# Horrible (?) system of PDEs

1. Dec 20, 2006

### Old Smuggler

Greetings to all. In a physics problem, I have come across a system of coupled PDEs
for 2 functions B(r,t) and V(r,t) on E^3 equipped with polar spherical coordinates (r,t,p).
(I write t for theta and p for phi.) With a comma denoting partial derivation and
D^2 denoting the Laplacian, the PDEs read

Equation 1

D^2 B = B*r^2*(sin^2 t)*( (V,r)^2 + (V,t)^2/r^2 ),

Equation 2

D^2 V + ( (2/r) + (B,r/B) )*V,r + ( 2*(cot t) + (B,t/B) )*V,t/r^2 = 0.

A trivial solution of this system is V = 0, B = 1 - m/r, where m is a constant.
However, I would want to find a nontrivial solution where V is not constant
and where said trivial solution represents a limit. Does anybody know how to
find such a solution?

2. Dec 21, 2006

### dextercioby

In the way written it's almost impossible to obtain a general analytic solution. However, since you mentioned it has to do with physics, perhaps you could spell out the problem that led you to this awful system.

Does it have to do with electromagnetism by any chance ?

Daniel.

3. Dec 21, 2006

### Old Smuggler

No, it hasn't anything to do with electromagnetism, but rather with gravity.

The system of PDEs comes from considering the axially symmetric line element

$$ds^2 = B [ -(1-V^2r^2{\sin}^2 {\theta})dt^2 + 2Vr^2{\sin}^2{\theta}dtd{\phi} + dr^2 + r^2( d{\theta}^2 + {\sin}^2{\theta}d{\phi}^2) ], (1)$$

combined with the equations

$$R_{{\perp}{\perp}} {\equiv} n^{\mu}n^{\nu}R_{{\mu}{\nu}}= 0, {\qquad} R_{r{\perp}} {\equiv} -n^{\mu}R_{r{\mu}} = R_{{\theta}{\perp}} = R_{{\phi}{\perp}} = 0$$

where $$R_{{\mu}{\nu}}$$ are the components of the Ricci tensor and $$n^{\mu}$$ are the
components of the normal unit vector field to the hypersurfaces t = constant in (1).

Last edited: Dec 21, 2006
4. Dec 21, 2006

### Chris Hillman

So are you trying to solve the EFE?

Hi, Old Smuggler,

Sounds like you are trying to find a solution in gtr or some other metric theory of gravity. If so, maybe the moderator should move this thread to the General and Special Relativity board? I have a lot of experience solving the EFE so there's a good chance I can say somthing helpful...

By the way, if you surround your pseudolatex with "tex" inside square brackets and "/tex" inside square brackets (sorry, I guess I don't know the metacode), this might suffice to obtain rendering of your equations, which would make them easier to read. (If your first try doesn't work, you can try tweaking with the "edit" button.)

Last edited: Dec 21, 2006
5. Dec 21, 2006

### Old Smuggler

Hello Dr. Hillman,

this problem does not come from GR; you see that the equations I wrote up are
different from the EFE. The question is, how relevant would your experience with
solving the EFE be for solving this problem?

Thank you for the tip, I have edited the last post to make it more readable.

6. Dec 25, 2006

### Chris Hillman

Hi again (btw, I go by "Chris"),

Er... a PDE is a PDE is a PDE? In particular, techniques or insights which apply to one geometrically motivated system of PDEs often apply to another. Even if not, this is in itself noteworthy.

(BTW, I have not tried to verify your claim about this not being the EFE; all such systems can be readily reformuated in many ways, so it is not always easy to tell at a glance whether you are regarding an old friend dressed in unfamiliar garb.)

Please forgive me if you have already considered all the "obvious questions" about your system. It can often be hard to gauge the level of sophistication of posters in public fora like this one, because the membership is so broad. It is all too easy to offend by "overshooting" or "undershooting", yet asking posters to summarize their background seems insulting to some...

I was pretty much simply planning to ask whether or not your system is hyperbolic/well-posed and whether you've computed its symmetry group or attempted an attack based on internal symmetries.

It probably would help if you told us a little more about the context. Where did you come across this system? In a paper or book?

Last edited: Dec 25, 2006
7. Dec 26, 2006

### Old Smuggler

Hi Chris,

I like that attitude. It means that you probably won't waste your time on this.
OK. Since $$2R_{{\perp}{\perp}}=G_{{\perp}{\perp}}+G^i_i$$
and $$R_{{\perp}j}=G_{{\perp}j}$$, where $${\bf G}$$
is the Einstein tensor, you see that in GR, the equations I wrote down would
correspond to a fluid with equation of state $$p=-{\rho}/3$$ and zero momentum density but with nonzero tension (since $$R_{ij}{\neq}0$$ from
the line element). This would not represent any acceptable physical solution in GR.
Sure. My background is physics; no sophisticated mathematics I'm afraid..
I believe that the system is well-posed since I calculated the first few terms
in a series solution; these terms fall off with distance so it seems that an
asymptotic flat solution should exist. Besides, the system has some similarities
with its counterpart in GR (that system can be found in Wald's book). However, the line element I wrote down does not take the Weyl canonical form since the Ricci tensor does not vanish. Therefore, a reformulation like that leading to the Ernst equation in GR does not seem too promising.

I do feel that the only sensible way to attack the system would be based on
symmetry, but I haven't calculated its symmetry group (unfortunately, I'm
unfamiliar with such techniques, but I could learn it if necessary).
Only a unpublished preprint I'm afraid.. I hope that won't act as a total turn-off for you. Thanks for your time.

Last edited: Dec 26, 2006
8. Dec 28, 2006

### Old Smuggler

Let me write down the first few terms of the series solution, valid in the weak field limit. This solution is

$$B(r,{\theta})=1-{\frac{r_s}{r}} + J_2{\frac{{\cal R}^2r_s}{2r^3}} (3{\cos}^2{\theta}-1)+{\frac{r_s^2a^2}{2r^4}}(3{\sin}^2{\theta}-1) +{\cdots}$$

$$V(r,{\theta})=-{\frac{r_sa}{r^3}}(1+{\frac{3r_s}{4r}}+{\cdots})$$

$$r_s{\equiv}{\frac{2MG}{c^2}}, {\qquad} M{\equiv}c^{-2}{\int}{\int}{\int}{\Big (}{\sqrt B}(T_{{\perp}{\perp}} +T^i_i)-2VT_{{\perp}{\phi}}{\Big )}{\sqrt h}d^3x,$$

$$a{\equiv}{\frac{J}{cM}}, {\qquad} J{\equiv}c^{-1}{\int}{\int}{\int}{\psi}^{\phi} T_{{\perp}{\phi}}{\sqrt h}d^3x,$$

$$h$$ is the determinant of the spatial 3-metric on the
t=constant hypersurfaces, $${\cal R}$$ is the mean coordinate radius of the source, $${\bf T}$$ is the (active) stress-energy tensor of the source and where the integration is taken over the source.
($$J_2$$ is the quadrupole-moment parameter, which is unspecified
in the solution, and $${\psi}{\equiv}{\frac{\partial}{{\partial}{\phi}}}$$ is a Killing vector field associated with the axial symmetry.)

We se that the solution is naturally expressed in terms of Komar integrals.

Last edited: Dec 28, 2006
9. Dec 29, 2006

### Chris Hillman

Do I have the right system?

Hi again, Old Smuggler,

I don't need to know where it comes from to attack your system, but let me make sure I understand correctly what your system is. If you are using a background polar spherical chart on euclidean three space, and assuming axisymmetry, then my understanding is that your system is:

$\Delta B = B \, r^2 \ \sin(\theta)^2 \, \left( V_r^2 + V_{\theta}^2/r^2 \right), \; \; \Delta V + \left( \frac{2}{r} + \frac{B_r}{B} \right) + \left( 2 \, \cot(\theta) + \frac{B_{\theta}}{B} \right) \, \frac{V_{\theta}}{r^2} = 0$

where B, V are unknown functions of $$r, \theta$$ only, where subscripts denote partials, and where

$\Delta = \frac{\partial^2}{\partial r^2} + \frac{2}{r} \, \frac{\partial}{\partial_r} + \frac{1}{r^2 \, \sin(\theta)^2} \, \frac{\partial^2}{\partial \theta^2}$

Do I have that right?

Actually, I was wondering if you have tried to set up a "nice" initial value formulation (nice: such that a standard result for second order hyperbolic systems would imply stability under small perturbations in initial conditions), as per Wald.

Actually, the Weyl-Papapetrou chart isn't key to obtaining the Ernst equation, and the Ernst equation isn't the only system with a large symmetry group, but never mind that for now.

If you can tell me that I have the right system, it should be a matter of a few minutes to compute the point symmetry group using Maple. If there is a verbatim environment at PF, I can paste in the commands. From this, plus Maple help, plus Peter Olver, Applications of Lie Groups to Differential Equations, you can learn enough to follow along and to do this yourself; with a bit of practice it rolls right along. Curiously enough, nonlinear equations are actually a bit easier, although their symmetry groups might be too small to be of much help. But in this case a group analysis can suggest special cases which are likely to lead to exact solutions by Lie's methods.

Last edited: Dec 29, 2006
10. Dec 29, 2006

### Old Smuggler

Hi Chris,

that is true. But I can understand that it would be more motivating to
know what lies behind this system. Since I'm not sure if that would be
appropriate to discuss here, I will send you an email if you are interested in
the wider context.
Yes, except for a couple of typos which I took the liberty to correct.

No, I haven't tried that.

I'm glad to hear that an exact solution may perhaps be found without too
much work. Anyway, I appreciate your efforts.

Last edited: Dec 29, 2006
11. Dec 29, 2006

### Chris Hillman

And the point symmetry group is...

Hi, Old Smuggler,

Oops, I forgot to write the $$\cot(\theta)$$ term in the Laplacian, didn't I? (That wasn't a trick question, I really did forget, but I am glad to see that you are on the ball!)

Note that I rewrote your $$B(r,t), V(r,t)$$ as $$b(r,\theta), v(r,\theta)$$ for convenience in what follows. I took the independent variables to be $$r, \theta$$ and the dependent variables to be $$b, v$$.

The point symmetry group only takes a minute or so to compute via Maple (using the liesymm package to obtain Lie's "determining equations" for the Lie algebra of the point symmetry group). I obtained a three-dimensional Lie algebra whose generators may be given as:

$\vec{X}_1 = b \, \partial_b, \; \vec{X}_2 = r \, \partial_r - v \, \partial_v, \; \vec{X}_3 = \partial_v$

In what follows I'll try to address a somewhat broader audience since I expect that symmetry methods should be of wide interest as the most general standard method of attack on systems of nonlinear partial differential equations. These methods are not a magic bullet--- they often don't yield any information you don't already know from more elementary reasoning, and sometimes they don't help at all, e.g. if the symmetry group is trivial. But this is a general attack which is not really very hard to try, so it is usually worth trying as a matter of course.

The general real linear combination of our three generators gives a multiscaling symmetry

$r \, \partial_r + k \, b \, \partial_b + (c- v) \, \partial_v$

where k, c are arbitrary real constants. From this we can read off a system of ODEs

$\dot{r} = r, \; \dot{\theta} = 0, \; \dot{b} = k \, b, \; \dot{v} = c-v$

which can be integrated to obtain the transformations in the group

$(r, \, \theta, \, b, \, v) \rightarrow (\lambda \, r, \, \theta, \, \lambda^k \, b, \, c+(v-c)/\lambda)$

where $$\lambda > 0$$ is the parameter of our uniparameter subgroup (for fixed k,c). (I should probably say that the transformations in the point symmetry group are the ones which take the system into a system having the same functional form. Also, the original parameter was the exponent in an exponential factor; for some reason I modified this to give an action by the positive reals under multiplication.) We also obtain the rational invariants of this flow, namely

$\theta = c[1], \; b/r^k = c[2], \; (c-v)/r = c[3]$

from which we obtain the symmetry Ansatz, which is the point of this exercise:

$b(r,\theta) = f(\theta) \, r^k, \; \; v(r,\theta) = c + r \, g(\theta)[/tex] Plugging this into the original system of equations, we obtain a system of two ODEs in $$\theta$$ only. (If this step fails, this indicates a computational error at an earlier stage; elimination is guaranteed if no errors have occured.) One gives f in terms of g; the other is a rather nasty third order nonlinear equation for g with no obvious symmetries. (Since no-one spoke up to tell me a verbatim environment, and since you didn't say whether you are a Maple user, I am not trying to include here the sequence of Maple commands I used, or to give the third order equation I came up with; to avoid transcription errors it is probably best for you to plug in the above Ansatz and find it for yourself.) In a special case we obtain the trivial solution [itex] b(r,\theta) = P_k(\cos(\theta)) \, r^k, \; v(r,\theta) = c$

where $$P_k$$ is the Legendre polynomial (this solution resembles an "interior mode" in an analogous situation in gtr), but you probably knew that.

Well, I'm not very surprised that I got stuck at a hard ODE; this often happens. But fear not, this method is very sensitive to small changes, so transforming to new variables can make all the difference. (In the above, it looks like I used up all the symmetries, so considering other symmetry assumptions probably won't help for the system expressed in the original variables.)

(By the way, one thing symmetry methods can help you determine is whether or not their exists a transformation to new (independent and dependent) variables in which the new system is linear. I have not tried that criterion in this example.)

If you often deal with PDEs, I'd strongly encourage you to learn symmetry methods (there are a number of excellent books out now), and to become familiar with Maple or Mathematica packages which automate the most time consuming step, solving the determining equations. These are always linear, so Groebner basis methods can "triangularize" and then solve (by a generalization of Gaussian reduction), and there are Maple and Mathematica routines which carry this out for differential rings. So the bottom line is that the determining equations are always solvable and you can quickly to fairly quickly determine the symmetry group (modulo computational limitations).

Once you have that, with a bit of experience you can read off a symmetry Ansatz as above and obtain a reduced system. In Lie's approach, each symmetry should enable you to reduce a system of PDEs to a system with one less variable (in a system of ODEs, each symmetry reduces the order by one). These symmetries "combine" in ways determined by the structure of the Lie algebra, but in principle if the symmetry group is sufficiently large you can often reduce to a single ODE and if that is solvable in closed form, you can often obtain complete solutions to the original system in closed form.

Lie's method is sometimes described as a generalization of "dimensional analysis", and in many cases (as here), the symmetry Ansatz method yields up an Ansatz which represents a multidimensional scaling which could in many cases have been guessed from more elementary reasoning. However, often one does obtain an Ansatz which might not easily have been guessed.

If there are boundary conditions, you can look at the subgroup of symmetries (acting on the indepedent variables) which preserve the boundary.

If your system includes some arbitrary functions (e.g. a forcing term) or parameters, you can use "group analysis" to determine special cases for which the symmetry group is enlarged. This is interesting because both because it can lead to solutions and because it can identify useful subcases. In your case, you probably have some "tuneable parameters" in your field equation, so you can apply a symmetry analysis to find out if there are any special values which yield a larger symmetry group.

In the case of a system which arises from a Lagrangian (as should happen with any "good" classical field theory), Noether's theorem relating variational symmetries (these form a subgroup of the point symmetrics) belongs to the subject of Lie analysis of differential equations, and gives a systematic way to determine conserved quantities (i.e. satisfying divergence identities). Those interested in soliton equations will be intrigued to note that Noether's notions of "generalized symmetries" (the terminology is unfortunately notoriously unstandardized) gives a systematic way to use elaborations of Lie's methods to find an infinite hierarchy of conservation laws.

Last edited: Dec 29, 2006
12. Dec 30, 2006

### Old Smuggler

Hi Chris,

great job! I will definitively learn this method when I get access to the
library (i.e., next week) and go through your post until I understand the
details. Meanwhile, I have some (undoubtedly naive) questions/comments.

This result should be checked against the weak field solution I wrote down
previously. The result for $$b(r,{\theta})$$ seems reasonable since
every term of the weak field solution takes this form (for $$k=-1,-3,-4,...$$; I presume that the restrictions of k to integers comes from putting
the symmetry Ansatz into the equations). So it seems that we have a series
solution of the type $$b(r,{\theta})=f^{(k)}({\theta})/r^k,$$,
right?

On the other hand, the form of $$v(r,{\theta})$$ is inconsistent
with the weak field solution, so something has apparently gone wrong here.
Yes, I'm a Maple user (but I haven't used any of the advanced stuff).
Since I'm not convinced that the Ansatz is correct, I have not tried to find
that nonlinear third order ODE.
Yes, the case v = constant is trivial since the system of PDEs then reduces
to the equation $${\triangle}b(r,{\theta})=0$$ combined with the
axisymmetric condition; which corresponds to a situation where the source
is static and axisymmetric. The solution is as for the Newtonian counterpart
(since we are only interested in asymptotically flat solutions)

$$b(r,{\theta})=P_k({\cos}{\theta})/r^k$$.

However, I'm interested in a situation where the source is "rotationally
deformed", i.e., a source which is spherically symmetric in absence of rotation; this would mean that the source is made of a material which does
not support shear forces, e.g., an ideal gas.
Does this mean that the system may be easier to solve if expressed in
cylindrical coordinates?
That's what I'm after, so I will try to learn this method as soon as possible.
Thank you for a very instructive lecture Chris!
In a slightly different version of the system of PDEs, the background
3-geometry is not $${\bf E}^3$$, but rather $${\bf S}^3$$
(actually half of it, so there are boundary conditions). This comment may
be relevant for that case.
Since the existence of tuneable parameters diminishes the predictive power
of any theory, I try to avoid them whenever possible; there are certainly
none in my field equations. (I will send you an email with references so you
can see for yourself.)

13. Dec 31, 2006

### Chris Hillman

Checking a symmetry group

Hi again, Old Smuggler,

First, let me emphasize again that I have not verified any of the claims you made about your system. I have not verified that the two versions you wrote down correspond as they should, nor have I thought about whether or not this is the EFE in disguise. I have not verified the weak-field solution you wrote down or even thought about whether this looks plausible. I simply took the explicit form you stated and computed the information I offered in a few minutes (it helps that I've done this before, of course!).

Somewhere in here I should probably have said that I am entirely self taught in PDEs (above the UG level) and in gtr. OTH, I do have a lot of experience computing symmetry groups and finding solutions from a suitable symmetry Ansatz deduced from a symmetry group, which probably wouldn't succeed if I had many any computational errors.

Well, very roughly, for a system of two second order PDEs in two dependent variables and two independent variables, according to Siklos (see my citation in thread on "how many solutions"), especially a hyperbolic system, you would in general expect the solution space to to be "parameterized" by a certain number of functions of one variable. So a subspace of the solution space which is "parameterized" by the solutions of a third order ODE is an infinitesimal snippet (nothing remotely approaching an open neighborhood of any solution, for example). As this suggests, in general you shouldn't expect symmetry methods to give anything like "the general solution" to a nonlinear systems of PDEs. (Which is in any case a much more difficult notion when dealing with nonlinear systems; for example, when are two parameters "independent"?) Indeed, you should expect symmetric solutions might be atypical in qualitative ways.

Indeed, in the case of say the KdV, various methods give different pieces of the solution space, and it is well known in the field of solitons that naive application as exemplified by my computation of the symmetry group and then a symmetry Ansatz give rather special solutions.

So, you shouldn't expect that your weak-field approximation should have any particular relation with the three-parameter family of solutions which I gave (up to solutions of the third order ODE).

I didn't say that, at least, I hope not. All three parameters are real valued.

You can check the symmetry group I gave by plugging in your system and computing the transformed system (rather laborious by hand, but Maple's dchange command will do this for you). You need only check that solutions of the original system are transformed to solutions of the transformed system.

The rational invariants are trivially determined by standard methods (see for example the books by Olver) from the vector field I wrote down (the generator of the multiscaling transformation with two fixed constants and one serving as the group parameter of a uniparameteric subgroup). You can write the general element of the symmetry group and verify that these do form a (nonabelian) transformation group.

I said a trivial special case is $b(r,\theta)= P_k(\cos(\theta)) \, r^k, \; v(r,\theta) = v_0$, but the "series" bit doesn't come from the symmetry group stuff. My comment was perhaps a bit misleading, since k can take any real value here. This is consistent with but not equivalent to what you get by separating variables in the usual Laplace equation (which I think of as more like an orthonormal basis of eigenfunctions).

If you assume v is constant, as you know, your system reduces to something you can attack by separation of variables, namely the "axisymmetric" Laplace equation with dependent variable b. But this is a "trivial case" holding little interest.

You can check the Ansatz by plugging it into the explicit system you gave and verifying that the variable r drops out. This is precisely what Lie's method is supposed to guarantee in this circumstance. If you play around, you will see that only very special assumptions allow you to eliminate a variable. As I said, these can often be determined by more elementary reasoning, but often Lie's methods lead you to an Ansatz which would not have been so obvious.

The rational invariants of the flow and the Ansatz are trivially determined from the vector field I wrote down (an element of the Lie algebra of the symmetry group). So determining generators for this symmetry group is the only laborious step. Fortunately, Lie's ideas are not very hard or even very "advanced", just unfamiliar to many who could, I think, benefit from making the effort to learn them. By the way, one of the books on this stuff is by Hans Stephani, who as he says in the forward was interested in gtr applications---which didn't pan out, but he was hooked regardless! That is not a bad outcome, I think.

Or oblate spheroidal, or prolate spheroidal, or parabolic--- there are infinitely many options, any of which might yield dramatic simplifications.

Lie's methods are very general, easy to try in most circumstances, and well worth learning IMHO.

Well, I've seen many people claim "there are no tuneable parameters in my field equations", only to be shot down in flames. No doubt you know why this statement seems a bit suspicious. However, if you submitted to GRG or CQG your referees will (or should) check that claim, so there's no need to argue about it here!

Again, I often compute symmetry groups just to stay in shape, as it were, I wasn't trying to get into discussions of a proposed alternative to gtr...

Last edited: Dec 31, 2006
14. Dec 31, 2006

### Old Smuggler

Well, if that's a general result, I must say it would be a real disappointment.
However, the reason for this may lie elsewhere, see below..
I should of course have checked the Ansatz right away, in my previous
post, sorry about that. Especially since it can be seen in a glance, that
the variable r does not drop out of the first equation, since the left hand
side is proportional to $$r^{k-2}$$ while the right hand side is
proportional to $$r^{k+2}$$. So it would seem that you have used
the wrong equation in your analysis (if the factor $$r^2$$ on the
right hand side is changed to $$r^{-2}$$ the Ansatz is correct).

15. Jan 1, 2007

### Chris Hillman

The Ansatz

There seems to be some confusion about notation/terminology which would probably be cleared up when you read Olver's book Applications of Lie Groups to Differential Equations. The factor you mention is the one which drops out if you start with the Ansatz I gave. It's also possible that as I feared might happen, there is some confusion about what the system of PDEs you wanted help with is. In any case, it seems you may no longer be interested in Lie's methods (if not you have enough information to start learning how to use his methods), so I propose that we let this drop.