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PARTIAL DIFFERENTIAL EQUATIONS Student Notes. ENGR 351 Numerical Methods for Engineers Southern Illinois University Carbondale College of Engineering Dr. L.R. Chevalier Dr. B.A. DeVantier. Photo Credit: Mr. Jeffrey Burdick. Partial Differential Equations. - PowerPoint PPT Presentation
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PARTIAL DIFFERENTIAL EQUATIONSStudent NotesENGR 351 Numerical Methods for EngineersSouthern Illinois University CarbondaleCollege of EngineeringDr. L.R. ChevalierDr. B.A. DeVantier
Photo Credit: Mr. Jeffrey Burdick
Partial Differential Equations An equation involving partial derivatives
of an unknown function of two or more independent variables
The following are examples. Note: u depends on both x and y
xyuxu
xuyu
yux
yxu
xyxu
xuu
yuxy
xu
2
2
2
22
2
33
2
2
2
2
2
2
58
612
Partial Differential Equations Because of their widespread application
in engineering, our study of PDE will focus on linear, second-order equations
The following general form will be evaluated for B2 - 4AC
A ux
B ux y
C uy
D
2
2
2 2
2 0
B2-4AC Category Example< 0 Elliptic Laplace equation (steady state with 2 spatial dimensions
= 0 Parabolic Heat conduction equation (time variablewith one spatial dimension
>0 Hyperbolic Wave equation (time-variable with onespatial dimension
2
2
2
2 0Tx
Ty
k Tx
Tt
2
2
2
2 2
2
2
1yx c
yt
Scope of Lectures on PDE Finite Difference: Elliptic
The Laplace Equation Finite difference solution Boundary conditions
Finite Difference: Parabolic Heat conduction Explicit method Simple implicit method
Specific Study Objectives Recognize the difference between elliptic,
parabolic, and hyperbolic PDE Recognize that the Liebmann method is
equivalent to the Gauss-Seidel approach for solving simultaneous linear algebraic equations
Recognize the distinction between Dirichlet and derivative boundary conditions
Know the difference between convergence and stability of parabolic PDE
Apply finite difference to parabolic and elliptic PDE using various boundary conditions and step sizes
Typically used to characterize steady-state boundary value problems
Before solving, the Laplace equation will be solved from a physical problem
Aux
Bux y
Cuy
D
ux
uy
2
2
2 2
2
2
2
2
2
0
0
Finite Difference: Elliptic EquationsB2- 4AC < 0
The Laplace Equation
Models a variety of problems involving the potential of an unknown variable
We will consider cases involving thermodynamics, fluid flow, and flow through porous media
2
2
2
2 0ux
uy
The Laplace Equation
Let’s consider the case of a plate heated from the boundaries
How is this equation derived from basic concepts of continuity?
How does it relate to flow fields?
2
2
2
2 0Tx
Ty
Consider the plate below, with thickness Dz.The temperatures are known at the boundaries.What is the temperature throughout the plate?
T = 200 T= 200
T = 400
T = 200
T = 200 T= 200
T = 400
T = 200
y
x
First, recognize how the shape can be setin an x-y coordinate system
T = 200 T= 200
T = 400
T = 200
y
x
Divide into a grid, with increments by Dx and Dy
T = 200 T= 200
T = 400y
x
What is the temperature here, if using a block centered scheme?
T= 200
T = 400
T = 200
y
x
T= 200
T = 400
T = 200
y
x
T = 200
What is the temperature here, if using a grid centered scheme?
y
xD x
D y
Consider the element shown below on the face of a plate Dz in thickness.
The plate is illustrated everywhere but at its edges or boundaries, where the temperature can be set.
q(y + D y)
q(x + D x)q(x)
q(y)
By continuity, the flow of heat in must equal the flow of heatout.
q x y z t q y x z tq x x y z t q y y x z tD D D D D D
D D D D D D D D
Consider the steady state heat flux qin and out of the elemental volume.
tzxyyqtzyxxq
tzxyqtzyxqDDDDDDDD
DDDDDD
0][
][DDDDDDD
DDDDDDDtzxyqtzxyyq
tzyxqtzyxxq
Rearranging terms
0D
D
DD
yyqyyq
xxqxxq
Dividing by Dx, Dy, Dz and D t :
0yq
xq yx
Again, this is our
continuity equation
As Dx & Dy approach zero, the equation reduces to:
qx
qy
0
The link between flux and temperature is provided by Fourier’s Law of heat conduction
q k CTii
where qi is the heat flux in the direction i.Substitute B into A to get the Laplace equation
Equation A
Equation B
qx
qy
0
q k C Tii
Equation A
Equation B
02
2
2
2
yT
xT
yT
Ckyx
TCk
xyq
xq
Consider Fluid FlowIn fluid flow, where the fluid is a liquid or a gas, the continuity equation is:
Vx
Vy
x y 0
The link here can by either of the following sets of equations:The potential function:
Stream function: yV
xV yx
xV
yV yx
2
2
2
2
2
2
2
20 0x y x y
The Laplace equation is then
Vx
Vy
x y 0
Vx
Vyx y
xV
yV yx
Flow in Porous Media
qx
qy
0
q K Hii Darcy’s Law
The link between flux and the pressure head is provided by Darcy’s Law
2
2
2
2 0hx
hy
2
2
2
2
ux
uy
f x y ( , )
For a case with sources and sinks within the 2-Ddomain, as represented by f(x,y), we have thePoisson equation.
Now let’s consider solution techniques.
Poisson Equation
(i,j)(i-1,j)
(i,j+1)
21,,1,
2
2
2,1,,1
2
2
2
2
yuuu
yu
xuuu
xu
jijiji
jijiji
D
D
(i+1,j)
(i,j-1)
Evaluate these equations based on the grid and central difference equations
022
21,,1,
2,1,,1
D
D
yuuu
xuuu jijijijijiji
(i,j)
(i+1,j)(i-1,j)
(i,j+1)
(i,j-1)
If D x = D y we can collect the termsto get:u u u u ui j i j i j i j i j 1 1 1 1 4 0, , , , ,
u u u u ui j i j i j i j i j 1 1 1 1 4 0, , , , ,
This equation is referredto as the Laplacian difference equation.
It can be applied to all interior points.
We must now considerwhat to do with the boundary nodes.
(i,j)
(i+1,j)(i-1,j)
(i,j+1)
(i,j-1)
Boundary Conditions Dirichlet boundary conditions: u is specified at
the boundary Temperature Head
Neumann boundary condition: the derivative is specified qi
Combination of both u and its derivative (mixed boundary condition)
ii xTor
xh
The simplest case is where the boundaries arespecified as fixed values.
This case is known as the Dirichlet boundaryconditions.
u1
u2
u3
u4
u1
u2
u3
u4
Consider how we can deal with the lower node shown, u1,1
-4u1,1 +u1,2+u2,1+u1 +u4 = 0
1,2
1,1 2,1
Note:This grid would resultin nine simultaneous equations.
Let’s consider how to model the Neumann boundary condition
ux
u ux
i j i j 1 1
2, ,
Dcentered finite divided differenceapproximation
suppose we wanted to consider this end grid point
(0,100)
(0,0) (200,0)
(200,100)h = 0.05x + 100
0xh
0
xh
0yh
xh
2
2
2
2
0yh
y
x
1,2
1,1 2,1
hx0
hy0
The two boundaries are consider to be symmetry lines due to the fact that the BC translates in the finite difference form to:
h i+1,j = h i-1,j
and
h i,j+1 = h i,j-1
1,2
1,1 2,1
hy0
h1,1 = (2h1,2 + 2 h2,1)/4
h1,2 = (h1,1 + h1,3+2h22)/42,2
hx0
ExampleThe grid on the next slide is designed to solve the LaPlace equation
02
2
2
2
yx
Write the finite difference equations for the nodes (1,1), (1,2), and (2,1). Note that the lower boundary is a Dirichlet boundary condition, the left boundary is a Neumann boundary condition, and Dx = Dy.
StrategyResolve the governing equation as
a finite difference equationResolve the boundary conditions
as finite difference equationsApply the governing equation at
each nodeAt boundary nodes, include the
finite difference estimation of the boundary
The Liebmann MethodMost numerical solutions of the
Laplace equation involves systems that are much larger that the general system we just evaluated
Note that there are a maximum of five unknown terms per line
This results in a significant number of terms with zero’s
The Liebmann Method In addition to the fact that they are
prone to round-off errors, using elimination methods on such sparse system waste a great amount of computer memory storing zeros
Therefore, we commonly employ approaches such as Gauss-Seidel, which when applied to PDEs is also referred to as Liebmann’s method.
The Liebmann Method In addition the equations will lead to a
matrix that is diagonally dominant. Therefore the procedure will converge
to a stable solution. Over relaxation is often employed to
accelerate the rate of convergence
u u u u u
u u ui j i j i j i j i j
i jnew
i jnew
i jold
1 1 1 1 4 0
1, , , , ,
, , ,
u u u u u
u u ui j i j i j i j i j
i jnew
i jnew
i jold
1 1 1 1 4 0
1, , , , ,
, , ,
As with the conventional Gauss Seidel method, the iterations are repeated until each point falls below a pre-specified tolerance:
si jnew
i jold
i jnew
u uu
, ,
,
100
Groundwater Flow Example
Modeling 1/2 of the system shown, we can develop the following schematic where Dx = Dy = 20 m
The finite difference equations can be solved using aa spreadsheet. This next example is part of the PDE example you can download from my homepage.
(0,100)
(0,0) (200,0)
(200,100)h = 0.05x + 100
0xh
0
xh
0yh
xh
2
2
2
2
0yh
y
x
100 =A1+0.05*20 =B1+0.05*20 =C1+0.05*20 =D1+0.05*20 =E1+0.05*20 =F1+0.05*20 =G1+0.05*20 =H1+0.05*20 =I1+0.05*20 =J1+0.05*20
=(A1+2*B2+A3)/4 =(B1+C2+B3+A2)/4 =(C1+D2+C3+B2)/4 =(D1+E2+D3+C2)/4 =(E1+F2+E3+D2)/4 =(F1+G2+F3+E2)/4 =(G1+H2+G3+F2)/4 =(H1+I2+H3+G2)/4 =(I1+J2+I3+H2)/4 =(J1+K2+J3+I2)/4 =(K1+K3+2*J2)/4
=(A2+2*B3+A4)/4 =(B2+C3+B4+A3)/4 =(C2+D3+C4+B3)/4 =(D2+E3+D4+C3)/4 =(E2+F3+E4+D3)/4 =(F2+G3+F4+E3)/4 =(G2+H3+G4+F3)/4 =(H2+I3+H4+G3)/4 =(I2+J3+I4+H3)/4 =(J2+K3+J4+I3)/4 =(K2+K4+2*J3)/4
=(A3+2*B4+A5)/4 =(B3+C4+B5+A4)/4 =(C3+D4+C5+B4)/4 =(D3+E4+D5+C4)/4 =(E3+F4+E5+D4)/4 =(F3+G4+F5+E4)/4 =(G3+H4+G5+F4)/4 =(H3+I4+H5+G4)/4 =(I3+J4+I5+H4)/4 =(J3+K4+J5+I4)/4 =(K3+K5+2*J4)/4
=(A4+2*B5+A6)/4 =(B4+C5+B6+A5)/4 =(C4+D5+C6+B5)/4 =(D4+E5+D6+C5)/4 =(E4+F5+E6+D5)/4 =(F4+G5+F6+E5)/4 =(G4+H5+G6+F5)/4 =(H4+I5+H6+G5)/4 =(I4+J5+I6+H5)/4 =(J4+K5+J6+I5)/4 =(K4+K6+2*J5)/4
=(2*A5+2*B6)/4 =(2*B5+C6+A6)/4 =(2*C5+D6+B6)/4 =(2*D5+E6+C6)/4 =(2*E5+F6+D6)/4 =(2*F5+G6+E6)/4 =(2*G5+H6+F6)/4 =(2*H5+I6+G6)/4 =(2*I5+J6+H6)/4 =(2*J5+K6+I6)/4 =(2*K5+2*J6)/4
100 =A1+0.05*20 =B1+0.05*20 =C1+0.05*20 =D1+0.05*20 =E1+0.05*20 =F1+0.05*20 =G1+0.05*20 =H1+0.05*20 =I1+0.05*20 =J1+0.05*20
=(A1+2*B2+A3)/4 =(B1+C2+B3+A2)/4 =(C1+D2+C3+B2)/4 =(D1+E2+D3+C2)/4 =(E1+F2+E3+D2)/4 =(F1+G2+F3+E2)/4 =(G1+H2+G3+F2)/4 =(H1+I2+H3+G2)/4 =(I1+J2+I3+H2)/4 =(J1+K2+J3+I2)/4 =(K1+K3+2*J2)/4
=(A2+2*B3+A4)/4 =(B2+C3+B4+A3)/4 =(C2+D3+C4+B3)/4 =(D2+E3+D4+C3)/4 =(E2+F3+E4+D3)/4 =(F2+G3+F4+E3)/4 =(G2+H3+G4+F3)/4 =(H2+I3+H4+G3)/4 =(I2+J3+I4+H3)/4 =(J2+K3+J4+I3)/4 =(K2+K4+2*J3)/4
=(A3+2*B4+A5)/4 =(B3+C4+B5+A4)/4 =(C3+D4+C5+B4)/4 =(D3+E4+D5+C4)/4 =(E3+F4+E5+D4)/4 =(F3+G4+F5+E4)/4 =(G3+H4+G5+F4)/4 =(H3+I4+H5+G4)/4 =(I3+J4+I5+H4)/4 =(J3+K4+J5+I4)/4 =(K3+K5+2*J4)/4
=(A4+2*B5+A6)/4 =(B4+C5+B6+A5)/4 =(C4+D5+C6+B5)/4 =(D4+E5+D6+C5)/4 =(E4+F5+E6+D5)/4 =(F4+G5+F6+E5)/4 =(G4+H5+G6+F5)/4 =(H4+I5+H6+G5)/4 =(I4+J5+I6+H5)/4 =(J4+K5+J6+I5)/4 =(K4+K6+2*J5)/4
=(2*A5+2*B6)/4 =(2*B5+C6+A6)/4 =(2*C5+D6+B6)/4 =(2*D5+E6+C6)/4 =(2*E5+F6+D6)/4 =(2*F5+G6+E6)/4 =(2*G5+H6+F6)/4 =(2*H5+I6+G6)/4 =(2*I5+J6+H6)/4 =(2*J5+K6+I6)/4 =(2*K5+2*J6)/4
100
=(A1+2*B2+A3)/4
=(A2+2*B3+A4)/4
=(A3+2*B4+A5)/4
=(A4+2*B5+A6)/4
=(2*A5+2*B6)/4
You will get anerror message inExcel that state thatit will not resolvea circular reference.
CAN USE EXCEL DEMONSTRATION
100 =A1+0.05*20 =B1+0.05*20 =C1+0.05*20 =D1+0.05*20 =E1+0.05*20 =F1+0.05*20 =G1+0.05*20 =H1+0.05*20 =I1+0.05*20 =J1+0.05*20
=(A1+2*B2+A3)/4 =(B1+C2+B3+A2)/4 =(C1+D2+C3+B2)/4 =(D1+E2+D3+C2)/4 =(E1+F2+E3+D2)/4 =(F1+G2+F3+E2)/4 =(G1+H2+G3+F2)/4 =(H1+I2+H3+G2)/4 =(I1+J2+I3+H2)/4 =(J1+K2+J3+I2)/4 =(K1+K3+2*J2)/4
=(A2+2*B3+A4)/4 =(B2+C3+B4+A3)/4 =(C2+D3+C4+B3)/4 =(D2+E3+D4+C3)/4 =(E2+F3+E4+D3)/4 =(F2+G3+F4+E3)/4 =(G2+H3+G4+F3)/4 =(H2+I3+H4+G3)/4 =(I2+J3+I4+H3)/4 =(J2+K3+J4+I3)/4 =(K2+K4+2*J3)/4
=(A3+2*B4+A5)/4 =(B3+C4+B5+A4)/4 =(C3+D4+C5+B4)/4 =(D3+E4+D5+C4)/4 =(E3+F4+E5+D4)/4 =(F3+G4+F5+E4)/4 =(G3+H4+G5+F4)/4 =(H3+I4+H5+G4)/4 =(I3+J4+I5+H4)/4 =(J3+K4+J5+I4)/4 =(K3+K5+2*J4)/4
=(A4+2*B5+A6)/4 =(B4+C5+B6+A5)/4 =(C4+D5+C6+B5)/4 =(D4+E5+D6+C5)/4 =(E4+F5+E6+D5)/4 =(F4+G5+F6+E5)/4 =(G4+H5+G6+F5)/4 =(H4+I5+H6+G5)/4 =(I4+J5+I6+H5)/4 =(J4+K5+J6+I5)/4 =(K4+K6+2*J5)/4
=(2*A5+2*B6)/4 =(2*B5+C6+A6)/4 =(2*C5+D6+B6)/4 =(2*D5+E6+C6)/4 =(2*E5+F6+D6)/4 =(2*F5+G6+E6)/4 =(2*G5+H6+F6)/4 =(2*H5+I6+G6)/4 =(2*I5+J6+H6)/4 =(2*J5+K6+I6)/4 =(2*K5+2*J6)/4
=B1+0.05*20
=(C1+D2+C3+B2)/4
=(C2+D3+C4+B3)/4
=(C3+D4+C5+B4)/4
=(C4+D5+C6+B5)/4
=(2*C5+D6+B6)/4
After selecting theappropriate command,EXCEL with performthe Liebmann methodfor you.
In fact, you will be ableto watch the iterations.
Table 2: Results of finite difference model.
A B C D E F G H I J K1 100 101 102 103 104 105 106 107 108 109 1102 101.6 102 102.6 103.4 104.2 105 105.8 106.6 107.4 108 108.43 102.5 102.7 103.1 103.7 104.3 105 105.7 106.3 106.9 107.3 107.54 103 103.1 103.4 103.9 104.4 105 105.6 106.1 106.6 106.9 1075 103.3 103.3 103.6 104 104.5 105 105.5 106 106.4 106.7 106.76 103.3 103.4 103.7 104 104.5 105 105.5 106 106.3 106.6 106.7
Table 2: Results of finite difference model.
A B C D E F G H I J K1 100 101 102 103 104 105 106 107 108 109 1102 101.6 102 102.6 103.4 104.2 105 105.8 106.6 107.4 108 108.43 102.5 102.7 103.1 103.7 104.3 105 105.7 106.3 106.9 107.3 107.54 103 103.1 103.4 103.9 104.4 105 105.6 106.1 106.6 106.9 1075 103.3 103.3 103.6 104 104.5 105 105.5 106 106.4 106.7 106.76 103.3 103.4 103.7 104 104.5 105 105.5 106 106.3 106.6 106.7
...end of problem.
• Assuming K = 5 m/day, we can calculate the q’s
• Use centered differences, for example @ cell B2
• qx = -K h/x = -(5m/d) (102.6m-101.6m) /(2*20m) = -0.125 m/d
• qy = -K h/y = -(5m/d) (101 m-102.7 m) /(2*20m) = 0.2125 m/d
Secondary Variables Because its distribution is described by
the Laplace equation, temperature is considered to be the primary variable in the heated plate problem
A secondary variable may also be of interest
In this case, the second variable is the rate of heat flux across the place surfaceq k C
Tii
Secondary Variables
yTT
kq
xTT
kq
iTCkq
jijiy
jijix
i
D
D
2'
2'
1,1.
,1.1
FINITE DIFFERENCE
APPROXIMATIONBASED ON RESULTSOF TEMPERATURE DISTRIBUTION
Secondary Variables
0
180tan
0
tan
1
1
22
x
x
y
x
x
y
yxn
qfor
qfor
qqq
The Resulting Flux Is A Vector With Magnitude And Direction
Note: is in degreesIf qx=0, is 90 or 270 depending on whether qy is positive or negative, respectively
Finite Difference: Parabolic Equations B2- 4AC = 0
Aux
Bux y
Cuy
D
2
2
2 2
2 0
These equations are used to characterizetransient problems.
We will first study this in one spatial directionthen we will discuss the results in 2-D.
kTx
Tt
2
2
Consider the heat-conduction equation
As with the elliptic PDEs, parabolic equations can be solved by substituting finite difference equations for the partial derivatives.
However we must now consider changes in time as well as space.
Finite Difference: Parabolic Equations B2- 4AC = 0
t
x
y
x
uil
spatial
{temporal
{
tTT
xTTTk
tT
xTk
li
li
li
li
li
D
D
1
211
2
2
2
Centered finite divided difference
Forward finite divided difference
2
111
1
211
2
2
xtk
whereTTTTT
tTT
xTTTk
li
li
li
li
li
li
li
li
li
li
DD
D
D
We can further reduce the equation:
NOTE:
Now the temperatureat a node is estimatedas a function of the temperature at the node, and surrounding nodes, but at a previous time
ExampleConsider a thin insulated rod 10 cm long with k = 0.835 cm2/s
Let D x = 2 cm and D t = 0.1 sec.
At t=0 the temperature of the rod is zero.
hot
cold
StrategyDraw the system and label nodesRewrite the governing equation
and boundaries in finite differenceRecognize the use of time and
space in the equations
Convergence and Stability Convergence means that as D x and D t
approach zero, the results of the numerical technique approach the true solution
Stability means that the errors at any stage of the computation are attenuated, not amplified, as the computation progresses
The explicit method is stable and convergent if
12
Derivative Boundary Conditions
T o
T L
In our previous example To and TL were constant values.
However, we may also have derivative boundary conditions
T T T T Ti i i i i0
10 1 0 12
Thus we introduce an imaginary point at i = -1This point provides the vehicle for providing the derivative BC
Derivative Boundary Conditions
q 0 =
0
T L
For the case of qo = 0, so T-1 = T1 .
In this case the balance at node 0 is:
llll010
10 T2T2TT
Derivative Boundary Conditions
q 0 =
10
T L
For the case of qo = 10, we need to know k’ [= k/(C)].Assuming k’ =1, then 10 = - (1) dT/dx,or dT/dx = -10
Derivative Boundary Conditions
D
D
D
llll
ll
ll
0101
0
11
11
T21x20T2TT
kx20TT
x2TTk10
Implicit Method
Explicit methods have problems relating to stability
Implicit methods overcome this but at the expense of introducing a more complicated algorithm
In this algorithm, we develop simultaneous equations
211 2
xTTT li
li
li
D
Explicit
grid point involved with space difference grid point involved with time difference
211 2
xTTT li
li
li
D
2
11
111 2
xTTT li
li
li
D
Explicit Implicit
With the implicit method, we develop a set of simultaneous equations at step in time
2
11
111 2
xTTT li
li
li
D
211 2
xTTT li
li
li
D
Explicit Implicit
k T T Tx
T Tt
T T T T
T f t
il
il
il
il
il
il
il
il
il
l l
11 1
11
2
1
11 1
11
01
01
2
1 2
D D
which can be expressed as:
For the case where the temperature level is given at the end by a function f0 i.e. x = 0
Implicit Method
T T T T
T f t
T T T f t
il
il
il
il
l l
il
il
il l
11 1
11
01
01
111
01
1 2
1 2
Substituting
In the previous example problem, we get a 4 x 4 matrixto solve for the four interior nodes for each time step
Implicit Method
Specific Study Objectives Recognize the difference between elliptic,
parabolic, and hyperbolic PDE Recognize that the Liebmann method is
equivalent to the Gauss-Seidel approach for solving simultaneous linear algebraic equations
Recognize the distinction between Dirichlet and derivative boundary conditions
Know the difference between convergence and stability of parabolic PDE
Apply finite difference to parabolic and elliptic PDE using various boundary conditions and step sizes