Introduction to Finite Di erence Methods · 2020. 1. 4. · Substituting Equation (6.c) into Equation (5) gives A n= 2 L Z L 0 sin ˇx L sin nˇx L dx: The sine function is orthogonal

Introduction to Finite Difference MethodsME 448/548 Notes

Gerald Recktenwald

Portland State University

Department of Mechanical Engineering

[email protected]

ME 448/548: Introduction to Finite Difference Approximation of the Heat Equation

Overview

1. Motivation for numerical solution to the heat equation

2. Finite difference notation

3. Finite difference approximations

ME 448/548: Introduction to Finite Difference Approximation of the Heat Equation page 1

Model Problem: 1D Heat Equation

∂u

∂t= α

∂2u

∂x2, 0 ≤ x ≤ L (1)

u(0, t) = 0, u(L, t) = 0 (2)

u(x, 0) = u0(x) (3)


Example: Place a hot pot on a table.

What is the transient temperature distribution in the table?

t < 0 t > 0

xL

x

0

L

TTL

T0

increasing time


Model Problem: 1D Heat Equation

Exact Solution

u(x, t) =

∞∑n=1

un(x, t) =

∞∑n=1

An exp(−αn2π2t/L2) sin(nπx

L

)(4)

An =2

L

∫ L0

u0(x) sin

(nπx

L

)dx. (5)


Example: Decay of u0(x) = sin(πx/L)

We will use the following problem to test our finite-difference solutions.

∂u

∂t= α

∂2u

∂x2, 0 ≤ x ≤ L, t ≥ 0 (6.a)

u(0, t) = u(L, t) = 0 (6.b)

u(x, 0) = sin

(πx

L

). (6.c)

This model problem has a simple solution that is convenient as a benchmark for numerical

schemes for the heat equation.


Example: Analytical solution to decay of u0(x) = sin(πx/L)

Substituting Equation (6.c) into Equation (5) gives

An =2

L

∫ L0

sin

(πx

L

)sin

(nπx

L

)dx.

The sine function is orthogonal in the following sense∫ L0

sin

(πx

L

)sin

(nπx

L

)dx =

{L/2 if n = 1

0 otherwise

so that

A1 = 1,

An = 0, n = 2, 3, . . .


Example: Analytical solution to decay of u0(x) = sin(πx/L)

Therefore, the analytical solution to the toy problem is

u(x, t) = exp

(−απ2t

L2

)sin

(πx

L

). (7)

The solution is just an exponential decay of the initial condition for this problem and

problems with Dirichlet BC and no source term.


Example: Analytical solution to decay of a square pulse

u0(x) =

0, if 0 ≤ x < xluc, if xl ≤ x ≤ xr0, if xr < x ≤ L.

uc

xl

xr

L0

An infinite number of An terms are needed to match the initial condition.

An =2

L

∫ xrxl

uc sin

(nπx

L

)dx

=2uc

nπ

[cos

(nπxl

L

)− cos

(nπxr

L

)]n = 0, 1, 2, . . .


Example: Analytical solution to decay of a square pulse

0 0.2 0.4 0.6 0.8 1−0.2

0

0.2

0.4

0.6

0.8

1

1.2

x

T(x

,t)

t = 1.0e−04

t = 1.0e−03

t = 1.0e−02

t = 5.0e−02


Numerical Model

ContinuousPDE for u(x,t)

Discretedifferenceequation

uik approximationto u(xi,tk)

Finitedifference

Algebraicsolution

orFinite volume

orFinite element

orBoundar element

or ...


Finite Difference Mesh

The finite difference solution is obtained at a finite set of points. These points are called

nodes and the network of these nodes is called a mesh or grid.

On a uniform mesh, the x-direction nodes are located at

xi = (i− 1)∆x, i = 1, 2, . . . , nx, ∆x =L

nx − 1(8)

where nx is the total number of spatial nodes, including those on the boundary.

i – 1 i + 1i

∆x

x = 0

∆x

i = 1 ...x = L

i = nx

xi – 1 xi xi + 1...



Analogous to the x-direction mesh, there are nodes at discrete times.

For uniform time steps

tk = (k − 1)∆t, k = 1, 2, . . . , nt, ∆t =tmax

nt − 1(9)

where nt is the number of time steps.



The x and t direction nodes form a finite version of a semi-infinite strip

t

i i+1i 1i=1 nx

k+1

k

k 1

. . .. . .

. . .. . .

. . .. . .

. . .

x=Lx=0

t=0, k=1

Interior node: u(x,t) is initially unknown.

Boundary node: u(0,t) and u(0,t) are knownor computable from additional data.

Initial condition: u(x,0) must be specified.


Nomenclature

For the one-dimensional mesh, we define

Symbol Meaning

u(x, t) Analytical solution (true solution).

u(xi, tk) Analytical solution evaluated at x = xi, t = tk.

uki Approximate numerical solution at x = xi, t = tk.


First order forward finite difference

Use a Taylor series expansion about the point xi

u(xi + δx) = u(xi) + δx∂u

∂x

∣∣∣∣xi

+δx2

2

∂2u

∂x2

∣∣∣∣∣xi

+δx3

3!

∂3u

∂x3

∣∣∣∣∣xi

+ · · ·

where δx is a small distance, and δx2 and δx3 are shorthand for (δx)2 and (δx)3.

Solve for ∂u/∂x|xi

∂u

∂x

∣∣∣∣xi

=u(xi + δx)− u(xi)

δx−δx

2

∂2u

∂x2

∣∣∣∣∣xi

−δx2

3!

∂3u

∂x3

∣∣∣∣∣xi

+ · · ·

In the limit as δx→ 0 the coefficients of the higher order derivative terms vanish, andthe first term on the right hand side becomes equal to the derivative on the left hand side.



Replace δx with ∆x = xi+1 − xi, and substitute ui ≈ u(xi)

∂u

∂x

∣∣∣∣xi

≈ui+1 − ui

∆x−

∆x

2

∂2u

∂x2

∣∣∣∣∣xi

−∆x2

3!

∂3u

∂x3

∣∣∣∣∣xi

+ · · · (10)

The mean value theorem can be used to replace the higher order derivatives (exactly)

∆x

2

∂2u

∂x2

∣∣∣∣∣xi

+(∆x)2

3!

∂3u

∂x3

∣∣∣∣∣xi

+ · · · =∆x

2

∂2u

∂x2

∣∣∣∣∣ξ

where xi ≤ ξ ≤ xi+1. Thus

∂u

∂x

∣∣∣∣xi

≈ui+1 − ui

∆x+

∆x

2

∂2u

∂x2

∣∣∣∣∣ξ

or∂u

∂x

∣∣∣∣xi

−ui+1 − ui

∆x≈

∆x

2

∂2u

∂x2

∣∣∣∣∣ξ

(11)



In general, ξ is not known. Since ∆x is the only parameter under the user’s control that

determines the error, the truncation error is simply written

∆x

2

∂2u

∂x2

∣∣∣∣∣ξ

= O(∆x)

This expression means that the left hand side is bounded by a product of an unknown

constant times ∆x. Although the expression does not give us the exact magnitude of

(∆x/2) (∂2u/∂x2)∣∣ξ, it indicates how quickly that term approaches zero as ∆x is

reduced.

Recall that the point of Big O notation is to look at the exponent of the truncation errorterm.

We conclude that the error in the first order forward difference formula is linear in the

mesh spacing.



Using big O notation, Equation (11) is written

∂u

∂x

∣∣∣∣xi

=ui+1 − ui

∆x+O(∆x). (12)

Equation (12) is called the forward difference formula for ∂u/∂x|xi because it involvesnodes xi and xi+1.


First order backward finite difference

Repeat the preceding analysis with a Taylor series expansion for u(xi − δx)

ui−1 = ui −∆x∂u

∂x

∣∣∣∣xi

+∆x2

2

∂2u

∂x2

∣∣∣∣∣xi

−(∆x)3

3!

∂3u

∂x3

∣∣∣∣∣xi

+ · · · (13)

Solve for ∂u/∂x|xi to get

∂u

∂x

∣∣∣∣xi

=ui − ui−1

∆x+

∆x

2

∂2u

∂x2

∣∣∣∣∣xi

−(∆x)2

3!

∂3u

∂x3

∣∣∣∣∣xi

+ · · ·

or

∂u

∂x

∣∣∣∣xi

=ui − ui−1

∆x+O(∆x). (14)

This is called the backward difference formula because it involves the values of u at xiand xi−1.


First order central difference

Combine Taylor Series expansions for ui+1 and ui−1 to get

∂u

∂x

∣∣∣∣xi

=ui+1 − ui−1

2∆x+O(∆x2) (15)


Summary of first order difference approximations

Forward:∂u

∂x

∣∣∣∣xi

=ui+1 − ui

∆x+O(∆x)

Backward:∂u

∂x

∣∣∣∣xi

=ui − ui−1

∆x+O(∆x)

Central:∂u

∂x

∣∣∣∣xi

=ui+1 − ui−1

2∆x+O(∆x2)


Second Order Central Difference

The Taylor series expansion for ui+1 is

ui+1 = ui + ∆x∂u

∂x

∣∣∣∣xi

+∆x2

2

∂2u

∂x2

∣∣∣∣∣xi

+(∆x)3

3!

∂3u

∂x3

∣∣∣∣∣xi

+ · · · (16)

Adding Equation (13) and Equation (16) yields

ui+1 + ui−1 = 2ui + ∆x2 ∂

2u

∂x2

∣∣∣∣∣xi

+2∆x4

4!

∂4u

∂x4

∣∣∣∣∣xi

+ · · ·

Notice that all odd derivatives cancel because of the alternating signs of terms on the

right hand side of Equation (13).


Second Order Central Difference

Solving for∂2u

∂x2

∣∣∣∣∣xi

gives

∂2u

∂x2

∣∣∣∣∣xi

=ui−1 − 2ui + ui+1

∆x2+

∆x2

6

∂4u

∂x4

∣∣∣∣∣xi

+ · · ·

or

∂2u

∂x2

∣∣∣∣∣xi

=ui−1 − 2ui + ui+1

∆x2+O(∆x2). (17)


Documents

Introduction to Finite Di erence Methods · 2020. 1. 4. · Substituting Equation (6.c) into Equation (5) gives A n= 2 L Z L 0 sin ˇx L sin nˇx L dx: The sine function is orthogonal