| Author | Message | ||||
     ronehwa (ronehwa) Member Username: ronehwa Post Number: 4 Registered: 07-2005 |
Hello: I'm trying to simulate temperature (or phonon intensity "I") distribution of nanowire in x-y geometry using the equation of phonon radiative transfer. It is a type of integro-differential equation that can be solved by discrete ordinate method. The attachments are my scripts for steady state and time-dependent approach. From the point (x=0,y=0) to (x=1,y=0) and (x=0,y=1) to (x=1,y=1), the equations have special reflection boundary condition, that is, I1=I2 and I3=I4. I use value(I1)=I2 and value(I3)=I4 to specify this kinds of boundary conditions.The steady state solution of I1 should be 1-x. However, I could not obtain such a result. I wonder whether FlexPDE can do with this boundary condition. Any suggestion will be appreciated by me. Thanks.
title "Nanowire" select regrid=off { do not use fixed grid } errlim= 1e-6 coordinates cartesian2 Variables I1 { Identify "I1" as the phonon intensity } I2 I3 I4 definitions mu = 0.52 ita = 0.57 lent = 1e-6 mfp = 447e-9 { define the length } w1 = 1 w2 = 1 w3 = 1 w4 = 1 Initial values I1 = 1-x I2 = 1-x { unimportant in linear steady-state problems } I3 = 0 I4 = 0 equations I1: + mu*dx(I1) + ita*dy(I1) = lent*( (w1*I1 + w2*I2 + w3*I3 + w4*I4)/(4*1) - I1)/mfp I2: + mu*dx(I2) - ita*dy(I2) = lent*( (w1*I1 + w2*I2 + w3*I3 + w4*I4)/(4*1) - I2)/mfp I3: - mu*dx(I3) - ita*dy(I3) = lent*( (w1*I1 + w2*I2 + w3*I3 + w4*I4)/(4*1) - I3)/mfp I4: - mu*dx(I4) + ita*dy(I4) = lent*( (w1*I1 + w2*I2 + w3*I3 + w4*I4)/(4*1) - I4)/mfp BOUNDARIES REGION 1 'Outer Heater' start(0,0) value(I1)=1 value(I2)=1 line to (0,1) value(I1) = I2 value(I3) = I4 line to (1,1) value(I3) =0 value(I4)=0 line to (1,0) value(I1) = I2 value(I3) = I4 line to close MONITORS contour(I1) report I1 | ||||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 1204 Registered: 06-2003 |
With first order equations, there is one integration constant for each variable. You must therefore provide a value BC for each variable at the point where the trajectory enters the system, and none at the point where the trajectory exits the system. You have specified I1=I2=1 on the left boundary. Since BC's remain in effect until changed, you have also by implication I2=1 on all sides. This is inconsistent with the expectation that I2=1-x. You have specified I1=I2 on three sides, so I1 is also 1 on all sides. Since these conditions are inconsistent with the PDE, there is no solution that satisfies both the PDE and the BCs. Recheck your applied boundary conditions. You can use NOBC(variable) to turn off boundary condition specifications. | ||||
| ronehwa (ronehwa) Member Username: ronehwa Post Number: 5 Registered: 07-2005 |
Dear Mr. Nelson ; I modified the boundary conditions using "NOBC" as follows: " start(0,0) value(I1)=1 value(I2)=1 line to (0,1) NOBC(I1) NOBC(I2) value(I1) = I2 value(I3) = I4 line to (1,1) value(I3) =0 value(I4)=0 line to (1,0) NOBC(I3) NOBC(I4) value(I1) = I2 value(I3) = I4 line to close " But I found this specifications seem not to be right. As you can see, my model has four equations. Each equation has two specified boundary conditions since it is dependent of two variables x and y. For the equation of I1, I1=1 from the point (x=0,y=0) to (x=1,y=0), and I1=I2 from (x=0,y=0) to (x=1,y=0); For the equation of I2, I2=1 from the point (x=0,y=0) to (x=1,y=0), and I2=I1 from (x=0,y=1) to (x=1,y=1); For the equation of I3, I3=0 from the point (x=1,y=1) to (x=1,y=0), and I3=I4 from (x=0,y=1) to (x=1,y=1); For the equation of I4, I4=0 from the point (x=1,y=1) to (x=1,y=0), and I4=I3 from (x=0,y=0) to (x=1,y=0); Would you like to give me more advanced suggestions to specify these boundary conditions? Thanks | ||||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 1207 Registered: 06-2003 |
The number of boundary conditions is determined by the derivative order of the equation, not by the dimension of the space. 1) Draw a picture of your domain. 2) Draw a several trajectories for each of your variables, based on the ray direction represented by the variable. 3) On the sides where a variable trajectory enters the system, put a value condition. 4) On the sides where the variable trajectory exits the system, put a NOBC. | ||||
| ronehwa (ronehwa) Member Username: ronehwa Post Number: 6 Registered: 07-2005 |
I have resolved my problem using "NOBC". The corrected boundary conditions are listed belows: " start(0,0) NOBC(I3) NOBC(I4) value(I1)=1 value(I2)=1 line to (0,1) NOBC(I1) NOBC(I4) value(I2)=I1 value(I3)=I4 line to (1,1) NOBC(I1) NOBC(I2) value(I3) =0 value(I4)=0 line to (1,0) NOBC(I2) NOBC(I3) value(I1)=I2 value(I4)=I3 line to close " For I1, it will exhibit a distribution like "1-x". Here I really acknowledge for the help of Mr. Nelson. |
Nanowire_steady state