A Computing first derivative based on second derivative

TheCanadian
Messages
361
Reaction score
13
I am trying to numerically solve a PDE, and just had a question as to the validity of a certain approach. For example, given the PDE:

$$ \frac {\partial ^2 E}{\partial t^2} = - k\frac {\partial E}{\partial t} + c^2 \frac {\partial ^2 E}{\partial z^2} - c\frac {\partial E}{\partial z}$$

If I was to apply an explicit Runge-Kutta method, I could make a substitution:

$$ u = \frac {\partial E}{\partial t} $$

and solve the following instead:

$$ \frac {\partial u}{\partial t} = - ku + c^2 \frac {\partial ^2 E}{\partial z^2} - c\frac {\partial E}{\partial z}
\\ \\
u = \frac {\partial E}{\partial t} $$

where the spatial derivatives are computed by a finite difference approximation (e.g. backward difference). I was just wondering: is solving the following a valid approach?

$$ \frac {\partial E}{\partial t} = \frac{1}{k}(-\frac {\partial ^2 E}{\partial t^2} + c^2 \frac {\partial ^2 E}{\partial z^2} - c\frac {\partial E}{\partial z})$$

Where now both the spatial derivatives and second time derivative are approximated by backward difference schemes, while I apply the Runge-Kutta method to the first time derivative of ##E##. I have not seen this done anywhere, but was just curious as to why (or why not) this is an unacceptable method? Mathematically, it seems odd that my first order derivative depends on higher order derivatives of itself, although I don't immediately see any reason why this is wrong numerically.
 
Physics news on Phys.org
I don't believe this makes sense. To the best of my knowledge, Runge-Kutta methods are specific to ordinary DEs, not partial DEs.
 
Mark44 said:
I don't believe this makes sense. To the best of my knowledge, Runge-Kutta methods are specific to ordinary DEs, not partial DEs.

Yes, that is correct. I would have to apply a method of lines first to discretize the ##z##-space. Once I have done this, I am applying the spatial derivatives to step ##E## forward in the ##z##-direction at a particular ##t##.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...

Similar threads

Replies
2
Views
2K
Replies
3
Views
3K
Replies
7
Views
3K
Replies
1
Views
2K
Replies
36
Views
4K
Replies
3
Views
3K
Back
Top