The Local Strong Solutions and Global Weak Solutions for a Nonlinear Equation

Kato Theorem (see [14].)Assume that conditions (i), (ii), and (iii) hold. If v₀ ∈ Y, there is a maximal T > 0 depending only on $$ and a unique solution v to the problem (7) such that

Moreover, the map v₀ → v(·, v₀) is a continuous map from Y to the following space:

Lemma 1. The operator A(u) = u²∂_x with u ∈ H^s(R), s > 3/2 belongs to G(H^s−1(R), 1, β).

Lemma 2. Assume that H(u) = 3u²∂_x with u ∈ H^s(R) and s > 3/2. Then, H(u) ∈ L(H^s(R), H^s−1(R)) for all u ∈ H^s(R). Moreover,

Lemma 3. For s > 3/2, u, z ∈ H^s(R) and w ∈ H^s−1, it holds that A(u) = [Λ^s, 3u²∂_x]Λ^−s ∈ L(H^s−1) for u ∈ H^s and

Lemma 4. Let u, z ∈ H^s with s > 3/2 and g(u) = (m − 1)Λ⁻²∂_x(u³). Then, g is bounded on bounded sets in H^s and satisfies

Proof. For s₀ > 1/2, we know that $$. Consequently, we have

Theorem 5. Assume that u₀ ∈ H^s(R) with s > 3/2. Then, there exists a T > 0 such that the system (5) or the problem (6) has a unique solution u(t, x) satisfying

Lemma 6. The solution of the problem (5) with m > 0 satisfies

where $$, and $$. Moreover, there exist two constants c₁ > 0 and c₂ > 0 depending only on m such that

Proof. Setting $$ and $$ and using the first equation of the problem (5), we obtain u = my − y_xx and

Using the Parseval identity and (21), we obtain (19) and (20).

Lemma 7. Let u₀ ∈ H^s, s > 3, and let T > 0 be the maximal existence time of the solution to the problem (6). Then, the problem (23) has a unique solution p ∈ C¹([0, T) × R, R). Moreover, the map p(t, ·) is an increasing diffeomorphism of R with p_x(t, x) > 0 for (t, x)∈[0, T) × R.

Proof. Using Theorem 5, we obtain u ∈ C¹([0, T); H^s−1(R)) and H^s−1 ∈ C¹(R). Therefore, we know that functions u(t, x) and u_x(t, x) are bounded, Lipschitz in space, and C¹ in time. Using the existence and uniqueness theorem for ordinary differential equations derives that the problem (23) has a unique solution p ∈ C¹([0, T) × R, R).

Differentiating (23) with respect to x gives rise to the following:

from which we obtain For every T^′ < T, using the Sobolev imbedding theorem yields that

It is inferred that there exists a constant K₀ > 0 such that $$ for (t, x)∈[0, T) × R. It completes the proof.

Lemma 8. Assume that u₀ ∈ H^s(R), s > 3/2. Let T be the maximal existence time of the solution u to the problem (6). Then, it has

where c > 0 is a constant independent of t.

Proof. Let ξ(x) = (1/2)e^−|x|, we have $$ for all g ∈ L²(R) and u = ξ⋆K₁(t, x). Using a simple density argument presented in [7], it suffices to consider s = 3 to prove this lemma. Let T be the maximal existence time of the solution u to the problem (6) with the initial value u₀ ∈ H³(R) such that u ∈ C([0, T), H³(R))∩C¹([0, T), H²(R)). From (6), we have

Since from (29), we have Using Lemma 6 and (30) derives that where c is a positive constant independent of t. Using (31) results in the following: Therefore, Using the Sobolev embedding theorem to ensure the uniform boundedness of u_x(s, η) for (s, η)∈[0, t] × R with t ∈ [0, T^′), from Lemma 7, for every t ∈ [0, T^′), we get a constant C(t) such that We deduce from (34) that the function p(t, ·) is strictly increasing on R with lim _x→±∞p(t, x) = ±∞ as long as t ∈ [0, T^′). It follows from (33) that Using the Gronwall inequality and (35) derives that (27) holds.

Definition 9 (weak solution). We call a function u : R₊ × R → R a weak solution of the Cauchy problem (5) provided that

• (i)

  u ∈ L^∞(R₊; L²(R));

• (ii)

  $$ in D^′([0, ∞) × R), that is, for all $$ there holds the following identity:

  $$ ()

Theorem 10. Let u₀(x) ∈ L²(R). Then, there exists a weak solution u(t, x) ∈ L^∞([0, ∞); L²(R)) to the problem (5).

Proof. Consider the problem (36). For an arbitrary T > 0, choosing a subsequence ε_n → 0, from (37), we know that $$ is bounded in L^∞ and $$ is uniformly bounded in L²(R). Therefore, we obtain that $$ is bounded in L²(R). Therefore, there exist subsequences $$ and $$, still denoted by $$ and $$, are weakly convergent to v in L²(R). Noticing (38) completes the proof.