1. Introduction
Let
Ω be a bounded domain in
Rn with the Lipschitz continuous boundary
∂Ω. In [
1] and in the author’s previous works [
2–
4], the following:
()
is investigated, which is in these works referred to as the equation of motion of vibrating membrane. Up to now, neither existence nor uniqueness of a solution to (
1) is obtained. In [
1–
3], we only have that
a sequence of approximate solutions to (
1)
converges to a function u in an appropriate function space, and that
if u satisfies the energy conservation law, it is a weak solution to (
1). In [
1], approximate solutions are constructed by the Ritz-Galerkin method and in [
2,
3] by Rothe’s method. In [
2], the boundary condition is not essentially discussed, and the observation is added in [
3]. In these works, the limit should satisfy the energy conservation law, and existence theorem of a global weak solution has not been established yet. Instead, in [
4], linear approximation for (
1) is established. On the other hand, the equation with the strong viscosity term −Δ
ut is investigated by several authors. For example, in [
5], it is investigated in the context of control theory, and it is asserted that if
and
ut(0) ∈
L2(
Ω), there exists a unique solution
for each
T > 0. Namely, the equation with strong viscosity term is well posed in
W1,2(
Ω), and since
W1,2 is a smaller class than the space of BV functions, this suggests that the influence of the term −Δ
ut is too strong.
In this paper, replacing the strong viscosity term −Δ
ut with
, we investigate it in the space of BV functions. Namely, our problem of this paper is as follows:
()
with initial and boundary conditions
()
()
We should note that the term “viscosity” probably means implying regularity. However, in this paper, we only investigate existence and uniqueness of (
2)–(
4), regularity is not investigated. This is the reason that in the title there is a quotation mark.
A function
u is said to be a
function of bounded variation or a BV
function in
Ω if the distributional derivative
Du is an
Rn valued finite Radon measure in
Ω. The vector space of all functions of bounded variation in
Ω is denoted by BV(
Ω). It is a Banach space equipped with the norm
(see, e.g., [
6–
8]). We should note that, for
u ∈ BV(
Ω), the operator
is multivalued. It is usually defined by the use of the subdifferential of the area functional. Namely, for a function
u ∈ BV(
Ω)∩
L2(
Ω), we regard
as
()
where
()
Here, readers should note that
J(
u) is not
. We are imposing (
4), and in the analysis in the space of BV functions, the most appropriate weak formulation of (
4) is to replace
with
(cf. [
3], see, also [
4, Appendix C]).
Now, we present our definition of a weak solution to (2)–(4).
Definition 1. A function u is a weak solution to (2)–(4) in [0, T) × Ω if u satisfies that
- (i)
u ∈ L∞((0, T); BV(Ω)∩L2(Ω)), ut ∈ L2((0, T) × Ω),
- (ii)
s-lim t↘0u(t, x) = u0(x) in L2(Ω),
- (iii)
there exist f0 ∈ ∂J(u0) and such that for ℒ1-a.e. t (ℒ1 denotes the one-dimensional Lebesgue measure), f1 ∈ ∂J(u) and, for any ,
()
If a function u ∈ L∞((0, ∞); BV(Ω)∩L2(Ω)) is a weak solution to (2)–(4) in [0, T) × Ω for each T > 0, then we say that u is a weak solution to (2)–(4) in [0, ∞) × Ω.
Our main theorem is as follows.
Theorem 2. Suppose that u0 ∈ BV(Ω)∩L2(Ω) and v0 ∈ L2(Ω). We further suppose that J(u0)∩L2(Ω) ≠ ∅. Then, there exists a unique weak solution to (2)–(4) in [0, ∞) × Ω.
Remark 3. If J(u0)∩L2(Ω) ≠ ∅, then the element is unique. Indeed, for each , J(u0 + εϕ) is differentiable at ε = 0 and (d/dε)J(u0 + εϕ)|ε=0 = (f0, ϕ) for each f0 ∈ J(u0)∩L2(Ω). Since f0 ∈ L2(Ω) and ϕ is arbitrary, f0 is uniquely determined.
2. Reduction of the Problem
In order to solve (
2)–(
4), we give a formal observation. Let us put
()
then (
2) becomes
ft = −
f −
g, which can be regarded as an ordinary differential equation to
f. By the variation-of-constants formula, we obtain that
. Noting that
g =
utt, we have
()
where
. Hence, formally, (
2) is reduced to
()
Definition of a weak solution to this equation is as follows.
Definition 4. Let f0 ∈ ∂J(u0), u0 ∈ BV(Ω)∩L2(Ω), and v0 ∈ L2(Ω). A function u is a weak solution to (10) with (3) and (4) in [0, T) × Ω if u satisfies that
- (i)
u ∈ L∞((0, ∞); BV(Ω)∩L2(Ω)), ut ∈ L2((0, T) × Ω),
- (ii)
s-lim t↘0u(t, x) = u0(x) in L2(Ω),
- (iii)
for any ϕ ∈ L2(Ω)∩BV(Ω) and for ℒ1-a.e. t,
()
Similar to the case of (2), we say that u is a weak solution to (10) with (3) and (4) in [0, ∞) × Ω if a function u ∈ L∞((0, ∞); BV(Ω)∩L2(Ω)) is a weak solution to (10) with (3) and (4) in [0, T) × Ω for each T > 0.
The previous observation is just formal. In the following proposition, we show it rigorously.
Proposition 5. Definitions 1 and 4 are equivalent.
Proof. It is sufficient to show that, for each T > 0, a function u is a weak solution to (2)–(4) in [0, T) × Ω if and only if it is a weak solution to (10) with (3) and (4) in [0, T) × Ω.
Suppose that u is a weak solution to (10) with (3) and (4) in [0, T) × Ω. Conditions (i) and (ii) of Definition 1 are the same as those of Definition 4. Thus, we only have to show (iii) of Definition 1. Let
()
Then, by (iii) of Definition
4, we have that
f1 ∈
∂J(
u) for
ℒ1-a.e.
t. Thus, by a direct calculation, we have that
u satisfies (iii) of Definition
1.
Next, we suppose that u is a weak solution to (2)–(4) in [0, T) × Ω. For each ϕ ∈ L∞((0, T); L2(Ω)∩BV(Ω)) and each , we put
()
Then
, and since
ψt(
t) =
ψ(
t) −
ρ(
t)
ϕ(
t), we have the following by (iii) of Definition
1:
()
By integration by parts, we have
()
Furthermore, we have the following by Fubini’s theorem:
()
Finally, noting that
, we have the following by (
14), (
15), and (
16):
()
Since
ϕ and
ρ are arbitrary, we have, for
ℒ1-a.e.
t,
()
which means that
u satisfies Definition
4, (iii).
Now, Theorem 2 is reduced to the following.
Theorem 6. Suppose that u0 ∈ BV(Ω)∩L2(Ω) and v0 ∈ L2(Ω). We further suppose that J(u0)∩L2(Ω) ≠ ∅ and let f0 ∈ J(u0)∩L2(Ω). Then, there exists a unique weak solution to (10), (3), and (4) in [0, ∞) × Ω.
Our strategy of proving Theorem
6 is the contracting mapping theorem. For this purpose, given that
, we solve
()
and show that the map
is a contraction. A weak solution to (
19) with (
3) and (
4) is defined as follows.
Definition 7. Let f0 ∈ ∂J(u0). A function u is a weak solution to (19) with (3) and (4) in [0, T) × Ω if u satisfies that
- (i)
u ∈ L∞((0, T); BV(Ω)∩L2(Ω)), ut ∈ L2((0, T) × Ω),
- (ii)
s-lim t↘0u(t, x) = u0(x) in L2(Ω),
- (iii)
for any ϕ ∈ L2(Ω)∩BV(Ω) and for ℒ1-a.e. t,
()
The proof of Theorem 6 consists of two parts. The first part is solving (19), and the second part is to show that the map is a contraction.
3. Existence and Uniqueness of a Solution to ()
Let u0 and v0 be as in Theorem 6. In this section, we show that there exists a unique solution to (19) with (3) and (4) in [0, T) × Ω for each T > 0.
Uniqueness is easy. Suppose that
u and
v are solutions to (
19) with (
3) and (
4) in [0,
T) ×
Ω, and inserting
ϕ =
v −
u to (iii) of Definition
7, integrating it from 0 to
T, obtaining another inequality by replacing
u and
v, and adding these two inequalities, we have
()
Since u(0) = v(0) = u0, we have the uniqueness of a solution to (19).
It is sufficient to show the existence in [0, T) × Ω for ℒ1-a.e. T. Approximate solutions are constructed by Rothe’s time semidiscretization method. In Rothe’s method, we should solve elliptic equations with respect to space variables. Here, we solve them by a direct variational method (namely, this is the method of discrete Morse semiflow, cf. [9] and references cited therein).
Suppose that
u0 ∈ BV(
Ω)∩
L2(
Ω) with
J(
u0)∩
L2(
Ω) ≠
∅ and
v0 ∈
L2(
Ω), and let
f0 ∈
J(
u0)∩
L2(
Ω). For a positive number
h, we construct a sequence
in the following way. For
l = 0, we let
u0 be as in (
3), and for
l ≥ 1, it is defined as a minimizer of the following functional:
()
in the class
L2(
Ω)∩BV(
Ω), where
()
Since
()
Fl is bounded from below, and hence, the existence of a minimizer of
ℱl follows.
Lemma 8 (energy inequality). consider the following:
()
Proof. Since ul is a minimizer of ℱl, we have
()
Hence, for each
l,
()
Thus, by induction on
l, we have the conclusion.
Next, we define approximate solutions
uh(
t,
x) and
for (
t,
x)∈(−
h,
∞) ×
Ω as follows: for (
l − 1)
h <
t ≤
lh,
()
Then Lemma
8 shows for each
T > 0
()
Now, we estimate the second term of the left hand side of (
29). Then,
()
where
C(
T) = 2
−1(
T + 2
−1e−2T), and thus, it is easy to see that
()
By (
29), we have, for each
ε > 0 and for each
T > 0,
()
where
()
Proposition 9. It holds that
- (1)
is uniformly bounded with respect to h;
- (2)
for any T > 0, is uniformly bounded with respect to h;
- (3)
for any T > 0, is uniformly bounded with respect to h;
-
Then there exist a sequence {hj} with hj → 0 as j → ∞ and a function u such that
- (4)
converges to ut as j → ∞ weakly in L2((0, ∞) × Ω);
- (5)
for any T > 0, converges to u as j → ∞ weakly star in L∞((0, T); L2(Ω));
- (6)
for any T > 0, converges to u as j → ∞ strongly in L∞((0, T); Lp(Ω)) for each 1 ≤ p < 1* = n/(n − 1);
- (7)
for any T > 0, converges to u as j → ∞ strongly in L∞((0, T); Lp(Ω)) for each 1 ≤ p < 1*;
- (8)
u ∈ L∞((0, ∞); BV(Ω));
- (9)
for ℒ1-a.e. t ∈ (0, ∞), converges to Du(t, ·) as j → ∞ in the sense of distributions;
- (10)
s- lim t↘0u(t) = u0 in L2(Ω).
Proof. Assertion (1) immediately follows from (32). Since we have
()
for each
t,
t′ ≥ 0, Assertion (1) implies that, for each
T > 0,
is uniformly bounded with respect to
h. Given that
t > 0, we let
l be an integer such that (
l − 1)
h <
t ≤
lh. Then,
()
By (
32),
()
Thus, we have
()
for each
l, where
. Hence,
()
Now, we have that
is uniformly bounded with respect to
h since
()
Since
C0 is increasing with respect to
T, Assertion (2) follows from (
32). Since
J is convex, we have
()
and Assertion (3) also holds.
Assertion (4) is a direct consequence of Assertion (1). Assertion (5) follows from Assertion (3). Furthermore, (34) and Assertion (1) imply that the function t ↦ uh(t, ·) ∈ L2(Ω) is equicontinuous with respect to h. By Sobolev’s theorem BV(Ω) ⊂ Lp(Ω) compactly for each 1 ≤ p < 1*. This means that, for any T > 0, {uh(t, ·)} is contained in a sequentially compact subset of Lp(Ω) which is independent of h and t ∈ [0, T]. Thus, by the Ascoli-Arzela theorem, we obtain Assertion (6).
Now, we have, for 1 ≤ p < 1*,
()
the right hand side of which converges to 0 as
h → 0 by (
38) and Assertion (6). Now, we have Assertion (7). Assertions (2) and (7) imply Assertions (8) and (9).
Letting t′ = 0 in (34), we have
()
Thus, by Assertion (1) the left hand side is uniformly bounded with respect to
h and, hence, passing to a subsequence if necessary, {
uh(
t) −
u0}
h>0 converges weakly in
L2(
Ω), and by Assertion (6), the weak limit is
u(
t) −
u0. Then, by the lower semicontinuity of
L2 norm, we have
()
which implies Assertion (10).
Now, our purpose is to show that u is a weak solution to (19). Proposition 9 implies that u satisfies (i) and (ii) of Definition 7.
Since
ul is a minimizer of
ℱl(
v), we have
()
Let us write, for (
l − 1)
h ≤
t <
lh,
. Then, for each
j and for
ℒ1-a.e.
t ∈ (0,
∞),
()
namely, for each
v ∈
L2(
Ω),
()
By Proposition
9 (7) we have that, for
ℒ1-a.e.
t ∈ (0,
∞),
()
strongly in
L1(
Ω) as
j →
∞. Let
T be a number
t such that (
46) and (
47) hold. We insert an arbitrary function
v ∈
L2((0,
T) ×
Ω) in (
46). Integrating it from 0 to
T, we have the following by Proposition
9 (4), (7), Fatou’s lemma, and the lower semicontinuity:
()
For a while, we write
hj =
h for simplicity. First, we note the following identity:
()
Let
L be the integer such that (
L − 1)
h <
T ≤
Lh. By (
49), we have
()
By (
42) and Proposition
9 (1),
is uniformly bounded with respect to
t ∈ [0,
T] and
h. On the other hand, since (
L − 1)
h <
T ≤
Lh,
()
Hence, we have by (
37)
()
and thus by (
37) again
()
as
h → 0. By Proposition
9 (1),
()
as
h → 0. Since
,
u(
T) ∈
L2(
Ω), (
47) implies that
in
L2(
Ω). In particular, we have
. Then, since
, we have
()
Summing up, we have
()
It is not difficult to show that
Gh →
G strongly in
L2((0,
T) ×
Ω). Hence, by (
48), we finally have
()
for each
v ∈
L2((0,
T) ×
Ω). By the convexity of
J, for each function
ψ ∈
L2(0,
T) with 0 ≤
ψ ≤ 1, we have
ψJ(
v) −
ψJ(
u) ≥
J(
u +
ψ(
v −
u)) −
J(
u). Thus, (
57) implies that
()
It is easy to extend this inequality to all nonnegative functions
ψ ∈
L2(0,
T). Hence, (iii) of Definition
7 holds for
ℒ1-a.e.
t ∈ (0,
∞).
4. Proof That Is a Contraction
Let
,
be functions in
L2((0,
T) ×
Ω). In this section, we write
()
Then,
()
Let
u,
v be a solution to (
19) with (
3) and (
4) for
,
, respectively. By (iii) of Definition
7,
()
Summing these, we have
()
Integrating from 0 to
t, we have the following by (
60) and by the fact that
u(0) =
v(0) =
u0:
()
We further integrate this from 0 to
T and write it
I +
II +
III. Then,
()
()
Here,
()
Thus,
()
These two terms are estimated as follows:
()
()
Summing up, we have
()
Hence, when 0 <
T < 1/2, putting
()
we have
()
As
T → 0,
K(
T) converges to 0. Thus, if
T is sufficiently small, 0 <
K(
T) < 1. This means that the map from
to
u is a contraction in
L2((0,
T) ×
Ω). Hence, there is a fixed point
and it is a solution to (
10) with (
3) and (
4) in [0,
T) ×
Ω.
End of the proof of Theorem 6.
Uniqueness of a Local Solution. Let
,
be solutions to (
10) with (
3) and (
4). Then, in the same calculus as before, we obtain
()
This implies the uniqueness.
Existence of a Time Global Solution. Suppose that
is a solution to (
10) with (
3) and (
4) in [0,
s]. First, we remark that
()
and the right hand side belongs to
L2(
Ω). Hence, we are able to solve (
10) with (
3) and (
4) from
t =
s. By the change of variable
, (
10) becomes
()
where
()
Hence, solving (
10) with (
4) and initial condition
,
, we obtain a function
u in [0,
T] ×
Ω which solves (
10). For
t ∈ [
s,
s +
T], put
. Then,
is a solution to (
10) with (
3) and (
4) in [0,
s +
T] ×
Ω. Repeating this process, we obtain a time global solution. Uniqueness of the local solution implies uniqueness of the global solution.
5. Uniform Estimates
Let
T be the small number presented in the previous section, and let
u be a solution to (
10) with (
3) and (
4) in [0,
T) ×
Ω. As we see in the previous section, it is obtained as a fixed point of the map
in
L2((0,
T) ×
Ω), where
u is a solution to (
19) with (
3) and (
4). The fixed point is obtained as in the following way. Let
u(0) be an arbitrary element of
L2((0,
T) ×
Ω) and put
τn(
u(0)) =
u(n). There is a constant
σ which is determined by
T such that 0 <
σ < 1 and
. Thus, we have
()
and hence,
()
namely, {
u(n)} is a Cauchy sequence in
L2((0,
T) ×
Ω) and it converges to a function
u. Since
u(n) =
τ(
u(n−1)), we have that
u =
τ(
u) by letting
n →
∞. Thus,
u is the fixed point of
τ.
Letting
m = 0 in (
78), we have
()
By the lower semicontinuity, an energy inequality for a solution to (
19) with (
3) and (
4) is obtained by letting
h → 0 in (
32):
()
Since
C0 is increasing with respect to
T, we have ess. sup
0<t≤TJ(
u(
t, ·)) ≤
C0(
T). Hence,
()
Recall that
C0 is presented as in (
33). By the proof of Proposition
9 (5), we have
()
Letting
n →
∞ in (
79), we finally have
()
Now, a solution
u to (
10) is a solution to (
19) for
, the fixed point of
τ. Hence, by (
81), we have
()
By (
83) and (
84), there exists a constant
C3 such that
()
Now, by (
83) and (
85), we obtain uniform estimates for
,
,
for the small
T.
Solving (
10) from
t =
s (0 <
s <
T), we have the following by (
83):
()
where
Us is as in (
76). By (
81) and Chebyshev’s inequality, for sufficiently large
R,
. Let
δ be an arbitrary small positive number and put
. Then, there exists an
s ∈ [
T −
δ,
T) = [
T − (1/
R2)(4
C0/(1 −
ε),
T)) such that
. Hereby, we have by (
83) and (
85) that, for such an
s, there exists a constant
C4 such that
()
Repeating this process, we have uniform estimates for
,
,
for each
. However, their upper bounds depend on
.