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Existence and Uniqueness of Weak Solutions to Parabolic IVP’s For T > 0, fixed, and U ⊂ R n a bounded open set, let U T = Then consider the following initial boundary value problem, ∂ t ux, t + Lux, t = fx, t in U T ux, 0 = u 0 x x ∈ U, ux, t = 0, x ∈ ∂U, 0 < t < T. n n i,j=1 j=1 x, t : x ∈ U, 0 < t < T . 1 Lux, t = − ∑ ∂ i a ij x, t ∂ j u + ∑ b j x, t ∂ j u + cx, tux, t where = − divax, t ∇u + ⃗ bx, t ⋅ ∇u + cx, tux, t with a ij , b j , c ∈ L ∞ U T for a 0 > 0, ⃗z ⋅ a ij x, t ⃗z ≥ a 0 | ⃗z | 2 and ∀z⃗ ∀x, t ∈ U T . Then L is said to be uniformly elliptic on U T . We recall that the assumptions on the coefficients a ij , b j , c, imply the existence of positive constants, a 1 , a 0 and μ 0 such that for all u, v ∈ H 1 U, and μ > μ 0 , |Bu, v, t| ≤ a 1 ‖u‖ 1 ‖v‖ 1 a. e. t ∈ 0, T Bu, u, t + μu, u 0 ≥ a 0 ‖u‖ 21 a. e. t ∈ 0, T where Bu, v, t = ∫ Then the mapping U n n i,j=1 j=1 ∑ ∂ i va ij x, t ∂ j u + v ∑ b j x, t ∂ j u + cx, t dx L μ = L + μI : H 10 U → H −1 U is an isomorphism, and the IBVP (1) is said to be a parabolic problem. For convenience, we will let V = H 10 U, H = H 0 U, and V ′ = H −1 U. Then V ⊂ H = H ′ ⊂ V ′ and ⟨f, v⟩ V ′ ×V = f, v H for f ∈ V ′ , v ∈ V. We recall that W0, T = u ∈ L 2 0, T : V : ∂ t u ∈ L 2 0, T : V ′ and that W0, T ⊂ C0, T : H. Then we define a weak solution of (1) to be a function u ∈ W0, T satisfying i ∂ t u, v H + Bu, v, t = f, v H ii u0 = u 0 . ∀v ∈ V a. e. t ∈ 0, T, 2 Equivalently, a weak solution satisfies T T T i ∫ 0 ∂ t u, v H dt + ∫ 0 Bu, v, tdt = ∫ 0 f, v H dt ii u0 = u 0 . ∀v ∈ W0, T We point out that in a parabolic problem, we can without loss of generality assume that 1 the bilinear form Bu, v is coercive. To see this, we simply observe that when L is uniformly elliptic, the associated bilinear form is at least V − H − coercive; i.e., B μ u, v = Bu, v + μu, v H is coercive for μ > μ 0 > 0. If u is a weak solution for (2) then it is easy to show that Ux, t = e μt ux, t solves i ∂ t U, v H + B μ U, v, t = e −μt f, v H ii U0 = u 0 . ∀v ∈ V a. e. t ∈ 0, T, Then existence of U is equivalent to existence of u and the problem for U involves a coercive bilinear form. Now we have, Theorem (Existence-uniqueness-continuous dependence) Suppose L is uniformly elliptic on U T with coefficients in L ∞ U T . Then for every f ∈ L 2 0, T : H −1 U and each u 0 ∈ H 0 U, there exists a unique u ∈ W0, T satisfying (2).Moreover, the linear mapping L 2 0, T : H −1 U × H 0 U : f, u 0 u ∈ W0, T is continuous. Proof- (uniqueness) Suppose u ∈ W0, T solves (2.2) in the case f = u 0 = 0. Then T T ∫ 0 ∂ t u, u H dt + ∫ 0 Bu, u, tdt = 0. But T T T ∫ 0 ∂ t u, u H dt = 1 ∫ 0 ∂ t u, u H dt = 1 d ∫ 0 u, u H dt 2 2 dt = 1 ‖uT‖ 2H − ‖u0‖ 2H . 2 Then 1 ‖uT‖ 2 + ∫ T Bu, u, tdt = 0, H 0 2 and, using the fact that B is coercive, 1 ‖uT‖ 2 + a 0 ∫ T ‖ut‖ 2 dt ≤ 1 ‖uT‖ 2 + ∫ T Bu, u, tdt = 0 H H H 0 0 2 2 Since u ∈ W0, T is continuous in t with values in H = H 0 U, it follows that ut = 0. This proves that the solution is unique, if it exists. The proof of existence will be carried out in steps. 1) Existence of Approximate Solutions Let w k denote an orthonormal basis for H which is, at the same time, an orthogonal basis for V. For N a positive integer, let N u N t = ∑ C jN t w j j=1 where the coefficients C jN t are required to satisfy i ∂ t u N , w k H + Bu N , w k , t = f, w k H k = 1, . . , N a. e. t ∈ 0, T, ii u N 0, w k H = u 0 , w k , . for k = 1, . . , N 3 2 Since w k denotes an orthonormal basis for H, this reduces to, N d C t + ∑ C jN tBw j , w k , t = f k t dt kN for k = 1, . . , N j=1 C kN 0 = u 0 , w k , . for k = 1, . . , N This is a system of linear ordinary differential equations in N unknowns where the coefficient matrix is positive definite on 0, T hence it follows that for every f ∈ L 2 0, T : H −1 U and each u 0 ∈ H 0 U, there exists a unique solution C kN t : for k = 1, . . , N ∈ C0, T N for the system ofODE’s and a corresponding approximate solution u N t for (3). 2) Energy Estimates We will show that there exist constants C 1 , C 2 , C 3 depending only on U, T and the coefficients in L, such that for each N, 1. a. ‖u N ‖ C0,T:H ≤ C 1 ‖u 0 ‖ H + ‖f‖ L 2 0,T:V ′ b. ‖u N ‖ L 2 0,T:V ≤ C 2 ‖u 0 ‖ H + ‖f‖ L 2 0,T:V ′ c. ‖u ′N ‖ L 2 0,T:V ′ ≤ C 3 ‖u 0 ‖ H + ‖f‖ L 2 0,T:V ′ We note first that N ‖u N 0‖ 2H = ∑u 0 , w j 2H ≤ ‖u 0 ‖ 2H j=1 and 2|f, u N H | = 2|⟨f, u N ⟩ V ′ ×V | ≤ 2‖ft‖ V ′ ‖u N t‖ V 1 ‖ft‖ 2 ′ + α‖u N t‖ 2 ≤ α V V Now it follows from (3) that ∂ t u N , u N H + Bu N , u N , t = f, u N H and we have that, ∂ t u N , u N H = 1 d u N , u N H = 1 d ||u N || 2H 2 dt 2 dt which leads to, i.e., d ||u || 2 + 2Bu N , u N , t ≤ 1 ‖ft‖ 2 ′ + α‖u N t‖ 2 V V α dt N H d ||u || 2 + 2α‖u N t‖ 2 ≤ 1 ‖ft‖ 2 ′ + α‖u N t‖ 2 V V V α dt N H Then, d ||u || 2 + α‖u N t‖ 2 ≤ 1 ‖ft‖ 2 ′ V V α dt N H and since ‖u N t‖ 2H ≤ ‖u N t‖ 2V , 3 d ||u || 2 + α‖u N t‖ 2 ≤ 1 ‖ft‖ 2 ′ H V α dt N H We rewrite this as, d ||u N || 2 e αt e −αt ≤ 1 ‖ft‖ 2 ′ H V α dt from which it follows by integrating from 0 to t, 1 ∫ T e αt ‖ft‖ 2 ′ dt ‖u N t‖ 2H ≤ e −αt ‖u N 0‖ 2H + α V 0 Since this holds for every t ∈ 0, T, it follows.that sup ‖u N t‖ 2H ≤ C 1 ‖u 0 ‖ 2H + ‖f‖ 2L 2 0≤t≤T 0,T:V ′ proving a. To prove b), we return to the estimate, d ||u || 2 + α‖u N t‖ 2 ≤ 1 ‖ft‖ 2 ′ V V α dt N H and we integrate with respect to t over 0, T to get T 1 ∫ T ‖ft‖ 2 ′ dt ‖u N T‖ 2H − ‖u N 0‖ 2H + α ∫ ‖u N t‖ 2V dt ≤ α V 0 0 Then T T ∫ 0 ‖u N t‖ 2V d ≤ α1 ‖u N 0‖ 2H + 12 ∫ 0 ‖ft‖ 2V ′ dt α which asserts that ‖u N ‖ L2 2 0,T:V ≤ C 2 ‖u 0 ‖ 2H + ‖f‖ 2L 2 0,T:V ′ Finally, to prove c), let V N = spanw 1 , … , w N . Then, for each v ∈ V with ‖v‖ V = 1, we can use the fact that w j is an ON basis in H to write v = v 1 + v 2 where v 1 ∈ V N and v 2 , w j H = 0 for j = 1, … , N. Since w j is also an orthogonal basis in V, we have N ‖v 1 ‖ 2V = v 1 , v 1 V = ∑v 1 , w j 2V ‖w j ‖ 2V ≤ ‖v‖ 2V ≤ 1 j=1 Now, it follows from (3) that ∂ t u N , v 1 H + Bu N , v 1 , t = f, v 1 H and ∂ t u N , v H = ∂ t u N , v 1 + v 2 H = ∂ t u N , v 1 H since N ∂ t u N t = ∑ C ′jN t w j and v 2 , w j H = 0 for j = 1, … , N. j=1 Then, 4 ∂ t u N , v H = ∂ t u N , v 1 H = f, v 1 H − Bu N , v 1 , t and |∂ t u N , v H | ≤ ‖f‖ V ′ ‖v 1 ‖ V + a 1 ‖u N ‖ V ‖v 1 ‖ V ≤ ‖f‖ V ′ + a 1 ‖u N ‖ V since ‖v 1 ‖ V ≤ 1 Then sup |⟨∂ t u N , v⟩ V ′ ×V | = ‖u ′N t‖ V ′ ≤ ‖f‖ V ′ + a 1 ‖u N ‖ V ‖v‖ V ≤1 Integrating this expression from 0 to T, and using b leads to c. 3) Existence of Weak Solutions u N t is bounded in L 2 0, T : V We see that b) implies c) implies ∂ t u N t is bounded in L 2 0, T : V ′ Then there exists a subsequence, u M t ⊂ u N t such that u M t ⇀ ut weakly in L 2 0, T : V weakly in L 2 0, T : V ′ ∂ t u M t ⇀ vt But then v must equal u ′ t since weak convergence in in L 2 0, T : V implies convergence in D ′ 0, T : V, which implies convergence in D ′ 0, T : V ′ and u M t ⇀ ut in D ′ 0, T : V ′ implies ∂ t u M t ⇀ u ′ t in D ′ 0, T : V ′ . Similarly, weak convergence in L 2 0, T : V ′ implies convergence in D ′ 0, T : V ′ so ∂ t u M t ⇀ vt in D ′ 0, T : V ′ and v = u ′ . Now we have, u M t ⇀ ut weakly in L 2 0, T : V ∂ t u M t ⇀ ∂ t ut weakly in L 2 0, T : V ′ which implies that u = w − lim u M belongs to W0, T. To see that this weak limit is a weak solution to the IVP, let p Vp = vt = ∑ d j t w j : d j ∈ C 1 0, T, 1 ≤ j ≤ p . j=1 Then it follows from (3) that for each M and all p ≤ M, T T T ∫ 0 ∂ t u M , vt H dt + ∫ 0 Bu M , vt, tdt = ∫ 0 f, vt H dt ∀v ∈ V p Now, the weak convergence results imply that we can let M tend to infinity to get, T T T ∫ 0 ∂ t u, vt H dt + ∫ 0 Bu, vt, tdt = ∫ 0 f, vt H dt and this holds for all v ∈ V p , for arbitrarily large p. But it is evident that ⋃ Vp is dense in L 2 0, T : V p>0 which means that the last result extends to all v ∈ L 2 0, T : V. Then u is a weak solution of the equation in (3). To see that u0 = u 0 , note that for arbitrary v ∈ C 1 0, T : V such that vT = 0, 5 T T T ∫ 0 −u, ∂ t vt H dt + ∫ 0 Bu, vt, tdt = ∫ 0 f, vt H dt + u0, v0 H Similarly, for each M, T T T ∫ 0 −u M , ∂ t vt H dt + ∫ 0 Bu M , vt, tdt = ∫ 0 f, vt H dt + u M 0, v0 H and if we subtract the latter equation from the former, we get T T ∫ 0 −u − u M , ∂ t vt H dt + ∫ 0 Bu − u M , vt, tdt = u0 − u M 0, v0 H . Now we let M tend to infinity, and obtain 0= u0 − lim u M 0, v0 M ∀v ∈ C 1 0, T : V. H But lim u M 0 = u 0 and C 1 0, T : V ⊂ C0, T : H and it follows that u0 = u 0 . Then u is a M weak solution of the IVP. Note that it now follows that every weakly convergent subsequence must converge to a weak solution of the IVP and since the weak solution has been shown to be unique, all subsequences must have the same weak limit. But in that case, the sequence u N itself must converge weakly to u ∈ W0, T. By using a v ∈ C 1 0, T : V such that v0 = 0, we can show in the same way that lim u M T = uT in H. Then we can show that u N converges strongly to u in L 2 0, T : V. M To see this, we write, T T T ∫ 0 ∂ t u, ut H dt + ∫ 0 Bu, ut, tdt = ∫ 0 f, ut H dt and T T ∫ 0 ∂ t u, ut H dt = 1 ∫ 0 d ‖ut‖ 2H dt = 1 ‖uT‖ 2H − 1 ‖u0‖ 2H 2 2 dt 2 Then T T ∫ 0 f, ut H dt − ∫ 0 But, ut, tdt = 1 ‖uT‖ 2H − 1 ‖u0‖ 2H 2 2 4 Next, we observe that T a 0 ‖u N − u‖ L2 2 0,T:V ≤ ∫ Bu N − u, u N − ut, tdt 0 T T T 0 0 0 ≤ ∫ Bu N , u N , tdt − ∫ Bu N , u, tdt − ∫ Bu, u N − u, tdt But, T T T ∫ 0 Bu N , u N , tdt = ∫ 0 f, u N t H dt − ∫ 0 ∂ t u N , u N H dt T = ∫ f, u N t H dt − 1 ‖u N T‖ 2H − ‖u N 0‖ 2H 0 2 and this leads to, T a 0 ‖u N − u‖ L2 2 0,T:V ≤ ∫ f, u N t H − Bu N , u, tdt 0 T − ∫ Bu, u N − u, tdt − 1 ‖u N T‖ 2H − ‖u N 0‖ 2H 0 2 Now, the weak convergence results imply we can allow N to tend to infinity on the right side 6 of this last expression and make use of (4) to obtain T T RHS ∫ f, ut H dt − ∫ But, ut, tdt − 1 ‖uT‖ 2H + 1 ‖u0‖ 2H = 0 0 0 2 2 from which it follows that ‖u N − u‖ L2 2 0,T:V 0 as N → ∞. Note it now follows that the energy inequalities b) and c) apply to the weak solution u and this implies that the solution depends continuously on the data. That is, u 0 , f ∈ H × L 2 0, T; V ′ u ∈ W0, T is continuous, where we recall that ‖u‖ 2W0,T = ∫ T 0 ‖ut‖ 2V + ‖∂ t ut‖ 2V ′ dt. 7