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