# How to decide PDE type for 1st order scalar PDE's

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 June 3, 2008, 05:34 How to decide PDE type for 1st order scalar PDE's #1 shantanu Guest   Posts: n/a Hello, can anyone please help me to decide the PDE type viz. elliptic, parabolic or hyperbolic if we have 1st order scalar PDE (eg. 1st order wave equation i.e. du/dt + c*du/dx=0). Thanks, Shantanu.

 June 4, 2008, 02:14 Re: How to decide PDE type for 1st order scalar PD #2 Vinayender Guest   Posts: n/a Hi, One method to find wether a pde is hyperbolic, ellliptic or parabolic as follows.... Hyperbolic problems are Initial value problems. Elliptic problems are Boundary value problems. Parabolic problems are Initial,Boundary value problmes. So take your equation :: du/dt + c*du/dx=0 ---------- 1 now write this equation can be compared with. du/dt *dt + du/dx *dx = du which is same as du/dt + du/dx * (dx/dt) = du/dt ----------- 2 comparing 1 and 2 we can say that along dx/dt is c du = 0, that is along the curve dx/dt = c pde is converted to ODE that is du = 0; which can be integrated and see the solution. There fore the solution is u = constant (along dx/dt = c) and the value of constant can be found by u(0) i.e, initial values. There fore final soution is u = u(0) (along dx/dt = c) and hence it is an initial value problem. And hence your given equation is an Hyperbolic.... Similarly if you want to solve an elliptic problem you need solution on a closed boundary ..for example you can solve (del**2)U =0 mathematically and see that you need solution on a closed by a boundary to solve such equations and hence will be elliptic .... similrly for parabolic... Also there is some mathematical procedure for finding a given pde is which one, that is if a pde admits wave like solution then the equation will be hyperbolic equation and the details of this procedure can be found in book by Hrish name is some thing like computational FD for external and internal flows........ Thanks, Vinayender.

 June 4, 2008, 05:51 Re: How to decide PDE type for 1st order scalar PD #3 Jed Guest   Posts: n/a I agree with the general message here, but it's not so clear cut. For instance, Hyberbolic problems usually need boundary conditions too unless the domain is infinite and many boundary value problems don't satisfy ellipticity. Especially for nonlinear problems, such classification is not particularly meaningful. What you are really interested in is what initial/boundary conditions and function spaces are necessary and sufficient to have well posedness (often a very hard problem). Finding coercive bilinear forms is also very relevant numerically since it tells you about what kind of preconditioning may be needed.

 June 4, 2008, 07:15 Re: How to decide PDE type for 1st order scalar PD #4 Vinayender Guest   Posts: n/a Hi Jed, First of all thank you very much for your comment on my message, it is always helps me to know wether my understanding is correct or not ? When I related initial value problem to Hyperbolic problem, I mean that to solve a hyperbolic problem initially we require the solution along some dataum line (which can be in time t=0 ,or space say a curve or surface) and we can find the solution to these pde by marching along (time or space)the characteristics lines (in the current example these lines are dx/dt = c). This explanation is in consistant with physics that a distrabance in a medium (governed by hyperbolic law)will propogate along charactersistic lines. And when I related boundary value problem to elliptic problem, I mean that to solve an elliptic pde we require solution all along the "CLOSED BOUNDARY" to solve for interior solution. Again this explanation is in consistant with the physics that for a disturbance in a medium (governed by hyperbolic law) will prapogate in all directions. Thanks, Vinayender.

 June 4, 2008, 07:58 Re: How to decide PDE type for 1st order scalar PD #5 John Guest   Posts: n/a shantanu: at others said you should specify IC and/ro BC first then determin type of eq., however, it is formal to take *t* as time and assume an IC, and take *x* as space, further you should determin solution domain, bounded, unbounded, etc. and define some BC for *u* there. your PDE seems to be both parabolic and hyperbolic. Jed: >Finding coercive bilinear forms is also ... Interesting, can you clarify, what you mean from *coercive*?

 June 4, 2008, 08:42 Re: How to decide PDE type for 1st order scalar PD #6 Vinayender Guest   Posts: n/a Hi Jhon, Above pde is Hyperbolic.........independent of wether t is time or spacial coordinate. But physically in that equation t is time and that equation represents "sclar linear convective equation". Jhon please read my 2nd post in this thread, where I told what I mean by Initial and boundary.. I should have used like this, prior represents knowing solution on some cure or surface which will not make a closed volume (area in case of 2d) and the later one is some curve or surface which makes a closed area or volume in 2d and 3d respectivly... Thanks, Vinayender.

 June 4, 2008, 09:20 Re: How to decide PDE type for 1st order scalar PD #7 John Guest   Posts: n/a i am not very expert but to my knowledge due to presence of time variable (and spatially hyperbolic), this PDE is always parabolic with respect to time, notice that in parabolic pde we have just one characteristic and have one sided propagation of informations along its direction, i.e., information never propagate backward in time, while in hyperbolic type information could propagate forward or backward, of course based on specific problem am i miss something?

 June 5, 2008, 06:04 Re: How to decide PDE type for 1st order scalar PD #8 Tom Guest   Posts: n/a First order scalar pdes in two variables are always hyperbolic. The difference comes when you go to systems and/or higher order derivatives where the characteristic equation is less trivial; e.g. for a second order pde or a system of two first order pdes the characteristic equation will be a quadratic which can have either real roots (hyperbolic), imaginary conjugate roots (elliptic) or a single repeated root (parabolic).

 June 5, 2008, 09:30 Re: How to decide PDE type for 1st order scalar PD #9 Vinayender Guest   Posts: n/a Hi John, You only said that "while in hyperbolic type information could propagate forward OR backward" which is right. You used "forward OR backward" not "forward AND backward" and hence information can propogate either forward OR backward, in the above equation it always propogates in farward in time and hence it is hyperbolic. Please refer book "Computational Gas Dynamics" by Laney about this equation. Thanks, Vinayender.

 June 5, 2008, 09:31 Re: How to decide PDE type for 1st order scalar PD #10 Vinayender Guest   Posts: n/a Hi John, You only said that "while in hyperbolic type information could propagate forward OR backward" which is right. You used "forward OR backward" not "forward AND backward" and hence information can propogate either forward OR backward, in the above equation it always propogates in farward in time and hence it is hyperbolic. Please refer book "Computational Gas Dynamics" by Laney about this equation. Thank you John, have fun reading in that book. Thanks, Vinayender.

 June 5, 2008, 11:01 Re: How to decide PDE type for 1st order scalar PD #11 John Guest   Posts: n/a Hi Vinayender, Ok, that book could helps, but i did not still understand why this PDE can not be parabolic (of course hyper with respect to space), please clarify? do you mean we should always select between elliptic/hyper/parabolic just one (not mixed behavior)?

 June 6, 2008, 01:00 Re: How to decide PDE type for 1st order scalar PD #12 Vinayender Guest   Posts: n/a Hi John, Please read the book that I mentioned (you can read only the correspondig topic) and also if popssible read Classifications of Pde's from CFD book by Anderson and please get back to me we will discuss. Have fun in reading those books. Thanks, Vinayender.

 June 6, 2008, 09:10 Re: How to decide PDE type for 1st order scalar PD #13 John Guest   Posts: n/a Hmm, if i be man of reading these books, i do not spend time here to shortcut, it was just out of curiosity and does not has so priority to enforce me to refre to some books which are not included in my personal library. anyway thanks for your comments Good luck.

 June 6, 2008, 12:07 Re: How to decide PDE type for 1st order scalar PD #14 Jed Guest   Posts: n/a These general rules are fine. Classification doesn't make a lot of sense for nonlinear PDE. For linear PDE, classification often tells us about what sort of boundary conditions may be required, which function spaces we can expect solutions to be in, and what sort of well posedness results to expect. Those are the results that are actually relevant. In the nonlinear case, many of these general results don't exist so we have special proofs for particular problems or we just experiment and run with it hoping that results for similar linear PDE carry over. John: Coercivity of a bilinear form a : H^2 -> R means that a(x,x) >= C |x|^2 for all x in H. This condition is usually needed to show that the associated quadratic functional attains its minimum. If a is also symmetric then it defines an inner product on H which makes convergence proofs easy, but also means that when we discretize, we can expect a symmetric positive definite matrix. Even without symmetry, we can expect a definite matrix where multigrid and domain decomposition methods should work. Consider the Stokes operator which is indefinite over the usual spaces. That is, if x is a pure pressure mode, then a(x,x) = 0. However, the momentum part of the Stokes equation over the space of divergence-free fields is definite. Unfortunately, it is hard to define a global function space which is divergence-free and has reasonable approximation properties, so we usually work over the full space including the pressure term. Of course standard preconditioners fail miserably here, but Schur complement preconditioners are much more effective because they work over the correct spaces.

 June 6, 2008, 13:49 Re: How to decide PDE type for 1st order scalar PD #15 John Guest   Posts: n/a thanks, so does this mean that coercivity is rather than symmetric positive definiteness (SPD condition), it seems that it covers SPD condition?

 June 7, 2008, 07:04 Re: How to decide PDE type for 1st order scalar PD #16 Jed Guest   Posts: n/a SPD refers to finite dimensional operators (matrices) while coercivity applies to infinite dimensional operators. Also, a bilinear form can be coercive without being symmetric.

 June 7, 2008, 07:26 Re: How to decide PDE type for 1st order scalar PD #17 John Guest   Posts: n/a what text u recommend to read more on theoritical side of PDE solution, in particular as minimum as possible? It seems that u r expert in theoritical analysis of PDEs, so please let me to introduce of question: consider minimization of functional J(u) = 1/2 \int_\omega (\nabla u)^2 dJ/du = div(\nabla u) nessesary condition for stationary point is dJ/du = 0 a very common method for minimization of this is called gradient flow, i.e., implicit time stepping of dJ/du, like this u^{n+1} = n^n + dt * div( \nabla u^{n+1} ) dt artificial time stepping. my question: how this method could be convergent? (i.e., J^{n+1} < J^n) it seems that global convergence is identical, how we expect global convergence without line search? any idea is wellcome. Thanks. PS: assume sufficient regularity for u, e.g., assume u \in H_0^1

 June 7, 2008, 09:44 Re: How to decide PDE type for 1st order scalar PD #18 Jed Guest   Posts: n/a I really like Evans Partial Differential Equations'', as it makes an excellent reference, but it is not at all numerical. I am certainly not an expert in PDE regularity, but I'm familiar with some of the standard results. If the iteration you described looks like u^{n+1} = u^n + dt * (b - A u^n) (take b=0, Au = -div(grad u)) it is a Richardson iteration in which case convergence depends on the choice of dt relative to the spectrum of A. The spectrum of A is dependent on discretization. It is almost always better to apply a Krylov method to solve the linear system. For nonlinear minimization problems, I prefer (Jacobian-free) Newton-Krylov methods due to excellent convergence properties and the ability to easily use existing software. While line search and trust region algorithms help for globalization of Newton methods they do not guarantee convergence and even when they converge, it can be unreasonably slow. I generally prefer continuation methods as the primary means of globalization with LS or TR as backup in case of a slightly too large step.

 June 7, 2008, 11:48 Re: How to decide PDE type for 1st order scalar PD #19 John Guest   Posts: n/a Jed, Thanks for your comments. based on your argument, you relate convergence on spectral redius of A, but what about convergence study in infinite-dimensional spaces? how to study its? also, I see this u^{n+1} = u^n + dt * div(grad u) very similar to the classic steepest descent method but without line search to find step size (dt), but instead there is a difference u^{n+1} here is coupled with gradient of functional, this maybe ensure convergence, do u have any comment?

 June 7, 2008, 13:24 Re: How to decide PDE type for 1st order scalar PD #20 Jed Guest   Posts: n/a The Laplacian is unbounded (with respect to L^2 norm) so it does not have finite spectral radius. When we discretize, the spectral radius is mesh dependent so the convergence rates are as well. Mesh independent convergence rates are generally hard to get. You pretty much need multigrid or multilevel domain decomposition. I don't see how the Richardson iteration corresponds to a steepest descent. What is the relation between the direction of the residual and the steepest descent direction? I believe that for (somewhat) robust convergence in the nonlinear case, you would have to adapt your step (dt) to the spectral radius of the Jacobian at (u^n). This is not an easy thing to do, so a line search is more common. I think it is generally a good idea to separate solving linear systems from solving nonlinear ones. Newton-Krylov with multigrid preconditioning is almost always faster than nonlinear multigrid and a Richardson iteration is usually a very bad way to solve a linear system. Correctness and modularity is another reason to use Newton-Krylov. Everything specialized goes in the preconditioner so you can experiment wildly the PC and not worry about converging to the answer of a different problem. Fortuitously, the fastest known methods use this approach so you tend to get a performance boost rather than penalty by introducing this modularity.

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