2.1 Heat Flow in a bar; Fourier’s law1. The units of κ are energy per length per time per temperature, for example, J / (cm s K) = W / (cmK).3. The units of Aρcu ∆ x are
cm2 gcm3
Jg K
Kcm = J .
5. ∂u∂x ( , t) = r
Aκ , t > t 0 .7. Measuring x in centimeters, the steady-state temperature distribution is u(x) = 0 .1x + 20, the solution of
−d2 udx 2 = 0 , u(0) = 20 , u(100) = 30 .
The heat ux is
−κ dudx
(x) = −0.802 ·0.1 = −0.0802 W/ cm2 .
Since the area of the bar’s cross-section is π cm2 , the rate at which energy is owing through the bar is 0 .0802π .=0.252 W (in the negative direction).
9. Consider the right endpoint ( x = ). Let the temperature of the bath be u , so that the difference in temperaturebetween the bath and the end of the bar is u( , t) −u . The heat ux at x = is, according to Fourier’s law,
−κ ∂u∂x
( , t),
so the statement that the heat ux is proportional to the temperature ow is written
−κ ∂u∂x
( , t) = α (u( , t) −u ) ,
where α > 0 is a constant of proportionality. This can be simplied to
αu ( , t) + κ ∂u∂x
( , t) = αu .
The condition at the right end is similar:
αu (0, t ) −κ ∂u∂x
(0, t ) = αu 0 .
(The sign change appears because, at the left end, a positive heat ux means that heat ows into the bar, whileat the right end the opposite is true.)
(b) The rate of mass transfer varies inversely with and directly with the square of r .
2.2 The hanging bar1. The sum of the forces on the cross-section originally at x = must be zero. As derived in the text, the internal
(elastic) force acting on this cross-section is
−Ak( ) dudx
( ),
while the external traction results in a total force of pA (force per unit area times area). Therefore, we obtain
−Ak( ) dudx
( ) + pA = 0 ,
or
k( )dudx ( ) = p.
The other boundary condition, of course, is u(0) = 0.3. (a) The stiffness is the ratio of the internal restoring force to the relative change in length. Therefore, if a bar
of length and cross-sectional area A is stretched to a length of 1 + p (0 < p < 1) times , then the bar willpull back with a force of 195 pA gigaNewtons. Equivalently, this is the force that must be applied to the endof the bar to stretch the bar to a length of (1 + p) .
(b) 196 / 195 m, or approximately 1 .005 m.(c) The BVP is
−195 ·109 d2 udx 2 = 0 , 0 < x < 1,
u(0) = 0 ,
195
·109 du
dx(1) = 10 9 .
It is easy to show by direct integration that the solution is u(x) = x/ 195, and therefore u(1) = 1 / 195. Sinceu is the displacement (in meters), the nal position of the end of the bar is 1+1 / 195 m, that is, the stretchedbar is 196 / 195 m.
5. The wave equation is
7.9 ·103 ∂ 2 u∂t 2 − 1.95 ·1011 ∂ 2 u
∂x 2 = f (x, t ), 0 < x < 1.
The units of f are N/ m3 (force per unit volume). The units of the rst term on the left are
kgm3
ms2 = kgm/ s2
m3 = Nm3 ,
and the units of the second term on the left areN
m21m
= Nm3 .
7. (a) We have∂ 2 u∂t 2 (x, t ) = −c2 θ2 u(x, t ),
while∂ 2 u∂x 2 (x, t ) = −θ2 u(x, t ).
Therefore,∂ 2 u∂t 2 −c2 ∂ 2 u
∂x 2 = −c2 θ2 u + c2 θ2 u = 0 .
(b) Regardless of the value of θ, u(0, t ) = 0 holds for all t . The only way that u( , t) = 0 can hold for all t is if
sin( θ ) = 0 ,
so θ must be one of the valuesθ = nπ , n = 0 , ±1, ±2, . . . .
1. Units of acceleration (length per time squared).3. Units of velocity (length per time).
5. The internal force acting on the end of the string at, say, x = , is
T ∂u∂x
( , t). (2.1)
If this end can move freely in the vertical direction, force balance implies that (2.1) must be zero.
7. If u(x, t ) = f (x −ct), then
∂v∂x
(x, t )= f (x −ct), ∂ 2 v∂x 2 (x, t )= f (x −ct),
∂v∂t
(x, t )= −cf (x −ct), ∂ 2 v∂t 2 (x, t )= c2 f (x −ct),
Therefore,∂ 2 u∂t 2 (x, t ) −c2 ∂ 2 u
∂x 2 (x, t ) = c2 f (x −ct) −c2 f (x −ct) = 0 ,
as desired. Similarly, if v(x, t ) = f (x + ct), then
∂v∂x
(x, t )= f (x + ct), ∂ 2 v∂x 2 (x, t )= f (x + ct),
∂v∂t
(x, t )= cf (x + ct), ∂ 2 v∂t 2 (x, t )= c2 f (x + ct),
and hence∂ 2 v∂t 2 (x, t ) −c2 ∂ 2 v
∂x 2 (x, t ) = c2 f (x + ct) −c2 f (x + ct) = 0 .
2.4 Advection; kinematic waves1. The solution u of
∂u∂t
+ c ∂u∂x
= 0 , 0 < x < , t > 0,
u(x, 0) = u0 (x), 0 < x < ,u(0, t ) = φ(t), t > 0.
can be found by reasoning similar to that used to solve the pure IVP (see (2.23) and following in the text). Thecross-section at point x at time t was ct units to the left at time t = 0. If x −ct > 0, then the value of u(x, t )is given by the initial condition: u(x, t ) = u 0 (x −ct). If x −ct < 0, then u(x, t ) is determined by the boundarycondition u(0, t ) = φ(t), and we must ask: At what time was the cross-section, now at point x at time t , found atx = 0? The answer is t 0 , where c(t −t0 ) = x, that is, t 0 = t −(1/c )x. Thus u (x, t ) = φ(t −(1/c )x) if x −ct < 0.
It is easy to verify by a direct calculuation that
u(x, t ) = u0 (x −ct), x −ct > 0,φ t − 1
c x , x −ct < 0
solves the given IBVP.
3. Suppose u solves
∂u∂t
+ c∂u∂x
= 0 , −∞< x < ∞, t > 0,
u(x, 0) = φ(x), −∞< x < ∞,
where c > 0, and suppose that φ(x) = 0 for all x ≤ a. The solution is u(x, t ) = φ(x − ct), so u(x, t ) = 0 if x −ct ≤a, that is, if t > (x −a)/c .
3.1 Linear systems as linear operator equations1. A function of the form f (x) = ax + b is linear if and only if b = 0. Indeed, if f : R →R is linear, then f (x) = ax ,
where a = f (1).3. Let f : R 2 →R 2 be dened by
f (x ) = x21 + x2
2x2 −x2
1
If we take x = (1 , 1), then f (x ) = (2 , 0), while 2x = (2 , 2) and hence f (2x ) = (8 , −2) = 2 f (x ). Thus f is notlinear.
5. (a) Vector space (subspace of C [0, 1]).(b) Not a vector space; does not contain the zero function. (Sub set of C [0, 1], but not a sub space .)(c) Vector space (subspace of C [0, 1]).
(d) Vector space.(e) Not a vector space; does not contain the zero polynomial. (Sub set of P n , but not a sub space .)
7. Suppose u∈C 1 [a, b] is any nonzero function. Then
L(2u) = 2 dudx
+ (2 u)3 = 2 dudx
+ 8 u3 ,
while2Lu = 2 du
dx + 2 u3 .
Since L(2u) = 2 Lu , L is not linear.9. Dene K : C 2 [a, b] →C [a, b] by
= α(a11 x1 + a12 x2 ) + β (a11 z 1 + a12 z 2 )α (a21 x1 + a22 x2 ) + β (a21 z 1 + a22 z 2 )
= α a11 x1 + a12 x2
a21 x1 + a22 x2+ β a11 z 1 + a12 z 2
a21 z 1 + a22 z 2= αAx + β Az .
This shows that the operator dened by A is linear.
(b) We have
(A (α x + β y )) i =n
j =1
a ij (αx j + βy j )
and
(α Ax + β Ay ) i = αj =1
a ij x j + β n
j =1
a ij yj .
Now,
(A (αx + β y ))i
=n
j =1
a ij (αx j + βy j )
=n
j =1
(αa ij x j + βa ij yj )
=n
j =1
αa ij x j +n
j =1
βa ij yj
= αn
j =1
a ij x j + β n
j =1
a ij yj
= ( αAx + β Ay )i ,
as desired. The third equality follows from the commutative and associative properties of addition, whilethe fourth equality follows from the distributive property of multiplication over addition.
3.2. EXISTENCE AND UNIQUENESS OF SOLUTIONS TO AX = B 9
3.2 Existence and uniqueness of solutions to Ax = b
1. The range of A isα(1, −1) : α ∈R ,
which is a line in the plane. See Figure 3.1.
−3 − 2 − 1 0 1 2 3−3
−2
−1
0
1
2
3
x1
x 2
Figure 3.1: The range of A (Exercise 3.2.1).
3. (a) A is nonsingular.(b) A is nonsingular.(c) A is singular. Every vector in the range is of the form
Ax = 1 11 1
x1
x2
= x1 + x2
x1 + x2.
That is, every y ∈R 2 whose rst and second components are equal lies in the range of A . Thus, for example,Ax = b is solvable for
b = 11 ,
but not forb = 1
2 .
5. The solution set is not a subspace, since it cannot contain the zero vector ( A0 = 0 = b ).7. The null space of L is the set of all rst-degree polynomials:
N (L) = u : [a, b] →R : u(x) = mx + c for some m, c ∈R .
9. (a) The only functions in C 2 [a, b] that satisfy
−d2 udx 2 = 0
are functions of the form u(x) = C 1 x + C 2 . The Neumann boundary condition at the left endpoint implies
that C 1 = 0, and the Dirichlet boundary condition at the right endpoint implies that C 2 = 0. Therefore,only the zero function is a solution to L m u = 0, and so the null space of L m is trivial.(b) Suppose f ∈C [a, b] is given and u∈C 2m [a, b] satises
−d2 udx2 (x) = f (x), a < x < b.
Integrating once yields, by the fundamental theorem of calculus,
A direct calculation (using Gaussian elimination) shows that this last system of equations has only the trivialsolution, and hence the three polynomials form a linearly independent set.
7. We know that P 2 has dimension 3, and therefore it suffices to show either that the given set of three vectors spans
P 2 or that it is linearly independent. If p ∈P 2 , then we write
c1 = p(x1 ), c3 = p(x3 ), c3 = p(x3 )
and dene the polynomialq (x) = c1 L1 (x) + c2 L2 (x) + c3 L3 (x).
and, similarly, q (x2 ) = p(x2 ), q (x3 ) = p(x3 ). But then p and q are second-degree polynomials that agree at threedistinct points, and three points determine a quadratic (just as two points determine a line). Therefore,
p(x) = c1 L1 (x) + c2 L2 (x) + c3 L3 (x),
and we have shown that every p ∈ P 2 can be written as a linear combination of L1 , L 2 , L 3 . This completes theproof.
9. Let L : C 2N [a, b] → C [a, b] be the second derivative operator. We wish to nd a basis for the null space of L . Afunction u
∈
C 2N [a, b] belongs to the null space of L if and only if it satises
d2 udx 2 (x) = 0 for all x∈[a, b], du
dx(0) = du
dx( ) = 0 .
The only functions satisfying the differential equation are rst-degree polynomials, u(x) = mx + c. Moreover, theboundary conditions are satised if and only if the slope m is zero. Thus u belongs to the null space of L if andonly if u(x) = c for some c ∈ R , that is, if and only if u is a constant function. A basis is therefore the set u1,where u1 (x) = 1 for all x∈[a, b].
3.4 Orthogonal bases and projections1. (a) The verication is a direct calculation of v i ·v j for the 6 combinations of i, j . For example,
v 1
·v 1 = 1
√ 31
√ 3+ 1
√ 31
√ 3+ 1
√ 31
√ 3= 1
3 + 1
3 + 1
3 = 1 ,
v 1 ·v 2 = 1√ 3
1√ 2 + 1
√ 3 0 + 1√ 3 −
1√ 2
= 1√ 6 −
1√ 6 = 0 ,
and so forth.
(b)
x = ( v 1 ·x )v 1 + ( v 2 ·x )v 2 + ( v 3 ·x )v 3
= 4√ 3 v 1 + 0 v 2 + − 2√ 6 v 3 .
3. We have
x + y 2 = ( x + y , x + y )= ( x , x + y ) + ( y , x + y )= ( x , x ) + ( x , y ) + ( y , x ) + ( y , y )= ( x , x ) + 2( x , y ) + ( y , y )= x 2 + 2( x , y ) + y 2 .
Therefore,x + y 2 = x 2 + y 2
if and only if ( x , y ) = 0.5. First of all, if
(y , z ) = 0 for all z∈W,
then since w i ∈W for each i, we have, in particular,
(y , w i ) = 0 , i = 1 , 2, . . . , n .
Suppose, on the other hand, that(y , w i ) = 0 , i = 1 , 2, . . . , n .
If z is any vector in W , then, since w 1 , w 2 , . . . , w n is a basis for W , there exist scalars α 1 , α 2 , . . . , α n such that
This completes the proof.7. The best approximation to the given data is y = 2 .0109x −0.0015151. See Figure 3.2.
1 1.5 2 2.5 32
2.5
3
3.5
4
4.5
5
5.5
6
x
y
Figure 3.2: The data from Exercise 3.4.7 and the best linear approximation.
9. The projection of g onto P 2 is
2π q 1 (x) + 0 q 2 (x) + 2√
5(π2
−12)π 3 q 3 (x)or
2π
+ 10(π 2 −12)π 3 −60 π 2 −12
π 3 x + 60 π 2 −12π 3 x2 .
See Figure 3.3.
0 0.2 0.4 0.6 0.8 1−0.5
0
0.5
1
y=sin(π x)quadratic approx.
Figure 3.3: The function g(x) = sin( πx ) and its best quadratic approximation over the interval [0 , 1] (seeExercise 3.4.9).
3.5 Eigenvalues and eigenvectors of a symmetric matrix1. The eigenvalues of A are λ 1 = 200 and λ 2 = 100, and the corresponding (normalized) eigenvectors are
3.5. EIGENVALUES AND EIGENVECTORS OF A SYMMETRIC MATRIX 13
respectively. The solution is
x = b ·u 1
λ1u 1 +
b ·u 2
λ 2u 2 = 1
1.
3. The eigenvalues and eigenvectors of A are λ 1 = 2 , λ2 = 1 , λ3 = −1 and
u 1 =
1√ 31√ 31√ 3
, u 2 =
1√ 20
− 1√ 2
, u 3 =
1√ 6
− 2√ 61√ 6
.
The solution of the Ax = b is
x = b ·u 1
λ 1u 1 +
b ·u 2
λ2u 2 +
b ·u 3
λ3u 3 = −3/ 2
1/ 25/ 2
.
5. The computation of u i
· b /λ i costs 2n operations, so computing all n of these ratios costs 2 n 2 operations.
Computing the linear combination is then equivalent to another n dot products, so the total cost is
2n 2 + n∗(2n −1) = 4 n 2 −n = O(4n 2 ).
7. (a) For k = 2 , 3, . . . , n −1, we have
Ls ( j )
k= 1
h2 (−sin(( k −1) jπh ) + 2sin ( kjπh ) −sin(( k + 1) jπh ))
= 1h2 (−sin( kjπh )cos( jπh ) + cos ( kjπh )sin( jπh ) + 2 sin ( kjπh )
− sin (kjπh )cos( jπh ) −cos(kjπh )sin( jπh ))
= 1h2 (2 −2cos( jπh ))sin( kjπh )
=
1
h2 (2 −2cos( jπh ))s ( j )
k .Thus
Ls ( j )
k= 1
h2 (2 −2cos( jπh )) s ( j )k , k = 2 , 3, . . . , n −1.
Similar calculations show that this formula holds for k = 1 and k = n also. Therefore, s ( j ) is an eigenvectorof L with eigenvalue
λ j = 2 −2cos( jπh )h2 .
(b) The eigenvalues λ j are all positive and are increasing with the frequency j .
(c) The right-hand side b can be expressed as
b = s (1) ·b s (1) + s (2) ·b s (2) + · · ·+ s ( n ) ·b s ( n ) ,
while the solution x of Lx = b is
x = s(1) ·x
λ1s (1) +
s(2) ·xλ2
s (2) + · · ·+ s( n ) ·x
λ ns ( n ) .
Since λj increases with j , this shows that, in producing x , the higher frequency components of b aredampened more than are the lower frequency components of b . Thus x is smoother than b .
Since W (t) = 0 for all t , we see that u1 , u 2 is linearly independent.
9. (a) Let u1 , u 2 be two functions dened on an interval, let t0 be a point in that interval, and suppose
u1 (t0 ) u2 (t0 )du 1dt (t0 ) du 2
dt (t0 ) = 0 .
We wish to prove that u1 , u 2 is linearly independent. We argue by contradiction, and suppose u1 , u 2 islinearly dependent. Then there exist c1 , c2 ∈R such that c1 u1 + c2 u2 = 0. This means that c1 u1 (t)+ c2 u2 (t) =0 for all t in the given interval, which in turn implies that c1
du 1dt (t ) + c2
du 2dt (t) = 0 for all t . But then, taking
t = t0 , we obtain
c1 u1 (t0 ) + c2 u2 (t0 ) = 0 ,
c1du 1
dt (t0 ) + c2
du 2
dt (t0 ) = 0 .
In matrix-vector form, this system is
u1 (t0 ) u2 (t0 )du 1dt (t0 ) du 2
dt (t0 ) c1
c2 = 0
0 .
Since (c1 , c2 ) = (0 , 0), this implies that the matrix
u1 (t0 ) u2 (t0 )du 1dt (t0 ) du 2
dt (t0 )
is singular, contradicting the assumption that its determinant is nonzero. This contradiction completes theproof.
We see that W (π/ 2) = −1 = 0, and hence, by part (a), u1 , u 2 is linearly independent. Nevertheless,W (0) = 0, which shows that the fact that u1 , u 2 is linearly independent does not imply that W (t) = 0 forall t , unless u1 and u2 are both solutions to the same second-order linear differential equation.
1. (a) We rst note that the zero function is a solution of (4.9), so S is nonempty. If u, v ∈ S and α, β ∈ R , thenw = αu + βv satises
a d2 wdt 2 + bdw
dt + cw = a d2
dt 2 [αu + βv ] + b ddt
[αu + βv ] + c(αu + βv )
= a α d2 udt 2 + β d
2 vdt 2 + b α du
dt + β dv
dt+ c(αu + βv )
= α a d2 udt 2 + bdu
dt + cu + β a d2 v
dt 2 + bdvdt
+ cv
= α ·0 + β ·0 = 0 .
Thus w is also a solution of (4.9), which shows that S is a subspace.(b) As explained in the text, no matter what the values of a,b,c (a = 0), the solution space is spanned by two
functions. Thus S is nite-dimensional and it has dimension at most 2. Moreover, it is easy to see that each
of the sets er 1 t
, er 2 t
(r 1 = r2 ), eµt
cos(λt ), eµt
sin( λt ) , and ert
, tert
is linearly independent, since aset of two functions is linearly dependent if and only if one of the functions is a multiple of the other. Thus,in every case, S has a basis with two functions, and so S is two-dimensional.
3. Suppose that the characteristic polynomial of (4.9) has a single, repeated root r = −b/ (2a) (so b2 −4ac = 0).Then, as shown in the text,
u(t) = c1 ert + c2 te rt
is a solution of (4.9) for each choice of c1 , c2 . We wish to show that, given k1 , k2 ∈R , there is a unique choice of c1 , c2 such that u(0) = k1 , du/dt (0) = k2 . We have
dudt
(t) = rc 1 ert + c2 ert + rc 2 te rt .
Therefore, the equations u(0) = k1 , du/dt (0) = k2 simplify to
1 0
r 1c = k.
Since the coefficient matrix is obviously nonsingular (regardless of the value of r ), there is a unique solution c1 , c2
for each k1 , k2 .5. (a) The only solution is u(x) = 0.
(b) The only solution is u(x) = 0.(c) The function u(x) = sin( πx ) is a nonzero solution. It is not unique, since any multiple of it is another
solution.7. By the product rule and the fundamental theorem of calculus,
ddt
t
t 0
ea ( t −s ) f (s) ds = ddt
eat t
t 0
e−as f (s) ds = aeat t
t 0
e−as f (s) ds + eat e−at f (t )
= a
t
t 0
ea ( t −s ) f (s) ds + f (t).
Therefore, with
u(t) = u0 ea ( t −t 0 ) + t
t 0
ea ( t −s ) f (s) ds,
we havedudt
(t ) = au 0 ea ( t −t 0 ) + a t
t 0
ea ( t −s ) f (s) ds + f (t)
= au (t) + f (t).
Thus u satises the differential equation. We also have
5. x 0 must be a multiple of the vector (1 , −1).7.
x (t) = 160
55
−3e−2t
−45e−t + 20 t
−7cos(t)
−11sin( t)
10 + 6 e−2t + 20 t −16cos(t) −8sin( t)
−5 −3e−2t + 45 e−t + 20 t −37cos( t) + 19 sin ( t)
9. (a) The solution is
x(t) = 32
e−t + 12
e3t , y(t) = 32
e−t − 12
e3t .
The population of the rst species ( x(t)) increases exponentially, while the population of the second species(y(t)) goes to zero in nite time ( y(t) = 0 at t = ln(3) / 4). Thus the second species becomes extinct, whilethe rst species increases without bound.
(b) If the initial populations are x(0) = r , y(0) = s, then the solution to the IVP is
x(t) = 12
(r + s)e−t + ( r −s)e3t , y(t) = 12
(r + s)e−t + ( s −r )e3t .
Therefore, if r = s, both populations decay to zero exponentially; that is, both species die out.
11. Assume v(t; s) satisesd2 vdt 2 + θ2 v = 0 , v(0; s) = 0 , dv
dt(0; s) = 0
for all s , and dene
u(t) = t
0v(t −s; s) ds.
Then
dudt
(t) = v(0; t) + t
0
dvdt
(t −s; s) ds = t
0
dvdt
(t −s; s) ds,
d2 udt 2 (t) = dv
dt(0; t) +
t
0
d2 vdt 2 (t −s; s) ds = f (t ) +
t
0
d2 vdt 2 (t −s; s) ds
(using the fact that v(0; t) = 0 and dvdt (0; t) = 0 for all t ). We then have
d2 udt 2 (t) + θ2 u(t) = f (t) +
t
0
d2 vdt 2 (t −s; s) ds + θ2
t
0v(t −s; s) ds
= f (t) + t
0
d2 vdt 2 (t −s; s) + θ2 v(t −s; s) ds
= f (t)
sinced2 vdt 2 (t −s; s) + θ2 v(t −s; s) = 0 for all s.
Also,
u(t) = 0
0v(t −s; s) ds = 0 , du
dt (t) =
0
0
dvdt
(t −s; s) ds = 0 ,
and thus u satises both the differential equations and the initial conditions.
The error denitely does not exhibit O(∆ t4 ) convergence to zero. The reason is the lack of smoothnessof the function 1 + |x − 1|; the rate of convergence given in the text only applies to ODEs dened bysmooth functions. When integrating from t = 0 to t = 0 .5, the nonsmoothness of the right-hand side is notencountered since x(t) < 1 on this interval. This explains why we observed good convergence in the rstpart of this problem.
4.5 Stiff systems of ODEs
1. (a) The exact solution isx (t) = 1
2 e−t + et
e−t −et .
(b) The norm of the error is approximately 0 .089103.(c) The norm of the error is approximately 0 .10658.
3. The largest value is ∆ t = 0 .04.5. (a) Applying the methods of Section 4.3, we nd the solution
x (t) = 12
3e−t −e−100 t
3e−t + e−100 t .
(b) By trial and error, we nd that ∆ t ≤ 0.02 is necessary for stability. Specically, integrating from t = 0 tot = 1, n = 49 (i.e. ∆ t = 1 / 49) yields
x 49 > x 0 ,while n ≥50 yields
x n ≤ x 0 .
(c) With x i +1 = x i + ∆ tAx i and y( i )1 = u 1 ·x i , we have
u 1 ·x i +1 = u 1 ·x i + ∆ tu 1 ·Ax i
⇒ y( i +1)1 = y( i )
1 + ∆ t(Au 1 ) ·x i
⇒ y( i +1)1 = y( i )
1 + ∆ tλ 1 u 1 ·x i
⇒ y( i +1)1 = y( i )
1 + ∆ tλ 1 y( i )1
⇒ y( i +1)1 = (1 + ∆ tλ 1 )y( i )
1 .
A similar calculation shows thaty( i +1)
2 = (1 + ∆ tλ 2 )y( i )2 .
For stability, we need
|1 + ∆ tλ 1 | ≤ 1,
|1 + ∆ tλ 2 | ≤ 1.
Since the eigenvalues of A are −1, −100, it is not hard to see that the upper bound for ∆ t is ∆ t ≤0.02, justas was determined by experiment.
(d) For the backward Euler method, a similar calculation shows that
5.1 The analogy between BVPs and linear algebraic systems
1. (a) Since M D is linear, it suffices to show that the null space of M D is trivial. If M D u = 0, then, du/dx = 0,that is, u is a constant function: u(x) = c. But then the condition u(0) = 0 implies that c = 0, and so u isthe zero function. Thus N (M D ) = 0.
(b) If u∈C 1D [0, ] and M D u = f , then we have
dudx
(x) = f (x), 0 < x < and u(0) = 0 ,
which imply that
u(x) =
x
0f (s) ds.
But we also must have u( ) = 0, which implies that
0f (s) ds = 0 .
If f ∈C [0, ] does not satisfy this condition, then it is impossible for M D u = f to have a solution.
3. (a) If v, w ∈ S are both solutions of Lu = f , then Lv = Lw, or (by linearity) L(v − w) = 0. Thereforev −w ∈N (L). But, since S is a subspace, v −w is also in S . If the only function in both N (L) and S is thezero function, then v −w must be the zero function, that is, v and w must be the same function. Therefore,if N (L) ∩S = 0, then Lu = f can have at most one solution for any f .
(b) We have already seen that Lu = f has a solution for any f ∈C [0, ] (see the discussion immediately precedingExample 5.1). We will use the result from the rst part of this exercise to show that the solution is unique.
The null space of L is the space of all rst degree polynomials:
N (L) = u : [0, ] →R : u(x) = ax + b for some a, b∈R .
i. Suppose u∈N (L) ∩S . Then u(x) = ax + b for some a, b∈R and
dudx 2
= 0 ⇒a = 0 .
Then u(x) = b, and
0u(x) dx = 0 ⇒b = 0 ⇒b = 0 .
Therefore u is the zero function, and so N (L) ∩S = 0. The uniqueness property then follows fromthe rst part of this exercise.
ii. Suppose u∈N (L) ∩S . Then u(x) = ax + b for some a, b∈R and
u(0) = 0 ⇒b = 0 .Then u(x) = ax , and
dudx
( ) = 0 ⇒a = 0 .
Therefore u is the zero function, and so N (L) ∩S = 0. The uniqueness property then follows fromthe rst part of this exercise.
5. The condition
0
dudx
(x)2
dx = 0 (5.1)
implies that du/dx is zero, and hence that u is constant. The boundary conditions on u would then imply thatu is the zero function. But, by assumption, u is nonzero (we assumed that ( u, u ) = 1). Therefore, (5.1) cannothold.
7. (a) If Lm u = 0, then u has the form u(x) = ax + b. The rst boundary condition, du/dx (0) = 0, implies thata = 0, and then the second boundary condition, u( ) = 0, yields b = 0. Therefore, u is the zero functionand N (Lm ) is trivial.
(b) For any f ∈C [0, ], the function
u(x) =
x z
0f (s) dsdz
belongs to C 2m [0, ] and satises
−d2 udx2 (x) = f (x), 0 < x < .
This shows that L m u = f has a solution for any f ∈C [0, ], and therefore R(Lm ) = C [0, ].
(c) Suppose u, v ∈C 2m [0, ]. Then
(Lm u, v ) =
−
0
d2 u
dx 2(x)v(x) dx
= −dudx
(x)v(x)0
+
0
dudx
(x) dvdx
(x) dx
=
0
dudx
(x) dvdx
(x) dx (since v( ) = 0, du/dx (0) = 0)
= u(x) dvdx
(x)0 −
0u(x) d2 v
dx2 (x) dx
= −
0u(x) d2 v
dx 2 (x) dx (since u( ) = 0, dv/dx (0) = 0)
= ( u, L m v).
Thus L m is symmetric.
(d) Suppose λ is an eigenvalue of Lm with corresponding eigenfunction u, and assume that u has been normalizedso that ( u, u ) = 1. Then
λ = λ(u, u ) = ( λu,u ) = ( Lm u, u )
= −
0
d2 udx 2 (x)u(x) dx
= −dudx
(x)u(x)0
+
0
dudx
(x)2
dx
=
0
dudx
(x)2
dx (since u( ) = 0, du/dx (0) = 0)
> 0.
The last step follows because every nonzero function in C 2m [0, ] has a nonzero derivative.
(b) If LR u = 0, then u must be a rst degree polynomial: u(x) = ax + b. The boundary conditions imply thata and b must satisfy the following equations:
−κa + αb = 0 ,(α + κ)a + αb = 0 .
The determinant is computed as follows:
−κ αα + κ α = −α(2κ + α ) < 0.
The only solution is a = b = 0, that is, u = 0, so N (LR ) is trivial.
5.2 Introduction to the spectral method; eigenfunctions1. If n
= m, then
(un , u m ) =
0sin nπx sin mπx dx
= 12
0cos (n −m)πx
−cos (n + m)πx dx
= 12
(n −m)π
sin (n −m)πx−
(n + m)π
sin (n + m)πx0
.
This last expression simplies to four terms, each including the sine function evaluated at an integer multiple of π , and hence each equal to zero. Thus ( un , u m ) = 0 for n = m.
5. The characteristic polynomial is r 2 −r + ( λ −5), so the characteristic roots are
r 1 = 1 −√ 21 −4λ2 , r2 = 1 + √ 21 −4λ2 . (5.2)
Case 1: λ = 21 / 4. In this case, the characteristic roots are r = 1 / 2, 1/ 2 and the general solution of the ODE is
u(x) = c1 ex/ 2 + c2 xe x/ 2 .
The boundary condition u(0) = 0 yields c1 = 0, and then the boundary condition u(1) = 1 implies that c2 = 0.Thus there is no nonzero solution to the BVP, and λ = 21 / 4 is not an eigenvalue.Case 2: λ < 21/ 4. In this case, the characteristic roots, given by (5.2), are real and distinct. The general solutionof the ODE is
u(x) = c1 er 1 x + c2 er 2 x .
The boundary conditions lead to the system
c1 + c2 = 0 ,c1 r 1 er 1 + c2 r 2 er 2 = 0 .
This system has the unique solution c1 = c2 = 0, so there is no nonzero solution for λ < 21/ 4. Hence no λ < 21/ 4is an eigenvalue.Case 3: λ > 21/ 4. In this case, the roots are complex conjugate:
r 1 = 12 −θi, r 2 = 1
2 + θi, θ =
√ 4λ −212
.
The general solution of the ODE is
u(x) = ( c1 cos (θx) + c2 sin (θx)) ex/ 2 .
The boundary condition u(0) = 0 yields c1 = 0, so any eigenfunction is of the form
u(x) = sin( θx)ex/ 2
.
The Neumann condition at x = 1 is equivalent to the equation
tan( θ) = −2θ, θ > 0.
Although this equation cannot be solved explicitly, a simple graph shows that there are innitely many solutions0 < θ 1 < θ 2 < · · ·, with
θk ∈2k −1
2 π, 2k + 1
2 π , k = 1 , 2, . . . .
Dene λ k by
θk =√ 4λk −21
2 ,
that is,
λk = θ2k +
214 , k = 1 , 2, . . . .
Then λ k , k = 1 , 2, . . . , are eigenvalues, and the corresponding eigenfunctions are
vk (x) = sin( θk x)ex/ 2 .
Using Newton’s method, we nd thatθ1
.= 1 .8366, θ2.= 4 .8158,
which yieldsλ1
.= 8 .6231, λ2.= 28 .4423.
The eigenfunctions are v1 , v2 , as given by the above formula.A direct calculation shows that
n 5 π 5 sin(nπx )The errors in approximating the original functions using 10 terms of the Fourier sine series are graphed in Figure5.1.
0 0.5 1−0.5
0
0.5
1
x0 0.5 1
−0.01
0
0.01
0.02
0.03
x
0 0.5 1−1
−0.5
0
0.5
1
1.5x 10
−3
x0 0.5 1
−5
0
5x 10
−5
x
Figure 5.1: Errors in approximating four functions by 10 terms of their Fourier sine series (see Exercise5.2.7). Top left: g(x) = x; Top right: h(x) = 1
2 − x − 12 ; Bottom left: m(x) = x − x3 ; Bottom right:
k(x) = 7 x − 10x3 + 3 x5 .
9. sin(3 πx ) (That is, all of the Fourier sine coefficients are zero, except the third, which is one.)
11. The series have the form∞
n =1
an cos (2n −1)πx2
,
where
(a) an = −4(2+( −1) n (2 n −1) π )π 2 (2 n −1) 2 ;
(b) an = 32cos ((2 n −1) π/ 4)sin 2 ((2 n −1) π/ 8)π 2 (2 n −1) 2 ;
(c) an = −8(24+12( −1) n (2 n −1) π +(2 n −1) 2 π 2 )
π 4 (2 n −1) 4 ;
(d) an = −8(5760+2880( −1) n (2 n −1) π +240(2 n −1) 2 π 2 +7(2 n −1) 4 π 4 )
π 6 (2 n −1) 6 .
The errors in approximating the original functions using 10 terms of the Fourier quarter-wave cosine series aregraphed in Figure 5.2.
Using integration by parts twice, almost exactly as in (5.19), we can express bn in terms of an :
−2T
0d
2u
dx 2 (x)sin (2n −1)πx2
dx
= −2T du
dx(x)sin (2n −1)πx
2 x =0
− (2n −1)π
2
0
dudx
(x)cos (2n −1)πx2
dx
= 2T (2n −1)π2 2
0
dudx
(x)cos (2n −1)πx2
dx
(since sin (0) = du/dx ( ) = 0)
= 2T (2n −1)π2 2 u(x)cos (2n −1)πx
2 x =0
+ (2n −1)π2
0u(x)sin (2n −1)πx
2dx
= T (2n −1)2 π 2
4 22
0u(x)sin (2n −1)πx
2dx
(since u(0) = cos ((2 n −1)π/ 2) = 0)
= T (2n −1)2 π 2
4 2 an .
This gives the desired result. (Actually, since it is known that the negative second derivative operator is symmetricunder the mixed boundary conditions, we can just appeal to (5.25), which is the above calculation writtenabstractly.)
9. Let a 1 , a 2 , a 3 , . . . be the Fourier quarter-wave sine coefficients of u ; then
−κ d2 udx 2 (x) = ∞
n =1
κ(2n −1)2 π 2
2002 an sin (2n −1)πx200
,
where κ = 3 / 2. We also have
0.001 =∞
n =1
0.004(2n −1)π
sin (2n −1)πx200
.
Setting the two series equal and solving for a n , we nd
u(x) =∞
n =1
3.2 ·102
3(2n −1)3 π 3 sin (2n −1)πx200
.
The temperature distribution is graphed in Figure 5.3.
0 20 40 60 80 1000
0.5
1
1.5
2
2.5
3
3.5
x
t e m p e r a t u r e
Figure 5.3: The temperature distribution (in degrees Celsius) in Exercise 5.3.9.
1. Suppose f , g ∈C 2D [0, ], so that f (0) = f ( ) = g(0) = g( ) = 0 .
Then
− ddx
k(x) df dx
, g
= −
0
ddx
k(x) df dx
(x) g(x) dx
= − k(x) df dx
(x)g(x)0
+
0k(x) df
dx(x) dg
dx(x) dx (integration by parts)
=
0k(x) df
dx(x) dg
dx(x) dx (using g(0) = g( ) = 0)
= k(x)f (x)dgdx (x) 0 −
0 f (x) ddx k(x)
dgdx (x) dx (integration by parts)
= −
0f (x) d
dxk(x) dg
dx(x) dx (using f (0) = f ( ) = 0)
= f, − ddx
k(x) dgdx
.
3. Take p(x) = ( x −c)3 (d −x)3 and v[c,d ] as suggested in the hint.5. Let v ∈C 2D [0, ] be a test function. We have
− ddx
k(x) dudx
(x) + p(x)u(x) = f (x), 0 < x <
⇒ − ddx
k(x) dudx
(x) v(x) + p(x)u(x)v(x) = f (x)v(x), 0 < x <
⇒ −
0
ddx
k(x) dudx
(x) v(x) dx +
0 p(x)u(x)v(x) dx =
0f (x)v(x) dx
⇒ −k(x) dudx
(x)v(x)0
+
0k(x) du
dx(x) dv
dx(x) dx +
0 p(x)u(x)v(x) dx =
0f (x)v(x) dx
⇒
0k(x) du
dx(x) dv
dx(x) dx +
0 p(x)u(x)v(x) dx =
0f (x)v(x) dx
⇒
0k(x) du
dx(x) dv
dx(x) + p(x)u(x)v(x) dx =
0f (x)v(x) dx.
The last step follows from the fact that v(0) = v( ) = 0. Thus the weak form of the given BVP is
nd u
∈
C 2D [a, b] such that
0
k(x) dudx
(x) dvdx
(x) + p(x)u(x)v(x) dx =
0
f (x)v(x) dx for all v
∈C 2D [a, b].
7. The calculation is the same as in Exercise 5; we integrate by parts only in the rst integral, and obtain thefollowing weak form: Find u∈C 2D [a, b] such that
0k(x) du
dx(x) dv
dx(x) + c(x) du
dx(x)v(x) + p(x)u(x)v(x) dx =
0f (x)v(x) dx for all v ∈C 2D [a, b].
Notice that now the left-hand side is not symmetric in u and v.9. Repeating the calculation beginning on page 167, we obtain
using F N as the approximating subspace. The weak form of the BVP is
nd u∈C 2D [a, b] such that
0k du
dx(x) dv
dx(x) dx =
0f (x)v(x) dx for all v ∈C 2D [a, b],
and the Galerkin method takes the form
nd vN =N
j =1
u j sin( jπx ) such that
0k dvn
dx (x) dv
dx(x) dx =
0f (x)v(x) dx for all v ∈F N .
We nd the coefficients u j , j = 1 , 2, . . . , N , by solving Ku = f , where
K ij =
0k dφ j
dx (x) dφi
dx (x) dx, F i =
0f (x)φi (x) dx
and φj (x) = sin( jπx ). We have
0k dφj
dx (x) dφ i
dx (x) dx =
0kijπ 2 cos( jπx )cos( iπx ) dx =
kj 2 π 2
2 , i = j,0, i = j.
This last step follows from the fact that the functions cos ( πx ), cos(2πx ), . . . , cos(Nπx ) are mutually orthogonalwith respect to the L2 (0, 1) inner product. Also, notice that
0f (x)φj (x) dx =
0f (x)sin( jπx ) dx = cj
2 ,
where cj is the Fourier sine coefficient of f . Since K is diagonal, it is easy to solve Ku = f :
u j = cj / 2kj 2 π 2 / 2
= cj
kj 2 π 2 .
Thus
vN (x) =N
j =1
cj
kj 2 π 2 sin ( jπx ).
This is exactly the solution obtained by the method of Fourier series in Section 5.3.2.
5. (a) The set S is the span of x, x 2 and therefore is a subspace.
(b) As discussed in the text, a(·, ·) automatically satises two of the properties of an inner product. Everyfunction p in S satises p(0) = 0, so a(·, ·) also satises the third property ( a(u, u ) = 0 implies that u = 0)for exactly the same reason as given in the text for the subspace V (see page 174).
(d) The bilinear form is not an inner product on P 2 , since a(1, 1) = 0 but 1 = 0 (a constant is “invisible” to theenergy inner product and norm). Therefore, every polynomial of the form (9
−3e)x2 + (4 e
−10)x + C
∈ S
is equidistant from f (x) = ex in the energy norm .
7. Write p(x) = x(1 −x) and q (x) = x(1/ 2 −x)(1 −x). The approximation will be v(x) = u1 p(x) + u2 q (x), whereKu = f . The stiffness matrix K is
K = 1
0 (1 + x) dpdx (x) dp
dx (x) dx 1
0 (1 + x) dqdx (x) dp
dx (x) dx
1
0 (1 + x) dpdx (x) dq
dx (x) dx 1
0 (1 + x) dqdx (x) dq
dx (x) dx
= 1
2 − 130
− 130
340
,
and the load vector f is
f = 1
0 xp(x) dx
1
0 xq (x) dx =
112
− 1120 .
Therefore,
u .= 0.16412
−0.038168 .
The exact solution is
u(x) = −14
x2 + 12
x − 14 ln2
ln (1 + x).
The exact and approximate solutions are graphed in Figure 5.4.
0 0.2 0.4 0.6 0.8 10
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
0.045
exactapproximate
Figure 5.4: The exact and approximate solutions from Exercise 5.5.7.
9. (a) It will take about 10 3 times as long, or 1000 seconds (almost 17 minutes), to solve a 1000
×1000 system,
and 100 3 times as long, or 10 6 seconds (about 11 .5 days) to solve a 10000 ×10000 system.
(b) Gaussian elimination consists of a forward phase, in which the diagonal entries are used to eliminate nonzeroentries below and in the same column, and a backward phase, in which the diagonal entries are used toeliminate nonzero entries above and in the same column. During the forward phase, at a typical step, thereis only 1 nonzero entry below the diagonal, and only 5 arithmetic operations are required to eliminate it (1division to compute the multiplier, and 2 multiplications and 2 additions to add a multiple of the currentrow to the next). Thus the forward phase requires O(5n) operations. A typical step of the backward phaserequires 3 arithmetic operations (a multiplication and an addition to adjust the right-hand side and a divisionto solve for the unknown). Thus the backward phase requires O(3n) operations. The grand total is O(8n)operations.
(c) It will take about 10 times as long, or 0 .1 seconds, to solve a 1000 ×1000 tridiagonal system, and 100 timesas long, or 1 second, to solve a 10000
6.2. PURE NEUMANN CONDITIONS AND THE FOURIER COSINE SERIES 41
The solution is
u(x, t ) =∞
n =1
10(1
−(
−1)n )
nπ e−
κn 2 π 2
100 2 ρct sin nπx
100.
The temperature at the midpoint after 20 minutes is
u(50, 1200) .= 1 .58 degrees Celsius .
11. (a) The steady-state temperature is u s (x) = x/ 20.
(b) The temperature u(x, t ) satises the IBVP
ρc ∂u∂t −κ ∂ 2 u
∂x 2 = 0 , 0 < x < 100, t > 0,
u(x, 0) = 5 , 0 < x < 100,u(0, t ) = 0 , t > 0,
u(100, t ) = 5 , t > 0.The solution is
u(x, t ) = x20
+∞
n =1
10nπ
e− κn 2 π 2
100 2 ρct sin nπx
100.
Considering the results of Exercise 6.1.9, it is obvious that at least several thousand seconds will elapsebefore the temperature is within 1% of steady state, so we can accurately estimate u(x, t ) using a singleterm of the Fourier series:
u(x, t ) .= x20
+ 10π
e− κπ 2
100 2 ρct sin πx
100.
We want to nd t large enough that
|u(x, t ) −u s (x)||u s (x)| ≤0.01
for all x∈[0, 100]. Using our approximation for u , this is equivalent to
200sin πx100
πxe− κπ 2
100 2 ρct
≤0.01.
A graph shows that200sin πx
100
πx ≤2
for all x , so we need
e− κπ 2
100 2 ρct
≤0.005.
This yields
t
≥ −1002 ρc ln 0.005
κπ 2
.= 4520 .
About 75 minutes and 20 seconds are required.
6.2 Pure Neumann conditions and the Fourier cosine series1. The solution is
Figure 6.3: The solution u(x, t ) from Exercise 6.2.4 at times 0, 0.02, 0.04, and 0.06, along with the steady-statesolution. These solutions were estimated using 10 terms in the Fourier series.
3. (a) If u, v
∈C 2
N [0, ], then
(LN u, v ) = −
0
d2 udx 2 (x)v(x) dx
= − dudx
(x)v(x)0
+
0
dudx
(x) dvdx
(x) dx
=
0
dudx
(x) dvdx
(x) dx (since dudx (0) = du
dx ( ) = 0)
= u(x) dvdx
(x)0 −
0u(x) d2 v
dx 2 (x) dx
= −
0u(x) d2 v
dx 2 (x) dx (since dvdx (0) = dv
dx ( ) = 0)
= ( u, L N v).
This shows that LN is symmetric.(b) Suppose λ is an eigenvalue of LN and u is a corresponding eigenvector, normalized so that ( u, u ) = 1. Then
λ = λ(u, u ) = ( λu,u )= ( LN u, u )
= −
0
d2 udx 2 (x)u(x) dx
= − dudx
(x)u(x)0
+
0
dudx
(x)2
dx
=
0
dudx
(x)2
dx (since dudx (0) = du
dx ( ) = 0)
≥ 0.Thus L N cannot have any negative eigenvalues.
6.2. PURE NEUMANN CONDITIONS AND THE FOURIER COSINE SERIES 43
has eigenpairs
λ 0 = 1 , γ 0 (x) = 1 , and λ n = 1 + κn2
π2
2 , γ n (x) = cos nπx , n = 1 , 2, 3, . . . .
The method of Fourier series can be applied to show that a unique solution exists for each f ∈C [0, ]. (Thekey is that 0 is not an eigenvalue of K , as it is of L N .)
7. As shown in this section, the solution to the IBVP is
u(x, t ) = d0 +∞
n =1
dn e−n 2 π 2 t/ 2
cos nπx ,
where d0 , d1 , d2 , . . . are the Fourier cosine coefficients of ψ. We have
∞
n =1
dn e−n 2 π 2 t/ 2
cos nπx≤
∞
n =1|dn |e−n 2 π 2 t/ 2
cos nπx
≤ ∞n =1 |dn |e−n
2π
2t/
2
≤ e−π 2 t/ 2 ∞
n =1|dn |e−( n 2 −1) π 2 t/ 2
.
This last series certainly converges (as can be proved, for example, using the comparison test), and
e−π 2 t/ 2
→0 as t → ∞.This shows that
d0 +∞
n =1
dn e−n 2 π 2 t/ 2
cos nπx→d0 as t → ∞.
The limit d0 is1
0ψ(x) dx,
the average of the initial temperature distribution.9. The steady-state temperature is about 0 .992 degrees Celsius.
11. (a) Suppose that the Fourier sine series of u(x, t ) on (0, ) is
u(x, t ) =∞
n =1
an (t)sin nπx , an (t) = 2
0u(x, t )sin nπx dx
and u satises∂u∂x
(0, t ) = ∂u∂x
( , t) = 0 for all t > 0.
The n th Fourier sine coefficient of −∂ 2 u/∂x 2 is computed as follows:
−2
0∂
2
u∂x 2 (x, t )sin nπx dx
= −2 ∂u
∂x(x, t )sin nπx
0 − nπ
0
∂u∂x
(x, t )cos nπx dx
= 2nπ2
0
∂u∂x
(x, t )cos nπx dx
= 2nπ2 u(x, t )cos nπx
0+ nπ
0u(x, t )sin nπx dx
= 2nπ2 ((−1)n u( , t) −u(0, t )) + n2 π 2
2 an (t).
Since the values u(0, t ) and u( , t) are unknown, we see that it is not possible to express the Fourier sinecoefficients of
and the formal calculation of the sine series of −∂ 2 u/∂x 2 yields
∞
n =1
2 (1 −(−1)n ) nπt sin( nπx ).
However,
−∂ 2 u∂x 2 (x, t ) = 0 ,
and so all of the Fourier sine coefficients of −∂ 2 u/∂x 2 should be zero. Thus the formal calculation is wrong.
6.3 Periodic boundary conditions and the full Fourier series1. The formula for the solution u is exactly the same as in Example 6.6, with a different value for κ (4.29 instead of
3.17). This implies that the amplitude of the solution is reduced by about 26%. Therefore, there is less variationin the temperature distribution in the silver ring as opposed to the gold ring.
3. (a) The IBVP is
ρc ∂u∂t −κ ∂ 2 u
∂x 2 = 0 , −5π < x < 5π, t > 0,
u(x, 0) = ψ(x), −5π < x < 5π,u(−5π, t ) = u(5π, t ), t > 0,
∂u∂x
(−5π, t ) = ∂u∂x
(5π, t ), t > 0.
(b) The solution is
u(x, t ) = c0 +∞
n =1
cn e− κn 225 ρc t cos nx
5,
where
c0 = 25 + π 4
9 ,
cn = 10(−1)n +1
n 4 , n = 1 , 2, 3, . . . .
(c) The steady-state temperature is the constant u s = 25 + π 4 / 9 degrees Celsius.
(d) We must choose t so that
|u s −u(x, t )||u s | ≤
0.01
holds for every x∈[−5π, 5π]. By trial and error, we nd that about 360 seconds (6 minutes) are required.
5. (a) To show that Lp is symmetric, we perform the now familiar calculation: we form the integral ( Lp u, v ) andintegrate by parts twice to obtain ( u, L p v). The boundary term from the rst integration by parts is
−dudx
(x)v(x)−
= dudx
(− )v(− ) − dudx
( )v( ).
Since both u and v satisfy periodic boundary conditions, we have
v(− ) = v( ), dudx
(− ) = dudx
( ),
so the boundary term vanishes. The boundary term from the second integration by parts vanishes for exactlythe same reason.
6.3. PERIODIC BOUNDARY CONDITIONS AND THE FULL FOURIER SERIES 45
(b) Suppose Lp u = λu , where u has been normalized: ( u, u ) = 1. Then
λ = λ(u, u ) = ( λu,u ) = ( Lp u, u ) = −
−d
2
udx2 (x)u(x) dx
= −dudx
(x)u(x)−
+
−
dudx
(x)2
dx
=
−
dudx
(x)2
dx
≥ 0.
The boundary terms vanishes because of the periodic boundary conditions:
dudx
(− )u(− ) = dudx
( )u( ).
7. (a) Since the ring is completely insulated, a steady-state temperature distribution cannot exist unless the netamount of heat being added to the ring is zero. This is exactly the same situation as a straight bar with theends, as well as the sides, insulated.
(b) Suppose u is a solution to (6.21). Then
−f (x) dx = −κ
−
∂ 2 u∂x 2 (x, t ) dx = −κ ∂u
∂x(x, t )
−= κ ∂u
∂x(− , t) −
∂u∂x
( , t) = 0 .
The last step follows from the periodic boundary conditions.(c) The negative second derivative operator, subject to boundary conditions, has a nontrivial null space, namely,
the space of all constant functions on ( − , ). In analogy to the Fredholm alternative for symmetric matrices,we would expect a solution to the boundary value problem to exist if and only if the right-hand-side functionis orthogonal to this null space. This condition is
−cf (x) dx = 0 for all c ∈R ,
or simply
−f (x) dx = 0 .
9. Consider the IBVP
ρc ∂u∂t −κ ∂ 2 u
∂x 2 + pu = f (x, t ), − < x < , t > t 0 ,
u(x, t 0 ) = ψ(x), − < x < ,u(− , t) = u( , t), t > t 0 ,
∂u∂x
(− , t) = ∂u∂x
( , t), t > t 0 .
Assume that p is a constant.
(a) We wish to derive the solution to the IBVP using the Fourier series method. The calculation is very similarto that carried out in Section 6.3.3 in the text, and we will use the same notation as in that section. Werepresent the solution u as the series
u(x) = a0 +∞
n =1
an cos nπx + bn sin nπx .
Substituting this series into the left side of the PDE yields
Following the derivation in Section 6.3.3, we nd a 0 , a 1 , . . . , b1 , b2 , . . . by solving the IVPs
ρcda 0
dt + pa0 = c0 (t), a0 (t0 ) = p0 ,
ρc da n
dt + κn 2 π 2
2 + p an = cn (t), an (t0 ) = pn , n = 1 , 2, . . . ,
ρc dbn
dt + κn 2 π 2
2 + p bn = dn (t), bn (t0 ) = q n , n = 1 , 2, . . . .
(b) The eigenvalues of the spatial operator are λ 0 = p and
λ n = κn 2 π 2
2 + p, n = 1 , 2, . . . .
Therefore, λ n ≥λ0 for all n , and all the eigevalues are positive if and only if p > 0.
11. Let f : (− , ) → R be an even function. We wish to prove that the full Fourier series of f reduces to the cosineseries of f , regarded as a function on (0 , ). The full Fourier series of f is
f (x) = a0 +∞
n =1
an cos nπx + bn sin nπx ,
where
a0 = 12
−f (x) dx,
an = 1
−f (x)cos nπx dx, n = 1 , 2, . . . ,
bn = 1
−f (x)sin nπx dx, n = 1 , 2, . . . .
For any function g : (
−, )
→R , if g is even, then
−g(x) dx = 2
0g(x) dx,
while if g is odd, then
−g(x) dx = 0 .
Since f is even,
a0 = 12
−f (x) dx = 1
0f (x) dx.
Also, f (x)cos(nπx/ ) is an even function of x (the product of even functions is even), and therefore
an = 1
−
f (x)cos nπx dx = 2
0f (x)cos nπx dx, n = 1 , 2, . . . .
Finally, f (x)sin( nπx/ ) is an odd function of x (the product of an even and an odd function is odd), and thus
bn = 1
−f (x)sin nπx dx = 0 , n = 1 , 2, . . . .
Thus the full Fourier series reduces to
f (x) = a0 +∞
n =1
an cos nπx ,
where
a0 = 1
0f (x) dx, a n = 2
0f (x)cos nπx dx, n = 1 , 2, . . . .
This is precisely the Fourier cosine series of f on the interval (0 , ) (cf. Section 6.2.2).
with k = 1 .5 and f (x) = 10 −7 x(25 −x)(100 −x)+1 / 240. The steady-state temperature is not unique; the solutionwith u(100) = 0 is shown in Figure 6.6.
0 20 40 60 80 100−1
0
1
2
3
4
5
6
x
t e m p e r a
t u r e
Figure 6.6: The computed steady-state temperature distribution from Exercise 6.5.1.
3. (a) The total amount of heat energy being added to the bar is 0 .51A W, where A is the cross-sectional area
(0.01A W through the left end and 0 .5A W in the interior). Therefore, 0 .51A W must be removed throughthe right end; that is, heat energy must be removed at a rate of 0 .51 W/ cm2 through the right end.
(b) The BVP is
−κ d2 udx 2 = 0 .005, 0 < x < 100,
k dudx
(0) = −0.01,
k dudx
(100) = −0.51.
The steady-state temperature is not unique; the temperature with u(100) = 0 is graphed in Figure 6.7.
0 20 40 60 80 1000
1
2
3
4
5
6
7
x
t e m p e r a t u r e
Figure 6.7: The computed steady-state temperature distribution from Exercise 6.5.3.
(b) Suppose v ∈ V and a (v, v ) = 0. By denition of a(·, ·), this is equivalent to
0κ(x) dv
dx(x)
2
dx = 0 .
Since the integrand is nonnegative, this implies that the integrand is in fact zero, and, since κ(x) is positive,we conclude that
dvdx
(x)
is the zero function. Therefore, v is a constant function.
(c) Thus, if K u = 0, it follows that v(x) = ni =0 u i φi (x) is a constant function, where u0 , u 2 , . . . , u n are the
components of u . But then there is a constant C such that v(x i ) = C , i = 0 , 1, 2, . . . , n . These nodal valuesof v are precisely the numbers u 0 , u 1 , . . . , u n , so we see that u = C u c , which is what we wanted to prove.
7. (a) Dene V = v ∈C 2 [0, ] : v(0) = 0 . Then the weak form of the BVP is
nd u∈ V such that a (u, v ) = ( f, v ) for all v ∈ V . (6.1)
(b) The fact that a solution of the strong form is also a solution of the weak form is proved by the usual argument:multiply the differential equation by an arbitrary test function v ∈ V , and then integrate by parts.
The proof that a solution of the weak form is also a solution of the strong form is similar to the argumentgiven in Section 6.5.2. Assuming that u satises the weak form (6.1), an integration by parts and somesimplication shows that
k( ) dudx
( )v( ) −
0
ddx
k(x) dudx
(x) + f (x) v(x) dx = 0 for all v ∈ V .
Since V ⊂ V , this implies that the differential equation
ddx
k(x) dudx
(x) + f (x) = 0 , 0 < x <
holds, and we then have
k( ) dudx
( )v( ) = 0 for all v ∈ V .
Choosing any v ∈ V with v( ) = 0 shows that the Neumann condition holds at x = .
9. The temperature distribution after 300 seconds is shown in Figure 6.8.
Figure 6.8: The temperature distribution from Exercise 6.5.9 (after 300 seconds).
11. Here is a sketch of the proof: If Ku = 0, then
u · Ku = 0 ⇒a(v, v ) = 0 ,
where v = n −1i =0 ui φi . But then, by the usual reasoning, v must be a constant function, and v(xn ) = 0 (since
φi (xn ) = 0 for i = 0 , 1, 2, . . . , n −1). Thus v is the zero function, which implies that the nodal values of v are allzero. Therefore u = 0, and so K is nonsingular.
7.1 The homogeneous wave equation without boundaries
1. The solution, by d’Alembert’s formula, is u(x, t ) = ( ψ(x −ct) + ψ(x + ct)). Figure 7.1 shows three snapshots of u.
−20 −15 −10 −5 0 5 10 15 20−5
0
5
10
15
20x 10
−4
Figure 7.1: Three snapshots of the solution of Exercise 7.1.1: t = 0 (dashed curve), t = 0 .01 (solid curve), andt = 0 .025 (dash-dotted curve).
3. The two waves join and add constructively, and then separate again. See Figure 7.2.
−40 − 30 − 20 − 10 0 10 20 30 400
0.5
1
1.5x 10
−3
x
t=0t=0.02
t=0.04
Figure 7.2: Constructive interference in Exercise 7.1.3. The “blip” in the center is the result of two wavescombining temporarily. (The velocity in this example is c = 1.)
where the support of ψ is [a, b]. The solution is u(x, t ) = ( ψ(x −ct) + ψ(x + ct)) / 2. Now, since ψ(x −ct) is aright-moving wave, we see that ψ(x1 −ct) = 0 if any of the following conditions hold:
• x1 ≤a;
• a < x 1 < b and t ≥ x 1 −ac (since ψ(x1 −ct) = 0 after the trailing edge of the wave passes x1 );
• x ≥ b and ( t ≤ x 1 −bc or t ≥ x 1 −a
c ) (since ψ(x1 −ct) = 0 before the leading edge of the wave reaches x1 andafter the trailing edge of the wave passes x1 ).
Similarly, ψ(x + ct) is a left-moving wave, and therefore ψ(x1 + ct) = 0 if any of the following is true:
• x1 ≥b;
• a < x 1 < b and t ≥ b−x 1c ;
• x1 ≤a and ( t ≤ a −x 1c or t ≥ b−x 1
c ).
Since u(x1 , t ) is guaranteed to be zero if both ψ(x1 −ct) and ψ(x1 + ct) are zero, we see that u(x1 , t ) = 0 if anyof the following is true:
• x1 ≤a and ( t ≤ a −x 1c or t ≥ b−x 1
c );
• a < x 1 < b and t ≥ x 1 −ac and t ≥ b−x 1
c ;
• x1 ≥b and ( t ≤ x 1 −bc or t ≥ x 1 −a
c ).
7. Consider the IBVP
∂ 2 u∂t 2 −c2 ∂ 2 u
∂x 2 = 0 , 0 < x < , t > 0,
u(x, 0) = ψ(x), 0 < x < ,∂u∂t
= 0 , 0 < x < ,
u(0, t ) = 0 , t > 0,u( , t) = 0 , t > 0,
where the support of ψ is [a, b] and 0 < a < b < . If we dene
u(x, t ) = 12
(ψ(x −ct) + ψ(x + ct)) ,
then we know that u satises the same PDE and initial condition as does the solution of the above IBVP. Also,
since the support of ψ is [a, b], u(0, t ) = 0 for all t ≤a/c (since u(0, t ) cannot be nonzero until the leading edge of the left-moving wave reaches x = 0), and similarly u( , t) = 0 for all t ≤ ( −b)/c (since u( , t) cannot be nonzerountil the leading edge of the right-moving wave reaches x = ). Therefore, if t < t f = min a/b, ( −b)/c , then usatises both boundary conditions, in addition to the initial conditions and the PDE, and hence is a solution of the above IBVP.
7.2 Fourier series methods for the wave equation1. The solution is found by setting cn = 0 in Example 7.4:
u(x, t ) =∞
n =1
bn cos cnπt25
sin nπx25
.
A graph analogous to Figure 7.6 from the text is given in Figure 7.3.
Fifty snapshots of the solution are shown in Figure 7.5. The solution gradually moves up; this is possible becauseboth ends are free to move vertically.
Figure 7.5: Fifty snapshots of the vibrating string from Example 7.4, with both ends free.
where
an (t) = bn − 4cn
c2 (2n −1)2 π 2 cos c(2n −1)πt2
+ 4cn
c2 (2n −1)2 π 2 ,
bn = 2(−1)n +1
125(2n −1)2 π 2 ,
cn = 4g(2n −1)π
.
Snapshots of the solution are graphed in Figure 7.6, which should be compared to Figure 7.9 from the text.
0 0.2 0.4 0.6 0.8 10
1
2x 10
−3
x
d i s p l a c e m e n t
t=0t=5e−05t=0.0001t=0.00015
Figure 7.6: Snapshots of the displacement of the bar in Example 7.6, taking into account the effect of gravity.
9. Consider a string whose (linear) density (in the unstretched state) is 0 .25 g/ cm and whose longitudinal stiffness is6500N. Suppose the unstretched (zero tension) length of the string is 40 cm. We wish to determine the length towhich the string must be stretched in order that its fundamental frequency be f 0 = 261 Hz (middle C). We will
write 0 = 40, k = 6 .5 ·108 (the stiffness of the string in g cm / s2 ), and T for the unknown tension. Note that themass of the string is m = 10 (g). We know that f 0 = c/ (2 ), where c is the wave speed and = 0 + ∆ . Also,c2 = T /ρ , where ρ = m/ . Finally, ∆ is related to T by Hooke’s law: k∆ / 0 = T . Putting all these equationstogether, we obtain
7.3. FINITE ELEMENT METHODS FOR THE WAVE EQUATION 57
7.3 Finite element methods for the wave equation
1. The fundamental period is 2 /c = 0 .2. Using h = 100 / 20 = 5, Dt = T / 60, and the RK4 method, we obtain thesolution shown in Figure 7.7.
0 0.2 0.4 0.6 0.8 1−0.2
0
0.2
x
d i s p l a c e m e n t
Figure 7.7: The computed solution of the wave equation in Exercise 7.3.1. Shown are 30 snapshots, plus theinitial displacement. Twenty subintervals in space and 60 time steps were used.
3. (a) Since the initial disturbance is 24cm from the boundary and the wave speed is 400 cm / s, it will take24/ 400 = 0 .06 s for the wave to reach the boundary.
(b) The IBVP is
∂ 2 u∂t 2 −c2 ∂ 2 u
∂x 2 = 0 , 0 < x < 50, t > 0,
u(x, 0) = 0 , 0 < x < 50,∂u∂t
(x, 0) = γ (x), 0 < x < 50,
u(0, t ) = 0 , t > 0,u(50, t ) = 0 , t > 0,
with c = 400 and γ given in the statement of the exercise.(c) Using piecewise linear nite elements and the RK4 method (with h = 50 / 80 and ∆ t = 6 ·10−4 ), we computed
the solution over the interval 0 ≤ t ≤0.06. Four snapshots are shown in Figure 7.8, which shows that it doestake 0 .06 seconds for the wave to reach the boundary. (The reader should notice the spurious “wiggles” inthe computed solution; these are due to the fact that the true solution is not smooth.)
0 10 20 30 40 50−0.1
0
0.1
x
d i s p
l a c e m e n t
Figure 7.8: The computed solution of the wave equation in Exercise 7.3.3.
5. It is not possible to obtain a reasonable numerical solution using nite elements. In Figure 7.9, we display theresult obtained using h = 2 .5 ·10−3 and ∆ t = 3 .75 ·10−4 . Three snapshots are shown.
Therefore, if ω does not equal cn/ (2 ) for any odd n, then the Fourier coefficients of u are uniformly boundedwith respect to t and go to zero like 1 /n 3 . Notice, though, that if ω is very close to cn/ (2 ) for some odd n , thenthe corresponding an (t) is larger (due to the factor of c2 n 2 −4 2 ω2 in the denominator), and thus that frequencyhas greater weight in the Fourier series.On the other hand, if ω = cn/ (2 ) for some odd n , then the corresponding Fourier coefficient an (t) grows withoutbound as t → ∞, and resonance is observed.
3. The experiment described in this problem is modeled by the IBVP
∂ 2 u∂t 2 −c2 ∂ 2 u
∂x 2 = 0 , 0 < x < 1, t > 0,
u(x, 0) = 0 , 0 < x < 1,
∂u∂t
(x, 0) = 0 , 0 < x < 1,
k ∂u∂x
(0, t ) = B sin(2 πωt ), t > 0,
u(1, t ) = 0 , t > 0.
Here c2 = k/ρ .= 3535 .53 m/ s.We transform this problem into one that we can analyze by the Fourier series method by shifting the data. Thefunction v(x, t ) = ( B/k )sin(2 πωt )(x −1) satises the given boundary conditions; if we dene w = u −v, then wsatises the IBVP
∂ 2 w∂t 2 −c2 ∂ 2 w
∂x 2 = 4π 2 ω2 Bk
sin(2πωt )( x −1), 0 < x < 1, t > 0,
w(x, 0) = 0 , 0 < x < 1,
∂w∂t
(x, 0) = −2πωBk
(x −1), 0 < x < 1,
k ∂w∂x
(0, t ) = 0 , t > 0,
w(1, t ) = 0 , t > 0.
We solve this IBVP by the usual method. We write
u(x, t ) =∞
n =1
an (t)sin( nπx ).
The Fourier coefficients a n are found by solving the IVP
d2 an
dt 2 + c2 n 2 π 2 an = cn (t), an (0) = 0 , dan
dt (0) = bn ,
where
cn (t) = 2 1
0
4π 2 ω2 Bk
sin(2πωt )( x −1)sin( nπx ) dx = −8πω 2 B
kn sin(2πωt )
and
bn = −4πωB
k 1
0(x −1)sin( nπx ) dx = 4ωB
kn .
It follows (see Section 4.2.3 from the text) that
an (t) = bn
cnπ sin(cnπt ) + 1
cnπ t
0sin( cnπ (t −s)) cn (s) ds = 4ωB
kcn 2 π sin (cnπt ) + 1
cnπ t
0sin( cnπ (t −s)) cn (s) ds.
If ω = cn/ 2, then
an (t) = 4ωBkcn 2 π
sin (cnπt ) − 8ω2 B (cn sin(2 πωt ) −2ω sin( cnπt ))
(a) From the above analysis, we see that the smallest resonant frequency is ωr = c/ 2, the fundamental frequencyof the wave equation modeling the problem.
(b) If ω = ωr , then a1 (t) is given by the second formula above, while an (t), n > 1, is given by the rst. Wecompute several snapshots of the displacement, using 20 terms in the Fourier series, on each of the timeintervals [0 , 0.005], [0.005, 0.01], [0.01, 0.015], and [0.015, 0.02]. These are shown in Figure 7.10. The graphsshow that the amplitude grows as t increase, and also that the shape of the displacement is dominated bythe rst fundamental mode.
0 0.5 1−0.5
0
0.5
0 0.5 1−0.5
0
0.5
0 0.5 1−0.5
0
0.5
0 0.5 1−0.5
0
0.5
Figure 7.10: Resonance in Exercise 7.3.3(a).
(c) With ω = ωr / 4, we compute the displacements at the same times as in the previous part of the exercise.The results are shown in Figure 7.11. Here the absence of resonance is clearly seen, in that the amplitudesremain bounded as t increases.
0 0.5 1−5
0
5x 10
−3
0 0.5 1−5
0
5x 10
−3
0 0.5 1−5
0
5x 10
−3
0 0.5 1−5
0
5x 10
−3
Figure 7.11: Absence of resonance in Exercise 7.3.3(b).
Solving this equation for the second derivative yields
d2 φdx 2 (x) = φ(x + ∆ x) −2φ(x) + φ(x −∆ x)
∆ x2 − 124
d4 φdx 4 (c1 ) + d4 φ
dx4 (c2 ) ∆ x2 .
Sinced4 φdx 4 (c1 ) + d4 φ
dx 4 (c2 ) →2 d4 φdx 4 (x)
as ∆ x → 0, it follows that the error in the central difference approximation to the second derivative is approxi-mately proportional to ∆ x2 .
5. With a Dirichlet condition at the left endpoint and a Neumann condition at the right, we must compute
u ( j )i
.= u(x i , t j ), i = 1 , 2, . . . , k, j = 1 , 2, . . . , n
(using the same notation as in the text). We apply the usual 2-2 scheme at each ( x i , t j ), i = 1 , 2, . . . , k − 1,resulting in equation (7.41) (page 301 in the text) for computing u( j +1)
i , i = 1 , 2, . . . , k −1. To compute u( j +1)k ,
we approximate the Neumann condition∂u∂x
(xk , t j ) = 0
byu ( j )
k +1 −u ( j )k−1
∆ x = 0 ⇒ u( j )
k +1 = u ( j )k−1 .
We then obtain equation (7.48) for u ( j +1)k . Finally, we use (7.44) to compute u (1)
i , i = 1 , 2, . . . , k −1, and (7.41)with i = k and ψ(x i +1 ) replaced by ψ(x i−1 ) to compute u (1)
k .To test the scheme, we try in on the given problem, using a time interval of [0 , 1] with initial values of k = 10 andn = 500 (this gives c∆ t/Dt = 0 .9). Doubling k and n at each iteration, we obtain the following maximum errors:
∆ x ∆ t maximum error0.1 0.002 1.9 ·10−2
0.05 0.001 4.8802 ·10−3
0.025 0.0005 1.2250 ·10−3
0.0125 0.00025 3.0792 ·10−4
These errors display the expected second-order behavior.
First-order PDEs and the Method of Characteristics
8.1 The simplest PDE and the method of characteristics
1. Consider the PDE∂u∂y
= 0 .
(a) Intuitively, since the solution is constant in y, the characteristics must be the vertical lines x = const. Wecan verify this formally as follows. Suppose we impose the initial condition u(f (s), g(s)) = h(s). Then thecharacteristics are determined by the equations
∂x
∂t = 0 , x(s, 0) = f (s),
∂y∂t
= 1 , y(s, 0) = g(s)
which yieldx(s, t ) = f (s), y(s, t ) = g(s) + t.
For each xed s, these equations parametrize the vertical line x = f (s).
(b) The IVP∂u∂y
= 0 , u(0, y) = φ(y)
is not well-posed. The general solution of the PDE is u(x, y ) = ψ(x). The initial condition u(0, y) = φ(y)then yields ψ(0) = φ(y), which is only possible if φ is a constant: φ(y) = c. Moreover, in this case, there areinnitely many solutions, one for each ψ satisfying ψ(0) = c. Therefore the given IVP either has no solution(if φ is not a constant function) or innitely many (if φ is a constant function).
3. Consider the PDE4 ∂u
∂x + 3 ∂u
∂t = 0 , 0 < x < 1, t > 0.
Given the initial condition u(f (s), g(s)) = h(s), the characteristics are determined by
∂x∂τ
= 4 , x(0, τ ) = f (s),
∂t∂τ
= 3 , t(0, τ ) = g(s),
which yieldx(s, τ ) = 4 τ + f (s), t(s, τ ) = 3 τ + g(s).
64 CHAPTER 8. FIRST-ORDER PDES AND THE METHOD OF CHARACTERISTICS
(a) Now consider the initial/boundary conditions
u(x, 0) = u0 (x), 0 < x < 1,u(0, t ) = φ(t), t > 0.
Characteristics passing through ( s, 0), 0 < s < 1, have the form
x = 4τ + s, t = 3τ ⇔ τ = t3
, s = x − 43
t.
This characteristic can also be written as t = (3 / 4)( x −s). On such characteristics,
u(x, t ) = v(s, τ ) = u0 (s) = u0 x − 43
t .
Characteristics through (0 , s ), s > 0, have the form
x = 4τ, t = 3τ + s ⇔ τ = x4
, s = t − 34
x.
This characteristic can also be written as t = s + (3 / 4)x. On such characteristics,
u(x, t ) = v(s, τ ) = φ(s) = φ t − 34
x .
The characteristic dividing the domain into two is t = (3 / 4)x. Thus
u(x, t ) = u0 x − 43 t , t ≤ 3
4 x,φ t − 3
4 x , t > 34 x.
(b) Now consider the initial/boundary conditionsu(x, 0) = u0 (x), 0 < x < 1,u(1, t ) = ψ(t), t > 0.
This set of conditions does not dene a well-posed problem, since every characteristic through a point(x0 , 0), 0 < x 0 < 1, also passes through a point (1 , t 0 ), 0 < t 0 < 3/ 4. These means that the solution isover-determined on part of the domain, and u0 , ψ cannot be specied independently of each other.
5. Consider the IBVP
∂u∂t
+ c∂u∂x
= 0 , 0 < x < 1, t > 0,
u(x, 0) = u0 (x), 0 < x < 1,
u(0, t ) = φ(t), t > 0.The solution is
u(x, t ) = u0 (x −ct), t < c −1 x,φ(t −c−1 x), t > c −1 x.
We assume that u0 and φ are both continuously differentiable.
(a) We wish to determine the conditions on u0 , φ that guarantee that u is continuous. Consider any ( x0 , t 0 )with t 0 = c−1 x0 . We have
u0 (x −ct) →u0 (x0 −ct0 ) = u0 (0) as ( x, t ) →(x0 , t 0 ),
φ(t −c−1 x) →φ(t0 −c−1 x0 ) = φ(0) as ( x, t ) →(x0 , t 0 ),
which shows that u is continuous if and only if u 0 (0) = φ(0).
Similar reasoning shows that this condition also ensures that ∂u/∂t is continuous.
7. Consider the IVP
∂u∂y
+ ∂ u∂x
= x + y, (x, y )∈R 2 ,
u(x, −x) = x2 , x∈R .
We solve the problem in characteristics variables as follows:
∂x∂t
= 1 , x(s, 0) = s,
∂y∂t
= 1 , y(s, 0) = −s,
∂v∂t
= x + y, v(s, 0) = s2 .
The rst two equations yield
x = s + t, y = −s + t ⇔ s = x −y2 , t = x + y2 ,
and therefore∂v∂t
= 2 t, v(s, 0) = s2
⇒ v(s, t ) = s2 + t2 .
Thus,
u(x, y ) = x −y2
2+ x + y
2
2= 1
2(x2 + y2 ).
8.2 First-order quasilinear PDEs1. Consider the following IVP:
x ∂u
∂x
+ y ∂u
∂y −(x + y)u = 0 ,
u(1, y) = φ(y).
(a) The characteristics are dened by the equations
∂x∂t
= x, x(s, 0) = 1 ,
∂y∂t
= y, y(s, 0) = s,
which yieldx = et , y = se t
⇔ t = ln x, s = yx
.
Notice that x must be positive. Also, since s represents y0 on the initial curve x = 1, we see that thecharacteristics can be written as y = y0 x. Thus the characteristic through (1 , y0 ) is the line with slope y0
passing through the origin. This also shows that there must be a discontinuity when x decreases to zero.
66 CHAPTER 8. FIRST-ORDER PDES AND THE METHOD OF CHARACTERISTICS
(b) In the characteristic variables, the solution v(s, t ) is dened by
∂v∂t = ( x + y)v = (1 + s)et v, v(s, 0) = φ(s).
The ODE is separable and therefore the IVP is easily solved:
v(s, t ) = φ(s)e(1+ s )( e t −1) .
Changing variables, we obtainu(x, y ) = φ y
xex + y−1−y/x .
(c) The formula for u shows that it is dened for all x > 0 (or for all x < 0, but the initial curve lies in the righthalf-plane). This is consistent with the analysis of the characteristics.
3. Consider the IVP
u ∂u∂x
+ u ∂u∂y
= 0 ,
u(x, 0) = φ(x).
The PDE is quasi-linear and can be solved by solving the characteristic equations:
∂x∂t
= v, x(s, 0) = s,
∂y∂t
= v, y(s, 0) = 0 ,
∂v∂t
= 0 , v(s, 0) = φ(s).
The equations for v yield v(s, t ) = φ(s); substituting this into the rst two equations yields x(s, t ) = s + φ(s)t,y = φ(s)t . From this we obtain s = x −y, t = y/φ (x −y), and the solution is
u(x, y ) = v(s, t ) = φ(s) = φ(x
−y),
that is, u(x, y ) = φ(x −y).If we test the initial curve using the Jacobian test, we obtain
∂x∂s (s, 0) ∂x
∂t (s, 0)∂y∂s (s, 0) ∂y
∂t (s, 0) = 1 φ(s)0 φ(s) = φ(s),
we see that the initial curve is noncharacteristic if and only if φ(s) = 0.
5. Consider the IVP
∂u∂x
+ u ∂u∂y
= 0 , x > 0,
u(x, 0) = x, x > 0.
(a) The IVP can be solved by solving the characteristic equations:
∂x∂t
= 1 , x(s, 0) = s,
∂y∂t
= v, y(s, 0) = 0 ,
∂v∂t
= 0 , v(s, 0) = s.
It is straightforward to obtain the solution:
x(s, t ) = s + t, y (s, t ) = st, v (s, t ) = s.
Solving x = s + t, y = st for s , t (and bearing in mind that s = x when t = 0), we obtain
(b) The characteristic indexed by s passes through the point ( s, 0), so we can regards s as x0 . Notice thatx = s + t , y = st implies that t = x −s and hence y = s(x −s). This shows that the characteristics are thelines y = x0 (x −x0 ).
(c) Since the slopes of the characteristics increase as x0 increases, it follows that, for x1 > x 0 , the characteristicsthrough ( x0 , 0) and ( x1 , 0) will intersect at some point ( x, y ) with x > x 1 . A simple calculation shows thaty = x0 (x −x0 ) and y = x1 (x −x1 ) intersect at the point ( x0 + x1 , x 0 x1 ). Taking the limit as x1 decreasestoward x0 reveals the rst point where the characteristic through ( x0 , 0) intersects another characteristic:(2x0 , x 2
0 ). Notice that this is a point on the curve y = x2 / 4, which is consistent with the results obtainedabove.
8.3 Burgers’s equation1. Consider the inviscid Burgers’s equation
∂u∂t
+ u ∂u∂x
= 0
with initial conditionu(x, 0) = u0 (x) = 1
1 + x2 .
We have seen that the characteristics are the lines
x = x0 + u0 (x0 )t ⇔ t = u0 (x0 )−1 (x −x0 ).
(a) We wish to show that the solution of the IVP is well-dened for at each point ( x, t ), x < 0, t > 0. To showthis, it suffices to prove that there is a unique characteristic passing through any such point. Since the slope(in the x, t plane) of every characteristic is positive, the characteristic through ( x0 , 0) passes through ( x, t )
only if x0 < x < 0 andt = (1 + x2
0 )( x −x0 ) ⇔ x30 −xx 2
0 + x0 + ( t −x) = 0 .
If we dene ψ(x0 ) = x30 − xx 2
0 + x0 + ( t − x), then it is easy to verify that ψ(0) > 0, ψ(x0 ) < 0 for x0
sufficiently large and negative, and ψ (x0 ) > 0 for all x0 < 0. It follows that ψ(x0 ) = 0 has a unique solutionx0 < 0, and hence there is a unique characteristic passing through ( x, t ).
(b) We now wish to describe the set of all ( x, t ), t > 0, for which the solution is uniquely dened. Referring tothe analysis in the text, culminating in Figure 8.9, we see that u(x, t ) is uniquely dened for all ( x, t ) outsidethe caustic, which is dened by the parametric equations
x = 1 + 3x20
2x0, t = (1 + x2
0 )2
2x0, x0 > 0.
The corner of the caustic can be found by setting the derivatives of x and t (with respect to x0 ) to zero and
solving, which yieldsx0 = 1
√ 3 , x = √ 3, t = 83√ 3 .
Moreover, we can solve
x = 1 + 3 x20
2x0
for x0 in terms of x to get
x0 = x ±√ x2 −33
.
Substituting into the equation for t yields the two branches of the caustic:
68 CHAPTER 8. FIRST-ORDER PDES AND THE METHOD OF CHARACTERISTICS
(where t 2 > t 1 for all x). Comparing to Figure 8.9, we see that u(x, t ) is uniquely dened if t > 0 and
x < √ 3 or x ≥√ 3 and ( t < t 1 or t > t 2 .
3. Consider the IVP
∂u∂t
+ (1 −u) ∂u∂x
= 0 , −∞< x < 0, t > 0,
u(x, 0) = 1x
, −∞< x < 0.
(a) The characteristic equations are
∂x∂τ
= 1 −v, x (s, 0) = s,
∂t∂τ
= 1 , t(s, 0) = 0 ,
∂v∂τ = 0 , v(s, 0) = 1s .
Solving yieldsx(s, t ) = s + s −1
s τ, t(s, t ) = τ, v(s, t ) = 1
s.
Setting s = x0 , we obtainx = x0 + x0 −1
x0t ⇔ t = x0
x0 −1(x −x0 ).
These are the equations of the characteristics.
(b) We can solve
x = s + s −1s
t
for s as follows:
x = s + s −1s t⇒s2 + ( t −x)s −t = 0
⇒s = x −t ± (t −x)2 + 4 t
2
⇒s = x −t − (t −x)2 + 4 t
2 ,
where the last step uses the fact that s must be negative. We then obtain u(x, t ) = v(s, τ ) = 1 /s , that is,
u(x, t ) = 2x −t − (t −x)2 + 4 t
.
(c) Notice that the slope of the characteristic decreases as x0 increases; therefore, the characteristics through(x0 , 0) and ( x1 , 0) will intersect at some point ( x, t ) with t < 0. Solving the equations
t = x0x0 −1 (x −x0 ), t = x1
x1 −1 (x −x1 )
for (x, t ) yields(x, t ) = ( x0 −x0 x1 + x1 , −x0 x1 ).
Notice that t < 0, as expected.
5. Consider the IVP
∂u∂t
+ a(u) ∂u∂x
= 0 , −∞< x < ∞, t > 0,
u(x, 0) = φ(x), x > 0.
The characteristics are dened by
x(s, τ ) = s + a(φ(s)) τ, t (s, τ ) = τ , v(s, τ ) = φ(s).
9.1 Green’s functions for BVPs in ODEs: Special cases1. u(x) = xe−x .
3. (a) This BVP models a bar whose top end (originally at x = 0) is free and whose bottom end is xed at x = .If we apply a unit force to the cross-section at x = s, then the part of the bar originally between x = s andx = will compress, and the part of the bar originally above x = s will just move rigidly with u(x) = u(s)for 0 ≤x < s . The compression of the bottom part of the bar will satisfy Hooke’s law:
k u(x)
−x = 1 ⇒u(x) = −x
k
(the uncompressed length of the part of the bar between x = s and x = is −x). Therefore we obtain
u(x) =
−sk , 0 < x < s,
−xk , s < x < .
The Green’s function is
G(x; s) = −s
k , 0 < x < s,
−xk , s < x < .
(b) The Green’s function is
G(x; s) =
max x,s
dz k(z )
.
(c) If k is constant, then
G(x; s) = −maxx, s k
= −s
k , 0 < x < s,
−xk , s < x < .
(d)u(x) = ln 2
2 − 14 −
x2
4 + x
2 − ln (1 + x)
2 .
5. By direct integration, we obtain the following solution to (9.11):
9.2. GREEN’S FUNCTIONS FOR BVPS IN ODES: THE SYMMETRIC CASE 73
Suppose G is symmetric in the sense that G(y, x ) = G(x, y ) for all x, y ∈[a, b]. We then have
(Mf,g )L 2 = b
a(Mf )( x)g(x) dx
= b
a b
aG(x, y )f (y) dy g(x) dx
= b
a b
aG(x, y )f (y)g(x) dydx
= b
a b
aG(x, y )f (y)g(x) dxdy (changing the order of integration)
= b
af (y)
b
aG(x, y )g(x) dx dy
= b
af (y)
b
aG(y, x )g(x) dx dy (since G is symmetric)
= b
af (y)( Mg)( y) dy = ( f ,Mg )L 2 .
Thus M is a symmetric operator with respect to the L 2 (a, b) inner product.
7. Dene H : R →R by
H (x) = 0, x < 0,1. 0 < x
We wish to prove that
∞
−∞δ (x)φ(x) dx = − ∞
−∞H (x) dφ
dx(x) dx
holds for all φ∈D(R ). That is, we wish to prove that if φ∈D(R ), then
∞
−∞H (x) dφ
dx(x) dx =
−φ(0) .
This is a direct calculation. Choose b > 0 sufficiently large that φ(x) = 0 for all x ≥b. Then
∞
−∞H (x) dφ
dx(x) dx = ∞
0
dφdx
(x) dx = b
0
dφdx
(x) dx = φ(b) −φ(0) = −φ(0) ,
as desired.
9. Let G be the Green’s function for the BVP
− ddx
P (x) dudx
+ R(x)u = f (x), a < x < b,
α 1 u(a) + α 2dudx
(a) = 0 ,
β 1u(b) + β
2
du
dx(b) = 0 .
Then
G(x; y) = −v 1 ( y ) v 2 ( x )P ( x ) W ( x ) , a ≤y ≤x,
−v 1 ( x ) v 2 ( y )P ( x ) W ( x ) , x ≤y ≤b,
where v1 and v2 are certain solutions of the homogeneous ODE
− ddx
P (x) dudx
+ R(x)u = 0 , a < x < b.
We wish to show that, for xed y ∈(a, b), u(x) = G(x; y) is the solution of the above BVP with the right-hand-sidefunction f (x) replaced by δ (x −y). To do this, we must interpret the BVP in its weak sense, which is
Here u must satisfy the given boundary conditions and the above variational equation must hold for all testfunctions v satisfying the same boundary conditions. Now, the Green’s function is continuous but its derivative
du/dx has a jump discontinuity at x = y. We will write
dudx
(y−) = limx→y −
dudx
(x) = −v1 (y)v2 (y)P (y)W (y)
,
dudx
(y+) = limx→y +
dudx
(x) = −v1 (y)v2 (y)P (y)W (y)
.
We now substitute u into the left-hand side of the variational equation to obtain
−P (x) dudx
(x)v(x)b
a+
b
aP (x) du
dx(x) dv
dx(x) + R(x)u(x)v(x) dx
= P (a) dudx
(a)v(a) −P (b) dudx
(b)v(b) +
y
aP (x) du
dx(x) dv
dx(x) + R(x)u(x)v(x) dx
+ b
yP (x) du
dx(x) dv
dx(x) + R(x)u(x)v(x) dx.
The integrands of both integrals on smooth on the intervals of integration, and hence integration by parts is valid:
−P (x) dudx
(x)v(x)b
a+
b
aP (x) du
dx(x) dv
dx(x) + R(x)u(x)v(x) dx
= P (a) dudx
(a)v(a) −P (b) dudx
(b)v(b) + P (y) dudx
(y−)v(y) −P (a) dudx
(a)v(a) + P (b) dudx
(b)v(b) −P (y) dudx
(y+ )v(y)
+ y
a − ddx
P (x) dudx
(x) v(x) + R(x)u(x)v(x) dx + b
y − ddx
P (x) dudx
(x) v(x) + R(x)u(x)v(x) dx
= P (y) du
dx(y−)v(y)
−P (y) du
dx(y+ )v(y) +
b
a −
d
dxP (x) du
dx(x) + R(x)u(x) v(x) dx
= P (y) dudx
(y−)v(y) −P (y) dudx
(y+ )v(y),
where the last step follows because, on both ( a, y ) and ( y, b), u(x) = G(x; y) is a smooth solution of the homoge-neous ODE
−v1 ( x ) v 2 ( y )P ( x ) W ( x ) , x ≤y ≤1,where v1 (x) = ex sin(2 x), v2 (x) = ex (sin (2x) −(tan (2)) cos (2 x)), and P (x)W (x) is the constant 2 tan (2). Thesolution to the BVP is
u(x) = 1
0G(x; y)f (y)w(y) dy,
where w(x) = e−2x .
5. The Green’s function isG(x; y) = ey 6 + 6 x + 3 x2 + x3 , 1
2 ≤y ≤x,ex 6 + 6 y + 3 y2 + y3 , x ≤y ≤1.
The solution to the BVP when f (x) = x is
u(x) = 232
+ 232
x + 6 x2 + 2 x3 −7ex−1 .
7. Let G be the Green’s function derived in this section, and let M be dened by ( Mf )( x) = b
a G(x; y)f (y) dy. (HereM maps complex C [a, b] into complex C 2 [a, b].) We wish to prove that ( Mf,g )w = ( f ,Mg )w for all f , g ∈C [a, b].In the following calculation, we use the facts that G is symmetric ( G(x; y) = G(y; x) for all x, y ∈[a, b]) and G isreal-valued, so that G(x; y) = G(x; y). We have
(Mf,g )w = b
a(Mf )(x)g(x)w(x) dx =
b
a b
aG(x; y)f (y)w(y) dy g(x)w(x) dx
= b
a b
aG(x; y)f (y)w(y)g(x)w(x) dydx
= b
a b
aG(x; y)f (y)w(y)g(x)w(x) dxdy
= b
a f (y) b
a G(x; y)g(x)w(x) dx w(y) dy
= b
af (y)
b
aG(x; y)g(x)w(x) dx w(y) dy
= b
af (y)
b
aG(y; x)g(x)w(x) dx w(y) dy
= b
af (y)( Mg)( y)w(y) dy = ( f ,Mg )w .
(Notice how the key step was the interchanging of the order of integration.) This proves the desired result.
provided t 0 < s . We notice that G(t, s ) = H (t −s)S (t, s ), where H is the Heaviside function. We will show that
∂G∂t (t, s ) = H (t −s) ∂S ∂t (t, s ) and ∂ 2
G∂t 2 (t, s ) = δ (t −s) + H (t −s) ∂ 2
S ∂t 2 (t, s )
(in the weak sense). Let φ be any smooth function having compact support in [ t0 , ∞), that is, any function forwhich φ(t) = 0 for all t∈(a, b), where t0 < a < b < ∞. We rst show that
− ∞t 0
G(t; s) dφdt
(t ) dt = ∞t 0
H (t −s) ∂S ∂t
(t, s )φ(t) dt,
which implies that∂G∂t
(t, s ) = H (t −s) ∂S ∂t
(t, s )
in the weak sense. We have
− ∞t 0
G(t; s) dφdt
(t) dt = − ∞t 0
H (t −s)S (t, s ) dφdt
(t ) dt
= − b
sS (t, s ) dφ
dt (t ) dt
= − S (t, s )φ(t)|bs + b
s
∂S ∂t
(t, s )φ(t) dt (by integration by parts)
= b
s
∂S ∂t
(t, s )φ(t) dt (since S (s, s ) = 0 and φ(b) = 0)
= ∞t 0
H (t −s) ∂S ∂t
(t, s )φ(t) dt.
This proves the desired formula for ∂G/∂t .Next, we will show that
− ∞t 0
∂G
∂t (t ; s) dφ
dt (t) dt = φ(s) +
∞t 0
H (t
−s) ∂ 2 S
∂t 2 ( t, s )φ(t) dt,
which implies that∂ 2 G∂t 2 (t, s ) = δ (t −s) + H (t −s) ∂ 2 S
∂t 2 (t, s )
in the weak sense. We have
− ∞t 0
∂G∂t
(t ; s) dφdt
(t ) dt = − ∞t 0
H (t −s) ∂S ∂t
(t, s ) dφdt
(t) dt
= − b
s
∂S ∂t
(t, s ) dφdt
(t) dt
= − ∂S ∂t
(t, s )φ(t)b
s+
b
s
∂ 2 S ∂t 2 (t, s )φ(t) dt (by integration by parts)
= φ(s) + b
s
∂ 2 S ∂t 2 (t, s )φ(t) dt (since ∂S∂t (s, s ) = 1 and φ(b) = 0)
= φ(s) + ∞t 0
H (t −s) ∂ 2 S ∂t 2 ( t, s )φ(t) dt.
This proves the desired formula for ∂G2 /∂t 2 .Since S (t, s ), as a function of t , satises the homogeneous version of the ODE, we see that
Making the change of variables u = x/ (2√ kt ), we obtain
∞
−∞S (x, t ) dx = 1
√ π ∞
−∞e−u 2
du = 1√ π √ π = 1 .
Also, for any x, y ∈R ,
∞
−∞S (x −y, t ) dx = ∞
−∞S (v, t ) dv = 1 .
3. For t = 660, six terms of the Fourier series are sufficient to give an accurate graph. For t = 61, even 100 termsare not sufficient to give a correct graph.
5. Following the example in Section 9.5.2, we obtain the following Green’s function:
G(x, t ; y, s ) = 2
ρc ∞n =1 e−k (2 n −1) 2 π 2 ( t −s ) / (4 2 ) sin (2 n −1) πy2 sin (2 n −1) πx
2 , 0 ≤s ≤ t,0, s > t.
Here we used the correct eigenfunctions for the given mixed boundary conditions and also adjust for the the factthat the constants appear differently in the PDE in this exercise than in the example in Section 9.5.2 (notice theextra factor of 1 / (ρc) in the formula for this Green’s function). As before, k = κ/ (ρc). Snapshots of G(x, t ; 75, 60)are shown in Figure 9.1.
0 20 40 60 80 1000
0.005
0.01
0.015
0.02
x
t e m p e r a t u r e
Graph of G(x,t;75,60) for various values of t
t=120t=240t=360t=480t=600
Figure 9.1: Snapshots of the Green’s function in Exercise 9.5.5 at times t = 120 , 240, 360, 480, 600 seconds.Twenty terms of the Fourier series were used to create these graphs.
φ(y) dy, n = 1 , 2, 3, . . .are the Fourier sine coefficients of the function φ. We recognize u as the solution of the IBVP
∂u∂t −k ∂ 2 u
∂x 2 = 0 , 0 < x < , t > 0,
u(x, 0) = φ(x), 0 < x < ,u(0, t ) = 0 , t > 0,u( , t) = 0 , t > 0
(see the derivation at the beginning of Section 6.1 in the text).
9.6 Green’s functions for the wave equation1. Let
u(x, t ) = 12c
t
0 x + c( t −s )
x −c( t −s )f (y, s ) dy ds.
We wish to show that u satises the inhomogeneous wave equation (with right-hand side f (x, t )) on the real line,subject to zero boundary conditions. We have
∂u∂t
(x, t ) = 12c
x+ c( t −t )
x −c( t −t )f (y, t ) dy + 1
2c t
0 cf (x + c(t −s), s ) + cf (x −c(t −s), s ) ds
= 12c
x
xf (y, t ) dy + 1
2c t
0 cf (x + c(t −s), s ) + cf (x −c(t −s), s ) ds
= 12c t
0 cf (x + c(t −s), s ) + cf (x −c(t −s), s ) ds,
∂ 2 u∂t 2 (x, t ) = 1
2c (cf (x + c(t −t), t ) + cf (x −c(t −t), t )) + 1
2c t
0c2 df
dx(x + c(t −s), s ) −c2 df
dx(x −c(t −s), s ) ds
= 12c
(cf (x, t ) + cf (x, t )) + 12c
t
0c2 df
dx(x + c(t −s), s ) −c2 df
dx(x −c(t −s), s ) ds
= f (x, t ) + c2
t
0
df dx
(x + c(t −s), s ) − df dx
(x −c(t −s), s ) ds,
∂u∂x
(x, t ) = 12c
t
0 f (x + c(t −s), s ) −f (x −c(t −s), s ) ds,
∂ 2 u∂x 2 (x, t ) = 1
2c t
0
df dx
(x + c(t −s), s ) − df dx
(x −c(t −s), s ) ds.
Thus
∂ 2 u∂t 2 (x, t ) −c2 ∂ 2 u
∂x 2 (x, t ) = f (x, t ) + c2
t
0
df dx
(x + c(t −s), s ) − df dx
(x −c(t −s), s ) ds
− c2
t
0
df dx
(x + c(t −s), s ) − df dx
(x −c(t −s), s ) ds
= f (x, t ),
as desired. It is obvious from the formulas for u and ∂u/∂t that both are identically zero when t = 0.
for all x,y,t, s ∈ R such that c(t −s) = |x −y|. (The two expressions differ when c(t −s) = |x −y| > 0, at leastwhen H is dened as in the text: H (t) = 1 if t > 0 and H (t) = 0 if t
≤0.) We have
H (c(t −s) − |x −y|) = 1 ⇔c(t −s) − |x −y| > 0
⇔c(t −s) > |x −y|⇔−c(t −s) < |x −y| < c (t −s) and t −s > 0
⇔x −y + c(t −s) > 0 and x −y −c(t −s) < 0 and t −s > 0
⇒H (x −y + c(t −s)) = 1 and H (x −y −c(t −s)) = 0 and H (t −s) = 1
⇔(H (( x −y) + c(t −s)) −H (( x −y) −c(t −s))) H (t −s) = 1 .
The reasoning can be reversed if we add the condition c(t −s) = |x −y|. This completes the proof.
5. Let u be the solution to
∂ 2 u
∂t 2 −c2 ∂ 2 u
∂x 2 = f (x, t ),
−∞< x <
∞, t > 0,
u(x, 0) = 0 , −∞< x < ∞,∂u∂t
(x, 0) = 0 , −∞< x < ∞,
where f is odd ( f (−x, t ) = −f (x, t )). We wish to show that u is also odd. We will do this by dening v(x, t ) =
−u(−x, t ) and proving that v satises the same IVP; then, by uniqueness, it will follow that v = u, that is,u(x, t ) = −u(−x, t ), as desired.We have
∂v∂t
(x, t ) = −∂u∂t
(−x, t ),
∂ 2 v
∂t2 (x, t ) =
−
∂ 2 u
∂t2 (
−x, t ),
∂v∂x
(x, t ) = ∂u∂x
(−x, t ),
∂ 2 v∂x 2 (x, t ) = −
∂ 2 u∂x 2 (−x, t ).
Therefore,
∂ 2 v∂t 2 (x, t ) −c2 ∂ 2 v
∂x 2 (x, t ) = −∂ 2 u∂t 2 (−x, t ) + c2 ∂ 2 u
∂x 2 (−x, t )
= −∂ 2 u∂t 2 (−x, t ) −c2 ∂ 2 u
∂x 2 (−x, t )
= −f (−x, t ) = f (x, t )
(since f is odd). Thus v satises the same wave equation as does u . Moreover, it is obvious that v also satisesthe same initial conditions; hence v = u and the proof is complete.
7. We outline the derivation of the Green’s function for
We dene f to be the even, periodic extension of f , and dene u to be the solution of
∂ 2
u∂t 2 −c2 ∂ 2
u∂x 2 = f (x, t ), −∞< x < ∞, t > 0,
u(x, 0) = 0 , −∞< x < ∞,∂ u∂t
(x, 0) = 0 , −∞< x < ∞.
From earlier results, we know that
u(x, t ) = t
0 ∞
−∞
12c
H (c(t −s) − |x −y|) f (y, s ) dyds = 12c
t
0 x + c( t −s )
x−c( t −s )f (y, s ) dy ds.
Then∂ u∂x
(x, t ) = t
0f (x + c(t −s), s ) − f (x −c(t −s), s ) ds.
It follows that
∂ u∂x
(x, 0) = t
0f (c(t −s), s ) − f (−c(t −s), s ) ds,
∂ u∂x
(x, ) = t
0f ( + c(t −s), s ) − f ( −c(t −s), s ) ds.
Since f is even, we have f (c(t −s), s ) = f (−c(t −s), s ), and hence
∂ u∂x
(x, 0) = 0 .
Also,
f ( −c(t −s), s ) = f (− + c(t −s), s ) (since f is even)
= ˜f ( + c(t −s), s ) (since
˜f is 2 -periodic) .
Therefore,∂ u∂x
(x, ) = 0 .
It follows that ˜u, restricted to the interval 0 < x < , solves the original IBVP.The derivation of the Green’s function for the IBVP is almost exactly the same as that given in the text (forDirichlet conditions). The result is
G(x, t ; y, s ) =∞
n = −∞
12c H (c(t −s) − |x −y −2n |) + H (c(t −s) − |x + y −2n |).
10.2 Properties of the Sturm-Liouville operator1. The verication that L is symmetric is exactly the same as that given (for the general case) on pages 389–390,
except for the treatment of the boundary term. We have
−P (x) dudx
(x)v(x)b
a= −P (b) du
du(b)v(b) + P (a) du
dx(a)v(a)
= −P (b) dudu
(b)v(b) (since v(a) = 0)
= β 1β 2
P (b)u(b)v(b) since dudx
(b) = −β 1β 2
u(b) .
Therefore,
(Lu,v )w = β 1β 2 P (b)u(b)v(b) + b
aP (x) dudx (x) dvdx (x) dx +
b
aR(x)u(x)v(x) dx.
The expression on the right is symmetric in u and v, which shows that
(u, L (v)) w = ( L(v), u )w = ( L(u), v)w ,
verifying that L is symmetric with respect to ( ·, ·)w .
3. Consider a Robin condition of the formα 1 u(a) + κ du
dx(a) = 0
(relative to the interval [ a, b]). We wish to determine the physically meaningful sign for α 1 in the case that thisboundary condition models heat ow through the left end of a bar. Recall that the heat ux at x = a (a vectorquantity) is given by
−κ du
dx(a)
(the direction of the vector is indicated by the sign). Since the normal direction to the left end of the bar is −1,the rate at which heat ows out of the bar at x = a is
−κ dudx
(a)(−1) = κ dudx
(a).
Thus consider the equation
κ dudx
(a) = γ (u(a) −T ),
where T is the temperature of the surroundings. This equation states that the rate at which heat ows out of thebar is proportional to the difference between the temperature at the left end of the bar and the temperature of the bar. The above boundary condition is equivalent to
which shows that we should take α1 = −γ in the Robin boundary condition. Thus α1 < 0 is the physicallymeaningful case for a Robin boundary condition at the left endpoint.
At the right endpoint x = b, the normal direction is 1, and therefore the rate at which heat ows out of the bar is
κ dudx
(b).
The same reasoning as above then shows that the boundary condition is
γu (b) + κ dudx
(b) = γT.
Therefore, we should take β 1 = γ , which shows that β 1 > 0 is the physically meaningful case.5. The eigenpairs are
λ n = 3 + nπln 2
2, un (x) = sin nπ ln x
ln 2, n = 1 , 2, 3, . . . .
7. (a) The eigenpairs are
λn = 4 + β + nπln 2
2, un (x) = x2 sin nπ ln x
ln 2, n = 1 , 2, 3, . . . .
(b) 0 is an eigenvalue if and only if
β = −4 − nπln 2
2
for some positive integer n .(c) There is exactly one negative eigenvalue if and only if
−4 − 4π 2
(ln2) 2 ≤β < −4 − π2
(ln2) 2 ,
while there are exactly k negative eigenvalues if and only if
−4 − (k + 1) 2 π 2
(ln2) 2 ≤β < −4 − k2 π 2
(ln2) 2 .
9. The eigenpairs are
λn = 2 + nπln 2
2, un (x) = x−1 nπ
ln 2 cos nπ ln x
ln 2+ sin nπ ln x
ln 2, n = 1 , 2, 3, . . . .
10.3 Numerical Methods for Sturm-Liouville problems1. We wish to apply the nite element method to
− ddx
P (x) dudx
+ R(x)u = λw (x)u, a < x < b,
u(a) = 0 ,
dudx
(b) = 0 .
Using the usual uniform mesh on the interval [ a, b], the nite element space of consists of all piecewise linearfunctions v (relative to the given mesh) satisfying v(a) = 0 (note that the Neumann condition is a naturalboundary condition). The standard basis for this space is φ1 , . . . , φ n (where each φj is the usual “hat” function).Applying the Galerkin method to the weak form (Equation (10.12) in the text), we obtain Au = λWu , where Aand W are the n ×n matrices dened by
Aij = b
aP (x) dφj
dx (x) dφ i
dx (x) + R(x)φj (x)φi (x) dx,
W ij = b
aw(x)φj (x)φi (x) dx.
(These are the same formulas derived in the text, except now i and j vary from 1 to n .)
3. The result is exactly the same as in Exercise 1, but now the basis for the nite element space is φ0 , φ1 , . . . , φ n ,the matrices A and W are (n + 1)
×(n + 1), and i, j vary from 0 to n in the formulas for A ij , W ij .
5. Let λ ∈ C and x ∈ C n satisfy Ax = λWx , and let ( x , y )W = ( Wx ) ·y . We can assume x satises ( x , x )W = 1.We then obtain
λ = λ(x , x )W = ( λWx ) ·x = ( Ax ) ·x = x ·(Ax ) = x ·(λWx ) = λ x ·(Wx ) = λ (( Wx ) ·x ) = λ.
(Notice how the symmetry of both A and W is used to move these matrices from one side of the inner productsto the other.) Since λ = λ , it follows that λ∈R .
10.4 Examples of Sturm-Liouville problems1. Let
u(x) =n
i =0
α i φi (x)
be a continuous piecewise linear function dened on [0 , ], relative to the usual uniform mesh. Here φ0 , φ1 , . . . , φ n is the standard basis. The average value of u on [0, ] is then
1
0u(x) dx = 1
0
n
i =0
α i φi (x) dx = 1 n
i =0
α i
0φi (x) dx.
Since the graph of each φi , i = 1 , . . . , n −1, forms a triangle of height 1 and base 2 h (h = /n ), it follows that
0φi (x) dx = h, i = 1 , 2, . . . , n −1;
similarly,
0φ0 (x) dx =
0φn (x) dx = h
2, i = 1 , 2, . . . , n −1.
Thus the average value of u on [0, ] is
h 12
α 0 + α 1 + · · ·+ α n −1 + 12
α n .
3. The fundamental frequency is about 3254Hz.
10.5 Robin boundary conditions1. The solution to the IBVP is
Figure 10.1: The temperature in the bar of Exercise 10.5.1 after 30 , 60, 90, 120, 150, 180 minutes.
Case 1 (λ > 0) The positive eigenvalues are the solutions of the equation
tan( √ λ ) = √ λαor
tan( s) = sα
, s > 0,
where s = √ λ . There are solutions sk.= (2 k + 1) π/ 2, k = 1 , 2, 3, . . . . The corresponding eigenpairs are
λ k = s2k2 , ψk (x) = sin sk x , k = 1 , 2, 3, . . . .
These eigenpairs exist regardless of the value of α . In addition, if α < 1, then there is an additional solutionof tan( s) = s/ (α ), namely, s0 ∈ (0, π/ 2), with corresponding eigenvalue λ0 = s2
0 / 2 and eigenfunctionψ0 (x) = sin( s0 x/ ).
Case 2 (λ = 0) If α = 1, then λ0 = 0 is an eigenvalue with eigenfunction ψ0 (x) = x. If α = 1, then 0 is not aneigenvalue.
Case 3 (λ < 0) Negative eigenvalues must satisfy the equation
tanh( √ −λ ) =√ −λ
αor
tanh( s) = sα
, s > 0,
where s = √ −λ . If α ≤1, then this equation has no solution and hence there are no negative eigenvalues.If α > 1, then there is a single solution s0 , leading to a single negative eigenvalue λ0 = −s2
0 / 2 andeigenfunction ψ0 (x) = sinh ( s0 x/ ).
Therefore, when α < 0, there is a sequence of eigenpairs
λ k = s2k2 , ψk (x) = sin sk x , k = 1 , 2, 3, . . . ,
where
λ k .= (2k + 1) 2 π 2
4 2 .There is also an eigenvalue λ 0 , which is positive if α < 1 (with eigenfunction ψ0 (x) = sin( √ λ0 x)), zero if α = 1(with eigenfunction ψ0 (x) = x), and negative if α > 1 (with eigenfunction ψ0 (x) = sinh ( √ −λ 0 x)).
10.6 Finite element methods for Robin boundary conditions1. The nite element formulation of
10.6. FINITE ELEMENT METHODS FOR ROBIN BOUNDARY CONDITIONS 87
where P (x) > 0 for all x ∈ [a, b], α,β > 0, is nearly the same as for the BVP (10.26) given in the text. TheDirichlet condition at x = a is an essential boundary condition (whereas the Neumann condition in (10.26) is a
natural boundary condition). This implies that we only consider nite element functions satisfying the Dirichletcondition, so we use the basis φ1 , φ2 , . . . , φ n instead of φ0 , φ1 , . . . , φ n . The weak form of the BVP is unchangedfrom (10.27), although now
P (a) dudx
(a)v(a)
vanishes because v(a) = 0 rather than because du/dx (a) = 0. The result is the linear system Au = f , where Aand f are given by the same formulas as in the text (page 420), except that the entries corresponding to φ0 areomitted. Thus A is now n ×n and f is now an n -vector.
3. The weak form of
− ddx
P (x) dudx
+ R(x)u = F (x), a < x < b,
−α 1 u(a) + α 2
du
dx(a) = 0 ,
β 1 u(b) + β 2dudx
(b) = 0
is
P (a) dudx
(a)v(a) −P (b) dudx
(b)v(b) + b
aP (x) du
dx(x) dv
dx(x) dx +
b
aR(x)u(x)v(x) dx =
b
aF (x)v(x) dx
for all test functions v. The boundary conditions imply
dudx
(a) = α1
α 2u(a), du
dx(b) = −
β 1β 2
u(b),
and therefore the weak form can be written as
α 1α 2
P (a)u(a)v(a) + β 1β 2 P (b)u(b)v(b) + b
aP (x) dudx (x) dvdx (x) dx +
b
aR(x)u(x)v(x) dx =
b
aF (x)v(x) dx.
Notice that the Robin conditions at the endpoints are natural boundary conditions, and hence no boundaryconditions are imposed on the test functions. We therefore take φ0 , φ1 , . . . , φ n as the basis for the nite elementspace (using the usual notation), and obtain the linear system Au = f , where A = K + M + G . The matrices Kand M are the same stiffness and mass matrices dened in the text (page 420), and G ∈R ( n +1) ×( n +1) is denedby
10.7 The theory of Sturm-Liouville problems: an outline
1. Let L : U → V be a compact linear operator, where U , V are normed linear spaces. We wish to show that L isbounded. We will argue by contradiction and assume L is unbounded. Then there exists a sequence un ⊂ U with un U = 1 for all n and Lu n V → ∞ as n → ∞. (Here we are using the alternate denition of the normof L: L = sup Lu V : u ∈ U, u U = 1.) Since L is compact and un is bounded in U , there must exist asubsequence un k of un such that Lu n k is convergent in V . But Lu n V → ∞ implies that Lu n k V → ∞,and hence that Lu n k is not convergent. This is a contradiction, which shows that the assumption that L isunbounded is not possible. Therefore, L must be bounded, which is what we wanted to prove.
3. Suppose that, for a given λ∈R , u = φ, u = ψ both satisfy
− ddx
P (x) dudx
+ R(x)u = λw (x)u, a < x < b,
α 1 u(a) + α 2dudx
(a) = 0 ,
β 1 u(b) + β 2dudx (b) = 0 ,
where P and w are assumed to be positive on the interval [ a, b]. Let
W (x) = φ(x) ψ(x)dφdx (x) dψ
dx (x) = φ(x) dψdx
(x) − dφdx
(x)ψ(x)
be the Wronskian of φ, ψ. Since φ, ψ satisfy the linear homogeneous ODE
− ddx
P (x) dudx
+ ( R(x) −λw (x)u = 0
on [a, b], we can show that φ, ψ is linearly independent by proving that W (x) = 0 for any given x ∈ [a, b]. Wehave
α 1 φ(a) + α 2dφ
dx
(a) = 0
⇒
dφ
dx
(a) =
−
α 1
α 2
φ(a),
and similarlydψdx
(a) = −α 1
α 2ψ(a).
Therefore,
W (a) = φ(a) dψdx
(a) − dφdx
(a)ψ(a) = −α 1
α 2φ(a)ψ(a) + α1
α 2φ(a)ψ(a) = 0 .
This shows that φ, ψ is linearly dependent, as desired.
90 CHAPTER 11. PROBLEMS IN MULTIPLE SPATIAL DIMENSIONS
By the divergence theorem,
Ω∇ ·
(k(
∇
u)v) =
∂ Ωk(
∇
u)v)
·n =
∂ Ωk ∂u
∂ nv.
The boundary term disappears if u and v both satisfy Dirichlet condition (since then v = 0 on ∂ Ω) and also if uand v both satisfy Neumann conditions (since then ∂u/∂ n = 0 on ∂ Ω). Thus, in either case,
(Lu,v ) = − Ω∇ ·(k∇u)v = Ωk∇u ·∇v.
Since the right-hand side is symmetric in u and v, so is the left-hand side; that is, ( Lu,v ) = ( Lv,u ) = ( u,Lv ),and we see that L is a symmetric operator.
7. We have
∇ ·F (u(x )) = ∂ ∂x 1
(F 1 (u(x ))) + ∂ ∂x 2
(F 2 (u(x ))) + ∂ ∂x 3
(F 3 (u(x )))
= F 1 (u(x
))
∂u
∂x 1 (x
) + F 2 (u(x
))
∂ u
∂x 2 (x
) + F 3 (u(x
))
∂u
∂x 3 (x
)= F (u(x )) ·∇u(x ).
9. Using the product rule for scalar functions, we obtain
∇ ·(v∇u) =3
i =1
∂ ∂x i
v ∂u∂x i
=3
i =1
∂v∂x i
∂u∂x i
+ v ∂ 2 u∂x 2
i =
3
i =1
∂v∂x i
∂u∂x i
+ v3
i =1
∂ 2 u∂x 2
i
= ∇v ·∇u + v∆ u.
11. Suppose u, v ∈C 2m (Ω). Then
(Lm u, v ) = − Ωv∆ u = Ω∇
u ·∇v − ∂ Ωv ∂u
∂ n (Green’s rst identity)
= Ω∇u ·∇v
= ∂ Ωu ∂v
∂ n − Ωu∆ v (Green’s rst identity)
= − Ωu∆ v = ( u, L m v).
The boundary terms vanish because the product that forms the integrand is zero over the entire boundary. Forexample,
∂ Ωv ∂u
∂ n = 0
since v = 0 on Γ 1 and ∂u/∂ n = 0 on Γ 2 .
11.2 Fourier series on a rectangular domain1. (a)
cmn = −4 (5 + 7( −1)m ) (−1 + ( −1)n )
m 3 n 3 π 6
(b) The graphs of the error are given in Figure 11.1.
3. The solution is given by
u(x ) =∞
m =1
∞
n =1
cmn
λmnsin( mπx 1 )sin( nπx 2 ),
where the cmn are the coefficients from Exercise 1 and
Figure 11.1: The error in approximating f (x, y) by the rst 4 terms (top) and the rst 25 terms (bottom) of
the double Fourier sine series. (See Exercise 11.2.1.)
5. (a) The IBVP is
ρc ∂u∂t −κ∆ u = 0 .02, x
∈Ω, t > 0,
u(x , 0) = 5 , x∈Ω,
u(x , t ) = 0 , x∈∂ Ω, t > 0.
The domain Ω is the rectangle x∈
R 2 : 0 < x 1 < 50, 0 < x 2 < 50 .
(b) The solution is
u(x , t ) =∞
m =1
∞
n =1
amn (t)sin mπx 1
50sin nπx 2
50,
where
amn (t) = bmn − cmn
κλ mne−κλ mn t/ ( ρc ) + cmn
κλ mn,
bmn = 20 (−1 + ( −1)m ) (−1 + ( −1)n )mnπ 2 ,
cmn = 2 (−1 + ( −1)m ) (−1 + ( −1)n )25mnπ 2 ,
λ mn = (m 2 + n 2 )π 2
2500 .
(c) The steady-state temperature u s (x ) satises the BVP
−κ∆ u = 0 .02, x∈Ω,
u(x ) = 0 , x∈∂ Ω.
The solution is
u s (x ) =∞
m =1
∞
n =1
cmn
κλ mnsin mπx 1
50sin nπx 2
50.
(d) The maximum difference between the temperature after 10 minutes and the steady-state temperature isabout 1 degree. The difference is graphed in Figure 11.2.
7. The minimum temperature in the plate reaches 4 degrees Celsius after 825 seconds.
9. The leading edge of the wave is initially 2/5 units from the boundary, and the wave travels at a speed of 261 √ 2units per second. Therefore, the wave reaches the boundary after √ 2/ 1305 .= 0 .00108 seconds. Figure 11.6 fromthe text shows the wave about to reach the boundary after 10 −3 seconds.
96 CHAPTER 11. PROBLEMS IN MULTIPLE SPATIAL DIMENSIONS
However, since φ(r, θ ) = r (1 −r )cos(θ)/ 5 (a function of r times cos ( θ)), all of the coefficients of φ are zero exceptfor cm 1 , m = 1 , 2, 3, . . . . Therefore,
u(r,θ,t ) =∞
m =1
am 1 (t)cos(θ)J 1 (α m 1 r ),
with am 1 (t) given above. After 30 seconds, the temperature distribution can be approximated accurately using asingle eigenfunction (corresponding to the largest eigenvalue, λ 11 ), so we need only
c11.= 9 .04433.
The temperature distribution after 30 seconds is graphed in Figure 11.8.
−10−5
05
10
−10
−5
0
5
10−0.04
−0.02
0
0.02
0.04
x1
x2
Figure 11.8: The approximation to solution u at t = 30, computed with one term of the (generalized) Fourierseries (see Exercise 11.3.7).
9. A circular drum of radius A has area πA2 , the same area as a square drum of side length √ πA . The fundamentalfrequency of such a square drum is
c π 2
(√ πA ) 2 + π 2(√ πA ) 2
2π = c
A1
√ 2π.= 0 .398942 c
A.
Comparing to Example 11.9, we see that the circular drum sounds a lower frequency than a square drum of equalarea.
11. s41.= 7 .588342434503804, s 42
.= 11 .06470948850118 , s 43.= 14 .37253667161759, s 44
.= 17 .61596604980483
11.4 Finite elements in two dimensions1. The mesh for this problem is shown in Figure 11.9, in which the free nodes are labeled. The stiffness matrix is
3. The mesh for this problem is shown in Figure 11.10, in which the free nodes are labeled. The stiffness matrix is
16 ×16 and neither it nor the load vector will be reproduced here. The matrix is singular, as is expected for aNeumann problem. The unique solution to KU = F with its last component equal to zero is
u .=
−0.071138
−0.047315
−0.0126560.001458
−0.067934
−0.047349
−0.0141450.001566
−0.065432
−0.046198
−0.015160
−0.000329
−0.062914
−0.046391
−0.0160650.000000
.
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
1
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
x1
x 2
Figure 11.10: The mesh for Exercise 11.4.3.
5. Let V n be the space of continuous piecewise linear functions relative to a given triangulation of the domain Ω, andlet φ1 , . . . , φ n be the standard basis for V n . Suppose that u : Ω→R belongs to L2 (Ω). We wish to nd v ∈V nthat minimizes w −u L 2 (Ω) over all w ∈ V n . We can write v =
ni =1 α i φi and use the projection theorem: v
must satisfy ( u
−v, w)L 2 (Ω) = 0 for all w
∈ V n , which is equivalent to ( u
−v, φ i )L 2 (Ω) = 0 for all i = 1 , 2, . . . , n .
98 CHAPTER 11. PROBLEMS IN MULTIPLE SPATIAL DIMENSIONS
We obtain the following equations:
(v, φ i ) = ( u, φ i ), i = 1 , 2, . . . , n
⇒
n
j =1
α j φj , φi = ( u, φ i ), i = 1 , 2, . . . , n
⇒
n
j =1
α j (φj , φi ) = ( u, φ i ), i = 1 , 2, . . . , n .
This last system of equations is equivalent to M α = f , where M is the usual mass matrix,
M ij = Ωφj φi , i , j = 1 , 2, . . . , n ,
and f is the vector dened by
f i =
Ω
vφ i , i = 1 , 2, . . . , n .
7. (a) A direct calculation shows that the inhomogeneous Dirichlet problem
−∇ ·(k(x )∇u) = f (x ), x∈Ω,
u(x ) = g(x ), x∈∂ Ω,
has weak formu∈G + V, a(u, v ) = ( f, v ) for all v ∈V = C 2D (Ω),
where G is any function satisfying the condition G(x ) = g(x ) for x ∈∂ Ω, and
G + V = G + v : v ∈V .
Substituting u = G + w, where w ∈V , into the weak form yields
w
∈V, a(w, v ) = ( f, v )
−a(G, v ) for all v
∈V.
In the Galerkin method, we replace V by a nite-dimensional subspace V n and solve
w ∈V n , a(w, v ) = ( f, v ) −a(G, v ) for all v ∈V n .
When using piecewise linear nite elements, we can satisfy the boundary conditions approximately by takingG to be a continuous piecewise linear function whose values at the boundary nodes agree with the givenboundary function g. For simplicity, we take G to be zero at the interior nodes. The resulting load vectoris then given by
f i = ( f, φ i ) −a(G, φ i ), i = 1 , 2, 3, . . . , n .
Since G is zero on interior nodes, the quantity a (G, φ i ) is nonzero only if (free) node i belongs to a triangleadjacent to the boundary.
(b) The regular triangulation of the unit square having 18 triangles has only four interior (free) nodes, and eachone belongs to triangles adjacent to the boundary. This means that every entry in the load vector is modied(which is not the typical case). Since f = 0, the load vector f is dened by
Multiplying the PDE by a test function v ∈ V = C 2 (Ω) and applying Green’s rst identity yields
Ωk∇u ·∇v = Ω
fv + ∂ Ωkvh for all v ∈ V .
We will apply Galerkin’s method with V m equal to the space of all continuous piecewise linear functionrelative to a given triangulation T . We label the standard basis of V m as φ1 , φ2 , . . . , φ n , and the nodes asz 1 , z 2 , . . . , z n . (Every node in the mesh is now free.) Thus V n = span φ1 , φ2 , . . . , φ n .We then dene the Galerkin approximation wn by
wn ∈ V n , a(wn , φ i ) = ( f, φ i ) +
∂ Ω
khφ i , i = 1 , 2, . . . , m .
Writing wn = mj =1 β j φj and
w =
β 1β 2...
β n
,
we obtain the system Kw = F . The stiffness matrix K∈
R m ×m is given by
K ij = a(φj , φi ), i , j = 1 , 2, . . . , n ,
and the load vector F∈
R m by
F i = ( f, φ i ) + ∂ Ωkhφ i , i = 1 , 2, . . . , n .
The boundary integral in the expression for F i is zero unless z i is a boundary node.(b) The compatibility condition for (11.1) is determined by the following calculation:
Ωf = − Ω∇ ·(k∇u) = − ∂ Ω
k ∂u∂ n
= − ∂ Ωkh.
(c) Now consider (11.1) with
k(x ) = 1 , f (x ) = x1 x2 + 34
, h(x ) = −3x2
1
5 ,
where Ω is the unit square. We will produce the nite element solution using the regular grid with 18triangles (16 nodes). The stiffness matrix K ∈ R 16 ×16 is singular, as should be expected, and there areinnitely many solutions. The unique solution u with last component equal to zero is
11.5. THE FREE-SPACE GREEN’S FUNCTION FOR THE LAPLACIAN 101
It follows that
R 2
∇
u(x )
·∇v(x ) dx =
2π
0 ∞0
∂u
∂r
∂v
∂r + 1
r 2
∂u
∂θ
∂v
∂θv rdrdθ,
as desired.
(b) We can derive the same result by starting with the PDE −∆ u = δ (x −y ) in polar coordinates, multiplyingby a test function, and integrating. The right-hand side becomes
R 2δ (x −y )v(x ) dx = v(y ).
On the left side, we have
2π
0 ∞0 −
∂ 2 u∂r 2 −
1r
∂u∂r −
1r 2
∂ 2 u∂θ 2 vrdrdθ
= − 2π
0 ∞0
∂ 2 u∂r 2 vrdr dθ −
2π
0 ∞0
∂u∂r vdrdθ −
∞0
1r 2
2π
0
∂ 2 u∂θ 2 v dθ r dr
= 2π
0 −r ∂u∂r
v∞0
+ ∞0
∂u∂r
∂v∂r
r dr + ∞0
∂u∂r
v dr dθ − 2π
0 ∞0
∂u∂r
vdrdθ −
∞0
1r 2
∂u∂θ
v2π
0 − 2π
0
∂u∂θ
∂v∂θ
dθ r dr
= 2π
0
∂u∂r
∂v∂r
rdrdθ + 2π
0 ∞0
1r 2
∂u∂θ
∂v∂θ
rdrdθ
= 2π
0
∂u∂r
∂v∂r
+ 1r 2
∂u∂θ
∂v∂θ
rdrdθ.
Once again, we have derived the desired result.7. The derivation is long and tedious. As of the time of this writing, it can be found at
102 CHAPTER 11. PROBLEMS IN MULTIPLE SPATIAL DIMENSIONS
11.6 The Green’s function for the Laplacian on a bounded domain
1. Suppose u satises
−∆ u = f (x ) in Ω,
α 1 u + α 2∂u∂ n
= 0 on ∂ Ω,
and let v be any test function in C 2 (Ω). Then
− Ω(∆ u)v = Ω
fv ⇒ − ∂ Ω
∂u∂n
v + Ω∇u ·∇v = Ω
fv.
The boundary condition yields
−∂u∂n
= α1
α 2u on ∂ Ω,
and hence we obtainα 1
α 2
∂ Ωuv +
Ω∇
u
·∇v =
Ωfv,
as desired.3. Let G be the free-space Green’s function for the Laplacian and dene
w(x ; y ) = α 1 G(x ; y ) + α ∂G∂ n x
(x ; y )
for all x∈∂ Ω and y ∈Ω. Assume that u(x ) = vΩ (x ; y ) satises
−∆ u = 0 in Ω
α 1 u + α 2∂u∂n
= w(x ; y ) on ∂ Ω,
wherew(x ; y ) = α 1 G(x ; y ) + α 2
∂G∂ n x
(x ; y ).
Deneu2 (x ) = Ω
vΩ (x ; y )f (y ) dy .
Then, for all v ∈C 2 (Ω),
− Ω(∆ x vΩ v = 0
⇒ − ∂ Ω
∂vΩ
∂ n x(x ; y )v(x ) dσx + Ω∇
x vΩ (x ; y ) ·∇v(x ) dx = 0
⇒ α1
α 2 ∂ ΩvΩ (x ; y )v(x ) dσx −
1α 2 ∂ Ω
w(x ; y )v(x ) dσx + Ω∇x vΩ (x ; y ) ·∇v(x ) dx = 0
⇒ α1
α 2 ∂ ΩvΩ (x ; y )v(x ) dσx + Ω∇
x vΩ (x ; y ) ·∇v(x ) dx = 1α 2 ∂ Ω
w(x ; y )v(x ) dσx .
Notice how we used the boundary condition satised by vΩ to substitute for ∂vΩ /∂ n x . We now multiply bothsides of this last equation by f (y ) and integrate over Ω to obtain
α 1
α 2 Ω ∂ ΩvΩ (x ; y )v(x ) dσx f (y ) dy + Ω Ω∇
x vΩ (x ; y ) ·∇v(x ) dx f (y ) dy
= 1α 2 Ω ∂ Ω
w(x ; y )v(x ) dσx f (y ) dy .
Changing the order of integration in the two integrals on the left yields
11.6. THE GREEN’S FUNCTION FOR THE LAPLACIAN ON A BOUNDED DOMAIN 105
Let v(x ) = GΩ (x ; y ), where GΩ is the Green’s function for the BVP
−∆ u = f (x ) in Ω,u = 0 on ∂ Ω.
Finally, dene Ω = x ∈ Ω : x −y > , where y ∈ Ω is given and > 0 is small enough that B (y ) ⊂ Ω.Notice that ∆ u = 0 in Ω, ∆ v = 0 in Ω , and v(x ) = 0 for all x ∈ ∂ Ω. Applying Green’s second identity to u andv yields
Ω(v∆ u −u∆ v) = ∂ Ω
v ∂u∂ n −u ∂v
∂ n dσx
⇒ 0 = ∂ Ωv(x ) ∂u
∂ n(x ) −u(x ) ∂v
∂ n(x ) dσx + S ( y )
v(x ) ∂u∂ n
(x ) −u(x ) ∂v∂ n
(x ) dσx
⇒ 0 = ∂ Ω0 ∂u
∂ n(x ) −φ(x ) ∂G Ω
∂ n x(x ; y ) dσx + S ( y )
GΩ (x ; y ) ∂u∂ n
(x ) −u(x ) ∂G Ω
∂ n x(x ; y ) dσx
⇒ ∂ Ωφ(x ) ∂G Ω
∂ n x(x ; y ) dσx = S ( y )
GΩ (x ; y ) ∂u∂ n
(x ) dσx − S ( y )u(x ) ∂G Ω
∂ n x(x ; y ) dσx .
Now, recall that GΩ = G −vΩ , where G is the free-space Green’s function for the Laplacian and vΩ (x ; y ) satises
−∆ x vΩ = 0 in Ω,vΩ (x ; y ) = G(x ; y ) on ∂ Ω.
We therefore have
∂ Ωφ(x ) ∂G Ω
∂ n x(x ; y ) dσx = S ( y )
G(x ; y ) ∂u∂ n
(x ) dσx − S ( y )u(x ) ∂ G
∂ n x(x ; y ) dσx −
S ( y )vΩ (x ; y ) ∂u
∂ n(x ) dσx + S ( y )
u(x ) ∂vΩ
∂ n x(x ; y ) dσx .
The integrands for the last two integrals are continuous, with no singularities as → 0+ . Therefore, these twointegrals tend to zero as →0+ . We have
G(x ; y ) = − 12π
ln ( x −y ) ,
and therefore G(x ; y ) is constant for x ∈S (y ). It follows that
S ( y )G(x ; y ) ∂u
∂ n(x ) dσx
is just a multiple of
S ( y )
∂u∂ n
(x ) dσx = B ( y )∆ u(x ) dx = 0
(applying the divergence theorem and using the fact that ∆ u = 0 in Ω). It remains only to calculate
S ( y )u(x ) ∂ G
∂ n x(x ; y ) dσx .
Now,
∇x G(x ; y ) = − x −y
2π x −y 2
and, on S (y ),n x = −
x −yx −y
(notice that this is the outward-pointing normal to Ω ). Therefore,
106 CHAPTER 11. PROBLEMS IN MULTIPLE SPATIAL DIMENSIONS
Thus
S ( y )
u(x ) ∂G
∂ n x
(x ; y ) dσx = 1
2π S ( y )
u(x ) dσx ,
which is the average value of u on S (y ), and therefore
S ( y )u(x ) ∂ G
∂ n x(x ; y ) dσx →u(y )
as →0+ . Therefore, it nally follows, by taking the limit as →0+ in Green’s second identity, that
∂ Ωφ(x ) ∂G Ω
∂ n x(x ; y ) dσx = −u(y ),
which, given the symmetry of GΩ , yields the desired result.
11. In the text it is shown that, if u solves
−∆ u = 0 in B R (0) ,
u = φ(x ) on S R (0) ,
thenu(x ) = S R (0)
R 2 − x 2
2πR x −y 2 φ(y ) dσy .
Now suppose that u is to represented as a function of polar coordinates ( r, θ ). Let the circle S R (0) be representedby y = ( R cos(ψ), R sin( ψ)), 0 ≤ψ ≤2π . Then
dσy = ∂y1
∂ψ
2
+ ∂y2
∂ψ
2
dψ = R 2 cos2 (ψ) + R 2 sin2 (ψ) dψ = Rdψ.
With x represented by the polar coordinates ( r, θ ), we have x 2 = r 2 and
x
−y 2 = ( r cos(θ)
−R cos(ψ)) 2 + ( r sin( θ)
−R sin( ψ)) 2 = r 2 + R 2
−2rR cos(ψ
−θ)
(where we used the trignometric identity cos ( θ)cos(ψ) + sin ( θ)sin( ψ) = cos ( ψ −θ)). Changing variables fromy to (r, θ ) thus yields
S R (0)
R 2 − x 2
2πR x −y 2 φ(y ) dσy = 12π
2π
0
R 2 −r 2
R 2 + r 2 −2Rr cos(ψ −θ)φ(ψ) dψ,
as desired.13. In Chapter 9, it was shown that the solution to
− ddx
P (x) dudx
+ R(x)u = 0 , a < x < b,
u(a) = ua ,u(b) = ub
isu(y) = P (a)ua
∂G∂y
(y; a) −P (b)ub∂G∂y
(y; b),
where G is the Green’s function for the inhomogeneous differential equation with homogeneous boundary condi-tions. If we take P (x) = 1, we obtain
u(y) = − −∂G∂y
(y; a)ua + ∂G∂y
(y; b)ub .
Thus u is obtained by “integrating” (that is, summing) the normal derivative of the Green’s function, times theboundary “function” over the boundary of Ω = ( a, b) (that is, the set a, b). Notice that the unit normal to Ωat x = b is n = 1, while at x = a it is n = −1, so that
108 CHAPTER 11. PROBLEMS IN MULTIPLE SPATIAL DIMENSIONS
We can follow the same reasoning as in 11.7.2. The solution of the same IVP in three dimensions is
u(x , t ) = t4π ∂B 1 (0)
ψ(x + cty ) dσy .
If, in this formula, ψ does not depend on x3 , then neither does u, and it is easy to see that u, regarded as afunction on R 2 , is the solution of the IVP given above. Therefore, it remains only to write u as an integral overa two-dimensional region, namely, B 1 (0). If S represents the upper hemisphere of ∂ B1 (0), then
∂B 1 (0)ψ(x + cty ) dσy = 2 S
ψ(x + cty ) dσy
= 2 1
−1 √ 1−x 2
−√ 1−x 2
ψ(x1 + cty1 , x 2 + cty2 )
1 −y21 −y2
2dy2 dy1
= 2 B 1 (0)
ψ(x + cty )
1 − y 2
dy .
It follows thatu(x , t ) = t
2π B 1 (0)
ψ(x + cty )
1 − y 2dy .
Performing the change of variables z = x + cty yields the alternate formula
u(x , t ) = 12πc B ct ( x )
ψ(z )
c2 t2 − z −x 2dy .
5. The solution of
∂ 2 u∂t 2 −c2 ∆ u = 0 , x
∈R 3 , t > 0,
u(x , 0) = 0 , x∈
R 3 ,∂u∂t
(x , 0) = ψ(x ), x
∈
R 3 .
isu(x , t ) = t
4π (ct)2 ∂B ct ( x )ψ(z ) dσz .
Letψ(x ) = 1, x < 1,
0, otherwise ,
and let u be the corresponding solution of the above IVP. Suppose rst that x ∈R 3 , x > R , and t < ( v −R)/c .This last condition implies that ct < x −R, and therefore, if z ∈S ct (x ), then
z −x = ct < x −R.
It is easy to see that this condition implies that z > R (since z is closer to x than x is to S R (0)). Therefore,
z
∈
S ct (x )
⇒ ψ(z ) = 0 ,
and it follows from the formula for u(x , t ) that u(x , t ) = 0.Now suppose t > ( x + R)/c , that is, ct > x + R. Then
z ∈S ct (x ) ⇒ z −x = ct > x + R,
which implies that z > R (since z −x ≤ z + x ). Therefore, in this case also,
z ∈S ct (x ) ⇒ ψ(z ) = 0 ,
and we see that u(x , t ) = 0.7. The causal Green’s function is
The errors in approximating f by a partial Fourier series are shown in Figure 12.1.
−1 −0.5 0 0.5 1−0.04
−0.02
0
0.020.04
x
−1 −0.5 0 0.5 1−0.04
−0.02
0
0.02
0.04
x
−1 −0.5 0 0.5 1−0.04
−0.02
0
0.02
0.04
x
Figure 12.1: The error in approximating f (x) = 1 − x2 by its partial complex Fourier series with N = 10 (top),N = 20 (middle), and N = 40 (bottom). (See Exercise 12.1.1.)
3. The complex Fourier coefficients of f are
cn = (−1)n +1 sin(1)nπ −1
.
The magnitudes of the error in approximating f by a partial Fourier series are shown in Figure 12.2.
using the formula for the sum of a nite geometric series:
m
n =0
r n = rm +1
−1r −1
.
Moreover, since eiθ is a 2π-periodic function of θ, we see that e2πij = 1 and hence that
N −1
n =0
eπijn/N 2= 0 .
On the other hand, by Euler’s formula,
N −1
n =0
eiπjn/N 2=
N −1
n =0
cos jnπN
+ i sin jnπN
2
=
N −1
n =0cos
2 jnπN −sin
2 jnπN + 2 i cos
jnπN sin
jnπN
=N −1
n =0
cos2 jnπN −
N −1
n =0
sin2 jnπN
+2 iN −1
n =0
cos jnπN
sin jnπN
.
Since this expression equals zero, we must have
N −1
n =0
cos2 jnπN −
N −1
n =0
sin2 jnπN
= 0 ,
N −1
n =0 cos jnπ
N sin jnπ
N = 0 .
The second equation is one of the results we set out to prove. The rst equation can be written, using thetrigonometric identity sin 2 (θ) = 1 −cos2 (θ), as
N −1
n =0
cos2 jnπN −
N −1
n =0
1 −cos2 jnπN
= 0 ,
which simplies to
2N −1
n =0
cos2 jnπN
= N,
orN −1
n =0cos
2 jnπN =
N 2 ,
which is the desired result. The resultN −1
n =0
sin2 jnπN
= N 2
then follows.
12.2 Fourier series and the FFT1. The graph of |F n | is shown in Figure 12.3.
Figure 12.3: The magnitude of the sequence F n from Exercise 12.2.1.
−1 −0.5 0 0.5 1−0.5
−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0.4
0.5
x
Figure 12.4: The function f (x) = x(1 − x2 ) (the curve) and the estimates of f (x j ) (the circles) computed fromthe complex Fourier coefficients of f (see Exercise 12.2.3.
where F n 15n = −16 is the DFT of f j 15
j = −16 ,
f j = f (x j ), x j = −1 + j16
, j = −16, −15, . . . , 15
(note that f (−1) = f (1)). Therefore, we can estimate f (x j ) by taking the inverse DFT (using the inverse FFT)of cn . The results are shown in Figure 12.4.
12.3. RELATIONSHIP OF SINE AND COSINE SERIES TO THE FULL FOURIER SERIES 115
while if j = m, then this sum is a nite geometric series:
N −1
n =0
e2πi ( j −m ) n/N =N −1
n =0
e2πi ( j −m ) /N n
=e2πi ( j −m ) /N
N
−1
e2πi ( j −m ) /N −1
= e2πi ( j −m ) −1e2πi ( j −m ) /N −1
= 0 .
Therefore,
am
N −1
n =0
e2πi ( j −m ) n/N = Na j , m = j,0, m = j,
and so we obtainN −1
n =0
An e2πijn/N = 1N
N −1
m =0
am
N −1
n =0
e2πi ( j −m ) n/N = 1N
Na j = a j ,
as desired.
7. Below we show the exact Fourier sine coefficients of f , the coefficients estimated by the DST, and the relativeerror. Since both a n and the estimated a n are zero for n even, we show only a n for n odd.
n a n estimated a n relative error1 2.5801 ·10−1 2.5801 ·10−1 6.2121 ·10−6
3 9.5560 ·10−3 9.5511 ·10−3 5.1571 ·10−4
5 2.0641
·10−3 2.0555
·10−3 4.1824
·10−3
7 7.5222 ·10−4 7.3918 ·10−4 1.7340 ·10−2
9 3.5393 ·10−4 3.3531 ·10−4 5.2608 ·10−2
11 1.9385 ·10−4 1.6778 ·10−4 1.3448 ·10−1
13 1.1744 ·10−4 8.0874 ·10−5 3.1135 ·10−1
15 7.6448 ·10−5 2.4279 ·10−5 6.8241 ·10−1
9. We have
2N −1
n =1
F n sin jnπN
= 4N −1
n =1
N −1
m =1
f m sin mnπN
sin jnπN
= 4
N −1
m =1f m
N −1
n =1sin
mnπN sin
jnπN .
By the hint,N −1
n =1
sin mnπN
sin jnπN
is either N/ 2 or 0, depending on whether m = j or not. It then follows immediately that
Figure 12.5: The partial Fourier sine series (50 terms) of f (x) = x.
12.3 Relationship of sine and cosine series to the full Fourier series
1. The partial Fourier sine series, with 50 terms, is shown in Figure 12.5. It appears that the series converges to thefunction F satisfying
F (x) = x −2k, −2k −1 < x < 2k + 1 , k = 0 , ±1, ±2, . . . .The period of this function is 2.
3. (a) The odd extension f odd is continuous on [ − , ] only if f (0) = 0. Otherwise, f odd has a jump discontinuityat x = 0.
(b) The even extension f even is continuous on [ − , ] for every continuous f : [0, ] →R .5. The quarter-wave sine series of f : [0, ] → R is the full Fourier series of the function f : [−2 ,2 ] → R obtained
by rst reecting the graph of f across the line x = (to obtain a function dened on [0 , 2 ]) and then taking theodd extension of the result. That is, f is dened by
f (x) =
f (x), 0 ≤x ≤ ,f (2 −x), < x ≤2 ,
−f (−x), − ≤x < 0,−f (x + 2 ), −2 ≤x < − .
Since this function is odd, its full Fourier series has only sine terms (all of the cosine coefficients are zero), andbecause of the other symmetry, the even sine terms also drop out.
12.4 Pointwise convergence of Fourier series1. The Fourier series of f converges pointwise to the function F dened by
F (x) = 0, x = ±(2k −1)π, k = 1 , 2, 3, . . . ,(x −2kπ )3 , (2k −1)π < x < (2k + 1) π, k = 0 , ±1, ±2, . . . .
3. There are many such functions f , but they all have one or more discontinuities. An example is
f (x) = x, 0 ≤x ≤ 1
2 ,x + 1 , 12 < x ≤1.
5. If f N converges uniformly to f on [a, b], then it obviously converges pointwise (after all, the maximum difference,over x ∈ [a, b], between f N (x) and f (x) converges to zero, so each individual difference must converge to zero).To prove that uniform convergence implies mean-square convergence, dene
7. Suppose f : [0, ] →R is continuous. Then f even , the even, periodic extension of f to R , is dened by
f even (x) = f (x −2k ), 2k ≤x < (2k + 1) , k = 0 , ±1, ±2, . . . ,f (−x + 2 k ), (2k −1) ≤x < 2k , k = 0 , ±1, ±2, . . . .
Obviously, then, f even is continuous on every interval (2 k ,(2k + 1) ) and ((2 k −1) ,2k ), that is, except possiblyat the points 2 k and (2 k −1) , k = 0 , ±1, ±2, . . . . We have
limx →2k − f even (x) = lim
x →2k − f (−x + 2 k ) = limx →0+
f (x) = f (0)
andlim
x →2k +f even (x) = lim
x →2k +f (x −2k ) = lim
x →0+f (x) = f (0) .
Since f even (2k ) = f (0) by denition, this shows that f even is continuous at x = 2κ .A similar calculation shows that
limx →(2 k−1) − f even (x) = lim
x→(2 k−1) +f even (x) = f even ((2 k −1) ) = f ( ),
and hence that f even is continuous at x = (2 k −1) .9. Given f : [0, ] →R , dene f : [−2 ,2 ] →R by
f (x) =
f (x), 0 ≤x ≤ ,f (2 −x), < x ≤2 ,
−f (−x), − ≤x < 0,
−f (x + 2 ), −2 ≤x < − .
Then dene F : R →R to be the periodic extension of f to R . Assuming f is piecewise smooth, the quarter-wavesine series of f converges to F (x) if F is continuous at x, and to
12
[F (x−) + F (x+)]
if F has a jump discontinuity at x. If f is continuous, then its quarter-wave sine series converges to f exceptpossibly at x = 0. If f is continuous and f (0) = 0, then the quarter-wave sine series of f converges to f at everyx∈[0, ].
12.5 Uniform convergence of Fourier series1. The function h(x) fails to satisfy h(−1) = h(1), so its Fourier coefficients decay like 1 /n and its Fourier series is
the slowest to converge. The function f satises f (−1) = f (1), but df/dx has a discontinuity at x = 0 (and thederivative of f per also has discontinuities at x = ±1). Therefore, the Fourier coefficients of f decay like 1/n 2 .Finally, gper and its rst derivative are continuous, but its second derivative has a jump discontinuity at x = ±1,so its Fourier coefficients decay like 1 /n 3 . The Fourier series of g is the fastest to converge.
3.5. Figure 12.6 shows the f , its partial Fourier series with 21, 41, 81, and 161 terms, and the line y = 1 .09. Zooming
in near x = 0 shows that the overshoot is indeed about 9%; see Figure 12.7.
12.6 Mean-square convergence of Fourier series1. (a) The innite series
∞
n =1
1n 2
converges to a nite value. 1 Therefore, 1/n ∈2 and so, by Theorem 12.36, the sine series converges to afunction in L2 (0, 1).
1 A standard result in the theory of innite series is that
Figure 12.6: The function f from Exercise 12.5.5, its partial Fourier series with 21, 41, 81, and 161 terms, andthe line y = 1 .09.
−0.02 0 0.02 0.04 0.06 0.08 0.11.075
1.08
1.085
1.09
1.095
1.1
1.105
x
y
21 terms41 terms81 terms161 terms
Figure 12.7: Zooming in on the overshoot in Figure 12.6.
(b) Figure 12.8 shows the sum of the rst 100 terms of the sine series. The graph suggests that the limit f is of the form f (x) = m(1 −x). The Fourier sine series of such an f are
cn = 2 1
0m(1 −x)sin( nπx ) = 2m
nπ , n = 1 , 2, 3, . . . .
Therefore, in order that cn = 1 /n , we must have m = π/ 2. Therefore,
f (x) = π2
(1 −x).
0 0.2 0.4 0.6 0.8 1−0.5
0
0.5
1
1.5
2
x
Figure 12.8: The partial sine series, with 100 terms, from Exercise 12.6.1.
converges if k is greater than 1. This can be proved, for example, by comparison with the improper integral
12.7. A NOTE ABOUT GENERAL EIGENVALUE PROBLEMS 119
3. The innite series∞
n =1
1
√ n2
=∞
n =1
1
nis the harmonic series , a standard example of a divergent series. Therefore, 1/ √ n ∈ 2 , and so the series doesnot converge to an L2 (0, 1) function.
5. (a) The function f (x) = xk belongs to L2 (0, 1) if and only if the integral 1
0 x2k dx is nite. Provided k = −1/ 2,we have
1
0x2k dx = lim
r →0+ 1
rx2k dx = lim
r →0+
1 −r 2k +1
1 + 2 k =
11+2 k , k > −1
2 ,
∞, k < −12 .
For k = −1/ 2,
1
0x2k dx =
1
0
dxx
= limr →0+
1
r
dxx
= limr →0+ −ln ( r ) = ∞.
Thus we see that f ∈L2 (0, 1) if and only if k > −1/ 2.(b) The rst few Fourier sine coefficients of f (x) = x−1/ 4 , as computed by the DST, are approximately
1.5816, 0.25353, 0.63033, 0.17859, 0.41060, 0.14157, . . . .(c) The graphs of f , the partial Fourier sine series, with 63 terms, and the difference between the two, are shown
in Figure 12.9.
0 0.2 0.4 0.6 0.8 10
2
4
6
x
y
y=x−1/4
sine series (63 terms)
0 0.2 0.4 0.6 0.8 1−0.1
0
0.1
x
Figure 12.9: The function f (x) = x− 1 / 4 , together with its partial Fourier sine series (rst 63 terms) (top), andthe difference between the two (bottom). See Exercise 12.6.5.
7. Since v n →v , there exists a positive integer N such that
n ≥N ⇒ v −v n < 2
.
Then, if n, m ≥N , we have
v n −v m ≤ v n −v + v −v m < 2
+ 2
= .
Therefore, v n is Cauchy.
12.7 A note about general eigenvalue problems1. The proof is a direct calculation, using integration by parts (Green’s identity): Suppose u, v ∈C 2D (Ω). Then
(K D u, v ) = − Ω∇ ·(k(x )∇u) v = Ωk(x )∇u ·∇v − ∂ Ω
3. If t is very large, then we can approximate u(x , t ) by the rst term in its generalized Fourier series. Using thesame notation as before for the eigenvalues and eigenfunctions of the negative Laplacian on Ω, we have
u(x , t ) .= c1 e−κλ 1 t/ ( ρc ) ψ1 (x ), t large .
The constant c1 is the rst generalized Fourier coefficient of the initial temperature 5:
c1 = Ω 5ψ1
Ω ψ21
.
The approximation is valid provided λ1 is a simple eigenvalue; that is, there is only one linearly independenteigenvector corresponding to λ 1 . Then all of the other terms in the generalized Fourier series decay to zero muchmore rapidly than does the rst term.
13.1 Implementation of nite element methods1. We must check that the one-point rule gives the exact values for the integral T R f when f (x ) = 1, f (x ) = x1 , or
f (x ) = x2 . With f (x ) = 1, we have
T Rf = (area of T R ) ·1 = 1
2
and12
f 13
, 13
= 12
.
With f (x ) = x1 , we have
T R
f =
1
0
1−x 1
0
x1 dx 2 dx1
= 1
0x1 −x2
1 dx 1 = 12 −
13
= 16
.
On the other hand,12
f 13
, 13
= 12 ·
13
= 16
,
so the rule is exact in this case as well. By symmetry, the rule must hold for f (x ) = x2 , which shows that therule has degree of precision at least 1. To show that the degree of precision is not greater than 1, we note that,for f (x ) = x2
g(z 1 , z 2 ) = f 1 + z 1 + 12 z 2 , z 2 = 1 + z 1 +
12 z 2 z 22 ,
we have
T f = T R
g = 1
0 1−z 1
01 + z 1 + 1
2z 2 z 22 dz 2 dz 1 .
This last integral also equals 1 / 8.5. Recall that the nodes in the mesh are enumerated from 1 to M . Let the constrained boundary nodes be those
with numbers c1 , c2 , . . . , c K . We dene a new K ×1 array CNodePtrs whose ith entry is ci . (Thus CNodePtrscontains pointers into the NodeList array, allowing one to nd the coordinates of any given constrained node.)We need the inverse of the mapping i →ci ; let us dene j →dj by
dj = i if j = ci ,0 if j = ci for all i = 1 , 2, . . . , K .
Now, under the original description of NodePtrs , the ith entry is zero if the ith node is constrained. We nowredene the ith entry to be −di if the ith node, while the i th entry is still gi if the ith node is free. (The use of the negative sign is just a trick to allow us to distinguish a free node from a constrained node; it avoids the needto store a separate ag.)Suppose we wish to solve a BVP with inhomogenous Dirichlet conditions (on all or part of the boundary). LetG be the piecewise linear function whose nodal value is zero everywhere except at the constrained boundarynodes. At the nonfree boundary nodes, the value of G is the assigned Dirichlet data. To solve the inhomogeneousDirichlet problem, we merely need to modify the load vector by replacing
F i = Ωfφ i
by
F i = Ωfφ i −a(G, φ i ).
The quantity a (G, φ i ) is nonzero only if the free node n f i belongs to a triangle that also contains a constrainedboundary node. The point here is simply that this can easily be determined from the information in the datastructure. As we loop over the triangles, we can determine, for each, whether it contains both a free node and aconstrained node. If it does, we modify the load vector accordingly.
13.2 Solving sparse linear systems1. Computing A = LU requires n −1 steps, and step i uses
(n −i)(1 + 2( n −i)) = 2( n −i)2 + ( n −i)
arithmetic operations (this is determined by simply counting the operations in the pseudo-code on page 604).The total number of operations required is
n −1
i =12(n −i)
2+ ( n −i) = 2
n −1
j =1 j
2+
n −1
j =1 j =
2n 3
3 − n2
2 − n6 .
This agrees with the O(2n 3 / 3) operation count given in the text.The computation of L −1 b requires n −1 steps, with 2( n −i) arithmetic operations per step. The total is
2n −1
i =1
(n −1) = 2n −1
j =1
j = n2 −n.
The nal step of back substitution requires n steps, with 2( n −i) + 1 operations per step, for a total of n
i =12(n −i) + 1 = 2
n −1
j =1
j + n = n 2 .
The total for the two triangular solves is 2 n 2
−n, which also agrees with the count given in the text.
The fact that A is symmetric allows the simplication
12
x ·Ay + 12
y ·Ax = y ·Ax ,
which was used above. We thus see that φ(x + y ) = φ(x ) + ∇φ(x ) ·y + O y 2 , with ∇φ(x ) = Ax −b .
5. An inner product has three properties:
(a) ( x , x ) ≥0 for all x , and ( x , x ) = 0 if and only if x = 0.
(b) ( x , y ) = ( y , x ) for all x , y .
(c) (α x + β y , z ) = α (x , z ) + β (y , z ) for all x , y , z and all α, β .
These properties hold for ( x , y )A = x · Ay . The third property would be true for any matrix A , the secondproperty obviously requires that A be symmetric, and the rst property requires that A be positive denite.
7. (a) With x (0) = (4 , 2), the negative gradient is −∇φ x (0) = (−2, −1), and so we must minimize
φ x (0)
−α
∇
φ x (0) = 7α 2
−5α
−18.
The minimizer is α = 5 / 14, and so
x (1) = 237
, 2314
.
(b) At x (1) = (23 / 7, 23/ 14), the negative gradient is ( −3/ 14, 3/ 7), so we must minimize
φ x (1) −α∇φ x (1) = 27196
α 2 − 45196
α − 529
28 .
The minimizer is α = 5 / 6, and so
x (2) = 8728
, 2 .
The desired solution is (3 , 2).
(c) With x (0) = 0 , the CG algorithm produces
x (1) .= (2 .67456, 2.34024),
with x (2) = (3 , 2), the exact solution.
9. If y = x −x (0) , then x = y + x (0) and so
Ax = b⇒
Ay + Ax (0) = b⇒
Ay = b −Ax (0) .
We can therefore replace b by b −Ax (0) , apply the CG algorithm to estimate y , and add x (0) to get the estimateof x .
13.3 An outline of the convergence theory for nite element meth-
ods1. Let the vector-valued function F be dened by F (x ) = ( f (x ), 0). Then
∇ ·(φF ) = φ∇ ·F + ∇φ ·F = φ ∂ f ∂x 1
+ ∂φ∂x 1
f,
so, by the divergence theorem,
Ωφ ∂f
∂x 1+ Ω
∂φ∂x 1
f = Ω∇ ·(φF ) = ∂ ΩφF ·n = ∂ Ω
φfn 1 .
If φ∈C ∞0 (Ω), then the boundary integral vanishes, and we obtain
Ωφ ∂ f
∂x 1= − Ω
∂φ∂x 1
f.
The derivation for ∂/∂x 2 is exactly analogous.
3. A direct computation shows thatf −g L 2 = 1
2 for all integers m,n,
whilef −g H 1 = 1
2 1 + m 2 π 2 + n 2 π 2 for all m,n.
Thus the L2 error is independent of m, n , while the H 1 error becomes arbitrarily large as m, n →∞.5. The following table shows the L2 and energy norm errors. The initial mesh has eight triangles, and the side
lengths of the triangles are divided by two each time the mesh is rened.h L 2 error energy error√ 2/ 2 1.7803 ·10−2 1.0190 ·10−1
√ 2/ 4 5.6705 ·10−3 5.3955 ·10−2
√ 2/ 8 1.5112 ·10−3 2.7318 ·10−2
√ 2/ 16 3.8399 ·10−4 1.3700 ·10−2
√ 2/ 32 9.6392 ·10−5
6.8554 ·10−3
√ 2/ 64 2.4123 ·10−5 3.4283 ·10−3
We see that the L2 error goes down by a factor of approximately four each time the mesh is rened which is
