Linear Stability of Solitary Waves for the Isothermal Euler–Poisson System

Abstract

We study the asymptotic linear stability of a two-parameter family of solitary waves for the isothermal Euler–Poisson system. When the linearized equations about the solitary waves are considered, the associated eigenvalue problem in $L^2$ space has a zero eigenvalue embedded in the neutral spectrum, i.e., there is no spectral gap. To resolve this issue, use is made of an exponentially weighted $L^2$ norm so that the essential spectrum is strictly shifted into the left-half plane, and this is closely related to the fact that solitary waves exist in the super-ion-sonic regime. Furthermore, in a certain long-wavelength scaling, we show that the Evans function for the Euler–Poisson system converges to that for the Korteweg–de Vries (KdV) equation as an amplitude parameter tends to zero, from which we deduce that the origin is the only eigenvalue on its natural domain with algebraic multiplicity two. We also show that the solitary waves are spectrally stable in $L^2$ space. Moreover, we discuss (in)stability of large amplitude solitary waves.

Appendix

1.1 Eigenvalue problems of the Euler–Poisson system and the KdV equation

We present a formal reduction of the eigenvalue problem of the Euler–Poisson system (1.10) to the eigenvalue problem of the KdV equation. By intoducing the scaling

$$\begin{aligned}&\xi := \varepsilon ^{1/2}x, \quad \lambda := \varepsilon ^{3/2}\Lambda , \quad (n_*,u_*,\phi _*)(\xi ):=\varepsilon ^{-1}(n_c,u_c,\phi _c)(x), \\&\quad (\dot{n}_*, \dot{u}_*, \dot{\phi }_*)(\xi ):=(\dot{n},\dot{u},\dot{\phi })(x), \end{aligned}$$

(1.10) becomes (letting $(\dot{n}_*,\dot{u}_*,\dot{\phi }_*)=(n,u,\phi )$ for notational simplicity)

By integrating (9.1)a–(9.1)b in $\xi$, we formally obtain that (recall that $c=\sqrt{1+K}+\varepsilon$)

Taking derivative of (9.1)c in $\xi$, and then subtracting the resulting equation from (9.1)b, $-\partial _\xi \phi$ term in the RHS of (9.1)b is canceled. Then, by applying the Taylor expansion, we obtain

$$\begin{aligned}&\varepsilon \Lambda u -(\sqrt{1+K}+\varepsilon ) \partial _\xi u + (K + 1 )\partial _\xi n - K \partial _\xi (\varepsilon n_*n) + \partial _\xi (\varepsilon u_*u)\nonumber \\&\quad + \varepsilon \partial _\xi ^3\phi - \partial _\xi (\varepsilon \phi _*\phi ) =O(\varepsilon ^2). \end{aligned}$$

(9.3)

Multiplying (9.1)a by $\mathsf {V}=\sqrt{1+K}$ and then adding the resulting equation to (9.3), we see that the terms $-\sqrt{1+K}\,\partial _\xi u$ and $(1+K)\partial _\xi n$ in (9.3) are canceled, and we have

$$\begin{aligned} \begin{aligned}&\mathsf {V}\Lambda n -\mathsf {V}\partial _\xi n + \mathsf {V}\partial _\xi (n_*u+u_*n) \\&+ \Lambda u - \partial _\xi u - K \partial _\xi (n_*n) + \partial _\xi (u_*u) + \partial _\xi ^3\phi - \partial _\xi (\phi _*\phi ) =O(\varepsilon ). \end{aligned} \end{aligned}$$

(9.4)

Using the relation (9.2) and Theorem 2.1, we obtain from (9.4) that

$$\begin{aligned} \Lambda n - \partial _\xi n + \mathsf {V}\partial _\xi (\Psi _\text {K} n) + \frac{1}{2\mathsf {V}}\partial _\xi ^3 n = O(\varepsilon ). \end{aligned}$$

(9.5)

1.2 Specific form of $A_*(\xi ,\Lambda ,\varepsilon )$

We denote $\partial _\xi$ by $'$ for simplicity. We let ${A_1}_*(\xi ,\varepsilon ):=A_1(x,\varepsilon )$, ${A_2}_*(\xi ,\varepsilon ):=A_2(x,\varepsilon )$, and $(n_*,u_*,\phi _*)(\xi ):=\varepsilon ^{-1}(n_c,u_c,\phi _c)(x)$ for $\xi =\varepsilon ^{1/2}x$. Then from (3.14) and (2.12), we have

$$\begin{aligned}&{A_1}_*= \small \begin{pmatrix} \frac{(c-\varepsilon u_*)\varepsilon ^{3/2} u_*'}{J} - \frac{K\varepsilon ^{3/2}n_*'}{J(1+\varepsilon n_*)} &{} \frac{(c-\varepsilon u_*)\varepsilon ^{3/2}n_*'}{J} + \frac{(1+\varepsilon n_*)\varepsilon ^{3/2} u_*'}{J} &{} 0 &{} \frac{1+\varepsilon n_*}{J} \\ \frac{K\varepsilon ^{3/2} u_*'}{J(1+\varepsilon n_*)} -\frac{K(c-\varepsilon u_*)\varepsilon ^{3/2}n_*'}{J(1+\varepsilon n_*)^2} &{} \frac{K\varepsilon ^{3/2} n_*'}{J(1+\varepsilon n_*)} + \frac{(c-\varepsilon u_*)\varepsilon ^{3/2} u_*'}{J} &{} 0 &{} \frac{(c-\varepsilon u_*)}{J} \\ 0 &{} 0 &{} 0 &{} 1 \\ -1 &{} 0 &{} e^{\varepsilon \phi _*} &{} 0 \end{pmatrix}, \\&\frac{1}{\sqrt{\varepsilon }}S^{-1}{A_1}_*S = \small \begin{pmatrix} \dfrac{a_{22}-\mathsf {V}a_{12}}{\varepsilon ^{1/2}} &{} \frac{a_{21} + \mathsf {V}a_{22} -\mathsf {V}a_{11} - \mathsf {V}^2a_{12} }{\varepsilon ^{3/2}} &{} \dfrac{a_{24}-\mathsf {V}a_{14} }{\varepsilon } &{} 0 \\ \varepsilon ^{1/2} \, a_{12} &{} \dfrac{a_{11} + \mathsf {V}a_{12}}{\varepsilon ^{1/2}} &{} a_{14} &{} 0 \\ 0 &{} \dfrac{e^{\varepsilon \phi _*}-1}{\varepsilon } &{} 0 &{} e^{\varepsilon \phi _*} \\ \dfrac{-a_{12}}{\varepsilon ^{1/2}} &{} -\dfrac{a_{11} + \mathsf {V}a_{12}}{\varepsilon ^{3/2}} &{} \dfrac{1-a_{14}}{\varepsilon } &{} 0 \end{pmatrix}, \\&\frac{\varepsilon ^{3/2}\Lambda }{\sqrt{\varepsilon }}S^{-1}{A_2}_*S= \frac{\Lambda }{J} \small \begin{pmatrix} \varepsilon ((c-\varepsilon u_*) - \mathsf {V}(1+\varepsilon n_*)) &{} \frac{K}{1+\varepsilon n_*} - \mathsf {V}^2(1+\varepsilon n_*) &{} 0 &{} 0 \\ \varepsilon ^2 (1+\varepsilon n_*) &{} \varepsilon (c-\varepsilon u_*)+\varepsilon \mathsf {V}(1+\varepsilon n_*) &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 \\ -\varepsilon (1+\varepsilon n_*) &{} -(c-\varepsilon u_*)-\mathsf {V}(1+\varepsilon n_*) &{} 0 &{} 0 \end{pmatrix}, \end{aligned}$$

where $a_{jk}$ is the (j, k)-entry of the matrix ${A_1}_*$. Here, we note that

$$\begin{aligned} \frac{1-a_{14}}{\varepsilon } = \frac{(c-\varepsilon u_*)^2- K - (1+ \varepsilon n_*)}{\varepsilon J} = \frac{2\sqrt{1+K} + \varepsilon - 2c u_*+ \varepsilon u_*- n_*}{J}. \end{aligned}$$

Hence, we may define $A_*(\xi ,\Lambda ,\varepsilon )$ for all $\varepsilon \in [0,\varepsilon _*]$, and using (2.12), it is straightforward to check that we have (3.58) at $\varepsilon =0$.

1.3 Proof of Lemma 3.9

Let $\mathcal {A}(\lambda ):=d/dx - A(x,\lambda ,\varepsilon )$. For a closed subspace $\text {Ran}(\lambda -\mathcal {L})$ of a Hilbert space $\mathcal {H}$, we have

$$\begin{aligned} \mathcal {H}=\text {Ran}(\lambda -\mathcal {L}) \oplus \text {Ran}(\lambda -\mathcal {L})^\perp = \text {Ran}(\lambda -\mathcal {L}) \oplus \text {Ker}\left( \lambda - \mathcal {L} \right) ^*\quad ( \,^*:= \text {Hermitian adjoint}) \end{aligned}$$

and a similar decomposition holds for a closed subspace $\text {Ran}\mathcal {A}(\lambda )$. We claim that

• C1. $\text {Ran}(\lambda -\mathcal {L})$ is closed if and only if $\text {Ran}(\mathcal {A}(\lambda ))$ is closed;

• C2. $\text {Ker}(\lambda -\mathcal {L})$ is isomorphic to $\text {Ker}(\mathcal {A}(\lambda ))$ ;

• C3. $\text {Ker}\left( \lambda - \mathcal {L} \right) ^*$ is isomorphic to $\text {Ker}(\mathcal {A}(\lambda )^*)$.

Then, Lemma 3.9 follows by the definition of the Fredholm index of an operator.

C2 and C3 are already checked in Section 3.2.⁷ We only prove the right direction of C1 since the converse is easier to check. Also, we only consider the case of the unweighted $L^2$-space for simplicity.

We suppose that $\text {Ran}(\lambda -\mathcal {L})$ is closed. For a sequence $\mathbf {f}_j=(f_j^1,f_j^2,f_j^3,f_j^4)^T \in \text {Ran}(\mathcal {A}(\lambda ))$ such that $\mathbf {f}_j \rightarrow \mathbf {f}=(f^1,f^2,f^3,f^4)^T$ in $(L^2)^4$ as $j \rightarrow \infty$, let $\mathbf {y}_j:=(n_j,u_j,\phi _j,\psi _j)^T\in (H^1)^4$ be a solution of $\mathcal {A}(\lambda )\mathbf {y}_j=\mathbf {f}_j$ for each $j\in \mathbb {Z}$.

We may decompose the last two components of $\mathcal {A}(\lambda )\mathbf {y}_j=\mathbf {f}_j$ (corresponding to the Poisson equation (3.12)) into two parts:

$$\begin{aligned} \left\{ \begin{array}{l l} \partial _x\phi _j^f - \psi _j^f = f_j^3, \\ \partial _x\psi _j^f - e^{\phi _c}\phi _j^f = f_j^4, \end{array} \right. \quad \left\{ \begin{array}{l l} \partial _x(\phi _j-\phi _j^f) - (\psi _j - \psi _j^f) = 0, \\ \partial _x(\psi _j - \psi _j^f) - e^{\phi _c}(\phi _j-\phi _j^f) + n_j = 0. \end{array} \right. \end{aligned}$$

(9.6)

Using the energy estimate, one can check that the solution $(\phi _j^f,\psi _j^f) \in (H^1)^2$ to the LHS of (9.6) exists for all $(f_j^3,f_j^4) \in (L^2)^2$, and $(\phi _j^f,\psi _j^f)$ converges to $(\phi ^f,\psi ^f)$ in $(H^1)^2$, where $(\phi ^f,\psi ^f)$ $\in$ $(H^1)^2$ satisfies

$$\begin{aligned} \partial _x\phi ^f - \psi = f^3, \quad \partial _x\psi ^f - e^{\phi _c}\phi = f^4. \end{aligned}$$

(9.7)

We rewrite the first system of (9.6) and the system (9.7) as

$$\begin{aligned} \mathcal {A}(\lambda )\begin{pmatrix} 0 \\ 0 \\ \phi _j^f \\ \psi _j^f \end{pmatrix} = \begin{pmatrix} L^{-1}\begin{pmatrix} 0 \\ \psi _j^f \end{pmatrix} \\ f_j^3 \\ f_j^4 \end{pmatrix} \quad \text {and} \quad \mathcal {A}(\lambda )\begin{pmatrix} 0 \\ 0 \\ \phi ^f \\ \psi ^f \end{pmatrix} = \begin{pmatrix} L^{-1}\begin{pmatrix} 0 \\ \psi ^f \end{pmatrix} \\ f^3 \\ f^4 \end{pmatrix}, \end{aligned}$$

(9.8)

respectively. Subtracting the LHS of (9.8) from $\mathcal {A}(\lambda )\mathbf {y}_j=\mathbf {f}_j$, we have

$$\begin{aligned} \mathcal {A}(\lambda )\begin{pmatrix} n_j \\ u_j \\ \phi _j-\phi _j^f \\ \psi _j-\psi _j^f \end{pmatrix} = \begin{pmatrix} \begin{pmatrix} f_j^1 \\ f_j^2 \end{pmatrix} -L^{-1}\begin{pmatrix} 0 \\ \psi _j^f \end{pmatrix} \\ 0 \\ 0 \end{pmatrix}, \end{aligned}$$

which implies that

$$\begin{aligned} (\lambda -\mathcal {L})\begin{pmatrix} n_j\\ u_j \end{pmatrix} =L\left( \begin{pmatrix} f_j^1 \\ f_j^2 \end{pmatrix} -L^{-1}\begin{pmatrix} 0 \\ \psi _j^f \end{pmatrix} \right) . \end{aligned}$$

Since $\text {Ran}(\lambda -\mathcal {L})$ is closed, there is $(n,u)\in (H^1)^2$ such that

$$\begin{aligned} (\lambda -\mathcal {L})\begin{pmatrix} n \\ u \end{pmatrix} =L\left( \begin{pmatrix} f^1 \\ f^2 \end{pmatrix} -L^{-1}\begin{pmatrix} 0 \\ \psi ^f \end{pmatrix} \right) , \end{aligned}$$

and this implies that

$$\begin{aligned} \mathcal {A}(\lambda )\begin{pmatrix} n \\ u \\ \tilde{\phi } \\ \tilde{\psi } \end{pmatrix} = \begin{pmatrix} \begin{pmatrix} f^1 \\ f^2 \end{pmatrix} -L^{-1}\begin{pmatrix} 0 \\ \psi ^f \end{pmatrix} \\ 0 \\ 0 \end{pmatrix}, \end{aligned}$$

(9.9)

where $(\widetilde{\phi },\widetilde{\psi })$ satisfies $\partial _x \widetilde{\phi } = \widetilde{\psi }$ and $\partial _x\widetilde{\psi } = e^{\phi _c}\widetilde{\phi } - n$. Now adding (9.9) and the RHS of (9.8), we conclude that $\text {Ran}(\mathcal {A}(\lambda ))$ is closed.

1.4 Non-negativity of the matrix $S_1$

We show that the symmetric matrix $S_1(x,\varepsilon )$ defined in (5.42) is non-negative for all sufficiently small $\varepsilon >0$.

Since $S_1$ is symmetric, it is enough to show that the eigenvalues of $S_1$,

$$\begin{aligned} 0, 0, 2\sqrt{K} R^{(1)}_{11} \pm \sqrt{2K^2( R^{(1)}_{12})^2 + 2( R^{(1)}_{21})^2}, \end{aligned}$$

are non-negative. Recalling the definition of the matrix $R^{(1)}$ (see (5.29)), $R_{11}^{(1)}$ is positive since

$$\begin{aligned} \begin{aligned} R_{11}^{(1)}&= \frac{c-u_c}{J} - \frac{c}{c^2-K} \\&= \frac{(c-u_c)(c^2-K) - c[(u_c-c)^2-K]}{J(c^2-K)} \\&= \frac{u_c[c(c-u_c)+K]}{J(c^2-K)} > 0, \end{aligned} \end{aligned}$$

where we have used the solitary wave identity (2.2a) for the inequality.

We check that $2\sqrt{K} R^{(1)}_{11} - \sqrt{2K^2( R^{(1)}_{12})^2 + 2( R^{(1)}_{21})^2}$ is positive. Since $R^{(1)}_{11}>0$, it is enough to show that

$$\begin{aligned} \begin{aligned}&4K(R^{(1)}_{11})^2 - 2K^2(R^{(1)}_{12})^2 - 2(R^{(1)}_{21})^2 \\&\quad = \frac{2K}{J_1}\left[ \underbrace{2\left( (c-u_c)(c^2-K)-cJ \right) ^2(1+n_c)^2}_{=:I_1=2(R^{(1)}_{11})^2J_1} \underbrace{- K\left( (c^2-K)(1+n_c)-J \right) ^2(1+n_c)^2}_{=:I_2} \right. \\&\quad \left. \underbrace{-K\left( c^2-K-(1+n_c)J\right) ^2}_{=:I_3}\right] \end{aligned} \end{aligned}$$

is positive, where $J_1:=J^2(c^2-K)^2(1+n_c)^2>0$. Using the solitary wave identity (2.2a) and the definition of $J=(c-u_c)^2-K$ (see (2.8)), we have

$$\begin{aligned} \begin{aligned} I_1&= 2\left( (c-u_c)(c^2-K)-c\left( (c-u_c)^2-K \right) \right) ^2(1+n_c)^2 \\&= 2\left( c(c^2-K)-c\left( c(c-u_c)-K(1+n_c) \right) \right) ^2 \\&= 2c^2 ( c u_c+K n_c )^2, \\ I_2&= - K\Big ((c^2-K)(1+n_c)-\left( (c-u_c)^2-K \right) \Big )^2(1+n_c)^2 \\&= - K\Big ((c^2-K)(1+n_c)^2-\big ( c(c-u_c)-K (1+n_c) \big ) \Big )^2 \\&= - K\left( n_c(c^2-K)(2+ n_c)+\left( cu_c+K n_c\right) \right) ^2 \\&= -K\left( cu_c+K n_c\right) ^2 -Kn_c^2(c^2-K)^2(2+n_c)^2 - 2Kn_c(c^2-K)(2+n_c)\left( cu_c+K n_c\right) \\&= -K\left( cu_c+K n_c\right) ^2 -4Kn_c^2(c^2-K)^2 - 4Kn_c(c^2-K)\left( cu_c+K n_c\right) \\&\quad + O\big (|n_c|^3 + |n_c|^2|u_c|\big ), \\ I_3&= -K\left( c^2-K-(1+n_c)\left( (c-u_c)^2-K \right) \right) ^2 \\&= -K\left( cu_c+K n_c\right) ^2. \end{aligned} \end{aligned}$$

Hence,

$$\begin{aligned} \begin{aligned} I_1+I_2+I_3&= 2(c^2-K) ( c u_c+K n_c )^2 -4Kn_c^2(c^2-K)^2 - 4Kn_c(c^2-K)\left( cu_c+K n_c\right) \\&\quad + O\big (|n_c|^3 + |n_c|^2|u_c|\big ) \\&= 2(c^2-K)( c u_c+K n_c )\left( c u_c -K n_c \right) -4Kn_c^2(c^2-K)^2 \\&\quad + O\big (|n_c|^3 + |n_c|^2|u_c|\big ) \\&= 2(c^2-K)( c^2 u_c^2 + K^2 n_c^2 - 2K c^2 n_c^2 ) +O\big (|n_c|^3 + |n_c|^2|u_c|\big ). \end{aligned} \end{aligned}$$

Since $\frac{cn_c}{1+n_c}=u_c$ from (2.2a), we obtain that

$$\begin{aligned} c^2 u_c^2 + K^2 n_c^2 - 2K c^2 n_c^2 = (c^2-K)^2n_c^2 \big ( 1+O(|n_c|) \big ). \end{aligned}$$

Therefore, we conclude that $I_1+I_2+I_3>0$ for all sufficiently small $\varepsilon >0$.