with mi IPl and nj =l we have Then,we have 2(u-u)Tv:+2(v-v:)Tu:+(u-u)T(v-v:) LP(U,VP,⊙，7)-G(U,VP,⊙，V,U,V) <2(u-u)Tv:+2(v-v:)Tul +l(u-u)T(v-v:) fP(U,VP)-gP(U,VP,n|U,VP) ≤u-l房+lv房+v-v3 =∑2(UVg）-s2,(U,vg,nU,V (,)Ep +lu房+专u-房+v-v) ≤u3+川v:房+362， For clarity,we denote U.,V[U and [Vl as u,v,ut and vt,respectively.Then we have Then we can prove u)=g-a-业+Pv-+wA ou,v)≤lo(u,v川≤uB+v.l层+362) ×(u-3+Iv-v2): +lu-u+u恨+lv-v+v:l恨 Because llu-url≤2andv-v:ll≤62，we can get =fij(ut,V)+Vufij(u,V)(u-u) fij(u,v)<fij(ut,vt)+Vfij(ut,vt)(u-ut) +7f.(u,v)(v-v) +7i(u,v)(w-v) +h(u,v)+o(u,v), +房+u房+郑2+ where VTfi.j(u,vt)=Aiut-(Ri.j-ui vt)vt, +R-vIu-u眼 V fij(ut,v)=A2v:-(Ri.j-ud vt)ut, +a房+眼+2+ h(u,v)contains all the second order terms,and the third +与R-vlv-v3. order and forth order terms are contained in o(u,v).Hence, we can get So if we let Auv)=岁u-w房+告v-尼 之wxim.v.房+ue+子+ -(Rij-uivt)(u-u)T(v-v:) +Rg-, +(v-)+(u-山)v)月， 2mu房+眼+2+0 1 and +R- o(u.v)=j[2(u-u)Tv:+2(v-v:)Tw we can prove that +(u-u)T(v-vt)](u-ut)T(v-vt). G(U,VP,⊙，7，U,V)≥LP(U,VP,⊙，7). Because That is to say,T should be smaller than some value which is dependent on the current U.and VP. -(Rij-uvi)(u-ut)T(v-vt) The second equation in Lemma 1 can be easily proved. ≤2R-ufv:(u-ua+v-v), B.PROOF OF LEMMA 2 and PROOF.From Lemma 1,we have j(wF(v-v)+(w-w)v) GP(U+1,VP+1,Vi,nJU:,V?) ≤lu(v-v)2+I(u-)Tv2 ≥LP(U+1,V8+1,Θ，V), GP(U,vP,,V,nU:,VP)=LP(U,v,,V.). ≤lu喉lv-v候+lv房lu-u匠， Furthermore,we have we have G(U+1,V+1,o,V,U,V) uv)s空+R-d+v.)u-u尼 ≤G(Ut,V,o,V,U,V). +(空+R-d+u)v-层 Hence,we can get LP(U+1,VR+1,⊙，V)≤LP(U,V,Θ，V) Because I(u-u)(v-ve)s (u-ui+llv-vi)/2,with mi = |Ω p i | and nj = |Ω˜ p j |. Then, we have L p (U, V p , Θ p t , Vt) − G p (U, V p , Θ p t , Vt, τt|Ut, V p t ) = f p (U, V p ) − g p (U, V p , τt|Ut, V p t ) = X (i,j)∈Ωp  ˆfi,j (U∗i, V p ∗j ) − g p i,j (U∗i, V p ∗j , τt|Ut, V p t )  , For clarity, we denote U∗i,V p ∗j ,[U∗i]t and [V p ∗j ]t as u,v,ut and vt, respectively. Then we have ˆfi,j (u, v) = 1 2  ￾ Ri,j − (u − ut + ut) T (v − vt + vt)
2 + λ1ku − ut + utk 2 F + λ2kv − vt + vtk 2 F  = ˆfi,j (ut, vt) + ∇ T u ˆfi,j (ut, vt)(u − ut) + ∇ T v ˆfi,j (ut, vt)(v − vt) + h(u, v) + o(u, v), where ∇ T u ˆfi,j (ut, vt) = λ1ut − (Ri,j − u T t vt)vt, ∇ T v ˆfi,j (ut, vt) = λ2vt − (Ri,j − u T t vt)ut, h(u, v) contains all the second order terms, and the third order and forth order terms are contained in o(u, v). Hence, we can get h(u, v) = λ1 2 ku − utk 2 F + λ2 2 kv − vtk 2 F − (Ri,j − u T t vt)(u − ut) T (v − vt) + 1 2 ￾ u T t (v − vt) + (u − ut) T vt
2 , and o(u, v) = 1 2 [2(u − ut) T vt + 2(v − vt) T ut + (u − ut) T (v − vt)](u − ut) T (v − vt). Because − (Ri,j − u T t vt)(u − ut) T (v − vt) ≤ 1 2 |Ri,j − u T t vt| ￾ ku − utk 2 F + kv − vtk 2 F
, and 1 2 ￾ u T t (v − vt) + (u − ut) T vt
2 ≤ ku T t (v − vt)k 2 F + k(u − ut) T vtk 2 F ≤ kutk 2 F kv − vtk 2 F + kvtk 2 F ku − utk 2 F , we have h(u, v) ≤ ( λ1 2 + 1 2 |Ri,j − u T t vt| + kvtk 2 F )ku − utk 2 F + (λ2 2 + 1 2 |Ri,j − u T t vt| + kutk 2 F )kv − vtk 2 F . Because |(u − ut) T (v − vt)| ≤ (ku − utk 2 F + kv − vtk 2 F )/2, we have |2(u − ut) T vt + 2(v − vt) T ut + (u − ut) T (v − vt)| ≤|2(u − ut) T vt| + |2(v − vt) T ut| + |(u − ut) T (v − vt)| ≤ku − utk 2 F + kvtk 2 F + kv − vtk 2 F + kutk 2 F + 1 2 [ku − utk 2 F + kv − vtk 2 F ] ≤kutk 2 F + kvtk 2 F + 3δ 2 , Then we can prove o(ut, vt) ≤ |o(ut, vt)| ≤ 1 4 ￾ kutk 2 F + kvtk 2 F + 3δ 2
× ￾ ku − utk 2 F + kv − vtk 2 F
. Because ||u − ut||2 F ≤ δ 2 and ||v − vt||2 F ≤ δ 2 , we can get ˆfi,j (u, v) ≤ ˆfi,j (ut, vt) + ∇ T u ˆfi,j (ut, vt)(u − ut) +∇ T v ˆfi,j (ut, vt)(v − vt) +( 5 4 kvtk 2 F + 1 4 kutk 2 F + 3 4 δ 2 + 1 2 λ1 + 1 2 |Ri,j − u T t vt|)ku − utk 2 F +( 5 4 kutk 2 F + 1 4 kvtk 2 F + 3 4 δ 2 + 1 2 λ2 + 1 2 |Ri,j − u T t vt|)kv − vtk 2 F . So if we let 1 τt ≥ max{2mi 5 4 kvtk 2 F + 1 4 kutk 2 F + 3 4 δ 2 + 1 2 λ1 + 1 2 |Ri,j − u T t vt| , 2nj 5 4 kutk 2 F + 1 4 kvtk 2 F + 3 4 δ 2 + 1 2 λ2 + 1 2 |Ri,j − u T t vt| }, we can prove that G p (U, V p , Θ p t , Vt, τt|Ut, V p t ) ≥ L p (U, V p , Θ p t , Vt). That is to say, τt should be smaller than some value which is dependent on the current Ut and V p t . The second equation in Lemma 1 can be easily proved. B. PROOF OF LEMMA 2 Proof. From Lemma 1, we have G p (Ut+1,V p t+1, Θ p t , Vt, τt|Ut, V p t ) ≥ L p (Ut+1, V p t+1, Θ p t , Vt), G p (Ut, V p t , Θ p t ,Vt, τt|Ut, V p t ) = L p (Ut, V p t , Θ p t , Vt). Furthermore, we have G p (Ut+1,V p t+1, Θ p t , Vt, τt|Ut, V p t ) ≤ G p (Ut, V p t , Θ p t , Vt, τt|Ut, V p t ). Hence, we can get L p (Ut+1, V p t+1, Θ p t , Vt) ≤ L p (Ut, V p t , Θ p t , Vt).