1
$\begingroup$

I am trying to solve a system of equations and have a question regarding the validity of my approach when implementing a fifth-order Cash-Karp Runge-Kutta (CKRK) embedded method with the method of lines. To give the questions some context, let me state the problem I am attempting to solve:

$$ \frac {\partial E}{\partial z} = - \frac {1}{c^2}\frac {\partial E}{\partial t} - \frac{1}{k} \frac {\partial ^2 E}{\partial z^2} - \frac{1}{kc^2} \frac {\partial^2 E}{\partial t^2} + iP \tag{1} $$ $$ \\ \frac {\partial P}{\partial t} = iNE^* \tag{2}\\ $$ $$ \frac {\partial N}{\partial t} = iPE \tag{3} $$

$$ E(z=0) = \frac{\partial E}{\partial z}(z=0) = E(t=0) = \frac{\partial E}{\partial t}(t=0) = 0,\\ P(t=0) = P_0e^{z/c}, N(t=0) = N_0e^{z/c} $$

where $ c = 3 \times 10^8, k = 1000, P_0 $ and $N_0$ are constants$ i=\sqrt{-1}$$; 0 \leq t \leq 1000, 0 \leq z \leq 1000 $

I am implementing CKRK on the above, and even though the first spatial derivative of $E$ depends on the second spatial derivative of $E$, the numerical method appears to work when solving (1)-(3) when I use the scheme of approximating the time and spatial derivatives of $E$ on the right hand side of (1) by a backward difference approximation (I am using an accuracy of 5).

To switch (1) above to a system of first order spatial derivatives in $z$, I could make the substitution:

$$ U = \frac {\partial E}{\partial z} $$

and solve the following equations for $E$ instead:

$$ U = \frac {\partial E}{\partial z} \tag{4}\\ $$ $$\frac {\partial U}{\partial z} = - \frac {k}{c^2}\frac {\partial E}{\partial t} - kU - \frac{1}{c^2}\frac {\partial^2 E}{\partial t^2} + kP \tag{5} $$

But when testing these same initial/boundary conditions using the same numerical method on the coupled equations (2) - (5), the code takes too long to finish (the step sizes required become extremely small). I believe it is due to the fact that the coefficients on the right hand side of (5) are very large and cause stability issues. I have tried to rescale the values for $z,t,P,N,$ and $E$, but doing so causes one of the other coupled equations to become unstable or has no effect (e.g. scaling $z$ does nothing to the value $ E = U\Delta z$ since both $U$ and $\Delta z$ would scale reciprocally and cancel any effect). It is due to similar reasons I am solving $E$ in the $z$-direction as opposed to doing the substitution $U = \frac {\partial E}{\partial t}$ and solving it in $t$ which is the standard method of lines approach (when I tried this method, the $\Delta t$ given by CKRK becomes very small).

So ultimately, instead of using equations (2) - (5), I was wondering if applying CKRK to (1) - (3) is still a valid approach where I approximate the derivatives of $E$ on the right-side of (1) by backward finite differences? It seems very odd to apply CKRK to a first order spatial derivative that depends on an approximation of the second order spatial derivative, but is this wrong? (I would be using stored intermediate values of $E$ to ensure the backward finite difference approximations are also following the Runge-Kutta method.)

$\endgroup$
  • $\begingroup$ This seems closely related to another question you posted. It would be courteous to at least link to that, since people have spent time reading and answering/commenting there. $\endgroup$ – David Ketcheson Oct 23 '16 at 5:37
  • $\begingroup$ @DavidKetcheson Thank you for the response. The main reason for not originally linking the previous post was because it focused on issues with using central finite difference approximations and the problem a user raised there is no longer an issue as far as I am aware (but the main question still stands). Although you are right as it may be helpful for users to read the previous question: scicomp.stackexchange.com/questions/25243/… $\endgroup$ – Mathews24 Oct 24 '16 at 1:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.