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Boundary Value Problems

Dr. Suresh Kumar, BITS Pilani

1

Chapter 7 Boundary Value Problems Note: This module is prepared from Chapter 7 of the text book (G.F. Simmons, Differential Equations with Applications and Historical Notes, TMH, 2nd ed., 1991) just to help the students. The study material is expected to be useful but not exhaustive. For detailed study, the students are advised to attend the lecture/tutorial classes regularly, and consult the text book.

Appeal: Please do not print this e-module unless it is really necessary.

Dr. Suresh Kumar, Department of Mathematics, BITS Pilani, Pilani Campus

Contents 0.1 0.2 0.3 0.4

One dimensional wave equation . . . . . . . . . . . One dimensional heat equation . . . . . . . . . . . The Laplace equation . . . . . . . . . . . . . . . . . Strum Liouville Boundary Value Problem (SLBVP) 0.4.1 Orthogonality of eigen functions . . . .

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Boundary Value Problems In this chapter, we shall discuss the solution of some boundary value problems.

0.1

One dimensional wave equation

Consider an elastic string of negligible mass and length π tied at the two ends along x-axis at the points (0, 0) and (π, 0). Suppose the string is pulled in the shape y = f (x) in the xy-plane and released. Then it can be shown that the vibrations of the string in the xy-plane are governed by the one dimensional wave equation ∂ 2y 1 ∂ 2y , = a2 ∂t2 ∂x2

(1)

where a is some positive constant, and y(x, t) is the displacement/vibration of the string along y-axis direction. The wave equation is subjected to the following four conditions. The first condition is y(0, t) = 0,

(2)

since the left end of the string is tied at (0, 0) for all the time, and hence it can not have displacement along the y-axis. The second condition is y(π, t) = 0

(3)

since the right end of the string is tied at (π, t) for all the time, and hence it can not have displacement along the y-axis. The third condition is ∂y = 0, ∂t

at t = 0,

(4)

since the string is in rest at t = 0. The fourth condition is y(x, 0) = f (x),

(5)

since the string is in the shape y = f (x) at t = 0. Once the string is released from the initial shape y(x, 0) = f (x), we are interested to find the distance or displacement of the string from the x-axis at any time t. It is equivalent to saying that we are interested to solve (1) for y(x, t) subject to the four conditions (2)-(5). 3

Boundary Value Problems

Dr. Suresh Kumar, BITS Pilani

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Assume that (1) possesses a solution of the form y(x, t) = u(x)v(t),

(6)

where u(x) and v(t) are to be determined. Plugging (6) into (1), we get 1 v ′′(t) u′′(x) = 2 = λ, a v(t) u(x)

(7)

where λ is some constant. This yields following two equations u′′(x) − λu(x) = 0,

(8)

v ′′(t) − λa2 v(t) = 0.

(9)

Now, let us first solve (8). Later, we shall look for the solution of (9). Considering (6), the condition y(0, t) = 0 in (2) gives u(0)v(t) = 0 or u(0) = 0. Similarly, y(π, t) = 0 in (3) gives u(π ) = 0. Further, we see that the nature of solution of (8) depends on the values of λ. √

√

(i) When λ > 0, the solution reads as u(x) = c1 e λx + c2 e− λx . Using the conditions u(0) = 0 and u(π) = 0, we get c1 = 0 = c2 , and hence u(x) = 0. This leads to the trivial solution y(x, t) = u(x)v(t) = 0, which is not of our interest. (ii) When λ = 0, the solution reads as u(x) = c1 x + c2 . Again, using the conditions u(0) = 0 and u(π) = 0, we get c1 = 0 = c2 , which leads to the trivial solution y(x, t) = u(x)v (t) = 0. (iii) When λ < 0, say, λ = −n2 , the solution reads as u(x) = c1 sin nx + c2 cos nx. Applying the condition u(0) = 0, we get c2 = 0. The condition u(π) = 0 then implies that c1 sin nπ = 0. Obviously, for a non-trivial solution we must have c1 6= 0. Then the condition c1 sin nπ = 0 forces n to be a positive integer. Thus, un (x) = sin nx,

(10)

is non-trivial solution of (8) for each positive integer n. Now, the solution of (9) with λ = −n2 reads as v(t) = c1 sin nat + c2 cos nat. The condition in (4) leads to u(x)v ′ (0) = 0 or v ′ (0) = 0, which in turn gives c1 = 0. So vn (t) = cos nat,

(11)

is non-trivial solution of (8). In view of (6), (10) and (11), we can say that yn (x, t) = un (x)vn (t) = sin nx cos nat,

(12)

is a solution of (1) for each positive integer n. It follows that y(x, t) =

∞ X n=1

bn yn (x, t) =

∞ X n=1

bn sin nx cos nat,

(13)

Boundary Value Problems

Dr. Suresh Kumar, BITS Pilani

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is also a solution of (1). To determine bn , we use the fourth condition y(x, 0) = f (x) given in (5). Then (13) gives f (x) =

∞ X

bn sin nx.

(14)

n=1

Notice that the series on right hand side in (14) is the Fourier sine series of f (x) in the interval [0, π]. So we have Z 2 π bn = f (x) sin nxdx. (15) π 0 Hence, y(x, t) =

∞ X

bn sin nx cos nat,

(16)

n=1

with bn from (15) is the solution of (1) subject to the four conditions (2)-(5).

0.2

One dimensional heat equation

Consider a uniform rod of length π aligned along x-axis from (0, 0) to (π, 0). Suppose that the two ends of the rod are kept at zero temperature all the time, and f (x) represents the temperature function at time t = 0. Then it can be shown that the temperature w(x, t) of the road is governed by the one dimensional heat equation ∂ 2w 1 ∂w , = 2 2 a ∂t ∂x

(17)

where a is some positive constant. The heat equation is subjected to the following three conditions. The first condition is w(0, t) = 0,

(18)

since the left end of the rod is kept at zero temperature for all t. The second condition is w(π, t) = 0

(19)

since the right end of the rod is kept at zero temperature for all t. The third condition is w(x, 0) = f (x),

(20)

since the temperature of the rod is given by f (x) at t = 0. Having known the temperature of the rod at t = 0, we are interested to find the temperature of the rod at any time t. It is equivalent to saying that we are interested to solve (17) for w(x, t) subject to the three conditions (18)-(20).

Boundary Value Problems

Dr. Suresh Kumar, BITS Pilani

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Assume that (17) possesses a solution of the form w(x, t) = u(x)v(t),

(21)

where u(x) and v(t) are to be determined. Plugging (21) into (17), we get 1 v ′ (t) u′′(x) = 2 = λ, u(x) a v(t)

(22)

where λ is some constant. This yields following two equations u′′(x) − λu(x) = 0,

(23)

v ′ (t) − λa2 v (t) = 0,

(24)

Following the strategy discussed in the previous section, the non-trivial solution of (23) subject to the conditions (18) and (19), reads as un (x) = sin nx,

(25)

where n a positive integer with λ = −n2 . Now, the solution of (24) with λ = −n2 reads as v(t) = c1 e−n vn (t) = e−n

2 a2 t

2 a2 t

. So (26)

,

is non-trivial solution of (24). In view of (21), (25) and (26), wn (x, t) = un (x)vn (t) = sin nxe−n

2 a2 t

(27)

,

is a solution of (17) for each positive integer n. It follows that w(x, t) =

∞ X

bn wn (x, t) =

n=1

∞ X

bn sin nxe−n

2 a2 t

,

(28)

n=1

is also a solution of (17). To determine bn , we use the third condition w(x, 0) = f (x) given in (20). Then (28) gives f (x) =

∞ X

bn sin nx.

(29)

n=1

Notice that the series on right hand side in (29) is the Fourier sine series of f (x) in the interval [0, π]. So we have Z 2 π f (x) sin nxdx. (30) bn = π 0 Hence,

w(x, t) =

∞ X

bn sin nxe−n

2 a2 t

,

n=1

with bn from (30) is the solution of (17) subject to the three conditions (18)-(20).

(31)

Boundary Value Problems

0.3

Dr. Suresh Kumar, BITS Pilani

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The Laplace equation

The steady state temperature w(x, y) (independent of time) in the two dimensional xy-plane is governed by ∂ 2w ∂ 2w = 0, + ∂y 2 ∂x2

(32)

known as the Laplace equation. With the transformations x = r cos θ and y = r sin θ, the polar form of (32) reads as ∂ 2w 1 ∂w 1 ∂ 2w = 0. + + r 2 ∂θ 2 ∂r 2 r ∂r

(33)

For, ∂w ∂x ∂w ∂y ∂w ∂w ∂w . = + = cos θ + sin θ ∂y ∂x ∂r ∂y ∂r ∂r ∂x 2 2 ∂ 2w ∂ 2w ∂ 2w 2 ∂ w 2 ∂ w = cos θ + cos θ sin θ + sin θ + sin θ cos θ ∂x∂y ∂r 2 ∂x2 ∂x∂y ∂y 2 ∂w ∂w ∂w ∂x ∂w ∂y ∂w = −r sin θ + r cos θ + = . ∂x ∂θ ∂y ∂y ∂θ ∂x ∂θ 2 2 ∂w ∂w 2 ∂ 2w ∂ 2w ∂ 2w 2 ∂ w 2 2 ∂ w 2 2 −r sin θ −r cos θ +r cos θ = r sin θ −r cos θ sin θ −r sin θ cos θ . 2 2 2 ∂x∂y ∂x∂y ∂x ∂θ ∂y ∂x ∂y 2

2

∂ w ∂w ∂ w Substituting the values of ∂r 2 , ∂r and ∂θ 2 into (33), we get (32). Suppose the steady state temperature is given on the boundary of a unit circle r = 1, say w(1, θ) = f (θ). Then the problem of finding the temperature at any point (r, θ) inside the circle is a Dirichlet’s problem for a circle. Now we shall solve (33) subject to the condition

w(1, θ) = f (θ).

(34)

Assume that (33) possesses a solution of the form w(r, θ) = u(r)v(θ),

(35)

where u(r) and v(θ) are to be determined. Plugging (35) into (33), we get r 2 u′′(r) + ru′ (r) v ′′(θ) = λ, =− v(θ) u(r)

(36)

where λ is some constant. This yields following two equations v ′′(θ) + λv(θ) = 0,

(37)

r 2 u′′(r ) + ru′ (r) − λu(r ) = 0.

(38)

The non-trivial solution of (37) reads as vn (θ) = an cos nθ + bn sin nθ,

(39)

Boundary Value Problems

Dr. Suresh Kumar, BITS Pilani

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where λ = n2 ; an , bn are constants such that both the terms on right hand side of (41) do not vanish together for n = 1, 2, 3, ...... Notice that (40) is a Cauchy-Euler DE with λ = n2 . So it transforms to d2 u − n2 u = 0, ds2

(40)

where r = es . Solution of this equation is u(z) = c1 ens + c2 e−ns

for n = 1, 2, 3, .....

where c1 and c2 are constants. In terms of r, the solution read as u(r ) = c1 r n + c2 r −n

for n = 1, 2, 3, .....

Since we are interested in solutions which are well defined inside the circle r = 1, we discard the second term r −n as it is not finite at r = 0. Thus, the solution of our interest is un (r) = r n , n = 1, 2, 3, .....

(41)

In view of (35), (40) and (41), wn (r, θ) = un (r)vn (θ) = r n (an cos nθ + bn sin nθ ),

(42)

is a solution of (33) for n = 1, 2, ....... It follows that ∞ X

wn (x, t) =

n=0

∞ X

r n (an cos nθ + bn sin nθ ),

(43)

n=1

is also a solution of (33). Since w(r, θ) =

a0 2

is also a solution of (33), so

∞ a0 X r n (an cos nθ + bn sin nθ ), + 2 n=1

(44)

is also a solution of (33). To determine a0 , an and bn , we use the third condition w(1, θ) = f (θ) given in (34). Then (44) gives f (θ) =

∞ a0 X (an cos nθ + bn sin nθ), + 2 n=1

(45)

Notice that the series on right hand side in (45) is the Fourier series of f (θ) in the interval [−π, π]. So we have Z 1 π an = f (φ) cos nφdφ, (n = 0, 1, 2, ....) (46) π −π 1 bn = π

Z

π

−π

f (φ) sin nφdφ, (n = 1, 2, 3, ......).

(47)

Boundary Value Problems

Dr. Suresh Kumar, BITS Pilani

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Thus, (44) with an from (46) and bn from (47) is the solution of (33) subject to the condition (34). Thus, the Dirichlet problem for the unit circle is solved. Now substituting an from (46) and bn from (47) into (44), we get " # Z ∞ 1 X n 1 π f (φ) r cos n(θ − φ) dφ. (48) w(r, θ) = + π −π 2 n=1 Let z = rei(θ−φ) = r[cos(θ − φ) + i sin(θ − φ)] so that z n = r n [cos n(θ − φ) + i sin n(θ − φ)]. Then we have # " ∞ ∞ 1 X n 1 X n + + z r cos n(θ − φ) = Re 2 n=1 2 n=1   z 1 + = Re 2 1−z   1+z = Re 2(1 − z)   (1 + z)(1 − z¯) = Re 2(1 − z)(1 − z¯)   (1 + z − z¯ − |z|2 ) = Re 2(1 − z − z¯ + |z|2 ) 1 − r2 = 2[1 − 2r cos(θ − φ) + r 2 ] So (48) becomes w(r, θ) =

1 2π

Z

π

−π

1 − r2 f (φ)dφ, 1 − 2r cos(θ − φ) + r 2

(49)

known as the Poission integral. It expresses the value of the harmonic function w(r, θ) at all points inside the circle r = 1 in terms of its values on the circumference of the circle. In particular, at r = 0, we have Z 1 π w(0, θ) = f (φ)dφ, (50) 2π −π which shows that the value of the harmonic function w at the center of the circle is the average of its values on the circumference.

0.4

Strum Liouville Boundary Value Problem (SLBVP)

Let p(x) 6= 0, p′ (x), q(x) and r(x) be continuous functions on [a, b]. Then the DE d [p(x)y ′ ] + [λq(x) + r (x)]y = 0, dx

(51)

with the boundary conditions c1 y(a) + c2 y ′ (a) = 0,

(52)

Boundary Value Problems

Dr. Suresh Kumar, BITS Pilani

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and d1 y(b) + d2 y ′ (b) = 0,

(53)

where neither both c1 and c2 nor both d1 and d2 are zero, is called a SLBVP. We see that y = 0 is trivial solution of (51). The values of λ for which (51) has non-trivial solutions, are known as its eigen values while the corresponding non-trivial solutions are known as eigen functions. Ex. 0.4.1. Find eigen values and eigen functions of the SLBVP y ′′ + λy = 0,

y(0) = 0, y (π) = 0.

Sol. 0.4.1. Eigen values are λ = n2 , where n is a positive integer. Eigen functions are yn = sin nx.

0.4.1

Orthogonality of eigen functions

Consider the SLBVP given by (51), (52) and (52). If ym and yn are any two distinct eigen functions corresponding to the eigen values λm and λn , then Z b q(x)ym (x)yn (x)dx = 0. a

In other words, any two distinct eigen functions ym and yn of the SLBVP are orthogonal with respect to the weight function q(x). Let us prove this result. Since ym and yn are eigen functions corresponding to the eigen values λm and λn , we have (pym′ )′ + (λm q + r)ym = 0

(54)

(pyn′ )′ + (λn q + r)yn = 0.

(55)

and

Multiplying (54) by yn and (55) by ym , and subtracting we get ′ ′ ) − ym (pyn′ )′ + (λm − λn )qym yn = 0 yn (pym

(56)

Moving the first two terms on right hand side, and then integrating from a to b, we have Z b Z b Z b ′ ′ ′ ′ qym yn dx = (λm − λn ) ) dx yn (pym ym (pyn ) dx − a a a Z b Z b b ′ ′ ′ ′ b yn′ (pym′ )dx ym (pyn )dx − [yn (pym )]a + = [ym (pyn )]a − a

a

′ (b)] − p(a)[ym (a)yn′ (a) − yn (a)ym′ (a)] = p(b)[ym (b)yn′ (b) − yn (b)ym = p(b)W (b) − p(a)W (a)

where W (x) = ym (x)yn′ (x) − yn (x)ym′ (x) is Wronskian of ym and yn . ∴ (λm − λn )

Z

a

b

qym yn dx = p(b)W (b) − p(a)W (a).

(57)

Boundary Value Problems

Dr. Suresh Kumar, BITS Pilani

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Notice that the eigen functions ym and yn are particular solutions of the SLBVP given by (51), (52) and (52). So we have ′ (a) = 0, c1 ym (a) + c2 ym

(58)

c1 yn (a) + c2 yn′ (a) = 0,

(59)

d1 ym (b) + d2 ym′ (b) = 0,

(60)

d1 yn (b) + d2 yn′ (b) = 0.

(61)

Now by the given, c1 and c2 are not zero together. So the homogeneous system given by (58) and (59) has a non-trivial solution. It follows that ym (a)yn′(a) − yn (a)ym′ (a) = W (a) must be zero. Likewise, (60) and (61) lead to ym (b)yn′ (b) − yn (b)ym′ (b) = W (b) = 0. So (57) becomes (λm − λn )

Z

b

qym yn dx = 0.

(62)

a

Also, λm 6= λn . So we get Z

b

qym yn dx = 0,

(63)

a

the desired result. Remark 0.4.1. The orthogonality property of eigen functions can be used to write a given function as the series expansion of eigen functions. Remark 0.4.2. A DE in the form d [p(x)y ′ ] + [λq(x) + r (x)]y = 0 dx is called in self adjoint form....