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Non-autonomous double phase eigenvalue problems with indefinite weight and lack of compactness

Tianxiang, Gou; Radulescu, Vicentiu

Abstract

In this paper, we consider eigenvalues to the following double phase problem with unbalanced growth and indefinite weight,-Delta pau-Delta qu=lambda m(x)|u|q-2uinRN,$$\begin{equation*} \hspace*{3pc}-\Delta _pa u-\Delta _q u =\lambda m(x)|u|{q-2}u \quad \mbox{in} \,\, \mathbb {R}<^>N, \end{equation*}$$where N > 2$N \geqslant 2$, 1{0, 1}(\mathbb {R}N, [0, +\infty))$, a not equivalent to 0$a \not\equiv 0$ and m:RN -> R$m: \mathbb {R}N \rightarrow \mathbb {R}$ is an indefinite sign weight which may admit non-trivial positive and negative parts. Here, Delta q$\Delta _q$ is the q$q$-Laplacian operator and Delta pa$\Delta _pa$ is the weighted p$p$-Laplace operator defined by Delta pau:=div(a(x)| backward difference u|p-2 backward difference u)$\Delta _pa u:=\textnormal {div}(a(x)|\nabla u|{p-2} \nabla u)$. The problem can be degenerate, in the sense that the infimum of a$a$ in RN$\mathbb {R}N$ may be zero. Our main results distinguish between the cases p

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Recei ed: 8 June 2023 Re ised: 3 Oc obe 2023 Accep ed: 27 Oc obe 2023 DOI: 10.1112/blms.12961 Bulle in o he London Ma hema ical Socie y RESEARCH ARTICLE Non-au onomous double phase eigen alue p oblems wi h inde ini e weigh and lack o compac ness Tianxiang Gou1Vicenţiu D. Rădulescu2,3,4,5,6 1School o Ma hema ics and S a is ics, Xi’an Jiao ong Uni e si y, Xi’an, Shaanxi, China 2Facul y o Applied Ma hema ics, AGH Uni e si y o Science and Technology, K akow, Poland 3Facul y o Elec ical Enginee ing and Communica ion, B no Uni e si y o Technology, B no, Czech Republic 4Depa men o Ma hema ics, Uni e si y o C aio a, C aio a, Romania 5Simion S oilow Ins i u e o Ma hema ics o he Romanian Academy, Bucha es , Romania 6School o Ma hema ics, Zhejiang No mal Uni e si y, Jinhua, China Co espondence Vicenţiu D. Rădulescu, Facul y o Applied Ma hema ics, AGH Uni e si y o Science and Technology, al. Mickiewicza 30, 30-059 K akow, Poland. Email: icen iu. adulescu@ima . o Funding in o ma ion Na ional Na u al Science Founda ion o China, G an /Awa d Numbe : 12101483; China Pos doc o al Science Founda ion, G an /Awa d Numbe : 2021M702620; Romanian Minis y o Resea ch, Inno a ion and Digi iza ion, G an /Awa d Numbe : PNRR-III-C9-2022-I8/22 Abs ac In his pape , we conside eigen alues o he ollow- ing double phase p oblem wi h unbalanced g ow h and inde ini e weigh , −Δ𝑎 𝑝𝑢−Δ 𝑞𝑢=𝜆𝑚(𝑥)|𝑢|𝑞−2𝑢in ℝ𝑁, whe e 𝑁⩾2,1<𝑝,𝑞<𝑁,𝑝≠𝑞,𝑎∈𝐶 0,1(ℝ𝑁, [0, +∞)),𝑎≢0and 𝑚∶ℝ𝑁→ℝis an inde ini e sign weigh which may admi non- i ial posi i e and nega i e pa s. He e, Δ𝑞is he 𝑞-Laplacian ope a o and Δ𝑎 𝑝is he weigh ed 𝑝-Laplace ope a o de ined by Δ𝑎 𝑝𝑢∶=di (𝑎(𝑥)|∇𝑢|𝑝−2∇𝑢).Thep oblemcanbe degene a e, in he sense ha he in imum o 𝑎in ℝ𝑁 may be ze o. Ou main esul s dis inguish be ween he cases 𝑝<𝑞and 𝑞<𝑝. In he i s case, we es ablish he exis ence o a con inuous amily o eigen alues, s a ing om he p incipal equency o a sui able single phase eigen alue p oblem. In he la e case, we p o e he exis ence o a disc e e amily o posi i e eigen alues, which di e ges o in ini y. MSC 2020 35P30 (p ima y), 35J70, 46E30, 47J10, 58C40, 58E05 (seconda y) © 2023 The Au ho s. Bulle in o he London Ma hema ical Socie y is copy igh © London Ma hema ical Socie y. This is an open access a icle unde he e ms o he C ea i e Commons A ibu ion License, which pe mi s use, dis ibu ion and ep oduc ion in any medium, p o ided he o iginal wo k is p ope ly ci ed. 734 wileyonlinelib a y.com/jou nal/blms Bull. London Ma h. Soc. 2024;56:734–755. NON-AUTONOMOUS DOUBLE PHASE EIGENVALUE PROBLEMS 735 1 INTRODUCTION In his pape , we in es iga e eigen alues o he ollowing double phase p oblem wi h unbalanced g ow h and inde ini e weigh , −Δ𝑎 𝑝𝑢−Δ 𝑞𝑢=𝜆𝑚(𝑥)|𝑢|𝑞−2𝑢in ℝ𝑁,(1.1) whe e 𝑁⩾2,1<𝑝,𝑞<𝑁,𝑝≠𝑞,𝑎∈𝐶 0,1(ℝ𝑁,[0,+∞)),𝑎≢0and 𝑚∶ℝ𝑁→ℝis an inde ini e sign weigh which may admi non- i ial posi i e and nega i e pa s. He e, Δ𝑞is he 𝑞-Laplacian ope a o and Δ𝑎 𝑝is he weigh ed 𝑝-Laplace ope a o de ined by Δ𝑎 𝑝𝑢∶=di (𝑎(𝑥)|∇𝑢|𝑝−2∇𝑢). Th oughou o his pape , we shall always assume ha he weigh unc ion 𝑚∶ℝ𝑁→ℝsa is ies he ollowing assump ion, (𝐻) 𝑚 = 𝑚1−𝑚 2, whe e 𝑚1,𝑚 2⩾0,𝑚1≢0,𝑚1∈𝐿 𝑁 𝑞(ℝ𝑁)∩𝐿 ∞(ℝ𝑁)and 𝑚2∈𝐿 ∞(ℝ𝑁). Rema k 1.1. In ou case, 𝑚2=0is allowable. P oblems like (1.1) a ise when one looks o he s a iona y solu ions o eac ion–di usion sys ems o he o m 𝑢𝑡=di [𝐷(𝑥, ∇𝑢)∇𝑢] + g(𝑥, 𝑢) (𝑥, 𝑡) ∈ ℝ𝑁× (0, ∞), whe e𝐷(𝑥,∇𝑢) = 𝑎(𝑥)|∇𝑢|𝑝−2 +|∇𝑢|𝑞−2. Thissys em hasa wide angeo applica ions inphysics and ela ed ields, such as biophysics, plasma physics and chemical eac ion design (see [7, 26]). In such applica ions, he unc ion 𝑢is a s a e a iable and desc ibes densi y o concen a ion o mul i-componen subs ances, di [𝐷(𝑥, ∇𝑢)∇𝑢] co esponds o he di usion wi h a di usion coe icien 𝐷(𝑥, ∇𝑢) and g(𝑥, 𝑢) is he eac ion and ela es o sou ce and loss p ocesses. Typically, in chemical and biological applica ions, he eac ion e m g(𝑥, 𝑢) has a polynomial o m wi h espec o he unknown concen a ion deno ed by 𝑢. The analysis o he double phase eigen alue p oblem (1.1) is closely associa ed wi h he ollowing single phase quasilinea eigen alue p oblem, −Δ𝑎 𝑟𝑢=𝜇𝑚(𝑥)|𝑢|𝑟−2𝑢in ℝ𝑁.(1.2) The i s pa o he pape is de o ed o he s udy o (1.2). The main esul s we es ablish ega ding (1.2) a e upcoming Theo em 3.1 and P oposi ion 3.1, which e eal ha he e exis a sequence o eigen alues o (1.2) and he i s eigen alue is simple. In he case o bounded domains and 𝑟=2, his p oblem is ela ed o he Riesz–F edholm heo y o sel -adjoin and compac ope a o s. The aniso opic linea case (i 𝑟=2and 𝑚(⋅)is non-cons an ) was i s conside ed in he pionee ing pape s o Boche [6], Hess and Ka o [17]andPleijel[25]. An impo - an con ibu ion in he case o unbounded domains is due o Alleg e o and Huang [1]and Szulkin and Willem [27]. In [27], he au ho s assumed ha weigh unc ion may ha e singula poin s. Equa ion (1.1) con ains he con ibu ion o wo di e en ial ope a o s in he le -hand side, so his p oblem is no homogeneous. In ac , he di e en ial ope a o 𝑢↦−Δ 𝑎 𝑝𝑢−Δ 𝑞𝑢is ela ed o 14692120, 2024, 2, Downloaded om h ps://londma hsoc.onlinelib a y.wiley.com/doi/10.1112/blms.12961 by B no Uni e si y O Technology, Wiley Online Lib a y on [13/05/2024]. See he Te ms and Condi ions (h ps://onlinelib a y.wiley.com/ e ms-and-condi ions) on Wiley Online Lib a y o ules o use; OA a icles a e go e ned by he applicable C ea i e Commons License 736 GOU and RĂDULESCU he ‘double-phase a ia ional unc ional de ined by 𝑢↦∫ℝ𝑁 𝑎(𝑥)|∇𝑢|𝑝+|∇𝑢|𝑞𝑑𝑥. The in eg and o his unc ional is he unc ion 𝜉(𝑥, 𝑡) = 𝑎(𝑥)𝑡𝑝+𝑡 𝑞 o all 𝑥∈ℝ𝑁and 𝑡⩾0. When 𝑎≡1, hen (1.1) becomes he so-called 𝑝&𝑞Laplacian p oblem, which was in es iga ed by Benouhiba and Belyacine [4, 5]. A ea u e o his pape is ha we do no assume ha he unc ion 𝑎(⋅)is bounded away om ze o, ha is, we do no equi e ha essin 𝑥∈ℝ𝑁𝑎(𝑥) > 0.Thisimplies ha he in eg and 𝜉(𝑥,𝑡) exhibi s unbalanced g ow h, namely he e holds ha 𝑡𝑞⩽𝜉(𝑥,𝑡) ⩽𝐶0(𝑡𝑝+𝑡 𝑞) o all 𝑥∈ℝ𝑁and 𝑡⩾0, (1.3) whe e 𝐶0>0 is a cons an . In his scena io, he s udy is ca ied ou in he amewo k o Musielak–O licz–Sobole spaces. Such unc ionals we e i s in es iga ed by Ma cellini [18–20] in he con ex o p oblems o he calculus o a ia ions and o non-linea elas ici y o s ongly aniso opic ma e ials. Fo such p oblems, he e is no global ( ha is, up o he bounda y) egula - i y heo y. The ea e only in e io egula i y esul s, which a e p ima ily due o Ba oni e al. [3]and Ma cellini [10, 20, 21]. In ac , mos o wo ks deal wi h double phase p oblems ha ing unbalanced g ow h in bounded domains o ℝ𝑁, we e e he eade s o [12–15, 22–24] and e e ences he ein. Howe e , he e exis ela i ely ew ones ea ing he p oblems in ℝ𝑁. The s udy o eigen alue p oblems like (1.1) is open un il now. Since (1.1) is se in he whole space ℝ𝑁, lack o compac ness is one o majo di icul ies we encoun e o discuss he eigen alue p oblem (1.1) in Musielak– O licz–Sobole spaces and mo e ca e ul analysis is needed in sui able weigh ed unc ions spaces. Indeed, his is mainly because he embedding 𝑊1,𝜉(ℝ𝑁)↪𝐿 𝑟(ℝ𝑁)is only con inuous o any 𝑞⩽𝑟⩽𝑞∗(see Lemma 2.3) and he weigh unc ion 𝑚∶ℝ𝑁→ℝis inde ini e, which cause ha he e i ica ion o he compac ness o he unde lying (minimizing and Palasi–Smale) sequences becomes di icul . Consequen ly, we manage o s udy he p oblem (1.1)inanewweigh edSobole space 𝐸de ined by he comple ion o 𝐶∞ 0(ℝ𝑁)unde he no m ‖𝑢‖𝐸∶= ‖∇𝑢‖𝜉+(∫ℝ𝑁|𝑢|𝑞max{𝑚2,𝜔}𝑑𝑥)1 𝑞,𝜔(𝑥)∶= 1 (1 + |𝑥|)𝑞,𝑥∈ℝ𝑁, whe e ‖⋅‖𝜉deno es he s anda d no m in 𝐷1,𝜉(ℝ𝑁). He e, 𝑊1,𝜉(ℝ𝑁)and 𝐷1,𝜉(ℝ𝑁)a e Musielak– O licz–Sobole spaces de ined in Sec ion 2. In his pape , when 𝑝<𝑞, we es ablish he exis ence o a con inuous amily o eigen alues o (1.1), s a ing om he p incipal equency o (1.2), see Theo ems 3.2 and 3.3.While𝑞<𝑝, we p o e he exis ence o a disc e e amily o posi i e eigen al- ues o (1.1), which di e ges o in ini y, see Theo em 3.4 and P oposi ion 3.2. The esul s we de i e e eal new ac s o eigen alues o double phase p oblems in ℝ𝑁. In bo h cases, we ac ually need o assume 𝑞<𝑞 ∗∶= 𝑁𝑞 𝑁−𝑞 , because o he unbalanced g ow h p ope y (1.3) wi h espec o he double phase ope a o and he dominance is he 𝑞-Laplacian e m. Thus, he p oblem unde con- side a ion is Sobole subc i ical and he ene gy unc ional 𝐽co esponding o (1.1) is well-de ined 14692120, 2024, 2, Downloaded om h ps://londma hsoc.onlinelib a y.wiley.com/doi/10.1112/blms.12961 by B no Uni e si y O Technology, Wiley Online Lib a y on [13/05/2024]. See he Te ms and Condi ions (h ps://onlinelib a y.wiley.com/ e ms-and-condi ions) on Wiley Online Lib a y o ules o use; OA a icles a e go e ned by he applicable C ea i e Commons License NON-AUTONOMOUS DOUBLE PHASE EIGENVALUE PROBLEMS 737 in he Sobole space 𝐸by Theo em 2.3, whe e 𝐽(𝑢) ∶= 1 𝑝∫ℝ𝑁 𝑎(𝑥)|∇𝑢|𝑝𝑑𝑥 + 1 𝑞∫ℝ𝑁|∇𝑢|𝑞𝑑𝑥 − 𝜆 𝑞∫ℝ𝑁 𝑚(𝑥)|𝑢|𝑞𝑑𝑥. Obse e ha 𝑝 𝑞<1+ 1 𝑁implies𝑝<𝑞 ∗.When double phasep oblemsa e se in boundeddomains in ℝ𝑁, hen he condi ion 𝑝 𝑞<1+ 1 𝑁can be applied o p o e he desi ed compac embedding esul s, o example [22, P oposi ion 4]. While double phase p oblems a e se in ℝ𝑁, he condi ion 𝑝 𝑞<1+ 1 𝑁can no longe be applicable o de i e he compac embedding esul s, which leads o lack o compac ness o he s udy. In his pape , such a condi ion is ac ually used o gua an ee he egula i y o solu ions o (1.1)(see[8, 9]), which along wi h he maximum p inciple de eloped in [23, 24] can lead o he simplici y o eigen alues, see P oposi ion 3.2. 2 PRELIMINARIES In he sec ion, we a e going o p esen some p elimina y esul s used o es ablish ou main heo ems. To deal wi h he eigen alue p oblem (1.1), we shall wo k in he co esponding Musielak–O licz–Sobole space. Fo he con enience o he eade s, le us i s p esen a ew de ini ions om [11, Sec ion 2] conce ning he main no ions and unc ion spaces used in his pape . De ini ion 2.1. A unc ion 𝜑 ∶ [0, +∞] → [0, +∞) is called a Φ- unc ion i 𝜑is con ex and le - con inuous on [0, +∞). In addi ion, 𝜑sa is ies ha 𝜑(0) = 0, lim 𝑡→0+𝜑(𝑡) = 0, lim 𝑡→+∞ 𝜑(𝑡) = +∞. De ini ion 2.2. A unc ion 𝜉∶ℝ𝑁× [0, +∞] → [0, +∞) is called a gene alized Φ- unc ion i i sa is ies he ollowing condi ions: (i) o almos e e y 𝑥∈ℝ𝑁,𝜉(𝑥,⋅)is a Φ- unc ion; (ii) o almos e e y 𝑡∈[0,+∞),𝜉(⋅,𝑡)is measu able. De ini ion 2.3. A gene alized Φ- unc ion 𝜉∶ℝ𝑁× [0, +∞] → [0, +∞) sa is ies Δ2-condi ion i he e exis s 𝐾⩾2such ha , o almos e e y 𝑥∈ℝ𝑁and 𝑡⩾0, 𝜉(𝑥,2𝑡) ⩾𝐾𝜉(𝑥, 𝑡). De ini ion 2.4. AΦ- unc ion𝜑 ∶ [0, +∞] → [0, +∞) issaid o be an𝑁- unc ion i i is con inuous and posi i e on [0, +∞). In addi ion, i sa is ies ha lim 𝑡→0+ 𝜑(𝑡) 𝑡=0, lim 𝑡→+∞ 𝜑(𝑡) 𝑡=+∞. A gene alized Φ- unc ion 𝜉∶ℝ𝑁× [0, +∞] → [0, +∞) is said o be a gene alized 𝑁- unc ion i , o almos e e y 𝑥∈ℝ𝑁,𝜉(𝑥,⋅)is an 𝑁- unc ion. 14692120, 2024, 2, Downloaded om h ps://londma hsoc.onlinelib a y.wiley.com/doi/10.1112/blms.12961 by B no Uni e si y O Technology, Wiley Online Lib a y on [13/05/2024]. See he Te ms and Condi ions (h ps://onlinelib a y.wiley.com/ e ms-and-condi ions) on Wiley Online Lib a y o ules o use; OA a icles a e go e ned by he applicable C ea i e Commons License 738 GOU and RĂDULESCU De ini ion 2.5. A gene alized 𝑁- unc ion 𝜉∶ℝ𝑁× [0, +∞] → [0, +∞) is called uni o mly con ex i , o any 𝜖>0, he e exis s 𝛿>0such ha , o almos e e y 𝑥∈ℝ𝑁, 𝜉(𝑥, 𝑠+𝑡 2)⩽(1 − 𝛿)𝜉(𝑥,𝑠)+𝜉(𝑥,𝑡) 2, whene e 𝑠,𝑡 ⩾0and |𝑥−𝑡|⩾𝜖max{|𝑠|,|𝑡|}. Wi h hese de ini ions in hand, we a e now eady o in oduce he double phase unc ion 𝜉∶ ℝ𝑁× [0, +∞) → [0, +∞) co esponding o (1.1)as 𝜉(𝑥,𝑡) ∶= 𝑎(𝑥)𝑡𝑝+𝑡 𝑞,𝑥∈ℝ𝑁,𝑡⩾0. (2.1) I is simple o check ha 𝜉is a gene alized 𝑁- unc ion. Mo eo e , 𝜉is uni o mly con ex and i sa is ies heΔ2-condi ion. Le us deno eby 𝑀(ℝ𝑁) he spaceconsis ing o all Lebesguemeasu able unc ion 𝑢∶ℝ𝑁→ℝ. The Musielak–O licz space 𝐿𝜉(ℝ𝑁)is de ined by 𝐿𝜉(ℝ𝑁)∶={𝑢∈𝑀(ℝ𝑁)∶𝜌 𝜉(𝑢) < +∞}, whe e 𝜌𝜉is he modula unc ion gi en by 𝜌𝜉(𝑢) ∶= ∫ℝ𝑁 𝜉(𝑥,|𝑢|)𝑑𝑥=∫ℝ𝑁 𝑎(𝑥)|𝑢|𝑝+|𝑢|𝑞𝑑𝑥. (2.2) He e, he space 𝐿𝜉(ℝ𝑁)is equipped wi h he Luxembu g no m gi en by ‖𝑢‖𝜉∶= in {𝜆>0∶𝜌 𝜉(𝑢 𝜆)⩽1}.(2.3) Using he abo e p ope ies sa is ied by 𝜉, we can easily check ha 𝐿𝜉(ℝ𝑁)is a Banach space, which is also sepa able and e lexi e. The Musielak–O licz–Sobole space 𝑊1,𝜉(ℝ𝑁)is de ined by 𝑊1,𝜉(ℝ𝑁)∶={𝑢∈𝐿 𝜉(ℝ𝑁)∶|∇𝑢|∈𝐿 𝜉(ℝ𝑁)}. He e, he space 𝑊1,𝜉(ℝ𝑁)is equipped wi h he no m ‖𝑢‖1,𝜉 ∶= ‖𝑢‖𝜉+‖∇𝑢‖𝜉, whe e ‖∇𝑢‖𝜉∶= ‖|∇𝑢|‖𝜉. Clea ly, 𝑊1,𝜉(ℝ𝑁)is a sepa able, e lexi e Banach space. Le us in oduce he associa ed homogeneous Musielak–O licz–Sobole 𝐷1,𝜉(ℝ𝑁)as he comple ion o 𝐶∞ 0(ℝ𝑁)unde he no m ‖∇𝑢‖𝜉. Nex , we a e going o show some ela ions be ween he no m in 𝐿𝜉(ℝ𝑁)and he modula unc ion 𝜌𝜉gi enby(2.2)and(2.3), espec i ely, p oo s o which can be comple ed by using he ing edien s p esen ed in [16, Sec ion 3.2]. 14692120, 2024, 2, Downloaded om h ps://londma hsoc.onlinelib a y.wiley.com/doi/10.1112/blms.12961 by B no Uni e si y O Technology, Wiley Online Lib a y on [13/05/2024]. See he Te ms and Condi ions (h ps://onlinelib a y.wiley.com/ e ms-and-condi ions) on Wiley Online Lib a y o ules o use; OA a icles a e go e ned by he applicable C ea i e Commons License NON-AUTONOMOUS DOUBLE PHASE EIGENVALUE PROBLEMS 739 Lemma 2.1. Le 𝜉∶ℝ𝑁× [0, +∞) → [0, +∞) be de ined by (2.1). Then, he ollowing asse ions hold. (i) ‖𝑢‖𝜉=𝜆i and only i 𝜌𝜉(𝑢 𝜆)=1. (ii) ‖𝑢‖𝜉< 1(= 1, > 1,𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦) i and only i 𝜌𝜉(𝑢) < 1(= 1,> 1,𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦). (iii) I ‖𝑢‖𝜉<1, hen‖𝑢‖max{𝑝,𝑞} 𝜉⩽𝜌𝜉(𝑢) ⩽‖𝑢‖min{𝑝,𝑞} 𝜉. (i ) I ‖𝑢‖𝜉>1, hen‖𝑢‖min{𝑝,𝑞} 𝜉⩽𝜌𝜉(𝑢) ⩽‖𝑢‖max{𝑝,𝑞} 𝜉. ( ) lim𝑛→+∞ ‖𝑢𝑛‖𝜉= 0(+∞,𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦) i and only i lim𝑛→+∞ 𝜌𝜉(𝑢𝑛)= 0(+∞,𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦). No e ha 𝑡𝑞⩽𝜉(𝑥,𝑡) o any 𝑥∈ℝ𝑁and 𝑡∈ℝ, by asse ion (ii)o Lemma 2.1, hen he e holds he ollowing embedding esul . Lemma 2.2. Le 𝜉∶ℝ𝑁× [0, +∞) → [0, +∞) be de ined by (2.1). Then, he embedding 𝐿𝜉(ℝ𝑁)↪ 𝐿𝑞(ℝ𝑁)is con inuous. As a consequence o Lemma 2.2 and Sobole ’s embeddings in 𝑊1,𝑞(ℝ𝑁)and 𝐷1,𝑞(ℝ𝑁) o 1< 𝑞<𝑁, we ha e he ollowing embedding esul . Lemma 2.3. Le 𝜉∶ℝ𝑁× [0, +∞) → [0, +∞) be de ined by (2.1). Then, he embedding 𝑊1,𝜉(ℝ𝑁)↪𝑊 1,𝑞(ℝ𝑁)↪𝐿 𝑟(ℝ𝑁)is con inuous o any 𝑞⩽𝑟⩽𝑞∗. Mo eo e , he embedding 𝐷1,𝜉(ℝ𝑁)↪𝐷 1,𝑞(ℝ𝑁)↪𝐿 𝑞∗(ℝ𝑁)is con inuous. 3 MAIN RESULTS In his sec ion, we shall conside he eigen alue p oblem (1.1) unde he assump ion (𝐻).The hypo hesis (𝐻) is always assumed o hold in wha ollows. Fi s , we shall p esen some esul s ela ed o he ollowing eigen alue p oblem, −Δ𝑎 𝑟𝑢=𝜇𝑚(𝑥)|𝑢|𝑟−2𝑢in ℝ𝑁.(3.1) Theo em 3.1. Assume (𝐻) holds, 𝑁⩾2,1<𝑟<𝑁,𝑎∈𝐶 0,1(ℝ𝑁,[0,+∞))and 𝑎≢0.Then, he e exis s a sequence o solu ions (𝜇𝑎,𝑟,𝑘,𝑢 𝑎,𝑟,𝑘)∈ℝ×𝐷 1,𝜂(ℝ𝑁) o (3.1) wi h 𝑢𝑎,𝑟,𝑘 ∈and 0<𝜇 𝑎,𝑟,1 <𝜇 𝑎,𝑟,2 ⩽⋯⩽𝜇𝑎,𝑟,𝑘 ⩽⋯,lim 𝑘→∞ 𝜇𝑎,𝑟,𝑘 →+∞ as 𝑘 → +∞, whe e 𝜂(𝑥, 𝑡) = 𝑎(𝑥)𝑡𝑟 o 𝑥∈ℝ𝑁and 𝑡⩾0, 𝑟∶= {𝑢∈𝐷 1,𝜂(ℝ𝑁)∶∫ℝ𝑁 𝑚(𝑥)|𝑢|𝑟𝑑𝑥 = 1}. P oo . De ine Ψ(𝑢) ∶= ∫ℝ𝑁 𝑎(𝑥)|∇𝑢|𝑟𝑑𝑥, 𝑀𝑟∶= 𝑟∩, 14692120, 2024, 2, Downloaded om h ps://londma hsoc.onlinelib a y.wiley.com/doi/10.1112/blms.12961 by B no Uni e si y O Technology, Wiley Online Lib a y on [13/05/2024]. See he Te ms and Condi ions (h ps://onlinelib a y.wiley.com/ e ms-and-condi ions) on Wiley Online Lib a y o ules o use; OA a icles a e go e ned by he applicable C ea i e Commons License 740 GOU and RĂDULESCU whe e he Sobole space is he comple ion o 𝐶∞ 0(ℝ𝑁)unde he no m ‖∇𝑢‖𝜂+(∫ℝ𝑁|𝑢|𝑟max{𝑚2,𝜔}𝑑𝑥)1 𝑟,𝜔(𝑥)= 1 (1 + |𝑥|)𝑟,𝑥∈ℝ𝑁. Reasoning as he p oo o [1, Lemma 1], we a e able o show Ψ(𝑢) es ic ed on 𝑀𝑟sa is ies he Palais–Smale condi ion. Then, by adap ing Ljus e nik–Schni elman heo y as he p oo o o hcomingTheo em 3.4,we can de i e he desi edconclusion. Thus, he p oo is comple ed. □ P oposi ion 3.1. Assume (𝐻) holds, 𝑁⩾2,1<𝑟<𝑁,𝑎∈𝐶 0,1(ℝ𝑁,[0,+∞))and 𝑎≢0.Then, he i s eigen alue 𝜇𝑎,𝑟,1 ob ained in Theo em 3.1 is simple and he eigen unc ion 𝑢𝑎,𝑟,1 has cons an sign. Mo eo e , i 𝑢∈𝐷 1,𝜂(ℝ𝑁)is a non- i ial solu ion o (3.1) co esponding o 𝜇>𝜇 𝑎,𝑟,1, hen𝑢 is sign-changing. Since he unc ion 𝑚is an inde ini e sign weigh , hen p oo o P oposi ion 3.1 is no s aigh o wa d. To p o e his, we need he ollowing auxilia y esul . Lemma 3.1. De ine 𝐼(𝑢,𝑣) ∶= −∫ℝ𝑁(Δ𝑎 𝑟𝑢)𝑢𝑟−𝑣 𝑟 𝑢𝑟−1 𝑑𝑥 − ∫ℝ𝑁(Δ𝑎 𝑟𝑣)𝑣𝑟−𝑢 𝑟 𝑣𝑟−1 𝑑𝑥, 𝑢,𝑣 ∈ 𝐷1,𝜂(ℝ𝑁), 𝑢, 𝑣 > 0. Then, 𝐼(𝑢,𝑣) ⩾0. Mo eo e , 𝐼(𝑢,𝑣) = 0 i and only i 𝑢=𝑘𝑣 o some 𝑘∈ℝ. P oo . Obse e ha ∇(𝑢𝑟−𝑣 𝑟 𝑢𝑟−1 )=(1+(𝑟−1) (𝑣 𝑢)𝑟)∇𝑢 − 𝑟(𝑣 𝑢)𝑟−1∇𝑣, ∇(𝑣𝑟−𝑢 𝑟 𝑣𝑟−1 )=(1+(𝑟−1) (𝑢 𝑣)𝑟)∇𝑣 − 𝑟(𝑢 𝑣)𝑟−1∇𝑢. Then, by he di e gence heo em, we see 𝐼(𝑢,𝑣) = ∫ℝ𝑁 𝑎(𝑥)((1+(𝑟−1) (𝑣 𝑢)𝑟)|∇𝑢|𝑟−𝑟 (𝑣 𝑢)𝑟−1|∇𝑢|𝑟−2(∇𝑣 ⋅∇𝑢))𝑑𝑥 +∫ℝ𝑁 𝑎(𝑥)((1+(𝑟−1) (𝑢 𝑣)𝑟)|∇𝑣|𝑟−𝑟 (𝑢 𝑣)𝑟−1|∇𝑣|𝑟−2(∇𝑢 ⋅∇𝑣))𝑑𝑥. (3.2) Using Young’s inequali y, we ha e 𝑟(𝑣 𝑢)𝑟−1|∇𝑢|𝑟−2(∇𝑣 ⋅∇𝑢)⩽𝑟(𝑣 𝑢)𝑟−1|∇𝑢|𝑟−1|∇𝑣|⩽(𝑟 − 1)(𝑣 𝑢)𝑟|∇𝑢|𝑟+|∇𝑣|𝑟, 𝑟(𝑢 𝑣)𝑟−1|∇𝑣|𝑟−2(∇𝑢 ⋅∇𝑣)⩽𝑟(𝑢 𝑣)𝑟−1|∇𝑣|𝑟−1|∇𝑢|⩽(𝑟 − 1)(𝑢 𝑣)𝑟|∇𝑣|𝑟+|∇𝑢|𝑟. As a consequence, coming back o (3.2), we can conclude 𝐼(𝑢,𝑣) ⩾0.I 𝐼(𝑢,𝑣) = 0, hen ∇𝑢 ⋅∇𝑣 = |∇𝑢||∇𝑣|,(𝑣 𝑢)𝑟|∇𝑢|𝑟=|∇𝑣|𝑟,(𝑢 𝑣)𝑟|∇𝑣|𝑟=|∇𝑢|𝑟. 14692120, 2024, 2, Downloaded om h ps://londma hsoc.onlinelib a y.wiley.com/doi/10.1112/blms.12961 by B no Uni e si y O Technology, Wiley Online Lib a y on [13/05/2024]. See he Te ms and Condi ions (h ps://onlinelib a y.wiley.com/ e ms-and-condi ions) on Wiley Online Lib a y o ules o use; OA a icles a e go e ned by he applicable C ea i e Commons License NON-AUTONOMOUS DOUBLE PHASE EIGENVALUE PROBLEMS 741 I hen ollows ha |𝑢∇𝑣 − 𝑣∇𝑢|=0. This implies ha he e exis s 𝑘∈ℝsuch ha 𝑢=𝑘𝑣and he p oo is comple ed. □ P oo o P oposi ion 3.1. No e i s ha 𝜇𝑎,𝑟,1 =in 𝑢∈𝑟 Ψ(𝑢). I 𝑢∈𝑟sa is ies Ψ(𝑢) = 𝑢𝑎,𝑟,1, hen |𝑢|∈𝑟and Ψ(|𝑢|)=𝑢 𝑎,𝑟,1. The e o e, wi hou es ic- ion, we may assume 𝑢𝑎,𝑟,1 is non-nega i e. Obse e ha 𝑢𝑎,𝑟,1 ∈𝐷 1,𝜂(ℝ𝑁)sa is ies he equa ion −Δ𝑎 𝑟𝑢𝑎,𝑟,1 +𝜇 𝑎,𝑟,1𝑚2(𝑥)|𝑢𝑎,𝑟,1|𝑟−2𝑢𝑎,𝑟,1 =𝜇 𝑎,𝑟,1𝑚1(𝑥)|𝑢𝑎,𝑟,1|𝑟−2𝑢𝑎,𝑟,1 ⩾0in ℝ𝑁. By maximum p inciple, 𝑢𝑎,𝑟,1 >0.Le 𝑢𝑎,𝑟,1 ∈𝑟and 𝑣𝑎,𝑟,1 ∈𝑟be wo posi i e eigen unc ions co esponding o 𝜇𝑎,𝑟,1, hen −Δ𝑎 𝑟𝑢𝑎,𝑟,1 =𝜇 𝑎,𝑟,1𝑚(𝑥)𝑢𝑟−1 𝑎,𝑟,1,−Δ 𝑎 𝑟𝑣𝑎,𝑟,1 =𝜇 𝑎,𝑟,1𝑚(𝑥)𝑣𝑟−1 𝑎,𝑟,1 in ℝ𝑁. I is simple o calcula e 𝐼(𝑢𝑎,𝑟,1,𝑣 𝑎,𝑟,1)=0.Asa esul o Lemma3.1,weha e𝑢𝑎,𝑟,1 =𝑘𝑣 𝑎,𝑟,1 o some 𝑘∈ℝ. This indica es ha 𝜇𝑎,𝑟,1 is simple. A guing by con adic ion, we suppose 𝑢∈𝐷 1,𝜂(ℝ𝑁)is a non-nega i e solu ion o (3.1) co esponding o 𝜇>𝜇 𝑎,𝑟,1. By he maximum p inciple, 𝑢>0. No ice ∫ℝ𝑁 𝑎(𝑥)|∇𝑢|𝑟𝑑𝑥 = 𝜇 ∫ℝ𝑁 𝑚(𝑥)|𝑢|𝑟𝑑𝑥 > 0. In addi ion, we know ha i 𝑢∈𝐷 1,𝜂(ℝ𝑁)is a solu ion o (3.1), hen 𝑘𝑢 ∈ 𝐷1,𝜂(ℝ𝑁)is also a solu ion o (3.1) o any 𝑘∈ℝ∖{0}. Then, by scaling, we may assume 0<∫ℝ𝑁 𝑚(𝑥)|𝑢|𝑟𝑑𝑥 < 1. (3.3) Le 𝑢𝑎,𝑟,1 ∈and 𝑢𝑎,𝑟,1 >0be an eigen unc ion o(3.1) co esponding o 𝜇𝑎,𝑟,1.Then, 𝑢𝑎,𝑟,1 sol es he equa ion −Δ𝑎 𝑟𝑢𝑎,𝑟,1 =𝜇 𝑎,𝑟,1𝑚(𝑥)|𝑢𝑎,𝑟,1|𝑟−2𝑢𝑎,𝑟,1 in ℝ𝑁. As a consequence o Lemma 3.1 and (3.3), we ha e 0⩽𝐼(𝑢,𝑢𝑎,𝑟,1)=𝜇∫ℝ𝑁 𝑚(𝑥)(𝑢𝑟−𝑢 𝑟 𝑎,𝑟,1)𝑑𝑥 + 𝜇𝑎,𝑟,1 ∫ℝ𝑁 𝑚(𝑥)(𝑢𝑟 𝑎,𝑟,1 −𝑢 𝑟)𝑑𝑥 =(𝜇−𝜇 𝑎,𝑟,1)∫ℝ𝑁 𝑚(𝑥)|𝑢|𝑟𝑑𝑥 − (𝜇 − 𝜇𝑎,𝑟,1)<0. This is impossible, hence 𝑢is sign-changing and he p oo is comple ed. □ Theo em 3.2. Assume (𝐻) holds, 𝑁⩾2,1<𝑝,𝑞<𝑁,𝑝≠𝑞,𝑎∈𝐶 0,1(ℝ𝑁,[0,+∞))and 𝑎≢0. Then, (1.1) has no non- i ial solu ions in 𝐷1,𝜉(ℝ𝑁) o any 0⩽𝜆⩽𝜇1,𝑞,1,whe e𝜇1,𝑞,1 >0is he i s eigen alue o (3.1) wi h 𝑎≡1and 𝑟=𝑞. 14692120, 2024, 2, Downloaded om h ps://londma hsoc.onlinelib a y.wiley.com/doi/10.1112/blms.12961 by B no Uni e si y O Technology, Wiley Online Lib a y on [13/05/2024]. See he Te ms and Condi ions (h ps://onlinelib a y.wiley.com/ e ms-and-condi ions) on Wiley Online Lib a y o ules o use; OA a icles a e go e ned by he applicable C ea i e Commons License 742 GOU and RĂDULESCU P oo . Le 𝑢∈𝐷 1,𝜉(ℝ𝑁)be a solu ion o (1.1) o some0⩽𝜆⩽𝜇1,𝑞,1. Obse e i s ha ∫ℝ𝑁 𝑎(𝑥)|∇𝑢|𝑝𝑑𝑥 + ∫ℝ𝑁|∇𝑢|𝑞𝑑𝑥 = 𝜆 ∫ℝ𝑁 𝑚(𝑥)|𝑢|𝑞𝑑𝑥. (3.4) This implies 𝑢=0i 𝜆=0. Le us assume 0<𝜆<𝜇 1,𝑞,1. Assume 𝑢≠0, i hen ollows om (3.4) ha ∫ℝ𝑁 𝑚(𝑥)|𝑢|𝑞𝑑𝑥 > 0. (3.5) In addi ion, since 𝜇1,𝑞,1 >0is he i s eigen alue o (3.1), hen ∫ℝ𝑁|∇𝑢|𝑞𝑑𝑥 ⩾𝜇1,𝑞,1 ∫ℝ𝑁 𝑚(𝑥)|𝑢|𝑞𝑑𝑥. (3.6) This along wi h (3.4) leads o 𝜇1,𝑞,1 ∫ℝ𝑁 𝑚(𝑥)|𝑢|𝑞𝑑𝑥 ⩽𝜆∫ℝ𝑁 𝑚(𝑥)|𝑢|𝑞𝑑𝑥. Using (3.5), we hen ge 𝑢=0. This is a con adic ion. Nex we assume 𝜆=𝜇 1. In his case, by combining (3.4)and(3.6), we ob ain ∫ℝ𝑁 𝑎(𝑥)|∇𝑢|𝑝𝑑𝑥 ⩽0, hence 𝑢=0. Thus, he p oo is comple ed. □ 3.1 Case 𝒑<𝒒 In his case, o es ablish he exis ence o solu ions o (1.1), we shall adap some ideas om [1]. Le us i s in oduce he weigh unc ion 𝜔(𝑥) = 1 (1 + |𝑥|)𝑞,𝑥∈ℝ𝑁. Le 𝐸be he comple ion o 𝐶∞ 0(ℝ𝑁)unde he no m ‖𝑢‖𝐸∶= ‖∇𝑢‖𝜉+(∫ℝ𝑁|𝑢|𝑞max{𝑚2,𝜔}𝑑𝑥)1 𝑞. I is s anda d o conclude ha 𝐸is a sepa able and e lexi e Banach space. In o de o p o e he exis ence o solu ions o (1.1), we shall de ine he associa ed ene gy unc ional 𝐽∶𝐸→ℝby 𝐽(𝑢) ∶= 1 𝑝∫ℝ𝑁 𝑎(𝑥)|∇𝑢|𝑝𝑑𝑥 + 1 𝑞∫ℝ𝑁|∇𝑢|𝑞𝑑𝑥 − 𝜆 𝑞∫ℝ𝑁 𝑚(𝑥)|𝑢|𝑞𝑑𝑥. Theo em 3.3. Assume (𝐻) holds, 𝑁⩾2,1<𝑝<𝑞<𝑁,𝑎∈𝐶 0,1(ℝ𝑁,[0,+∞))and 𝑎≢0.Then, he e exis posi i e solu ions o (1.1) o any𝜆>𝜇 1,𝑞,1. 14692120, 2024, 2, Downloaded om h ps://londma hsoc.onlinelib a y.wiley.com/doi/10.1112/blms.12961 by B no Uni e si y O Technology, Wiley Online Lib a y on [13/05/2024]. See he Te ms and Condi ions (h ps://onlinelib a y.wiley.com/ e ms-and-condi ions) on Wiley Online Lib a y o ules o use; OA a icles a e go e ned by he applicable C ea i e Commons License NON-AUTONOMOUS DOUBLE PHASE EIGENVALUE PROBLEMS 749 whe e he second ac holds because o 𝑚1∈𝐿 𝑁 𝑞(ℝ𝑁) om he assump ion (𝐻). Combining (3.23), (3.24)and(3.25), by (3.22), we hen ob ain ∫ℝ𝑁(𝑎(𝑥)(|∇𝑢𝑛|𝑝−2∇𝑢𝑛−|∇𝑢|𝑝−2∇𝑢)+(|∇𝑢𝑛|𝑞−2∇𝑢𝑛−|∇𝑢|𝑞−2∇𝑢))⋅(∇𝑢𝑛−∇𝑢 )𝑑𝑥 = 𝑜𝑛(1). Obse e ha |𝑧1−𝑧 2|𝑟⩽𝐶((|𝑧1|𝑟−2𝑧1−|𝑧2|𝑟−2𝑧2)⋅(𝑧1−𝑧 2))𝜃 2(|𝑧1|𝑟+|𝑧2|𝑟)1− 𝜃 2,∀𝑧 1,𝑧 2∈ℝ𝑁,(3.26) whe e 𝜃=𝑟i 1<𝑟<2and 𝜃=2i 𝑟⩾2. Then, we see ∫ℝ𝑁 𝑎(𝑥)(|∇𝑢𝑛−∇𝑢|𝑝)𝑑𝑥 + ∫ℝ𝑁|∇𝑢𝑛−∇𝑢|𝑞𝑑𝑥 ⩽𝐶(∫ℝ𝑁 𝑎(𝑥)(|∇𝑢𝑛|𝑝−2∇𝑢𝑛−|∇𝑢|𝑝−2∇𝑢)⋅(∇𝑢𝑛−∇𝑢 )𝑑𝑥)𝜃 2(∫ℝ𝑁 𝑎(𝑥)(|∇𝑢𝑛|𝑝+|∇𝑢|𝑝)𝑑𝑥)1− 𝜃 2 +𝐶 (∫ℝ𝑁(|∇𝑢𝑛|𝑞−2∇𝑢𝑛−|∇𝑢|𝑞−2∇𝑢)⋅(∇𝑢𝑛−∇𝑢 )𝑑𝑥)𝜃 2(∫ℝ𝑁|∇𝑢𝑛|𝑞+|∇𝑢|𝑞𝑑𝑥)1− 𝜃 2=𝑜 𝑛(1). This immedia ely indica es ha 𝑢𝑛→𝑢in 𝐷1,𝜉(ℝ𝑁)as 𝑛→∞. Taking ad an age o (3.20)and (3.21), we hen ge ∫ℝ𝑁 𝑚(𝑥)|𝑢𝑛|𝑞𝑑𝑥 = ∫ℝ𝑁 𝑚(𝑥)|𝑢|𝑞𝑑𝑥 + 𝑜𝑛(1), because o 𝜆𝑛=𝜆+𝑜 𝑛(1) and 𝜆≠0. In iew o (3.18), ∫ℝ𝑁 𝑚2(𝑥)|𝑢𝑛|𝑞𝑑𝑥 = ∫ℝ𝑁 𝑚2(𝑥)|𝑢|𝑞𝑑𝑥 + 𝑜𝑛(1). Since 𝑢𝑛→𝑢in 𝐷1,𝑞(ℝ𝑁)as 𝑛→∞, by Ha dy’s inequali y, ∫ℝ𝑁|𝑢𝑛−𝑢|𝑞 (1 + |𝑥|)𝑞𝑑𝑥 ⩽(𝑝 𝑁−𝑝)𝑝∫ℝ𝑁|∇𝑢𝑛−∇𝑢|𝑞𝑑𝑥 = 𝑜𝑛(1). Consequen ly, we de i e ha 𝑢𝑛→𝑢in 𝐸as 𝑛→∞. Thus, he p oo is comple ed. □ Theo em 3.4. Assume (𝐻) holds, 𝑁⩾2,1<𝑞<𝑝<𝑁,𝑎∈𝐶 0,1(ℝ𝑁,[0,+∞))and 𝑎≢0.Then, he e exis s a sequence o solu ions (𝜆𝑘,𝑢 𝑘)∈ℝ×𝐸wi h 𝑢𝑘∈and 0<𝜆 1<𝜆 2⩽⋯⩽𝜆𝑘⩽⋯,lim 𝑘→∞ 𝜆𝑘→+∞as 𝑘 → +∞. P oo . To es ablish he exis ence o a sequence o eigen alues o (1.1), we shall ake in o accoun Ljus e nik–Schni elman heo y in [2]. De ine Σ∶={𝐴⊂∶𝐴is compac and 𝐴=−𝐴 }. 14692120, 2024, 2, Downloaded om h ps://londma hsoc.onlinelib a y.wiley.com/doi/10.1112/blms.12961 by B no Uni e si y O Technology, Wiley Online Lib a y on [13/05/2024]. See he Te ms and Condi ions (h ps://onlinelib a y.wiley.com/ e ms-and-condi ions) on Wiley Online Lib a y o ules o use; OA a icles a e go e ned by he applicable C ea i e Commons License 750 GOU and RĂDULESCU Fo a se 𝐴∈Σ, he genus o 𝐴is de ined by 𝛾(𝐴) ∶= min{𝑛∈ℕ∶exis s a unc ion 𝜑∈𝐶(𝐴,ℝ𝑛∖{0}) sa is ying 𝜑(−𝑥) = −𝜑(𝑥)}. I such a minimum does no exis , we se 𝛾(𝐴) = +∞. Le us now de ine Σ𝑘∶= {𝐴∈Σ∶𝛾(𝐴)⩾𝑘},∀𝑘∈ℕ+. Fi s we see ha , o any 𝑘∈ℕ+,Σ𝑘≠∅. Indeed, le 𝑋𝑘be a 𝑘-dimensional subspace o 𝐸,by Bo suk–Ulam’s heo em, hen 𝛾(∩𝑋 𝑘)⩾𝑘.De ine  𝜆𝑘∶= in 𝐴∈Σ𝑘 sup 𝑢∈𝐴 Φ(𝑢). Since Σ𝑘+1 ⊂Σ 𝑘, 𝜆𝑘⩽ 𝜆𝑘+1 o any 𝑘∈ℕ+. F om Lemma 3.2, 𝜆𝑘is a c i ical poin o 𝐽 es ic ed on  o any 𝑘∈ℕ+. Then, we de i e ha 0<  𝜆1< 𝜆2⩽⋯⩽ 𝜆𝑘⩽ 𝜆𝑘+1 ⩽⋯. Nex we p o e  𝜆𝑘→+∞as 𝑘→+∞.Le {𝑒𝑖}⊂𝐸be such ha 𝐸=span{𝑒1,𝑒 2,…,𝑒 𝑖,…}.Le {𝑒′ 𝑖}⊂ 𝐸be such ha 𝐸′=span{𝑒′ 1,𝑒′ 2,…,𝑒′ 𝑖,…}, whe e 𝐸′deno es he dual space o 𝐸.De ine𝑋𝑖∶= span{𝑒𝑖}and 𝑌𝑘∶= 𝑘 ⨁ 𝑖=1 𝑋𝑖,𝑍 𝑘∶= ∞ ⨁ 𝑖=𝑘 𝑋𝑖,∀𝑘∈ℕ+. Le 𝐴∈Σ 𝑘sa is y 𝛾(𝐴) ⩾𝑘. By basic p ope ies o he genus, we ha e 𝐴∩𝑍 𝑘≠∅.De ine 𝛽𝑘∶= in 𝐴∈Σ𝑘 sup 𝑢∈𝐴∩𝑍𝑘 𝐽(𝑢), ∀ 𝑘 ∈ ℕ+. Then, 𝛽𝑘→+∞as 𝑘→∞. O he wise, we may assume {𝛽𝑘}⊂ℝis bounded. Thus, he e exis s a sequence {𝑢𝑘}⊂𝐴∩𝑍 𝑘such ha {Φ(𝑢𝑘)} ⊂ ℝis bounded. I hen ollows ha {𝑢𝑘}is bounded in 𝐸. Fu he , he e exis s 𝑢∈𝐸such ha 𝑢𝑘⇀𝑢in 𝐸as 𝑛→∞. Obse e ha ⟨𝑒′ 𝑖,𝑢⟩=⟨𝑒′ 𝑖,𝑢 𝑘⟩+ 𝑜𝑘(1) = 𝑜𝑘(1), because o 𝑢𝑘∈𝑍 𝑘. The e o e, we ha e 𝑢=0and 𝑢𝑘⇀0in 𝐸as 𝑘→∞.This along wi h he assump ion ha 𝑚1∈𝐿 𝑁 𝑞(ℝ𝑁) om he assump ion (𝐻) leads o ∫ℝ𝑁 𝑚1(𝑥)|𝑢𝑘|𝑞𝑑𝑥 = 𝑜𝑘(1). Since 𝑚2⩾0 om he assump ion (𝐻), ∫ℝ𝑁 𝑚(𝑥)|𝑢𝑘|𝑞𝑑𝑥 = ∫ℝ𝑁 𝑚1(𝑥)|𝑢𝑘|𝑞𝑑𝑥 − ∫ℝ𝑁 𝑚2(𝑥)|𝑢𝑘|𝑞𝑑𝑥 ⩽𝑜𝑘(1), which is impossible due o 𝑢𝑘∈. Consequen ly, we ge ha 𝛽𝑘→+∞as 𝑘→∞.Thanks o  𝜆𝑘⩾𝛽𝑘 o any 𝑘∈ℕ+, 𝜆𝑘→+∞as 𝑘→∞.Since𝑢𝑘∈𝐸is a c i ical poin o 𝐸 es ic ed on , he e exis s 𝜆𝑘∈ℝsuch ha −Δ𝑎 𝑝𝑢𝑘−Δ 𝑞𝑢𝑘=𝜆 𝑘𝑚(𝑥)|𝑢𝑘|𝑞−2𝑢𝑘in ℝ𝑁, 14692120, 2024, 2, Downloaded om h ps://londma hsoc.onlinelib a y.wiley.com/doi/10.1112/blms.12961 by B no Uni e si y O Technology, Wiley Online Lib a y on [13/05/2024]. See he Te ms and Condi ions (h ps://onlinelib a y.wiley.com/ e ms-and-condi ions) on Wiley Online Lib a y o ules o use; OA a icles a e go e ned by he applicable C ea i e Commons License NON-AUTONOMOUS DOUBLE PHASE EIGENVALUE PROBLEMS 751 whe e 𝜆𝑘=1 𝑞∫ℝ𝑁 𝑎(𝑥)|∇𝑢𝑘|𝑝𝑑𝑥 + 1 𝑞∫ℝ𝑁|∇𝑢𝑘|𝑞𝑑𝑥 > Φ(𝑢𝑘)=  𝜆𝑘,∀𝑘∈ℕ+. Thus, he p oo is comple ed. □ Lemma 3.3. De ine 𝐼(𝑢,𝑣) ∶ = −∫ℝ𝑁(Δ𝑎 𝑝𝑢)𝑢𝑞−𝑣 𝑞 𝑢𝑞−1 𝑑𝑥 − ∫ℝ𝑁(Δ𝑞𝑢)𝑢𝑞−𝑣 𝑞 𝑢𝑞−1 𝑑𝑥 −∫ℝ𝑁(Δ𝑎 𝑝𝑣)𝑣𝑞−𝑢 𝑞 𝑣𝑞−1 𝑑𝑥 − ∫ℝ𝑁(Δ𝑞𝑣)𝑣𝑞−𝑢 𝑞 𝑣𝑞−1 𝑑𝑥, (3.27) whe e 𝑢,𝑣 ∈ 𝐷1,𝜉(ℝ𝑁),𝑢, 𝑣 > 0 and 1<𝑞<𝑝.Then,𝐼(𝑢,𝑣) ⩾0. Mo eo e , 𝐼(𝑢,𝑣) = 0 i and only i 𝑢=𝑘𝑣 o some 𝑘∈ℝ. P oo . Le us i s show 𝐼1(𝑢,𝑣)∶=−∫ℝ𝑁(Δ𝑎 𝑝𝑢)𝑢𝑞−𝑣 𝑞 𝑢𝑞−1 𝑑𝑥 − ∫ℝ𝑁(Δ𝑎 𝑝𝑣)𝑣𝑞−𝑢 𝑞 𝑣𝑞−1 𝑑𝑥 ⩾0, 𝑢, 𝑣 ∈ 𝐷1,𝜉(ℝ𝑁), 𝑢, 𝑣 > 0. I is s aigh o wa d o compu e ∇(𝑢𝑞−𝑣 𝑞 𝑢𝑞−1 )=(1+(𝑞−1) (𝑣 𝑢)𝑞)∇𝑢 − 𝑞(𝑣 𝑢)𝑞−1∇𝑣, (3.28) ∇(𝑣𝑞−𝑢 𝑞 𝑣𝑞−1 )=(1+(𝑞−1) (𝑢 𝑣)𝑞)∇𝑣 − 𝑞(𝑢 𝑣)𝑞−1∇𝑢. (3.29) The e o e, by he di e gence heo em, we de i e ha 𝐼1(𝑢, 𝑣) = ∫ℝ𝑁 𝑎(𝑥)((1+(𝑞−1) (𝑣 𝑢)𝑞)|∇𝑢|𝑝−𝑞 (𝑣 𝑢)𝑞−1|∇𝑢|𝑝−2∇𝑢 ⋅∇𝑣)𝑑𝑥 +∫ℝ𝑁 𝑎(𝑥)((1+(𝑞−1) (𝑢 𝑣)𝑞)|∇𝑣|𝑝−𝑞 (𝑢 𝑣)𝑞−1|∇𝑣|𝑝−2∇𝑣 ⋅∇𝑢)𝑑𝑥. Using Young’s inequali y, we know ha 𝑞(𝑣 𝑢)𝑞−1|∇𝑢|𝑝−2|∇𝑢 ⋅∇𝑣|⩽𝑞(𝑣 𝑢)𝑞−1|∇𝑢|𝑝−1|∇𝑣| ⩽𝑞(𝑝 − 1) 𝑝(𝑣 𝑢)𝑝(𝑞−1) 𝑝−1 |∇𝑢|𝑝+𝑞 𝑝|∇𝑣|𝑝 =𝑞(𝑝 − 1) 𝑝(𝑣 𝑢)𝑝(𝑞−1) 𝑝−1 |∇𝑢|𝑝2(𝑞−1) 𝑞(𝑝−1) |∇𝑢|𝑝(𝑝−𝑞) 𝑞(𝑝−1) +𝑞 𝑝|∇𝑣|𝑝 ⩽(𝑞 − 1)(𝑣 𝑢)𝑞|∇𝑢|𝑝+𝑝−𝑞 𝑝|∇𝑢|𝑝+𝑞 𝑝|∇𝑣|𝑝. 14692120, 2024, 2, Downloaded om h ps://londma hsoc.onlinelib a y.wiley.com/doi/10.1112/blms.12961 by B no Uni e si y O Technology, Wiley Online Lib a y on [13/05/2024]. See he Te ms and Condi ions (h ps://onlinelib a y.wiley.com/ e ms-and-condi ions) on Wiley Online Lib a y o ules o use; OA a icles a e go e ned by he applicable C ea i e Commons License 752 GOU and RĂDULESCU Simila ly, we can ge 𝑞(𝑢 𝑣)𝑞−1|∇𝑣|𝑝−2|∇𝑣 ⋅∇𝑢|⩽𝑞(𝑢 𝑣)𝑞−1|∇𝑣|𝑝−1|∇𝑢|⩽(𝑞 − 1)(𝑢 𝑣)𝑞|∇𝑣|𝑝+𝑝−𝑞 𝑝|∇𝑣|𝑝+𝑞 𝑝|∇𝑢|𝑝. I hen ollows ha 𝐼1(𝑢, 𝑣) ⩾0. Nex , we p o e ha 𝐼2(𝑢,𝑣)∶=−∫ℝ𝑁(Δ𝑞𝑢)𝑢𝑞−𝑣 𝑞 𝑢𝑞−1 𝑑𝑥 − ∫ℝ𝑁(Δ𝑞𝑣)𝑣𝑞−𝑢 𝑞 𝑣𝑞−1 𝑑𝑥 ⩾0, 𝑢, 𝑣 ∈ 𝐷1,𝜉(ℝ𝑁), 𝑢, 𝑣 > 0. In iew o (3.28)and(3.29), by he di e gence heo em, 𝐼2(𝑢, 𝑣) = ∫ℝ𝑁(1+(𝑞−1) (𝑣 𝑢)𝑞)|∇𝑢|𝑞−𝑞 (𝑣 𝑢)𝑞−1|∇𝑢|𝑞−2∇𝑢 ⋅∇𝑣 𝑑𝑥 +∫ℝ𝑁(1+(𝑞−1) (𝑢 𝑣)𝑞)|∇𝑣|𝑞−𝑞 (𝑢 𝑣)𝑞−1|∇𝑣|𝑞−2∇𝑣 ⋅∇𝑢 𝑑𝑥. Using again Young’s inequali y, we ob ain 𝑞(𝑣 𝑢)𝑞−1|∇𝑢|𝑞−2|∇𝑢 ⋅∇𝑣|⩽𝑞(𝑣 𝑢)𝑞−1|∇𝑢|𝑞−1|∇𝑣|⩽(𝑞 − 1)(𝑣 𝑢)𝑞|∇𝑢|𝑞+|∇𝑣|𝑞, 𝑞(𝑢 𝑣)𝑞−1|∇𝑣|𝑞−2|∇𝑣 ⋅∇𝑢|⩽𝑞(𝑢 𝑣)𝑞−1|∇𝑣|𝑞−1|∇𝑢|⩽(𝑞 − 1)(𝑢 𝑣)𝑞|∇𝑣|𝑞+|∇𝑢|𝑞. The e o e, we ha e 𝐼2(𝑢, 𝑣) = 0. Acco dingly, he e holds ha 𝐼(𝑢,𝑣) ⩾0 o any 𝑢,𝑣 ∈ 𝐷1,𝜉(ℝ𝑁) and 𝑢,𝑣 > 0.I 𝐼(𝑢,𝑣)=0, hen 𝐼2(𝑢, 𝑣) = 0. This leads o ∇𝑢 ⋅∇𝑣 = |∇𝑢||∇𝑣|,(𝑣 𝑢)𝑞|∇𝑢|𝑞=|∇𝑣|𝑞,(𝑢 𝑣)𝑞|∇𝑣|𝑞=|∇𝑢|𝑞, As a consequence, we see ha |𝑢∇𝑣 − 𝑣∇𝑢|=0. This implies ha he e exis s 𝑘∈ℝsuch ha 𝑢=𝑘𝑣and he p oo is comple ed. □ Rema k 3.1. In ac , Lemma 3.3 is es ablished o he double phase ope a o unde he assump ion 𝑞<𝑝, which is no a di ec consequence o Lemma 3.1. I is unknown o us i Lemma 3.3 emains alid o he case 𝑝<𝑞. F om he p oo o Lemma 3.3, one can see ha he assump ion 𝑞<𝑝is c ucial, which is he p emise o he use o Young’s inequali y. P oposi ion 3.2. Assume (𝐻) holds, 𝑁⩾2,1<𝑞<𝑝<𝑁,𝑝 𝑞<1+ 1 𝑁,𝑎∈𝐶 0,1(ℝ𝑁,[0,+∞)) and 𝑎≢0. Assume ha any eigen unc ion o (1.1) co esponding o 𝜆is non-nega i e. Then, 𝜆 is simple. P oo . Le 𝑢∈𝐸be a non-nega i e eigen unc ion o (1.1) co esponding o 𝜆. I ollows om [8]and[23, P oposi ion 3] o [24, P oposi ion 2.3] ha 𝑢>0.Le 𝑢>0 and 𝑣>0 be wo eigen unc ions o (1.1) co esponding o 𝜆. Then, we see ha −Δ𝑎 𝑝𝑢−Δ 𝑞𝑢 = 𝜆𝑚(𝑥)𝑢𝑞−1,−Δ 𝑎 𝑝𝑣−Δ 𝑞𝑣 = 𝜆𝑚(𝑥)𝑣𝑞−1 in ℝ𝑁. 14692120, 2024, 2, Downloaded om h ps://londma hsoc.onlinelib a y.wiley.com/doi/10.1112/blms.12961 by B no Uni e si y O Technology, Wiley Online Lib a y on [13/05/2024]. See he Te ms and Condi ions (h ps://onlinelib a y.wiley.com/ e ms-and-condi ions) on Wiley Online Lib a y o ules o use; OA a icles a e go e ned by he applicable C ea i e Commons License NON-AUTONOMOUS DOUBLE PHASE EIGENVALUE PROBLEMS 753 As a esul , he e holds ha 𝐼(𝑢,𝑣)=𝜆∫ℝ𝑁 𝑚(𝑥)(𝑢𝑞−𝑣 𝑞)𝑑𝑥 + 𝜆 ∫ℝ𝑁 𝑚(𝑥)(𝑣𝑞−𝑢 𝑞)𝑑𝑥 = 0. I hen ollows om Lemma 3.3 ha he desi ed conclusion holds. This comple es he p oo . □ P oposi ion 3.3. Assume (𝐻) holds, 𝑁⩾2,1<𝑝,𝑞<𝑁,𝑝≠𝑞,𝑎∈𝐶 0,1(ℝ𝑁,[0,+∞))and 𝑎≢0. Then, 𝜇1,𝑞,1 =in ⎧ ⎪ ⎨ ⎪ ⎩ 1 𝑝∫ℝ𝑁𝑎(𝑥)|∇𝑢|𝑝𝑑𝑥 + 1 𝑞∫ℝ𝑁|∇𝑢|𝑞𝑑𝑥 1 𝑞∫ℝ𝑁𝑚(𝑥)|𝑢|𝑞𝑑𝑥 ∶𝑢∈𝐸∖{0},∫ℝ𝑁 𝑚(𝑥)|𝑢|𝑞𝑑𝑥 > 0⎫ ⎪ ⎬ ⎪ ⎭ . P oo . Since 𝜇1,𝑞,1 is he i s eigen alue o (3.1)and𝐸⊂𝐷 1,𝑞(ℝ𝑁), 𝜇1,𝑞,1 =in {∫ℝ𝑁|∇𝑢|𝑞𝑑𝑥 ∫ℝ𝑁𝑚(𝑥)|𝑢|𝑞𝑑𝑥 ∶𝑢∈𝐷 1,𝑞(ℝ𝑁)∖{0}, ∫ℝ𝑁 𝑚(𝑥)|𝑢|𝑞𝑑𝑥 > 0} ⩽in ⎧ ⎪ ⎨ ⎪ ⎩ 1 𝑝∫ℝ𝑁𝑎(𝑥)|∇𝑢1,𝑞,1|𝑝𝑑𝑥 + 1 𝑞∫ℝ𝑁|∇𝑢|𝑞𝑑𝑥 1 𝑞∫ℝ𝑁𝑚(𝑥)|𝑢|𝑞𝑑𝑥 ∶𝑢∈𝐸∖{0},∫ℝ𝑁 𝑚(𝑥)|𝑢|𝑞𝑑𝑥 > 0⎫ ⎪ ⎬ ⎪ ⎭ . Le 𝑢1,𝑞,1 ∈𝐸be an eigen unc ion o (3.1) co esponding o 𝜇1,𝑞,1 and 𝑝<𝑞, hen in ⎧ ⎪ ⎨ ⎪ ⎩ 1 𝑝∫ℝ𝑁𝑎(𝑥)|∇𝑢|𝑝𝑑𝑥 + 1 𝑞∫ℝ𝑁|∇𝑢|𝑞𝑑𝑥 1 𝑞∫ℝ𝑁𝑚(𝑥)|𝑢|𝑞𝑑𝑥 ∶𝑢∈𝐸∖{0},∫ℝ𝑁 𝑚(𝑥)|𝑢|𝑞𝑑𝑥 > 0⎫ ⎪ ⎬ ⎪ ⎭ ⩽ 𝑛𝑝 𝑝∫ℝ𝑁𝑎(𝑥)|∇𝑢1,𝑞,1|𝑝𝑑𝑥 + 𝑛𝑞 𝑞∫ℝ𝑁|∇𝑢1,𝑞,1|𝑞𝑑𝑥 𝑛𝑞 𝑞∫ℝ𝑁𝑚(𝑥)|𝑢1,𝑞,1|𝑞𝑑𝑥 =𝜇 1,𝑞,1 +𝑜 𝑛(1) as 𝑛→∞. Simila ly, i 𝑞<𝑝, hen in ⎧ ⎪ ⎨ ⎪ ⎩ 1 𝑝∫ℝ𝑁𝑎(𝑥)|∇𝑢|𝑝𝑑𝑥 + 1 𝑞∫ℝ𝑁|∇𝑢|𝑞𝑑𝑥 1 𝑞∫ℝ𝑁𝑚(𝑥)|𝑢|𝑞𝑑𝑥 ∶𝑢∈𝐸∖{0},∫ℝ𝑁 𝑚(𝑥)|𝑢|𝑞𝑑𝑥 > 0⎫ ⎪ ⎬ ⎪ ⎭ ⩽ 1 𝑝𝑛𝑝∫ℝ𝑁𝑎(𝑥)|∇𝑢1,𝑞,1|𝑝𝑑𝑥 + 1 𝑞𝑛𝑞∫ℝ𝑁|∇𝑢1,𝑞,1|𝑞𝑑𝑥 1 𝑞𝑛𝑞∫ℝ𝑁𝑚(𝑥)|𝑢1,𝑞,1|𝑞𝑑𝑥 =𝜇 1,𝑞,1 +𝑜 𝑛(1) as 𝑛→∞. Thus, he desi ed esul ollows and he p oo is comple ed. □ 14692120, 2024, 2, Downloaded om h ps://londma hsoc.onlinelib a y.wiley.com/doi/10.1112/blms.12961 by B no Uni e si y O Technology, Wiley Online Lib a y on [13/05/2024]. See he Te ms and Condi ions (h ps://onlinelib a y.wiley.com/ e ms-and-condi ions) on Wiley Online Lib a y o ules o use; OA a icles a e go e ned by he applicable C ea i e Commons License 754 GOU and RĂDULESCU Rema k 3.2. Unde he assump ions o Theo em 3.4, by Theo em 3.2 and P oposi ion 3.3,weha e 𝜆1>in ⎧ ⎪ ⎨ ⎪ ⎩ 1 𝑝∫ℝ𝑁𝑎(𝑥)|∇𝑢|𝑝𝑑𝑥 + 1 𝑞∫ℝ𝑁|∇𝑢|𝑞𝑑𝑥 1 𝑞∫ℝ𝑁𝑚(𝑥)|𝑢|𝑞𝑑𝑥 ∶𝑢∈𝐸∖{0},∫ℝ𝑁 𝑚(𝑥)|𝑢|𝑞𝑑𝑥 > 0⎫ ⎪ ⎬ ⎪ ⎭ . Rema k 3.3. The a gumen sde elopedin hispape allow oob ainsimila esul si hehypo hesis (𝐻) is eplaced by he ollowing condi ion in oduced by Szulkin and Willem [27], ()𝑚∈𝐿 1 𝑙𝑜𝑐(ℝ𝑁),𝑚+=𝑚 1+𝑚 2≠0,𝑚1∈𝐿 𝑁 𝑞(ℝ𝑁), o e e y 𝑦∈ℝ𝑁,lim𝑥→𝑦 |𝑥− 𝑦|𝑞𝑚2(𝑥) = 0 and lim|𝑥|→∞ |𝑥|𝑞𝑚2(𝑥) = 0, whe e 𝑚+∶= max{𝑚(𝑥), 0}. ACKNOWLEDGEMENTS T. Gou was suppo ed by he Na ional Na u al Science Founda ion o China (No. 12101483) and he Pos doc o al Science Founda ion o China (No. 2021M702620). V.D. Rădulescu was suppo ed by he g an “Nonlinea Di e en ial Sys ems in Applied Sciences” o he Romanian Minis y o Resea ch, Inno a ion and Digi iza ion, wi hin PNRR-III-C9-2022-I8/22. The au ho s would like o hank wa mly he anonymous e e ees o hei e y p ecise eading o ou pape and o gi ing cons uc i e commen s and sugges ions. JOURNAL INFORMATION The Bulle in o he London Ma hema ical Socie y is wholly owned and managed by he London Ma hema ical Socie y, a no - o -p o i Cha i y egis e ed wi h he UK Cha i y Commission. All su plus income om i s publishing p og amme is used o suppo ma hema icians and ma hema ics esea ch in he o m o esea ch g an s, con e ence g an s, p izes, ini ia i es o ea ly ca ee esea che s and he p omo ion o ma hema ics. ORCID VicenţiuD.Rădulescu h ps://o cid.o g/0000-0003-4615-5537 REFERENCES 1. W. Alleg e o and Y. X. Huang, Eigen alues o he inde ini e-weigh p-Laplacian in weigh ed spaces,Funkcial. Ek ac 38 (1995), no. 2, 233–242. 2. A. Amb ose i and A. 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