Nonlocal eigenvalue problems arising in a generalized phase-field-type system

Abstract

We deal with a generalized phase-field-type system that arises as a transformed system of reaction-diffusion equations with a conservation law. We consider the stationary problem which is reduced to a scalar elliptic equation with a nonlocal term, and study the linearized eigenvalue problem. We first prove by the spectral comparison argument that the number of unstable eigenvalues for the problem coincides with the one of the linearized eigenvalue problem for the original system. We next show a limiting behavior of eigenvalues for the scalar problem as the coefficient of the nonlocal term goes to infinity.

Appendix

1.1 Proof of Lemma 4

We put $X^1:=H^1(\varOmega )$. $\Vert \cdot \Vert _{X^1}$ and $(\cdot ,\cdot )_{X^1}$ stand for the norm and the inner product in $H^1$ respectively. In order to make clarity, we also use the notation $\Vert \cdot \Vert _{L^2}$ for the $L^2$ norm throughout the section. Recall

$$\begin{aligned} \phi _j\in L^2(\varOmega ), \quad (\phi _i,\phi _j)_{L^2}=\delta _{ij}~\quad (1\le i,~ j\le N), \end{aligned}$$

and V in (45)

$$\begin{aligned} V=\{\varphi \in L^2(\varOmega ) : (\varphi ,\phi _j)_{L^2}=0~~(1\le j\le N)\}. \end{aligned}$$

First step: We define linear functionals on $X^1$ as

$$\begin{aligned} T_j(u):=(u,\phi _j)_{L^2}\quad (u\in X^1),\qquad 1\le j\le N. \end{aligned}$$

Then $T_j\in (X^1)'$ (the dual space of $X^1$). By the Riesz representation theorem there is a unique $v_j\in X^1$ such that

$$\begin{aligned} T_j(u)=(u,v_j)_{X^1}\qquad (u\in X^1). \end{aligned}$$

We show

• (i) $\Vert v_j\Vert _{L^2}>0, \quad v_j\notin V~(1\le j\le N)$.

• (ii) $\{v_1, v_2,\ldots , v_N\}$ is linearly independent.

For $\phi _j$ there is a sequence $\{u_m^{(j)}\}_{m=1,2,\ldots }\subset X^1$ such that

$$\begin{aligned} \Vert u_m^{(j)}-\phi _j\Vert _{L^2}\rightarrow 0\quad (m\rightarrow \infty ), \end{aligned}$$

since $H^1(\varOmega )$ is dense in $L^2(\varOmega )$. We may assume $1/2\le \Vert u_m^{(j)}\Vert _{L^2} \le 2$ because of $\Vert \phi _j\Vert _{L^2}=1$. On the other hand for each j

$$\begin{aligned} \lim _{m\rightarrow \infty }T_j(u_m^{(j)})=\lim _{m\rightarrow \infty }(u_m^{(j)},\phi _j)_{L^2}=1. \end{aligned}$$

This yields

$$\begin{aligned} \Vert T_j\Vert _{(X^1)'}:=\sup _{u\in X^1}\frac{|T_j(u)|}{\Vert u\Vert _{X^1}} \ge \frac{|T_j(u_m^{(j)})|}{\Vert u_m^{(j)}\Vert _{X^1}}>0, \end{aligned}$$

for some m. The Riesz representation theorem tells

$$\begin{aligned} \Vert v_j\Vert _{X^1}=\Vert T_j\Vert _{(X^1)'}>0\qquad (1\le j\le N), \end{aligned}$$

hence

$$\begin{aligned} T_j(v_j)=(v_j,\phi _j)_{L^2}=(v_j, v_j)_{X^1}>0. \end{aligned}$$

This implies $v_j\notin V~(1\le j\le N)$ and (i) is concluded.

Next we set

$$\begin{aligned} \sum _{j=1}^N{c_jv_j}=0,\qquad c_j\in \mathbb {C}~(1\le j\le N). \end{aligned}$$

Arbitrarily given $u\in X^1$, we have

$$\begin{aligned} 0=\left( u, \sum _{j=1}^N c_jv_j\right) _{X^1} =\sum _{j=1}^N\overline{c_j}(u, v_j)_{X^1} = \sum _{j=1}^N\overline{c_j}(u, \phi _j)_{L^2}. \end{aligned}$$

(55)

We use the sequence $\{u_m^{(i)}\}$ in the proof of (i). Putting $u=u_m^{(i)}$ in (55) and taking $m\rightarrow \infty$ yields

$$\begin{aligned} 0=\sum _{j=1}^N\overline{c_j}(\phi _i, \phi _j)_{L^2}=\overline{c_i}. \end{aligned}$$

Hence $c_j=0~(1\le j\le N)$ by which we obtain the desired assertion (ii).

Second step: Put $Y_1:=X^1\cap V$ and

$$\begin{aligned} Y_2:=L.H.[v_1, v_2, \ldots , v_N]. \end{aligned}$$

Then $\mathrm{dim}Y_2=N$. Apply the Gram-Schmidt orthonormalization to the family $\{v_j\}_{j=1,\ldots ,N}$. Then we obtain a family $\{\hat{v}_j\}_{j=1,\ldots ,N}$ such that

$$\begin{aligned} (\hat{v}_i, \hat{v}_j)_{X^1}=\delta _{i,j}, \quad L.H.[\hat{v}_1, \hat{v}_2,\ldots , \hat{v}_N]=Y_2. \end{aligned}$$

We have a transformation from $(v_1, \ldots , v_n)$ into $(\hat{v}_1,\ldots , \hat{v}_N)$, indeed there is a regular matrix P such that P is triangular and

$$\begin{aligned} (\hat{v}_1,\ldots ,\hat{v}_N)=(v_1,\ldots , v_N)P. \end{aligned}$$

By this P we define

$$\begin{aligned} (\hat{\phi }_1, \ldots , \hat{\phi }_N):=(\phi _1,\ldots ,\phi _N)P. \end{aligned}$$

It is clear that $\hat{\phi }_j\in L^2(\varOmega )$ and $\{\hat{\phi }_1,\ldots ,\hat{\phi }_N\}$ is linearly independent. By the relations

$$\begin{aligned} (u,v_j)_{X^1}=(u,\phi _j)_{L^2}\quad (1\le j\le N), \end{aligned}$$

we have

$$\begin{aligned} (u, \hat{v}_j)_{X^1}=(u,\hat{\phi }_j)_{L^2}\qquad (\forall u\in H^1(\varOmega )), \quad 1\le j\le N. \end{aligned}$$

For $u\in X^1$, we set

$$\begin{aligned}&u=u'+u'',\\&u':=u-\sum _{j=1}^N(u, \hat{v}_j)_{X^1}\hat{v}_j,\quad u'':=\sum _{j=1}^N(u, \hat{v}_j)_{X^1}\hat{v}_j. \end{aligned}$$

Clearly $u''\in Y_2$. We prove $u'\in V$.

$$\begin{aligned}&(u',\hat{\phi }_j)_{L^2}=\left( u-\sum _{k=1}^N(u,\hat{v}_k)_{X^1}\hat{v}_k, \hat{\phi }_j\right) _{L^2} \\&=(u,\hat{\phi }_j)_{L^2}-\sum _{k=1}^N(u,\hat{v}_k)_{X^1}(\hat{v}_k, \hat{\phi }_j)_{L^2} \\&=(u,\hat{v}_j)_{X^1}-\sum _{k=1}^N(u,\hat{v}_k)_{X^1}(\hat{v}_k, \hat{v}_j)_{X^1}=0. \end{aligned}$$

In the sequel we obtain $u'\in Y_1=X^1\cap V$ and $u=u'+u''\in Y_1\oplus Y_2~(Y_1\bot Y_2)$.

Third step: We can replace the basis $\{v_j\}_{j=1,\ldots ,N}$ of $Y_2$ by $\{w_j\}_{j=1,\ldots ,N}$ so that

$$\begin{aligned} (w_i, \phi _j)_{L^2}=\delta _{i,j}\quad (1\le i, j\le N) \end{aligned}$$

holds. Indeed define the matricies,

$$\begin{aligned} J:=((v_i, v_j)_{X^1})_{1\le i, j\le N},\qquad A:=(a_{ij})_{1\le i,j\le N}:=J^{-1}. \end{aligned}$$

Then

$$\begin{aligned} w_j=\sum _{\ell }^N a_{\ell j}v_{\ell } (1\le j\le N) \end{aligned}$$

create the desired basis, since

$$\begin{aligned} (w_i,\phi _j)_{L^2}=(w_i, v_j)_{X^1}=\sum _{\ell =1}^N a_{\ell i}(v_\ell , v_j)_{X^1}=\delta _{ij}. \end{aligned}$$

We then complete the proof of Lemma 4. $\square$

1.2 Proof of Theorem 3 (II)

Given sequence $\{\tau (m)\}_{m=1,2,\ldots },~\tau (m)\rightarrow \infty ~(m\rightarrow \infty )$, there is a subsequence $\{\tau (m_p)\}_{p=1,2,\ldots }$ and a orthonormal family of functions $\{\varPhi _k\}_{k=1,2,\ldots }$ such that

$$\begin{aligned}&\lim _{p\rightarrow \infty }\varPhi _k(\cdot ; \tau (m_p))=\varPhi _k(\cdot )~~\mathrm{strongly} ~\mathrm{in}~~L^2(\varOmega ), \\&\lim _{p\rightarrow \infty }\varPhi _k(\cdot ; \tau (m_p))=\varPhi _k(\cdot )~~\mathrm{weakly}~\mathrm{in}~~H^1(\varOmega ). \end{aligned}$$

Then

$$\begin{aligned} P(-d\varDelta \varPhi _k - a_\infty (\cdot )\varPhi _k)=\lambda _k\varPhi _k, \end{aligned}$$

namely,

$$\begin{aligned} d\varDelta \varPhi _k + a_\infty (x)\varPhi _k+\lambda _k\varPhi _k =\sum _{j=1}^N\frac{(d\varDelta \varPhi _k + a_\infty (\cdot )\varPhi _k, \phi _j)_{L^2}}{\Vert \phi _j\Vert _{L^2}^2}\, \phi _j. \end{aligned}$$

(56)

We construct approximate eigenfunction. Define

$$\begin{aligned} \tilde{\varPhi }_k(x;\tau ):= & {} \varPhi _k(x)+\frac{1}{\tau }\sum _{j=1}^N\frac{(\varvec{A}\varPhi _k,\phi _j)_{L^2}}{\Vert \phi _j\Vert _{L^2}^4}\, v_j(x), \\ \varvec{A}:= & {} d\varDelta + a_\infty (\cdot ), \end{aligned}$$

where $v_j\in H^1(\varOmega )~(1\le j\le N)$ and

$$\begin{aligned} (v_i,\phi _j)_{L^2}=\delta _{ij}\Vert \phi _i\Vert _{L^2}^2\quad (1\le i,j \le N). \end{aligned}$$

(57)

For simplicity of notation we write $\varPhi _{k,\tau }=\varPhi _k(\cdot ;\tau )$ and $\tilde{\varPhi }_{k,\tau }=\tilde{\varPhi }_k(\cdot ;\tau )$.

Multiply by $\tilde{\varPhi }_{k,\tau }$ the equation

$$\begin{aligned} -d\varDelta \varPhi _{k,\tau } - a(\cdot ;\tau )\varPhi _{k,\tau }+\tau \sum _{j=1}^N(\varPhi _{k,\tau }, \phi _j)_{L^2}\, \phi _j-\mu _k(\tau )\varPhi _{k,\tau }=0~~(x\in \varOmega ), \end{aligned}$$

and integrate it over $\varOmega$ and we obtain

$$\begin{aligned}&((-d\varDelta - a(\cdot ;\tau ))\varPhi _{k,\tau },\varPhi _k)_{L^2} +\tau \sum _{j=1}^N(\varPhi _{k,\tau }, \phi _j)_{L^2}(\phi _j, \varPhi _k)_{L^2} -\mu _k(\tau )(\varPhi _{k,\tau }, \varPhi _k)_{L^2}\\&\quad +\frac{1}{\tau }\sum _{j=1}^N\frac{(\varvec{A}\varPhi _k,\phi _j)_{L^2}}{\Vert \phi _j\Vert _{L^2}^4} \left\{ ((-d\varDelta - a(\cdot ;\tau ))\varPhi _{k,\tau }, v_j)_{L^2} +\tau \sum _{\ell =1}^N(\varPhi _{k,\tau },\phi _\ell )_{L^2}(\phi _\ell ,v_j)_{L^2}\right\} \\&\quad -\frac{1}{\tau }\sum _{j=1}^N\frac{(\varvec{A}\varPhi _k,\phi _j)_{L^2}}{\Vert \phi _j\Vert _{L^2}^4}\, \mu _k(\tau )(\varPhi _{k,\tau },v_j)_{L^2}=0. \end{aligned}$$

Utilizing (57), $(\varPhi _k,\phi _j)_{L^2}=0$ and taking integration by part, the above equations turns to be

$$\begin{aligned}&(\varPhi _{k,\tau }, (-d\varDelta - a(\cdot ;\tau ))\varPhi _k)_{L^2} -\mu _k(\tau )(\varPhi _{k,\tau }, \varPhi _k)_{L^2}\\&\quad +\frac{1}{\tau }\sum _{j=1}^N\frac{(\varvec{A}\varPhi _k,\phi _j)_{L^2}}{\Vert \phi _j\Vert _{L^2}^4} \Big \{(d\nabla \varPhi _{k,\tau }, \nabla v_j)_{L^2} - (a(\cdot ;\tau )\varPhi _{k,\tau }, v_j)_{L^2} +\tau (\varPhi _{k,\tau },\phi _j)_{L^2}\Vert \phi _j\Vert _{L^2}^2 \Big \}\\&\quad -\frac{1}{\tau }\sum _{j=1}^N\frac{(\varvec{A}\varPhi _k,\phi _j)_{L^2}}{\Vert \phi _j\Vert _{L^2}^4}\, \mu _k(\tau )(\varPhi _{k,\tau },v_j)_{L^2}=0. \end{aligned}$$

Applying (56) to the above equality leads to

$$\begin{aligned}&\tau (\lambda _k-\mu _k(\tau ))(\varPhi _{k,\tau },\varPhi _k)_{L^2} - (\varPhi _{k,\tau }, (a(\cdot ;\tau )-a_\infty (\cdot ))\varPhi _k)_{L^2} \\&\quad +\sum _{j=1}^N\frac{(\varvec{A}\varPhi _k,\phi _j)_{L^2}}{\Vert \phi _j\Vert _{L^2}^4} \Big \{(d\nabla \varPhi _{k,\tau },\nabla v_j)_{L^2}- (a(\cdot ;\tau )\varPhi _{k,\tau },v_j)_{L^2}\Big \}\\&\quad -\sum _{j=1}^N\frac{(\varvec{A}\varPhi _k,\phi _j)_{L^2}}{\Vert \phi _j\Vert _{L^2}^4}\, \mu _k(\tau )(\varPhi _{k,\tau },v_j)_{L^2}=0. \end{aligned}$$

Putting $\tau =\tau (m_p)$ and taking $p\rightarrow \infty$, we have

$$\begin{aligned}&\lim _{p\rightarrow \infty }(\varPhi _{k,\tau (m_p)}, \varPhi _k)_{L^2}=\Vert \varPhi _k\Vert _{L^2}^2=1,\\&\quad \lim _{p\rightarrow \infty }\mu _k(\tau (m_p))(\varPhi _{k,\tau (m_p)}, v_j)=\lambda _k(\varPhi _k,v_j),\\&\quad \lim _{p\rightarrow \infty }\Big \{(d\nabla \varPhi _{k,\tau (m_p)}, \nabla v_j)_{L^2} - (a(\cdot ;\tau )\varPhi _{k,\tau (m_p)}, v_j)_{L^2}\Big \} \\&\qquad =(d\nabla \varPhi _k,\nabla v_j)_{L^2} - (a_\infty (\cdot )\varPhi _k, v_j)_{L^2}. \end{aligned}$$

With the aid of these facts we obtain

$$\begin{aligned}&\lim _{p\rightarrow \infty }\tau (m_p)(\mu _k(\tau (m_p))-\lambda _k)\\&\quad =\sum _{j=1}^N\frac{(\varvec{A}\varPhi _k, \phi _j)_{L^2}}{\Vert \phi _j\Vert _{L^2}^4}\Big \{(d\nabla \varPhi _k,\nabla v_j)_{L^2} - (a_\infty (\cdot )\varPhi _k, v_j) -\lambda _k(\varPhi _k,v_j)_{L^2}\Big \} \\&\quad =-\sum _{j=1}^N\frac{(\varvec{A}\varPhi _k, \phi _j)_{L^2}}{\Vert \phi _j \Vert _{L^2}^4}((d\varDelta + a_\infty (\cdot )+\lambda _k)\varPhi _k, v_j)_{L^2} \\&\quad =-\sum _{j=1}^N\frac{(\varvec{A}\varPhi _k, \phi _j)_{L^2}}{\Vert \phi _j\Vert _{L^2}^4}\left( \sum _{\ell =1}^N\frac{(\varvec{A}\varPhi _k, \phi _\ell )_{L^2}}{\Vert \phi _\ell \Vert _{L^2}^2}\phi _\ell , v_j\right) _{L^2} \quad (\text{ by } (56))\\&\quad =-\sum _{j=1}^N\frac{(\varvec{A}\varPhi _k, \phi _j)_{L^2}^2}{\Vert \phi _j\Vert _{L^2}^4} =-\sum _{j=1}^N\frac{(\varvec{A}\varPsi _k, \phi _j)_{L^2}^2}{\Vert \phi _j\Vert _{L^2}^4}, \end{aligned}$$

where in the last equality we used $(\phi _\ell , v_j)_{L^2}=\delta _{\ell ,j} \Vert \phi _j\Vert _{L^2}^2$ and the fact that $\varPhi _k$ and $\varPsi _k$ are linearly dependent because of the simplicity of $\lambda _k$. Since $\{\tau (m)\}_{m=1,2,\ldots }$ is arbitrary, we can get the assertion of Theorem 3 (II). $\square$

1.3 On the convergence of solutions as $\delta \rightarrow +0$

In general the following proposition holds:

Proposition 3

Assume that $\partial \varOmega$ is of class $C^2$ and that $f\in C^1(\mathbb {R})$ satisfies

$$\begin{aligned} \xi \,f(\xi )\leqq -c_1\xi ^2+c_2\quad (\xi \in \mathbb {R}), \end{aligned}$$

where $c_1>0$ and $c_2>0$ are constants. Then for solutions $\{u_\delta \}_{\delta >0}$ of (11) and an arbitrary sequence $\{\delta _p\}_{p=1}^\infty$ with $\lim _{p\rightarrow \infty } \delta _p=0$, there exist a subsequence $\{\zeta _p\}_{p=1}^\infty$ $\subset \{\delta _p\}_{p=1}^\infty$ and a solution $u_0$ of (15) such that

$$\begin{aligned} \lim _{p\rightarrow \infty }\Vert u_{\zeta _p}-u_0\Vert _{C^1(\overline{\varOmega })}=0. \end{aligned}$$

It is easy to see that $f(u)=u-u^3$ satisfies the condition in the proposition. When $f(u)=u-u^3$ and the domain $\varOmega$ is a one-dimensional interval, say (0, 1), minimizers of (12) must be monotone [9, 30]. We can easily check that given $m\in (0, 1/\sqrt{3})$, there are $\delta _m$ and $d_m$ such that for $\delta <\delta _m$ a unique constant solution $u_c$ of (11) satisfies $f'(u_c)=1-3u_c^2>0$ and it is unstable for $d<d_m$. This implies the minimizers $u^*_\delta$ are strictly decreasing/increasing. On the other hand, the equation (15) has a unique strictly decreasing/increasing solution $u_0^*$ under the same condition for m and d [13]. Combining these facts and (16), we can assert that $\lim _{\delta \rightarrow +0}\Vert u^*_\delta -u^*_0\Vert _{L^\infty }=0$. Once we obtained the convergence for the strictly monotone solutions, by reflection and appropriate scaling, we obtain the convergence for multi-mode solutions. In order to establish the convergence for the general f enjoying the condition of Proposition 3 or higher-dimensional domains we need a further investigation, which will be a future work. In the rest of this subsection we give a sketch of the proof of Proposition 3.

Sketch of Proof

Putting $v=u-m$ and $g(\xi )=f(\xi +m)$, we rewrite (11) as

$$\begin{aligned} \varDelta v+g(v)-\frac{1}{\delta }\langle v\rangle =0\quad \text {in}\quad \varOmega , \quad \frac{\partial v}{\partial \nu }=0\quad \text {on}\quad \partial \varOmega . \end{aligned}$$

(58)

Then $v_\delta =u_\delta -m$ is a solution of this equation. We show that $\{v_{\delta }\}_{\delta >0}$ is relatively compact in $C^1(\overline{\varOmega })$.

Drop the subscript $\delta$ of $v_\delta$ for a while. We first obtain by (58)

$$\begin{aligned} \int _\varOmega g(v) dx-\frac{1}{\delta } \vert \varOmega \vert \langle v\rangle =0, \end{aligned}$$

(59)

and

$$\begin{aligned} \int _\varOmega \vert \nabla v\vert ^2 dx+\int _\varOmega (-v g(v)) dx +\frac{1}{\delta } \vert \varOmega \vert \langle v\rangle ^2=0. \end{aligned}$$

(60)

Then we can verify that

$$\begin{aligned} -\xi \, g(\xi )>0\quad \text {for}\quad \vert \xi \vert \geqq L \end{aligned}$$

holds for a positive number L thanks to the assumption on f. From (60)

$$\begin{aligned} \int _\varOmega \vert \nabla v\vert ^2 dx+\int _{\vert v\vert \geqq L} (-v g(v)) dx +\frac{1}{\delta } \vert \varOmega \vert \langle v\rangle ^2=\int _{\vert v\vert < L} (v g(v)) dx \end{aligned}$$

(61)

follows, where we simply write $\{x\in \varOmega :\vert v(x)\vert \geqq L\}$ by $|v|\geqq L$ and so on. Since the right hand side (61) is bounded, there is a positve M such that

$$\begin{aligned} 0\leqq \int _\varOmega \vert \nabla v\vert ^2 dx\leqq M,\quad 0\leqq \int _{\vert v\vert \geqq L} (-v g(v)) dx \leqq M,\quad 0\leqq \frac{1}{\delta } \vert \varOmega \vert \langle v\rangle ^2\leqq M. \end{aligned}$$

(62)

We estimate $|\langle v\rangle |/\delta$. By (59)

$$\begin{aligned} \frac{1}{\delta } \vert \varOmega \vert \,\, \vert \langle v\rangle \vert= & {} \vert \int _\varOmega g(v) dx\vert \leqq \int _{v \geqq L} (-g(v)) dx+ \int _{ v \leqq -L} g(v) dx+ \int _{\vert v\vert \le L} \vert g(v)\vert dx\\\leqq & {} \frac{1}{L} \int _{\vert v\vert \geqq L} (-v\,g(v)) dx+ \int _{\vert v\vert<L} \vert g(v)\vert dx\leqq \frac{M}{L}+ \vert \varOmega \vert \sup _{\vert \eta \vert <L} \vert g(\eta )\vert , \end{aligned}$$

where we used (62). Therefore there is a constant $K>0$ such as $(1/\delta ) \vert \langle v\rangle \vert \le K$.

We next verify that there is a number $R>0$ such that

$$\begin{aligned} -R\le \inf _{x\in \varOmega }v_\delta (x)\le \sup _{x\in \varOmega } v_\delta (x)\le R, \end{aligned}$$

(63)

by applying to (58) the fact

$$\begin{aligned} g(\xi )<-K \quad (\xi \geqq R), \qquad g(\xi )\geqq K\quad (\xi \leqq -R), \end{aligned}$$

for a number R and $(1/\delta ) \vert \langle v_\delta \rangle \vert \le K$. Namely, the supremum and the infimum of $v_\delta$ cannot takes in the range $|v|>R$.

Finally, applying the Schauder estimate to (58) with the above estimate, we obtain that $\{v_\delta \}_{\delta >0}$ is bounded in $C^{1,\kappa }(\overline{\varOmega })$ for $0<\kappa <1$ and consequently, it is compact in $C^1(\overline{\varOmega })$. Hence, since $u_\delta =v_\delta +m$, we have the assertion of Proposition 3 up to the convergence of the subsequence. In conclusion, the limit function, denoted by $u_0$, satisfies (15). $\square$