**3. Asymptotic stability with convergence rates**

6 Will-be-set-by-IN-TECH

<sup>|</sup>*φ*(*x*1)| ≤ *<sup>C</sup>*(<sup>1</sup> <sup>+</sup> *<sup>x</sup>*1)−1/*q*, *<sup>x</sup>*<sup>1</sup> <sup>&</sup>gt; 0,

In this chapter we only treat the stationary solutions *φ*(*x*1) with *φx*<sup>1</sup> > 0 and discuss their stability; however, we must get the similar stability result of the monotone decreasing stationary solutions by using the same argument introduced in this chapter. (We refer the

In this subsection we reformulate our problem by the anti-derivative method. To this end we

 ∞ *x*1

Here, we assume the integrability of *v*(*x*, *t*) over **R**+. This transformation is motivated by the argument in Liu-Nishihara [9]. By using (2.4), we can reformulate (1.8)–(1.10) in terms of

*φx*<sup>1</sup> *f* ��

*zx*<sup>1</sup> (0, *x*�

*v*(*y*, *x*�

<sup>∗</sup> (*φ*)=(*<sup>f</sup>* ��

Once we obtain the solution for the problem (2.5)–(2.7), the differentiation *v* = *zx*<sup>1</sup> is the solution for (1.8)–(1.10). Namely, we will apply the weighted energy method in the partial Fourier space and try ot derive the global solution in time to the reformulated problem (2.5)–(2.7). We will discuss this reformulated problem in Section 3 to prove our main theorems.

We introduce the weight function employed in the weighted energy method. Our weight

*<sup>j</sup>*(*φ*)*zx*<sup>1</sup> , *hj* =

*<sup>w</sup>*(*u*)=(−*eAu* <sup>+</sup> <sup>1</sup>)/ *<sup>f</sup>*1(*u*) for *<sup>u</sup>* <sup>∈</sup> [*ub*, 0], (2.8)

, *t*) *dy*. (2.4)

<sup>∗</sup> (*φ*) · ∇*x*� *z dy* = −*g*<sup>1</sup> + ∇*x*� · *<sup>h</sup>*∗, (2.5)

, *t*) = 0, (2.6)

*gj dy*.

*<sup>n</sup>* (*φ*)), and *g*1, ∇*x*� · *h*<sup>∗</sup> are

*z*(*x*, 0) = *z*0(*x*), (2.7)

 ∞ *x*1

<sup>2</sup> (*φ*), ··· , *f* ��

*z*(*x*, *t*) = −

 ∞ *x*1

) − *φ*(*y*))*dy*, *f* ��

*gj* = *fj*(*φ* + *zx*<sup>1</sup> ) − *fj*(*φ*) − *f* �

(*φ*) · ∇*z* +

*solution φ*(*x*1) *if and only if ub* < 0*. The solution verifies φx*<sup>1</sup> > 0 *and*

*where q is the degeneracy exponent of f*<sup>1</sup> *and C is a positive constant.*

<sup>1</sup>(0) = 0*: In this case the problem* (1.6)*–*(1.7) *admits a unique smooth*

*where C and c are positive constants.*

(ii) *Degenerate case where f* �

**2.3 Reformulated problem**

introduce a new function *z*(*x*, *t*) as

*zt* − Δ*z* + *f* �

*<sup>x</sup>*<sup>1</sup> (*u*0(*y*, *x*�

nonlinear terms defined by *h*<sup>∗</sup> = (*h*2, ··· , *hn*) and

reader to [2].)

*z*(*x*, *t*) as

where *<sup>z</sup>*0(*x*) = <sup>−</sup> <sup>∞</sup>

**2.4 Weight function**

function is defined as

In the final section, we apply our weighted energy method in the partial Fourier space to the linearized problem. We consider the linearized problem corresponding to the half space problem (2.5)–(2.7). Namely, we consider (2.5) with *gj* = 0 for *j* = 1, ··· , *n*. For this linearized equation, we treat the special situation that

$$\text{"}\,^r f\_j(\mu) \text{ are linear in } \mu \in [\mathfrak{u}\_b, 0] \text{ for } j = 2, \dots, n. \text{"}\,^r$$

Then our initial value problem of the linearized equation is written as

$$z\_t - \Delta z + f'(\phi) \cdot \nabla z = 0 \tag{3.1}$$

together with (2.6) and (2.7). Taking the Fourier transform with respect to *<sup>x</sup>*� <sup>∈</sup> **<sup>R</sup>***n*−<sup>1</sup> for the linearized problem (3.1), (2.6), (2.7), we obtain

$$\begin{aligned} \dot{z}\_t - \hat{z}\_{\mathbf{x}\_1 \mathbf{x}\_1} + |\mathring{\xi}|^2 \hat{z} + f\_1'(\phi)\hat{z}\_{\mathbf{x}\_1} + i\tilde{\xi} \cdot f\_\*'(\phi)\hat{z} &= \mathbf{0}, \\ \dot{z}\_{\mathbf{x}\_1}(0, \xi, t) &= 0, \\ \dot{z}(\mathbf{x}\_1, \xi, 0) &= \hat{z}\_0(\mathbf{x}\_1, \xi), \end{aligned} \tag{3.2}$$

where *<sup>ξ</sup>* = (*ξ*2, ··· , *<sup>ξ</sup>n*) <sup>∈</sup> **<sup>R</sup>***n*−<sup>1</sup> *<sup>ξ</sup>* is the Fourier variable corresponding to *x*� = (*x*2, ··· , *xn*) ∈ **R***n*−1, *f* � <sup>∗</sup>(*φ*)=(*<sup>f</sup>* � <sup>2</sup>(*φ*), ··· , *f* � *<sup>n</sup>*(*φ*)), and *ξ* · *f* � <sup>∗</sup>(*φ*) = <sup>∑</sup>*<sup>n</sup> <sup>j</sup>*=<sup>2</sup> *ξ<sup>j</sup> f* � *<sup>j</sup>*(*φ*). This is the formulation of our linearized problem in the partial Fourier space **R**ˆ *<sup>n</sup>* <sup>+</sup> <sup>=</sup> **<sup>R</sup>**<sup>+</sup> <sup>×</sup> **<sup>R</sup>***n*−<sup>1</sup> *<sup>ξ</sup>* .

(ii) *Degenerate case where f* �

*yields that* (3.7)*–*(3.9) *with α* = 1*.*

we see the following properties.

*c*|·|*L*<sup>2</sup>

*w*(*φ*)¯

where

*c*|·|*L*<sup>2</sup> ≤|·|*L*<sup>2</sup>

<sup>1</sup> ≤|·|*L*<sup>2</sup>

*z*ˆ and take the real part, obtaining

<sup>1</sup>(0) = <sup>0</sup>*: Suppose that z*<sup>0</sup> <sup>∈</sup> *<sup>L</sup>*<sup>2</sup>

where *w* is the weight function defined by (2.8). For this weighted norm, by using Lemma 2.2,

<sup>257</sup> Application of the Weighted Energy Method in the Partial Fourier Space to Linearized Viscous Conservation Laws with Non-Convex Condition

We prove (i) and (ii) in Theorem 3.1 in parallel. We first derive (3.4). We multiply (3.2)1 by

2|*z*ˆ|

<sup>2</sup> <sup>−</sup> *<sup>w</sup>*(*φ*)Re(¯

<sup>2</sup> <sup>+</sup> <sup>|</sup>*ξ*<sup>|</sup>

2|*z*ˆ| 2 *L*2 *<sup>w</sup>* + |

(*c*1*D*<sup>ˆ</sup> <sup>1</sup> <sup>−</sup> *<sup>κ</sup>*|*ξ*<sup>|</sup>

Here, by virtue of (2.9)1, the last term of the left-hand side of (3.13) is positive. We multiply

<sup>2</sup>) <sup>−</sup> <sup>1</sup> 2

2|*z*ˆ|

(*w f*1)��(*φ*)*φx*<sup>1</sup> |*z*ˆ|

*z*ˆ*z*ˆ*x*<sup>1</sup> ),

(*ub*)|*z*ˆ(0, *ξ*, *t*)|

<sup>2</sup>) + *<sup>c</sup>φx*<sup>1</sup> <sup>|</sup>*z*ˆ<sup>|</sup>

*φx*<sup>1</sup> *<sup>z</sup>*ˆ<sup>|</sup> 2

> 2|*z*ˆ| 2 *L*2 *w*

, we choose *κ* > 0 such that *κ* < *c*<sup>1</sup> and integrate (3.15) over [0, *t*].

<sup>2</sup>*τD*<sup>ˆ</sup> <sup>1</sup>(*ξ*, *<sup>τ</sup>*)*d<sup>τ</sup>* <sup>≤</sup> *<sup>C</sup>*|*z*ˆ0(·, *<sup>ξ</sup>*)<sup>|</sup>

*<sup>w</sup>* ≤ *C*|·|*L*<sup>2</sup> for the non-degenerate case : *f* �

<sup>1</sup> for the degenerate case : *<sup>f</sup>* �

*Proof of Theorem 3.1.* Throughout this proof, we use the weighted *L*<sup>2</sup> norm:

 <sup>∞</sup> 0 *w φ*(*x*1) |*a*(*x*1)|


*<sup>w</sup>* ≤ *C*|·|*L*<sup>2</sup>

1 2

D<sup>1</sup> = *w*(*φ*)(|*z*ˆ*x*<sup>1</sup> |

<sup>F</sup><sup>1</sup> <sup>=</sup> <sup>1</sup> 2 (*w f*1)�

> *∂ ∂t* |*z*ˆ| 2 *L*2

with a positive constant *c*1, where

<sup>2</sup>*<sup>t</sup>* (*κ* > 0) to get

*∂ ∂t* (*e κ*|*ξ*| 2*t* |*z*ˆ| 2 *L*2 *w* ) + *e κ*|*ξ*| 2*t*

2|*z*ˆ| 2 *L*2 *w*


2 *L*2 *<sup>w</sup>* + *t* 0 *e κ*|*ξ*|

*e κ*|*ξ*| 2*t*

(3.13) by *eκ*|*ξ*<sup>|</sup>

This yields

Noting that *<sup>D</sup>*<sup>ˆ</sup> <sup>1</sup> ≥ |*ξ*<sup>|</sup>

*<sup>w</sup>*(*φ*) *<sup>∂</sup> ∂t* |*z*ˆ| <sup>2</sup> + *∂ ∂x*<sup>1</sup>

<sup>2</sup> <sup>+</sup> <sup>|</sup>*ξ*<sup>|</sup>

and *w* is a weight function defined by (2.8). By virtue of (2.9)2 in Lemma 2.2, we have

where *c* is a some positive constant. Therefore, integrating (3.11) in *x*<sup>1</sup> ∈ **R**+, we get

*<sup>w</sup>* <sup>+</sup> *<sup>c</sup>*1*D*<sup>ˆ</sup> <sup>1</sup> <sup>−</sup> (*w f*1)�

2 *L*2 *<sup>w</sup>* + |*ξ*|

(*φ*)|*z*ˆ|

D<sup>1</sup> ≥ *cw*(*φ*)(|*z*ˆ*x*<sup>1</sup> |

*<sup>D</sup>*<sup>ˆ</sup> <sup>1</sup> <sup>=</sup> <sup>|</sup>*z*ˆ*x*<sup>1</sup> <sup>|</sup>

<sup>1</sup>(*L*1) *and* (*z*0)*x*<sup>1</sup> <sup>∈</sup> *<sup>H</sup>*1(*L*1)*. Then this*

(0) < 0,

(3.10)

(0) = 0.

F<sup>1</sup> + D<sup>1</sup> = 0, (3.11)

2,

2, (3.12)

<sup>2</sup> <sup>≤</sup> <sup>0</sup> (3.13)

*<sup>L</sup>*<sup>2</sup> . (3.14)

) ≤ 0. (3.15)

, (3.16)

2 *L*2 *w*

<sup>2</sup>*dx*<sup>1</sup> 1/2 ,

Furthermore, we sometimes use the differentiated problem. We differentiate the problem (3.1), (2.6), (2.7) with respect to *x*1. Then this yields our problem (1.15) together with (1.9) and (1.10), and the corresponding problem in the partial Fourier space:

$$\begin{aligned} \mathfrak{d}\_t - \mathfrak{d}\_{\mathbf{x}\_1 \mathbf{x}\_1} + |\mathfrak{f}|^2 \mathfrak{d} + (f\_1'(\mathfrak{g}) \mathfrak{d})\_{\mathbf{x}\_1} + i \mathfrak{f} \cdot f\_\*'(\mathfrak{g}) \mathfrak{d} &= \mathbf{0}, \\ \mathfrak{d}(\mathfrak{d}\_\bullet \mathfrak{f}, t) &= \mathbf{0}, \\ \mathfrak{d}(\mathbf{x}\_1, \mathfrak{f}, \mathbf{0}) &= \mathfrak{d}\_0(\mathbf{x}\_1, \mathfrak{f}). \end{aligned} \tag{3.3}$$

Here we note that *v* = *zx*<sup>1</sup> . By applying the weighted energy method to the above problems, we obtain the pointwise estimate of solutions.

#### **3.1 Energy method**

We apply the energy method to the problems (3.2) and (3.3) formulated in the partial Fourier space and derive pointwise estimates of solutions to (3.2). We use *L*<sup>2</sup> space for the variable *x*<sup>1</sup> ∈ **R**<sup>+</sup> in the normal direction. The result is given as follows.

**Theorem 3.1** (Pointwise estimate)**.** *Let φ*(*x*1) *be a stationary solution with φx*<sup>1</sup> > 0*. Then the solution to the problem* (3.2) *verifies the following pointwise estimate.*

(i) *Non-degenerate case where f* � <sup>1</sup>(0) <sup>&</sup>lt; <sup>0</sup>*: Suppose that <sup>z</sup>*ˆ0(·, *<sup>ξ</sup>*) <sup>∈</sup> *<sup>H</sup>*2(**R**+) *for each <sup>ξ</sup>* <sup>∈</sup> **<sup>R</sup>***n*−<sup>1</sup> *<sup>ξ</sup> . Then it holds*

$$|\sharp(\cdot,\xi,t)|\_{L^{2}\_{a}} \leq \mathsf{C}e^{-\kappa|\xi|^{2}t}|\sharp\_{0}(\cdot,\xi)|\_{L^{2}\_{a'}}\tag{3.4}$$

$$|\sharp\_{\mathbf{x}\_{1}}(\cdot,\xi,t)|\_{L^{2}} \leq \mathsf{C}e^{-\kappa|\xi|^{2}t} (|\sharp\_{0}(\cdot,\xi)|\_{L^{2}\_{u}} + |(\sharp\_{0})\_{\mathbf{x}\_{1}}(\cdot,\xi)|\_{L^{2}})\,\mathrm{}\tag{3.5}$$

$$|\mathfrak{z}\_{\mathbf{x},\mathbf{x}\_{1}}(\cdot,\xi,t)|\_{L^{2}} \leq \mathsf{C}e^{-\mathsf{x}|\xi|^{2}t} \left( |\mathfrak{z}\_{0}(\cdot,\xi)|\_{L^{2}\_{\mathfrak{a}}} + |(\mathfrak{z}\_{0})\_{\mathbf{x}\_{1}}(\cdot,\xi)|\_{H^{1}} \right) \tag{3.6}$$

*with <sup>α</sup>* <sup>=</sup> <sup>0</sup>*, for <sup>ξ</sup>* <sup>∈</sup> **<sup>R</sup>***n*−<sup>1</sup> *<sup>ξ</sup> and t* ≥ 0*, where* |·|*L*<sup>2</sup> *<sup>α</sup> denotes the L*<sup>2</sup> *<sup>α</sup> norm with respect to x*<sup>1</sup> ∈ **R**+*, and C and κ are positive constants.*

(ii) *Degenerate case where f* � <sup>1</sup>(0) = <sup>0</sup>*: Suppose that <sup>z</sup>*ˆ0(·, *<sup>ξ</sup>*) <sup>∈</sup> *<sup>L</sup>*<sup>2</sup> <sup>1</sup>(**R**+) *and* (*z*ˆ0)*x*<sup>1</sup> (·, *<sup>ξ</sup>*) <sup>∈</sup> *<sup>H</sup>*1(**R**+) *for each <sup>ξ</sup>* <sup>∈</sup> **<sup>R</sup>***n*−<sup>1</sup> *<sup>ξ</sup> . Then it holds that* (3.4)*–*(3.6) *with α* = 1*.*

As a simple corollary we have the following decay estimate.

**Corollary 3.2** (Decay estimate)**.** *Assume the same conditions of Proposition 3.1. Then the solution to the problem* (3.1)*,* (2.6)*,* (2.7) *satisfies the following decay estimate.*

(i) *Non-degenerate case where f* � <sup>1</sup>(0) <sup>&</sup>lt; <sup>0</sup>*: Suppose that z*<sup>0</sup> <sup>∈</sup> *<sup>H</sup>*2(*L*1)*. Then this yields*

$$\|\partial\_{\mathbf{x}'}^k z(t)\|\_{\mathcal{L}\_\mathbf{a}^2} \le \mathsf{C} t^{-(n-1)/4 - k/2} \|z\_0\|\_{L\_\mathbf{a}^2(L^1)'} \tag{3.7}$$

$$\|\partial\_{\mathbf{x}'}^k z\_{\mathbf{x}\_1}(t)\|\_{\mathcal{L}^2} \le \mathcal{C} t^{-(n-1)/4 - k/2} (\|z\_0\|\_{L^2\_a(L^1)} + \|(z\_0)\_{\mathbf{x}\_1}\|\_{L^2(L^1)}),\tag{3.8}$$

$$\|\|\partial\_{\mathbf{x}'}^k z\_{\mathbf{x}\_1 \mathbf{x}\_1}(t)\|\|\_{\mathcal{L}^2} \le \mathcal{C} t^{-(n-1)/4 - k/2} \left(\|z\_0\|\_{L^2\_\mathbf{a}(L^1)} + \|(z\_0)\_{\mathbf{x}\_1}\|\_{H^1(L^1)}\right) \tag{3.9}$$

*with α* = 0*, for t* > 0*, where k* ≥ 0 *is an integer and C is a positive constant.*

(ii) *Degenerate case where f* � <sup>1</sup>(0) = <sup>0</sup>*: Suppose that z*<sup>0</sup> <sup>∈</sup> *<sup>L</sup>*<sup>2</sup> <sup>1</sup>(*L*1) *and* (*z*0)*x*<sup>1</sup> <sup>∈</sup> *<sup>H</sup>*1(*L*1)*. Then this yields that* (3.7)*–*(3.9) *with α* = 1*.*

*Proof of Theorem 3.1.* Throughout this proof, we use the weighted *L*<sup>2</sup> norm:

$$|a|\_{L^2\_w} = \left(\int\_0^\infty w\left(\phi(x\_1)\right) |a(x\_1)|^2 d\mathbf{x}\_1\right)^{1/2}.$$

where *w* is the weight function defined by (2.8). For this weighted norm, by using Lemma 2.2, we see the following properties.

> *c*|·|*L*<sup>2</sup> ≤|·|*L*<sup>2</sup> *<sup>w</sup>* ≤ *C*|·|*L*<sup>2</sup> for the non-degenerate case : *f* � (0) < 0, *c*|·|*L*<sup>2</sup> <sup>1</sup> ≤|·|*L*<sup>2</sup> *<sup>w</sup>* ≤ *C*|·|*L*<sup>2</sup> <sup>1</sup> for the degenerate case : *<sup>f</sup>* � (0) = 0. (3.10)

We prove (i) and (ii) in Theorem 3.1 in parallel. We first derive (3.4). We multiply (3.2)1 by *w*(*φ*)¯ *z*ˆ and take the real part, obtaining

$$\frac{1}{2}w(\phi)\frac{\partial}{\partial t}|\mathbf{\hat{z}}|^2 + \frac{\partial}{\partial \mathbf{x}\_1}\mathcal{F}\_1 + \mathcal{D}\_1 = 0,\tag{3.11}$$

where

8 Will-be-set-by-IN-TECH

Furthermore, we sometimes use the differentiated problem. We differentiate the problem (3.1), (2.6), (2.7) with respect to *x*1. Then this yields our problem (1.15) together with (1.9) and (1.10),

Here we note that *v* = *zx*<sup>1</sup> . By applying the weighted energy method to the above problems,

We apply the energy method to the problems (3.2) and (3.3) formulated in the partial Fourier space and derive pointwise estimates of solutions to (3.2). We use *L*<sup>2</sup> space for the variable

**Theorem 3.1** (Pointwise estimate)**.** *Let φ*(*x*1) *be a stationary solution with φx*<sup>1</sup> > 0*. Then the*

2*t*

2*t*

2*t* 

<sup>1</sup>(0) = <sup>0</sup>*: Suppose that <sup>z</sup>*ˆ0(·, *<sup>ξ</sup>*) <sup>∈</sup> *<sup>L</sup>*<sup>2</sup>

**Corollary 3.2** (Decay estimate)**.** *Assume the same conditions of Proposition 3.1. Then the solution*


(|*z*ˆ0(·, *ξ*)|*L*<sup>2</sup>


*<sup>α</sup> denotes the L*<sup>2</sup>

<sup>1</sup>(0) <sup>&</sup>lt; <sup>0</sup>*: Suppose that z*<sup>0</sup> <sup>∈</sup> *<sup>H</sup>*2(*L*1)*. Then this yields*

�*z*0�*L*<sup>2</sup>

�*z*0�*L*<sup>2</sup>

*α*

<sup>1</sup>(0) <sup>&</sup>lt; <sup>0</sup>*: Suppose that <sup>z</sup>*ˆ0(·, *<sup>ξ</sup>*) <sup>∈</sup> *<sup>H</sup>*2(**R**+) *for each <sup>ξ</sup>* <sup>∈</sup> **<sup>R</sup>***n*−<sup>1</sup>

*<sup>α</sup>* + |(*z*ˆ0)*x*<sup>1</sup> (·, *ξ*)|*L*<sup>2</sup>

*<sup>α</sup>* + |(*z*ˆ0)*x*<sup>1</sup> (·, *ξ*)|*H*<sup>1</sup>

*<sup>α</sup>*(*L*<sup>1</sup>) + �(*z*0)*x*<sup>1</sup> �*L*<sup>2</sup>(*L*<sup>1</sup>)

*<sup>α</sup>*(*L*<sup>1</sup>) + �(*z*0)*x*<sup>1</sup> �*H*<sup>1</sup>(*L*<sup>1</sup>)

<sup>1</sup>(*φ*)*v*ˆ)*x*<sup>1</sup> + *iξ* · *f* �

<sup>∗</sup>(*φ*)*v*<sup>ˆ</sup> = 0,

(3.3)

*<sup>ξ</sup> . Then*

, (3.5)

(3.6)

, (3.4)

*<sup>α</sup> norm with respect to x*<sup>1</sup> ∈ **R**+*, and*

<sup>1</sup>(**R**+) *and* (*z*ˆ0)*x*<sup>1</sup> (·, *<sup>ξ</sup>*) <sup>∈</sup> *<sup>H</sup>*1(**R**+)

*α*(*L*<sup>1</sup>), (3.7)

, (3.8)

(3.9)

<sup>2</sup>*v*ˆ + (*f* �

and the corresponding problem in the partial Fourier space:

*v*ˆ*<sup>t</sup>* − *v*ˆ*x*<sup>1</sup> *<sup>x</sup>*<sup>1</sup> + |*ξ*|

*v*ˆ(*x*1, *ξ*, 0) = *v*ˆ0(*x*1, *ξ*).

*x*<sup>1</sup> ∈ **R**<sup>+</sup> in the normal direction. The result is given as follows.

*solution to the problem* (3.2) *verifies the following pointwise estimate.*

<sup>|</sup>*z*ˆ*x*<sup>1</sup> (·, *<sup>ξ</sup>*, *<sup>t</sup>*)|*L*<sup>2</sup> <sup>≤</sup> *Ce*−*κ*|*ξ*<sup>|</sup>

<sup>|</sup>*z*ˆ*x*<sup>1</sup> *<sup>x</sup>*<sup>1</sup> (·, *<sup>ξ</sup>*, *<sup>t</sup>*)|*L*<sup>2</sup> <sup>≤</sup> *Ce*−*κ*|*ξ*<sup>|</sup>

As a simple corollary we have the following decay estimate.

*to the problem* (3.1)*,* (2.6)*,* (2.7) *satisfies the following decay estimate.*

*<sup>x</sup>*� *zx*<sup>1</sup> (*t*)�L<sup>2</sup> <sup>≤</sup> *Ct*−(*n*−1)/4−*k*/2

*<sup>x</sup>*� *zx*<sup>1</sup> *<sup>x</sup>*<sup>1</sup> (*t*)�L<sup>2</sup> <sup>≤</sup> *Ct*−(*n*−1)/4−*k*/2

*with α* = 0*, for t* > 0*, where k* ≥ 0 *is an integer and C is a positive constant.*

*<sup>ξ</sup> and t* ≥ 0*, where* |·|*L*<sup>2</sup>

*<sup>ξ</sup> . Then it holds that* (3.4)*–*(3.6) *with α* = 1*.*

*<sup>α</sup>* <sup>≤</sup> *Ct*−(*n*−1)/4−*k*/2�*z*0�*L*<sup>2</sup>

*<sup>α</sup>* <sup>≤</sup> *Ce*−*κ*|*ξ*<sup>|</sup>


*v*ˆ(0, *ξ*, *t*) = 0,

we obtain the pointwise estimate of solutions.

**3.1 Energy method**

*it holds*

(i) *Non-degenerate case where f* �

*with <sup>α</sup>* <sup>=</sup> <sup>0</sup>*, for <sup>ξ</sup>* <sup>∈</sup> **<sup>R</sup>***n*−<sup>1</sup>

*for each <sup>ξ</sup>* <sup>∈</sup> **<sup>R</sup>***n*−<sup>1</sup>

*C and κ are positive constants.* (ii) *Degenerate case where f* �

(i) *Non-degenerate case where f* �

�*∂k*

�*∂k*

�*∂k*

*<sup>x</sup>*� *<sup>z</sup>*(*t*)�L<sup>2</sup>

$$\begin{aligned} \mathcal{D}\_1 &= w(\phi)(|\mathfrak{z}\_{\mathfrak{x}\_1}|^2 + |\mathfrak{z}|^2|\mathfrak{z}|^2) - \frac{1}{2}(wf\_1)''(\phi)\phi\_{\mathfrak{x}\_1}|\mathfrak{z}|^2, \\ \mathcal{F}\_1 &= \frac{1}{2}(wf\_1)'(\phi)|\mathfrak{z}|^2 - w(\phi)\text{Re}(\tilde{\mathfrak{z}}\mathfrak{z}\_{\mathfrak{x}\_1}). \end{aligned}$$

and *w* is a weight function defined by (2.8). By virtue of (2.9)2 in Lemma 2.2, we have

$$\mathcal{D}\_1 \ge cw(\phi)(|\mathfrak{k}\_{\ge 1}|^2 + |\mathfrak{f}|^2|\mathfrak{k}|^2) + c\phi\_{\ge 1}|\mathfrak{k}|^2,\tag{3.12}$$

where *c* is a some positive constant. Therefore, integrating (3.11) in *x*<sup>1</sup> ∈ **R**+, we get

$$\frac{\partial}{\partial t} |\hat{z}|\_{L^2\_{\psi}}^2 + c\_1 \hat{D}\_1 - (wf\_1)'(u\_b) |\hat{z}(0, \xi, t)|^2 \le 0 \tag{3.13}$$

with a positive constant *c*1, where

$$\hat{D}\_1 = |\mathfrak{z}\_{\mathfrak{x}\_1}|\_{L^2\_w}^2 + |\mathfrak{f}|^2 |\mathfrak{z}|\_{L^2\_w}^2 + |\sqrt{\mathfrak{g}\_{\mathfrak{x}\_1}} \mathfrak{z}|\_{L^2}^2. \tag{3.14}$$

Here, by virtue of (2.9)1, the last term of the left-hand side of (3.13) is positive. We multiply (3.13) by *eκ*|*ξ*<sup>|</sup> <sup>2</sup>*<sup>t</sup>* (*κ* > 0) to get

$$\frac{\partial}{\partial t} (e^{\kappa |\tilde{\xi}|^2 t} |\hat{z}|\_{L^2\_w}^2) + e^{\kappa |\tilde{\xi}|^2 t} (c\_1 \mathcal{D}\_1 - \kappa |\tilde{\xi}|^2 |\hat{z}|\_{L^2\_w}^2) \le 0. \tag{3.15}$$

Noting that *<sup>D</sup>*<sup>ˆ</sup> <sup>1</sup> ≥ |*ξ*<sup>|</sup> 2|*z*ˆ| 2 *L*2 *w* , we choose *κ* > 0 such that *κ* < *c*<sup>1</sup> and integrate (3.15) over [0, *t*]. This yields

$$e^{\kappa |\tilde{\boldsymbol{\xi}}|^{2}t} |\boldsymbol{\varepsilon}(\cdot, \boldsymbol{\xi}, t)|\_{L^{2}\_{\nu}}^{2} + \int\_{0}^{t} e^{\kappa |\tilde{\boldsymbol{\xi}}|^{2}\tau} \hat{D}\_{1}(\tilde{\boldsymbol{\xi}}, \tau) d\tau \leq \mathsf{C} |\boldsymbol{\varepsilon}\_{0}(\cdot, \boldsymbol{\xi})|\_{L^{2}\_{\nu}}^{2} \tag{3.16}$$

Integrating (3.21) in *x*<sup>1</sup> ∈ **R**+, we have

2 *<sup>L</sup>*<sup>2</sup> + |*ξ*|


2 *<sup>L</sup>*<sup>2</sup> + *t* 0 *e κ*|*ξ*|

≤ *C*|(*v*ˆ0)*x*<sup>1</sup> (·, *ξ*)|


Then this yields the desired estimate:

(1 + *t*)*γe*

*κ*|*ξ*| 2*t*

> *κ*|*ξ*| 2*t*


2 *<sup>L</sup>*<sup>2</sup> + *t* 0

2 *<sup>L</sup>*<sup>2</sup> + *C*


2 *<sup>L</sup>*<sup>2</sup> + *t* 0

≤ *C*|*z*ˆ0(·, *ξ*)|


≤ *C*|*v*ˆ0(·, *ξ*)|

 *t* 0

(1 + *t*)*γe*

where *<sup>D</sup>*<sup>ˆ</sup> <sup>3</sup> <sup>=</sup> <sup>|</sup>*v*ˆ*x*<sup>1</sup> *<sup>x</sup>*<sup>1</sup> <sup>|</sup>

*∂ ∂t* (*e κ*|*ξ*| 2*t* |*v*ˆ*x*<sup>1</sup> | 2 *<sup>L</sup>*<sup>2</sup> ) + *e*

*e κ*|*ξ*| 2*t*

(1 + *t*)*γe*

*κ*|*ξ*| 2*t*

≤ *C*|(*v*ˆ0)*x*<sup>1</sup> (·, *ξ*)|

*e κ*|*ξ*| 2*t*

(*κ* > 0) to get

*∂ ∂t* |*v*ˆ*x*<sup>1</sup> | 2

<sup>2</sup>|*v*ˆ*x*<sup>1</sup> <sup>|</sup> 2

> *κ*|*ξ*| 2*t*

> > 2 *<sup>L</sup>*<sup>2</sup> + *t* 0 *e κ*|*ξ*|

2 *<sup>L</sup>*<sup>2</sup> + *C*

Thus, employing (3.16) and (3.20) to the above inequality, we obtain

*<sup>L</sup>*<sup>2</sup> <sup>+</sup> <sup>2</sup>*D*<sup>ˆ</sup> <sup>3</sup> <sup>≤</sup> *<sup>C</sup>*(|*v*ˆ<sup>|</sup>

<sup>2</sup>|*v*ˆ*x*<sup>1</sup> <sup>|</sup> 2

<sup>259</sup> Application of the Weighted Energy Method in the Partial Fourier Space to Linearized Viscous Conservation Laws with Non-Convex Condition

<sup>2</sup>*τD*<sup>ˆ</sup> <sup>3</sup>(*ξ*, *<sup>τ</sup>*)*d<sup>τ</sup>*

<sup>2</sup>*τ*(|*v*ˆ(·, *<sup>ξ</sup>*, *<sup>τ</sup>*)<sup>|</sup>

<sup>2</sup>*τD*<sup>ˆ</sup> <sup>3</sup>(*ξ*, *<sup>τ</sup>*)*d<sup>τ</sup>* <sup>≤</sup> *<sup>C</sup>*(|*z*ˆ0(·, *<sup>ξ</sup>*)<sup>|</sup>

Finally, we apply (3.10) to the estimates (3.16), (3.20) and (3.24). Then this gives the desired

Here, for later use, we derive the corresponding time weighted estimate. We multiply (3.15) (or (3.19), (3.23)) with 0 <sup>&</sup>lt; *<sup>κ</sup>* <sup>≤</sup> *<sup>c</sup>*<sup>1</sup> (or 0 <sup>&</sup>lt; *<sup>κ</sup>* <sup>&</sup>lt; 2) by (<sup>1</sup> <sup>+</sup> *<sup>t</sup>*)*<sup>γ</sup>* (*<sup>γ</sup>* <sup>≥</sup> 0) and integrate over [0, *<sup>t</sup>*].

(1 + *τ*)*γe*

 *t* 0

(1 + *τ*)*γe*

 *t* 0

*κ*|*ξ*|

*κ*|*ξ*| <sup>2</sup>*τ* *κ*|*ξ*|

(1 + *τ*)*γ*−1*e*

*κ*|*ξ*|

(1 + *τ*)*γe*

<sup>2</sup>*τD*<sup>ˆ</sup> <sup>3</sup>(*ξ*, *<sup>τ</sup>*)*d<sup>τ</sup>*


<sup>2</sup>*τD*<sup>ˆ</sup> <sup>1</sup>(*ξ*, *<sup>τ</sup>*)*d<sup>τ</sup>*

*κ*|*ξ*|

<sup>2</sup>*τD*<sup>ˆ</sup> <sup>2</sup>(*ξ*, *<sup>τ</sup>*)*d<sup>τ</sup>*

*κ*|*ξ*|

2 *<sup>H</sup>*<sup>1</sup> + |*ξ*|

<sup>2</sup>*τ*|*z*ˆ(·, *<sup>ξ</sup>*, *<sup>τ</sup>*)<sup>|</sup>

<sup>2</sup>*τ*|*v*ˆ(·, *<sup>ξ</sup>*, *<sup>τ</sup>*)<sup>|</sup>

2 *<sup>L</sup>*<sup>2</sup> *dτ*,

2 *<sup>L</sup>*<sup>2</sup> *dτ*,

<sup>2</sup>|*v*ˆ(·, *<sup>ξ</sup>*, *<sup>τ</sup>*)<sup>|</sup>

2 *L*2 *dτ*,

estimates (3.4)–(3.6) with *α* = 0, 1. Hence the proof of Theorem 3.1 is completed.

2 *<sup>L</sup>*<sup>2</sup> + *γC*

> 2 *<sup>L</sup>*<sup>2</sup> + *C*

(1 + *τ*)*γe*

(1 + *τ*)*γe*

where *γ* ≥ 0, and *C* is a positive constant. These will be used in the next subsection.

2 *<sup>L</sup>*<sup>2</sup> + *t* 0

(2*D*<sup>ˆ</sup> <sup>3</sup> <sup>−</sup> *<sup>κ</sup>*|*ξ*<sup>|</sup>

 *t* 0 *e κ*|*ξ*|

Then we choose *κ* > 0 such that *κ* < 2 and integrate (3.23) over [0, *t*]. This yields

2

*<sup>H</sup>*<sup>1</sup> + |*ξ*||*v*ˆ|

*<sup>L</sup>*<sup>2</sup> ) <sup>≤</sup> *Ceκ*|*ξ*<sup>|</sup>

2 *<sup>H</sup>*<sup>1</sup> + |*ξ*|

2

*<sup>L</sup>*<sup>2</sup> and *<sup>C</sup>* is a positive constant. We multiply (3.22) by *<sup>e</sup>κ*|*ξ*<sup>|</sup>

2*t* (|*v*ˆ| 2

> 2 *L*2

*<sup>L</sup>*<sup>2</sup> ), (3.22)

2

*<sup>L</sup>*<sup>2</sup> ). (3.23)

*<sup>H</sup>*<sup>1</sup> ). (3.24)

(3.25)

(3.26)

(3.27)

*<sup>H</sup>*<sup>1</sup> + |*ξ*||*v*ˆ|

<sup>2</sup>|*v*ˆ(·, *<sup>ξ</sup>*, *<sup>τ</sup>*)<sup>|</sup>

*<sup>w</sup>* + |*v*ˆ0(·, *<sup>ξ</sup>*)|

2 *<sup>L</sup>*<sup>2</sup> )*dτ*.

2

2*t*

where *C* is a positive constant.

We next prove (3.5). Multiplying (3.3)1 by ¯ *v*ˆ and taking the real part, then we have

$$\frac{1}{2}\frac{\partial}{\partial t}|\mathcal{O}|^2 + \frac{\partial}{\partial x\_1}\mathcal{F}\_2 + \mathcal{D}\_2 = 0,\tag{3.17}$$

where

$$\mathcal{D}\_2 = |\mathfrak{d}\_{\mathbb{X}\_1}|^2 + |\tilde{\mathfrak{e}}|^2 |\mathfrak{d}|^2 + \frac{1}{2} f\_1^{\prime\prime}(\phi) \phi\_{\mathbb{X}\_1} |\vartheta|^2 \Big|$$

$$\mathcal{F}\_2 = \frac{1}{2} f\_1^{\prime}(\phi) |\vartheta|^2 - \text{Re}(\bar{\vartheta}\vartheta\_{\mathbb{X}\_1}).$$

We integrate (3.17) in *x*<sup>1</sup> ∈ **R**<sup>+</sup> to obtain

$$\frac{\partial}{\partial t} |\boldsymbol{\vartheta}|\_{L^2}^2 + 2\boldsymbol{\hat{D}}\_2 \le \mathbb{C} |\boldsymbol{\vartheta}|\_{L^2}^2 \tag{3.18}$$

where *<sup>D</sup>*<sup>ˆ</sup> <sup>2</sup> <sup>=</sup> <sup>|</sup>*v*ˆ*x*<sup>1</sup> <sup>|</sup> 2 *<sup>L</sup>*<sup>2</sup> + |*ξ*| <sup>2</sup>|*v*ˆ<sup>|</sup> 2 *<sup>L</sup>*<sup>2</sup> and *<sup>C</sup>* is a positive constant. We multiply (3.18) by *<sup>e</sup>κ*|*ξ*<sup>|</sup> <sup>2</sup>*<sup>t</sup>* (*κ* > 0) to get

$$\frac{\partial}{\partial t}(e^{\kappa|\xi|^2 t}|\vartheta|^2\_{L^2}) + e^{\kappa|\xi|^2 t}(2\mathcal{D}\_2 - \kappa|\xi|^2|\vartheta|^2\_{L^2}) \le \mathsf{C} e^{\kappa|\xi|^2 t}|\vartheta|^2\_{L^2}.\tag{3.19}$$

Then we choose *κ* > 0 such that *κ* < 2 and integrate (3.19) over [0, *t*]. This yields

$$e^{\kappa |\tilde{\xi}|^2 t} |\vartheta(\cdot, \tilde{\xi}, t)|\_{L^2}^2 + \int\_0^t e^{\kappa |\tilde{\xi}|^2 \tau} \hat{D}\_2(\tilde{\xi}, \tau) d\tau \le \mathbb{C} |\vartheta\_0(\cdot, \tilde{\xi})|\_{L^2}^2 + \mathbb{C} \int\_0^t e^{\kappa |\tilde{\xi}|^2 \tau} |\vartheta(\cdot, \tilde{\xi}, \tau)|\_{L^2}^2 d\tau.$$

Noting that *v* = *zx*<sup>1</sup> and |*v*ˆ|*L*<sup>2</sup> ≤ *C*|*v*ˆ|*L*<sup>2</sup> *<sup>w</sup>* , we apply (3.16) to the above inequality. Then we get

$$\|e^{\kappa|\xi|^2 t} |\vartheta(\cdot,\widetilde{\xi},t)|\_{L^2}^2 + \int\_0^t e^{\kappa |\widetilde{\xi}|^2 \tau} \dot{D}\_2(\widetilde{\xi},\tau) d\tau \le \mathbb{C} (|\mathfrak{z}\_0(\cdot,\xi)|\_{L^2\_w}^2 + |\mathfrak{z}\_0(\cdot,\xi)|\_{L^2}^2). \tag{3.20}$$

We shall show (3.6). Multiplying (3.3)1 by <sup>−</sup> ¯ *v*ˆ*x*<sup>1</sup> *<sup>x</sup>*<sup>1</sup> and taking the real part, then we have

$$\frac{1}{2}\frac{\partial}{\partial t}|\vartheta\_{\mathbf{x}\_{1}}|^{2} + \frac{\partial}{\partial \mathbf{x}\_{1}}\mathcal{F}\_{3} + \mathcal{D}\_{3} = \mathbf{0},\tag{3.21}$$

where

$$\begin{split} \mathcal{D}\_{3} &= |\mathfrak{H}\_{\mathbf{x}\_{1}\mathbf{x}\_{1}}|^{2} + |\xi|^{2} |\mathfrak{H}\_{\mathbf{x}\_{1}}|^{2} + \frac{3}{2} f\_{1}^{\prime\prime}(\boldsymbol{\phi}) \boldsymbol{\phi}\_{\mathbf{x}\_{1}} |\mathfrak{H}\_{\mathbf{x}\_{1}}|^{2} \\ & \quad - \frac{1}{2} \left( (f\_{1} f\_{1}^{\prime\prime})^{\prime} f\_{1} \right)^{\prime}(\boldsymbol{\phi}) \boldsymbol{\phi}\_{\mathbf{x}\_{1}} |\hat{v}|^{2} - \frac{1}{2} i \tilde{\xi} (f\_{1} f\_{\ast}^{\prime\prime})^{\prime}(\boldsymbol{\phi}) \boldsymbol{\phi}\_{\mathbf{x}\_{1}} |\hat{v}|^{2} \, \, \, \, \, \, \\ \mathcal{F}\_{3} &= \frac{1}{2} f\_{1}^{\prime}(\boldsymbol{\phi}) |\mathfrak{H}\_{\mathbf{x}\_{1}}|^{2} + \frac{1}{2} \{ (f\_{1} f\_{1}^{\prime\prime})^{\prime}(\boldsymbol{\phi}) + i \tilde{\xi} f\_{\ast}^{\prime\prime}(\boldsymbol{\phi}) \} \boldsymbol{\phi}\_{\mathbf{x}\_{1}} |\boldsymbol{\phi}|^{2} \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,$$

Integrating (3.21) in *x*<sup>1</sup> ∈ **R**+, we have

10 Will-be-set-by-IN-TECH

*v*ˆ and taking the real part, then we have

<sup>1</sup> (*φ*)*φx*<sup>1</sup> |*v*ˆ|

2,

F<sup>2</sup> + D<sup>2</sup> = 0, (3.17)

*<sup>L</sup>*<sup>2</sup> , (3.18)

<sup>2</sup>*τ*|*v*ˆ(·, *<sup>ξ</sup>*, *<sup>τ</sup>*)<sup>|</sup>

2

<sup>2</sup>*<sup>t</sup>* (*κ* > 0)

*<sup>L</sup>*<sup>2</sup> . (3.19)

2 *<sup>L</sup>*<sup>2</sup> *dτ*.

*<sup>L</sup>*<sup>2</sup> ). (3.20)

where *C* is a positive constant.

where

where *<sup>D</sup>*<sup>ˆ</sup> <sup>2</sup> <sup>=</sup> <sup>|</sup>*v*ˆ*x*<sup>1</sup> <sup>|</sup>

to get

*e κ*|*ξ*| 2*t*

where

We next prove (3.5). Multiplying (3.3)1 by ¯

We integrate (3.17) in *x*<sup>1</sup> ∈ **R**<sup>+</sup> to obtain

2 *<sup>L</sup>*<sup>2</sup> + |*ξ*|


*e κ*|*ξ*| 2*t*

*∂ ∂t* (*e κ*|*ξ*| 2*t* |*v*ˆ| 2 *<sup>L</sup>*<sup>2</sup> ) + *e*

2 *<sup>L</sup>*<sup>2</sup> + *t* 0 *e κ*|*ξ*|

Noting that *v* = *zx*<sup>1</sup> and |*v*ˆ|*L*<sup>2</sup> ≤ *C*|*v*ˆ|*L*<sup>2</sup>


We shall show (3.6). Multiplying (3.3)1 by <sup>−</sup> ¯

D<sup>3</sup> = |*v*ˆ*x*<sup>1</sup> *<sup>x</sup>*<sup>1</sup> |

<sup>F</sup><sup>3</sup> <sup>=</sup> <sup>1</sup> 2 *f* � 2 *<sup>L</sup>*<sup>2</sup> + *t* 0 *e κ*|*ξ*|

> 1 2 *∂ ∂t* |*v*ˆ*x*<sup>1</sup> | <sup>2</sup> + *∂ ∂x*<sup>1</sup>

<sup>2</sup> <sup>+</sup> <sup>|</sup>*ξ*<sup>|</sup>

− 1 2 (*f*<sup>1</sup> *f* �� <sup>1</sup> )� *f*<sup>1</sup> �

<sup>1</sup>(*φ*)|*v*ˆ*x*<sup>1</sup> |

<sup>−</sup> Re(*v*ˆ*<sup>t</sup>* ¯

<sup>2</sup>|*v*ˆ*x*<sup>1</sup> <sup>|</sup> <sup>2</sup> + 3 2 *f* ��

<sup>2</sup> + 1 2 {(*f*<sup>1</sup> *f* �� <sup>1</sup> )�

*v*ˆ*x*<sup>1</sup> ) − (|*ξ*|

1 2 *∂ ∂t* |*v*ˆ| <sup>2</sup> + *∂ ∂x*<sup>1</sup>

D<sup>2</sup> = |*v*ˆ*x*<sup>1</sup> |

*∂ ∂t* |*v*ˆ| 2

> *κ*|*ξ*| 2*t*

Then we choose *κ* > 0 such that *κ* < 2 and integrate (3.19) over [0, *t*]. This yields

<sup>F</sup><sup>2</sup> <sup>=</sup> <sup>1</sup> 2 *f* � <sup>1</sup>(*φ*)|*v*ˆ|

<sup>2</sup>|*v*ˆ<sup>|</sup> 2 <sup>2</sup> <sup>+</sup> <sup>|</sup>*ξ*<sup>|</sup>

2|*v*ˆ| <sup>2</sup> + 1 2 *f* ��

<sup>2</sup> <sup>−</sup> Re(¯

*<sup>L</sup>*<sup>2</sup> <sup>+</sup> <sup>2</sup>*D*<sup>ˆ</sup> <sup>2</sup> <sup>≤</sup> *<sup>C</sup>*|*v*ˆ<sup>|</sup>

(2*D*<sup>ˆ</sup> <sup>2</sup> <sup>−</sup> *<sup>κ</sup>*|*ξ*<sup>|</sup>

<sup>2</sup>*τD*<sup>ˆ</sup> <sup>2</sup>(*ξ*, *<sup>τ</sup>*)*d<sup>τ</sup>* <sup>≤</sup> *<sup>C</sup>*|*v*ˆ0(·, *<sup>ξ</sup>*)<sup>|</sup>

*v*ˆ*v*ˆ*x*<sup>1</sup> ).

2

*<sup>L</sup>*<sup>2</sup> and *<sup>C</sup>* is a positive constant. We multiply (3.18) by *<sup>e</sup>κ*|*ξ*<sup>|</sup>

2|*v*ˆ| 2

<sup>2</sup>*τD*<sup>ˆ</sup> <sup>2</sup>(*ξ*, *<sup>τ</sup>*)*d<sup>τ</sup>* <sup>≤</sup> *<sup>C</sup>*(|*z*ˆ0(·, *<sup>ξ</sup>*)<sup>|</sup>

<sup>1</sup> (*φ*)*φx*<sup>1</sup> |*v*ˆ*x*<sup>1</sup> |

<sup>2</sup> <sup>−</sup> <sup>1</sup> 2 *iξ*(*f*<sup>1</sup> *f* �� ∗ )�

(*φ*) + *iξ f* ��

<sup>1</sup> (*φ*)*φx*<sup>1</sup> + *iξ f* �

(*φ*)*φx*<sup>1</sup> |*v*ˆ|

<sup>2</sup> + *f* ��

2

<sup>∗</sup> (*φ*)}*φx*<sup>1</sup> |*v*ˆ|

2 *<sup>L</sup>*<sup>2</sup> + *C*

*<sup>L</sup>*<sup>2</sup> ) <sup>≤</sup> *Ceκ*|*ξ*<sup>|</sup>

2*t* |*v*ˆ| 2

 *t* 0 *e κ*|*ξ*|

*<sup>w</sup>* , we apply (3.16) to the above inequality. Then we get

2 *L*2

*v*ˆ*x*<sup>1</sup> *<sup>x</sup>*<sup>1</sup> and taking the real part, then we have

*<sup>w</sup>* + |*v*ˆ0(·, *<sup>ξ</sup>*)|

F<sup>3</sup> + D<sup>3</sup> = 0, (3.21)

(*φ*)*φx*<sup>1</sup> |*v*ˆ|

2

<sup>∗</sup>(*φ*))Re(*v*<sup>ˆ</sup> ¯

2,

*v*ˆ*x*<sup>1</sup> ).

$$2\frac{\partial}{\partial t}|\boldsymbol{\vartheta}\_{\boldsymbol{x}\_{1}}|\_{L^{2}}^{2} + 2\boldsymbol{\hat{D}}\_{3} \leq \mathcal{C}(|\boldsymbol{\vartheta}|\_{H^{1}}^{2} + |\boldsymbol{\xi}| |\boldsymbol{\vartheta}|\_{L^{2}}^{2}),\tag{3.22}$$

where *<sup>D</sup>*<sup>ˆ</sup> <sup>3</sup> <sup>=</sup> <sup>|</sup>*v*ˆ*x*<sup>1</sup> *<sup>x</sup>*<sup>1</sup> <sup>|</sup> 2 *<sup>L</sup>*<sup>2</sup> + |*ξ*| <sup>2</sup>|*v*ˆ*x*<sup>1</sup> <sup>|</sup> 2 *<sup>L</sup>*<sup>2</sup> and *<sup>C</sup>* is a positive constant. We multiply (3.22) by *<sup>e</sup>κ*|*ξ*<sup>|</sup> 2*t* (*κ* > 0) to get

$$\frac{\partial}{\partial t} (e^{\kappa |\xi|^2 t} |\vartheta\_{\mathbf{x}\_1}|\_{L^2}^2) + e^{\kappa |\xi|^2 t} (2\hat{D}\_3 - \kappa |\xi|^2 |\vartheta\_{\mathbf{x}\_1}|\_{L^2}^2) \le \mathbb{C} e^{\kappa |\xi|^2 t} (|\vartheta|\_{H^1}^2 + |\xi| |\vartheta|\_{L^2}^2). \tag{3.23}$$

Then we choose *κ* > 0 such that *κ* < 2 and integrate (3.23) over [0, *t*]. This yields

$$\begin{split} & \|e^{\kappa |\tilde{\xi}|^2 t} |\mathfrak{d}\_{\mathbf{x}\_1}(\cdot, \tilde{\xi}, t)|\_{L^2}^2 + \int\_0^t e^{\kappa |\tilde{\xi}|^2 \tau} \hat{D}\_3(\tilde{\xi}, \tau) d\tau \\ & \leq C |(\hat{v}\_0)\_{\mathbf{x}\_1}(\cdot, \tilde{\xi})|\_{L^2}^2 + C \int\_0^t e^{\kappa |\tilde{\xi}|^2 \tau} (|\hat{v}(\cdot, \tilde{\xi}, \tau)|\_{H^1}^2 + |\tilde{\xi}|^2 |\hat{v}(\cdot, \tilde{\xi}, \tau)|\_{L^2}^2) d\tau. \end{split}$$

Thus, employing (3.16) and (3.20) to the above inequality, we obtain

$$\left|e^{\kappa|\xi|^2 t} \langle \vartheta\_{\mathbf{x}}(\cdot, \widetilde{\xi}\_{\prime} t) \rangle\_{\mathbf{L}^2}^2 + \int\_0^t e^{\kappa|\xi|^2 \tau} \hat{D}\_3(\widetilde{\xi}, \tau) d\tau \right| \le \mathcal{C}(|\mathfrak{z}\_0(\cdot, \widetilde{\xi})|\_{\mathbf{L}^2\_w}^2 + |\mathfrak{z}\_0(\cdot, \widetilde{\xi})|\_{\mathbf{H}^1}^2). \tag{3.24}$$

Finally, we apply (3.10) to the estimates (3.16), (3.20) and (3.24). Then this gives the desired estimates (3.4)–(3.6) with *α* = 0, 1. Hence the proof of Theorem 3.1 is completed.

Here, for later use, we derive the corresponding time weighted estimate. We multiply (3.15) (or (3.19), (3.23)) with 0 <sup>&</sup>lt; *<sup>κ</sup>* <sup>≤</sup> *<sup>c</sup>*<sup>1</sup> (or 0 <sup>&</sup>lt; *<sup>κ</sup>* <sup>&</sup>lt; 2) by (<sup>1</sup> <sup>+</sup> *<sup>t</sup>*)*<sup>γ</sup>* (*<sup>γ</sup>* <sup>≥</sup> 0) and integrate over [0, *<sup>t</sup>*]. Then this yields the desired estimate:

(1 + *t*)*γe κ*|*ξ*| 2*t* |*z*ˆ(·, *ξ*, *t*)| 2 *<sup>L</sup>*<sup>2</sup> + *t* 0 (1 + *τ*)*γe κ*|*ξ*| <sup>2</sup>*τD*<sup>ˆ</sup> <sup>1</sup>(*ξ*, *<sup>τ</sup>*)*d<sup>τ</sup>* ≤ *C*|*z*ˆ0(·, *ξ*)| 2 *<sup>L</sup>*<sup>2</sup> + *γC t* 0 (1 + *τ*)*γ*−1*e κ*|*ξ*| <sup>2</sup>*τ*|*z*ˆ(·, *<sup>ξ</sup>*, *<sup>τ</sup>*)<sup>|</sup> 2 *<sup>L</sup>*<sup>2</sup> *dτ*, (3.25) (1 + *t*)*γe κ*|*ξ*| 2*t* |*v*ˆ(·, *ξ*, *t*)| 2 *<sup>L</sup>*<sup>2</sup> + *t* 0 (1 + *τ*)*γe κ*|*ξ*| <sup>2</sup>*τD*<sup>ˆ</sup> <sup>2</sup>(*ξ*, *<sup>τ</sup>*)*d<sup>τ</sup>* ≤ *C*|*v*ˆ0(·, *ξ*)| 2 *<sup>L</sup>*<sup>2</sup> + *C t* 0 (1 + *τ*)*γe κ*|*ξ*| <sup>2</sup>*τ*|*v*ˆ(·, *<sup>ξ</sup>*, *<sup>τ</sup>*)<sup>|</sup> 2 *<sup>L</sup>*<sup>2</sup> *dτ*, (3.26) (1 + *t*)*γe κ*|*ξ*| 2*t* |*v*ˆ*x*<sup>1</sup> (·, *ξ*, *t*)| 2 *<sup>L</sup>*<sup>2</sup> + *t* 0 (1 + *τ*)*γe κ*|*ξ*| <sup>2</sup>*τD*<sup>ˆ</sup> <sup>3</sup>(*ξ*, *<sup>τ</sup>*)*d<sup>τ</sup>* ≤ *C*|(*v*ˆ0)*x*<sup>1</sup> (·, *ξ*)| 2 *<sup>L</sup>*<sup>2</sup> + *C t* 0 (1 + *τ*)*γe κ*|*ξ*| <sup>2</sup>*τ* |*v*ˆ(·, *ξ*, *τ*)| 2 *<sup>H</sup>*<sup>1</sup> + |*ξ*| <sup>2</sup>|*v*ˆ(·, *<sup>ξ</sup>*, *<sup>τ</sup>*)<sup>|</sup> 2 *L*2 *dτ*, (3.27)

where *γ* ≥ 0, and *C* is a positive constant. These will be used in the next subsection.

*for t* > 0*, where k* ≥ 0 *is an integer and C is a positive constant.*

12, 14] and divide into four steps.

*κ*|*ξ*| 2*t*

for *γ* ≥ 0 and 0 ≤ *β* ≤ *α*, where

*∂ ∂t* |*z*ˆ| 2 *L*2 *β*,*w*

for 0 ≤ *β* ≤ *α*, where we define

small that *αC�* ≤ *c*1. This yields

2 *L*2 *β* + *γC*

> 1 2 *∂ ∂t*


> *∂ ∂t* |*z*ˆ| 2 *L*2 *β*,*w*

(3.33) is positive. We now observe that


2 *L*2 *β* + *t* 0

 *t* 0

(1 + *t*)*γe*

have

≤ *C*|*z*ˆ0(·, *ξ*)|

The proof of Corollary 3.4 is completely same as the proof of Corollary 3.2 and omitted here. *Proof of Theorem 3.3.* We use the weighted energy method to the problems (3.2) and (3.3) formulated in the partial Fourier space. Our computation is similar to the one used in [4, 10,

<sup>261</sup> Application of the Weighted Energy Method in the Partial Fourier Space to Linearized Viscous Conservation Laws with Non-Convex Condition

> *κ*|*ξ*| <sup>2</sup>*τ*

To prove (3.31), we use the equality (3.11). Notice that, by virtue of (2.9)1 in Lemma 2.2, we

<sup>2</sup> <sup>−</sup> *<sup>C</sup>*|*z*ˆ*x*<sup>1</sup> <sup>|</sup>

(<sup>1</sup> <sup>+</sup> *<sup>x</sup>*1)*β*F<sup>1</sup>

+ (<sup>1</sup> <sup>+</sup> *<sup>x</sup>*1)*β*D<sup>1</sup> <sup>+</sup> *<sup>β</sup>*(<sup>1</sup> <sup>+</sup> *<sup>x</sup>*1)*β*−1(−F1) = 0.

− (*w f*1)�

 *φ*(*x*1) |*v*(*x*1)|

and *C* is a positive constant. Here, by virtue of (2.9)1, the last term of the left-hand side of

for any *�* > 0, where *C�* is a constant depending on *�*. We apply this inequality to the term on

+ *C�*|*a*| 2 *L*2

> 2 *L*2 *β*−1

≤ *βC*|*z*ˆ*x*<sup>1</sup> |

2

≤ *�*|*a*| 2 *L*2 *β*

<sup>+</sup> *<sup>c</sup>*3**D**<sup>ˆ</sup> *<sup>β</sup>* <sup>+</sup> <sup>2</sup>*βc*2|*z*ˆ<sup>|</sup>

<sup>2</sup>*τ*|*z*ˆ(·, *<sup>ξ</sup>*, *<sup>τ</sup>*)<sup>|</sup>

2 *L*2 *β dτ*

*<sup>φ</sup>x*<sup>1</sup> *<sup>z</sup>*ˆ<sup>|</sup> 2 *L*2 *β* ,

> *x*1

(*ub*)|*z*ˆ(0, *ξ*, *t*)|

<sup>2</sup>*dx*<sup>1</sup> 1/2 ,

**<sup>D</sup>**<sup>ˆ</sup> *<sup>β</sup>*(*ξ*, *<sup>τ</sup>*) + *<sup>β</sup>*|*z*ˆ(·, *<sup>ξ</sup>*, *<sup>τ</sup>*)<sup>|</sup>

2 *L*2 *β*−1 *dτ*

<sup>2</sup> (3.32)

<sup>2</sup> <sup>≤</sup> *<sup>β</sup>C*|*z*ˆ*x*<sup>1</sup> <sup>|</sup>

2 *L*2 *β* , we choose *�* > 0 so

*<sup>L</sup>*<sup>2</sup> (3.34)

2 *L*2 *β*−1 (3.31)

(3.33)

**Step 1.** First, we show the following space-time weighted energy inequality:

(1 + *τ*)*γ*−1*e*

2 *L*2 *β* + |*ξ*| 2|*z*ˆ| 2 *L*2 *β* + |

and *C* and *κ* are positive constants. Notice that **D**ˆ <sup>0</sup> coincides with *D*ˆ <sup>1</sup> in (3.14).

− F<sup>1</sup> ≥ *c*2|*z*ˆ|

2 +

We integrate this equality over *x*<sup>1</sup> ∈ **R**<sup>+</sup> and use (3.12) and (3.32), obtaining


the right-hand side of (3.33) by taking *<sup>a</sup>* <sup>=</sup> *<sup>z</sup>*ˆ*x*<sup>1</sup> . Noting that **<sup>D</sup>**<sup>ˆ</sup> *<sup>β</sup>* ≥ |*z*ˆ*x*<sup>1</sup> <sup>|</sup>

2 *L*2 *β*−1

(1 + *x*1)*βw*

with positive constants *<sup>c</sup>*<sup>2</sup> and *<sup>C</sup>*. Now we multiply (3.11) by (<sup>1</sup> <sup>+</sup> *<sup>x</sup>*1)*<sup>β</sup>* (0 <sup>≤</sup> *<sup>β</sup>* <sup>≤</sup> *<sup>α</sup>*) to get

**<sup>D</sup>**<sup>ˆ</sup> *<sup>β</sup>* <sup>=</sup> <sup>|</sup>*z*ˆ*x*<sup>1</sup> <sup>|</sup>

(<sup>1</sup> <sup>+</sup> *<sup>x</sup>*1)*β*|*z*ˆ<sup>|</sup>

<sup>+</sup> *<sup>c</sup>*1**D**<sup>ˆ</sup> *<sup>β</sup>* <sup>+</sup> <sup>2</sup>*βc*2|*z*ˆ<sup>|</sup>

(1 + *τ*)*γe*

*κ*|*ξ*|

*Proof of Corollary 3.2.* By virtue of the Plancherel theorem, we have (2.3). Substituting (3.4) into (2.3), we obtain

$$\begin{split} \|\partial\_{\mathbf{x}'}^k z(t)\|\_{\mathcal{L}\_a^2}^2 &\leq C \int\_{\mathbf{R}\_\xi^{n-1}} |\xi|^{2k} e^{-\kappa |\xi|^2 t} |\hat{z}\_0(\cdot, \xi)|\_{L\_a^2}^2 d\xi \\ &\leq C \sup\_{\substack{\mathfrak{F} \in \mathbf{R}\_\xi^{n-1} \\ \mathfrak{F} \in \mathbf{R}\_\xi^{n-1}}} |\hat{z}\_0(\cdot, \xi)|\_{L\_a^2}^2 \int\_{\mathbf{R}\_\xi^{n-1}} |\xi|^{2k} e^{-\kappa |\xi|^2 t} d\xi \\ &\leq C t^{-(n-1)/2 - k} \|z\_0\|\_{L\_a^2(L^1)'}^2 \end{split}$$

where *C* is a positive constant. Here we used (2.2) and the simple inequality

$$\int\_{\mathbb{R}^{n-1}\_{\xi}} |\tilde{\xi}|^{2k} e^{-\kappa |\tilde{\xi}|^2 t} d\xi \le C t^{-(n-1)/2 - k}$$

with a constant *C*. By applying the same argument to the pointwise estimates (3.5) and (3.6) we can derive the decay estimate (3.8) and (3.9), respectively. Thus this completes the proof.

#### **3.2 Weighted energy method**

In the last subsection, we restrict to the non-degenerate case *f* � <sup>1</sup>(0) < 0 and apply the weighted energy method to the problems (3.2) and (3.3). This yields sharp pointwise estimates of solutions to (3.2). We use the weighted space *L*<sup>2</sup> *<sup>α</sup>*(**R**+) (*α* ≥ 0) for *x*<sup>1</sup> ∈ **R**<sup>+</sup> in the normal direction and this gives the additional decay (1 + *t*)−*<sup>α</sup>*/2. The result is stated as follows.

**Theorem 3.3** (Pointwise estimate)**.** *Let f* � <sup>1</sup>(0) < 0 *and let φ*(*x*1) *be a stationary solution with <sup>φ</sup>x*<sup>1</sup> <sup>&</sup>gt; <sup>0</sup>*. Let <sup>α</sup>* <sup>≥</sup> <sup>0</sup> *and suppose that <sup>z</sup>*ˆ0(·, *<sup>ξ</sup>*) <sup>∈</sup> *<sup>L</sup>*<sup>2</sup> *<sup>α</sup>*(**R**+) *and* (*z*ˆ0)*x*<sup>1</sup> (·, *<sup>ξ</sup>*) <sup>∈</sup> *<sup>H</sup>*1(**R**+) *for each <sup>ξ</sup>* <sup>∈</sup> **<sup>R</sup>***n*−<sup>1</sup> *<sup>ξ</sup> . Then the solution to the problem* (3.2) *verifies the pointwise estimate*

$$|\mathfrak{z}(\cdot,\mathfrak{z},t)|\_{L^2} \le \mathcal{C}(1+t)^{-a/2} e^{-\kappa |\mathfrak{z}|^2 t} |\mathfrak{z}\_0(\cdot,\mathfrak{z})|\_{L^2\_{\mathfrak{a}}} \tag{3.28}$$

$$|\mathfrak{f}\_{\mathbf{x}\_{1}}(\cdot,\xi,t)|\_{L^{2}} \leq \mathfrak{C}(1+t)^{-n/2}e^{-\kappa|\xi|^{2}t} (|\mathfrak{f}\_{0}(\cdot,\xi)|\_{L^{2}\_{a}} + |(\mathfrak{f}\_{0})\_{\mathbf{x}\_{1}}(\cdot,\xi)|\_{L^{2}}),\tag{3.29}$$

$$\left|\sharp\_{\mathbf{x}\_{1}\mathbf{x}\_{1}}(\cdot,\xi,t)\right|\_{L^{2}} \leq \mathsf{C}(1+t)^{-\kappa/2}e^{-\kappa|\xi|^{2}t}\left(\left|\sharp\_{0}(\cdot,\xi)\right|\_{L^{2}\_{\mu}} + \left|(\sharp\_{0})\_{\mathbf{x}\_{1}}(\cdot,\xi)\right|\_{H^{1}}\right) \tag{3.30}$$

*for <sup>ξ</sup>* <sup>∈</sup> **<sup>R</sup>***n*−<sup>1</sup> *<sup>ξ</sup> and t* ≥ 0*, where the norms* |·|*L*<sup>2</sup> *,* |·|*H*<sup>1</sup> *and* |·|*L*<sup>2</sup> *<sup>α</sup> are with respect to x*<sup>1</sup> ∈ **R**+*, and C and κ are positive constants.*

As an easy consequence, we have the following decay estimate.

**Corollary 3.4** (Decay estimate)**.** *Assume the same conditions of Theorem 3.3. Let z*<sup>0</sup> <sup>∈</sup> *<sup>L</sup>*<sup>2</sup> *<sup>α</sup>*(*L*1) *and* (*z*0)*x*<sup>1</sup> <sup>∈</sup> *<sup>H</sup>*1(*L*1) *for <sup>α</sup>* <sup>≥</sup> <sup>0</sup>*. Then the solution to the problem* (3.1)*,* (2.6)*,* (2.7) *satisfies the decay estimate*

$$\begin{aligned} \|\partial\_{\mathbf{x}'}^k z(t)\|\_{\mathcal{L}^2} &\leq \mathcal{C} (1+t)^{-a/2} t^{-(n-1)/4 - k/2} \|z\_0\|\_{L^2\_a(\mathcal{L}^1)}, \\ \|\partial\_{\mathbf{x}'}^k z\_{\mathbf{x}\_1}(t)\|\_{\mathcal{L}^2} &\leq \mathcal{C} (1+t)^{-a/2} t^{-(n-1)/4 - k/2} \left(\|z\_0\|\_{L^2\_a(\mathcal{L}^1)} + \|(z\_0)\_{\mathbf{x}\_1}\|\_{L^2(\mathcal{L}^1)}\right), \\ \|\partial\_{\mathbf{x}'}^k z\_{\mathbf{x}\_{\mathbf{1}}\mathbf{x}\_1}(t)\|\_{\mathcal{L}^2} &\leq \mathcal{C} (1+t)^{-a/2} t^{-(n-1)/4 - k/2} \left(\|z\_0\|\_{L^2\_a(\mathcal{L}^1)} + \|(z\_0)\_{\mathbf{x}\_1}\|\_{H^1(\mathcal{L}^1)}\right). \end{aligned}$$

*for t* > 0*, where k* ≥ 0 *is an integer and C is a positive constant.*

The proof of Corollary 3.4 is completely same as the proof of Corollary 3.2 and omitted here.

*Proof of Theorem 3.3.* We use the weighted energy method to the problems (3.2) and (3.3) formulated in the partial Fourier space. Our computation is similar to the one used in [4, 10, 12, 14] and divide into four steps.

**Step 1.** First, we show the following space-time weighted energy inequality:

$$\begin{split} &(1+t)^{\gamma}e^{\kappa|\tilde{\xi}|^{2}t}|\boldsymbol{\mathring{\varepsilon}}(\cdot,\boldsymbol{\mathring{\xi}},t)|\_{L^{2}\_{\tilde{\beta}}}^{2} + \int\_{0}^{t}(1+\tau)^{\gamma}e^{\kappa|\tilde{\xi}|^{2}\tau} \left(\mathbf{\hat{D}}\_{\tilde{\beta}}(\boldsymbol{\mathring{\xi}},\tau) + \beta|\boldsymbol{\mathring{\varepsilon}}(\cdot,\boldsymbol{\mathring{\xi}},\tau)|\_{L^{2}\_{\tilde{\beta}-1}}^{2}\right) d\tau \\ & \leq \mathsf{C} |\boldsymbol{\mathring{\varepsilon}}\_{0}(\cdot,\boldsymbol{\mathring{\xi}})|\_{L^{2}\_{\tilde{\beta}}}^{2} + \gamma\mathsf{C} \int\_{0}^{t}(1+\tau)^{\gamma-1}e^{\kappa|\tilde{\xi}|^{2}\tau} |\boldsymbol{\mathring{\varepsilon}}(\cdot,\boldsymbol{\mathring{\xi}},\tau)|\_{L^{2}\_{\tilde{\beta}}}^{2} d\tau \end{split} \tag{3.31}$$

for *γ* ≥ 0 and 0 ≤ *β* ≤ *α*, where

12 Will-be-set-by-IN-TECH

*Proof of Corollary 3.2.* By virtue of the Plancherel theorem, we have (2.3). Substituting (3.4)


with a constant *C*. By applying the same argument to the pointwise estimates (3.5) and (3.6) we can derive the decay estimate (3.8) and (3.9), respectively. Thus this completes the proof.

energy method to the problems (3.2) and (3.3). This yields sharp pointwise estimates of

direction and this gives the additional decay (1 + *t*)−*<sup>α</sup>*/2. The result is stated as follows.

−*κ*|*ξ*| 2*t*

−*κ*|*ξ*| 2*t*

> −*κ*|*ξ*| 2*t*

(*z*0)*x*<sup>1</sup> <sup>∈</sup> *<sup>H</sup>*1(*L*1) *for <sup>α</sup>* <sup>≥</sup> <sup>0</sup>*. Then the solution to the problem* (3.1)*,* (2.6)*,* (2.7) *satisfies the decay*

<sup>−</sup>(*n*−1)/4−*k*/2�*z*0�*L*<sup>2</sup>

−(*n*−1)/4−*k*/2

−(*n*−1)/4−*k*/2

**Corollary 3.4** (Decay estimate)**.** *Assume the same conditions of Theorem 3.3. Let z*<sup>0</sup> <sup>∈</sup> *<sup>L</sup>*<sup>2</sup>


(|*z*ˆ0(·, *ξ*)|*L*<sup>2</sup>


*α*(*L*<sup>1</sup>),

�*z*0�*L*<sup>2</sup>

�*z*0�*L*<sup>2</sup>

*α*

*<sup>ξ</sup> . Then the solution to the problem* (3.2) *verifies the pointwise estimate*

<sup>≤</sup> *Ct*−(*n*−1)/2−*k*�*z*0�<sup>2</sup>

≤ *C* sup *<sup>ξ</sup>*∈**R***n*−<sup>1</sup> *ξ*

where *C* is a positive constant. Here we used (2.2) and the simple inequality

 **R***n*−<sup>1</sup> *ξ* |*ξ*| 2*ke* −*κ*|*ξ*| 2*t*

In the last subsection, we restrict to the non-degenerate case *f* �

solutions to (3.2). We use the weighted space *L*<sup>2</sup>

*<sup>φ</sup>x*<sup>1</sup> <sup>&</sup>gt; <sup>0</sup>*. Let <sup>α</sup>* <sup>≥</sup> <sup>0</sup> *and suppose that <sup>z</sup>*ˆ0(·, *<sup>ξ</sup>*) <sup>∈</sup> *<sup>L</sup>*<sup>2</sup>

<sup>|</sup>*z*ˆ(·, *<sup>ξ</sup>*, *<sup>t</sup>*)|*L*<sup>2</sup> <sup>≤</sup> *<sup>C</sup>*(<sup>1</sup> <sup>+</sup> *<sup>t</sup>*)−*α*/2*<sup>e</sup>*

<sup>|</sup>*z*ˆ*x*<sup>1</sup> (·, *<sup>ξ</sup>*, *<sup>t</sup>*)|*L*<sup>2</sup> <sup>≤</sup> *<sup>C</sup>*(<sup>1</sup> <sup>+</sup> *<sup>t</sup>*)−*α*/2*<sup>e</sup>*

<sup>|</sup>*z*ˆ*x*<sup>1</sup> *<sup>x</sup>*<sup>1</sup> (·, *<sup>ξ</sup>*, *<sup>t</sup>*)|*L*<sup>2</sup> <sup>≤</sup> *<sup>C</sup>*(<sup>1</sup> <sup>+</sup> *<sup>t</sup>*)−*α*/2*<sup>e</sup>*

*<sup>x</sup>*� *<sup>z</sup>*(*t*)�L<sup>2</sup> <sup>≤</sup> *<sup>C</sup>*(<sup>1</sup> <sup>+</sup> *<sup>t</sup>*)−*α*/2*<sup>t</sup>*

*<sup>x</sup>*� *zx*<sup>1</sup> (*t*)�L<sup>2</sup> <sup>≤</sup> *<sup>C</sup>*(<sup>1</sup> <sup>+</sup> *<sup>t</sup>*)−*α*/2*<sup>t</sup>*

*<sup>x</sup>*� *zx*<sup>1</sup> *<sup>x</sup>*<sup>1</sup> (*t*)�L<sup>2</sup> <sup>≤</sup> *<sup>C</sup>*(<sup>1</sup> <sup>+</sup> *<sup>t</sup>*)−*α*/2*<sup>t</sup>*

*<sup>ξ</sup> and t* ≥ 0*, where the norms* |·|*L*<sup>2</sup> *,* |·|*H*<sup>1</sup> *and* |·|*L*<sup>2</sup>

As an easy consequence, we have the following decay estimate.

**Theorem 3.3** (Pointwise estimate)**.** *Let f* �


*L*2 *α*(*L*<sup>1</sup>),

*<sup>d</sup><sup>ξ</sup>* <sup>≤</sup> *Ct*−(*n*−1)/2−*<sup>k</sup>*

<sup>1</sup>(0) < 0 and apply the weighted

, (3.28)

*<sup>α</sup> are with respect to x*<sup>1</sup> ∈ **R**+*, and C*

 ,

, (3.29)

(3.30)

*<sup>α</sup>*(*L*1) *and*

*<sup>α</sup>*(**R**+) (*α* ≥ 0) for *x*<sup>1</sup> ∈ **R**<sup>+</sup> in the normal

*<sup>α</sup>*(**R**+) *and* (*z*ˆ0)*x*<sup>1</sup> (·, *<sup>ξ</sup>*) <sup>∈</sup> *<sup>H</sup>*1(**R**+) *for each*

<sup>1</sup>(0) < 0 *and let φ*(*x*1) *be a stationary solution with*

*<sup>α</sup>* + |(*z*ˆ0)*x*<sup>1</sup> (·, *ξ*)|*L*<sup>2</sup>

*<sup>α</sup>* + |(*z*ˆ0)*x*<sup>1</sup> (·, *ξ*)|*H*<sup>1</sup>

*<sup>α</sup>*(*L*<sup>1</sup>) + �(*z*0)*x*<sup>1</sup> �*L*<sup>2</sup>(*L*<sup>1</sup>)

*<sup>α</sup>*(*L*<sup>1</sup>) + �(*z*0)*x*<sup>1</sup> �*H*<sup>1</sup>(*L*<sup>1</sup>)

into (2.3), we obtain

**3.2 Weighted energy method**

*<sup>ξ</sup>* <sup>∈</sup> **<sup>R</sup>***n*−<sup>1</sup>

*for <sup>ξ</sup>* <sup>∈</sup> **<sup>R</sup>***n*−<sup>1</sup>

*estimate*

*and κ are positive constants.*

�*∂k*

�*∂k*

�*∂k*

�*∂k <sup>x</sup>*� *<sup>z</sup>*(*t*)�<sup>2</sup> L2 *<sup>α</sup>* ≤ *<sup>C</sup>* **R***n*−<sup>1</sup> *ξ* |*ξ*| 2*ke* −*κ*|*ξ*| 2*t*

$$\mathbf{D}\_{\beta} = |\mathfrak{z}\_{\mathfrak{x}\_1}|\_{L^2\_{\beta}}^2 + |\mathfrak{f}|^2 |\mathfrak{z}|\_{L^2\_{\beta}}^2 + |\sqrt{\mathfrak{g}\_{\mathfrak{x}\_1}} \mathfrak{z}|\_{L^2\_{\beta}}^2$$

and *C* and *κ* are positive constants. Notice that **D**ˆ <sup>0</sup> coincides with *D*ˆ <sup>1</sup> in (3.14).

To prove (3.31), we use the equality (3.11). Notice that, by virtue of (2.9)1 in Lemma 2.2, we have

$$-\mathcal{F}\_1 \ge c\_2|\mathfrak{z}|^2 - \mathbb{C}|\mathfrak{z}\_{\mathfrak{x}\_1}|^2 \tag{3.32}$$

with positive constants *<sup>c</sup>*<sup>2</sup> and *<sup>C</sup>*. Now we multiply (3.11) by (<sup>1</sup> <sup>+</sup> *<sup>x</sup>*1)*<sup>β</sup>* (0 <sup>≤</sup> *<sup>β</sup>* <sup>≤</sup> *<sup>α</sup>*) to get

$$\begin{split} \frac{1}{2} \frac{\partial}{\partial t} \left\{ (1+\varkappa\_1)^{\beta} |\mathfrak{z}|^{2} \right\} &+ \left\{ (1+\varkappa\_1)^{\beta} \mathcal{F}\_1 \right\}\_{\varkappa\_1} \\ &+ (1+\varkappa\_1)^{\beta} \mathcal{D}\_1 + \beta (1+\varkappa\_1)^{\beta-1} (-\mathcal{F}\_1) = 0. \end{split}$$

We integrate this equality over *x*<sup>1</sup> ∈ **R**<sup>+</sup> and use (3.12) and (3.32), obtaining

$$\frac{\partial}{\partial t} \left| \hat{z} \right|\_{L^2\_{\beta, \nu}}^2 + c\_1 \hat{\mathbf{D}}\_{\beta} + 2\beta c\_2 \left| \hat{z} \right|\_{L^2\_{\beta - 1}}^2 - (wf\_1)'(u\_b) \left| \hat{z}(0, \xi, t) \right|^2 \le \beta \mathbb{C} \left| \hat{z}\_{\mathbf{x}\_1} \right|\_{L^2\_{\beta - 1}}^2 \tag{3.33}$$

for 0 ≤ *β* ≤ *α*, where we define

$$\|v\|\_{L^2\_{\beta,\nu}} = \left(\int\_0^\infty (1+\mathfrak{x}\_1)^\beta w(\mathfrak{\phi}(\mathfrak{x}\_1)) |v(\mathfrak{x}\_1)|^2 d\mathfrak{x}\_1\right)^{1/2}.$$

and *C* is a positive constant. Here, by virtue of (2.9)1, the last term of the left-hand side of (3.33) is positive. We now observe that

$$\left|a\right|\_{L^2\_{\beta-1}}^2 \le \epsilon \left|a\right|\_{L^2\_{\beta}}^2 + \mathcal{C}\_{\epsilon} \left|a\right|\_{L^2}^2$$

for any *�* > 0, where *C�* is a constant depending on *�*. We apply this inequality to the term on the right-hand side of (3.33) by taking *<sup>a</sup>* <sup>=</sup> *<sup>z</sup>*ˆ*x*<sup>1</sup> . Noting that **<sup>D</sup>**<sup>ˆ</sup> *<sup>β</sup>* ≥ |*z*ˆ*x*<sup>1</sup> <sup>|</sup> 2 *L*2 *β* , we choose *�* > 0 so small that *αC�* ≤ *c*1. This yields

$$\frac{\partial}{\partial t} \left| \hat{z} \right|\_{L^2\_{\beta, \nu}}^2 + c\_3 \hat{\mathbf{D}}\_{\beta} + 2\beta c\_2 \left| \hat{z} \right|\_{L^2\_{\beta - 1}}^2 \le \beta \mathbf{C} \left| \hat{z}\_{\mathbf{x}\_1} \right|\_{L^2}^2 \tag{3.34}$$

for *γ* > *α*, where *D*ˆ <sup>1</sup> is defined in (3.12), and *C* and *κ* are positive constants. Notice that (3.38)

<sup>263</sup> Application of the Weighted Energy Method in the Partial Fourier Space to Linearized Viscous Conservation Laws with Non-Convex Condition

To prove (3.38), we recall the inequality (3.25) which is the same as (3.31) with *β* = 0. We need to estimate the second term on the right-hand side of (3.25). This can be done by applying the technique due to Nishikawa in [10]. When *α* > 0 is not an integer, we have from (3.36) with


where *C* is a positive constant. Now, using a simple interpolation inequality |*a*|*L*<sup>2</sup> ≤

*<sup>θ</sup> <sup>t</sup>* 0

*θ* = *α* − [*α*], *λ* = *γ* − 1, *μ* = [*α*] and the corresponding *ν* determined by *λ* = *μθ* + *ν*(1 − *θ*).

*dτ* <sup>1</sup>−*<sup>θ</sup>*

where we have used the fact that (*ν* − [*α*] + 1)(1 − *θ*) = *γ* − *α*. Substituting this estimate into

**Step 4.** Finally, we prove (3.29) and (3.30). Employing (3.38), we can estimate the last term

<sup>2</sup>*τD*<sup>ˆ</sup> <sup>2</sup>(*ξ*, *<sup>τ</sup>*)*d<sup>τ</sup>*

*<sup>α</sup>* <sup>≤</sup> *<sup>C</sup>*(<sup>1</sup> <sup>+</sup> *<sup>t</sup>*)*γ*−*α*(|*z*ˆ0(·, *<sup>ξ</sup>*)<sup>|</sup>

(1 + *τ*)*γe*

*<sup>α</sup>* <sup>+</sup> <sup>|</sup>*v*ˆ0(·, *<sup>ξ</sup>*)<sup>|</sup>

*κ*|*ξ*|

2 *<sup>H</sup>*<sup>1</sup> )

*κ*|*ξ*|

2 *L*2

> 2 *<sup>L</sup>*<sup>2</sup> + *t* 0

for *γ* > *α*. This means that (3.30). Hence this completes the proof of Theorem 3.3.

2 *L*2

2 *<sup>L</sup>*<sup>2</sup> *dτ*

<sup>2</sup>*τ*|*z*ˆ(·, *<sup>ξ</sup>*, *<sup>τ</sup>*)<sup>|</sup>

2 *L*2 *α*−[*α*]

2 *L*2 *α*−[*α*]−1 ≤ *C*|*z*ˆ0(·, *ξ*)|

≤ *C*|*z*ˆ0(·, *ξ*)|

(<sup>1</sup> <sup>+</sup> *<sup>τ</sup>*)*ν*|*a*(*τ*)<sup>|</sup>

2 *L*2 *α* ,

2 *L*2 *α* ,

2 *L*2 *θ dτ* <sup>1</sup>−*<sup>θ</sup>* ,

<sup>≤</sup> *<sup>C</sup>*(<sup>1</sup> <sup>+</sup> *<sup>t</sup>*)*γ*−*α*|*z*ˆ0(·, *<sup>ξ</sup>*)<sup>|</sup>

2 *L*2

<sup>2</sup>*τD*<sup>ˆ</sup> <sup>3</sup>(*ξ*, *<sup>τ</sup>*)*d<sup>τ</sup>*

*<sup>α</sup>* <sup>+</sup> <sup>|</sup>*v*ˆ0(·, *<sup>ξ</sup>*)<sup>|</sup>

2 *L*2 ) (3.41)

2 *L*2 *α* , (3.39)

(3.40)

<sup>2</sup>*t*/2*z*ˆ(*x*1, *<sup>ξ</sup>*, *<sup>t</sup>*),

gives the desired estimate (3.28) even if *α* > 0 is not an integer.

(1 + *t*)[*α*]

*e κ*|*ξ*|

> 2 *<sup>L</sup>*<sup>2</sup> *dτ*

(<sup>1</sup> <sup>+</sup> *<sup>τ</sup>*)*μ*|*a*(*τ*)<sup>|</sup>

*κ*|*ξ*|

2 *L*2 *α <sup>t</sup>* 0

of the right-hand side of (3.26). Namely, we obtain

2 *<sup>L</sup>*<sup>2</sup> + *t* 0

*<sup>L</sup>*<sup>2</sup> <sup>+</sup> *<sup>C</sup>*(<sup>1</sup> <sup>+</sup> *<sup>t</sup>*)*γ*−*α*|*z*ˆ0(·, *<sup>ξ</sup>*)<sup>|</sup>

*κ*|*ξ*| 2*t*

On the other hand, by applying (3.38) and (3.41) to (3.27), we get

(1 + *τ*)[*α*]

(<sup>1</sup> <sup>+</sup> *<sup>τ</sup>*)*λ*|*a*(*τ*)<sup>|</sup>

 *t* 0

 *t* 0

≤ *<sup>t</sup>* 0

 *t* 0

(1 + *t*)*γe*

≤ *C*|*v*ˆ0(·, *ξ*)|

*κ*|*ξ*| 2*t*

2

for *γ* > *α*. Thus this yields (3.29).

Then, using (3.39), we arrive at the estimate

≤ *C*|*z*ˆ0(·, *ξ*)|

(3.25), we get the desired estimate (3.38).


(1 + *t*)*γe*

(1 + *τ*)*γ*−1*e*

*e κ*|*ξ*| 2*t*

(0 ≤ *θ* ≤ 1) and the Hölder inequality, we see that

2 *L*2 *θ*−1 *dτ*

provided that *<sup>λ</sup>* <sup>=</sup> *μθ* <sup>+</sup> *<sup>ν</sup>*(<sup>1</sup> <sup>−</sup> *<sup>θ</sup>*) with 0 <sup>≤</sup> *<sup>θ</sup>* <sup>≤</sup> 1. We use (3.40) for *<sup>a</sup>* <sup>=</sup> *<sup>e</sup>κ*|*ξ*<sup>|</sup>

<sup>2</sup>*τ*|*z*ˆ(·, *<sup>ξ</sup>*, *<sup>τ</sup>*)<sup>|</sup>

(1 + *τ*)*ν*−[*α*]

(1 + *τ*)*γe*


<sup>≤</sup> *<sup>C</sup>*(<sup>1</sup> <sup>+</sup> *<sup>t</sup>*)*γ*−*α*(|*z*ˆ0(·, *<sup>ξ</sup>*)<sup>|</sup>

*l* = [*α*] that


for constants *c*<sup>3</sup> and *C*. Multiplying (3.34) by *eκ*|*ξ*<sup>|</sup> <sup>2</sup>*<sup>t</sup>* (*κ* > 0), we obtain

$$\frac{\partial}{\partial t} \left\{ e^{\kappa |\tilde{\xi}|^2 t} |\mathcal{Z}|^2\_{L^2\_{\beta, \mu}} \right\} + e^{\kappa |\tilde{\xi}|^2 t} \left( c\_3 \mathbf{\hat{D}}\_{\tilde{\beta}} - \kappa |\tilde{\xi}|^2 |\mathcal{Z}|^2\_{L^2\_{\tilde{\beta}}} \right) + 2 \beta c\_2 e^{\kappa |\tilde{\xi}|^2 t} |\mathcal{Z}|^2\_{L^2\_{\tilde{\beta}-1}} \le \beta C e^{\kappa |\tilde{\xi}|^2 t} |\mathcal{Z}\_{x\_1}|^2\_{L^2}.$$

As in (3.15), we choose *κ* > 0 such that *κ* ≤ *c*3. Then we multiply the resulting inequality by (<sup>1</sup> <sup>+</sup> *<sup>t</sup>*)*<sup>γ</sup>* (*<sup>γ</sup>* <sup>≥</sup> 0) and integrate over [0, *<sup>t</sup>*]. This yields

$$\begin{split} &(1+t)^{\gamma}e^{\kappa|\boldsymbol{\xi}|^{2}t}|\boldsymbol{\varepsilon}(\cdot,\boldsymbol{\xi},t)|\_{L^{2}\_{\beta}}^{2} + \int\_{0}^{t}(1+\tau)^{\gamma}e^{\kappa|\boldsymbol{\xi}|^{2}\tau} \left(\hat{\mathbf{D}}\_{\beta}(\boldsymbol{\xi},\tau) + \beta|\boldsymbol{\varepsilon}(\cdot,\boldsymbol{\xi},\tau)|\_{L^{2}\_{\beta-1}}^{2}\right)d\tau \\ & \qquad \leq C|\boldsymbol{\varepsilon}(\cdot,\boldsymbol{\xi})|\_{L^{2}\_{\beta}}^{2} + \gamma C \int\_{0}^{t}(1+\tau)^{\gamma-1}e^{\kappa|\boldsymbol{\xi}|^{2}\tau}|\boldsymbol{\varepsilon}(\cdot,\boldsymbol{\xi},\tau)|\_{L^{2}\_{\beta}}^{2}d\tau \\ & \qquad \quad + \beta C \int\_{0}^{t}(1+\tau)^{\gamma}e^{\kappa|\boldsymbol{\xi}|^{2}\tau}|\boldsymbol{\varepsilon}\_{1}(\cdot,\boldsymbol{\xi},\tau)|\_{L^{2}}^{2}d\tau, \end{split} \tag{3.35}$$

where *C* is a positive constant. Here the last term on the right-hand side of (3.35) is already estimated in (3.25) because |*z*ˆ*x*<sup>1</sup> | 2 *<sup>L</sup>*<sup>2</sup> <sup>≤</sup> **<sup>D</sup>**<sup>ˆ</sup> <sup>0</sup> <sup>=</sup> *<sup>D</sup>*<sup>ˆ</sup> 1. Therefore the proof of (3.31) is complete.

**Step 2.** Next we show the following estimate for *α* ≥ 0:

$$\begin{split} & (1+t)^{l} e^{\kappa |\tilde{\xi}|^{2}t} |\mathfrak{z}(\cdot,\tilde{\xi},t)|\_{L^{2}\_{a-l}}^{2} \\ & + \int\_{0}^{t} (1+\tau)^{l} e^{\kappa |\tilde{\xi}|^{2}\tau} \left( \mathbf{\hat{D}}\_{a-l}(\tilde{\xi},\tau) + (a-l) |\mathfrak{z}(\cdot,\tilde{\xi},\tau)|\_{L^{2}\_{a-l-1}}^{2} \right) d\tau \leq \mathsf{C} |\mathfrak{z}\_{0}(\cdot,\tilde{\xi})|\_{L^{2}\_{a}}^{2} \end{split} \tag{3.36}$$

for each integer *l* with 0 ≤ *l* ≤ [*α*], where *C* and *κ* are positive constants. Note that if *α* ≥ 0 is an integer, then (3.36) with *l* = *α* gives the desired estimate (3.28).

We prove (3.36) by induction with respect to the integer *l* with 0 ≤ *l* ≤ [*α*]. First we put *γ* = 0 and *β* = *α* in (3.31). This shows that (3.36) holds true for *l* = 0. Now, let 1 ≤ *j* ≤ [*α*] (for *α* ≥ 1) and suppose that (3.36) holds true for *l* = *j* − 1. In particular, we suppose that

$$\int\_0^t (1+\tau)^{j-1} e^{\kappa |\xi|^2 \tau} |\mathfrak{z}(\cdot, \xi, \tau)|\_{L^2\_{a-j}}^2 d\tau \le \mathbb{C} |\mathfrak{z}\_0(\cdot, \xi)|\_{L^2\_a}^2. \tag{3.37}$$

Then we prove (3.36) for *l* = *j*. To this end, we put *γ* = *j* and *β* = *α* − *j* in (3.31). This gives

$$\begin{split} & (1+t)^{j} e^{\kappa |\boldsymbol{\xi}|^{2}} |\boldsymbol{\varepsilon}(\cdot,\boldsymbol{\xi},t)|\_{L^{2}\_{a-j}}^{2} \\ & \quad + \int\_{0}^{t} (1+\tau)^{j} e^{\kappa |\boldsymbol{\xi}|^{2}\tau} \left( \mathsf{D}\_{a-j}(\boldsymbol{\xi},\tau) + (a-j) |\boldsymbol{\varepsilon}(\cdot,\boldsymbol{\xi},\tau)|\_{L^{2}\_{a-j-1}}^{2} \right) d\tau \\ & \quad \leq \mathsf{C} |\boldsymbol{\varepsilon}\_{0}(\cdot,\boldsymbol{\xi})|\_{L^{2}\_{a-j}}^{2} + j\mathsf{C} \int\_{0}^{t} (1+\tau)^{j-1} e^{\kappa |\boldsymbol{\xi}|^{2}\tau} |\boldsymbol{\varepsilon}(\cdot,\boldsymbol{\xi},\tau)|\_{L^{2}\_{a-j}}^{2} d\tau \leq \mathsf{C} |\boldsymbol{\varepsilon}\_{0}(\cdot,\boldsymbol{\xi})|\_{L^{2}\_{a-j}}^{2} \end{split}$$

where we used (3.37) in the last estimate. This shows that (3.36) holds true also for *l* = *j* and therefore the proof of (3.36) is complete.

**Step 3.** Next, when *α* > 0 is not an integer, we show that

$$\begin{split} & (1+t)^{\gamma} e^{\kappa |\boldsymbol{\xi}|^{2}t} |\boldsymbol{\varepsilon}(\cdot,\boldsymbol{\xi},t)|\_{L^{2}}^{2} + \int\_{0}^{t} (1+\tau)^{\gamma} e^{\kappa |\boldsymbol{\xi}|^{2}\tau} \hat{D}\_{1}(\boldsymbol{\xi},\tau) d\tau \\ & \leq \mathsf{C} (1+t)^{\gamma-\mathsf{R}} |\boldsymbol{\varepsilon}\_{0}(\cdot,\boldsymbol{\xi})|\_{L^{2}\_{a}}^{2} \end{split} \tag{3.38}$$

for *γ* > *α*, where *D*ˆ <sup>1</sup> is defined in (3.12), and *C* and *κ* are positive constants. Notice that (3.38) gives the desired estimate (3.28) even if *α* > 0 is not an integer.

To prove (3.38), we recall the inequality (3.25) which is the same as (3.31) with *β* = 0. We need to estimate the second term on the right-hand side of (3.25). This can be done by applying the technique due to Nishikawa in [10]. When *α* > 0 is not an integer, we have from (3.36) with *l* = [*α*] that

$$\begin{aligned} (1+t)^{[a]} e^{\kappa |\tilde{\xi}|^2 t} |\hat{\varepsilon}(\cdot, \tilde{\xi}, t)|\_{L^2\_{a-[a]}}^2 &\leq \mathbb{C} |\hat{\varepsilon}\_0(\cdot, \tilde{\xi})|\_{L^2\_{a'}}^2 \\ \int\_0^t (1+\tau)^{[a]} e^{\kappa |\tilde{\xi}|^2 \tau} |\hat{\varepsilon}(\cdot, \tilde{\xi}, \tau)|\_{L^2\_{a-[a]-1}}^2 &\leq \mathbb{C} |\hat{\varepsilon}\_0(\cdot, \tilde{\xi})|\_{L^2\_{a'}}^2 \end{aligned} \tag{3.39}$$

where *C* is a positive constant. Now, using a simple interpolation inequality |*a*|*L*<sup>2</sup> ≤ |*a*| *θ L*2 *θ*−1 |*a*| 1−*θ L*2 *θ* (0 ≤ *θ* ≤ 1) and the Hölder inequality, we see that

$$\begin{split} &\int\_{0}^{t} (1+\tau)^{\lambda} |a(\tau)|\_{L^{2}}^{2} d\tau \\ &\leq \left( \int\_{0}^{t} (1+\tau)^{\mu} |a(\tau)|\_{L^{2}\_{\theta-1}}^{2} d\tau \right)^{\theta} \left( \int\_{0}^{t} (1+\tau)^{\nu} |a(\tau)|\_{L^{2}\_{\theta}}^{2} d\tau \right)^{1-\theta} \end{split} \tag{3.40}$$

provided that *<sup>λ</sup>* <sup>=</sup> *μθ* <sup>+</sup> *<sup>ν</sup>*(<sup>1</sup> <sup>−</sup> *<sup>θ</sup>*) with 0 <sup>≤</sup> *<sup>θ</sup>* <sup>≤</sup> 1. We use (3.40) for *<sup>a</sup>* <sup>=</sup> *<sup>e</sup>κ*|*ξ*<sup>|</sup> <sup>2</sup>*t*/2*z*ˆ(*x*1, *<sup>ξ</sup>*, *<sup>t</sup>*), *θ* = *α* − [*α*], *λ* = *γ* − 1, *μ* = [*α*] and the corresponding *ν* determined by *λ* = *μθ* + *ν*(1 − *θ*). Then, using (3.39), we arrive at the estimate

$$\begin{aligned} &\int\_0^t (1+\tau)^{\gamma-1} e^{\kappa |\boldsymbol{\xi}|^2 \tau} |\boldsymbol{\varepsilon}(\cdot, \boldsymbol{\xi}, \tau)|\_{L^2}^2 d\tau \\ &\leq C |\boldsymbol{\varepsilon}\_0(\cdot, \boldsymbol{\xi})|\_{L^2\_x}^2 \left( \int\_0^t (1+\tau)^{\nu-[\boldsymbol{\alpha}]} d\tau \right)^{1-\theta} \leq C(1+t)^{\gamma-a} |\boldsymbol{\varepsilon}\_0(\cdot, \boldsymbol{\xi})|\_{L^2\_x}^2. \end{aligned}$$

where we have used the fact that (*ν* − [*α*] + 1)(1 − *θ*) = *γ* − *α*. Substituting this estimate into (3.25), we get the desired estimate (3.38).

**Step 4.** Finally, we prove (3.29) and (3.30). Employing (3.38), we can estimate the last term of the right-hand side of (3.26). Namely, we obtain

$$\begin{split} &(1+t)^{\gamma} e^{\kappa |\boldsymbol{\xi}|^{2}t} |\boldsymbol{\vartheta}(\cdot,\boldsymbol{\xi},t)|\_{L^{2}}^{2} + \int\_{0}^{t} (1+\tau)^{\gamma} e^{\kappa |\boldsymbol{\xi}|^{2}\tau} \hat{D}\_{2}(\boldsymbol{\xi},\tau) d\tau \\ & \leq C |\hat{v}\_{0}(\cdot,\boldsymbol{\xi})|\_{L^{2}}^{2} + \mathcal{C}(1+t)^{\gamma-a} |\hat{z}\_{0}(\cdot,\boldsymbol{\xi})|\_{L^{2}\_{x}}^{2} \leq \mathcal{C}(1+t)^{\gamma-a} (|\hat{z}\_{0}(\cdot,\boldsymbol{\xi})|\_{L^{2}\_{x}}^{2} + |\hat{v}\_{0}(\cdot,\boldsymbol{\xi})|\_{L^{2}}^{2}) \end{split} \tag{3.41}$$

for *γ* > *α*. Thus this yields (3.29).

14 Will-be-set-by-IN-TECH

2|*z*ˆ| 2 *L*2 *β* 

As in (3.15), we choose *κ* > 0 such that *κ* ≤ *c*3. Then we multiply the resulting inequality by

*κ*|*ξ*| <sup>2</sup>*τ*

*κ*|*ξ*|

<sup>2</sup>*τ*|*z*ˆ*x*<sup>1</sup> (·, *<sup>ξ</sup>*, *<sup>τ</sup>*)<sup>|</sup>

where *C* is a positive constant. Here the last term on the right-hand side of (3.35) is already

for each integer *l* with 0 ≤ *l* ≤ [*α*], where *C* and *κ* are positive constants. Note that if *α* ≥ 0 is

We prove (3.36) by induction with respect to the integer *l* with 0 ≤ *l* ≤ [*α*]. First we put *γ* = 0 and *β* = *α* in (3.31). This shows that (3.36) holds true for *l* = 0. Now, let 1 ≤ *j* ≤ [*α*] (for *α* ≥ 1)

> 2 *L*2 *α*−*j*

<sup>2</sup>*τ*|*z*ˆ(·, *<sup>ξ</sup>*, *<sup>τ</sup>*)<sup>|</sup>

Then we prove (3.36) for *l* = *j*. To this end, we put *γ* = *j* and *β* = *α* − *j* in (3.31). This gives

**<sup>D</sup>**<sup>ˆ</sup> *<sup>α</sup>*−*j*(*ξ*, *<sup>τ</sup>*)+(*<sup>α</sup>* <sup>−</sup> *<sup>j</sup>*)|*z*ˆ(·, *<sup>ξ</sup>*, *<sup>τ</sup>*)<sup>|</sup>

*κ*|*ξ*|

where we used (3.37) in the last estimate. This shows that (3.36) holds true also for *l* = *j* and

(1 + *τ*)*j*−1*e*

2 *<sup>L</sup>*<sup>2</sup> + *t* 0

> 2 *L*2 *α*

*<sup>c</sup>*3**D**<sup>ˆ</sup> *<sup>β</sup>* <sup>−</sup> *<sup>κ</sup>*|*ξ*<sup>|</sup>

(1 + *τ*)*γe*

(1 + *τ*)*γ*−1*e*

*κ*|*ξ*|

**<sup>D</sup>**<sup>ˆ</sup> *<sup>α</sup>*−*l*(*ξ*, *<sup>τ</sup>*)+(*<sup>α</sup>* <sup>−</sup> *<sup>l</sup>*)|*z*ˆ(·, *<sup>ξ</sup>*, *<sup>τ</sup>*)<sup>|</sup>

and suppose that (3.36) holds true for *l* = *j* − 1. In particular, we suppose that

*κ*|*ξ*|

<sup>2</sup>*<sup>t</sup>* (*κ* > 0), we obtain

*κ*|*ξ*| 2*t* |*z*ˆ| 2 *L*2 *β*−1

**<sup>D</sup>**<sup>ˆ</sup> *<sup>β</sup>*(*ξ*, *<sup>τ</sup>*) + *<sup>β</sup>*|*z*ˆ(·, *<sup>ξ</sup>*, *<sup>τ</sup>*)<sup>|</sup>

2 *L*2 *β dτ*

*<sup>L</sup>*<sup>2</sup> <sup>≤</sup> **<sup>D</sup>**<sup>ˆ</sup> <sup>0</sup> <sup>=</sup> *<sup>D</sup>*<sup>ˆ</sup> 1. Therefore the proof of (3.31) is complete.

2 *L*2 *α*−*l*−1 

*dτ* ≤ *C*|*z*ˆ0(·, *ξ*)|

2 *L*2 *α*−*j*−1 *dτ*

2 *L*2 *α*−*j*

*κ*|*ξ*|

<sup>2</sup>*τD*<sup>ˆ</sup> <sup>1</sup>(*ξ*, *<sup>τ</sup>*)*d<sup>τ</sup>*

<sup>2</sup>*τ*|*z*ˆ(·, *<sup>ξ</sup>*, *<sup>τ</sup>*)<sup>|</sup>

(1 + *τ*)*γe*

<sup>≤</sup> *<sup>β</sup>Ceκ*|*ξ*<sup>|</sup>

2 *L*2 *β*−1 *dτ*

*dτ* ≤ *C*|*z*ˆ0(·, *ξ*)|

2 *L*2 *α*

*dτ* ≤ *C*|*z*ˆ0(·, *ξ*)|

2 *L*2 *α*

. (3.37)

2 *L*2 *α* ,

(3.38)

2*t* |*z*ˆ*x*<sup>1</sup> | 2 *L*2 .

(3.35)

(3.36)

+ 2*βc*2*e*

<sup>2</sup>*τ*|*z*ˆ(·, *<sup>ξ</sup>*, *<sup>τ</sup>*)<sup>|</sup>

2 *<sup>L</sup>*<sup>2</sup> *dτ*,

for constants *c*<sup>3</sup> and *C*. Multiplying (3.34) by *eκ*|*ξ*<sup>|</sup>

(<sup>1</sup> <sup>+</sup> *<sup>t</sup>*)*<sup>γ</sup>* (*<sup>γ</sup>* <sup>≥</sup> 0) and integrate over [0, *<sup>t</sup>*]. This yields

2 *L*2 *β* + *t* 0

> *t* 0

(1 + *τ*)*γe*

2

an integer, then (3.36) with *l* = *α* gives the desired estimate (3.28).

(1 + *τ*)*j*−1*e*

2 *L*2 *α*−*j*


**Step 3.** Next, when *α* > 0 is not an integer, we show that

*κ*|*ξ*| 2*t*

<sup>≤</sup> *<sup>C</sup>*(<sup>1</sup> <sup>+</sup> *<sup>t</sup>*)*γ*−*α*|*z*ˆ0(·, *<sup>ξ</sup>*)<sup>|</sup>


**Step 2.** Next we show the following estimate for *α* ≥ 0:

2 *L*2 *α*−*l*


2 *L*2 *β* + *γC*

+ *βC t* 0


 *t* 0

(1 + *τ*)*<sup>j</sup> e κ*|*ξ*| <sup>2</sup>*τ*

therefore the proof of (3.36) is complete.

2 *L*2 *α*−*j* + *jC t* 0

(1 + *t*)*γe*

≤ *C*|*z*ˆ0(·, *ξ*)|

*∂ ∂t e κ*|*ξ*| 2*t* |*z*ˆ| 2 *L*2 *β*,*w* + *e κ*|*ξ*| 2*t* 

(1 + *t*)*γe*

(1 + *t*)*<sup>l</sup> e κ*|*ξ*| 2*t*

+ *t* 0

*κ*|*ξ*| 2*t*

≤ *C*|*z*ˆ0(·, *ξ*)|

estimated in (3.25) because |*z*ˆ*x*<sup>1</sup> |

(1 + *τ*)*<sup>l</sup> e κ*|*ξ*| <sup>2</sup>*τ*

(1 + *t*)*<sup>j</sup> e κ*|*ξ*| 2*t*

+ *t* 0

On the other hand, by applying (3.38) and (3.41) to (3.27), we get

$$\begin{aligned} &(1+t)^{\gamma}e^{\kappa|\xi|^2t}|\widehat{v}\_{\mathfrak{X}\_1}(\cdot,\xi,t)|\_{L^2}^2 + \int\_0^t (1+\tau)^{\gamma}e^{\kappa|\xi|^2\tau} \widehat{D}\_3(\xi,\tau)d\tau \\ &\leq C(1+t)^{\gamma-\alpha}(|\widehat{z}\_0(\cdot,\xi)|\_{L^2\_{\mathfrak{X}}}^2 + |\vartheta\_0(\cdot,\xi)|\_{H^1}^2) \end{aligned}$$

for *γ* > *α*. This means that (3.30). Hence this completes the proof of Theorem 3.3.

#### **4. References**


**1. Introduction**

applications.

Committee of University of Macau.

Since Bachelier, a French mathematician, first tried to give a mathematical definition for Brownian motion and used it to model the dynamics of stock process in 1900, financial mathematics has developed a lot. Black & Scholes (1973) and Merton (1973) respectively used the geometric Brownian motion (GBM) to model the underlying asset's price process so that they opened the gate of easy ways to compute option prices, which led one of the major breakthroughs of modern finance. Many good ideas have been proposed to model the stock pricing processes since then. Merton (1976) first introduced the jumps into the asset price processes in his seminar paper. More recently, a lot of exponential Lévy models, including Kou's model, Variance Gamma (VG) model, Inverse Gaussian (IG) model, Normal Inverse Gaussian (NIG) model, and CGMY model, etc., were proposed to add jumps in the financial models so that they can describe the statistical properties of financial time series better [e.g. see Cont & Tankov (2003) and references there in]. Also, serval stochastic volatility modes were presented [e.g. see Heston (1993), Bates (1996), and Duffie *et al* (2000), etc.]. Empirical financial data indicate that these models are usually more consist with financial markets.

**Fourier Transform Methods for Option Pricing\*** 

*Department of Mathematics, University of Macau, Macao,* 

**11**

Deng Ding

*China* 

Under assuming that the price of an underlying asset follows a GBM, Black & Scholes (1973) showed that the value of a European option satisfies a boundary problem of heat equation (Black-Scholes equation) so that they derived an explicit formula (Black-Scholes formula) for the value. However, this nice analytic tractability in option pricing can not be carried over to the most exponential Lévy models or the stochastic volatility modes for asset returns. Thus, many new approaches, including efficient numerical methods, for option pricing have been proposed. These approaches and methods can be classified into four major groups: The partial integro-differential equation (PIDE) and various numerical methods for such equations; The Monte Carlo methods via the stochastic simulation techniques for underlying asset price processes; Directly numerical integration and various numerical methods via differential integral transforms; Backward stochastic differential equation and its numerical methods. Each of them has its advantages and disadvantages for different financial models and specific

Since Stein & Stein (1991) first used Fourier inversion method to find the distribution of the underlying asset in a stochastic volatility model, the Fourier transform methods have become a very active field of financial mathematics. Heston (1993) applied the characteristic function

\*The work was supported by the research grant MYRG136(Y1-L2)-FST11-DD from the Research
