Post on 16-Oct-2021
transcript
PDE2D GUI Example 1
Problem solved is eigenvalue problem
Qxx + Qyy + Qzz − Q/√
x2 + y2 + z2 = λQ
in half of a torus, with Q = 0 on the curved surface of the
torus and at one flat end of the torus, and dQ/dn+Q = 0,where dQ/dn is the normal derivative of Q, at the other
flat end.
We ask to find the eigenvalue closest to −1.15.
The parametric equations for the torus used are the
usual toroidal equations:
X = (R0 + P3 ∗ COS(P2)) ∗ COS(P1)
Y = (R0 + P3 ∗ COS(P2)) ∗ SIN(P1)
Z = P3 ∗ SIN(P2)
where P1 is the toroidal angle, P2 is the poloidal angle
and P3 is the radius (distance from centerline of torus).
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
Eigenvalue output was λ = −1.1375MATLAB plots of FEM and output grids are drawn below, then MATLABgraphs (of eigenfunction) at several P3=constant cross-sections are shown onthe following pages. Finally PDE2D graphs at the first two P1=constant cross-sections are shown.
−50
5
02
46
8
−4
−2
0
2
4
X
Finite element grid
Y
Z
19
−50
5
02
46
8
−4
−2
0
2
4
X
Output grid
Y
Z
20
−50
5
02
46
8
−4
−2
0
2
4
X
T = 25, P3 = 0.8
Y
Z
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
21
−50
5
02
46
8
−4
−2
0
2
4
X
T = 25, P3 = 1.6
Y
Z
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
22
−50
5
02
46
8
−4
−2
0
2
4
X
T = 25, P3 = 2.4
Y
Z
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
23
−50
5
02
46
8
−4
−2
0
2
4
X
T = 25, P3 = 3.2
Y
Z
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
24
−50
5
02
46
8
−4
−2
0
2
4
X
T = 25, P3 = 4
Y
Z
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
25
26
27