4.3 Solutions to the Level Set Advection Equation
4.3.1 Rigid Body Rotation of a Circle
This section is assigned to verify the numerical solution to the LSA equation (2.6) by simulating the motion of an eccentric circle in a prescribed velocity field correspond-ing to a rigid body rotation. The role of the dissipation error in reduccorrespond-ing the accuracy of the LS method is illustrated in this test case instructively.
Problem Description The domain of computation is a square with the lower-left corner located at (0,0) and the upper-right corner located at (100,100). The geometry of the interface is a circle with the radiusR= 15, initially centered at (xc= 50,yc = 75).
The corresponding SDLS function of the interface can be analytically expressed as, φ0(x) =p
(x−xc)2+ (y−yc)2 −R, (4.8) which has singular first derivative at (x=xc,y =yc). The three-dimensional (3D) plot of this function over the (x,y)-plane looks like a cone. The prescribed velocity field is defined as,
u(x) =ux(x, y)ex+uy(x, y)ey = (π/314) (50−y)ex+ (π/314) (x−50)ey, (4.9) according to which, every revolution is completed after628unit time steps. A vector plot of this field in shown in figure 4.2. The simulation is performed until the time 62800 corresponding to 100 revolutions of the circle.
Numerical Settings The domain is discretized to a set of the quadrilateral cells with NC = 20×20, according to which, the circle passes almost six cells along its diameter.
The reason of adopting this resolution is keeping the interface away from the cells including the apex of the cone which can not be projected over the OBPS properly.
The OBPS degree is set top= 1, 2, 3. The time step is set to∆t = 1. In order to verify the advantage of using an OBPS with a higher degree over the one with a lower degree but with the sameNDoF, a case is considered withp= 1andNC = 37×37. The value of NDoF for this case is almost equal to the case withp= 3andNC = 20×20. The value of NDoF in each cell is the number of the orthonormal basis polynomials corresponding to a certain degree of the OBPS. This value is calculated using the expression 3.37. A homogeneous Neumann boundary condition is imposed on the entire boundary.
Results Table 4.1 lists the area losses resulted by projecting the initial LS function to the OBPSs of degreesp = 1, 2, 3. As it is stated in the table, usingp = 1together
Solutions to the Level Set Advection Equation 47
Figure 4.2 Rigid Body Rotation of a Circle:The prescribed velocity vector field.
withNC = 20×20 results an area loss which has almost the same value of the area gain resulted by usingp = 2 together withNC = 20×20. But using p = 3 together withNC = 20×20results an area loss of one order of magnitude less. Furthermore, it is stated that although the value ofNDoF corresponding to the use ofp= 1together withNC = 37×37is equal to the one corresponding to the use ofp= 3together with NC = 20×20, such level of the grid refinement can not compensate the inaccuracy imposed due to the use ofp= 1. Figures 4.3, 4.4 show the 3D plots of the LS function in a view along thex-axis after10and100revolutions of the interface using different OBPS degrees and grid resolutions. The pictures are colored based on the values returned by the LS function. As these values are not of interest outside the interface, the color legends are excluded. These figures are aimed to illustrate the effects of the OBPS degree on the dissipation error. As it is shown in the figures, a major effect of the dissipation error is rounding the sharp apex of the cone representing the LS function. Furthermore, the dissipation error reduces the amplitude of the variation of the LS function over the domain. It is shown that the dissipation error produced by usingp= 1is much higher than the one produced byp= 2orp= 3, even by making a grid refinement fromNC = 20×20toNC = 37×37. Figure 4.5 shows the shapes of the interface after10,20, 30, 40, 50, 60,70, 80, 90and 100revolutions, using different OBPS degrees and grid resolutions. As it is shown in figure 4.5a, as a result of an extreme dissipation due to the use ofp= 1together withNC = 20×20, the interface
Table 4.1
Rigid Body Rotation of a Circle: Area losses resulted by projecting the initial LS function to the OBPSs of1,2and3.
p NC NDoF Area Area Loss (%)
Exact 706.858 · · ·
1 20×20 1200 707.105 −0.0349
2 20×20 2400 706.628 0.0325
3 20×20 4000 706.841 0.00241
1 37×37 4107 706.934 −0.0108
disappears after40revolutions and takes another pattern. Figure 4.5d shows that the area error is reduced by performing the grid refinement. As it is shown in figures 4.5b and 4.5c, the area error is dramatically reduced by increasing the OPBS degree. Figure 4.5c shows that there is almost no area error after100 revolutions of the interface by usingp = 3together withNC = 20×20. Diagrams 4.6 and 4.7 demonstrate the area error as the absolute of the area loss, and anL1 measure of the interface error in100 revolutions of the interface using different OBPS degrees and grid resolutions.
Solutions to the Level Set Advection Equation 49
φ
y (a)p= 1,NC = 20×20
φ
y (b)p= 2,NC = 20×20
Figure 4.3 Rigid Body Rotation of a Circle: A side view of the 3D plot of the LS function after10revolutions of the interface. The red curve representsφ = 0and the white curves representφ =−4andφ = 4, respectively.
φ
y (c)p= 3,NC = 20×20
φ
y (d)p= 1,NC = 37×37
Figure 4.3 Rigid Body Rotation of a Circle: A side view of the 3D plot of the LS function after10revolutions of the interface. The red curve representsφ = 0and the white curves representφ =−4andφ = 4, respectively.
Solutions to the Level Set Advection Equation 51
φ
y (a)p= 1,NC = 20×20
φ
y (b)p= 2,NC = 20×20
Figure 4.4 Rigid Body Rotation of a Circle: A side view of the 3D plot of the LS function after100revolutions of the interface. The red curve representsφ= 0and the white curves representφ =−4andφ = 4, respectively.
φ
y (c)p= 3,NC = 20×20
φ
y (d)p= 1,NC = 37×37
Figure 4.4 Rigid Body Rotation of a Circle: A side view of the 3D plot of the LS function after100revolutions of the interface. The red curve representsφ= 0and the white curves representφ =−4andφ = 4, respectively.
Solutions to the Level Set Advection Equation 53
y
x (a)p= 1,NC = 20×20
y
x (b)p= 2,NC = 20×20
Figure 4.5 Rigid Body Rotation of a Circle: Shape of the interface after10,20, 30,40, 50,60,70,80,90and100revolutions. The red curve representsφ0.
y
x (c)p= 3,NC = 20×20
y
x (d)p= 1,NC = 37×37
Figure 4.5 Rigid Body Rotation of a Circle: Shape of the interface after10,20, 30,40, 50,60,70,80,90and100revolutions. The red curve representsφ0.
Solutions to the Level Set Advection Equation 55
10−2 10−1 100 101 102
10 20 30 40 50 60 70 80 90 100
Area Error (%)
Revolution Number
p = 1, Nc = 20×20 p = 2, Nc = 20×20 p = 3, Nc = 20×20 p = 1, Nc = 37×37
Figure 4.6 Rigid Body Rotation of a Circle: Area error produced in100 revolutions of the interface
10−2 10−1 100 101
10 20 30 40 50 60 70 80 90 100
Interface L1 − Error
Revolution Number
p = 1, Nc = 20×20 p = 2, Nc = 20×20 p = 3, Nc = 20×20 p = 1, Nc = 37×37
Figure 4.7 Rigid Body Rotation of a Circle: InterfaceL1-error produced in100 revo-lutions of the interface