't be used solely. Is there some way to eliminate the solutions given by (***) when solving the system numerically, and only obtain solutions given by (*)? Thanks in advance.
• August 30th 2012, 06:18 PM
TriKri
Re: Eliminating dual solutions?
No one knows? :(
• August 31st 2012, 05:31 AM
TriKri
Re: Eliminating dual solutions?
I read in these course notes (on page 31) that "if a crazy looking operator like $\sqrt{-\nabla^2_H}$ is okay, then the exact problem can be recast into a canonical problem". What is a canonical problem, and does that have anything to do with my question? Can that help me in some way? Thanks.

P.S. The linear operator $\hat{C}$ that I'm using in my problem has the property that if it's applied to a function $e^{i\,\vec{k}\,\vec{x}}$, the outcome will be $-|\vec{k}|\,e^{i\,\vec{k}\,\vec{x}},$ which I guess means that $\hat{C}$ can be expressed as $-\sqrt{-\nabla^2}$, which looks very similar to the one used in the slide show.

Edit: Oh, and I forgot to say that the PDE is only supposed to be solved numerically, not analytically.