Proof.Since the coefficients of (3) are locally Lipschitz continuous, then ∀X0,Y0,Z0∈D∗, there exists a local solution Xt,Yt,Zt on t∈0,λe quasi surely (q.s.), where λe represents the explosion time. To show λe=+∞ q.s., we prove Xt,Yt,Zt does not explode to infinity in a finite time. Suppose k0>1 is large enough such that (s.t) X0,Y0,Z0 lies in the interval 1/k0,k03. For k≥k0, define(5)λk=inft∈0,λe:Xt∉1k,k or Yt∉1k,k or Zt∉1k,k,where λk is increasing as k⟶∞. We have λ∞=limk⟶∞λk, therefore λ∞≤λe quasi surely. Suppose we guarantee that λ∞=∞ q.s., then λe=∞ and vω:Xt,Yt,Zt∈R+3=1 q.s. If we assume 0<Vλ∞<+∞, then there exists a pair of constants χ>0 and ε∈0,1 s.t.(6)Vλ∞≤χ≥vλ∞≤χ≥ε.
Then, ∃k1≥k0 s.t.(7)VΠk:=Vλk≤χ≥ε, for all k≥k1.
Set a function U1:R+3⟶R+ by(8)U1X,Y,Z=X−ln X+Y−ln Y+Z−ln Z−3.
We note the function gx:=x−ln x−1≥0 for any x>0. Using the G-Ito lemma for the function U1, we get(9)dU1=∂U1∂xdX+∂U1∂ydY+∂U1∂zdZ+12∂2U1∂x2dX2+∂2U1∂y2dY2+∂2U1∂z2dZ2=1−1XdX+1−1YdY+1−1ZdZ+12σ¯12+σ¯22+σ¯32dBt=:LU1dBt+ΘX,Y,ZdBt,where(10)LU1=X−1rK−rXK−βY1+αY+δX−1ZX+Y−1βX1+αY−ρ+ϑ+γ+Z−1γYZ−Z−1μ+δ+12∑i=13σ¯i2=r+rKX−rKX2−βX1+αY+βY1+αY−ρY−ϑY−γZY−δXZ−μZ−r+ρ+ϑ+γ+μ+δ+12∑i=13σ¯i2,ΘX,Y,Z=σ¯12X−1+σ¯22Y−1+σ¯32Z−1.
We note that the region D∗=X+Y+Z≤K and all the parameters are positive, then we have(11)LU1≤r+rKX+βY1+αY+ρ+ϑ+γ+μ+δ+12∑i=13σ¯i2≤r+rKX+βY+ρ+ϑ+γ+μ+δ+12∑i=13σ¯i2≤ΛK+ρ+ϑ+γ+μ+δ+12∑i=13σ¯i2,where Λ=:maxr+r/K,β. We denote(12)C=ΛK+ρ+ϑ+γ+μ+δ+12∑i=13σ¯i2.
Therefore,(13)dU1=:LU1dBt+ΘX,Y,ZdBt≤CdBt+ΘX,Y,ZdBt.
Integrate (13) from 0 to λk∧χ,(14)U1Xλk∧χ,Yλk∧χ,Zλk∧χ≤U1X0,Y0,Z0+C⋅Bλk∧χ+∫0λk∧χΘXt,Yt,ZtdBt,and take the G-expectation,(15)^U1Xλk∧χ,Yλk∧χ,Zλk∧χ≤U1X0,Y0,Z0+C⋅^Bλk∧χ=U1X0,Y0,Z0+C⋅π¯2⋅λk∧χ.
Note the set Πkω:=ω:λkω≤χ and (7), then VΠkω≥ε for all k≥k1. We see that the definition of λk, then for everyω∈Πkω, there exist at least Xλkω or Yλkω or Zλkω equals to k or 1/k. For example, if Xλkω=k or Xλk=1/k, then U1Xλk,Yλk,Zλk=k−1−ln k+Yλk−1−ln Yλk+Zλk−1+ln Zλk≥k−1−ln k, or U1Xλk,Yλk,Zλk=1/k−1+ln k+Yλk−1−ln Yλk+Zλk−1+ln Zλk≥1/k−1+ln k. Thus,(16)U1Xλk,Yλk,Zλk≥min1k−1+ln k,k−1−ln k.
From (7) and (14)–(16), we have(17)^IΠkωU1Xλk,Yλk,Zλk=^IΠkωU1Xλk∧χ,Yλk∧χ,Zλk∧χ≤U1X0,Y0,Z0+C⋅^IΠkωBλk∧χ≤U1X0,Y0,Z0+C⋅π¯2⋅χ<+∞,(18)^IΠkωU1Xλk,Yλk,Zλk≥^1k+ln k−1∧k−1−ln k⋅IΠkω=1k+ln k−1∧k−1−ln k⋅^IΠkω=1k+ln k−1∧k−1−ln k⋅VΠkω≥1k+ln k−1∧k−1−ln k⋅ε.
Therefore, from inequalities (17) and (18), we have(19)1k+ln k−1∧k−1−ln k⋅ε≤^IΠkωU1Xλk,Yλk,Zλk<+∞.
Letting k⟶∞, we find out inequality (19) is a contradiction. Thus, Vλ∞<+∞=0, namely, vλ∞=+∞=1 and vω:Xt,Yt,Zt∈R+3,t≥0=1.