Hyperbolic Partial Differential Equations
Evolution equations associated with irreversible physical processes like diffusion and heat
conduction lead to parabolic partial differential equations. When the equation is a model for
a reversible physical process like propagation of acoustic or electromagnetic waves, then
the evolution equation is generally hyperbolic.
The mathematical models usually begin with a conservation statement that is some
version of Newton’s second law
d mV
⃗
dt
⃗
=F
which, when applied to an arbitrary ball B inside the domain U ⊂ R n where the process is
occuring, often takes the form
∂ t ∫ ρ∂ t u dx =
∫∂B F⃗ x, t ⋅ ⃗n dS +
momentum
surface tractions
B
∫B f dx
body forces
In the usual way, this leads to
∫B ∂ t ρ∂ t u − divF⃗ − f dx = 0
∀B ⊂ U
and, since B is arbitrary,
⃗ x, t = fx, t
∂ t ρ∂ t ux, t − divF
for x, t ∈ U × 0, T
If u here represents, say, the deflection from the equilibrium state for an elastic membrane,
then we will have as a constituitive relation,
⃗ x, t = A∇ux, t
F
and then
ρ∂ tt ux, t = ∇A∇ux, t + fx, t
for x, t ∈ U × 0, T.
To complete the specifications for a well posed problem, we would add two initial conditions
ux, 0 = u 0 x
and
∂ t ux, 0 = u 1 x
for x ∈ U,
and boundary conditions of some sort, say
ux, t = 0
for
x ∈ ∂U, t ∈ 0, T.
We will represent this problem, in general, in the form
∂ tt ux, t + Lux, t = fx, t
in U T
1
ux, 0 = u 0 x in U
∂ t ux, 0 = u 1 x in U
u=0
on ∂U × 0, T
where L denotes a second order operator which is uniformly elliptic on U T with coefficients
which we will suppose to be in C 1 U T . In this case, the initial boundary value problem is said
to be "hyperbolic". Now we are going to try to devise a framework in which this problem can
be viewed as if it behaves in more or less the same way as an ordinary differential equation
1
for a real, vector valued function of t.
Because of what we already know about the elliptic operator L, if our solution u were to
′
belong to L 2 0, T : V, where V = H 10 U, (and H = H 0 U, V = H −1 U ) then
′
′
Lux, t ∈ L 2 0, T : V and if f ∈ L 2 0, T : H = L 2 U T  ⊂ L 2 0, T : V , then this imposes
′
the condition ∂ tt u ∈ L 2 0, T : V . Then this creates the following situation with regard to the
spaces where u and its time derivatives are found,
u ∈ L 2 0, T : V
∂ tt u ∈ L 2 0, T : V
′
∂ t u ∈ L 2 0, T : H;
which suggests
⟨∂ tt u, u⟩ V ′ ×V = ∂ tt u, u H = −∂ t u, ∂ t u H = −‖∂ t u‖ 2H .
i.e.,
Then, since L 2 0, T : V ⊂ L 2 0, T : H ⊂ L 2 0, T : V
u, ∂ t u ∈ L 2 0, T : H
∂ t u, ∂ tt u ∈ L 2 0, T : V
′
′
, we have
hence
u ∈ C0, T : H
hence
∂ t u ∈ C 0, T : V
′
Here we made use of the fact that if a function and its first derivative are both square
integrable on 0, T then the function must, in fact, be Holder continuous on 0, T (of order
1
).This permits the following interpretation of the initial conditions,
2
u⋅, t  u 0 ∈ H
∂ t u⋅, t  u 1 ∈ V
′
as t  0
as t  0.
Then a reasonable weak formulation of (1) might be the following:
given
f ∈ L 2 0, T : H , u 0 ∈ H,
and
u1 ∈ V′
u ∈ L 2 0, T : V, with ∂ t u ∈ L 2 0, T : H and ∂ tt u ∈ L 2 0, T : V
find
⟨∂ tt u, v⟩ V ′ ×V + Bu, v, t = ⟨f, v⟩ V ′ ×V
such that
u0 = u 0
′
∀v ∈ V a. e. in0, T
∂ t u0 = u 1
In fact, we are going to assume the initial data is such that u 0 ∈ V, and u 1 ∈ H and then it
remains to be seen if we can show that this weak problem has a unique solution for all
admissible data.
Theorem For all data f ∈ L 2 0, T : H and u 0 ∈ V,
u 1 ∈ H, there exists a unique
′
u ∈ L 2 0, T : V, with ∂ t u ∈ L 2 0, T : H and ∂ tt u ∈ L 2 0, T : V such that
i)
ii)
⟨∂ tt u, v⟩ V ′ ×V + Bu, v, t = f, v H
u0 = u 0
∀v ∈ V a. e. in0, T
2
∂ t u0 = u 1
Moreover, the solution mapping
L 2 0, T : H × H × V ′ : f, u 0 , u 1   u, ∂ t u ∈ L 2 0, T : V × L 2 0, T : H
is continuous.
Proof- (uniqueness) We will first prove that the solution is unique if it exists. Let u denote a
2
solution of the weak problem (2) corresponding to data, u 0 = u 1 = f = 0. We would like to
proceed by writing
T
∫ 0 ⟨u ′′ , u ′ ⟩ V ′ ×V + Bu, u ′ , tdt = 0.
However, since u ′ belongs to L 2 0, T : H, and not to L 2 0, T : V, neither of the terms
⟨u ′′ , u ′ ⟩ V ′ ×V , Bu, u ′ , t is defined. Instead, we must define for a fixed s ∈ 0, T,
s
∫ t uτ dτ
vt =
if 0 ≤ t ≤ s
0
s≤t≤T
if
Then v ∈ V for all t, 0 ≤ s ≤ T, and
s
∫ 0 ⟨∂ tt u, v⟩ V ′ ×V + Bu, v, tdt = 0
Now u ′ 0 = 0, and vs = 0 so,
s
∫ 0 − ⟨u ′ , v ′ ⟩ V ′ ×V + Bu, v, t dt = 0
Next, we observe that v ′ t = −ut for 0 ≤ t ≤ s, and hence
s
∫ 0 ⟨u ′ , u⟩ V ′ ×V − Bv ′ , v, tdt = 0
This implies that
s
∫ 0 1 d ‖ut‖ 2H − Bv, v, t dt =
2 dt
s
= − ∫ Cu, v, t + Dv, v, tdt
0
where
⃗ ⋅ ∇v +
Cu, v, t = ∫ ub
Du, v, t =
U
1
2
uv∇ ⋅ ⃗
bdx
∫ ∇v ⋅ ∂ t A∇u + v∂ t⃗b ⋅ ∇u + uv∂ t cdx.
1
2
U
This leads to
1
2
‖us‖ 2H + Bv0, v0, t
‖us‖ 2H + ‖v0‖ 2V ≤ C ∫
and
Now write
s
0
s
= − ∫ Cu, v, t + Dv, v, tdt
0
‖ut‖ 2H + ‖vt‖ 2V dt
t
v0 = ∫ uτ dτ := wt
0
and express this last estimate in terms of w,
‖us‖ 2H + ‖ws‖ 2V ≤ C ∫
s
0
‖ut‖ 2H + ‖wt − ws‖ 2V dt
But
‖wt − ws‖ 2V ≤ 2‖wt‖ 2V + 2‖ws‖ 2V
hence
3
‖us‖ 2H + 1 − 2sC‖ws‖ 2V ≤ C ∫
s
0
‖ut‖ 2H + ‖wt‖ 2V dt
If we choose T 1 > 0 sufficiently small that 1 − 2T 1 C ≥
‖us‖ 2H + ‖ws‖ 2V ≤ C ∫
If we let
Us = ∫
s
0
s
0
1
2
, then we will have
‖ut‖ 2H + ‖wt‖ 2V dt
for 0 ≤ s ≤ T 1
‖ut‖ 2H + ‖wt‖ 2V dt, this estimate asserts
U ′ s ≤ C Us;
i.e.,
Us ≤ U0 e Cs .
But U0 = 0 so Us = 0 for 0 ≤ s ≤ T 1 and this implies us = 0 for 0 ≤ s ≤ T 1 . We can
repeat this argument on T 1 , 2T 1 , 2T 1 , 3T 1 , …, n − 1T 1 , nT 1  for nT 1 > T in order to
eventually conclude that ut = 0 for 0 ≤ t ≤ T. This proves the uniqueness of the weak
solution.
Observe that weakening the notion of the solution to the IBVP enlarges the class of
admissible solutions so that proving existence becomes easier in general. At the same time,
however, proving uniqueness in a larger class usually becomes more difficult, as this proof
illustrates.
The existence proof, like the proof for the existence of a solution to the parabolic
problem will proceed in a series of steps.
1) Existence of Approximate SolutionsLet w k denote an orthonormal basis for H that is, simultaneously, an orthogonal basis
for V. Then for each positive integer N, define
N
u N t =
∑ C j,N t w j
j=1
where the C j,N are required to satisfy, for each k, 1 ≤ k ≤ N,
i)
ii)
u N " t, w k  H + Bu N , w k , t = f, w k  H
C k,N 0 = u 0 , w k  H
3
C ′k,N 0 = u 1 , w k  H
Here we are using the fact that ⟨u, v⟩ V ′ ×V = u, v H . Now (3) is equivalent to,
N
C ′′k,N t
+ ∑ Bw j , w k , tC k,N t = f k t
j=1
C k,N 0 = u 0 , w k  H
C ′k,N 0 = u 1 , w k  H
which is a system of second order linear ODE’s of the form,
⃗ ′′ t + B jk tC
⃗ N t = ⃗ft,
C
N
⃗ N 0 = U
⃗ 0,
C
⃗ ′ 0 = U
⃗ 1,
C
N
where the coefficient matrix, B jk t is uniformly positive definite on 0, T. It is well known
⃗ N t, for each N, hence there exists for
that such a system has a unique global solution, C
each N, a unique approximate solution u N t for the weak boundary problem.
4
2) Energy Estimates
2
a) max ‖u N t‖ 2V + ‖u ′N t‖ H
0≤t≤T
≤ C 1 ‖u 0 ‖ V + ‖u 1 ‖ H + ‖f‖ L 2 U T 
4
‖u ′′N ‖ L 2 0,T:V ′  ≤ C 2 ‖u 0 ‖ V + ‖u 1 ‖ H + ‖f‖ L 2 U T 
b)
It follows from (3) that
u N " t, u ′N  H + Bu N , u ′N , t = f, u ′N  H
and
2
u N " t, u ′N  H = 1 d u ′N t, u ′N  H = 1 d ‖u ′N t‖ H
2 dt
2 dt
Bu N , u ′ , t = ∫ ∇u ′ A∇u N dx + ∫ u ′ ⃗
b ⋅ ∇u N + c u ′ u N dx
N
U
N
U
N
N
=: B 1 u N , u ′N , t + B 2 u N , u ′N , t
We can assume WLOG that the matrix A is symmetric, which leads to
B 1 u N , u ′N , t = ∫ ∇u ′N A∇u N dx
U
= 1 d ∫ ∇u N A∇u N dx − 1 ∫ ∇u N A ′ ∇u N dx
2 dt U
2 U
and
B 1 u N , u ′N , t ≥ 1 d ∫ ∇u N A∇u N dx − C‖u N ‖ 2V
2 dt U
In addition,
B 2 u N , u ′N , t ≤ C ‖u N ‖ 2V + ‖u ′N ‖ 2H
and, combining these leads to
1 d ‖u ′ t‖ 2 + ∫ ∇u N A∇u N dx
N
H
U
2 dt
≤ C ‖u ′N ‖ 2H + ‖u N ‖ 2V + ‖f‖ 2H
Note here that in order to estimate f, u ′N  H in terms of ‖u ′N ‖ H , we need f in L 2 0, T : H and
not in L 2 0, T : V ′  where we might have thought it should be. Next we observe that by the
ellipticity assumption, we have
∫U ∇u N A∇u N dx ≥ a 0 ‖u N ‖ 2V
which implies that
1 d ‖u ′ t‖ 2 + ∫ ∇u N A∇u N dx
N
H
U
2 dt
≤ C ′ ‖u ′N ‖ 2H + ∫ ∇u N A∇u N dx + ‖f‖ 2H .
U
If we let,
2
Ut := ‖u ′N t‖ H + ∫ ∇u N A∇u N dx
U
then
d Ut − C Ut = d e −Ct Ute Ct ≤ C ′′ ‖f‖ 2 .
H
dt
dt
and it follows that Ut satisfies,
5
Ut ≤
t
U0 + C ′′ ∫ ‖fs‖ 2H ds e Ct
0
But,
2
U0 = ‖u ′N 0‖ H + ∫ ∇u N 0 A∇u N 0dx
U
≤ C ‖u 1 ‖ 2H + ‖u 0 ‖ 2V
and this leads to
‖u ′N t‖ 2H + ∫ ∇u N t A∇u N tdx ≤ C ‖u 1 ‖ 2H + ‖u 0 ‖ 2V + ‖f‖ L2 2 U T  .
U
Finally, we can make use of the ellipticity assumption together with the Poincare inequality
to arrive at the conclusion
‖u ′N t‖ 2H + ‖u N t‖ 2V ≤ C ‖u 1 ‖ 2H + ‖u 0 ‖ 2V + ‖f‖ L2 2 U T 
5
since this last estimate holds for all t ∈ 0, T,
max ‖u ′N t‖ 2H + ‖u N t‖ 2V
0≤t≤T
≤ C ‖u 1 ‖ 2H + ‖u 0 ‖ 2V + ‖f‖ L2 2 U T  .
This proves (4a). Alternatively, we could integrate (5) to obtain
‖u ′N ‖ L2 2 0,T,H + ‖u N ‖ L2 2 0,T,V ≤ CT ‖u 1 ‖ 2H + ‖u 0 ‖ 2V + ‖f‖ L2 2 U T  .
6
Next, fix v ∈ V with ‖v‖ V ≤ 1, and write v = v 1 + v 2 where v 1 ∈ spanw 1 , … , w N  = M N and
v 2 M N . Then ‖v 1 ‖ V ≤ ‖v‖ V ≤ 1 and
⟨u ′′N , v⟩ V ′ ×V = u ′′N , v 1  H = f, v 1  H − Bu N , v 1 .
It follows then, using the Cauchy-Schwartz inequality and the boundedness of B, that
⟨u ′′N , v⟩ V ′ ×V ≤ C‖ft‖ H + ‖u N t‖ V 
hence
‖u ′′N t‖ V ′ = sup ⟨u ′′N , v⟩ V ′ ×V ≤ C‖ft‖ H + ‖u N t‖ V 
‖v‖ V ≤1
and
T
T
∫ 0 ‖u ′′N t‖ 2V ′ dt ≤ C ∫ 0 ‖ft‖ 2H + ‖u N t‖ 2V dt
≤ C ‖f‖ L2 2 U T  + ‖u 1 ‖ 2H + ‖u 0 ‖ 2V
and this is equivalent to (4b).
3. Existence of Weak Solutions
The energy estimates (4) imply that
u N t is bounded in L 2 0, T : V
u ′N t is bounded in L 2 0, T : H
u ′′N t is bounded in L 2 0, T : V
′
6
and it follows that there exists a subsequence u n t ⊂ u N t such that
u n t converges weakly to u in L 2 0, T : V
u ′n t converges weakly to v in L 2 0, T : H
u ′′n t converges weakly to w in L 2 0, T : V
′
In the usual way, making use of the fact that
′
L 2 0, T : V ⊂ L 2 0, T : H ⊂ L 2 0, T : V
⊂ D ′ 0, T : V
′
we get that u ′ = v, u ′′ = v ′ = w. It remains now to show that this weak limit point is a weak
solution, i.e., that it satisfies (2). Let
m
Vm =
vt =
∑ d j t w j : d j t ∈ C 2 0, T 1 ≤ j ≤ m
.
j=1
For m ≤ n,
T
T
∫ 0 u n " t, v H + Bu n , v, tdt = ∫ 0 f, v H dt
for all v ∈ V m
For m = n, let n tend to infinity and use the weak convergence results to get
T
T
∫ 0 u" t, v H + Bu, v, tdt = ∫ 0 f, v H dt
for all v ∈
⋃ Vm
m>0
Since w k  is a basis for V, it follows that ⋃ V m = L 2 0, T : V and therefore,
m>0
T
T
∫ 0 u" t, v H + Bu, v, tdt = ∫ 0 f, v H dt
for all v ∈ L 2 0, T : V.
′
In addition, u ∈ C0, T : H, and u ′ ∈ C0, T : V  hence ut  u0 in H and u ′ t  u ′ 0
′
in V as t → 0 + .
Finally, for any v ∈ C 2 0, T : V such that vT = v ′ T = 0, we have
T
∫ 0 u, v ′′  H + Bu, v, t − f, v H dt = ⟨u ′ 0, v0⟩ − ⟨u0, v ′ 0⟩
and
T
∫ 0 u n , v ′′  H + Bu n , v, t − f, v H dt = ⟨u ′n 0, v0⟩ − ⟨u n 0, v ′ 0⟩
Then
T
∫ 0 u − u n , v ′′  H + Bu − u n , v, tdt = ⟨u ′ 0 − u ′n 0, v0⟩ − ⟨u0 − u n 0, v ′ 0⟩
and the weak convergence of the subsequence u n  implies that the left side of this
equation tends to zero as n tends to infinity. On the right side, recalling (3ii), we have
u n 0 converges in V to u 0 and u ′n 0 converges in H to u 1 , which implies
u0 = u 0
and
u ′ 0 = u 1 .
Then it follows that u is a weak solution of the IBVP. But then every subsequence of the
sequence of approximate solutions, u N , must converge to a weak solution. Since the weak
solution has been shown to be unique, it follows that all subsequences have the same weak
7
limit. But in this case, the sequence u N , must itself converge, weakly, to the weak
solution.
Note that the estimate (5) applies to u = lim u N which shows that the mapping
L 2 0, T : H × H × V ′ : f, u 0 , u 1   u, ∂ t u ∈ L 2 0, T : V × L 2 0, T : H
is continuous.
Finally notice that if the initial conditions in (1),
ux, 0 = u 0 x and ∂ t ux, 0 = u 1 x,
were replaced by final conditions, ux, T = u 0 x and ∂ t ux, T = u 1 x, then the new
problem still admits a unique weak solution which depends continuously on the data. This is
in contrast to the parabolic problem where the final value problem is not well posed.
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