MA342H: Homework #2 solutions
1. Let A be an open, bounded subset of R2 and suppose that u(x, y) satisfies
uxx + uyy = f (x, y) in A
u(x, y) = g(x, y) on ∂A
for some given functions f, g. Show that there exists a constant C > 0 such that
max |u| ≤ sup |g| + C sup |f |.
A∪∂A
A
∂A
Let L(u) = uxx + uyy for convenience. We seek a function w(x, y) = a + bx2 such that
L(w) ≥ L(u) ≥ −L(w) in A
.
w ≤ u ≤ −w
on ∂A
The first condition reads 2b ≥ f ≥ −2b, so we need to have b ≥ 0 and
|f | ≤ 2b in A
⇐⇒
b≥
1
sup |f |.
2 A
The second condition reads w ≤ g ≤ −w, so we need to have w ≤ 0 and
|g| ≤ −w on ∂A
⇐⇒
|g| ≤ −a − bx2 on ∂A.
This means that both conditions will be satisfied, if we define the constants a, b by
1
−a = sup |g| + bx2 .
b = sup |f |,
2 A
∂A
Using the comparison principle, we now get |u| ≤ −w at all points, hence also
|u(x, y)| ≤ −a − bx2 ≤ −a = sup |g| + bx2
∂A
≤ sup |g| + C · sup |f |.
A
∂A
2. Show that the maximum principle does not hold for the wave equation utt = c2 uxx ,
where c > 0. Hint: there are separable solutions of the form u(x, t) = sin(ax) sin(bt).
We note that u(x, t) = sin(ax) sin(bt) satisfies the wave equation if and only if
utt = c2 uxx
⇐⇒
−b2 u = −a2 c2 u
⇐⇒
b = ±ac.
Consider the case a = 1 and b = c, for instance. Then u(x, t) = sin(x) sin(ct) satisfies the
wave equation in the open, bounded set A = (0, 2π) × (0, 2π/c) and u(x, t) vanishes on ∂A.
In particular, the maximum value of u(x, t) is not attained on the boundary of A.
3. Solve the second-order equation utt + 2uxx + 3uxt − ut − 2u = 0.
One may factor this equation in the same way that one factors the quadratic
f (x, t) = t2 + 2x2 + 3xt − t − 2 = t2 + (3x − 1)t + 2(x2 − 1).
In this case, the discriminant of the quadratic is
∆ = (3x − 1)2 − 8(x2 − 1) = x2 − 6x + 9 = (x − 3)2 ,
so the two roots of the quadratic are given by
1 − 3x ± (x − 3)
=⇒ t = −x − 1 or t = −2x + 2.
2
This gives the factorisation f (x, t) = (t + x + 1)(t + 2x − 2) and one may similarly factor
2
∂2
∂ ∂
∂
∂
+2 2 +3
−
−2 u
0=
∂t2
∂x
∂x ∂t ∂t
∂
∂
∂
∂
=
+
+1
+2
− 2 u.
∂t ∂x
∂t
∂x
t=
In order to solve this equation, we introduce variables α, β such that
∂
∂
∂
=
+
,
∂α
∂t ∂x
∂
∂
∂
=
+2 .
∂β
∂t
∂x
According to the chain rule, we need to have tα = xα = tβ = 1 and xβ = 2, so we let
α = 2t − x
t=α+β
.
⇐⇒
β =x−t
x = α + 2β
In terms of these new variables α and β, the given equation takes the form
∂
∂
∂
+1
−2 u=0
⇐⇒
+ 1 w = 0,
∂α
∂β
∂α
where w = uβ − 2u. Solving this first-order linear equation, one finds that
wα + w = 0
=⇒
(eα w)α = 0
=⇒
eα w = f (β)
for some arbitrary function f . Since w = uβ − 2u, we conclude that
uβ − 2u = e−α f (β)
=⇒
=⇒
=⇒
(e−2β u)β = e−α−2β f (β)
Z
−2β
e u = e−α−2β f (β) dβ + G(α)
u = e−α F (β) + e2β G(α)
for some arbitrary functions F, G. Recalling the definitions of α and β, we finally get
u(x, t) = ex−2t F (x − t) + e2x−2t G(2t − x)
e − x).
= e−t Fe(x − t) + ex G(2t
4. Let f, g, h, ϕ, ψ be some given functions and consider the equation
utt − 2uxx + uxt + ut = f (x, t), where 0 ≤ x ≤ 1 and t ≥ 0.
Show that there is at most one solution u(x, t) which satisfies the conditions
u(0, t) = g(t),
u(1, t) = h(t),
u(x, 0) = ϕ(x),
ut (x, 0) = ψ(x).
First, we consider the case f = g = h = 0. Multiplying the given equation by ut , we get
0 = ut utt − 2ut uxx + ut uxt + u2t
1 2
1 2
ut t − 2(ut ux )x + 2utx ux +
u
+ u2t
=
2
2 t x
1 2
1 2
+ u2t .
ut t − 2(ut ux )x + u2x t +
u
=
2
2 t x
Define an energy function E(t) by integrating the perfect time derivatives, namely
Z 1
1
2
2
E(t) =
ut (x, t) + ux (x, t) dx.
2
0
In view of our computation above, we must then have
1 Z 1
1 2
′
E (t) = 2ut ux − ut −
u2t dx.
2
0
0
Since u vanishes at the endpoints x = 0, 1 by assumption, the same is true for ut and this
means that the energy E(t) is decreasing.
Next, we consider the general case and let f, g, h, ϕ, ψ be arbitrary. Suppose that the
given problem has two solutions u, v. Then their difference w = u − v satisfies
wtt − 2wxx + wxt + wt = 0, where 0 ≤ x ≤ 1 and t ≥ 0
subject to the initial and boundary conditions
w(0, t) = w(1, t) = w(x, 0) = wt (x, 0) = 0.
According to the previous paragraph, the energy of such a function is decreasing, so
Z 1
1
2
2
wt (x, t) + wx (x, t) dx
E(t) =
2
0
Z 1
1
2
2
≤
wt (x, 0) + wx (x, 0) dx = 0.
2
0
This implies that wt = wx = 0 at all points, so w is constant. In particular, w = 0 at all
points, so u = v at all points. That is, the given problem has at most one solution.