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**Example text**

Therefore, λi (φ(y)) ≈ λi− , and the solution indeed is about H (x − y, t − τ ; λi− ). For the more complicated case y < 0 and x > 0, G i is about H (xi− − y, t − τ ; λi− ). Here we still solve the dual equation in the left quarter plane. The initial point (x, t), however, is in the right quarter plane. This means that initially (τ ≈ t) the solution is about H (x − y, t − τ ; λi+ ), restricted to the left quarter plane y < 0. As τ goes backward (becomes smaller), the center of the heat kernel may or may not enter the shock layer, depending on whether i > p or i < p.

29) is the right one. 14. For that we will need the following lemma. 6. 36) where |λ| > 0 is a parameter. Proof. 36) is obtained by direct calculation. Next, we consider the transversal fields i = p. 28) is roughly H (x − y, t − τ ; λi− ). 11) in the left quarter plane with the initial point (x, t) in it. Therefore, λi (φ(y)) ≈ λi− , and the solution indeed is about H (x − y, t − τ ; λi− ). For the more complicated case y < 0 and x > 0, G i is about H (xi− − y, t − τ ; λi− ). Here we still solve the dual equation in the left quarter plane.

Liu, Y. Zeng When we apply Duhamel’s principle in the next section, the above lemmas are not sufficient to handle the leading terms in the nonlinear ources, and we need to take cancellation into account. 6 in the next section. 4. Let the constants 1 ≤ α < 3, µ > 0, µ > 0, λ = λ , and C¯ ≥ |λ − λ |. Let k ≥ 0 be an integer. 24) where µ∗ > max(µ, µ ) is arbitrarily chosen, and char is the characteristic function. Next we discuss the dissipation of damping waves. Define ∞ t2 Kα,β (x, t; t1 , t2 ; ε, λ, µ) ≡ t1 −∞ (t − τ )− α−α −1 2 α (t − τ + 1)− 2 β H (x − y, t − τ ; λ, µ)(τ + 1)− 2 e−ε|y|/µ dydτ, ∞ t Lα,β (x, t; ε, µ) ≡ −∞ 0 (t − τ ) − α−α2 −1 (t − τ + 1) β H (x − y, t − τ ; −ε, µ)(τ + 1)− 2 e−ε|y|/µ dydτ, t Mα,β (x, t; ε, µ) ≡ 0 ∞ −∞ (t − τ ) − α−α2 −1 (t − τ + 1) β ⎧ ⎪ ⎨1 β (t) ≡ log(t + 1) ⎪ 2−β ⎩ (t + 1) 2 if β > 2 if β = 2 .