Section 3.3 Wave Equation, the Method of Separation of Variables 123

already gives a very good picture of how the string moves. This can be justified byobserving that the coefficients in the series in (12) decrease rapidly to zero, and sothe contributions of additional terms become small.

For problems with nonzero initial displacement and velocity, we haveonly to work them as in Examples 1 arid 3 and put the results together, asspecified by equations (8) and (9). That is, from the initial displacementf(x) we find the ba’s as in Example 1, from the initial velocity g(x) we findthe b ‘s as in Example 3, and then we put these results into (8). We notetoo that, if we have the Fourier sine series of f(x) and/or g(x) at hand, ourtask is considerably reduced, as illustrated by Example 3.

Exercises 3.3In Exercises 1 10, (a) solve the boundary value problem (1) (3) for a string of ‘unitlcngth,s’ubect to the given co’aditzons.(b) Illustrate the motion of the string by piott’ng a partial sum of your seriessolution at. various value of t. 7b decide how many terms to include in yourpa’rtia.l sum, compare the graph at t = 0 and the (J’raph of 1(x). The !Ir’o.pIIs should‘match when you have enough terms iii. your partial sum.1. f(x) = .D5siiiir:u, g(x) = U, c =

2. f(.r) sin r:r cos ir.r, g(a) = 0, c =

3. 1(x) siii ir.r + 3 sin 2yr.r — sin 5ir.r, g(x) = 0, c = 1.4. f(x) = sin irx + sin 3icr + 3 sin 7irx, g(:r) = sin 2ir.i. c = 1.5. g(•) 0, c = 4,

f( — J 2x if 0 < .c <.ei 2(1—c) if<.r’<i.

— 6. g(x) = 2, c =

0 if0<x<

= 3(.i — 1/3) if . < .r <

1(1) if<x<1.

7. g(.r) = 1, 1,

if0<:r<

f(.) = I if - < i <

4(1—.r) if<.r<l.

8. ,f(x) = .i; sin icr. g(r) = 0. c -

9. 1(x) = x(1 — •r), g(x) = siuirx, c = 1

124 Chapter 3 Partial Diflérential Equations in Rectangular Coordinates

10. g(r) = 0, c = 1,

4x if0<x,

J(;r) — 1/2) if <x <

( 4(x—1) if<x<l.

11. Time period of motion. (a) Show that the nth normal mode (11) is periodicin time with period 2L/nc. Conclude that for any n, a period of u,, is 2L/c.(b) Show that any superposition of normal modes is periodic in time with period2L/c. Conclude that the string vibrates with a time period 2L/c.(c) Shape of the string at half a time period. Using (8), show that for all :rand t, u(x, t + L/c) = —u(L — .r, t). What does this imply about the shape of thestring at half a time period’?

Project Problem: Solve a case of the wave equation with dainpi ig in Exercise 12and then apply your solution to a specific problem by domg any one of Exercises 1315.12. Damped vibrations of a string. In the presence of resistance proportionalto velocity, the one dimensional wave equation becomes

2 a2 u+ 2k (‘dt2 th2

(see (7), Section 3.2). We will solve this equation subject to conditiojis (2) and (3)by following the method of this section.(a) Assume a product solution of the form u(z, t) = X(.r)T(i), and derive thefollowing equations for X and T:

X” + 2X = 0, X(0) = 0, X(L) = 0,

T” + 2kT’ +(1w)2T = 0,

where is the separation constant.(h) Show that

flit . flit= ,,, and X = = sin ii = 1, 2

(c) To deterimne the solutions in T we have to solve T” + 2k T’ + ( e)2T = 0.Review the general solution of the second order linear differential equation withconstant coefficients (Appendix A.2), and explain why three possible cases are tohe treated separately:’n < , a

= ,and n > . The respective solutions for

T areT, = (,kt (a cosli X,,t + b,, sinh A

Tjj ae + btc_kt,

T = e._kt(aL cos .Xt + bTl sin A,t),

where

______________

Section 3.3 Wave Equation, the Method of Separation of Variables 125

(d) Conclude that when is not a positive integer, the solution is

u(x, t) = 0—ktsin cosh A7t + b,, sinh Ant)

sin x(a1,cos A, + b, sin Ant),<n < CO

where these smns run over integers only, and where

2 1 nita,,L J 1(x) sm dx. n 1, 2.

and the h,, are determined from the equation

—ku,, + X,,b,,= 2 f (x) sin dx. n = 1,2

,.e) Conclude that when is a positive integer, the soiutiou is as in (ci) with theone additional term

sin(x)(ae’ + bte)

with a. and h,, as in (ci), except that b is determined from the equation

2 i’ k—ka + hkl.= L J g(x) sin —x dx.

— 13. Solve‘9 .9c)’u chi. d— ulit2 +

ii(O, t.) = u(i-, t) = 0,

nCr, 0) = sin .r, (.r, 0) = 0.

IHint: Since k .5 and L it, we have ii. > for all n. So only one case fromthe solution of Exercise 12 needs to be considered.]14. Solve

li2fi DI! 1)2tt

‘ti(O, t) = ‘fL(ir, t) = 0,

u(x,0)= .rsiu.r, ‘(x,0) =0.

15. (a) Solve

thc d2u

ii(O, t) = ‘u(ir, t) = 0.Du

u(x, 0) 0, -—(x, 0) = 10.(h) Illustrate graphically the fact that the solution tends to zero as t tends toinfinity.

126 Chapter 3 Partial Differential Equations in Rectangular Coordinates

3.4 D’Alembert’s MethodAs promised earlier, we will show in this sect;ion how the Fourier series

solution of the boiuiclary value prohleiii associated with the vibrating string,

‘2 ‘9‘n 2(1) C)2C 2’

O<x<L, t>O,

(2) ‘u(O, t) = 0 and ‘u(L, t) = 0 for all t > 0,

(3) u(x, 0) = f(x) and (i, 0) = g(.r) ft)r 0 <x <L,

has a simpler expression in terms of the initial data f and g More precisely,

we will show that the solution of (1) (3) is given by

1 1 d+(t

(4) u(x, ) — [f*C1;— ct) + f*(x + Ct)] +

— J g*() ds,2 2c i—d

where f and g denote the odd extensions of J and g. This is called

d’Alembert’s solution of tile vibrating string probleni, and it has an in

teresting interpretation 111 terms of traveling waves. The derivation of (4)

from the solution of the previous section ((8), Section 3.3) involves using

trigonometric identities, as illustrated by the following example.

EXAMPLE 1 From Fourier series to d’Alembert’s solutionWhen f(:i:) = sin and g(:s) = 0, the Fourier series method of the previous

section yields the following solution of (1) (3):

mxuC, t) = sin —t-— cos

(See Example 2, Section 3.3.) On the other hand, cl’Aleinbert’x solution (4) yieldsthe following form of the solution:

1 . ‘nor . mit‘u(:r, t) = [sin j(:r — ct) + sin _L_(v + ct)]

Recalling the trigonometric identity sin a cos b = - [sin(a + b) + sin(a — b)], we see

that the two solutions are the same.

The derivation of d’Aleinbert’s solution (4) from the Fourier series solu

tion of the previous section is based on similar ideas and is outlined in the

exercises. It is more instructive at this point to check the validity of (4) by

verifying that it satisfies the equations (1)—(3). To simplify the notation, we

drop the * and use the same notation for a function and its odd extension.

Section 3.4 D’Alembert’s Method 127

In addition, we assume that all derivatives encountered in the following computations exist. We begin by showing that (4) satisfies (1). Differentiating(4) with respect to t, using the chain rule and the fundamental theorem ofcalculus, we get

thL L) 1 1= {[.f(x_ct)+f(x+ct)i+f g(s)ds}

= [—cf’(x — et) + cf’(x + et)] + [g( + Ct) + g(x — ct)].

Differentiating a second time with respect to t, we get

=

[f”( — Ct) + f”(i; + et)] + [g’(x + ct) g’(r —

Likewise, differentiating (4) with respect to x we get

I 1= [.1 (x — et) + j (x + et)] + — [g(:r: + ct) — g(x

—

and

32’tm 1 ,, 1dT2 =

[.1 (x — ct) + f”(x + rt)] + — [q’(x + ct) g’(x — ct)].

It follows that = c2, and so (4) satisfies the wave equation (1).To check that (4) satisfies (2) and (3), we use the fact that f* and g* are

odd and 2L—periodic. For example, to check the boundary condition (2) at= 0, we plug c = 0 into (4) arid get

1 1u(0, t) = — {f4(—ct) + f*(ct)] +

— / g*(s) ds = 0,2 2J—ct

because ,f* is odd, so f*(_ct) = _f*(ct), and g* is oH, so its integral overa symmetric interval is 0. We leave the verification of the second conditioniii (2) arid (3) to Exercise 14.

Geometric Interpretation of D’Alembert’s SolutionWhemi the initial velocity is zero, d’Alenibert’s solution takes on the simplerform

(5) u(x, t) = ‘[f*(x- ct) + f*( + ct)]

This has an interesting geometric interpretation. For fixed t, the graph off*

(x — ct) (as a function of x) is obtained by translating the graph of f* (x)

128 Chapter 3 Partial Differential Equations in Rectangular Coorclinates

by ct units to the right. As t increases, the graph represents a wave traveling

to the right with velocity c. Similarly, the graph of f* (x + Ct) is a wave

traveling to the left with velocity c. VTe see froni (5) that this solution of

the wave equation is an average of two waves traveling in opposite directions

with shapes determined from the initial shape of the string.

The general form of d’Alembert’s solution (4) is harder to interpret ge

ornetrically. It does tell us, however, that the displacement at the point x

at time t > 0 is determined entirely by the initial displacements at positions

x — ct and x + ct and by the initial velocity on the interval between x — ct and

x + ct. To understand the contribution of the initial velocity to the motion,

let G denote an antiderivative of g*• Hence

G(x)=

*(Z) dz,

for some fixed number a. Note that

G(x+2L)—G(x)=] g*(z)dz=J g*(z)dz=0,-L

where the second equality follows from Theorem 1, Section 2.1, and the last

equality follows because g is odd. Hence G is 2L periodic. (Antiderivatives

of periodic functions were discussed in Exercises 1516, Section 2.1.) The

solution (4) may be rewritten in terms of f* and G as follows:

1 1u(x, t) = [f’(x — Ct) + f*(x + ct)] + — [G(x + Ct) — G(x — ct)j

(6)1 1 1 1

= _[f*(x_Ct)__G(x_Ct)1+_[f*(x+Ct)+_G(x+Ct)1.

showing that, in general, the solution still consists of right- and left-moving

waveforms. The main difference from the case represented by (5) is that

here the two waveforms need no longer have the same shape. The functions[f*(x) — G(x)j and [f*(x) + G(z)] give the shapes of the right- and

left-moving waves, respectively.

EXAMPLE 2 D’Alembert’s solution with zero initial velocity

Consider the wave problem of Example 1, Section 3.3, where L = 1, c =

( x if0,f(x) =

3(1—x) ifx1,

and g(x) = 0. (a) Use d’Alernbert’s solution to determine the shape of the string

at times t = and f.(b) Determine the first time when the string returns to its initial shape.

Section 3.4 D’Alembert’s Method 129

Solution (a) Since g(x) = 0, we use (5) and get the shape of the string at timet as the graph of [f*(x— 1/3) + f*(x + 1/3)]. To plot this graph, we firstplot f* (x), the 2-periodic odd extension of f. By translating this graph to the leftby 1/3 unit, we obtain the graph of f* (x + 1/3); translating it to the right by thesame amount, we obtain the graph of f*(x

— 1/3). Now the shape of the string attime t = ir/3 is obtained by averaging (adding and dividing by two) the graphs off*(x + 1/3) and .1 *(:r— 1/3). Since we are only interested in the shape of the string,we restrict the graphs to the interval 0 < x < 1. See Figure 1, where we have alsoplotted the graphs for = 27r/3.

(h) The string returns to its initial shape when

- t/) + f(x + t/)] = ,f* (x).

Since f* is 2—periodic, tins happens when f/ir = 2, or t = 2ir. I

EXAMPLE 3 D’Alembert’s solution with nonzero initial velocityUse d’Alcmbert’s nietliod to solve the wave problem (1) (3) with c 1, L = 1,f(x) = 0, and g(x) = .r for 0< x < 1.Solution We use (4) with ,f*

= 0. Hence

1 x+t1(7) u(x,t)

=

g*(s)ds = (G(x+t) —G(x—t)),

where g4 is the odd 2-periodic extension of g and G is an antiderivative of g. Letus take

0(x)= / g*(z)dz

(a).1 f(x)

l2/1/1/3

1/2fm(x+ 1/3)

f”(x)

1/3 2/3

.05

—0.1

2

u(x, it/3)

x

l/2t(x + 2/3).1

.05- - - - - -

-

l/2/*(x— 2/3)

2/3-2/3X

0.05-

—.05-

1/21*(x— 1/3)

(b)—.1 (c) u(x, 2it/3)

liure 1 (a) Initial shape of the string, f(x), 0 < .r < 1, and its odd extension f*(x). (b) Sllapshot of the stringat tulle t = ir/3, obtainedi by averaging the translates ,f*(x + 1/3) and ,f*(:r

— 1/3): u(x, ir/3) = 1/2(f*(.T + 1/3) +i — 1/3)). (c) Snapshot of the string at time / = 2ir/3: u(.i:, 2ir/3) 1/2 (f* (x + 2/3) + f* (.z: — 2/3)).

130 Chapter 3 Partial Differential Equations in Rectangular Coordinates

To complete the solution, we imist determine G. From our discussion preceding (6),

we know that C is 2—periodic. Thus, it suffices to deternune G on any interval of

length 2. Since g (a) = a on the interval (—1, 1), we obtain

br all i in (— 1, 1). Hence

1 iGCr)

=

z dz = .r — —

1,1.2_I jf —1<a<i,G(.i)

= { ãr + ) otherwise.

F’igure 2 (a) Graph of q, the

a—periodic 0(1(1 extension of g.

(b) Graph of C, an antideriva—

ive of g. Note that c; is 2-

periodic.

The graph of C is shown in Figure 2(b). According to (7), to get a snapshot of

the string at a given time t, it suffices to take the difference of the left and right

translates of the graph of C by t units and (livide by 2. In this case we have two

half—length waves still, but one is inverted. We then just superpose theni again.

Note how this yields the zero initial position when t = 0.

Characteristic Lines

Figure 3

Here we discuss some interesting properties of the wave equation and its

solution. Up until now, in order to interpret ‘u(x, t.) as the shape of the

string at a given time t, we have been thinking of ‘u(x, t) as a function of x,

for a fixed value of t. For the sake of our discussion, it would help to change

the way we think of ‘u and consider u(x, t) as a function of (a;, t), where

a; and t vary simultaneously in the xt-plane. Because the string has finite

length L, we are particularly interested in the semi—infinite strip S, which

consists of the points (as, t) with 0 x L and t 0.

Tue lines with slopes ±‘ in the xt-plane play an important role.. These

are called the characteristic lines of the wave equation and can he written

in the form

(8) x — ct = :o — ct0 (slope = > 0, through the point (x0, to));

(9) x + ct = xo + ct0 (slope = — < 0, through the point (x0, to)).

The characteristic lines consists of two families of parallel lines that cover the

xt-plane (Figure 3). Each point (x0, t0) is the vertex of an isosceles triangle

formed by two characteristic lines, which intersect the x-axis at the points

— ct0 and x0 + ct0 (Figure 4). The base of this triangle, [x0 — ct0, x0 + ctol,

(a)

//2

(b)

Y

— P2

-I

G(x)

01 3x

LX— ctx0— Cl0 x+ ct=x0+Cl0

Section 3.4 DAleinbert’s IVlothod 131

is called the interval of dependence of the point (x0, to) (Figure 4). Tounderstand this terminology, from d’Alcmberts solution (4), we see thatu(xO, t0) is deternuned by the values of f* at the emiponits of the intervalof dependence awl by integrating g on that itterval. The values of f andg outside the interval of dependence of (x0. t)) do not affect in ally waythe value of ‘u(xo, to); consequently, any pertiubation outside this intervalis not felt at the point (:ro, to). From Figure 4 it is clear that the intervalof dependeiice [.r1) — ct, T0 + cto] lies entirely inside the interval [0, L] ifand only if the point (co to) belongs to the triangular regioll I hounded bythe interval [0, Lj and the characteristic lines :r — et = 0 (through the originwith slope ) and u + et = f, (through the (0. L) with slope -b). Thus, for

J’igure 4 Interval of ‘_ (.r, t) inside the region I, (.r, 1) depends only on the values of f and g overa tidence, centered at :r0, the interval [0, U; awl so iii order to compute u (.r, 1) We (10 not need the

no hus ct0 J)eiiodic extensions of f and q.

EXAMPLE 4 Using intervals of dependenceCotisider the wave J)rOhleil i

= 10.,.r. 1) < i < 1, 1 > 1).

u(0, 1) =r 0, u(1, 1) 0.u(:i, 0) = a( 1 — a), u (a, 0) = Ri.

Find u(a. 1) for (a. I) iii region i iii Figure l . with L 1 awl 2.Solution Applying (lAleniherts solution and using the fact that the interval ofdependence of a point in region I lies entuely in [0, 1], we obtain

u(a, 1) [[(u — 21) + [(a + 21)] (i.A constant 2 .1 .

Bcotant 1[(— 2t)(1 — (a — 2/)) + (a + 21)(i — (.r f 2/fl] + s2

A constan 22

= —4/2+.r--.i+(a+21)2—(.r—21)2

P2 _42 +.r —a2 + 8/a.

ligure 5 Characteristic par— We next (lescrihe a iiiethiod for finding the values of a at potts out—dhlogram.side the region I. We need the following interesting identity. Let us call acharacteristic parallelogram one whose sides lie on characteristic lines.

PROPOSITION 1 Let P1, P2, Qi , Q2 (1(1.0 it ( I h1(’ \‘Cttic(’s Of L (h1iuHt (9isti( )a.rahlelugram, with

CHARACTERISTIC P1 thiagonallv oJ)l)usite to lb (Figure 5). ‘linniPARALLELOGRAM

(10) u(P) + u(J’2) = u(Qi ) + i1(Q2).

Proof Write

(11) A(x, t) = [f*(x— ci) — G(a: — Ct)], B(x, t) = [f*(a. + ci) + G(:u + ci)],

132 Chapter 3 Partial Differential Equations in Rectangular Coordinates

where G is an antiderivative of q. Then for all (:r, t), we have from (6)

‘u(x, t) = A(x, ) + B(x, t).

We note from (11) that A is constant on the characteristic lilies :5 — ct =

— ct, while B is constant on the characteristic lines :1; + et = ro + ct0.

Hence A(Pi) = A(Q1), B(Pi) = B(Q2), B(P2) = B(Qi), A(P2) = A(Q2)

(Figure 5), aiid so

u(Pi) + ‘u(P2) = A(P1)+ B(P1)+ A(P2)+ B(P2) = u(Qi) + ‘u(Q2).

With the help of (10) we can deternmie u in the strij) S in Figure 6 by

using geometric constructions to reduce to the region I. For thus pirpose,

we (hvide S into triangular and polygonal regions bounded by characteristic

lines, as follows. Start with two characteristic lines that enanate froni the x—

axis at x = 0 and n: = L, and reflect on the boumlary of S along characteristic

lines with opposite slopes. Label these regions by II, III, IV, and so on, as

shown in Figure 6.

Figure 6 I)ividing the

______________________________________________________________________________

ctrip by reHecting charac- EXAMPLE 5 Using characteristic parallelograms

-eristic lines. Refer to Example 1. Deterunne the values of u in the region II.

Solution Let P2 = (x0, t0) he an arbitrary point in the region II. Form a character

istic parallelogram with vertices P1, F2, Qi , Q2, as shown iii Figure 7. The vertices

P2 and arc on the characteristic line .i: — 2t = 0, and the vertex Qi is on the

boundary line :5 = 0. From Proposition 1, we have u(Pi ) = u(Qi ) + n(Q2)— u(P2).

We will find ‘u(P2) and v(Q2) by using the formula o(.r, t) = —4t2 + 5 — x2 + 8tz

from Example 4, because P.2 and Q2 ale in the region I. Also, ?I(Qi) = 0, because

of the boundary condition u(0, I) = 0 for all t > 0.

To determi me the coordinates of P2 and Q2, we use simple geometric considera

tions using the equations of characteristic lines as labeled in Figure 7. For example,

Q2 is the intersection point of the characteristic lines z+2t z0+2t0 and i: — 2f. = 0.

Adding the equations, we get 2.r so + 2t0 or z = 21 From .r = 2t, we obtain=

, and so Q2= (ici+2I O)+2t() ). The coordinates of Q and P2 are coin—

puted similarly. We have Q = (0, —‘i’ + to) and P ( —x+2tn —xn+2f0). Let us

simplify time notation and write P1 (z, t) instead of (xe, t0). Then, for Pi (z, t)

in the region II.

u(a’, t) = uV2)— u(P2)z+2t.’s+21r —x+2t —z+2f

= ‘‘ 2 ‘ 4)—u(

2 ‘ 4u(Q2)

Pigure 7 — 4(X + 2t2+

:r + 2t — 1x + 2t2+8z + 2t x +2t

— 4’ 2 2 / 2 4-v(P2)

4( —x + 2t9 — —x + 2t (X + 2t12 — 8z + 2t —x + 2

4’ 2 2/ 2 4

Characteristic lines

x+4tx.

Section 3.4 D’Alernbert’s Method 133

Note how this formula for u(x, t) satisfies the wave equation and the boundary

condition at x = 0. The other conditions iii the wave problem do not concern the

points in the region II and thus should not he checked.

In the exercises, you are asked to find the values of u in the regions II, III,

and IV, by using techniques similar to the ones of Examples 4 and 5.

Exercises 3.4

In &e’rcises 1 8, use d ‘AleTnbert ‘s formula (4) to solve the boundary value problem

(1) — (3) for a string of unit length, subject to the given coridjtzons. In each case,

describe completely ,f’ and C (an anticle’rzvatzve of’ g*) (see Ezarnples 2 and 3 for

hints).1. f(:r) = sin r:r, g(x) = 0, C =

V 2. f(.c) = sinirxcosirx, g(:r) = 0, r =

Y’ 3 f() = Sill ltX + 3 i1n ii(i) iii C’ = •1.

4.f(:r)=0, g(x)=1,c=1

V 5. f(x) as in Exorcise 5, Section 3.3, g(:r) = :r, e = 1.

6. f(:r) 0, q(i) = cosirx, c 1.

7. ,f(:r) = 0, ij(x) = —10, e 1.

8. f(.r) = 0, g(r) = sinir.r, 1.

V 9. Determine the first time the string returns to its initial shape in Exercises 1

aiid 5.

10. Plot the solution in Exercise 1 for t = 1/2 and 1.

11. Plot the solution in Exercise 4 for t = 1/2 and 1.

12. Time period of motion. Prove the results of Exercise 11(b), Section 3.3,

using d’Alcnibert’s solution.

13. Suppose that both .f and q are symmetric about r = 4; that is, /‘(L — i) = j(:r)and g(L — i) = g(.r). Show that

Lu(,x’. I + —) = -o(’, 1)

for all (I < .r < L and P > 0.

14. Check that d’Alembc’rt.’s solution (4) satisfies (a) the boundary condition u(L, I) =

0 for all 1 > 0: and (h) the initial conditions (3). [1-lint: For (a), use the oddness

and the 2L—poriodicity of ,f and .]15. Consider the boundary value problem (1)--(3) with c = 1, L = 1 , q(z) = 0, and

( 4,r if 0< i: <

f(;i’)= 4(—:z’) if<:r<,

10 if<x1.

(a) Use d’Aheinbert’s method to plot the string at times t = 0, , .(b) For t , identify the points on the string that are still in rest position.

134 Chapter 3 Partial Differential Equations in Rectangular Coordinates

(c) Take a point r on the string with zero initial displacement ( < a; < 1). Howlong does it take before the point :r starts to vibrate?(d) What is the answer in (c) for an arbitrary string constant c > 0?

16. D’Alembert’s solution for zero initial velocity. (a) Starting from (8),Section 3.3, show that, if the initial velocity g(x) = 0, then

1 ‘nit . nitu(x, t) = b, [sin —T (:i: — rt) + sin (a: + ct)].

n= 1

(b) Derive d’Alembert’s solution (4) using (a). [Hint: ,f*(s) = h,, sin s.j

17. Project Problem: D’Alembert’s solution (the general case). Let, g*

denote the odd 2L—periodic extension of g, and let

= / g4 (s) d,s.

(a) Show that C is even and 2L—periodic, and

GCr) = B,, (cos a’ —

n=1

where

B1, =

— / gr)sin xd.z; = —cb (ii = 1,2, ..

[Hint: Exercise 33. Section 2.3.1(h) Use (a) to show that

717r ‘nitG(a + Ct) — C(x — ct) B.[cos _-j__(x + ct) — cos 7—(x — ct)].

i,=1

i ‘.r+rI ‘Dc * . n-i(c) Usmg (b), show that L-_ g (s) ds=

b, sm —L—x Sm(cl) Use (c), (8) of Section 3.3, and Exercise 16 to derive d’Alembert’s solution (4).

..J 18. Project Problem: Conservation of energy. The energy at time t of avibrating string is given by

L

E(t)=f (u+c2u.)dx.

(a) By differentiating under the integral sign, show that

dE 1L2

= j (UtUtt + c uu) dx.a

(b) Use the wave equation (1) to replace fl by c2u and obtain

dE 1L

= c (U,Ut) dx.U,’ J

Section 3.5 The One Dimensional Heat Equation 135

(c) Using (2), show that u(O, t) = u(L, t) = 0 for all t > 0.(ci) Prove the principle of conservation of energy, which states that the energy duringthe free vibrations of a string is constant for all time. [Hint: Prove that dE/dt = 0,using (b) and (c).119. Refer to Example 3.(a) What are the characteristic lines?(b) Find the intervals of dependence of the points (.5, .2) and (.3, 2).(c) Describe the region I in this case. Which one of the points in (b) belongs to theregion I?(ci) Find u(.x, t) for all points in the region I.

20. Refer to Example 3. Find ‘u(x, t) for all points in the region II.

21. Refer to Example 4. Find u(.r, t) for all points in the region III (see Figure 8).

22. Refer to Example 4. Find ‘u(x, t) for all points in the region IV (see Figure 8).

I ‘[‘he One Dimensional Heat EquationIn this and the following section we study the temperature distribution ina. uniform bar of length L with insulated lateral surface and flO internalsources of heat, subject to certain boundary and initial conditions. Todescribe the problem, let u(x, t) (0 < x < L, t > 0) represent the temperatureof the point :i: of the bar at time t. (Figure 1). Given that the initial

‘( ‘ temperature distribution of the bar is u(x, 0) = f(x), and given that theends of the bar are held at constant temperature 0, we ask, What is u(x, t)for 0 < :r < L, t > 0? As you would expect, to answer this question, wemust solve a boundary value problem. We will show in the appendix at the

0 x end of this section that u satisfies the one dimensional heat equation

0 L On—=r2- 0<x<L, t>0.

I Insulated bar with$)k kepi. it 00.

In addition, ‘ii. satisfies tlxc’ boundary conditions

•

. u(0,t) = 0 and n(L,t) = 0 for all t >0• c I problem is first or—• I, wi only need one mi—

il ion, unlike the wave and the initial conditionI where two conditions

*rII’ IWI’(h’dl. u(x. 0) = f(.c) for 0 < .r < L.

We solve this problem Using the method of separation of variables. Afterdoing so, we will introduce the notion of steady-state temperature and use

it to solve a related heat problem with nonzero boundary data. Interestingand important variations on these problems are presented in time followingsection.

S