Solucionario Del Capitolu 29 De Chapra

CHAPTER 27
27.1 The solution can be assumed to be T = eλx. This, along with the second derivative T” = λ2eλx, can be substituted into the differential equation to give
λ2 e λx − 0.1e λx = 0

which can be used to solve for
λ2 − 0.1 = 0 λ = ± 0.1

Therefore, the general solution is
T = Ae
0.1x

+ Be −

0.1x

The constants can be evaluated by substituting each of the boundaryconditions to generate two equations with two unknowns,
200 = A + B 100 = 23.62434 A + 0.042329 B

which can be solved for A = 3.881524 and B = 196.1185. The final solution is, therefore,
T = 3.881524 e
0.1x

+ 196.1185e −

0.1x

which can be used to generate the values below:
x 0 1 2 3 4 5 6 7 8 9 10 T 200 148.2747 111.5008 85.97028 69.10864 59.21565 55.29373 56.94741 64.34346 78.22764 100200

100

0 0 2 4 6 8 10

27.2 Reexpress the second-order equation as a pair of ODEs:

dT =z dx dz = 0.1T dx

The solution was then generated on the Excel spreadsheet using the Heun method (without iteration) with a step-size of 0.01. An initial condition of z = −55 was chosen for the first shot. The first few calculation results are shown below.
x 0 0.1 0.2 0.3 0.4 0.5 T z 200.000194.600 189.395 184.378 179.547 174.894 -55.000 -53.028 -51.108 -49.240 -47.420 -45.649 k11 k12 Tend zend k21 k22 phi1 phi2 -55.000 20.000 194.500 -53.000 -53.000 19.450 -54.000 19.725 -53.028 19.460 189.297 -51.082 -51.082 18.930 -52.055 19.195 -51.108 18.939 184.284 -49.214 -49.214 18.428 -50.161 18.684 -49.240 18.438 179.454 -47.396 -47.396 17.945 -48.318 18.192 -47.420 17.955 174.805 -45.625-45.625 17.480 -46.523 17.718 -45.649 17.489 170.330 -43.900 -43.900 17.033 -44.774 17.261

The resulting value at x = 10 was T(10) = 315.759. A second shot using an initial condition of z (0) = −70 was attempted with the result at x = 10 of T(10) = −243.249. These values can then be used to derive the correct initial condition,
z( 0) = −55 + − 70 + 55 (100 − 315.759 ) = −60.79 − 243.249 − 315.759The resulting fit, along with the two “shots” are displayed below:
400 300 200 100 0 -100 0 -200 -300 2 4 6 8 10

27.3 A centered finite difference can be substituted for the second derivative to give,
Ti −1 − 2Ti + Ti +1 h2 − 0.1Ti = 0

or for h = 1,
− Ti −1 + 2.1Ti − Ti +1 = 0

The first node would be
2.1T1 − T2 = 200

and the last node would be

− T9 + 2.1T10 = 100

Thetridiagonal system can be solved with the Thomas algorithm or Gauss-Seidel for (the analytical solution is also included)
x 0 1 2 3 4 5 6 7 8 9 10 T 200 148.4838 111.816 86.32978 69.47655 59.57097 55.62249 57.23625 64.57365 78.3684 100 Analytical 200 148.2747 111.5008 85.97028 69.10864 59.21565 55.29373 56.94741 64.34346 78.22764 100

27.4 The second-order ODE can be expressed as the followingpair of first-order ODEs,
dy =z dx dz 2 z + y − x = dx 8

These can be solved for two guesses for the initial condition of z. For our cases we used z(0) −1 −0.5 y(20) −6523.000507 7935.937904 Clearly, the solution is quite sensitive to the initial conditions. These values can then be used to derive the correct initial condition,
z( 0) = −1 + − 0.5 + 1 (8 + 6523.000507 ) = −0.774154 7935.937904 +6523.000507

The resulting fit is displayed below:

12 8 4 0 0 5 10 15 20

27.5 Centered finite differences can be substituted for the second and first derivatives to give,

8

yi +1 − 2 y i + yi −1 ∆x
2

−2

y i +1 − y i −1 − yi + x i = 0 ∆x

or substituting ∆x = 2 and collecting terms yields
2.5 y i +1 − 5 yi + 1.5 yi −1 + x i = 0

This equation can be written for each nodeand solved with either the Gauss-Seidel method or a tridiagonal solver to give
x 0 2 4 6 8 10 12 14 16 18 20 T 5 4.287065 4.623551 5.600062 6.960955 8.536414 10.18645 11.72749 12.78088 12.39044 8

12 8 4 0 0 5 10 15 20

27.6 The second-order ODE can be expressed as the following pair of first-order ODEs,
dT =z dx dz 7 4 = 1.2 × 10 (T + 273) − 5(150 − T ) dx

The solution was then...