HOMEWORK 4
SHUANGLIN SHAO
1. P45. # 1.
Proof. By the maximum principle, u(x, t) = 1 − x2 − 2kt attains the maximum at the bottom or on the two sides. When t = 0, 1 − x2 − 2kt = 1 − x2
attains the maximum at x = 0, i.e., 1 − x2 = 1.
When x = 0, 1 − x2 − 2kt = 1 − 2kt attains the maximum 1 in the interval
0 ≤ t ≤ T.
When x = 1, 1 − x2 − 2kt = −2kt attains the maximum 0 in the interval
0 ≤ t ≤ T.
Thus the maximum of u is 1 in the closed rectangle {0 ≤ x ≤ 1, 0 ≤ t ≤
T }.
2. P45. # 2.
Proof. (a). Let M (T ) = the maximum of u(x, t) in the closed rectangle
{0 ≤ x ≤ l, 0 ≤ t ≤ T }. M (T ) is a decreasing function of T . Indeed, when
T1 ≤ T2 , the rectangle R1 = {0 ≤ x ≤ l, 0 ≤ t ≤ T1 } is contained in the
rectangle R2 = {0 ≤ x ≤ l, 0 ≤ t ≤ T2 }; the bottom and the two lateral
sides of R1 are contained in those of R2 . On the other hand, the uniqueness
of solutions to the diffusion equation shows that the solution u on R2 is an
extension of u on R1 . So by the maximum principle, M (T1 ) ≤ M (T2 ).
(b). Similarly as in proving (a)., m(T ) is an increasing function of T . 3. P45. #3.
Proof. (a). By the strong maximum principle, u(x, t) > 0 in the interior
points 0 < x < 1, 0 < t < ∞ because the minimum value of u, 0, is attained
at the boundary point, and at the two lateral sides.
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(b). We follow the hint. Let µ(t) = the maximum of u(x, t) over 0 ≤ x ≤ 1.
Let X(t) ∈ [0, 1] such that
µ(t) = u(X(t), t).
On the two lateral sides, the value of u is 0. By part (a) , µ(t) > 0. So
X(t) ∈ (0, 1). On the point (X(t), t), following the proof of the maximum
principle in the book,
ux (X(t), t) = 0, ut (X(t), t) ≤ 0.
So we differentiate µ in t,
µ0 (t) = ux X 0 (t) + ut = ut ≤ 0.
We can establish it by a different method. For each t > 0, let µ(t) = the
maximum of u(x, t) over 0 ≤ x ≤ 1. By part (a), u(x, t) ≥ 0 for all x, t.
Then µ(t) ≥ 0 for t > 0.
Suppose that 0 < t1 < t2 . Define R(t1 , t2 ) = {0 ≤ x ≤ 1, t1 ≤ t ≤ t2 }. Since
the value of u is 0 on the two lateral sides, by the maximum principle, the
maximum of u on R(t1 , t2 ) is µ(t1 ). This implies that
µ(t2 ) ≤ µ(t1 ).
So µ is decreasing in t > 0.
4. P46. # 4.
Proof. (a).
On the two lateral sides and on the bottom, the minimum
and the maximum of u is 0 and 41 , respectively. So by the strong maximum
principle, 0 < u(x, t) < 1.
(b). Both u(x, t) and u(1−x, t) satisfy the equation ut = kuxx and the two
lateral side conditions, and the initial condition. By the uniqueness theorem
for the diffusion equation,
u(x, t) = u(1 − x, t).
(c). Let
Z
E(t) =
(u(x, t))2 dx.
2
We differentiate it in t,
dE(t)
=2
dt
Z
uut dx
Z
= 2k
=
uuxx dx
uux |x=∞
x=−∞
Z
= −2k
Z
− 2k
(ux )2 dx
(ux )2 dx ≤ 0.
So E(t) is decreasing in t.
5. P46. # 6.
Proof. We prove it by considering the difference w = v − u. The function
w(x, t) satisfies the equation
wt = kwxx ,
with w being nonnegative when either t = 0, orx = 0 or x = l. By the
maximum principle, the minimum value of w is attained when either t = 0,
or x = 0 or x = l. So
w ≥ 0.
Thus
u≤v
for 0 ≤ t < ∞ and 0 ≤ x ≤ l.
6. P52. # 1.
Proof. The function φ satisfies that
φ(x) = 1, |x| < l; φ(x) = 0, |x| > l.
So the solution u is
Z l
(x−y)2
1
e− 4kt φ(y)dy
u(x, t) = √
4πkt −l
Z l
Z −l
(x−y)2
(x−y)2
1
1
=√
e− 4kt dy − √
e− 4kt dy
4πkt −∞
4πkt −∞
3
For the first integral,
1
√
4πkt
Z
l
e
−
y−x
√
4kt
2
dy
−∞
Z √l−x
4kt −z 2 √
1
e
=√
4ktdz
4πkt −∞
Z √l−x
4kt −z 2
1
e dz
=√
π −∞
=
Erf ( √l−x
)
4kt
2
.
Similarly for the second integral,
Z −l y−x 2
1
− √
4kt
√
e
dy
4πkt −∞
Z − √l+x
4kt −z 2 √
1
=√
4ktdz
e
4πkt −∞
Z − √l+x
4kt −z 2
1
=√
e dz
π −∞
=
)
Erf (− √l+x
4kt
2
.
Hence
u(x, t) =
Erf ( √l−x
)
4kt
2
−
Erf (− √l+x
)
4kt
2
.
7. P52. # 2.
Proof. The function φ satisfies that
φ(x) = 1, for x > 0; φ(x) = 3, for x < 0.
So the solution u is
Z ∞
(x−y)2
1
e− 4kt φ(y)dy
4πkt −∞
Z ∞
Z 0
(x−y)2
(x−y)2
1
3
=√
e− 4kt dy + √
e− 4kt dy
4πkt 0
4πkt −∞
u(x, t) = √
4
For the first integral,
Z ∞ x−y 2
1
− √
4kt
e
dy
4πkt 0
Z −∞
1
2√
e−z 4kt(−dz)
=√
x
4πkt √4kt
Z √x
4kt −z 2
1
e dz
=√
π −∞
x
Erf ( √4kt
)
=
.
2
√
Similarly for the second integral,
2
Z 0
y−x
3
− √
4kt
√
e
dy
4πkt −∞
Z − √x
4kt −z 2 √
3
=√
4ktdz
e
4πkt −∞
Z − √x
4kt −z 2
3
=√
e dz
π −∞
x
3Erf (− √4kt
)
=
.
2
Hence
u(x, t) =
x
Erf ( √4kt
)
2
+
x
3Erf (− √4kt
)
2
.
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8. P52. # 3.
Proof.
Z ∞
(x−y)2
1
√
e− 4kt φ(y)dy
4πkt −∞
Z ∞
(x−y)2
1
√
e− 4kt +3y dy
4πkt −∞
Z ∞
x2 −2xy+y 2 +12kyt
1
4kt
√
e−
dy
4πkt −∞
Z ∞
(y−x+6kt)2 +12kxt−36k2 t2
1
4kt
√
e−
dy
4πkt −∞
Z ∞
(y−x+6kt)2
1
√
e− 4kt +3x−9kt dy
4πkt −∞
Z ∞ y−x+6kt 2
3x−9kt
e
√
−
4kt
√
e
dy
4πkt −∞
Z
e3x−9kt ∞ −y2
√
e dy
π
−∞
u(x, t) =
=
=
=
=
=
=
= e3x−9kt .
9. P52. # 6.
Proof. Let A =
R∞
0
2
e−x dx =
A2 =
=
=
=
=
=
=
1
2
R∞
−∞ e
−x2 dx
. Thus
1 ∞ ∞ −x2 −y2
e
dxdy
4 −∞ −∞
Z
1
2
2
e−(x +y ) dxdy
4 R2
Z
1 ∞ −r2
e 2πrdr
4 0
Z
π ∞ −r2 2
e d(r )
4 0
Z
π ∞ −x
e dx
4 0
π
(1 − 0)
4
π
.
4
Z
Z
6
So
√
π
.
2
A=
10. P52. #7.
Proof. From Exercise # 6,
Z ∞
Z
2
e−p dp = 2
−∞
If setting p =
√x ,
4kt
√
∞
2
e−x dx =
√
π.
0
we have
∞
Z
2
π=
e
x
− 4kt
−∞
√
dx
√
= π
4kt
Z
∞
S(x, t)dx,
−∞
which implies that
Z
∞
S(x, t)dx = 1.
−∞
11. P53. # 8.
Proof. Let
x2
1
S(x, t) = √
e− 4kt .
2 πkt
Since S is even function in x,
S(x, t) = S(−x, t).
Thus S(x, t) = S(|x|, t).
max S(x, t) = max S(x, t).
δ≤x<∞
δ≤|x|<∞
For each t > 0, the function S(x, t) is decreasing on the interval [δ, ∞).
Hence
δ2
1
max S(x, t) = √
e− 4kt .
δ≤x<∞
2 πkt
7
√
δ2
We will apply the L’Hospital rule to show that e− 4kt / 4πkt goes to zero as
t goes to zero.
t−1/2
1
√ lim δ2
2 πk t→0 e 4kt
1
−1/2t−3/2
= √ lim
δ2
−δ 2
2 πk t→0 e 4kt
2
4kt
δ2
2k
= 2 t1/2 e− 4kt
δ
δ2
which goes to zero as both t1/2 and e− 4kt go to zero when t goes to zero.
This proves the claim.
12. P53. # 15.
Proof. Let u, v be two solutions to the diffusion equation with Neumann
boundary conditions. Let w = u − v. Then w satisfies the following system
of equations:

wt − kwxx = 0.



w(x, 0)
= 0,
wx (0, t)
= 0,



wx (l, t)
= 0.
We multiply the equation by w to obtain
w(wt − kwxx ) = 0.
Thus
d(w2 )
− k(wx w)x + kwx2 = 0.
2dt
Integrating both sides and using the boundary conditions, we see that
Z
Z
1d
w2 dx + k wx2 dx = 0.
2 dt
R
Since k ≥ 0, we see that the quantity w2 dx is decreasing for t > 0. Thus
Z
Z
2
(u(x, t) − v(x, t)) dx ≤ w2 (x, 0)dx = 0.
This implies that
u(x, t) = v(x, t) for all x, t.
This proves the uniqueness.
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13. P 54. # 16.
Proof. We set u(x, t) = e−bt v(x, t). We substitute u into the equation
0 = ut − kuxx + bu = e−bt (−bv + vt − kvxx + bv) .
Thus
vt − kvxx = 0.
The initial condition changes to v(x, 0) = φ(x). Hence by the solution
formula for the homogeneous diffusion equation,
Z ∞
v(x, t) =
S(x − y, t)φ(y)dy,
−∞
where
S(x, t) = √
Therefore
−bt
Z
x2
1
e− 4kt .
4πkt
∞
S(x − y, t)φ(y)dy.
u(x, t) = e
−∞
This is the solution.
Department of Mathematics, KU, Lawrence, KS 66045
E-mail address: [email protected]
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