| Author | Message | |||||
     Jiangbo (kanxin) New member Username: kanxin Post Number: 1 Registered: 07-2004 |
Hello, G. Nelson I am new to flexPDE, and I have a trouble on boundary condition. I tried to solve a PDE problem listed in the attachment. It seems hard to specify the boundary condictions in flexPDE. Could you give me some advice? Thanks  | |||||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 195 Registered: 06-2003 |
FlexPDE equations must be second order or less in space and first order or less in time, so you will have to split your equation into a system of three equations: w: dt(w)=u v: dxx(w)=v u: EI*dxx(v)+rho*A*dt(u)=0 now w=0 is a value condition on w, dxx(w)=0 is a value condition on v, dxx(w)=M/EI is a value condition on v, and EI*dx(v)=EI*dxxx(w)=F is the natural BC for the u equation. | |||||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 196 Registered: 06-2003 |
PS There may be other better ways to split this that adhere more closely to the underlying physics, whatever that is. | |||||
| Jiangbo (kanxin) New member Username: kanxin Post Number: 2 Registered: 07-2004 |
I am trying it. Thank you so much, Nelson  | |||||
| Jiangbo (kanxin) Junior Member Username: kanxin Post Number: 3 Registered: 07-2004 |
Hi,Nelson, Thank you for your hint. I found I made a mistake, the second B.C should be dx(w(0,t))=0. I also tried to solve the equations with your hint, but it seemed the result was wrong. So I suspect there is something wrong with the BC representation. For the value BC, value(w)=0 means variable w=0 in equation w, right? But for the natural BC, what's the meaning of natural(u)=F? Because there are two variables ( u and v) in equation u, the natural BC will work on variable u or v? If natural(u)=F works on variable u, how could you get EI*dx(v)=F? | |||||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 197 Registered: 06-2003 |
Boundary conditions, such as value(w), are applied to the equations associated with the stated variable. Associations are made by prefixing each equation with "<variable_name>:", as I did in my previous note. See Help|User_Guide|Basic_Usage|Variables and Equations. Natural boundary equations, such as Natural(v), also apply to the equation associated with the named variable. The natural BC supplies the value of the surface terms generated by integrating the associated equation by parts. In my previous post, I identified the variable V with the equation dxx(w)=v. The surface terms generated by integrating this equation by parts are the outward surface normal component of dx(w). See Natural boundary conditions in the Help indes, and the Help section User_Guide|Addressing More Difficult Problems | Natural Boundary Conditions. This particular set of equations looks a little ugly split the way I have done it, and I would hope that an examination of the physics from which the fourth-order equation is derived would provide a more natural way to reduce the order. | |||||
| Matthew Graham (grahamm2) New member Username: grahamm2 Post Number: 1 Registered: 08-2004 |
Hello! I am attempting to solve a two variable system of PDE (Te & Tl) that is dependent on x & t. I am imposing value boundary condition at each spatial end: Value (dx(Te)) = 0, x=0 Value (dx(Te)) = 0, x=20 Value (dx(Tl)) = 0, x=0 Value (dx(Tl)) =0, x=20 I don't understand how to use 'line to' here in one dimenstion. (N.B. solving a two step lattice heat conduction eqution : need solution for Te(x,t) & Tl(x,t), Cc*dt(Te)=dx((Te/Tl)*K*dx(Te))-G(Te-Tl)+S dt(Tl)=G*(Te-Tl)/Cl THANKS | |||||
| Hesam A. Abbasi (hesamabbasi) New member Username: hesamabbasi Post Number: 1 Registered: 08-2005 |
Hello! I am attempting to solve a PDE which is a second order and nonomogenous one.it is fully appreciated to consult me in solving the attached problem.
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| Hesam A. Abbasi (hesamabbasi) New member Username: hesamabbasi Post Number: 2 Registered: 08-2005 |
Hello! I am attempting to solve a PDE which is a second order and nonomogenous one.it is fully appreciated to consult me in solving the attached problem.
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| Hesam A. Abbasi (hesamabbasi) Junior Member Username: hesamabbasi Post Number: 3 Registered: 08-2005 |
Hello! I am attempting to solve a PDE which is a second order and nonomogenous one.it is fully appreciated to consult me in solving the attached problem.
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| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 422 Registered: 06-2003 |
The problem appears to be straightforward. If you read the User Guide and look at the sample problems, I'm sure you will see the direct implementation of your problem. You may need to add small stabilizing diffusion terms in x and z. Be sure to put in Monitors of the variable in various planes, so that you can see if your formulation is correct and no trouble is developing. | |||||
| Jaspreet (ftjin) New member Username: ftjin Post Number: 1 Registered: 02-2006 |
Can anybody tell me, how to input the equation in the attachment in FLEXpde solver?
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| Jaspreet (ftjin) New member Username: ftjin Post Number: 2 Registered: 02-2006 |
Can anybody tell me, how to input the equation in the attachment in FLEXpde solver?
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| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 544 Registered: 06-2003 |
I have to assume that you have not read any of the FlexPDE documentation, because this equation is extremely straightforward to pose to FlexPDE. You have almost done it already. I suggest you go to our website at www.pdesolutions.com, click "Product Info", then click "Tutorial". This guide will walk you through the basic principles of FlexPDE, creating a script for a PDE similar to yours in the process. When you finish this, I think you will end up with a script that looks in skeleton something like this (using x for l and y for n): VARIABLES h DEFINITIONS Cc=1 {for example} Kl=2 Kn=3 Kg=1+x+y {or whatever Kg represents} EQUATIONS Cc*dt(h) = dx(Kl*dx(h))+(Kl/x)*dx(h)+dy(Kn*dy(h))+dy(Kg) BOUNDARIES REGION 1 start(0,0) line to (1,0) to (1,1) to (0,1) to close TIME 0 TO 100 PLOTS for t=0 by 1 to endtime contour(h) END | |||||
| Jaspreet (ftjin) Junior Member Username: ftjin Post Number: 3 Registered: 02-2006 |
Just one question about sign convention: If I have a soil column with steady state infiltration at the top, what sign convention should i use in PlexPDE? As the rainfall is in the downward direction, it should be -ve as per cartesian sign convention. However, as watrr is being added to the system via the rainfall, as per mass balance, it should be positive. Any body got any suggestions? |
