Solution

Math 1470
Homework #2
Math 1470 Homework #2 Solutions
P38: 2 By d’Alembert’s formula,
Z
1
1 x+ct
2
2
u(x, t) =
log[1 + (x + ct) ] + log[1 + (x − ct) ] +
(4 + s) ds
2
2c x−ct
1
=
log[1 + (x + ct)2 ] + log[1 + (x − ct)2 ] + (4 + x)t.
2
P38: 9 Factoring the operators yields (∂x +∂t )(∂x −4∂t )u = 0. So the similar argument
as for the standard wave equation (i.e. a change of variable argument) implies that
u(x, t) = f (x − t) + g(4x + t).
Plugging the initial conditions we have
f (x) + g(4x) = x2
− g 0 (x) + g 0 (4x) = ex .
Solving the above and plugging the results into the expression of u we find
u(x, t) =
4 x+t/4
1
e
− ex−t + x2 + t2 .
5
4
P46: 6 Consider the difference w = u − v. Then wt − kwxx = 0, and w ≤ 0 on
{t = 0}, {x = 0} and {x = l}. So Maximum Principle applied to w implies that w ≤ 0
for 0 ≤ t < ∞ and 0 ≤ x ≤ l.
√
P52: 1 Letting p = (y − x)/ 4kt gives
Z l
1
2
u(x, t) = √
e−(y−x) /4kt dy
4πkt −l
Z (l−x)/√4kt
1
2
e−p dp
=√
√
π −(l+x)/ 4kt
1
l−x
1
l+x
+ Erf √
.
= Erf √
2
2
4kt
4kt
Problem 1.
Use energy method to show that the following PDE

 ρ(x)utt = [k(x)ux ]x − q(x)u, 0 < x < l, t > 0,
u(0, x) = ut (0, x) = 0,
0 ≤ x ≤ l,

u(t, 0) = u(t, l) = 0,
t>0
has a unique solution u = 0. Here we assume that ρ(x) ≥ ρ0 > 0, k(x) ≥ k0 > 0, and
q(x) ≥ 0, where ρ0 and k0 are constants.
1
Math 1470
Homework #2
Proof. Multiplying the PDE by ut and integrating over [0, l] we get
Z
Z l
Z
d 1 l 2
d 1 l 2
(kux )x ut dx −
ρut dx =
qu dx
dt 2 0
dt 2 0
0
Z
Z
l
d 1 l 2
d 1 l 2
= kux ut −
kux dx −
qu dx
dt 2 0
dt 2 0
0
Plugging in the initial and boundary conditions it follows that
Z
d 1 l
2
2
2
ρut + kux + qu dx = 0.
dt 2 0
Therefore
Z l
Z l
1
1
2
2
2
2
2
2
ρut + kux + qu dx (t) =
ρut + kux + qu dx (0) = 0.
2 0
2 0
Using the bounds of ρ, k, q it infers that it must be
u ≡ 0.
Problem 1.
Following the same notations as used in class, let ΩT = {0 < t < T, 0 < x < l} and
Γ1 = {t = 0, 0 ≤ x ≤ l}, Γ2 = {0 ≤ t ≤ T, x = 0} ∪ {0 ≤ t ≤ T, x = l}. Let
u, ut , uxx ∈ C(ΩT ), and moreover
ut − kuxx ≤ 0 in ΩT ,
k > 0.
Show that
max u = max u.
Γ1 ∪Γ2
ΩT
Proof. The proof goes almost identical as for the case of ut − kuxx = 0. We let
v = u + εx2 for ε > 0. So v satisfies
vt − kvxx = ut − kuxx − 2ε < 0.
Thus from the argument in class we know
max v = max v.
Γ1 ∪Γ2
ΩT
From the definition of v we obtain
max u ≤ max v ≤ max v ≤ max u + εl2 .
ΩT
ΩT
Γ1 ∪Γ2
Letting ε → 0 finishes the proof.
2
Γ1 ∪Γ2
Math 1470
Homework #2
Problem 2.
Using the same notation as in the previous problem. Now consider the damped diffusion equation
ut = kuxx + cu, k > 0, c ≥ 0.
If |u(0, x)| ≤ M on Γ1 and |u(t, x)| ≤ B on Γ2 , prove that
|u(t, x)| ≤ max{M, B}ect ,
in ΩT .
Proof. Let v = e−ct u, then the PDE for v is
vt − kvxx = 0,
with estimates on the boundary as
|v(0, x)| = |u(0, x)| ≤ M on Γ1 ,
|v(t, x)| ≤ Be−ct ≤ B on Γ2 .
Hence an application of Maximum Principle infers that
max |v| ≤ max |v| ≤ max{M, B},
ΩT
Γ1 ∪Γ2
and thus
|u(t, x)| ≤ max{M, B}ect ,
3
in ΩT .

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