Boundary-Value Problems in Rectangular Coordinates

CHAPTER 13 Boundary-Value Problems in Rectangular Coordinates

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

13.1 Separable Partial Differential Equations • Linear PDEIf we let udenote 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.

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

Example 1 Find product solution of SolutionLet u = X(x)Y(y) and thenWe introduce a real separation constant as −.

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

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) areX = c4 cosh 2x + c5 sinh 2x and Thus (7)where A2 = c4c6, B2 = c5c6.

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

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

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

Example 2 Classify the following equations: Solution (a)

Example 2 (2)

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.

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

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

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

13.3 Heat Equation • IntroductionThe heat equation can be described by the following (1) (2) (3)

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

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

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)

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

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

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)

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

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

Solution of the BVP • Assuming u(x, t) = X(x)T(t), then (1) givesso that (4) (5)

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) isX(0) = 0 and X(L) = 0 imply that c1= 0 and c2sin L = 0. Thus we have  = n/L, n = 1, 2, 3, …

The eigenvalues and eigenfunctions are

Let An = c2c3,Bn = c2c4, solutions that satisfy (1) and (2) are (7)and (8)

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

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

Standing Wave • It is easy to transform (8) into

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.

Fig 13.9

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

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.

Thus we have the following equation (6)For  = 0, (6) becomes X” = 0, X’(0) = 0, X’(a) = 0The solution is X = c1 + c2x. X’(0) = 0 implies c2 = 0 and X = c1also satisfies the condition X’(a) = 0. Thus X = c1, c1  0 is a nontrivial solution. • For  = −2 < 0,  > 0, (6) possesses no nontrivial solutions.

For  = 2 > 0,  > 0, (6) becomes X” + 2X = 0, X’(0) = 0, X’(a) = 0Applying 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) areFor Y” – Y = 0, when 0 = 0, the solution is Y = c3 +c4y. Y(0) = 0 implies c3 = 0 and so Y = c4y.

For n = (n/a)2, n = 1, 2, …, the solution isY = 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

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

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

Dirichlet Problem • Please verify that the solution of the following Dirichlet Problem

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.

Problem 1:

Problem 2:

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 13.15.

Fig 13.15

It is an exercise that the solution of problem 1 is

The solution of problem 2 is

Boundary-Value Problems in Rectangular Coordinates