Hi guys, I'm hoping one of you will be able to help with an issue I'm having. I'm trying to find the solution to an ordinary differential equation. Specifically, it's the equation of motion of a D brane that comes from the Born-Infeld action. It's not a particularly nice equation so I'm solving it numerically. The background is AdS so the radial coordinate has a boundary at \(z=0\). One can show that the asymptotic expansion at the boundary goes like \(\phi = m z + v z^3 + k z^3 \log z + \ldots \). I am interested in looking at v as a function of m - effectively m is the dial and v changes. I can also evaluate the action as a function of m as well. The equation has 2 solutions - one that extends to the centre and where the derivative of the function is zero at the centre. The other one ends at some finite \(z_0\). \(\phi(z_0) = \frac{\pi}{2}\) and \(\phi'(z_0)\) diverges. I showed all this numerically by providing the boundary conditions in the interior of AdS: ie \(\phi (z = \text{centre}) = \text{constant}<\frac{\pi}{2}\) and \(\phi'(z = \text{centre})= 0\) or \(\phi (z_0) =\frac{\pi}{2}\) and \(\phi'(z_0)= \text{big\ number}\). Now I want to do something different, and for technical reasons I need to provide one boundary condition at the boundary and one in the interior or both at the boundary (really I need to put m in as a boundary condition), so I changed my code so that the boundary conditions are \(\phi(\epsilon) = m \epsilon + v \epsilon^3 + k \epsilon^3 \log \epsilon\) and \(\phi'(\epsilon) = m + 3 v \epsilon^2 + k \epsilon^2 ( 3 \log \epsilon+1)\). m,v and k I knew from the other method so I was expecting to get the same pictures and results as before. I didn't. I've since tried various combinations of boundary conditions at each end and nothing seems to work. The solution blows up at multiple points with this method when it didn't with the previous method. What I want to know is are there any specific numerical techniques that I can use to deal with boundary conditions at both ends of the solution. I'm using mathematica which essentially defaults to no brain mode and picks the method it thinks will work best. It's not always that reliable. Cheers.
Can you please show what equation you are trying to solve and what are the boundary conditions? Do you have any fall-off condition at infinity? What is the difference between the first and the second problem, namely, is the second problem simply the first one shifted by epsilon?
The EOM is long and unpleasant. The action is \(S = \int dz \frac{(z-1)^3 (z+1) \cos^3 \phi(z) }{z^5} \sqrt{1+z^2 \phi'(z)^2}\) The z coordinate runs between 0 at the boundary and 1 at the centre of the space. Originally the solutions I got by using the BC's \(\phi (1) = \text{constant}<\frac{\pi}{2}\) and \(\phi'(1)= 0\) or \(\phi (z_0) =\frac{\pi}{2}\) and \(\phi'(z_0)= \text{big\ number}\) which are at the "interior end" of the solution, rather than the boundary. There is a numerical issue here: if I provide the BC's exactly at z=1 or \(z=z_0\) the program doesn't find any solution. Numerically I put in \(1-\epsilon\) or \(z_0-\epsilon\) where \(\epsilon\) is some small number (of order \(10^{-6}\) does the trick.) The BC's above determine m, v and k so to start with I tried solve the same problem as a check of the method, putting in the values of m,v and k I produced from the previous method, assuming this would reproduce the previous solution. The numerical issue above also affects the BC's at the boundary, so instead of supplying \(\phi(0)\) and derivative I supply \(\phi(\epsilon)\) and derivative. I swapped the BC's for m, v and k and tried to reproduce \(\phi(1)\) and \(\phi'(1)\) (I picked the first solution that extends all the way to the centre to start. I should also point out that the solution that gets "chosen" from the point of view of the boundary depends only on m. Below some \(m = m_{\text{crit}}\) you get the solution that extends all the way to the centre. Above \(m=m_{\text{crit}}\) and you have the solution that ends at \(z=z_0\). Thanks. Please Register or Log in to view the hidden image!
I tried to get the equation from your action. Please check my errors. Does the equation look something like this? \( z^3(z^2-1)\dot\phi(z\ddot\phi+\dot\phi) = (z^2\dot\phi^2+1)[z(2z^2+2z+3) + 6(z+1)(z^2\dot\phi^2+2)\tan\phi], \) where dot denotes the derivative with respect to z.
Hi Temur, I must admit, I've got a line of my mathematica code to work out the equation of motion and the result it gives is far lass pretty than that. It looks like in your equation you have a \(\dot{\phi}^4\) term that I don't have and no \(\dot{\phi}^3\) term that I do have. I spent today working on this problem and I discovered the BDF method that seems to work a bit better. I also had a chat with my supervisor about it and it looks like the solutions are unstable, ie they always want to diverge away from the true solution when the parameters are not exactly tuned to the correct values. I don't know if you are familiar with the numerical solution of the Schrodinger equation for Charmonium, which quite often gets done by undergrads as a project, but the eigenfunctions there essentially behave like that. We suspect / hope that we can use some iteration type program to get the solutions to converge as far as possible and that will be sufficient. Thanks for your help. If you're interested I'd be happy to takl about this some more. Please Register or Log in to view the hidden image!
Thanks for your reply. Assuming that my equations are at least a correct sketch of the true equations, the equation you are dealing with is degenerate, because the coefficients in front of the second derivative of phi becomes zero at various points (namely, z=0 and z=1). To solve it stably, one approach would be to do some clever transformation of the equation so that degeneracy does not occur.
Hi Temur, I've been thinking a bit about this problem today while I've been working on some other things. You're quite right of course, the EOM is badly behaved at z = 1 and z = 0, but unfortunately, I've chosen a coordinate system that has nice properties for the physics of the problem and I suspect that I'd lose that if I were to massage the system so that the EOM was a bit better behaved. I'll let you know how the progress, if any, goes. Please Register or Log in to view the hidden image!
