22.520 Numerical Methods for PDEs : Video 9: 2D Finite Di ...

22.520 Numerical Methods for PDEs : Video 9: 2DFinite Difference Equations

David J. Willis

February 14, 2015

David J. Willis 22.520 Numerical Methods for PDEs : Video 9: 2D Finite Difference EquationsFebruary 14, 2015 1 / 29

Thought Experiment

Let’s extend the string equation to a membrane:

∇2u =∂2u

∂x2+

∂2u

∂y2= f (x , y) (1)

Figure : The membrane problem domain.

With boundary conditions on both ends of:

uL = uR = 0 (2)


Thought Experiment

Goal: Satisfy the governing equation at each location on themembrane


Step 1: Discretization

Break membrane into a series of equally sized ”chunks”.Let’s break the membrane into N nodes in the x−direction and Mnodes in the y−direction:

Total number of nodes = N ∗M.∆x = ∆y → N = M.


Step 1: Discretization

Recall, we would like to satisfy the PDE at each node (equation pernode)To simplify life, let’s number the nodes

Relating the node number (NN) to m and n index:

NN = N ∗ (m − 1) + n

This node numbering scheme will become quite useful later.


Step 2: Discretize the Governing Equation

Governing equation: At each node of the domain, we wish to ensure:(∂2u

∂x2+

∂2u

∂y2

)i

= fi (x , y)

We can write the discrete finite difference equation:

un−1 − 2un + un+1

(∆x)2+

um−1 − 2um + um+1

(∆y)2' fn,m(x , y)

Finite Differences: Determine the solution u that satisfies theapproximate difference equations at each of the nodes of the discretedomain.



Let’s look at node 12 to begin:(un−1,m − 2un,m + un+1,m

(∆x)2

)+

(un,m−1 − 2un,m + un,m+1

(∆y)2

)' fn,m

u1,3 − 2u2,3 + u3,3

(∆x)2+

u2,2 − 2u2,3 + u2,4

(∆y)2' f2,3



Consider now, how the equation at node 12 looks when written usingthe node numbers (not m, n locations) :

u1,3 − 2u2,3 + u3,3

(∆x)2+

u2,2 − 2u2,3 + u2,4

(∆y)2' f2,3

u11 − 2u12 + u13

(∆x)2+

u7 − 2u12 + u17

(∆y)2' f12



So, according to the node numbering scheme used here, the equationapproximating the PDE at node 12 is:

u11 + u13 + u7 + u17 − 4u12

(∆x)2' f12



Let’s try this for node 13:

So, for node 13, the equation approximating the PDE is:

Node 13 :u12 + u14 + u8 + u18 − 4u13

(∆x)2' f13



Let’s try this for node 18:

So, for node 18, the equation approximating the PDE is:

Node 18 :u17 + u19 + u13 + u23 − 4u18

(∆x)2' f18



Summary: for the internal nodes

Node 12 :u11 + u13 + u7 + u17 − 4u12

(∆x)2' f12

Node 13 :u12 + u14 + u8 + u18 − 4u13

(∆x)2' f13

Node 18 :u17 + u19 + u13 + u23 − 4u18

(∆x)2' f18


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 amatrix and a vector):

Node 12 :1

(∆x)2(u11 + u13 + u7 + u17 − 4u12) ' f12

1

(∆x)2

: . . . : : : : : : : . . . :0 . . . 1 . . . 0 . . . 1 −4 1 . . . 0 . . . 1 . . . :: . . . : : : : : : : . . . :

u1

:u7

:u11

u12

u13

:u17

:uN

=

f1

::::

f12

::::

uN



Node 13 :1

(∆x)2(u12 + u15 + u8 + u18 − 4u13) ' f13

1

(∆x)2

: . . . : : : : : : : . . . :0 . . . 1 . . . 0 . . . 1 −4 1 . . . 0 . . . 1 . . . :: . . . : : : : : : : . . . :

u1

:u8

:u12

u13

u14

:u18

:uN

=

:::::

f13

::::

uN



Node 18 :1

(∆x)2(u17 + u19 + u13 + u23 − 4u18) ' f18

1

(∆x)2

: . . . : : : : : : : . . . :0 . . . 1 . . . 0 . . . 1 −4 1 . . . 0 . . . 1 . . . :: . . . : : : : : : : . . . :

u1

:u13

:u17

u18

u19

:u23

:uN

=

f1

::::

f18

::::

uN



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 nodesThe value 1 is in the off diagonal entries



Don’t forget to setup the RHS vector

In this example f = 1 at each of the internal nodes

f =

f1

:f6

f7

f8

f9

f10

f11

f12

f13

f14

f15

:

=

0:0111001110:


Step 4: Incorporate Boundary Conditions

Introduce boundary conditions.We know that u = 0 at all boundary nodes

bottom : u1 = u2 = u3 = u4 = u5 = 0

top : u21 = u22 = u23 = u24 = u25 = 0

left : u6 = u11 = u16 = 0

right : u10 = u15 = u20 = 0


Step 4: Incorporate Boundary Conditions

Need to incorporate these boundary conditions into the A-matrix:Example: u22 = 0 = f22:

Make the 22nd-row of the A-matrix = 0Insert a value of 1 on the diagonal (A22,22 = 1)Set the 22nd-entry of the f-vector = 0 (f22 = 0)

When applied to all boundary conditions, the A-Matrix takes the form:


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 =

f1

:f6

f7

f8

f9

f10

f11

f12

f13

f14

f15

:

=

0:0111001110:


Examine the A-Matrix Structure

The banded A-Matrix Structure (1, zeros,1,-4,1,zeros,1):


Examine the A-Matrix Structure

The A-Matrix Structure (21 x 21 nodes: Use, spy(A)):


Solve using matlab

Solve the linear system of equations using Matlab.

We will use the ’backslash’ operator in Matlab:

Solution = A\f (3)


Solution

Figure : Solution for N = M = 5,NumNodes = 25, Max Deflection = -0.06667


Solution

Figure : Solution for N = M = 11,NumNodes = 121, Max Deflection =-0.073010


Solution



Solution



Starting to think about the Error

Figure : Convergence of the Max Deflection Value (compared withdeflection for N = 101)


What have we learned?

Finite difference expressions in 2D follow the same structure as 1D:

Approximate the PDE expression at nodes in the domainDetermine the solution at those nodes

The general steps for a finite difference problem are:

Step 1: Discretize the domain (in 1D: N − 1 intervals, N points).Step 2: Write finite difference equations that approximate governingequation at each internal node of the problemStep 3: Form a linear system (Ax = b)Step 4: Enforce the boundary conditions where relevantStep 5: Solve the system of equations


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22.520 Numerical Methods for PDEs : Video 9: 2D Finite Di ...

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