California State University, Northridge
[pic]
|College of Engineering and Computer Science
Mechanical Engineering Department
Engineering Analysis Notes | |
| |Larry Caretto March 14, 2009 |
Numerical Analysis of Partial Differential Equations
Solution Properties for Finite-difference Equations
A numerical solution to an ordinary or a partial-differential equation should satisfy various properties. These are listed below.
Consistency. A finite-difference expression is consistent with the differential equation if the truncation error approaches zero (ignoring roundoff) as the discrete steps (in space and time) approach zero. This is usually the case for finite-difference expressions. However, there are some algorithms, such as the DuFort-Frankel algorithm, in which the truncation error depends on the ratio (Δt/Δx)2, that are only conditionally consistent.
Stability. A finite-difference equation should be stable. The errors in a stable finite-difference equation will not grow without bound as the solution progresses.
Convergent. A convergent finite-difference expression tends to the exact solution of the differential equation as the grid size tends towards zero. According to the Lax Equivalence Theorem, a properly posed linear initial value problem and a finite difference approximation to it that satisfies the consistency condition will be convergent if it is stable.
Physical reality. Solutions should produce physically realistic results. Densities should be positive. Temperature changes should not violate the second law of thermodynamics. This should be true for each node in the solution. This requirement applies not only to the numerical method, but to physical models for complex flow phenomena such as turbulence, combustion, gaseous radiation, and multiphase flow.
Accuracy. There are many sources of error in numerical solutions. We have discussed truncation errors caused by the numerical approaches. Additional errors, known as iteration errors, are possible when approximate solutions to the finite-difference equations are obtained. Furthermore, inaccuracies can be introduced by poor physical models or assumptions. For example, the solution of a potential flow problem will have possible errors from truncation and iteration error. However, the assumption of potential flow can introduce errors as compared to the actual physical problem. However, these errors may be acceptable in some cases.
Finite-difference methods and stability for the diffusion equation
The notes on the entitled “Introduction to Numerical Calculus”, referred to here as “introductory notes”, we applied finite-difference and finite-element methods to a simple ordinary differential equation. The extension of the finite-difference approach used there to partial differential equations is fairly straightforward. As an example of this, consider the following differential equation, known as the diffusion equation or the heat conduction equation.
[pic] [3-1]
The quantity α represents the thermal diffusivity in heat transfer and the diffusion coefficient in diffusion problems. We will call this term the diffusivity. The dependent variable T can be a general potential, although here we are using the usual symbol for temperature. Equation [3-1] has an open boundary in time; we do not have to specify the conditions at a future time. This equation is formally classified as a parabolic partial differential equation.
We need an initial condition for the temperature at all points in the region. We can write a general initial condition as T(x,0) = T0(x). Similarly we can write arbitrary boundary conditions for the boundaries at xmin and xmax: T(xmin,t) = TL(t) and T(xmax,t) = TR(t). These conditions, as well as values for the geometry and the diffusivity must be specified before we can solve the differential equation.
We can construct finite-difference grids in space and time. For simplicity, we will assume constant step sizes, Δt and Δx. We define our time and space coordinates on this grid by the integers n and i, so that we can write
[pic] [3-2]
We can obtain a solution process, known as the explicit method, if we use a forward difference for the time derivative and a central difference for the space derivatives. This method is also called the forward-time-central-space (FTCS) method to emphasize the nature of the finite-difference approximations to the time and space derivatives in the diffusion equation. The forward-time and central-space approximations are given by the following equations.
[pic] [3-3]
These equations are modifications of equations [19] and [29] from the introductory notes on numerical calculus. In those notes we dealt with ordinary derivatives and needed only one subscript for the dependent grid variable. Here, in a multidimensional case, we have dependent variable, T, as a function of time and distance, T(x,t). Thus we define Tin = T(xi, tn). We use differences between variables like Tin, T ni+1, and Tni-1 to give differences in the x-direction that are used to compute spatial derivatives. The constant n superscript in the space derivative is an indication that we are holding time constant and varying distance. Similarly, the time derivatives are reflected in the finite-difference form by using variations in the superscript n while holding the subscript i constant.
The forward difference for the time derivative is chosen over the more accurate central difference to provide a simple algorithm for solving this problem. There are other problems with the use of the central difference expression for the time derivative that we will discuss later.
If we substitute the finite-difference expressions in [3-3] into equation [3-1], we get the following result.
[pic] [3-4]
We have written the combined error term to show the order of the error for both Δx and Δt. We have not written these as separate terms since O() notation simply shows how the error depends on the power of the step size. If we neglect the truncation error, we can obtain a simple equation for the value of the temperature at the new time step.
[pic] [3-5]
We see that the grid spacings and the thermal diffusivity appear in a single term. We can define this as a single term to save writing.
[pic] [3-6]
With this definition, equation [3-5] may be written as follows.
[pic] [3-7]
With this finite difference equation, each value of T at the new time step is related only to T values at the previous time step.[1] These are the temperature at the same spatial location and the two neighboring T values in the grid. Equation [3-7] is illustrated below.
|[pic] | | |[pic] | | |[pic] |
| |times f |+ |times (1 – 2f) |+ |times f | |
| | | |= | | | |
| | | |[pic] | | | |
From the initial conditions, we know all the T values at the first time step. That is, Ti0 equals the given initial conditions at x, T0(xi). Similarly, the T values at i = 0, and i = imax are known from the boundary conditions. We can illustrate this problem for a simple case where the initial conditions and the boundary conditions are constants. We define the boundaries of our one-dimensional region as x = 0 and x = L. The initial value, denoted as Tinit, is the same for all values of x. The constant boundary value at x = 0, for all times greater than zero, will be called T0, and the boundary value at x = L will be denoted as TL. We can compare our numerical results for this problem with the analytical solution for both the value of T and its gradient. The analytical solution to equation [3-1] for the constant boundary conditions used here, is given by the following equations.[2]
[pic] [3-8]
This analytical solution can be used to obtain the value of T and its gradient, at any time and space coordinate, for any given values of (, L, Tinit, T0, and TL. Note that the gradient of T usually represents a physical flux such as the heat flux or the diffusion flux.
In order to obtain a numerical solution, which is only obtained at discrete points, we must specify numerical values of (, L, Tinit, T0, and TL. For simplicity let us choose ( = 1, L = 1, Tinit = 1000 and T0 = TL = 0. We will start with a small computational grid using only five points in the x direction. This gives Nx = 4 and Δx = 0.25. We also pick a time step, Δt = 0.01 and calculate to a maximum time of 0.2. This requires 20 time steps.
Table 3-1 below shows the results of the numerical calculations with the spatial variation in columns and the time variation in rows. There are two initial condition rows. The “t = 0” row shows the initial conditions. The “t = 0+” row shows the temperatures immediately after t = 0 when the boundary conditions are applied. This second row is the starting time point, n = 0, for the numerical calculations. These two initial condition rows illustrate the discontinuity in the T values at the two points (0,0) and (L,0) in the solution domain. This discontinuity does not affect our ability to obtain an analytical solution. We must treat it explicitly in the numerical solution.
In addition to the modified initial condition row, where we know all the values of T, we have the boundary condition columns. For all time points at x = 0 and x = L = 1, we set the table entries to the known boundary values. Although the initial and boundary conditions are particularly simple in this example, we could handle initial conditions that were arbitrary functions of x and boundary conditions that were arbitrary conditions of time by placing the appropriate results for the initial and boundary temperatures at the correct places in the table.
Once the modified initial condition row and the boundary condition columns are set, the solution for the remaining temperatures, using equation [3-7] is straightforward. We first use equation [3-6] to compute the value of f.
[pic] [3-9]
Once the value of f is known, we can apply equation [3-7] to each node, in successive rows. The calculations below are for the first time step.
[pic] [3-10]
These are the results shown in the “n = 1” row of the table. The numerical results are symmetric in x as would be expected from the uniform initial condition and the symmetric boundary condition. The process illustrated in equation [3-10] is used to calculate the remaining results in the table. The final two entries in the table are the exact results, calculated for the different values of x at the final time t = 0.2, and the error. The latter is the absolute value of the difference between the numerical and analytical solution.
The numerical calculation of the final values for an end time of t = 0.2 can be repeated for time steps of 0.02 and 0.04. These results are shown in the Tables 3-2 and 3-3 below. When the time step is doubled from 0.01 to 0.02 the solution results are similar. However, the error in the temperatures at the final time step has increased by about a factor of three.
When the time step is increases to 0.04, the results are completely different, and these results do not match our ideas of physical reality. We expect that the region will eventually reach a constant value of T equal to the two boundary temperatures of zero. As this diffusion process takes place, we do not expect to see any T values outside the range bounded by the initial value of 1000 and the boundary value of 0.
|Table 3-1 |
|Numerical solution of the Diffusion Equation |
|( = 1, Δx = 0.25, Δt = 0.01, tmax = 0.20 |
| |i = 0 |i = 1 |i = 2 |i = 3 |i = 4 |
| |x = 0.00 |x = 0.25 |x = 0.50 |x = 0.75 |x = 1.00 |
| |t = 0 |1000 |1000 |1000 |1000 |1000 |
|n = 0 |t = 0+ |0 |1000 |1000 |1000 |0 |
|n = 1 |t = 0.01 |0 |840 |1000 |840 |0 |
|n = 2 |t = 0.02 |0 |731.2 |948.8 |731.2 |0 |
|n = 3 |t = 0.03 |0 |649.0 |879.2 |649.0 |0 |
|n = 4 |t = 0.04 |0 |582.0 |805.5 |582.0 |0 |
|n = 5 |t = 0.05 |0 |524.6 |734.0 |524.6 |0 |
|n = 6 |t = 0.06 |0 |474.2 |667.0 |474.2 |0 |
|n = 7 |t = 0.07 |0 |429.2 |605.3 |429.2 |0 |
|n = 8 |t = 0.08 |0 |388.7 |548.9 |388.7 |0 |
|n = 9 |t = 0.09 |0 |352.1 |497.7 |352.1 |0 |
|n = 10 |t = 0.10 |0 |319.1 |451.1 |319.1 |0 |
|n = 11 |t = 0.11 |0 |289.2 |408.9 |289.2 |0 |
|n = 12 |t = 0.12 |0 |262.0 |370.5 |262.0 |0 |
|n = 13 |t = 0.13 |0 |237.5 |335.8 |237.5 |0 |
|n = 14 |t = 0.14 |0 |215.2 |304.4 |215.2 |0 |
|n = 15 |t = 0.15 |0 |195.0 |275.8 |195.0 |0 |
|n = 16 |t = 0.16 |0 |176.8 |250.0 |176.8 |0 |
|n = 17 |t = 0.17 |0 |160.2 |226.5 |160.2 |0 |
|n = 18 |t = 0.18 |0 |145.2 |205.3 |145.2 |0 |
|n = 19 |t = 0.19 |0 |131.6 |186.1 |131.6 |0 |
|n = 20 |t = 0.20 |0 |119.2 |168.6 |119.2 |0 |
|Exact |t = 0.20 |0 |125.1 |176.9 |125.1 |0 |
|Error |t = 0.20 |0 |5.8 |8.2 |5.8 |0 |
|Table 3-2 |
|Numerical solution of the Conduction Equation |
|( = 1, Δx = 0.25, Δt = 0.02, tmax = 0.20 |
| |i = 0 |i = 1 |i = 2 |i = 3 |i = 4 |
| |x = 0.00 |x = 0.25 |x = 0.50 |x = 0.75 |x = 1.00 |
| |t = 0 |1000 |1000 |1000 |1000 |1000 |
|n = 0 |t = 0+ |0 |1000 |1000 |1000 |0 |
|n = 1 |t = 0.02 |0 |680 |1000 |680 |0 |
|n = 2 |t = 0.04 |0 |564.8 |795.2 |564.8 |0 |
|n = 3 |t = 0.06 |0 |457.9 |647.7 |457.9 |0 |
|n = 4 |t = 0.08 |0 |372.1 |526.2 |372.1 |0 |
|n = 5 |t = 0.10 |0 |302.3 |427.6 |302.3 |0 |
|n = 6 |t = 0.12 |0 |245.7 |347.4 |245.7 |0 |
|n = 7 |t = 0.14 |0 |199.6 |282.3 |199.6 |0 |
|n = 8 |t = 0.16 |0 |162.2 |229.4 |162.2 |0 |
|n = 9 |t = 0.18 |0 |131.8 |186.4 |131.8 |0 |
|n = 10 |t = 0.20 |0 |107.1 |151.4 |107.1 |0 |
|Exact |t = 0.20 |0 |125.1 |176.9 |125.1 |0 |
|Error |t = 0.20 |0 |18.0 |25.4 |18.0 |0 |
|Table 3-3 |
|Numerical solution of the Conduction Equation |
|( = 1, Δx = 0.25, Δt = 0.04, tmax = 0.20 |
| |i = 0 |i = 1 |i = 2 |i = 3 |i = 4 |
| |x = 0.00 |x = 0.25 |x = 0.50 |x = 0.75 |x = 1.00 |
| |t = 0 |1000 |1000 |1000 |1000 |1000 |
|n = 0 |t = 0+ |0 |1000 |1000 |1000 |0 |
|n = 1 |t = 0.04 |0 |360 |1000 |360 |0 |
|n = 2 |t = 0.08 |0 |539.2 |180.8 |539.2 |0 |
|n = 3 |t = 0.12 |0 |-35.3 |639.6 |-35.3 |0 |
|n = 4 |t = 0.16 |0 |419.2 |-224.2 |419.2 |0 |
|n = 5 |t = 0.20 |0 |-260.9 |599.3 |-260.9 |0 |
|Exact |t = 0.20 |0 |125.1 |176.9 |125.1 |0 |
|Error |t = 0.20 |0 |385.9 |422.5 |385.9 |0 |
The solutions with a time step of 0.04 have some negative values of T in the solution. You can verify that these unexpected negative values are correctly computed from the finite difference equation. There is something obviously wrong with these results, but there is no obvious problem. We have used the same finite difference equation that had previously produced correct results. The problem that we are facing here results from the translation of the differential equation into a finite difference equation. It is an issue known as stability. Numerical algorithms that are not stable can produce erroneous results whose errors grow without bound.
We can do a heuristic analysis of stability for the algorithm considered here. This is a heuristic analysis because it uses physical reasoning rather than examining the error growth of the finite difference equations. This argument proceeds by considering the physical implications of the finite-difference equation, [3-7].
[pic] [3-7]
This equation shows that the T value at the new time step, for any given value of xi depends on the value of T at the previous time step, at the same value of xi, multiplied by the factor (1 – 2f). Depending on the value of f, the (1 – 2f) factor may be positive, negative or zero. If the factor is negative, we have a physically unrealistic result. For given values of T at the neighboring nodes, the new value of T at xi will be smaller, if the T value at the previous time step is larger. This is not correct. We expect that larger values of T at xi at one time step should lead to larger values of T at xi at the next time step. Thus, we do not want to use negative values for the factor (1 – 2f). This sets the following limit on our numerical calculations.
[pic] [3-11]
In the three sets of calculations above the value of f was 0.16, 0.32, and 0.64. Thus, the third set of calculations violated the stability limit. Because we used an inadmissible value of f, we obtained incorrect results from our calculation. The stability limit becomes a more serious concern when we are dealing with very small values of Δx. We would have to choose a very small value of Δt to satisfy the stability condition.
The stability limit on Δt differs from the need to choose a small Δt for accuracy. The stability limit places a significant restriction on Δt. If we were to half the step size, Δx, to obtain a more accurate solution, we would have to cut Δt by ¼. Thus, doubling the number of space steps requires a factor of four increase in the time steps or an overall factor of eight in the total nodes in the calculation. This requirement for stability is separate from the requirement for accuracy. Accuracy depends on small sizes for Δt and Δx. We may want to use a larger step size for Δt to obtain a desired accuracy, however we cannot use a step size that violates the stability condition in equation [3-11] if use the explicit method.
The appendix to this section of notes contains a derivation for the truncation error for this approach. There are two equivalent equations for this truncation error. Both of these are written as an infinite series in equation [3-12].
[pic] [3-12]
We see that the truncation error depends in a complex way on the two finite-difference grid spacings, Δt and Δx. Choosing f = 1/6 eliminates the first (k = 2) term in the truncation error and gives a higher order truncation error.
The method just described is called the explicit method because all temperatures at the new time step can be written as an explicit function of temperatures at the old time step. Alternative, implicit approaches use more than one value of the temperature at the new time step in the finite-difference equation. The Crank-Nicholson method is the most popular implicit method. This method uses a finite-difference representation of the conduction equation at a time point midway between the two specified time grid lines. Although there is no computational point there, we can write this time derivative as follows, using a central-difference approximation. In this expression the step size is Δt/2, the distance from the location of the derivative to the location of each of the two points used in its computation.
[pic] [3-13]
Although this midpoint location gives us an easy way to compute the time derivative, there is no such easy way to compute the space derivative. We can use an average of the finite difference expression for the space derivative at the two time points, n and n+1. Such an average produces a truncation error that can be seen by adding the Taylor series for fi+1 and fi-1 from equations [2-15] and [2-18] in the notes on basics.
[pic] [3-14]
When we solve the last equation above for fi, we get the truncation error that arises when we use the arithmetic average of two neighboring points (fi+1 and fi-1) to approximate the value at the midpoint.
[pic] [3-15]
Applying this result to the [pic]term gives the following result.
[pic] [3-16]
In this case the truncation error is proportional to [(Δt)2] because we are taking an average of this derivative at two time points to represent the derivative at the midpoint. We can use the second-order equation for the second derivative for each term in this average. Doing this and combining equations [3-16] and [3-13] gives
[pic] [3-17]
The finite difference that we obtain from equation [3-17] contains three temperatures at the new time point. We can rewrite this equation, introducing the factor f defined in equation [3-6] as follows.
[pic] [3-18]
This equation describes a tridiagonal system of simultaneous linear algebraic equations. As we saw in the notes on basics, this set of equations is easily solved by the Thomas algorithm.
A final approach, known as the fully implicit approach, uses we use a backward difference for the time derivative and a central difference for the space derivatives at the point (xi,, tn+1). In this approach we write the two derivatives as follows.
[pic] [3-19]
Substituting these finite difference expressions into the differential equation gives the following result.
[pic] [3-20]
Rearranging the finite-difference terms (ignoring the truncation error) and using our usual definition of f = αΔt/(Δx)2, gives the following result, after some algebra.
[pic] [3-21]
We can use our heuristic approach to the analysis of stability for equation [3-21]. Here we see that the signs of the coefficients of the Ti terms, at both time steps, have the same signs for any value of f. (The value of f must always be positive. The diffusivity is positive and we will always choose Δt > 0.) Because Tin and Tin+1 both have the same sign, any increase in Tin will give an increase in Tin+1. Thus, we expect the fully implicit method to be stable for all values of f.
We cannot apply the heuristic analysis to the Crank-Nicholson method in equation [3-18], because we cannot clearly isolate the effect of the current and future T values at the same node. A formal stability analysis shows that the Crank-Nicholson method is stable for any step size. However, it produces physically unrealistic results if the time step is too large.
Accuracy of the methods
The truncation error analysis in the appendix to this section shows that the accuracy of the various methods for the solution of the conduction equation depends in a complex way on the interactions of the step sizes Δx and Δt. We expect that the Crank-Nicholson method will be more accurate because it has a higher order truncation error. However, we did see that the truncation error of the explicit method would be the same as that of the Crank-Nicholson method if we chose f = 1/6. A comparison of the three methods outlined here would have to consider the computer time required for their implementation. The explicit method would require less computer time per time step in the calculation because it does not require the use of the Thomas algorithm.
The implicit method and the Crank-Nicholson method should require about the same amount of computer time. (The Crank-Nicholson method will be slightly longer because of its more complicated right-hand-side term in the tridiagonal set of equations.) The accuracy of these two methods can be compared for an example problem using the same geometry, thermal diffusivity, and boundary and initial conditions as used in previous sample calculations. (0 ≤ x ≤ L; α = 1;Tinit = 1000; T0 = TL = 0.)
Table 3-4 shows a small set of results for the Crank-Nicholson method. These calculations are done for Δx = 0.01 and Δt = 0.0005. The value of f in this case is (1)(0.005)/(0.01)2 = 5. With f > 0.5, an explicit calculation is not possible. All time steps, up to a final time of 0.0125, are in Table 3-4, but only nodes 1-4 in the x direction are shown in the table. The results for nodes 96 through 100 are mirror images of the results for nodes one through four.
For the first time step in Table 3-4, the node nearest the border has a negative temperature. This leads to severe oscillations in the temperatures at this spatial node. These die out by the time step, n = 25. There are slight oscillations in the second node in from the border as well. The third node in from the border does not have any oscillations, but the temperature change between the time steps n = 2 and n = 3 is very small compared to the temperature changes for the previous and subsequent time steps. This is a sign that the effects of the temperature oscillations near the border are penetrating into the body of the Crank-Nicholson calculation. (These same results are also present in the symmetric border at x = L.) The Crank-Nicholson calculation is stable because these initial oscillations die out. If the method were not stable, these oscillations would grow without bound.
Table 3-5 shows the partial results for the fully implicit method for the same set of nodes used in Table 3-4. Here there are no physically unrealistic oscillations. Although these results are more realistic, the error in this method is greater than the error in the Crank-Nicholson method, even for a small maximum time of 0.0125. The error in the boundary gradient at the final time step (n = 25) is not shown in the table. However, the error in this term, using the fully implicit method, is about 1.7 times the error using the Crank-Nicholson method.
|Table 3-4 |
|Crank-Nicholson Solution of the Conduction Equation |
|( = 1, Δx = 0.01, Δt = 0.0005, tmax = 0.0125 |
| |i = 0 |i = 1 |i = 2 |i = 3 |i = 4 |
| |x = 0 |x = 0.01 |x = 0.02 |x = 0.03 |x = 0.04 |
| |t = 0 |1000 |1000 |1000 |1000 |1000 |
|n = 0 |t = 0+ |0 |1000 |1000 |1000 |1000 |
|n = 1 |t = 0.0005 |0 |-73.35 |423.96 |690.85 |834.09 |
|n = 2 |t = 0.001 |0 |352.75 |305.27 |440.73 |599.81 |
|n = 3 |t = 0.0015 |0 |25.70 |320.81 |439.19 |533.34 |
|n = 4 |t = 0.002 |0 |203.86 |209.57 |347.52 |473.02 |
|n = 5 |t = 0.0025 |0 |56.79 |252.91 |334.12 |422.43 |
|n = 6 |t = 0.003 |0 |141.46 |177.47 |298.20 |397.48 |
|n = 7 |t = 0.0035 |0 |66.73 |209.02 |279.22 |363.26 |
|n = 8 |t = 0.004 |0 |109.40 |160.30 |263.81 |347.29 |
|n = 9 |t = 0.0045 |0 |68.71 |179.63 |245.68 |324.49 |
|n = 10 |t = 0.005 |0 |90.79 |148.20 |237.92 |311.75 |
|n = 11 |t = 0.0055 |0 |67.50 |159.07 |222.68 |296.08 |
|n = 12 |t = 0.006 |0 |78.99 |138.51 |217.76 |285.25 |
|n = 13 |t = 0.0065 |0 |65.08 |144.07 |205.56 |273.92 |
|n = 14 |t = 0.007 |0 |70.94 |130.31 |201.68 |264.62 |
|n = 15 |t = 0.0075 |0 |62.29 |132.69 |192.04 |255.97 |
|n = 16 |t = 0.008 |0 |65.10 |123.21 |188.58 |247.99 |
|n = 17 |t = 0.0085 |0 |59.50 |123.75 |180.95 |241.06 |
|n = 18 |t = 0.009 |0 |60.65 |117.00 |177.71 |234.21 |
|n = 19 |t = 0.0095 |0 |56.86 |116.50 |171.59 |228.43 |
|n = 20 |t = 0.01 |0 |57.10 |111.53 |168.52 |222.53 |
|n = 21 |t = 0.0105 |0 |54.43 |110.47 |163.53 |217.57 |
|n = 22 |t = 0.011 |0 |54.19 |106.68 |160.64 |212.45 |
|n = 23 |t = 0.0115 |0 |52.22 |105.35 |156.49 |208.11 |
|n = 24 |t = 0.012 |0 |51.73 |102.36 |153.78 |203.64 |
|n = 25 |t = 0.0125 |0 |50.21 |100.93 |150.27 |199.78 |
|Exact |t = 0.0125 |0 |50.43 |100.66 |150.48 |199.72 |
|Error |t = 0.0125 |0 |0.216 |0.272 |0.212 |0.061 |
In addition to providing an accuracy comparison, these example calculations in Table 3-4 show that there is no inherent link between stability and physical reality. In these notes we derived the stability criterion for the explicit method by using arguments of physical reality. However, we see that the Crank-Nicholson method can produce intermediate results that are unrealistic physically, but the method still is stable. The fully implicit method does not produce such unrealistic results, but its accuracy is less than that of the Crank-Nicholson method.
The differences in the accuracy of the two methods increase as the calculation is run for a longer maximum time. Table 3-6 shows some overall results for the final time step when the calculations are done for a maximum time of 1. The errors in the fully implicit method, with its truncation error that is O[Δt, (Δx)2] are significantly higher than those in the Crank-Nicholson calculation, which has a truncation error that is O[(Δt)2, (Δx)2]. Both calculations took 2000 time steps with f = 5.0. Table 3-6 also shows the results of an explicit calculation that used 20,000 time steps to achieve the same maximum time. These time steps were the minimum allowed with the maximum value of f = 0.50. (If we wanted to use f = 1/6 to get the higher order truncation error from the explicit method, we would need 60,000 time steps to reach the same maximum time.)
|Table 3-5 |
|Fully-implicit Solution of the Conduction Equation |
|( = 1, Δx = 0.01, Δt = 0.0005, tmax = 0.0125 |
| |i = 0 |i = 1 |i = 2 |i = 3 |i = 4 |
| |x = 0 |x = 0.01 |x = 0.02 |x = 0.03 |x = 0.04 |
| |t = 0 |1000 |1000 |1000 |1000 |1000 |
|n = 0 |t = 0+ |0 |1000 |1000 |1000 |1000 |
|n = 1 |t = 0.0005 |0 |358.26 |588.17 |735.71 |830.39 |
|n = 2 |t = 0.001 |0 |218.22 |408.43 |562.69 |682.35 |
|n = 3 |t = 0.0015 |0 |166.26 |322.13 |460.74 |578.96 |
|n = 4 |t = 0.002 |0 |139.05 |272.65 |396.35 |507.18 |
|n = 5 |t = 0.0025 |0 |121.84 |240.25 |352.17 |455.26 |
|n = 6 |t = 0.003 |0 |109.75 |217.08 |319.77 |415.99 |
|n = 7 |t = 0.0035 |0 |100.65 |199.49 |294.81 |385.13 |
|n = 8 |t = 0.004 |0 |93.50 |185.57 |274.85 |360.14 |
|n = 9 |t = 0.0045 |0 |87.68 |174.19 |258.43 |339.38 |
|n = 10 |t = 0.005 |0 |82.82 |164.67 |244.62 |321.81 |
|n = 11 |t = 0.0055 |0 |78.69 |156.56 |232.81 |306.69 |
|n = 12 |t = 0.006 |0 |75.13 |149.54 |222.55 |293.50 |
|n = 13 |t = 0.0065 |0 |72.00 |143.38 |213.53 |281.87 |
|n = 14 |t = 0.007 |0 |69.24 |137.93 |205.52 |271.52 |
|n = 15 |t = 0.0075 |0 |66.77 |133.05 |198.35 |262.22 |
|n = 16 |t = 0.008 |0 |64.55 |128.66 |191.88 |253.82 |
|n = 17 |t = 0.0085 |0 |62.54 |124.67 |186.01 |246.17 |
|n = 18 |t = 0.009 |0 |60.70 |121.03 |180.64 |239.17 |
|n = 19 |t = 0.0095 |0 |59.02 |117.70 |175.71 |232.74 |
|n = 20 |t = 0.01 |0 |57.47 |114.62 |171.16 |226.79 |
|n = 21 |t = 0.0105 |0 |56.03 |111.78 |166.95 |221.28 |
|n = 22 |t = 0.011 |0 |54.70 |109.13 |163.04 |216.16 |
|n = 23 |t = 0.0115 |0 |53.46 |106.67 |159.38 |211.37 |
|n = 24 |t = 0.012 |0 |52.30 |104.36 |155.96 |206.88 |
|n = 25 |t = 0.0125 |0 |51.21 |102.20 |152.76 |202.67 |
|Exact |t = 0.0125 |0 |50.43 |100.66 |150.48 |199.72 |
|Error |t = 0.0125 |0 |0.779 |1.542 |2.273 |2.956 |
Table 3-6 also shows the execution time for calculations done on a 700 MHz Pentium III computer using a C++ program. The measured execution times are so short that they may not be meaningful. In general, solutions for this problem took about 0.5 to 0.6 microseconds per node per time step for the implicit and Crank Nicholson method and about 0.06 to 0.08 microseconds per node per time step for the explicit method. For example, a Crank Nicholson solution with Δx = 0.001 (1000 space nodes) and Δt = 0.00001 (100,100 time steps for a maximum time of 1) took 63.39 seconds. This is equivalent to (63.39 seconds) / (1000 space steps) / (100,000 time steps) ● (1,000,000 μsec/s) = 0.6449 microseconds per node per time step. To get a similar accuracy (rms error in temperature at the final time of 1x10-12) with the same Δx required a time step of 0.00000002 and an execution time of 4,110.4 seconds. This corresponds to a computation time of (4,110.4)(0.001)( 0.00000002) = 0.0822 microseconds per node per time step.
|Table 3-6 |
|Comparison of Results and Errors at time = tmax for Various Methods |
|( = 1, Δx = 0.01, Δt = 0.0005(CN and Implicit), = 0.00005 for explicit, tmax = 1 |
|Data Item in Each Row |Exact Value |Crank Nicholson |Fully Implicit |Explicit with smaller|
| | | | |Δt |
|Maximum Value of T |0.065865 |0.065903 | |0.065728 |
|Boundary Gradient at x = 0 |0.20689 |0.29711 |0.21220 |0.20676 |
|Time step Δt | |0.0005 |0.0005 |0.00005 |
|Execution time (seconds) | |0.11 |0.05 |0.11 |
|Maximum error in T | |4.7x10-5 |167x10-5 |8.1x10-5 |
|RMS error in T | |3.3x10-5 |118x10-5 |1.3x10-5 |
|Gradient error | |2.1x10-5 |531x10-5 |13.2x10-5 |
Numerical solutions of Laplace’s Equation
We start the solution of Laplace’s equation in the same way as we started the solution of the diffusion equation. We assume that the differential equation is defined over a rectangular region.
[pic] [3-21]
We can construct a rectangular finite difference grid in the two-dimensional region. For simplicity, we will assume constant step sizes, Δx and Δy. We define our time and space coordinates on this grid by the integers i and j, so that we can write
[pic] [3-22]
The location of a particular value T(xi,yj) is denoted as Tij. We can use the same second-order central-difference equation for the second derivative that we used in equation [3-3] for the diffusion equation with this Tij notation.
[pic] [3-23]
Substituting these finite-difference representations into Laplace’s equation and ignoring the truncation error, which is second order in each independent variable, gives
[pic] [3-24]
Multiplying through by (Δx)2 and defining β = Δx/Δy gives the following finite-difference equation.
[pic] [3-25]
We have an equation like this for each node in our two-dimensional grid. The set of equations represents a set of simultaneous linear equations that we have to solve simultaneously. If we have a problem in which the values of u are specified at all the boundaries (called the Dirichlet boundary conditions) we will have (N – 1)(M – 1) nodes at which we will have to solve an equation like [3-25]. We can illustrate a typical set of equations by considering a very small grid with N = 6 and M = 5. This sample grid is shown in Figure 1. In this case we have (6 – 1)(5 – 1) = 20 nodes
Figure 1. Example of small finite-difference grid.
[pic]
We see that the form of equation [3-25] is to relate each node to its four nearest neighbors. The nodes related by the typical finite-difference equation are said to form a computational molecule as shown in figure 1. We can write this set of equations in a matrix form if we first choose an order for the equations. For example we can start in the lower left-hand corner and solve the equations for all the nodes in one row before moving on to the next row. In this case we start solving in the following order u11, u21, u31, u41, u51, u12, u22, u32, etc. When solving an equation near a boundary such as the equation for u12, the value of u02, is a known boundary condition. Since it is known, it can be moved to the right-hand side of the equation giving the following result.
[pic] [3-26]
Each of the four computational nodes located in a corner of the grid would have two known boundary values that would be placed on the right hand side of the equation. The equation for the computational molecule centered at node 11 would be written as follows.
[pic] [3-27]
The entire set of matrix equations for the grid shown above is presented on the next page. For simplicity, the matrix equations use β = 1 giving the general equation for Δx = Δy.
[pic] [3-28]
Figure 2. Matrix Representation of Second-Order Finite-Difference Equations for Laplace’s Equation with Δx = Δy
[pic]
The coefficient matrix is a 20 by 20 matrix with 400 coefficients. Most of the coefficients are zero. All the nonzero coefficients are shown, but the large blank areas contain zeros that are not explicitly shown in this diagram. For Laplace’s equation the right hand side matrix contains only the boundary conditions where the value of u on a boundary has been specified.
A matrix such as the one shown in figure 2, where most of the coefficients are zero, is called a sparse matrix. Special solution techniques are used to solve such matrix equations and special computer data structures are used so that it is not necessary to store the large number of zero coefficients in the left-hand-side array.
The main approach to solving the equations is to use iteration. In this approach an initial guess is made for all the unknowns on the grid and the values of each grid node are updated in one iteration steps. These iteration steps continue until the errors in the equation are sufficiently small. For simple approaches, we can consider the solution of equation [3-28] on a two-dimensional grid. The grid numbering system for this two-dimensional problem runs from 1 to Nx in the x direction and 1 to Ny in the y direction. The nodes along i = 1, j = 1, i = Nx, and j = Ny are known values from the specified boundary potentials. We thus have (Nx – 2)(Ny – 2) interior points at which we have to solve equation [3-28] for the potential. If Nx = Ny = 1002, we will have 106 unknown values of potential that are obtained from 106 versions of equation [3-28], one for each node.
To solve this system of 106 equations by iteration, we would make an initial guess for each unknown potential. We denote the initial guess of potential at node (i,j) as uij(0); subsequent iterations of this potential are denoted as uij(n). In particular, we can rearrange equation [3-28] to solve for the new iteration of uij in terms of the previous iterations.
[pic] [3-29]
Equation [3-29], in which the new value of ui,j is computed in terms of the old values is known as Jacobi iteration. An alternative approach uses the most recent values of potential in the iteration process. If we adjust potentials starting at the lowest values of i and j and increasing these indices, the most recent iteration for ui,j-1 and ui-1,j will be available when we compute ui,j. We could use these most recent iterations as follows:
[pic] [3-30]
The only difference between equations [3-29] and [3-30] is in the n+1 superscript for the first two potentials in equation [3-30] Equation [3-30] defines Gauss-Seidel iteration. There is an improvement to this iteration process known as overrelaxation.[3] We can regard equation [3-30] as providing a correction to the previous potential value. This interpretation of equation [3-30] is illustrated below.
[pic] [3-31]
Overrelaxation is based on the principle that if a little correction is good, then more is better. Instead of simply adding ΔTi,j we add ωΔTi,j where the constant ω is selected to provide faster convergence. This method, called simultaneous overrelaxation or SOR, is given by equation [3-32].
[pic] [3-32]
Equation [3-32] can be simply coded as follows. As usual we assume that all the variables have been properly defined and initialized prior to this code. The variables Nx, Ny, and omega are used to denote the x and y grid nodes and the relaxation coefficient, ω. The unknown potentials, uij, are represented by the U[i][j] array.
for ( i = 2; i ................
................
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