Numerical Methods for PDEs : Video 9: 2D Finite Difference February 14, Equations / 29

Transcription

1 Numerical Methods for PDEs Video 9 2D Finite Difference Equations February 4, Numerical Methods for PDEs Video 9 2D Finite Difference February 4, Equations 205 / 29

2 Thought Experiment Let s extend the string equation to a membrane 2 u = 2 u x u = f (x, y) () y 2 With boundary conditions on both ends of u L = u R = 0 (2) Numerical Methods for PDEs Video 9 2D Finite Difference February 4, Equations / 29

3 Thought Experiment Goal Satisfy the governing equation at each location on the membrane Numerical Methods for PDEs Video 9 2D Finite Difference February 4, Equations / 29

4 Step Discretization Break membrane into a series of equally sized chunks. Let s break the membrane into N nodes in the x direction and M nodes in the y direction Total number of nodes = N M. x = y N = M Numerical Methods for PDEs Video 9 2D Finite Difference February 4, Equations / 29

5 Step Discretization Recall, we would like to satisfy the PDE at each node (equation per node) To simplify life, let s number the nodes Relating the node number (NN) to m and n index NN = N (m ) + n This node numbering scheme will become quite useful later Numerical Methods for PDEs Video 9 2D Finite Difference February 4, Equations / 29

6 Step 2 Discretize the Governing Equation Governing equation At each node of the domain, we wish to ensure ( 2 ) u x u y 2 = f i (x, y) We can write the discrete finite difference equation u n 2u n + u n+ ( x) 2 + u m 2u m + u m+ ( y) 2 f n,m (x, y) Finite Differences Determine the solution u that satisfies the approximate difference equations at each of the nodes of the discrete domain. i Numerical Methods for PDEs Video 9 2D Finite Difference February 4, Equations / 29

7 Step 2 Discretize the Governing Equation Let s look at node 2 to begin ( ) ( ) un,m 2u n,m + u n+,m un,m 2u n,m + u n,m+ ( x) 2 + ( y) 2 f n,m u,3 2u 2,3 + u 3,3 ( x) 2 + u 2,2 2u 2,3 + u 2,4 ( y) 2 f 2, Numerical Methods for PDEs Video 9 2D Finite Difference February 4, Equations / 29

8 Step 2 Discretize the Governing Equation Consider now, how the equation at node 2 looks when written using the node numbers (not m, n locations) u,3 2u 2,3 + u 3,3 ( x) 2 + u 2,2 2u 2,3 + u 2,4 ( y) 2 f 2,3 u 2u 2 + u 3 ( x) 2 + u 7 2u 2 + u 7 ( y) 2 f Numerical Methods for PDEs Video 9 2D Finite Difference February 4, Equations / 29

9 Step 2 Discretize the Governing Equation So, according to the node numbering scheme used here, the equation approximating the PDE at node 2 is u + u 3 + u 7 + u 7 4u 2 ( x) 2 f Numerical Methods for PDEs Video 9 2D Finite Difference February 4, Equations / 29

10 Step 2 Discretize the Governing Equation Let s try this for node 3 So, for node 3, the equation approximating the PDE is Node 3 u 2 + u 4 + u 8 + u 8 4u 3 ( x) 2 f Numerical Methods for PDEs Video 9 2D FiniteFebruary Difference 4, Equations / 29

11 Step 2 Discretize the Governing Equation Let s try this for node 8 So, for node 8, the equation approximating the PDE is Node 8 u 7 + u 9 + u 3 + u 23 4u 8 ( x) 2 f Numerical Methods for PDEs Video 9 2D FiniteFebruary Difference 4, Equations 205 / 29

12 Step 2 Discretize the Governing Equation Summary for the internal nodes Node 2 Node 3 Node 8 u + u 3 + u 7 + u 7 4u 2 ( x) 2 f 2 u 2 + u 4 + u 8 + u 8 4u 3 ( x) 2 f 3 u 7 + u 9 + u 3 + u 23 4u 8 ( x) 2 f Numerical Methods for PDEs Video 9 2D FiniteFebruary Difference 4, Equations / 29

13 Step 3 Form a Linear System of Equations Each node s finite difference equation has 5-unknowns This will become a system of equations (recall multiplication of a matrix and a vector) Node 2 ( x) 2 (u + u 3 + u 7 + u 7 4u 2 ) f ( x) u u 7 u u 2 u 3 u 7 u N = f f 2 u N Numerical Methods for PDEs Video 9 2D FiniteFebruary Difference 4, Equations / 29

14 Step 3 Form a Linear System of Equations Node 3 ( x) 2 (u 2 + u 5 + u 8 + u 8 4u 3 ) f 3 ( x) u u 8 u 2 u 3 u 4 u 8 u N = f 3 u N Numerical Methods for PDEs Video 9 2D Finite Difference Equations February 4, / 29

15 Step 3 Form a Linear System of Equations Node 8 ( x) 2 (u 7 + u 9 + u 3 + u 23 4u 8 ) f 8 ( x) u u 3 u 7 u 8 u 9 u 23 u N = f f 8 u N Numerical Methods for PDEs Video 9 2D Finite Difference Equations February 4, / 29

16 Step 3 Form a Linear System of Equations This results in a linear system of equations, Au = f Looking only at the A-matrix for right now Examining the Structure of A (for internal nodes) The value 4 is on the diagonal entry for equations with internal nodes The value is in the off diagonal entries Numerical Methods for PDEs Video 9 2D FiniteFebruary Difference 4, Equations / 29

17 Step 3 Form a Linear System of Equations Don t forget to setup the RHS vector In this example f = at each of the internal nodes f = f f 6 f 7 f 8 f 9 f 0 f f 2 f 3 f 4 f 5 = Numerical Methods for PDEs Video 9 2D Finite Difference Equations February 4, / 29

18 Step 4 Incorporate Boundary Conditions Introduce boundary conditions. We know that u = 0 at all boundary nodes bottom u = u 2 = u 3 = u 4 = u 5 = 0 top u 2 = u 22 = u 23 = u 24 = u 25 = 0 left u 6 = u = u 6 = 0 right u 0 = u 5 = u 20 = Numerical Methods for PDEs Video 9 2D FiniteFebruary Difference 4, Equations / 29

19 Step 4 Incorporate Boundary Conditions Need to incorporate these boundary conditions into the A-matrix Example u 22 = 0 = f 22 Make the 22nd-row of the A-matrix = 0 Insert a value of on the diagonal (A 22,22 = ) Set the 22nd-entry of the f-vector = 0 (f 22 = 0) When applied to all boundary conditions, the A-Matrix takes the form Numerical Methods for PDEs Video 9 2D FiniteFebruary Difference 4, Equations / 29

20 Step 4 RHS Vector Boundary Nodes Don t forget to setup the RHS vector In this example RHS = 0 at each of the boundary nodes f = f f 6 f 7 f 8 f 9 f 0 f f 2 f 3 f 4 f 5 = Numerical Methods for PDEs Video 9 2D Finite Difference Equations February 4, / 29

21 Examine the A-Matrix Structure The banded A-Matrix Structure (, zeros,,-4,,zeros,) Numerical Methods for PDEs Video 9 2D FiniteFebruary Difference 4, Equations / 29

22 Examine the A-Matrix Structure The A-Matrix Structure (2 x 2 nodes Use, spy(a)) Numerical Methods for PDEs Video 9 2D FiniteFebruary Difference 4, Equations / 29

23 Solve using matlab Solve the linear system of equations using Matlab. We will use the backslash operator in Matlab Solution = A\f (3) Numerical Methods for PDEs Video 9 2D FiniteFebruary Difference 4, Equations / 29

24 Solution Figure Solution for N = M = 5, NumNodes = 25, Max Deflection = Numerical Methods for PDEs Video 9 2D FiniteFebruary Difference 4, Equations / 29

25 Solution Figure Solution for N = M =, NumNodes = 2, Max Deflection = Numerical Methods for PDEs Video 9 2D FiniteFebruary Difference 4, Equations / 29

26 Solution Figure Solution for N = M = 2, NumNodes = 44, Max Deflection = Numerical Methods for PDEs Video 9 2D FiniteFebruary Difference 4, Equations / 29

27 Solution Figure Solution for N = M = 4, NumNodes = 68, Max Deflection = Numerical Methods for PDEs Video 9 2D FiniteFebruary Difference 4, Equations / 29

28 Starting to think about the Error Figure Convergence of the Max Deflection Value (compared with deflection for N = 0) Numerical Methods for PDEs Video 9 2D FiniteFebruary Difference 4, Equations / 29

29 What have we learned? Finite difference expressions in 2D follow the same structure as D Approximate the PDE expression at nodes in the domain Determine the solution at those nodes The general steps for a finite difference problem are Step Discretize the domain (in D N intervals, N points). Step 2 Write finite difference equations that approximate governing equation at each internal node of the problem Step 3 Form a linear system (Ax = b) Step 4 Enforce the boundary conditions where relevant Step 5 Solve the system of equations Numerical Methods for PDEs Video 9 2D FiniteFebruary Difference 4, Equations / 29