HOMEWORK 3
SHUANGLIN SHAO
1. P38. # 1.
Proof. By the solution formula, we have
1 x+ct
u(x, t) =
e
+ ex−ct +
2
1 x+ct
e
+ ex−ct −
=
2
Z
1 x+ct
sin ydy
2c x−ct
1
(cos(x + ct) − cos(x − ct))
2c
2. P38. # 2.
Proof. By the solution formula,
Z
1
1 x+ct
2
2
u(x, t) =
log(1 + (x + ct) ) + log(1 + (x − ct) ) +
(4 + y)dy
2
2c x−ct
1
1
=
log(1 + (x + ct)2 ) + log(1 + (x − ct)2 ) +
(4 + x + ct)2 − (4 + x − ct)2
2
4c
1
= log(1 + (x + ct)2 )(1 + (x − ct)2 ) + t(4 + x).
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3. P38. # 5.
See the class notes of Section 2.2.
4. P38. # 6.
Proof. The greatest displacement is ac .
u(x, t) =
1
2a
a
|[x − ct, x + ct] ∩ (−a, a)| ≤
= .
2c
2c
c
1
So the possible maximum value is ac . When t =
u(x,
2a
c ,
2a
1
) = |[x − 2a, x + 2a] ∩ (−a, a)|.
c
2c
When x = 0,
u(0,
Hence the maximum value
2a
a
2a
)=
= .
c
2c
c
a
can be achieved.
c
5. P38. # 8.
Proof. (a). Let v = ru. Then
vtt = rutt , vrr = 2ur + rurr .
Then
utt =
vrr
2ur
vtt
, urr =
−
.
r
r
r
So
vtt = c2 vrr .
(b). By the solution formula,
v(r, t) = f (r + ct) + g(r − ct).
Hence
u=
v
f (r + ct) g(r − ct)
=
+
.
r
r
r
(c). If u(r, 0) = φ(r) and ut (r, 0) = ψ(r),
v(r, 0) = ru(r, 0) = rφ(r), vt (r, 0) = rut (r, 0) = rψ(r).
Hence
1
1
(rφ(r + ct) + rφ(r − ct)) +
2
2c
Z
1
1
u(r, t) = (φ(r + ct) + φ(r − ct)) +
2
2cr
Z
v(r, t) =
r+ct
yψ(y)dy.
r−ct
Since u = vr ,
r+ct
yψ(y)dy.
r−ct
2
6. P38. #9.
Proof. We use the coordinate method at the beginning of this section. We
look for solutions for equations:
(1)
(a∂x + b∂t )(c∂x + d∂t )u = 0,
where ad − bc 6= 0. We choose new coordinates:
ξ = bx − at, η = dx − ct.
By the chain rule,
∂x u = b∂ξ u + d∂η u,
∂t u = −a∂ξ u − c∂η u.
Then we have
c∂x + d∂t = (bc − ad)∂ξ ,
a∂x + b∂t = (ad − bc)∂η .
Then the equation (1) can be rewritten as
−(ad − bc)2 ∂ξη u = 0.
Therefore ∂ξη u = 0. The solution to this equation is
u(ξ, η) = f (ξ) + g(η)
where f and g are functions of one variable. Therefore the general solution
to the equation (1) is
u(x, t) = f (bx − at) + g(dx − ct),
where f, g are functions of one variable. We will use it in Ex. 9, Ex. 10,
Ex. 11.
To Equation (1), we impose the initial conditions
(2)
u(x, 0) = φ(x), ut (x, 0) = ψ(x).
Then
(3)
(4)
f (bx) + g(dx) = φ(x),
0
−af (bx) − cg 0 (dx) = ψ(x).
Thus by integrating the second equation above, we see that
Z x
a
c
(5)
− f (bx) − g(dx) =
ψ(y)dy + A,
b
d
0
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where A = − ab f (0) − dc g(0). Therefore combining Equation (2) and (3),
Z x
b
bc
−bd
x
A
ψ(y)dy −
f (x) =
φ( ) +
,
ad − bc 0
ad − bc b
ad − bc
Z x
d
ad
x
A
bd
ψ(y)dy −
φ( ) −
.
g(x) =
ad − bc 0
ad − bc b
ad − bc
Thus
(6)
u(x, t) = f (bx − at) + g(dx − ct)
!
Z x− a t
Z x− c t
d
b
bd
ad
c
bc
a
=
ψ(y)dy −
ψ(y)dy +
φ(x − t) −
φ(x − t).
ad − bc
ad − bc
d
ad − bc
b
0
0
Now for this equation in Ex. 9, uxx − 3uxt − 4utt = 0. We factorize it
(∂x − 4∂t )(∂x + ∂t )u = 0.
Hence
a = 1, b = −4, c = 1, d = 1.
By using the solution formula (6),
!
Z x−t
Z x+ t
4
4
t
1
4
ψ(y)dy −
ψ(y)dy + φ(x − t) + φ(x + ).
u(x, t) = −
5
5
5
4
0
0
By the initial conditions,
φ(x) = x2 , ψ(x) = ex .
Therefore
u(x, t) = −
1
t
4 x−t
4
t
e
− ex+ 4 + (x − t)2 + (x + )2 .
5
5
5
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6.1. Remark. To conclude from this problem, we formulate a general lemma:
6.2. Lemma. If u satisfies the following wave equation:
(a∂x + b∂t )(c∂x + d∂t )u = 0,
u(x, 0) = φ(x), ut (x, 0) = ψ(x).
ad − bc 6= 0.
Then
bd
u(x, t) =
ad − bc
Z
0
x− dc t
Z
ψ(y)dy −
0
x− ab t
!
ψ(y)dy +
ad
c
bc
a
φ(x− t)−
φ(x− t)
ad − bc
d
ad − bc
b
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7. P38. # 10.
Proof. The equation can be written as
(∂x + 5∂t )(∂x − 4∂t )u = 0.
With the initial conditions
u(x, 0) = φ(x), ψ(x) = ψ(x).
Then in Lemma 6.2,
a = 1, b = 5, c = 1, d = −4.
Thus
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u(x, t) =
9
x+ 4t
Z
0
Z
ψ(y)dy −
x− 5t
!
ψ(y)dy
0
4
t
5
t
+ φ(x + ) + φ(x − ).
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4
9
5
8. P38. # 11.
Proof. For this equation, we have a particular solution
1
u(x, t) = − sin(x + t).
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Then we look for solutions to the homogeneous equation
(3∂x + ∂t )(∂x + 3∂t )v = 0.
Then in Lemma 6.2,
a = 3, b = 1, c = 1, d = 3.
Then by (6),
v(x, t) = f (x − 3t) + g(3x − t),
where f, g are functions of one variable. Therefore
1
u(x, t) = f (x − 3t) + g(3x − t) −
sin(x + t),
16
where f, g are functions of one variable.
9. P41. # 2.
Proof. (a). The solution u satisfies
utt = uxx .
Since e =
1 2
2 (ut
+
u2x )
and p = ut ux ,
∂t e = ut utt + ux uxt ,
∂x p = ux utx + ut uxx .
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Then
∂t e = ∂x p.
On the other hand
∂x e = ut utx + ux uxx ,
∂t p = utt ux + ut uxt .
Then
∂x e = ∂t p.
(b). From (a), we have
∂t2 e = ∂tx p = ∂x2 e.
Thus e satisfies the linear homogeneous wave equation. Similarly p satisfies
the linear homogeneous wave equation.
10. P41. # 3.
Proof. (a). The solution u satisfies utt = uxx . For the translate of u =
u(x − y, t), we differentiate both sides in t and x,
utt (x − y, t) = uxx (x − y, t).
(b). Differentiate both sides of utt = uxx in x in any order,
(∂xk u)tt = (∂xk u)xx .
Differentiate both sides of utt = uxx in t in any order,
(∂tk u)tt = (∂tk u)xx .
Therefore any derivative of u satisfies the wave equation.
(c). For the dilation u → u(ax, at), differentiate u in x and t twice:
a2 uxx (ax, at) = a2 utt (ax, at).
If u satisfies the wave equation, u(ax, at) satisfies the wave equation.
11. P41. # 4.
Proof. The equation utt = uxx has a solution
u(t, x) = f (x + t) + g(x − t),
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where f and g are functions of one variable. Thus
u(x + h, t + k) = f (x + t + h + k) + g(x − t + h − k),
u(x − h, t − k) = f (x + t − h − k) + g(x − t − h + k),
u(x + k, t + h) = f (x + t + h + k) + g(x − t + k − h),
u(x − k, t − h) = f (x + t − h − k) + g(x − t + h − k).
Therefore
u(x + h, t + k) + u(x − h, t − k) = u(x + k, t + h) + u(x − k, t − h).
12. P41. # 6.
Proof. (a). We know that u(r, t) = α(r)f (t − β(r)). We compute
ut = α(r)f 0 (t − β(r)),
00
utt = α(r)f (t − β(r)),
ur = α0 (r)f (t − β(r)) + α(r)f 0 (t − β(r))(−β 0 (r)).
00
urr = α (r)f (t − β(r)) + 2α0 (r)f 0 (t − β(r))(−β 0 (r))
00
00
+ α(r)f (t − β(r))(β 0 (r))2 + α(r)f 0 (t − β(r))(−β (r)).
Equating utt = c2 urr + n−1
r ur implies that
00
α(r) − c2 α(r)(β 0 (r))2 f (t − β(r))
n−1
00
2
0
0
0
+ c 2α (r)β (r) + α(r)β (r) +
α(r)β (r) f 0 (t − β(r))
r
n−1 0
00
2
− c α (r) +
α (r) f (t − β(r)) = 0.
r
(b). Thus setting the coefficients of f , f 0 and f 00 equal to zero, we have
α(r) − c2 α(r)(β 0 (r))2 = 0,
n−1
00
2α0 (r)β 0 (r) + α(r)β (r) +
α(r)β 0 (r) = 0
r
n−1 0
00
α (r) +
α (r) = 0.
r
00
(c). When n = 1, α (r) = 0 implies that α(r) = ar + b, where a, b ∈ R.
The equation β 0 (r) = 1c or − 1c , we have
r
β(r) = ± + c1 , where c1 ∈ R.
c
00
0
0
The equation α (r)β (r) + α(r)β (r) = 0 implies that ac = 0, thus a = 0.
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Hence α(r) = b, β(r) = ± rc + c1 , where b, c1 ∈ R. This solution of α(r)
establishes (d).
When n = 3 from β 0 (r) = ± 1c , β(r) = ± rc + c1 , c1 ∈ R. We are left with two
equations
α(r)
= 0,
α0 (r) +
r
2α0 (r)
00
α (r) +
= 0.
r
Thus
α(1)
α(r) =
.
r
Department of Mathematics, KU, Lawrence, KS 66045
E-mail address: [email protected]
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