# Re-writing higher order spatial derivatives as lower order system

• Jul 26th 2010, 08:23 PM
nkinar
Re-writing higher order spatial derivatives as lower order system
I've been working with the following PDE:

$\displaystyle $\nabla ^2 p = \frac{{\partial ^2 p}}{{\partial x^2 }} + \frac{{\partial ^2 p}}{{\partial y^2 }} = A\frac{{\partial ^2 p}}{{\partial t^2 }} + B\frac{{\partial p}}{{\partial t}}$$

What I would like to do is to re-write the second-order spatial derivatives in this PDE as first order derivatives.

This is what I have attempted, but I am uncertain as to whether this is correct. I introduce another variable $\displaystyle q$, and then:

$\displaystyle $\frac{{\partial q}}{{\partial t}} = \nabla \cdot \vec p$$

$\displaystyle $\nabla \cdot \vec q = A\frac{{\partial p}}{{\partial t}} + Bp$$

I reason that this is correct since

$\displaystyle $\nabla \cdot \frac{{\partial q}}{{\partial t}} = \nabla \cdot \nabla \cdot \vec p = \nabla ^2 p$$

$\displaystyle $\frac{\partial }{{\partial t}}\left( {\nabla \cdot \vec q} \right) = \frac{\partial }{{\partial t}}\left( {A\frac{{\partial p}}{{\partial t}} + Bp} \right)$$

Now is it reasonable to claim that the LHS of the two equations above are the same?
• Jul 27th 2010, 03:47 AM
Jester
I see an immediate problem. You have both $\displaystyle p$ and $\displaystyle \vec{p}$. These are not same! Similar for $\displaystyle q$ and $\displaystyle \vec{q}.$
• Jul 27th 2010, 03:52 AM
Jester
Second $\displaystyle \nabla \cdot \nabla \cdot \vec{p}$ makes no sense.
• Jul 27th 2010, 07:32 AM
nkinar
Danny: Thank you very much for checking this over. Are the problems that you see issues with notation, or can the two equations be written in a more precise fashion?

Thus

$\displaystyle $\nabla \cdot \frac{{\partial q}}{{\partial t}} = \nabla \cdot \left( {\nabla \cdot \vec p} \right)$$

Essentially what I would like to do is to write the PDE in terms of first order derivatives.
• Jul 27th 2010, 07:37 AM
nkinar
So given

$\displaystyle $\frac{{\partial ^2 p}}{{\partial x^2 }} + \frac{{\partial ^2 p}}{{\partial y^2 }} = A\frac{{\partial ^2 p}}{{\partial t^2 }} + B\frac{{\partial p}}{{\partial t}}$$

is there a way to split this into a system involving only first order derivatives?
• Jul 27th 2010, 07:45 AM
nkinar
So to put this in another way:

Is

$\displaystyle $\frac{{\partial ^2 p}}{{\partial x^2 }} + \frac{{\partial ^2 p}}{{\partial y^2 }} = A\frac{{\partial ^2 p}}{{\partial t^2 }} + B\frac{{\partial p}}{{\partial t}}$$

Also equivalent to the following?

$\displaystyle $\frac{{\partial q}}{{\partial t}} = \frac{{\partial p}}{{\partial x}} + \frac{{\partial p}}{{\partial y}}$$

$\displaystyle $\frac{{\partial q}}{{\partial x}} + \frac{{\partial q}}{{\partial y}} = A\frac{{\partial p}}{{\partial t}} + Bp$$
• Jul 27th 2010, 08:24 AM
Jester
These are not equivalent since elimiating the q gives

$\displaystyle Ap_{tt} + Bp _{t} = p_{xx} + 2 p_{xy} + p_{yy}$.
• Jul 27th 2010, 08:47 AM
nkinar
Thanks, Danny: I had suspected that something was amiss!

$\displaystyle $\nabla \cdot \left( {\nabla p} \right) = \nabla ^2 p = \frac{{\partial ^2 p}}{{\partial x^2 }} + \frac{{\partial ^2 p}}{{\partial y^2 }} = A\frac{{\partial ^2 p}}{{\partial t^2 }} + B\frac{{\partial p}}{{\partial t}}$$

This is re-written as:

$\displaystyle $A\frac{{\partial p}}{{\partial t}} + Bp = \nabla \cdot \vec q$$

$\displaystyle $\frac{{\partial \vec q}}{{\partial t}} = \nabla p$$

Where q is a vector. I believe that this is correct since by taking the time derivative of the first equation:

$\displaystyle $A\frac{{\partial ^2 p}}{{\partial t^2 }} + B\frac{{\partial p}}{{\partial t}} = \frac{\partial }{{\partial t}}\left( {\nabla \cdot \vec q} \right)$$

Applying the del operator to the following:

$\displaystyle $\nabla \cdot \left( {\frac{{\partial \vec q}}{{\partial t}}} \right) = \frac{\partial }{{\partial t}}\left( {\nabla \cdot \vec q} \right) = \nabla \cdot \left( {\nabla p} \right) = \nabla ^2 p$$

We can thus say that both sides of the above are equal.
• Jul 27th 2010, 08:51 AM
Jester
Ya - I'll agree that that!
• Jul 27th 2010, 08:53 AM
nkinar
Thanks, Danny! I think that it looks good as well! Should I go ahead and mark the thread as solved?
• Jul 27th 2010, 08:54 AM
Jester
Sure - If you're happy with the answer.
• Jul 27th 2010, 08:56 AM
nkinar
Yes, I am happy with the answer. Once again, thank you very much! I'll go ahead and close the thread.