Correction to: Eigenvalue bifurcation in doubly nonlinear problems with an application to surface plasmon polaritons

Correction to: Nonlinear Differ. Equ. Appl. (2021) 28:9 https://doi.org/10.1007/s00030-020-00668-2

1 Introduction

In [1] we proved a result on bifurcation from simple isolated eigenvalues for a class of generally non-self-adjoint eigenvalue problems which are nonlinear in the solution as well as in the eigenvalue parameter. The general result was then applied in the context of surface plasmon polaritons (SPPs). Although the proof of the main result is correct, the definition of the isolatedness for an eigenvalue is not suitable for our purposes and does not lead to the linearized operator being Fredholm. As a consequence, wrong assumptions were checked in the application to SPPs. In this erratum we provide the correct assumptions and verify that they are satisfied in our examples.

In doing so we also remove some minor errors and inconsistencies. Section 2 lists and explains all the necessary corrections. For readers’ convenience we have incorporated all the corrections in the arXiv version of this paper [2].

Notation. The numbering of theorems and equations refers to the one used in [1]. New equations are numbered with (1’), (2’), etc.

2 Corrections

• 1. The definition of the isolatedness of the linear eigenvalue $\omega _0$ has to be modified. The original definition was the isolatedness of $\omega _0$ in the complex $\omega -$plane. However, for the Fredholm property of $L(\cdot ,\omega _0)$ (used in the proof of Theorem 2.1) the isolatedness of the zero eigenvalue of $L(\cdot ,\omega _0)$ (i.e. for $\omega =\omega _0$ fixed) is needed. The new assumptions at the beginning of Section 2 (to replace items (i) and (ii) on p. 3) are as follows

  • (E1) $\omega _0$ is algebraically simple in the sense that $\kappa =0$ is an algebraically simple eigenvalue of the standard eigenvalue problem $L(\cdot ,\omega _0)u=\kappa u$, i.e. $\ker (L(\cdot ,\omega _0)^2)=\ker (L(\cdot ,\omega _0))=\langle \varphi _0 \rangle$,

  • (E2) $\omega _0$ is isolated in the sense that $\kappa =0$ is an isolated eigenvalue of the problem $L(\cdot ,\omega _0)u=\kappa u$.

• 2. We check these new assumptions (E1) and (E2) for the examples from Section 4.1.2 in Appendix A of this erratum. The text of the appendix is to be understood as an addition to Section 4.2.

• 3. Equation (7) in assumption (f4) has to be corrected (weakened to an estimate in $L^2$):

  $$\begin{aligned} \left\| f(\cdot ,\omega ,\varphi )-g_\omega (\cdot )|\varphi |^\frac{1}{\alpha }\varphi \right\| ={\mathcal {O}}\left( \left\| |\varphi |^{\frac{1}{\alpha }+1+\beta }\right\| \right) \quad \text{ as }\quad {\mathbb {C}}\ni \varphi \rightarrow 0. \end{aligned}$$

  (7)

• 4. The statement of the main theorem should be formulated more clearly regarding the uniqueness statement:

Theorem 2.1

Suppose that (E1), (E2) hold, i.e. $\omega _0$ is an algebraically simple and isolated eigenvalue of L with eigenfunction $\varphi _0$ and that A, W and f satisfy assumptions (A1)–(f4). Let also $\tau \in (0,\min \{1,\alpha \beta \}]$. Then there is a unique branch bifurcating from $(\omega _0,0)$. There exists $\varepsilon _0>0$ s.t. for any $\varepsilon \in (0,\varepsilon _0)$ the solution $(\omega ,\varphi )$ normalized to satisfy $\langle \varphi ,\varphi _0^*\rangle =\varepsilon ^\alpha$ has the form

$$\begin{aligned} \omega =\omega _0+\varepsilon \nu +\varepsilon ^{1+\tau }\sigma ,\qquad \quad \varphi =\varepsilon ^\alpha \varphi _0+\varepsilon ^{\alpha +1}\phi +\varepsilon ^{\alpha +1+\tau }\psi , \end{aligned}$$

(8)

with $\nu ,\sigma \in {\mathbb {C}}$ and $\phi ,\psi \in D(A)\cap \langle \varphi _0^*\rangle ^\perp$.

• 5. Equation (26): a constant C was missing in the last estimate:

  $$\begin{aligned} \begin{aligned} \Vert I(\cdot ,\omega )\Vert _\infty&\le \frac{2M}{r^3}|\omega -\omega _0|^3=\frac{2M}{r^3}\varepsilon ^3|\nu +\varepsilon ^\tau \sigma |^3\le \frac{2MC}{r^3}\varepsilon ^3(1+\varepsilon ^\tau r_1). \end{aligned} \end{aligned}$$

• 6. Correction of a sign in the proof of Proposition 3.1. We have on p. 17, l. 2–3

  $$\begin{aligned}P_0(\nu \partial _\omega W(\cdot ,\omega _0)\varphi _0)=-P_0(g_{\omega _0}(\cdot )|\varphi _0|^\frac{1}{\alpha }\varphi _0) \end{aligned}$$

  and hence on p. 17, l. 11-12

  $$\begin{aligned} \begin{aligned} {\mathcal {B}}P_0(\nu \partial _\omega W(\cdot ,\omega _0)\varphi _0)&=-{\mathcal {B}}P_0(g_{\omega _0}(\cdot )|\varphi _0|^\frac{1}{\alpha }\varphi _0)=-P_0\big ({\mathcal {B}}(g_{\omega _0}(\cdot )){\mathcal {B}}(|\varphi _0|^\frac{1}{\alpha }\varphi _0)\big )\\&=-P_0(g_{\omega _0}(\cdot )|\varphi _0|^\frac{1}{\alpha }\varphi _0). \end{aligned} \end{aligned}$$

• 7. Correction of a typo in the proof of Proposition 3.1. On p. 17, l. 17 “a unique solution in $P_0D(A)$” should be replaced by “a unique solution in $Q_0 D(A)$”.

• 8. Figure 1 (d) displays the plot of ${\tilde{d}}_1$ and not ${\tilde{d}}_{-1}$. The caption for (d) and the text discussing this plot has to be modified accordingly.

• 9. In equation (62) the numerator 1 has to be replaced by $2\pi$.

Checking assumptions (E1) and (E2)

For a fixed $\omega _0$ we denote the corresponding linear eigenfunction $\varphi (\cdot ,\omega _0)$ by

$$\begin{aligned} {{\widetilde{\varphi }}}_0:=\varphi (\cdot ,\omega _0), \end{aligned}$$

where $\varphi$ is defined by (58)-(59). Clearly, $\varphi _0={{\widetilde{\varphi }}}_0/\Vert {{\widetilde{\varphi }}}_0\Vert$.

Verification of (E1): $\omega _0$ is simple. Because in (58)-(59) the constants A, B, C, D are unique (up to normalization of ${{\widetilde{\varphi }}}_0$), it is clear that $\ker L(\cdot ,\omega _0)=\langle \widetilde{\varphi }_0\rangle$, so 0 as eigenvalue of $L(\cdot ,\omega _0)$ is geometrically simple. To prove the algebraic simplicity, suppose by contradiction that there exists a Jordan chain associated to $\omega _0$. This means, there exists $u\in D(A)$ (see (54)) such that

$$\begin{aligned} L(\cdot ,\omega _0)u=-u''-W(\cdot ,\omega _0)u={{\widetilde{\varphi }}}_0\qquad \text{ in } \,{\mathbb {R}}. \end{aligned}$$

(1')

Solving (1’) explicitly using the variation of constants, one finds

$$\begin{aligned} u(x)={\left\{ \begin{array}{ll} c_1e^{\lambda _-x}+\frac{1}{2\lambda _-}\left( e^{\lambda _-x}\int _x^0e^{-\lambda _-t}{{\widetilde{\varphi }}}_0(t)\,{\textrm{d}}t+e^{-\lambda _-x}\int _{-\infty }^xe^{\lambda _-t}{{\widetilde{\varphi }}}_0(t)\,{\textrm{d}}t\right) &{} \text {for}\, x<0, \\ c_2e^{\mu x}+c_3e^{-\mu x}+\frac{1}{2\mu }\left( e^{\mu x}\int _x^de^{-\mu t}{{\widetilde{\varphi }}}_0(t)\,{\textrm{d}}t+e^{-\mu x}\int _0^xe^{\mu t}{{\widetilde{\varphi }}}_0(t)\,{\textrm{d}}t\right) &{} \text {for}\, x\in (0,d), \\ c_4e^{-\lambda _+x}+\frac{1}{2\lambda _+}\left( e^{\lambda _+x}\int _x^{\infty }\!e^{-\lambda _+t}{{\widetilde{\varphi }}}_0(t)\,{\textrm{d}}t+e^{-\lambda _+x}\int _d^xe^{\lambda _+t}{{\widetilde{\varphi }}}_0(t)\,{\textrm{d}}t\right) &{} \text {for}\, x>d. \end{array}\right. } \end{aligned}$$

(2')

In order to belong to D(A), u must satisfy the $C^1$-matching at the interfaces $x=0$ and $x=d$. This implies that the constants $c_1$, $c_2$, $c_3$, $c_4$ have to solve the linear system

$$\begin{aligned} T\left( c_1,c_2,c_3,c_4\right) ^{{\textsf{T}}}=b, \end{aligned}$$

where

$$\begin{aligned} T:=\left( \begin{matrix} -1 &{} 1 &{} 1 &{} 0 \\ -\lambda _- &{} \mu &{} -\mu &{} 0 \\ 0 &{} e^{\mu d} &{} e^{-\mu d} &{} -e^{-\lambda _+d} \\ 0 &{} \mu e^{\mu d} &{} -\mu e^{-\mu d} &{} \lambda _+e^{-\lambda _+d} \end{matrix}\right) \end{aligned}$$

and

$$\begin{aligned} 2b:=\left( \begin{array}{c} \frac{1}{\lambda _-}\int _{-\infty }^0e^{\lambda _-t}{{\widetilde{\varphi }}}_0(t)\,{\textrm{d}}t-\frac{1}{\mu }\int _0^de^{-\mu t}{{\widetilde{\varphi }}}_0(t)\,{\textrm{d}}t\\ -\int _{-\infty }^0e^{\lambda _-t}{{\widetilde{\varphi }}}_0(t)\,{\textrm{d}}t-\int _0^de^{-\mu t}{{\widetilde{\varphi }}}_0(t)\,{\textrm{d}}t\\ - \frac{e^{-\mu d}}{\mu }\int _0^de^{\mu t}{{\widetilde{\varphi }}}_0(t)\,{\textrm{d}}t+\frac{e^{\lambda _+ d}}{\lambda _+}\int _d^{\infty }\!e^{-\lambda _+t}{{\widetilde{\varphi }}}_0(t)\,{\textrm{d}}t\\ e^{-\mu d}\int _0^de^{\mu t}{{\widetilde{\varphi }}}_0(t)\,{\textrm{d}}t+e^{\lambda _+ d}\int _d^{\infty }\!e^{-\lambda _+t}{{\widetilde{\varphi }}}_0(t)\,{\textrm{d}}t \end{array}\right) \,. \end{aligned}$$

Note that T is singular since

$$\begin{aligned} \det T = e^{-(\lambda _++\mu )d}\left( (\lambda _+-\mu )(\lambda _--\mu )- e^{2\mu d}(\lambda _++\mu )(\lambda _-+\mu )\right) =0 \end{aligned}$$

by our choice of d in (60). In order to find a contradiction and exclude the existence of a solution $u\in D(A)$ of (1’), we now prove that b is not orthogonal to the kernel of ${{\overline{T}}}^{{\textsf{T}}}$. Standard computations show that $\ker {{\overline{T}}}^{{\textsf{T}}}$ is one-dimensional and given by

$$\begin{aligned} \ker {{\overline{T}}}^{{\textsf {T}}}=\langle p\rangle :={{\,\text {span}\,}}\,\left( \overline{\lambda _-}\left( {{\overline{\mu }}}-\overline{\lambda _+}\right) e^{-{{\overline{\mu }}} d}, \left( {{\overline{\mu }}}-\overline{\lambda _+}\right) e^{-{{\overline{\mu }}} d}, \overline{\lambda _+}\left( \overline{\lambda _-}+{{\overline{\mu }}}\right) , \overline{\lambda _-}+{{\overline{\mu }}}\right) ^{{\textsf {T}}}. \end{aligned}$$

The scalar product $\left( b,p\right)$ then reads

$$\begin{aligned} \begin{aligned} \left( b,p\right)&=-2(\lambda _+-\mu )e^{-\mu d}\int _{-\infty }^0e^{\lambda _-t}{{\widetilde{\varphi }}}_0(t)\,{\textrm{d}}t+2e^{\lambda _+ d}(\lambda _-+\mu )\int _d^\infty e^{-\lambda _+t}{{\widetilde{\varphi }}}_0(t)\,{\textrm{d}}t\\&\quad +\frac{e^{-\mu d}(\lambda _+-\mu )}{\mu }\left[ (\lambda _--\mu )\int _0^de^{-\mu t}{{\widetilde{\varphi }}}_0(t)\,{\textrm{d}}t-(\lambda _-+\mu )\int _0^de^{\mu t}{{\widetilde{\varphi }}}_0(t)\,{\textrm{d}}t\right] \,. \end{aligned} \end{aligned}$$

After evaluating the integrals for ${{\widetilde{\varphi }}}_0$ given by (58)–(59), and some algebraic computations in which the identity $e^{2\mu d}=\frac{(\mu -\lambda _+)(\mu -\lambda _-)}{(\mu +\lambda _+)(\mu +\lambda _-)}$ is frequently used, we get

$$\begin{aligned} \begin{aligned} \left( b,p\right)&=\frac{e^{-\lambda _+ d}}{2(\lambda _--\mu )}\bigg [\frac{\lambda _-^2-\mu ^2}{\lambda _+}+\frac{\lambda _+^2-\mu ^2}{\lambda _-}\\&\quad -\frac{ d(\lambda _+^2-\mu ^2)(\lambda _-^2-\mu ^2)+(\lambda _-+\lambda _+)(\lambda _-\lambda _+-\mu ^2)}{\mu ^2}\bigg ]\,. \end{aligned} \end{aligned}$$

Finally, we check that $\left( b,p\right)$ is non-zero for the values of $\lambda _\pm$, $\mu$, and d obtained numerically in Sect. 4.1.2. We obtain $\left( b,p\right) \approx -19.38 -46.36 {\textrm{i}}$ for Case 2 and $\left( b,p\right) \approx 0.82 + 1.58 {\textrm{i}}$ for Case 3.

Verification of (E2): $\omega _0$ is isolated. For $L(x,\omega _0):=-\frac{d^2}{dx^2}-W(x,\omega _0)$ the essential spectrum is given by

$$\begin{aligned} \sigma _\text {ess}\left( L(\cdot ,\omega _0)\right) =\bigcup _\pm \left\{ \lambda -W_\pm (\omega _0): \lambda \in [0,\infty )\right\} \,. \end{aligned}$$

Since $W_\pm (\omega _0)\in {\mathbb {C}}\setminus {\mathbb {R}}$ in both Case 2 and Case 3, we have $0\notin \sigma _\text {ess}\left( L(\cdot ,\omega _0)\right)$. As the essential spectrum is closed, 0 is isolated from $\sigma _\text {ess}\left( L(\cdot ,\omega _0)\right)$.

Next, we show the isolatedness of 0 from other eigenvalues of $L(\cdot ,\omega _0)$. If $\kappa \in {\mathbb {C}}\setminus \{0\}$ is an eigenvalue of $L(\cdot ,\omega _0)$, it must be

$$\begin{aligned} \begin{aligned}&\frac{1}{\mu (0)}\left[ \log \left( \frac{(\mu (0)-\lambda _-(0))(\mu (0)-\lambda _+(0))}{(\mu (0)+\lambda _-(0))(\mu (0)+\lambda _+(0))}\right) \!+2\pi {\textrm{i}}m\right] \\&\quad =\frac{1}{\mu (\kappa )}\!\left[ \log \left( \frac{(\mu (\kappa )-\lambda _-(\kappa ))(\mu (\kappa )-\lambda _+(\kappa ))}{(\mu (\kappa )+\lambda _-(\kappa ))(\mu (\kappa )+\lambda _+(\kappa ))}\right) \!+2\pi {\textrm{i}}{{\tilde{m}}}\right] \end{aligned} \end{aligned}$$

for some ${{\tilde{m}}}\in {\mathbb {Z}}$, where $\mu (\kappa ):=\sqrt{-W_*(\omega _0)-\kappa }$ and $\lambda _\pm (\kappa ):=\sqrt{-W_\pm (\omega _0)-\kappa }$.

Suppose there exists a sequence $(\kappa _j)_j\subset {\mathbb {C}}$ of such eigenvalues of $L(\cdot ,\omega _0)$ which converges to 0. Then, by continuity of the maps $\kappa \mapsto \mu (\kappa )$ and $\kappa \mapsto \lambda _\pm (\kappa )$, we infer ${{\tilde{m}}}=m$ and therefore

$$\begin{aligned} f(\kappa _j):=\mu (\kappa _j)\left( \log g(0)+2\pi {\textrm{i}}m\right) -\mu (0)\left( \log g(\kappa _j)+2\pi {\textrm{i}}m\right) =0 \end{aligned}$$

must hold, where

$$\begin{aligned} g(\kappa ):=\frac{(\mu (\kappa )-\lambda _-(\kappa ))(\mu (\kappa )-\lambda _+(\kappa ))}{(\mu (\kappa )+\lambda _-(\kappa ))(\mu (\kappa )+\lambda _+(\kappa ))}. \end{aligned}$$

Since $\kappa \mapsto f(\kappa )$ is differentiable at 0, a necessary condition for $f(\kappa _j)=0$ with $\kappa _j\rightarrow 0$ is $f'(0)=0$, that is

$$\begin{aligned} f'(0)=\mu '(0)\left( \log g(0)+2\pi {\textrm{i}}m\right) -\mu (0)\frac{g'(0)}{g(0)}=0. \end{aligned}$$

(3')

By simple computations one gets

$$\begin{aligned} g'(0)=\frac{g(0)}{\mu (0)}\left( \frac{1}{\lambda _+(0)}+\frac{1}{\lambda _-(0)}\right) , \end{aligned}$$

from which, by (3’) and the definition of d in (60), one infers

$$\begin{aligned} -d-\left( \frac{1}{\lambda _+(0)}+\frac{1}{\lambda _-(0)}\right) =0. \end{aligned}$$

This contradicts the fact that $d>0$ and that $\lambda _\pm$ are chosen with positive real part. As a consequence, we deduce that $\omega _0$ is isolated in the sense of (E2).