1. Boundary-Value Problems in Rectangular Coordinates
CHAPTER 13
Boundary-Value Problems in Rectangular Coordinates

2. Contents 13.1 Separable Partial Differential Equations
13.2 Classical Equations and Boundary-Value Problems
13.3 Heat Equation
13.4 Wave Equation
13.5 Laplace’s Equation
13.6 Nonhomogeneous Equations and Boundary Conditions
13.7 Orthogonal Series Expansions
13.8 Fourier Series in Two Variable

3. 13.1 Separable Partial Differential Equations
Linear PDE If we let u denote the dependent variable and x, y are independent variables, the general form of a linear second-order PDE is given by (1) When G(x, y) = 0, (1) is homogeneous; otherwise it is nonhomogeneous.

4. Separation of Variables
If we assume that u = X(x)Y(y), then

5. Example 1 Find product solution of
Solution Let u = X(x)Y(y) and then We introduce a real separation constant as −.

6. Example 1 (2)
Thus
For the three cases:  = 0: X” = 0, Y’ = 0 (3)  = −2 > 0,  > X” – 42X = 0, Y’ − 2Y = 0 (4)  = 2 > 0,  > X” + 42X = 0, Y’ + 2Y = 0 (5)

7. Example 1 (3)
Case I: ( = 0) The solutions of (3) are X = c1 + c2x and Y = c3. Thus (6) where A1 = c1c3 , B1 = c2c3.
Case II: ( = −2) The solutions of (4) are X = c4 cosh 2x + c5 sinh 2x and Thus (7) where A2 = c4c6, B2 = c5c6.

8. Example 1 (4)
Case III: ( = 2) The solutions of (5) are X = c7 cos 2x + c8 sin 2x and Thus (8) where A3 = c7c9, B3 = c8c9.

9. If u1, u2, …, uk are solution of a homogeneous linear
THEOREM 13.1
If u1, u2, …, uk are solution of a homogeneous linear
partial differential equation, then the linear combination
u = c1u1 + c2u2 + … + ckuk
where the ci = 1, 2, …, k are constants, is also a solution.
Superposition Principles

10. If linear second-order differential equation
DEFINITION 13.1
If linear second-order differential equation
where A, B, C, D, E, and F are real constants, is said
to be
hyperbolic if
parabolic if
elliptic if
Classification of Equations

11. Example 2
Classify the following equations:
Solution (a)

12. Example 2 (2)

13. 13.2 Classical Equations and Boundary-Value Problems
Introduction Typical second-order PDEs: (1) (2) (3) They are known as one-dimensional heat equation, one-dimensional wave equation, and Laplace’s equations in two dimensions, respectively.

14. Note:
Laplace’s equation is abbreviated 2u = 0, where is called the two-dimensional Laplacian of u. In three dimension the Laplacian of u is

15. Boundary-Value Problems
Solve: Subject to: (BC) (11) (IC)

16. and Solve: Subject to: (BC) (12)

17. 13.3 Heat Equation
Introduction The heat equation can be described by the following (1) (2) (3)

18. Solution of the BVP
Using u(x, t) = X(x)T(t), and − as the separation constant: (4) (5) (6)

19. Now the boundary conditions in (2) become u(0, t) = X(0)T(t) = 0 and u(L, t) = X(L)T(t) = 0. Then we can have X(0) = X(L) = 0 and (7) From the previous discussions, we have

20. When the boundary conditions X(0) = X(L) = 0 are applied to (8) and (9), these solutions are only X(x) = 0. Applying the first condition to (10) gives c1 = 0. Therefore X(x) = c2 sin x. The condition X(L) = 0 implies that (11) We have sin L = 0 for c2  0 and  = n/L, n = 1, 2, 3, … The values n = n2 = (n/L)2, n = 1, 2, 3, … and the corresponding solutions (12)

21. are the eigenvalues and eigenfunctions, respectively
are the eigenvalues and eigenfunctions, respectively. The general solution of (6) is and so (13) where An = c2c3.

22. Now using the initial conditions u(x, 0) = f(x), 0 < x < L, we have (14) By the superposition principle the function (15) must satisfy (1) and (2). If we let t = 0, then

23. It is recognized as the half-range expansion of f in a sine series
It is recognized as the half-range expansion of f in a sine series. If we let An = bn, n = 1, 2, 3, … thus (16) We conclude that the solution of the BVP described by (1), (2) and (3) is given by infinite series (17)

24. For example, u(x, 0) = 100, L = , and k = 1, then

25. 13.4 Wave Equation
Introduction Consider the wave equations (1) (2) (3)

26. Solution of the BVP
Assuming u(x, t) = X(x)T(t), then (1) gives so that (4) (5)

27. Using X(0) = 0 and X(L) = 0, we have
Using X(0) = 0 and X(L) = 0, we have (6) Only  = 2 > 0,  > 0 leads to nontrivial solutions. Thus the general solution of (4) is X(0) = 0 and X(L) = 0 imply that c1= 0 and c2 sin L = 0. Thus we have  = n/L, n = 1, 2, 3, …

28. The eigenvalues and eigenfunctions are

29. Let An = c2c3, Bn = c2c4, solutions that satisfy (1) and (2) are

30. Setting t = 0 in (8) and using u(x, 0) = f(x) gives Since it is a half-range expansion of f in a sine series, we can write An = bn: (9)

31. To determine Bn we differentiate (8) w. r. t
To determine Bn we differentiate (8) w.r.t. t and set t = 0: Thus we obtain (10)

32. Standing Wave
It is easy to transform (8) into

33. When n = 1, u1(x, t) is called the first standing wave, the first normal mode or the fundamental mode of vibration. The frequency f1 = a/2L of the first normal mode is called the fundamental frequency or first harmonic. See Fig 13.9.

34. Fig 13.9

35. 13.5 Laplace’s Equation
Introduction Consider the following boundary-value problem (1) (2) (3)

36. Solution of the BVP
With u(x, y) = X(x)Y(y), (1) becomes The three homogeneous boundary conditions in (2) and (3) translate into X’(0) = 0, X’(a) = 0, Y(0) = 0.

37. Thus we have the following equation. (6) For  = 0, (6) becomes
Thus we have the following equation (6) For  = 0, (6) becomes X” = 0, X’(0) = 0, X’(a) = 0 The solution is X = c1 + c2x. X’(0) = 0 implies c2 = 0 and X = c1 also satisfies the condition X’(a) = 0. Thus X = c1, c1  0 is a nontrivial solution.
For  = −2 < 0,  > 0, (6) possesses no nontrivial solutions.

38. For  = 2 > 0,  > 0, (6) becomes
For  = 2 > 0,  > 0, (6) becomes X” + 2X = 0, X’(0) = 0, X’(a) = 0 Applying X’(0) = 0 to the solution X = c1 cos x + c2 sin x, implies c2 = 0 and so X = c1 cos x . The condition X’(a) = 0 gives −c1  sin a = 0, and we must have  = n/a, n = 1, 2, 3, …. The eigenvalues of (6) are n = (n/a)2, n = 1, 2, …
By corresponding 0 with n = 0, the eigenfunctions of (6) are For Y” – Y = 0, when 0 = 0, the solution is Y = c3 + c4y. Y(0) = 0 implies c3 = 0 and so Y = c4y.

39. For n = (n/a)2, n = 1, 2, …, the solution is
For n = (n/a)2, n = 1, 2, …, the solution is Y = c3 cosh (ny/a) + c4 sinh (ny/a) Y(0) = 0 implies c3 = 0 and so Y = c4 sinh (ny/a).
The solutions un = XY are

40. The superposition principle yields
The superposition principle yields (7) Set y = b, then is a half-range expansion of f in a Fourier cosine series.

41. If we let A0b = a0/2 and An sin (nb/a)= an, n = 1, 2, …., we have

42. Dirichlet Problem
Please verify that the solution of the following Dirichlet Problem

43. is

44. Superposition Principle
We want to break the following problem (11) into two problems, each of which has homogeneous boundary conditions on parallel boundaries, as shown in the following tables.

45. Problem 1:

46. Problem 2:

47. Suppose that u1 and u2 are solutions of problem 1 and problem 2, respectively. If we define u = u1 + u2, then and so on. See Fig

48. Fig 13.15

49. It is an exercise that the solution of problem 1 is

50. The solution of problem 2 is

51. 13.6 Nonhomogeneous BVPs
Introduction A typical nonhomogeneous BVP for the heat equation is (1) When heat is generated at a constant rate r within a rod, the heat equation in (1) takes the form (2) Equation (2) is shown not to be separable.

52. Change of Dependent variables
u = v + ,  is a function to be determined.

53. Time Independent PDE and BCs
Time Independent PDE and BCs First consider the heat source F and the boundary conditions are time-independent: (3)

54. In (3), u0 and u1 denotes constants
In (3), u0 and u1 denotes constants. If we let u(x, t) = v(x, t) + (x), (3) cane be reduced to two problems:

55. Example 1 Solve (2) subject to
Solution If we let u(x, t) = v(x, t) + (x), then (4) since t = 0.

56. Example 1 (2)
Substituting (4) into (3) gives (5) Equation (5) reduces to a homogeneous PDE if we demand that  be a function satisfying the ODE Thus we have (6)

57. Example 1 (3)
Furthermore, We have v(0, t) = 0 and v(1, t) = 0, provided we choose (0) = 0 and (1) = u0 Applying these conditions to (6) implies c2 = 0, c1 = r/2k + u0.

58. Example 1 (4)
Thus Finally the initial condition u(x,0) = v(x, 0) + (x) implies v(x,0) = u(x, 0) − (x) = f(x) – (x). We have the new homogeneous BVP:

59. Example 1 (5)
In the usual manner we find

60. Example 1 (6)
A solution of the original problem is (8) Observe that

61. Time Dependence PDE and BCs
Under this situation, a new form of solution is u(x, t) = v(x, t) + (x, t) Since (9) (1) becomes (10)

62. The BCs on v in (10) become homogeneous if we demand that
The BCs on v in (10) become homogeneous if we demand that (11) We now construct a function  that satisfies both conditions in (11). One such function is (12) Please note that xx = 0. If we substitute (13) the problems in (1) become

63. (14) where G(x, t) = F(x, t) – t.

64. Before solving (14), we outline the basic strategy:
Before solving (14), we outline the basic strategy: Make the assumption that time-dependent coefficients vn(t) and Gn(t) can be found such that both v(x, t) and G(x, t) in (14) can be expanded in the series (15) where sin(nx/L), n = 1, 2, … are the eigenfunctions of X”+ X = 0, X(0) = 0, X(L) = 0 corresponding to the eigenvalues n = n2 = n22/L2

65. Example 2
Solve
Solution We match this problem with (1) to get k = 1, L = 1, F(x, t) = 0, u0(t) = cos t, u1(t) = 0, f(x) = 0. From (12) we get and then as indicated in (13), we use the substitution (16)

66. Example 2 (2)
to obtain the BVP for v(x, t): (17) The eigenvalues and eigenfunctions of the Sturm-Liouville problem X +X = 0, X(0) = 0, X(1) = 0 are n = n2 = n22 and sin nx, n = 1, 2, ….

67. Example 2 (3)
With G(x, t) = (1 – x) sin t, we assume from (15) and for fixed t, v and G can be written as Fourier sine series: (18) and (19)

68. Example 2 (4)
By treating t as a parameter, then Hence (20)

69. Example 2 (5)
From (18), we have (21) The PDE becomes

70. Example 2 (6)
For each n, the general solution of the above ODE: where Cn denotes the arbitrary constant. Thus (22)

71. Example 2 (7)
The Cn can be found by applying the initial condition v(x, 0) to (22). From the Fourier series

72. Example 2 (8)
Therefore

73. 13.7 Orthogonal Series Expansions
Example 1 The temperature in a rod of unit length is determined from solve for u(x, t).

74. Example 1 (2)
Solution If we let u(x, t) = X(x)T(t) and −  as the separation constant, we have (1) (2) (3)

75. Example 1 (3)
(1) and (3) comprise the regular Sturm-Liouville problem (4) As in Example 2 of Sec 12.5, (4) possesses nontrivial solutions only for  = 2 > 0,  > 0. The general solution is X = c1 cos x + c2 sin x. X(0) = 0 implies c1 = 0. Applying the second condition in (4) to X = c2 sin x implies (5)

76. Example 1 (4)
Because the graph of y = tan x and y = −x/h, h > 0, have an infinite number of points of intersections for x > 0, (5) has an infinite number of roots. If the consecutive positive roots are denoted by n, n = 1, 2, …, then the eigenvalues n = n2 and the corresponding eigenfunctions X(x) = c2 sin nx, n = 1, 2, …. The solution of (2) is

77. Example 1 (5)
Now at t = 0, u(x, 0) = 1, 0 < x < 1, so that (6) (6) is an expansion of u(x, 0) = 1 in terms of the orthogonal functions arising from the Sturm-Liouville problem (4). The set {sin nx} is orthogonal w.r.t. the weight function p(x) = 1. From (8) of Sec 12.1, we have (7)

78. Example 1 (6)
We found that (8)

79. Example 1 (7)
Thus (7) becomes

80. Example 2
See Fig The PDE is described by

81. Fig 13.19

82. Example 2 (2)
Solution Similarly we have (9) (10) (11) (9) together with the homogeneous boundary conditions in (11), (12) is a regular Sturm-Liouville problem.

83. Example 2 (3)
For  = 0 and  = −2,  > 0, the only solution is X = 0. For  = 2,  > 0, applying X(0) = 0 and X(1) = 0 to the solution X = c1 cos x + c2 sin x implies c1 = 0, c2 cos  = 0. Thus n = (2n – 1)/2 and the eigenvalues are n = n2 = (2n – 1)22/4, and the corresponding eigenfunctions are

84. Example 2 (4)
The initial condition t(x, 0) = 0 implies X(x)T(0) = 0 or T(0) = 0. When applied to T(t) = c3 cos ant + c4 sin ant of (10) implies c4 = 0, T(t) = c3 cos ant = c3 cos a((2n – 1)/2)t. Thus

85. Example 2 (5) When t = 0, we must have, for 0 < x < 1, (14)
As in Example 1, the set {sin((2n – 1)/2)x} is orthogonal w.r.t. the weight function p(x) = 1 on [0, 1]. We have

86. Example 2 (6)
Finally

87. 13.8 Fourier Series in Two Variables
Heat and Wave Equation in Two Dimensions Two-dimensional heat equation: (1) Two-dimensional wave equation: (2)

88. Fig 13.21

89. Example 1
Find the temperature u(x, y, t) in the plate if the initial temperature is f(x, y) and if the boundary conditions are held at temperature zero for time t > 0.
Solution We must solve

90. Example 1 (2)
If we let u = XYT, we get (3) Similarly, we can obtain and so (4) (5)

91. Example 1 (3)
By the same reason, we introduce another separation constant − in (5) then
Now the homogeneous conditions

92. Example 1 (4)
Thus we have two problems, one in x (7) and the other in y (8) Similarly we have two independent sets of eigenvalues and eigenfunctions defined by sin b = 0 and sin c = 0. That is (9)

93. Example 1 (5)
(10) After substituting the values in (9) into (6), its general solution is

94. Example 1 (6) Using the superposition principle in a double sum (11)
At t = 0, we have (12) and (13)

95. Equation (11) is called a sine series in two variables
Equation (11) is called a sine series in two variables. The cosine series in two variables is given by

97. Thank You !