numerical method for engineers-chapter 23
TRANSCRIPT
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CHAPTER 23
23.1
x f(x)xi2 0.261799388 0.965925826xi1 0.523598776 0.866025404xi 0.785398163 0.707106781xi+1 1.047197551 0.5xi+2 1.308996939 0.258819045
true = sin(/4) = 0.70710678
The results are summarized as
first-order second-orderForward 0.79108963 0.72601275
11.877% 2.674%Backward 0.60702442 0.71974088
14.154% 1.787%Centered 0.69905703 0.70699696
1.138% 0.016%
23.2
x f(x)xi2 21 1.322219295xi1 23 1.361727836xi 25 1.397940009xi+1 27 1.431363764xi+2 29 1.462397998
The results are summarized as
first-order second-orderForward 0.016711878 0.017309258
3.799% 0.360%Backward 0.018106086 0.017281994
4.227% 0.517%Centered 0.017408982 0.017371197
0.214% 0.003%
23.3
x f(x)
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xi2 1.8 6.049647464xi1 1.9 6.685894442xi 2 7.389056099xi+1 2.1 8.166169913xi+2 2.2 9.025013499
Both the first and second derivatives have the same value,
The results are summarized as
first-order second-orderFirst derivative 7.401377351 7.389031439
-0.166750% 0.000334%Second derivative 7.395215699 7.389047882
-0.083361% 0.000111%
23.4 The true value is sin(/4) = 0.70710678.
23.5 The true value is 1/x = 1/5 = 0.2.
23.6 The true value
Equation (23.9) can be used to compute the derivative as
x0 = 0.5 f(x0) = 1.125
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x1 = 1 f(x1) = 24x2 = 2 f(x2) = 48
Centered difference:
23.7 At x = xi, Eq. (23.9) is
For equispaced points that are h distance apart, this equation becomes
23.8 (a)
x f(x)xi2 -0.5 -17.125xi1 -0.25 -16.0156xi 0 -15xi+1 0.25 -13.9844xi+2 0.5 -12.875
(b)
x f(x)
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xi2 0.2 0.039203xi1 0.3 0.08598xi 0.4 0.14737xi+1 0.5 0.219396xi+2 0.6 0.297121
(c)
x f(x)xi2 2 0.786843xi1 2.5 1.100778xi 3 1.557408xi+1 3.5 2.338254xi+2 4 4.131729
(d)
x f(x)xi2 0.6 0.62948xi1 0.8 0.540569xi 1 0.479426xi+1 1.2 0.433954xi+2 1.4 0.398355
(e)
x f(x)xi2 1.6 6.553032xi1 1.8 7.849647xi 2 9.389056xi+1 2.2 11.22501xi+2 2.4 13.42318
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23.9 The first forward difference formula of O(h2) from Fig. 23.1 can be used to estimate the velocity for the first point at t = 0,
The acceleration can be estimated with the second forward difference formula of O(h2) from Fig. 23.1
For the interior points, centered difference formulas of O(h2) from Fig. 23.3 can be used to estimate the velocities and accelerations. For example, at the second point at t = 25,
For the final point, backward difference formulas of O(h2) from Fig. 23.2 can be used to estimate the velocities and accelerations. The results for all values are summarized in the following table.
t y v a0 0 1.40 -0.0096
25 32 1.16 -0.009650 58 0.92 -0.009675 78 0.68 -0.0096
100 92 0.44 -0.0096125 100 0.20 -0.0096
23.10 Here is a VBA program to implement a Romberg algorithm to estimate the derivative of a given function. It is set up to evaluate the derivative from Example 23.1.
Option Explicit
Sub RhombTest()Dim maxit As IntegerDim es As Double, x As Doublex = 0.5maxit = 3es = 0.001MsgBox RhomDiff(x, maxit, es)End Sub
Function RhomDiff(x, maxit, es)
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Dim n As Integer, j As Integer, k As Integer, iter As IntegerDim i(10, 10) As Double, ea As Double, del As Doublen = 1i(1, 1) = DyDx(x, n)iter = 0Do iter = iter + 1 n = 2 ^ iter i(iter + 1, 1) = DyDx(x, n) For k = 2 To iter + 1 j = 2 + iter - k i(j, k) = (4 ^ (k - 1) * i(j + 1, k - 1) - i(j, k - 1)) / (4 ^ (k - 1) - 1) Next k ea = Abs((i(1, iter + 1) - i(1, iter)) / i(1, iter + 1)) * 100 If (iter >= maxit Or ea <= es) Then Exit DoLoopRhomDiff = i(1, iter + 1)End Function
Function DyDx(x, n)Dim a As Double, b As Doublea = x - x / nb = x + x / nDyDx = (f(b) - f(a)) / (b - a)End Function
Function f(x)f = -0.1 * x ^ 4 - 0.15 * x ^ 3 - 0.5 * x ^ 2 - 0.25 * x + 1.2End Function
When the program is run, the result is
23.11 Here is a VBA program uses Eq. 23.9 to obtain first-derivative estimates for unequally spaced data.
Option Explicit
Sub TestDerivUnequal()Dim n As Integer, i As IntegerDim x(100) As Double, y(100) As Double, dy(100) As DoubleRange("a5").Selectn = ActiveCell.RowSelection.End(xlDown).Selectn = ActiveCell.Row - nRange("a5").SelectFor i = 0 To n x(i) = ActiveCell.Value ActiveCell.Offset(0, 1).Select y(i) = ActiveCell.Value ActiveCell.Offset(1, -1).SelectNext i
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For i = 0 To n dy(i) = DerivUnequal(x, y, n, x(i))Next iRange("c5").SelectFor i = 0 To n ActiveCell.Value = dy(i) ActiveCell.Offset(1, 0).SelectNext iEnd Sub
Function DerivUnequal(x, y, n, xx)Dim ii As IntegerIf xx < x(0) Or xx > x(n) Then DerivUnequal = "out of range"Else If xx < x(1) Then DerivUnequal = DyDx(xx, x(0), x(1), x(2), y(0), y(1), y(2)) ElseIf xx > x(n - 1) Then DerivUnequal = _ DyDx(xx, x(n - 2), x(n - 1), x(n), y(n - 2), y(n - 1), y(n)) Else For ii = 1 To n - 2 If xx >= x(ii) And xx <= x(ii + 1) Then If xx - x(ii - 1) < x(ii) - xx Then 'If the unknown is closer to the lower end of the range, 'x(ii) will be chosen as the middle point DerivUnequal = _ DyDx(xx, x(ii - 1), x(ii), x(ii + 1), y(ii - 1), y(ii), y(ii + 1)) Else 'Otherwise, if the unknown is closer to the upper end, 'x(ii+1) will be chosen as the middle point DerivUnequal = _ DyDx(xx, x(ii), x(ii + 1), x(ii + 2), y(ii), y(ii + 1), y(ii + 2)) End If Exit For End If Next ii End IfEnd If
End Function
Function DyDx(x, x0, x1, x2, y0, y1, y2)DyDx = y0 * (2 * x - x1 - x2) / (x0 - x1) / (x0 - x2) _ + y1 * (2 * x - x0 - x2) / (x1 - x0) / (x1 - x2) _ + y2 * (2 * x - x0 - x1) / (x2 - x0) / (x2 - x1)End Function
When the program is run, the result is shown below:
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The results can be compared with the true derivatives which can be calculated with analytical solution, f(x) =5e–2x – 10xe–2x. The results can be displayed graphically below where the computed values are represented as points and the true values as the curve.
-0.8
-0.6
-0.4
-0.2
0
0.2
0 1 2 3 4 5
An even more elegant approach is to put cubic splines through the data (recall Sec. 20.2 and the program used for the solution to Prob. 20.10) to evaluate the derivatives. Here is the result of applying that program to this problem.
-0.8
-0.6
-0.4
-0.2
0
0.2
0 1 2 3 4 5
23.12 (a) Create the following M function:
function y=f(t)y=9.81*70/12*(1-exp(-12/70*t));
Then implement the following MATLAB session:
>> Q=quad(@f,0,10)
Q = 298.5546
(b)
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(c) Implement the following MATLAB session:
>> x=[9.99 10.01];>> y=f(x);>> d=diff(y)./diff(x)
d = 1.7667
(d)
23.13 (a) Create the following M function:
function y=fn(x)y=1/sqrt(2*pi)*exp(-(x.^2)/2);
Then implement the following MATLAB session:
>> x=-2:.1:2;>> y=fn(x);>> Q=quad(@fn,-1,1)
Q = 0.6827
>> Q=quad(@fn,-2,2)
Q = 0.9545
Thus, about 68.3% of the area under the curve falls between –1 and 1 and about 95.45% falls between –2 and 2.
(b)>> x=-2:.1:2;>> y=fn(x);>> d=diff(y)./diff(x);>> x=-1.95:.1:1.95;>> d2=diff(d)./diff(x);>> x=-1.9:.1:1.9;>> plot(x,d2,'o')
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Thus, inflection points (d2y/dx2 = 0) occur at –1 and 1.
23.14 (a) Create the following M function:
function y=fn(x)y=1/sqrt(2*pi)*exp(-(x.^2)/2);
Then implement the following MATLAB session:
>> x=-1:.5:1;>> y=fn(x);>> I=trapz(x,y)
I = 0.6725
>> x=-2:.5:2;>> y=fn(x);>> I=trapz(x,y)
I = 0.9500
Note that there is another MATLAB function called trap that can also be used to solve this problem. To learn more about this function type
>> help trap
The resulting description will be displayed,
I = trap(func,a,b,n): multiple-application trapezoidal rule. input: func = name of function to be integrated a, b = integration limits n = number of segments
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output: I = integral estimate
Here are the results of using this function to solve the problem:
>> I=trap(@fn,-1,1,4)
I = 0.6725
>> I=trap(@fn,-2,2,8)
I = 0.9500
Thus, about 67.25% of the area under the curve falls between –1 and 1 and about 95% falls between –2 and 2.
(b)>> d=diff(y)./diff(x);>> x=-1.75:.5:1.75;>> d2=diff(d)./diff(x);>> x=-1.5:.5:1.5;>> plot(x,d2,'o')
Thus, inflection points (d2y/dx2 = 0) occur at –1 and 1.
23.15Program IntegrateUse imslImplicit NoneInteger::irule=1Real::a=-1.,b=1,errabs=0.0,errrel=0.001Real::errest,res,fExternal f
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Call QDAG(f,a,b,errabs,errrel,irule,res,errest)Print '('' Computed = '',F8.4)',resPrint '('' Error estimate ='',1PE10.3)',errestEnd Program
Function f(x)Implicit NoneReal:: x , fReal::piParameter(pi=3.1415927)f=1/sqrt(2.*pi)*exp(-x**2/2.)End Function
Answers:
x = -1 to 1: Computed = 0.6827 Error estimate = 4.069E-06x = -2 to 2: Computed = 0.9545 Error estimate = 7.975E-06x = -3 to 3: Computed = 0.9973 Error estimate = 5.944E-06
23.16 MATLAB Script:
% Prob2316 Integration programa=0;b=pi/2;integral=quad(@ff,a,b)
function y=ff(x)y=cos(cos(x));
>> prob2316
integral = 1.2020
23.17 MATLAB Script saved as prob2317.m:
%Numerical Integration of sin(t)/t = function sint(t)%Limits: a=0, b=2pi%Using the "quad" and "quadl" function for numerical integration%Plot of functiont=0.01:0.01:2*pi;y=ff2(t);plot(t,y); grid%Integrationformat longa=0.01;b=2*pi;Iquad=quad('ff2',a,b)Iquadl=quadl('ff2',a,b)
function y=ff2(t)y=sin(t)./t;
MATLAB execution:
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>> prob2317
Iquad = 1.40815163720125
Iquadl = 1.40815163168846
23.18%Centered Finite Difference First & Second Derivatives of Order O(dx^2)%Using diff(y)
dx=1.;y=[1.4 2.1 3.3 4.8 6.8 6.6 8.6 7.5 8.9 10.9 10];dyf=diff(y);
% First Derivative Centered FD using diffn=length(y);for i=1:n-2
dydxc(i)=(dyf(i+1)+dyf(i))/(2*dx);end
%Second Derivative Centered FD using diffdy2dx2c=diff(dyf)/(dx*dx);
fprintf('first derivative \n'); fprintf('%f\n', dydxc)fprintf('second derivative \n'); fprintf('%f\n', dy2dx2c)
first derivative 0.9500001.3500001.7500000.9000000.9000000.4500000.1500001.7000000.550000second derivative 0.500000
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0.3000000.500000-2.2000002.200000-3.1000002.5000000.600000-2.900000
23.19% Finite Difference Approximation of slope% For f(x)=exp(-2*x)-x% f'(x)=-2*exp(-2*x)-1% Centered diff. df/dx=(f(i+1)-f(i-1))/2dx + O(dx^2)% Fwd. diff. df/dx=(-f(i+2)+4f(i+1)-3f(i))/2dx + O(dx^2)% Bkwd. diff. df/dx=(3f(i)-4f(i-1)+f(i-2))/2dx + O(dx^2)
x=2;fx=exp(-2*x)-x;dfdx2=-2*exp(-2*x)-1;
%approximationdx=0.5:-0.01:.01;for i=1:length(dx) %x-values at i+-dx and +-2dx xp(i)=x+dx(i); x2p(i)=x+2*dx(i); xn(i)=x-dx(i); x2n(i)=x-2*dx(i); %f(x)-values at i+-dx and +-2dx fp(i)=exp(-2*xp(i))-xp(i); f2p(i)=exp(-2*x2p(i))-x2p(i); fn(i)=exp(-2*xn(i))-xn(i); f2n(i)=exp(-2*x2n(i))-x2n(i); %Finite Diff. Approximations Cdfdx(i)=(fp(i)-fn(i))/(2*dx(i)); Fdfdx(i)=(-f2p(i)+4*fp(i)-3*fx)/(2*dx(i)); Bdfdx(i)=(3*fx-4*fn(i)+f2n(i))/(2*dx(i));enddx0=0;plot(dx,Fdfdx,'--',dx,Bdfdx,'-.',dx,Cdfdx,'-',dx0,dfdx2,'*')gridtitle('Forward, Backward, and Centered Finite Difference approximation - 2nd Order Correct')xlabel('Delta x')ylabel('df/dx')gtext('Centered'); gtext('Forward'); gtext('Backward')
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23.20 First, we will use forward expansions. The Taylor series expansion about a = xi and x = xi+2 (2x steps forward) can be written as:
(1)
Taylor series expansion about a = xi and x = xi+1 (x steps forward):
(2)
Multiply Eq. 2 by 2 and subtract the result from Eq. 1 to yield
(3)
Next, we will use backward expansions. The Taylor series expansion about a = xi and x = xi2 (2x steps backward) can be written as:
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(4)
Taylor series expansion about a = xi and x = xi1 (x steps backward):
(5)
Multiply Eq. 5 by 2 and subtract the result from Eq. 4 to yield
(6)
Add Eqs (3) and (6)
(7)
Equation 7 can be solved for
23.21 (a)
(b)
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(c)
23.22
rad/s
m/s
rad/s2
m/s2
23.23 Use the same program as was developed in the solution of Prob. 23.11
Option Explicit
Sub TestDerivUnequal()Dim n As Integer, i As IntegerDim x(100) As Double, y(100) As Double, dy(100) As DoubleRange("a5").Selectn = ActiveCell.RowSelection.End(xlDown).Selectn = ActiveCell.Row - nRange("a5").SelectFor i = 0 To n x(i) = ActiveCell.Value ActiveCell.Offset(0, 1).Select y(i) = ActiveCell.Value ActiveCell.Offset(1, -1).SelectNext iFor i = 0 To n dy(i) = DerivUnequal(x, y, n, x(i))Next iRange("c5").SelectFor i = 0 To n ActiveCell.Value = dy(i) ActiveCell.Offset(1, 0).SelectNext i
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End Sub
Function DerivUnequal(x, y, n, xx)Dim ii As IntegerIf xx < x(0) Or xx > x(n) Then DerivUnequal = "out of range"Else If xx < x(1) Then DerivUnequal = DyDx(xx, x(0), x(1), x(2), y(0), y(1), y(2)) ElseIf xx > x(n - 1) Then DerivUnequal = _ DyDx(xx, x(n - 2), x(n - 1), x(n), y(n - 2), y(n - 1), y(n)) Else For ii = 1 To n - 2 If xx >= x(ii) And xx <= x(ii + 1) Then If xx - x(ii - 1) < x(ii) - xx Then 'If the unknown is closer to the lower end of the range, 'x(ii) will be chosen as the middle point DerivUnequal = _ DyDx(xx, x(ii - 1), x(ii), x(ii + 1), y(ii - 1), y(ii), y(ii + 1)) Else 'Otherwise, if the unknown is closer to the upper end, 'x(ii+1) will be chosen as the middle point DerivUnequal = _ DyDx(xx, x(ii), x(ii + 1), x(ii + 2), y(ii), y(ii + 1), y(ii + 2)) End If Exit For End If Next ii End IfEnd If
End Function
Function DyDx(x, x0, x1, x2, y0, y1, y2)DyDx = y0 * (2 * x - x1 - x2) / (x0 - x1) / (x0 - x2) _ + y1 * (2 * x - x0 - x2) / (x1 - x0) / (x1 - x2) _ + y2 * (2 * x - x0 - x1) / (x2 - x0) / (x2 - x1)End Function
The result of running this program is shown below:
23.24
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23.25 The flow rate is equal to the derivative of volume with respect to time. Equation (23.9) can be used to compute the derivative as
x0 = 1 f(x0) = 1x1 = 5 f(x1) = 8x2 = 8 f(x2) = 16.4
Therefore, the flow is equal to 2.95 cm3/s.
23.26 The velocity at the surface can be computed with Eq. (23.9) as
x0 = 0 f(x0) = 0x1 = 0.002 f(x1) = 0.287x2 = 0.006 f(x2) = 0.899
Therefore, the shear stress can be computed as
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23.27 The first forward difference formula of O(h2) from Fig. 23.1 can be used to estimate the velocity for the first point at t = 10,
For the interior points, centered difference formulas of O(h2) from Fig. 23.3 can be used to estimate the derivatives. For example, at the second point at t = 20,
For the final point, backward difference formulas of O(h2) from Fig. 23.2 can be used to estimate the derivative. The results for all values are summarized in the following table.
t c dc/dt log c log(dc/dt)10 3.52 0.1195 0.546543 -0.9226320 2.48 0.0885 0.394452 -1.0530630 1.75 0.0625 0.243038 -1.2041240 1.23 0.044 0.089905 -1.3565550 0.87 0.031 -0.06048 -1.5086460 0.61 0.021 -0.21467 -1.67778
A log-log plot can be developed
y = 0.9946x - 1.4527
R2 = 0.9988
-2
-1
0
-0.4 -0.2 0 0.2 0.4 0.6
The resulting best-fit equation can be used to compute k = 10–1.45269 = 0.035262 and n = 0.994579.
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