On the Cauchy Problem for a Class of Weakly Dissipative One-Dimensional Shallow Water Equations

Theorem 1. Given u₀ ∈ H^s(ℝ), s > 3/2, then there exist a T and a unique solution u to (2) such that

Theorem 2. Let a − 2b > 0, b > 0: suppose that u₀ ∈ H²(ℝ), and there exists a x₀ ∈ ℝ such that $$,

Then the corresponding solution u(x, t) to (2) with u₀ as the initial datum blows up in finite time.

Proof. Suppose that the solution exists globally. From (8) and initial condition (10), we have y(q(x₀, t), t) = 0 and

for all t ≥ 0. Due to u(x, t) = G*y(x, t), we can write u(x, t) and u_x(x, t) as Therefore, for all t > 0.

By direct calculation, for x ≤ q(x₀, t), we have

Similarly, for x ≥ q(x₀, t), we have So for any fixed t, combination of (15) and (16), we obtain for all x ∈ ℝ.

From the expression of u_x(x, t) in terms of y(x, t), differentiating u_x(q(x₀, t), t) with respect to t, we have

where we have used (17), and the inequality $$. In addition, we also used the equation $$, which is obtained by differentiating equation (3).

For (11), we know that

Claim. u_x(q(x₀, t), t) < 0 is decreasing. (u(q(x₀, t), t) + λ/b) ² < (u_x(q(x₀, t), t) + λ/b) ² and (u(q(x₀, t), t) − λ/b) ² < (u_x(q(x₀, t), t) + λ/b) ², for all t ≥ 0.

Suppose that there exists a t₀ such that (u(q(x₀, t), t) + λ/b) ² < (u_x(q(x₀, t), t) + λ/b) ² and (u(q(x₀, t), t) − λ/b) ² < (u_x(q(x₀, t), t) + λ/b) ² on [0, t₀); then (u(q(x₀, t₀), t₀) + λ/b) ² = (u_x(q(x₀, t₀), t₀) + λ/b) ² or (u(q(x₀, t₀), t₀) − λ/b) ² = (u_x(q(x₀, t₀), t₀) + λ/b) ².

Now, let

Firstly, differentiating I(t), we have Secondly, by the same argument, we obtain Therefore, it follows from (21), (22), and the continuity property of ODEs that for all t > 0. This implies that t₀ can be extended to the infinity.

Moreover, using (21) and (22) again, we have the following equation for [2(u_x + λ/b) ² − (u + λ/b) ² − (u−λ/b)²](q(x₀, t), t):

where we use u_x(q(x₀, t), t) = −I(t) + II(t).

Now, recalling (18), we have

Putting (25) into (24), it yields

Before finishing the proof, we need the following technical lemma.

Lemma  3 (see [15]). Suppose that Ψ(t) is twice continuously differential satisfying

Then ψ(t) blows up in finite time. Moreover the blow-up time can be estimated in terms of the initial datum as

Let $$; then (26) is an equation of type (27) with C₀ = b²/4. The proof is complete by applying Lemma 3.

Remark 4. When b = 1, Theorem 2 reduces to the result in [19].

Theorem 5. Let a = 2b > 0. Suppose that u₀ ∈ H²(ℝ) and there exists a x₀ ∈ ℝ such that $$,

Then the corresponding solution u(x, t) to (2) with u₀ as the initial datum blows up in finite time.

Proof. We easily obtain

Differentiating u_x at the point (q(x₀, t), t) with respect to t, we get

Process of the proof is similar to Theorem 2. Thus to be concise, we omit the detailed proof.

Theorem 6. When a = 2b > 0 and λ = 0, then the nonlinear wave equation (2) breaks if and only if some portion of the positive part of y₀(x) lies to the left of some portion of its negative part.

Proof. As studied in [1], when a = 2b > 0 and λ = 0, rewriting (2) yields

Recalling Mckean’s theorem in [7], (32) breaks if and only if some portion of the positive part of $$ lies to the left of some portion of its negative part.

So (34) breaks if and only if some portion of the positive part of $$ lies to the left of some portion of its negative part.

This completes the proof.

Remark 7. Mckean’s theorem [7] is for the special case a = 2, b = 1. Condition a = 2b here is more general. However, the necessary and sufficient condition for (2) is still a challenging problem for us at present.

Theorem 8. Suppose that u₀ ∈ H³(ℝ), and $$ is one sign. Then the corresponding solution to (2) exists globally.

Proof. We can assume that y₀ ≥ 0. It is sufficient to prove that u_x(x, t) has a lower and upper bound for all t. In fact,

Therefore, we have This completes the proof.

Theorem 9. Let 0 < a ≤ 3b. Assume that for some T ≥ 0 and s ≥ 5/2, u ∈ C([0, T); H^s(ℝ)) is a strong solution of (2). If u₀(x) = u(x, 0) has compact support [a,c], then for t ∈ (0, T), one has

where L(t) and l(t) denote continuous nonvanishing functions, with L(t) > 0 and l(t) < 0 for t ∈ (0, T). Furthermore, L(t) is strictly increasing function, while l(t) is strictly decreasing function.

Proof. Since u₀ has compact support in x in [a, c], from (8), so does y(, t) has compact support in x in [q(a, t), q(c, t)] in its lifespan. Hence the following functions are well-defined:

with Thus, for x > q(c, t), we obtain Similarly, for x < q(a, t), we have Hence, as consequences of (40) and (41), we get On the other hand, It is easy to get Putting the identity (44) into dE(t)/dt, we have where we have used (42).

Therefore, in the lifespan of the solution, we get

By the same argument, one can check that the following identity for F(t) is true:

In order to complete the proof, it is sufficient to let L(t) = (1/2)E(t) and l(t) = (1/2)F(t), respectively.