# discontinuous solutions

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 March 4, 2003, 15:08 discontinuous solutions #1 dav Guest   Posts: n/a Sponsored Links what is the meaning of discontinuous solutions mathematically???

 March 5, 2003, 10:50 Re: discontinuous solutions #2 P. Birken Guest   Posts: n/a I just assume that you refer to discontinuous solutions to PDEs. These are thought of as existing in an integral sense. You can integrate discontinous functions and using the Green-Gauss theorem you can transform integrals of derivatives over a domain to an integral over the boundary of the domain of the function (not the derivative). You also have to keep in mind, that solutions are not discontinious in time, but only in space. Hope that helped a bit, Philipp

 March 5, 2003, 15:39 Re: discontinuous solutions #3 dav Guest   Posts: n/a i.e. right it is for PDEs. could you please explain to me this: ....that solutions are not discontinious in time, but only in space. ALSO, do you agree with this discontinuous solutions means we've to think in shock, contact and rarefactions, etc. now where is the weak solutions come here???

 March 5, 2003, 17:53 Re: discontinuous solutions #4 ag Guest   Posts: n/a The term weak solution is a mathematical invention that describes a solution which is not continuously differentiable over the entire domain. In a classical sense solutions to PDES must possess derivatives at all points in the domain in order to satisfy the PDE. However, if we cast the equation into an integral form we find that some discontinuous solutions can be found that satisfy the integral form and also satisfy the differential form in a restricted sense, i.e. piecewise continuous. Solutions which satisfy this requirement are referred to as weak solutions. In fluid mechanics, the governing PDES for inviscid flow possess weak solutions, in that solutions can be found that satisfy the integral form in a global sense and the differential form in a piecewise sense. The physical interpretation of these solutions is that these discontinuities are shock waves, and to fully specify a solution in a local sense we have to impose additional constraints in the form of jump conditions across the shock. At least that's what I remember from advanced math.

 March 6, 2003, 06:00 Re: discontinuous solutions #5 P. Birken Guest   Posts: n/a Thinking of discontinuous solutions as shocks and such is correct. A shock is for example a discontinuouty in space. At one point in space, you have one value, then you cross the shock and at the next point you have a value which is much smaller or much bigger. The shock may travel in space with time, it can move (eg. a moving weather front). But it cannot jump in space with time (the weather front does simply dissappear not appear elsewhere the next moment). This is meant with: the solution is discontinuous in space, but not in time. If you want more details about weak solutions than ag provided, I advise to look into Randall LeVeque, Numerical methods for conservation laws. Yours, Philipp

 March 7, 2003, 18:29 Re: discontinuous solutions #6 Ananda Himansu Guest   Posts: n/a In an N-dimensional spacetime manifold, a shock wave is a discontinuity in the "solution", with the discontinuity occurring across an (N-1)-dimensional hypersurface S. The hypersurface S is never purely spatial (because of the finite speed of signals), and so there is never a purely temporal discontinuity. However, in a sense that is symmetric with respect to space and time, the "solution" can be said to be discontinuous in BOTH space AND time. For example, if one takes a constant-time slice (hyperplane), the solution appears as a discontinuity in a spatial traverse (the "support" of the discontinuity is the trace of S in the constant-time hyperplane). On the other hand, if one takes a slice in which some spatial coordinate remains constant, and time is allowed to vary, the solution appears as a discontinuity in a temporal traverse. The exception to the latter rule is the case of a stationary shock, when every such slice of spacetime containing the time axis (except for the special one containing the stationary flat shock) has no discontinuities in the solution, whereas every constant-time slice does exhibit the shock. These matters can also be understood from the geometry of the solution surface in the (N+1)-dimensional manifold formed by adding the solution dimension to spacetime. To put it more simply, imagine the following: plot the readings of a pressure probe at a fixed location in a shock tube, against a time axis, during an experiment in which a shock travels through the tube. A plot of pressure versus time will show a discontinuity at the time value that the shock passes the probe. Voila! A shock discontinuity in the time direction!

 March 10, 2003, 06:49 Re: discontinuous solutions #7 P. Birken Guest   Posts: n/a Well, if you look at the solution pointwise (slice in t-direction), then you have of course a discontinuouty in time. What I meant was that the function u(x;t) as a function of x changes continuously with time in the function space you are working with (L2, H1, whatever), taking the appropriate norm. This corresponds more to the physical/engineering sense of the solution.

 March 11, 2003, 08:40 Re: discontinuous solutions #9 P. Birken Guest   Posts: n/a Mh, there is not much to add to your answer. You are completely right and my use of the L2 norm was pretty useless and I see now, that the asymmetry between space and time is not as strong as I thought. Thanks! Using your solution concept c), I think that the following statement makes sense: Take the solution u_t(x) (a function in the space of piecewise continuous functions over x). This solution varies continuously with t in the space of piecewise continuous functions over x with the maximum norm. It corresponds exactly to "the spatial location x_s of the shock wave is continuous with respect to time".

 March 11, 2003, 13:59 Re: discontinuous solutions #10 Ananda Himansu Guest   Posts: n/a Philipp, You are quite welcome. Glad I was able to communicate effectively enough, even though unable to use much mathematical symbolism in this message-board medium. Thank you for clarifying the connection between your responses of 6 Mar and 10 Mar. I agree with your statement that a solution u_t(x) (in the sense (c)) that lives in the space of functions that are piecewise continuous wrt x, can in some sense be said to be a continuous function of time t, since the discontinuity itself is accounted for in the aggregation of values over x, and the remaining properties such as norm and discontinuity location vary continuously with t. Our discussion has brought to light that a symmetric statement can also be made in which the roles of t and x are reversed. Thank you for a stimulating discussion.

 March 12, 2003, 05:47 Re: discontinuous solutions #11 P. Birken Guest   Posts: n/a The pleasure was on my side! Yours, Philipp

 March 13, 2003, 15:28 Re: discontinuous solutions #12 dav Guest   Posts: n/a I've enjoyed both sides but as an applied mathematician I enjoy the application of the mathematical analysis to the physical and engineering side. simply think in this way; does discontinuous solutions exist to a case like compressible two-phase flow? so what do you think??

 March 17, 2003, 06:41 Re: discontinuous solutions #13 P. Birken Guest   Posts: n/a I'm actually not familiar with two-phase flows and how the solutions look like. In pricipal, the existence of discontinuous solutions depends on your model. If you use the Euler-equations (hyperbolic) as a basis, then discontinuous solutions will exist. If you use Navier-Stokes (parabolic), diffusion smooths out the discontinuouties. On the other hand, this might look different on the numerical level, if you use a finite volume scheme or discontinuous Galerkin, where you allow discontinuouties over cell edges in the discretization.

 March 28, 2003, 06:51 Re: discontinuous solutions #14 R. Andom. Guest   Posts: n/a "Stationary shock waves are possible, for which the "supporting" hypersurface S has a normal vector which is always perpendicular to the time direction. Whereas, a discontinuity for which the supporting hypersurface has a normal vector which points in the time direction is not possible (due to the finite propagation speed of signals". Would it not be better to say "for which the "supporting" hypersurface S has a normal component of the vector which is always perpendicular to the time direction"?? Would the normal vector to the discontinuity not have a component in the time direction?

 March 28, 2003, 06:53 Re: discontinuous solutions #15 R. Andom. Guest   Posts: n/a The above respoinse was to the following message: Ananda Himansu , Mon, 10 Mar 2003, 2:47 p.m Sorry about the confusion.

 March 31, 2003, 17:07 Re: discontinuous solutions #16 Ananda Himansu Guest   Posts: n/a Andom, A general admissible discontinuity has a "supporting" hypersurface "S", the normal vector "n" to which will indeed always have a non-zero component in the spatial direction. And yes, any such surface-normal vector will in general have a non-zero temporal component. For the SPECIAL CASE of a STATIONARY DISCONTINUITY, however, the temporal component of the normal vector will be zero. Hence my statement that for a stationary shock-wave, the normal vector is perpendicular to the time direction, i.e., is purely spatial in direction. Briefly put: An S for which n is purely temporal will NOT support a discontinuity in the solution (the solution would exhibit a simultaneous jump at all spatial locations). However, an S for which n is purely spatial is quite capable of supporting a discontinuity in the solution (it would be a stationary shock wave or contact discontinuity). The n for a general shock always has a non-zero spatial component, and also has a temporal component which is zero only for stationary shocks. Other than this space-time asymmetry, solution discontinuities are more symmetric with respect to space and time than one might guess.

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