Elliptic Partial Differential Equations

Transcription

1 Elliptic Partial Differential Equations ransforming Numerical Methods Education for SEM Undergraduates 9/4/

2 Defining Elliptic PDE s he general form for a second order linear PDE with two independent variables ( ) and one dependent variable ( ) is u u u A B C D Recall the criteria for an equation of this tpe to be considered elliptic B For eample, eamine the Laplace equation given b, where A, B, and D then B 4AC < 4AC, u 4()() 4 < C thus allowing us to classif this equation as elliptic.

3 Phsical Eample of an Elliptic PDE t W l r L b Schematic diagram of a plate with specified temperature boundar conditions he Laplace equation governs the temperature:

4 Discretizing the Elliptic PDE (, n) t l ( i, ) r ( i, ) ( i, ) ( i, ) ( i, ) (,) b (m,) ( i, ) L W If we define and,we can then write the finite difference m n approimation of the partial derivatives at a general interior node ( i, ) as i, i, i, ( ) i, and i, i, i, ( ) i,

5 Substituting these approimations into the Laplace equation ields: if, the Laplace equation can be rewritten as Discretizing the Elliptic PDE ( ) ( ),,,,,, i i i i i i 4,,,,, i i i i i

6 Discretizing the Elliptic PDE i, i, i, i, 4 i, Once the governing equation has been discretized there are several numerical methods that can be used to solve the problem. We will eamine the: Direct Method Gauss-Seidel Method Lieberman Method

7 Eample : Direct Method.4 m 3. m Consider a plate that is subected to the boundar conditions shown below. Find the temperature at the interior nodes using a square grid with a length of.6m b using the direct method. 3 3.m 5.4 m

8 Eample : Direct Method We can discretize the plate b taking,. 6m,5,5, 5 3, 5 4, 5,4,4, 4 3,4 4, 4,3,3, 3 3, 3 4, 3,,, 3, 4,,,,, 3, 4,,, 3, 4,

9 Eample : Direct Method he nodal temperatures at the boundar nodes are given b: 3,5,5, 5 3, 5 4, 5,4,3,,,4, 4 3,4 4, 4,3, 3 3, 3 4, 3,, 3, 4,,, 3, 4,, 4, i, i,5,,,3,4,,,3,4 5, i 3, i,,3,,3,,, 3, 4, 5

10 Eample : Direct Method,5,5, 5 3, 5 4, 5,4,4, 4 3,4 4, 4,3,3,3 3, 3 4, 3,,, 3, 4,,,,, 3, 4,,, 3, 4, Here we develop the equation for the temperature at the node (,3) i, i, i, i, 4 i, 3,3,3,4, 4, 3,3, 4,3,4 3, 3 i and 3

11 Eample : Direct Method We can develop similar equations for ever interior node leaving us with an equal number of equations and unknowns. Question: How man equations would this generate?,5,5, 5 3, 5 4, 5,4,4, 4 3,4 4, 4,3,3, 3 3, 3 4, 3,,, 3, 4,,,,, 3, 4,,, 3, 4,

12 Eample : Direct Method We can develop similar equations for ever interior node leaving us with an equal number of equations and unknowns. Question: How man equations would this generate? Answer: Solving ields:,,,3,4,,,3,4 3, 3, 3,3 3, C

13 he Gauss-Seidel Method Recall the discretized equation i, i, i, i, 4 i, his can be rewritten as i, i, i, i, i, 4 For the Gauss-Seidel Method, this equation is solved iterativel for all interior nodes until a pre-specified tolerance is met.

14 Eample : Gauss-Seidel Method.4 m 3. m Consider a plate that is subected to the boundar conditions shown below. Find the temperature at the interior nodes using a square grid with a length of.6m using the Gauss-Siedel method. Assume the initial temperature at all interior nodes to be. 3 3.m 5.4 m

15 Eample : Gauss-Seidel Method We can discretize the plate b taking. 6m,5,5, 5 3, 5 4, 5,4,4, 4 3,4 4, 4,3,3, 3 3, 3 4, 3,,, 3, 4,,,,, 3, 4,,, 3, 4,

16 Eample : Gauss-Seidel Method he nodal temperatures at the boundar nodes are given b: 3,5,5, 5 3, 5 4, 5,4,3,,,4, 4 3,4 4, 4,3, 3 3, 3 4, 3,, 3, 4,,, 3, 4,, 4, i, i,5,,,3,4,,,3,4 5, i 3, i,,3,,3,,, 3, 4, 5

17 Eample : Gauss-Seidel Method Now we can begin to solve for the temperature at each interior node using i, i, i, i, 4 Assume all internal nodes to have an initial temperature of zero. i,,5,4,3,,5, 5 3, 5 4, 5,4, 4 3,4 4, 4,3, 3 3, 3 4, 3,, 3, 4, Iteration # i and i and,,,,, ,, 4,3,,,,,, 3, 4,,, 3, 4,

18 Eample : Gauss-Seidel Method After the first iteration, the temperatures are as follows. hese will now be used as the nodal temperatures for the second iteration

19 Eample : Gauss-Seidel Method Iteration # i and,,,,, present previous,, present, % ε a,

20 Eample : Gauss-Seidel Method he figures below show the temperature distribution and absolute relative error distribution in the plate after two iterations: emperature Distribution Absolute Relative Approimate Error Distribution % 34 % 6% 54 % 83 % 58% 3 % 6 % 3% 7 % 45 % 4%

21 Eample : Gauss-Seidel Method emperature Distribution in the Plate () Node Number of Iterations Eact,,,3,4,,,3,4 3, 3, 3,3 3,

22 he Lieberman Method Recall the equation used in the Gauss- Siedel Method, i, i, i, i, i, 4 Because the Guass-Siedel Method is guaranteed to converge, we can accelerate the process b using over- relaation. In this case, relaed new old i, λi, ( λ) i,

23 Eample 3: Lieberman Method.4 m 3. m Consider a plate that is subected to the boundar conditions shown below. Find the temperature at the interior nodes using a square grid with a length of.6m. Use a weighting factor of.4 in the Lieberman method. Assume the initial temperature at all interior nodes to be. 3 3.m 5.4 m

24 Eample 3: Lieberman Method We can discretize the plate b taking. 6m,5,5, 5 3, 5 4, 5,4,4, 4 3,4 4, 4,3,3, 3 3, 3 4, 3,,, 3, 4,,,,, 3, 4,,, 3, 4,

25 Eample 3: Lieberman Method We can also develop equations for the boundar conditions to define the temperature of the eterior nodes. 3,5,5, 5 3, 5 4, 5,4,3,,,4, 4 3,4 4, 4,3, 3 3, 3 4, 3,, 3, 4,,, 3, 4,, 4, i, i,5,,,3,4,,,3,4 5, i 3, i,,3,,3,,, 3, 4, 5

26 Eample 3: Lieberman Method Now we can begin to solve for the temperature at each interior node using the rewritten Laplace equation from the Gauss-Siedel method. Once we have the temperature value for each node we will appl the over relaation equation of the Lieberman method Assume all internal nodes to have an initial temperature of zero. Iteration # Iteration #,,,,, i and i and, λ ( λ) relaed new old,,,.4(3.5) (.4) 43.,,, λ ( λ) relaed new old,,,,.4(9.68) (.4) 4.565

27 Eample 3: Lieberman Method After the first iteration the temperatures are as follows. hese will be used as the initial nodal temperatures during the second iteration

28 Eample 3: Lieberman Method Iteration # i and,,,,, ε a, present previous,, present, % λ ( λ) relaed new old,,,.4( ) (.4)

29 Eample 3: Lieberman Method he figures below show the temperature distribution and absolute relative error distribution in the plate after two iterations: emperature Distribution Absolute Relative Approimate Error Distribution % 4 % % 53 % 8 % 57% 9 % 55 % 3% 6 % 39 % 7.5%

30 Eample 3: Lieberman Method Node,,,3,4,,,3,4 3, 3, 3,3 3,4 emperature Distribution in the Plate () Number of Iterations 9 Eact

31 Alternative Boundar Conditions In Eamples -3, the boundar conditions on the plate had a specified temperature on each edge. What if the conditions are different? For eample, what if one of the edges of the plate is insulated. In this case, the boundar condition would be the derivative of the temperature. Because if the right edge of the plate is insulated, then the temperatures on the right edge nodes also become unknowns. 3 3.m Insulated 5.4 m

32 Alternative Boundar Conditions he finite difference equation in this case for the right edge for the nodes for,3,.. n m, m, m, m, 4 m, However the node ( m, ) is not inside the plate. he derivative boundar condition needs to be used to account for these additional unknown nodal temperatures on the right edge. his is done b approimating the derivative at the edge node as ( m, ) m ( m, ) m,, ( ) m, 3 3. m Insulated.4m 5

33 Alternative Boundar Conditions Rearranging this approimation gives us, We can then substitute this into the original equation gives us, Recall that is the edge is insulated then, Substituting this again ields, m m m,,, ) ( 4 ) (,,,,, m m m m m, m 4,,,, m m m m

34 Eample 3: Alternative Boundar Conditions A plate.4m 3. m is subected to the temperatures and insulated boundar conditions as shown in Fig.. Use a square grid length of.6m. Assume the initial temperatures at all of the interior nodes to be. Find the temperatures at the interior nodes using the direct method. 3 3.m Insulated 5.4 m

35 Eample 3: Alternative Boundar Conditions We can discretize the plate taking,. 6m,5,5, 5 3, 5 4, 5,4,4, 4 3,4 4, 4,3,3, 3 3, 3 4, 3,,, 3, 4,,,,, 3, 4,,, 3, 4,

36 Eample 3: Alternative Boundar Conditions We can also develop equations for the boundar conditions to define the temperature of the eterior nodes. 3,5,5, 5 3, 5 4, 5, ;,,3,4,4,4, 4 3,4 4, 4 i, 5; i,,3,4,3,3, 3 3, 3 4, 3 Insulated i,5 3; i,,3,4,,,, 3, 4,,, 3, 4, 4, ;,,3,4,,, 3, 4, 5

37 Eample 3: Alternative Boundar Conditions,5,5, 5 3, 5 4, 5,4,4, 4 3,4 4, 4,3,3,3 3, 3 4, 3,,, 3, 4,,,,, 3, 4,,, 3, 4, Here we develop the equation for the temperature at the node (4,3), to show the effects of the alternative boundar condition. i4 and 3 4 3,3 4, 4,4 4, 3 3,3 4, 44,3 4, 4

38 Eample 3: Alternative Boundar Conditions he addition of the equations for the boundar conditions gives us a sstem of 6 equations with 6 unknowns. Solving ields:,,,3,4,,,3,4 3, 3, 3,3 3,4 4, 4, 4,3 4, C

39 HE END