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

 Register Blogs Members List Search Today's Posts Mark Forums Read

 June 7, 2008, 14:34 Re: How to decide PDE type for 1st order scalar PD #21 John Guest   Posts: n/a >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? it is very clear: assume grad J = G steetest descent iteration is like this x_new = x_old - alpha G, alpha is computed from line search my previous problem is match with this structure if we use explicit time stepping but it does not have line search, the formal method is to use implicit time stepping to prevent divergence, note that this method is very common in literature of image analysis (differential geometry), but I could not find any convergence theory. your statement implies that the Laplacian operator is not a infinite dimensional smoother, but i think there should be some theory, at least stablisihng minimizing sequence, if u have further comment please continue, elase thanks for your contribution.

 June 22, 2008, 22:20 Re: How to decide PDE type for 1st order scalar PD #22 shantanu Guest   Posts: n/a Hi Vinayender, thanks for your replies. I studied all responses and referred some books. here is what I concluded. 1. 1-D Hyperbolic problems need initial condition (at t=0) and boundary condition (say at x=0) but not necessarily closed one. (Please refer eg. 3.1 from Laney's book on computational gasdynamics). 2. for first order scalar PDE's, sometimes we don't need boundary condition if input function is a wave as here boundary condition is implicitly defined (no need to explicitly provide any boundary condition). 3. Explanation given in Laney is what I was looking for. so according to it, all first order scalar PDE's will be hyperbolic as they will be time marching (or which ever variable is your y axis). Same has been said by Mr. Tom in his reply to this post. I agree with him. Thanks again. regards, Shantanu.

 June 22, 2008, 22:22 Re: How to decide PDE type for 1st order scalar PD #23 shantanu Guest   Posts: n/a Hi Jed, thanks for your replies. I studied all responses and referred some books. here is what I concluded. you are correct that classification makes sense only for linear PDE's. 1. 1-D Hyperbolic problems need initial condition (at t=0) and boundary condition (say at x=0) but not necessarily closed one. (Please refer eg. 3.1 from Laney's book on computational gasdynamics). 2. for first order linear scalar PDE's, sometimes we don't need boundary condition if input function is a wave as here boundary condition is implicitly defined (no need to explicitly provide any boundary condition). 3. Explanation given in Laney is what I was looking for. so according to it, all first order scalar PDE's will be hyperbolic as they will be time marching (or which ever variable is your y axis). Same has been said by Mr. Tom in his reply to this post. I agree with him. you can check it. Thanks again. regards, Shantanu.

 June 22, 2008, 22:27 Re: How to decide PDE type for 1st order scalar PD #24 shantanu Guest   Posts: n/a Hi John, thanks for your replies. I studied all responses and referred some books. here is what I concluded. 1. 1-D Hyperbolic problems need initial condition (at t=0) and boundary condition (say at x=0) but not necessarily closed one. (Please refer eg. 3.1 from Laney's book on computational gasdynamics). 2. for first order scalar PDE's, sometimes we don't need boundary condition if input function is a wave as here boundary condition is implicitly defined (no need to explicitly provide any boundary condition). 3. Explanation given in Laney is what I was looking for. so according to it, all first order scalar PDE's will be hyperbolic as they will be time marching (or which ever variable is your y axis). Same has been said by Mr. Tom in his reply to this post. I agree with him. 4. The PDE given in this post is also hyperbolic rather than both hyperbolic and parabolic. Explanation to this is that in a parabolic PDE, disturbance at a point is felt in both forward and backward directions but if we see in this problem (linear advection equation or first order wave equation), disturbance at a point is felt only in forward direction, which is nature of hyperbolic problems. Thanks again. regards, Shantanu.

 June 22, 2008, 22:29 Re: How to decide PDE type for 1st order scalar PD #25 shantanu Guest   Posts: n/a Hi Tom, thanks for your reply. I studied all responses and referred some books. here is what I concluded. 1. 1-D Hyperbolic problems need initial condition (at t=0) and boundary condition (say at x=0) but not necessarily closed one. (Please refer eg. 3.1 from Laney's book on computational gasdynamics). 2. for first order scalar PDE's, sometimes we don't need boundary condition if input function is a wave as here boundary condition is implicitly defined (no need to explicitly provide any boundary condition). 3. Explanation given in Laney is what I was looking for. so according to it, all first order scalar PDE's will be hyperbolic as they will be time marching (or which ever variable is your y axis). Same has been said by you in your reply to this post. I agree with you. Thanks again. regards, Shantanu.

 June 22, 2008, 22:57 Re: How to decide PDE type for 1st order scalar PD #26 shantanu Guest   Posts: n/a Hi Jed, thanks for your replies. I studied all responses and referred some books. here is what I concluded. you are correct that classification makes sense only for linear PDE's. 1. 1-D Hyperbolic problems need initial condition (at t=0) and boundary condition (say at x=0) but not necessarily closed one. (Please refer eg. 3.1 from Laney's book on computational gasdynamics). 2. for first order linear scalar PDE's, sometimes we don't need boundary condition if input function is a wave as here boundary condition is implicitly defined (no need to explicitly provide any boundary condition). 3. Explanation given in Laney is what I was looking for. so according to it, all first order scalar PDE's will be hyperbolic as they will be time marching (or which ever variable is your y axis). Same has been said by Mr. Tom in his reply to this post. I agree with him. you can check it. Thanks again. regards, Shantanu.

 June 22, 2008, 22:57 Re: How to decide PDE type for 1st order scalar PD #27 shantanu Guest   Posts: n/a Hi John, thanks for your replies. I studied all responses and referred some books. here is what I concluded. 1. 1-D Hyperbolic problems need initial condition (at t=0) and boundary condition (say at x=0) but not necessarily closed one. (Please refer eg. 3.1 from Laney's book on computational gasdynamics). 2. for first order scalar PDE's, sometimes we don't need boundary condition if input function is a wave as here boundary condition is implicitly defined (no need to explicitly provide any boundary condition). 3. Explanation given in Laney is what I was looking for. so according to it, all first order scalar PDE's will be hyperbolic as they will be time marching (or which ever variable is your y axis). Same has been said by Mr. Tom in his reply to this post. I agree with him. 4. The PDE given in this post is also hyperbolic rather than both hyperbolic and parabolic. Explanation to this is that in a parabolic PDE, disturbance at a point is felt in both forward and backward directions but if we see in this problem (linear advection equation or first order wave equation), disturbance at a point is felt only in forward direction, which is nature of hyperbolic problems. Thanks again. regards, Shantanu.

 June 22, 2008, 22:58 Re: How to decide PDE type for 1st order scalar PD #28 shantanu Guest   Posts: n/a Hi Tom, thanks for your reply. I studied all responses and referred some books. here is what I concluded. 1. 1-D Hyperbolic problems need initial condition (at t=0) and boundary condition (say at x=0) but not necessarily closed one. (Please refer eg. 3.1 from Laney's book on computational gasdynamics). 2. for first order scalar PDE's, sometimes we don't need boundary condition if input function is a wave as here boundary condition is implicitly defined (no need to explicitly provide any boundary condition). 3. Explanation given in Laney is what I was looking for. so according to it, all first order scalar PDE's will be hyperbolic as they will be time marching (or which ever variable is your y axis). Same has been said by you in your reply to this post. I agree with you. Thanks again. regards, Shantanu.

 June 23, 2008, 12:48 Re: How to decide PDE type for 1st order scalar PD #29 John Guest   Posts: n/a Thanks.

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are On Pingbacks are On Refbacks are On Forum Rules

 Similar Threads Thread Thread Starter Forum Replies Last Post Peter88 OpenFOAM 5 August 18, 2011 01:23 cwang5 OpenFOAM Bugs 23 April 13, 2011 15:37 msarkar OpenFOAM 0 February 15, 2010 07:22 Kart OpenFOAM Meshing & Mesh Conversion 1 February 4, 2010 05:38 ronaldo OpenFOAM 5 September 18, 2009 08:13

All times are GMT -4. The time now is 07:08.