Dirichlet problem with mixed singularities

In this section, we will study problems having the so-called mixed singularities, that is, both time and space ones. Moreover, in some theorems we omit the assumption that the nonlinearity f in the differential equation is positive. In literature we can find results

about the solvability of singular Dirichlet problems with sign-changing nonlinearities which mostly concernw-solutions. Here, we will prove the existence of solutions to prob- lem (7.1) provided f has mixed singularities. We assume thatA1,A2are closed intervals containing 0 and

f ∈Car(0,T)×D , whereD =A1\ {0} ×

A2\ {0} , f has time singularities att=0 and att=T,

and space singularities atx=0 and aty=0,

(7.67)

that is, there exists (x,y)∈Dsuch that

_ε 0 _{f(t,x,y)dt= ∞,} T T−ε _{f(t,x,y)dt= ∞forε∈0,}T 2 , lim sup x→0

_{f(t,x,y)= ∞ for a.e.t∈[0,T] and somey∈A2\ {0},}

lim sup y→0

_{f(t,x,y)= ∞ for a.e.t∈[0,T] and somex∈A1\ {0}.}

Since problem (7.1) containsφ-Laplacian and has mixed singularities, we cannot use theorems of Sections1.2 and1.3. Hence, we will prove their new generalized version. In order to do it we will consider the sequence of regular problems

φ(u) + fk(t,u,u)=0, u(0)=ak, u(T)=bk, (7.68) where fk∈Car([0,T]×R2_{),ak,bk∈R,k∈N.}

Theorem 7.44 (principle forφ-Laplacian and mixed singularities). Let (7.67) hold, let εk>0,ηk>0 fork∈Nand assume that

lim k→∞εk=0, klim→∞ηk=0; (7.69) fk(t,x,y)= f(t,x,y) for a.e.t∈1 k,T− 1 k , for eachk > 2 T and for each (x,y)∈A1×A2,|x| ≥εk,|y| ≥ηk;

(7.70)

there exists a bounded setΩ⊂C1_{[0,T] such that} the regular problem (7.68) has a solutionuk∈Ω anduk(t),u_k(t) ∈A1×A2fort∈[0,T], k > 2

T.

(7.71)

Then there existu∈C[0,T] and a subsequence{uk} ⊂ {uk}such that lim

Further assume that there is a finite setS= {s1,. . .,sν} ⊂(0,T) such that the sequenceφu_k is equicontinuous

on each interval [a,b]⊂(0,T)\S. (7.72) Thenu∈C1_{((0,T)\S) and} lim →∞u k(t)=u(t) locally uniformly on (0,T)\S. Assume, in addition, that limk→∞ak=0, limk→∞bk=0 and that

S=s∈(0,T) :u(s)=0 oru(s)=0 oru(s) does not exist. (7.73) Thenφ(u)∈ACloc((0,T)\S) anduis aw-solution of problem (7.1).

Denotes0=0 andsν+1=T. Moreover, let there beη∈(0,T/2),λ0,μ0,λ1,μ1,. . .,λν+1, μν+1∈ {−1, 1},0∈Nandψ∈L1[0,T] such that

λifk

t,uk(t),uk(t) signuk(t)≥ψ(t)

for a.e.t∈si−η,si ∩(0,T), and alli∈ {0,. . .,ν+ 1},≥0, (7.74) μifk

t,uk(t),uk(t) signuk(t)≥ψ(t)

for a.e.t∈si,si+η ∩(0,T), and alli∈ {0,. . .,ν+ 1},≥0. (7.75) Thenφ(u) ∈ AC[0,T] and uis a solution of problem (7.1). Moreover, (u(t),u(t)) ∈

A1×A2holds fort∈[0,T]. Proof

Step 1. Convergence of the sequence{uk}.

Assume that conditions (7.67), (7.70), and (7.71) hold. By (7.71) there existsr >0 such that the sequence{uk}of solutions to problem (7.68) satisfies

_uk

C1≤r fork >

2 T.

Hence, the sequence{uk}is bounded and equicontinuous on [0,T]. Due to the Arzel`a- Ascoli theorem, this yields the existence of a functionu ∈ C[0,T] and a subsequence {uk} ⊂ {uk}such that lim→∞uk(t)=u(t) uniformly on [0,T].

Step 2. Convergence of the sequence{u

k}.

Assume, in addition to step 1, that condition (7.72) holds and choose an arbitrary interval [a,b] ⊂(0,T)\S. Then{φ(u

k)}and consequently{uk}is equicontinuous on [a,b]. Since{u

k}is also bounded on [a,b], we can use the Arzel`a-Ascoli theorem and choose a subsequence{uk}such that it uniformly converges on [0,T] and lim→∞uk(t)= u(t) uniformly on [a,b]. Using the diagonalization theorem we deduce that we can

choose the uniformly converging on [0,T] subsequence{uk}so that lim →∞u k(t)=u(t) locally uniformly on (0,T)\S. Therefore,u∈C1_((0,T)\S).

Step 3. Convergence of the approximate nonlinearities{fk}.

Assume, in addition to step 2, that limk→∞ak=0, limk→∞bk=0, and that condition (7.73) holds. Thenu(0)=u(T)=0. DefineU=V1∪V2∪S, where

V1=t∈(0,T) : f(t,·,·) :D →Ris not continuous,

V2=

t∈(0,T) : the equality in condition (7.70) is not fulfilled.

Choose an arbitraryt∈(0,T)\U. Then there exists0 ∈Nsuch that for all ≥0we havet∈[1/k,T−1/k],|uk(t)| ≥εk,|uk(t)| ≥ηkand fk t,uk(t),uk(t) = f t,uk(t),u_k(t) . Sincetis an arbitrary element in (0,T)\Uand meas(U)=0, we get

lim →∞fk t,uk(t),uk(t) = f t,u(t),u(t) a.e. on [0,T]. (7.76) Step 4. The functionuis aw-solution.

Now, choose an arbitrary interval [a,b]⊂(0,T)\S. Then there exist∗∈N,ε∗>0, andη∗>0 such that for all≥∗

_fk

t,uk(t),u_k(t) ≤h(t) for a.e.t∈[a,b], where

h(t)=supf(t,x,y):ε∗≤ |x| ≤r,η∗≤ |y| ≤r∈L1[a,b].

Therefore, we can apply the Lebesgue dominated convergence theorem and get f(t,u(t), u(t))∈L1[a,b] and lim →∞ b a fk s,uk(s),u_k(s) ds= b a f s,u(s),u(s) ds. Integrating the equality

φu_k(t) + fkt,uk(t),u_k(t) =0 for a.e.t∈[0,T], (7.77) we get φu_k(t) −φu_k(a) + t a fk s,uk(s),u_k(s) ds=0 fort∈[a,b], which for→ ∞leads to

φu(t) −φu(a) +

t a f

Since [a,b] can be an arbitrary interval in (0,T)\S, we deduce thatφ(u)∈ACloc((0,T)\

S) anduis aw-solution of problem (7.1). Step 5. The functionuis a solution.

Assume, in addition to step 3, that there existη∈(0,T/2),λ0,. . .,λν+1,μ0,. . .,μν+1∈ {−1, 1},0 ∈ N, and ψ ∈ L1[0,T] such that conditions (7.74) and (7.75) are valid. Since u is aw-solution of problem (7.1), it remains to prove thatφ(u) ∈ AC[0,T]. By step 3, f(t,u(t),u(t)) ∈L1[a,b] for each [a,b]⊂(0,T)\S. So, it suffices to prove f(t,u(t),u(t)) ∈L1[ci,di] fori =0,. . .,ν+ 1, where (ci,di)= (si−η,si+η)∩(0,T). Choose an arbitraryi ∈ {0,. . .,ν+ 1}and t ∈ (ci,di)\S. Thenu(t) = 0. If we use equality (7.76) and the fact that{u_k}locally uniformly converges touon (0,T)\S, we obtain

lim →∞fk

t,uk(t),u_k(t) signu_k(t)= ft,u(t),u(t) signu(t)

for a.e.t∈ [ci,di]. If we multiply equality (7.77) by signu_k(t) and then integrate over [ci,di], we get for≥0

di

ci fk

s,uk(s),u_k(s) signu_k(s)ds≤φu_kdi +φu_kci ≤2φ(r). Therefore, the Fatou lemma yields f(t,u(t),u(t))∈L1[ci,di], by conditions (7.74) and (7.75). Hence, f(t,u(t),u(t))∈L1[0,T] andφ(u)∈AC[0,T]. Remark 7.45. (i)Theorem 7.44guarantees the existence of a solutionuwhich can change its sign.

(ii) According to Step 4 of the proof ofTheorem 7.44, we can claim thatTheorem 7.44

remains valid if we replace (7.75) with

fkt,uk(t),u_k(t) ≥ψ(t)

for a.e.t∈si−η,si+η ∩(0,T) and alli∈ {0,. . .,ν+ 1},≥0.

(7.78)

(iii) If f has no singularity aty=0, then we putηk =0 fork∈NinTheorem 7.44. Moreover, due to step 3 of the proof ofTheorem 7.44, the setSin (7.73) consists only of the zeros ofu. This will be accounted for in the next theorem. We will assume that

f ∈Car(0,T)×D can change its sign, D=(0,∞)×R,

and f has mixed singularities att=0, t=T, x=0. (7.79)

Theorem 7.46. Let (7.79) hold. Letσ1andσ2be a lower function and an upper function of