General Asymptotic Supnorm Estimates for Solutions of One-Dimensional Advection-Diffusion Equations in Heterogeneous Media

Theorem 1. If$$ solves problem (1a), (1b), then u(·, t) ∈ C⁰([0, T_*[, L^q(ℝ)) for each p₀ ≤ q < ∞, and

for all 0 < t < T_*.

Proof. The proof is standard, so we will only sketch the basic steps. Taking S ∈ C¹(ℝ) such that S^′(v) ≥ 0 for all v, S(0) = 0,  S(v) = sgn⁡(v) for   | v | ≥ 1, let (given δ > 0)$$, so that  L_δ(u)→|u| as δ → 0, uniformly in  u. Let  Φ_δ(u) = L_δ(u) ^q. Given R > 0, 0 < ϵ ≤ 1, let ζ_R(·) be the cut-off function ζ_R(x) = 0 for |x| ≥ R, $$ for |x| < R. Multiplying (1a) by $$ if q ≠ 2, or u(x, t) · ζ_R(x) if q = 2, and integrating the result on ℝ × [0, t], we obtain, letting δ → 0 and then R → ∞, since u ∈ L^∞(ℝ × [0, t]),

where $$, $$+ϵ, and By Gronwall’s lemma, (14a) and (14b) give$$, from which we obtain (13) by simply letting ϵ → 0. This shows, in particular, that$$ if p₀ ≤ q < ∞. Now, to get u(·, t) ∈ C⁰([0, T_*[, L^q(ℝ)), it is sufficient to show that, given ɛ > 0 and 0 < T < T_* arbitrary, we can find R   = R(ɛ, T) ≫ 1 large enough so that we have $$ for any 0 ≤ t ≤ T. Taking ψ ∈ C²(ℝ) with 0 ≤ ψ ≤ 1 and ψ(x) = 0 for all x ≤ 0, ψ(x) = 1 for all x ≥ 1, let Ψ_R,M ∈ C²(ℝ) be the cut-off function given by Ψ_R,M(x) = 0 if |x | ≤ R − 1, Ψ_R,M(x) = ψ(|x | − R + 1) if R − 1 < |x| < R, and Ψ_R,M(x) = 1 if R≤|x | ≤ R + M, Ψ_R,M(x) = ψ(R + M + 1−|x|) if R + M<|x | < R + M + 1, Ψ_R,M(x) = 0 if |x | ≥ R + M + 1, where R > 1, M > 0 are given. Multiplying (1a) by $$ if q ≠ 2, or u(x, t) · Ψ_R,M(x) if q = 2, and integrating the result on ℝ × [0, t], 0 < t ≤ T, we obtain, as in (14a) and (14b), by letting δ → 0, M → ∞, that $$ for all 0 ≤ t ≤ T, provided that we take R > 1 sufficiently large. This gives the continuity result, and the proof is complete.

Theorem 2. Let q ≥ 2p₀. If $$ is such that $$, then

Proof. Consider (20a) first. From (5), (17), and (18), we have

This gives or, in terms of $$ defined by $$ if q > 2, $$ if q = 2, Using (19), we then get $$, which is equivalent to (20a). Similarly, (20b) can be obtained, using (9).

Theorem 3. Let q ≥ 2p₀. For each 0 ≤ t₀ < T_*, we have

for all t₀ ≤ t < T_*.

Proof. Set $$. There are three cases to consider.

Case I. $$ for all t₀ ≤ τ ≤ t. By (20a), Theorem 2, we must then have $$ for all τ ∈ [t₀, t]∖E_q, so that $$ is monotonically decreasing in [t₀, t]. In particular, $$ in this case, and (26) holds.

Case II. $$ and $$ for some t₁ ∈   ]t₀, t]. In this case, let t₂ ∈   ]t₀, t] be such that we have $$ for all t₀ ≤ τ < t₂, while $$. We claim that $$ for every t₂ ≤ τ ≤ t: in fact, if this were not true, we could then find t₃, t₄ with t₂ ≤ t₃ < t₄ ≤ t such that $$ for all t₃ < τ ≤ t₄, $$. By (20a), Theorem 2, this would require $$ for all τ ∈  ]t₃, t₄]∖E_q, so that $$ could not increase anywhere on [t₃, t₄]. This contradicts $$, and so we have $$ for every t₂ ≤ τ ≤ t, as claimed. On the other hand, by (20a), $$ has to be monotonically decreasing on [t₀, t₂], just as in Case I. Therefore, we have $$ in this case again, which shows (26).

Case III. Consider $$. This gives $$ for every t₀ ≤ τ ≤ t, by repeating the argument used on the interval [t₂, t] in Case II. It follows that we must have 𝕌_q(t₀; t) ≤ λ_q(t) in this case, and the proof of Theorem 3 is complete.

Theorem 4. Let p₀ ≤ p < ∞, 0 ≤ t₀ < T_*. Then

for any t₀ ≤ t < T_*, where 𝔹(t₀; t) and 𝕌_p(t₀; t) are given in (24) and (25) above.

Proof. Let k ∈ ℤ, k ≥ 2. Applying (26) successively with q = 2p, 4p, …, 2^kp, we obtain

where Now, for 1 ≤ ℓ ≤ k − 1, by Young’s inequality (see, e.g., [11, page 622]); in particular, we get, from (28a) and (28b), since K(k, ℓ) ≤ 2p for all 0 ≤ ℓ ≤ k − 1. Letting k → ∞, (27) is obtained.

Theorem 5. Let q ≥ 2p₀, and ℬ   ≥ 0 be as defined in (35). Then

where $$ is the constant in the Nash inequality (19).

Proof. We set p = q/2 and assume that 𝒰_p is finite. As in the proof of Theorem 2, we take v ∈ L^∞(ℝ × [0, ∞[) given by v(x, t) = |u(x, t)|^p if p > 1, v(x, t) = u(x, t) if p = 1. It follows that

Therefore, from (18), we have, for some null set E_2p ⊂ [0, ∞[, for all t ∈ [0, ∞[∖E_2p, and so, by (19), This gives, by Young’s inequality ([11, page 622]), for all   t ∈ [0, ∞[∖E_2p, Setting we claim that In fact, let us argue by contradiction. If (42) is false, we can pick 0 < η ≪ 1 and a sequence (t_j) _j≥0, t_j → ∞, such that $$ (for all j ≥ 0) and g(t) ≤ λ_p + η/2 for all t ≥ t₀. From (20a), Theorem 2, it will then follow that In fact, suppose that (43) were false, so that we had $$ for some $$. Taking j ≫ 1 with $$, we could then find $$ such that $$ for all $$, while $$, and so there would exist $$ with $$ positive at t = t_*. By (20a), we would have $$, but this would contradict the fact that $$ everywhere on $$. Thus, we conclude that (43) cannot be false, as claimed. We then obtain, from (19), (40), and (43), for all t ∈ [t₀, ∞[∖E_2p. Recalling that $$, g(t)   ≤ λ_p   + η/2, for  all t ≥ t₀, this gives for some constant K(η) > 0 independent of t, which cannot be, since this implies This contradiction shows (42), which is equivalent to (36), and the proof is complete.

Theorem 6. Let p ≥ p₀. Then

where C₂, C_∞ are the constants given in (19) and (9).

Proof. Again, assuming 𝒰_p finite (otherwise, (49) is obvious; cf. endnote⁴), we introduce, as in the previous proof, v ∈ L^∞(ℝ × [0, ∞[)   given by v(x, t) = |u(x, t)|^p if p > 1, and v(x, t) = u(x, t) if p = 1. Thus, (40) is valid, and setting λ_p ∈ ℝ, g ∈ L^∞([0, ∞[) by

we have that (49) is obtained if we show that We argue by contradiction and assume that (51) is false. Taking then 0 < η ≪ 1, t₀ ≫ 1 so that $$ and g(t) ≤ λ_p + η/2 hold for all t ≥ t₀, we get, by (9) and (40), for all t ∈ [t₀, ∞[∖E_2p. Since $$, g(t) ≤ λ_p + η/2, this gives for some constant K(η) > 0 independent of t. As before, this implies that $$ for all t ≥ t₀, which is impossible because $$ is finite. This contradiction establishes (51) above, completing the proof of Theorem 6.

Theorem 7. Let p ≥ p₀. Assuming b ∈ L^∞(ℝ × [0, ∞[), then (6), (48a), and (48b) hold.