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Finite difference method

From Wikipedia, the free encyclopedia

In mathematics, finite-difference methods (FDM) are numerical methods for solving differential equations by approximating them with difference equations, in which finite differences approximate the derivatives. FDMs are thus discretization methods. FDMs convert a linear (non-linear) ODE (Ordinary Differential Equations) /PDE (Partial differential equations) into a system of linear (non-linear) equations, which can then be solved by matrix algebra techniques. The reduction of the differential equation to a system of algebraic equations makes the problem of finding the solution to a given ODE ideally suited to modern computers, hence the widespread use of FDMs in modern numerical analysis[1].

Today, FDMs are the dominant approach to numerical solutions of partial differential equations.[1]

YouTube Encyclopedic

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  • ✪ Finite Difference Method for Solving ODEs: Example: Part 1 of 2
  • ✪ Topic 7a -- One-dimensional finite-difference method
  • ✪ MIT Numerical Methods for Partial Differential Equations Lecture 1: Convection Diffusion Equation
  • ✪ Shift Operator (E) II Finite Differences
  • ✪ Application of Finite Differences in Newton-Raphson's Method | Programming Numerical Methods (2018)


. . In this segment we're going to take an example of Finite Difference Method of solving boundary value ordinary differential equations. So let's go and see that how we can use the theory behind the finite difference methods to solve ordinary differential equation boundary value problems. And we are going to look at an example of doing that. So let's go and take a simple, so this is the problem statement which is given to us for the example. You have a second order differential equation which is of this particular form. . So you are given this and you are given the boundary conditions here that the value of u at 2 is 0.008 and the value of u at 6.5 is 0.003. So what you want to be able to do is, you want to be able to find out what u is as a function of r because u is the dependent variable and r is the independent variable and you want to be able to find what u is at the different points. In order to keep the problem simple, so somebody's asking so basically the problem statement is find u as a function of, function of r. But since we are given the numerical methods we cannot find u as a function of r at every data point, so somebody might say hey, use 4 nodes. Use 4 nodes to do the problem. Let me say 4, equidistant. Equidistant meaning that the nodes have the same distance between each other. So what that implies is that we need to break up our interval which is going from 2 to 6.5 into 4 nodes to be able to solve the problem. So I am here, so let's suppose if I am here, at r equal to 2 and then here I am at r equal to 6.5. So if I am going to break it up into 4 nodes I'll have another node here and another node right here so that means that three segments here. So the segment width will be divided by 3, the difference between the two. So if I want to calculate delta r it's just (6.5-2)/3 which is 1.5. And I have four nodes here so this will become the first, second node will be at r + delta r which will be at 3.5 and the next node will be at 3.5 plus delta r which will be at 5.0 so that's where the next node is going to be at 5.0. And what I am going to do is I'm going to number these nodes. I am going to number this node as 1, number this node as 2, and number this node as 3, and number this node as 4 and that's how I'll be designating that. So I already know what the value of u at 1 is because it is already given to me. The value of u is already given to me as 0.008 and I already know what the value at 4 is which is given to me as the value at r=6.5 is 0.003. So what that implies that I have to find out what the value of the u is, the dependent variables at the second node and the third node and if I am able to do that, that means that I have been able to solve the problem. That I can draw the profile of u, then I can do things like interpolation or spline interpolation to be able to find the value of u at some other point approximately. Even the values which I get at 2 and 3 are known approximately because I will be using approximations for my derivatives. So let's go and see that how do we go about doing this problem here by using Finite Difference Methods is by, what I am going to do is I am going to say hey, d2u/dr2 which is one of the terms in the ordinary differential equation which was given to me, that I can approximate it at any node i. So if I have a node i because I have to approximate this second derivative at each node, what is this approximation equal to? So this approximation will be equal to u_1+1 - 2(u_i) + u_i-1, so you're basically taking the information of the dependent variable at the node ahead and the node where you are and the node behind, divided by delta r squared. And that gives you the approximation of the second derivative. So, again, there are different approximations that you can choose, this is the central divided difference approximation of the second derivative of the function and then also what I am going to do is, another derivative; du/dr which I need to approximate at node i. And let's suppose I am going to use a forward divided difference scheme for that, so I am going to say u_i+1 - u_i divided by delta r. I am going to use that as my approximation to do that. So what that implies is that what I am trying to do is to change these derivative into unknown still, I still have unknowns of u_i, u_i-1 and things like that but now there are unknowns at specific points and that is what numerical methods is all about, being able to convert a problem into, into basic operations of multiplication, addition, division, and subtraction, that's what you are doing, that you are setting these up so that you can get simultaneous linear equations. So let's go and see how we end up doing that. So the ordinary differential equation woudl turn out to be [u_i+1 - 2(u_i) + u_i-1]/(delta r)^2 [u_i+1 - 2(u_i) + u_i-1]/(delta r)^2, because that is the approximation of the first derivative, plus (1/r), so since you are writing the equation at a particular node it will be r_i, that's what I will get there, times (u_i+1 - u_i)/(delta r) and then you have minus u/r^2, so since we are writing the equation at node i would be u_i and since we are writing equation node i, it will be the value of r which you have at that point equal to 0. So that's what you have reduced your ordinary differential equation into but you can see that it's not just solving one equation, one unknown at a time because you have, let's suppose, in this case at node i you have 3 unknowns. You have this, this, and this so you cannot solve three equations, three unknowns at this, three equations, so three unknowns in one equation in a single instance. You will need multiple equations to be able to, to be able to do that. So that's why what we are going to do is we are going to write down this equation at node 2 and 3 because we already know what the value at node 1 and node 4 is because those are the boundary conditions which are given to us. So if we write down this equation for node 2 and 3 we will be able to set up two equations and 4 unknowns but since we know the value of the function at the first node and the, and the last node, that will set us up a four equation, four unknowns. So let's go and see how we go about doing that. So I am going to write down this equation at, what I am going to do is I am going to write it one at a time, so node i, so which is node 1, so let me look at node 1. Node 1, I have u_1 is same as u at, at 2, is the value of the displacement of u at 2, it is 0.008. So that's my first equation. Now if I write down the equation at node 2, let me see what do I get from there, I get, I'll put i equal to 2 in there. I'll get u_3 minus 2(u_2) + u_1, I am substituting i=2 now, and that's what I will get when I subtitute it there, divided by delta r. Delta r is nothing but 1.5, squared, plus 1 divided by r_i which is r_2, (u_3 - u_2)/(delta r), so maybe I should just keep it as delta r for the time being here, divided by delta r, minus u_i which is u_2 divided by (r_i)^2 which is (r_2)^2 equal to, equal to 0. So since I am getting that, what that means is that I got to substitute the values of delta r and r_2 which I know, so I will get (u_3 - 2(u_2) + u_1)/(delta r)^2 which is what, 1.5 squared. plus (1/r_2) which is what, r_2 is 3.5, times (u_3 - u_2) divided by delta r, which is 1.5 squared, minus u_2 divided by r_2 squared which is (3.5)^2 equal to 0. If I expand this I'll be getting 0.444 for u_1, so combining all the u_1 terms, combining all the u_2 terms, and then combining all the u_3 terms and that's what I'll get as my equation number 2. And that's the end of this segment. . . .


Derivation from Taylor's polynomial

First, assuming the function whose derivatives are to be approximated is properly-behaved, by Taylor's theorem, we can create a Taylor series expansion

where n! denotes the factorial of n, and Rn(x) is a remainder term, denoting the difference between the Taylor polynomial of degree n and the original function. We will derive an approximation for the first derivative of the function "f" by first truncating the Taylor polynomial:

Setting, x0=a we have,

Dividing across by h gives:

Solving for f'(a):

Assuming that is sufficiently small, the approximation of the first derivative of "f" is:

Accuracy and order

The error in a method's solution is defined as the difference between the approximation and the exact analytical solution. The two sources of error in finite difference methods are round-off error, the loss of precision due to computer rounding of decimal quantities, and truncation error or discretization error, the difference between the exact solution of the original differential equation and the exact quantity assuming perfect arithmetic (that is, assuming no round-off).

The finite difference method relies on discretizing a function on a grid.
The finite difference method relies on discretizing a function on a grid.

To use a finite difference method to approximate the solution to a problem, one must first discretize the problem's domain. This is usually done by dividing the domain into a uniform grid (see image to the right). This means that finite-difference methods produce sets of discrete numerical approximations to the derivative, often in a "time-stepping" manner.

An expression of general interest is the local truncation error of a method. Typically expressed using Big-O notation, local truncation error refers to the error from a single application of a method. That is, it is the quantity if refers to the exact value and to the numerical approximation. The remainder term of a Taylor polynomial is convenient for analyzing the local truncation error. Using the Lagrange form of the remainder from the Taylor polynomial for , which is

, where ,

the dominant term of the local truncation error can be discovered. For example, again using the forward-difference formula for the first derivative, knowing that ,

and with some algebraic manipulation, this leads to

and further noting that the quantity on the left is the approximation from the finite difference method and that the quantity on the right is the exact quantity of interest plus a remainder, clearly that remainder is the local truncation error. A final expression of this example and its order is:

This means that, in this case, the local truncation error is proportional to the step sizes. The quality and duration of simulated FDM solution depends on the discretization equation selection and the step sizes (time and space steps). The data quality and simulation duration increase significantly with smaller step size.[2] Therefore, a reasonable balance between data quality and simulation duration is necessary for practical usage. Large time steps are useful for increasing simulation speed in practice. However, time steps which are too large may create instabilities and affect the data quality.[3][4]

The von Neumann and Courant-Friedrichs-Lewy criteria are often evaluated to determine the numerical model stability.[3][4][5][6]

Example: ordinary differential equation

For example, consider the ordinary differential equation

The Euler method for solving this equation uses the finite difference quotient

to approximate the differential equation by first substituting it for u'(x) then applying a little algebra (multiplying both sides by h, and then adding u(x) to both sides) to get

The last equation is a finite-difference equation, and solving this equation gives an approximate solution to the differential equation.

Example: The heat equation

Consider the normalized heat equation in one dimension, with homogeneous Dirichlet boundary conditions

(boundary condition)
(initial condition)

One way to numerically solve this equation is to approximate all the derivatives by finite differences. We partition the domain in space using a mesh and in time using a mesh . We assume a uniform partition both in space and in time, so the difference between two consecutive space points will be h and between two consecutive time points will be k. The points

will represent the numerical approximation of

Explicit method

The stencil for the most common explicit method for the heat equation.
The stencil for the most common explicit method for the heat equation.

Using a forward difference at time and a second-order central difference for the space derivative at position (FTCS) we get the recurrence equation:

This is an explicit method for solving the one-dimensional heat equation.

We can obtain from the other values this way:


So, with this recurrence relation, and knowing the values at time n, one can obtain the corresponding values at time n+1. and must be replaced by the boundary conditions, in this example they are both 0.

This explicit method is known to be numerically stable and convergent whenever .[7] The numerical errors are proportional to the time step and the square of the space step:

Implicit method

The implicit method stencil.
The implicit method stencil.

If we use the backward difference at time and a second-order central difference for the space derivative at position (The Backward Time, Centered Space Method "BTCS") we get the recurrence equation:

This is an implicit method for solving the one-dimensional heat equation.

We can obtain from solving a system of linear equations:

The scheme is always numerically stable and convergent but usually more numerically intensive than the explicit method as it requires solving a system of numerical equations on each time step. The errors are linear over the time step and quadratic over the space step:

Crank–Nicolson method

Finally if we use the central difference at time and a second-order central difference for the space derivative at position ("CTCS") we get the recurrence equation:

This formula is known as the Crank–Nicolson method.

The Crank–Nicolson stencil.
The Crank–Nicolson stencil.

We can obtain from solving a system of linear equations:

The scheme is always numerically stable and convergent but usually more numerically intensive as it requires solving a system of numerical equations on each time step. The errors are quadratic over both the time step and the space step:

Usually the Crank–Nicolson scheme is the most accurate scheme for small time steps. The explicit scheme is the least accurate and can be unstable, but is also the easiest to implement and the least numerically intensive. The implicit scheme works the best for large time steps.


The figures below present the solutions given by the above methods to approximate the heat equation

with the boundary condition

The exact solution is

Comparison of Finite Difference Methods
Explicit method (not stable)
Implicit method (stable)
Crank-Nicolson method (stable)

Example: The Laplace operator

The (continuous) Laplace operator in -dimensions is given by . The discrete Laplace operator depends on the dimension .

In 1D the Laplace operator is approximated as

This approximation is usually expressed via the following stencil

and which represents a symmetric, tridiagonal matrix. For an equidistant grid one gets a Toeplitz matrix.

The 2D case shows all the characteristics of the more general nD case. Each second partial derivative needs to be approximated similar to the 1D case

which is usually given by the following stencil


Consistency of the above-mentioned approximation can be shown for highly regular functions, such as . The statement is

To proof this one needs to substitute Taylor Series expansions up to order 3 into the discrete Laplace operator.



Similar to continuous subharmonic functions one can define subharmonic functions for finite-difference approximations

Mean value

One can define a general stencil of positive type via

If is (discrete) subharmonic then the following mean value property holds

where the approximation is evaluated on points of the grid, and the stencil is assumed to be of positive type.

A similar mean value property also holds for the continuous case.

Maximum principle

For a (discrete) subharmonic function the following holds

where are discretizations of the continuous domain , respectively the boundary .

A similar maximum principle also holds for the continuous case.

See also


  1. ^ a b Christian Grossmann; Hans-G. Roos; Martin Stynes (2007). Numerical Treatment of Partial Differential Equations. Springer Science & Business Media. p. 23. ISBN 978-3-540-71584-9.
  2. ^ Arieh Iserles (2008). A first course in the numerical analysis of differential equations. Cambridge University Press. p. 23. ISBN 9780521734905.
  3. ^ a b Hoffman JD; Frankel S (2001). Numerical methods for engineers and scientists. CRC Press, Boca Raton.
  4. ^ a b Jaluria Y; Atluri S (1994). "Computational heat transfer". Computational Mechanics. 14: 385–386. doi:10.1007/BF00377593.
  5. ^ Majumdar P (2005). Computational methods for heat and mass transfer (1st ed.). Taylor and Francis, New York.
  6. ^ Smith GD (1985). Numerical solution of partial differential equations: finite difference methods (3rd ed.). Oxford University Press.
  7. ^ Crank, J. The Mathematics of Diffusion. 2nd Edition, Oxford, 1975, p. 143.

Various lectures and lecture notes

This page was last edited on 7 August 2019, at 14:39
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