Numerical Analysis of a Second-Order Algorithm for the Time-Dependent Natural Convection Problem

A Proof of Theorem 3.5

At time t n + 1 , the true solution ( u , p , T ) satisfies (3.2). We decompose errors as follows:

Subtracting (3.2) from (3.1), and taking the difference of the obtained equation for two consecutive time-steps, we have

Setting

and adding them together, multiplying it by Δ ⁢ t , and using (2.5), (2.6), we can obtain

We bound the other terms on the right of (A.2) as follows. For all ε 2 , ε 3 > 0 ,

where θ n + 1 ∈ ( t n , t n + 1 ) . Similarly, we can obtain

(A.7) T 5 = − Δ ⁢ t ⁢ c ̄ ⁢ ( S u ⁢ 1 n + 1 , T n + 1 , ζ T ⁢ 1 n + 1 − ζ T ⁢ 1 n Δ ⁢ t ) ≤ C ⁢ Δ ⁢ t ⁢ ∥ T ∥ ∞ , 1 2 ⁢ ∥ ∇ η u ⁢ 1 n + 1 − η u ⁢ 1 n Δ ⁢ t ∥ 2 + C ⁢ Δ ⁢ t ⁢ ∥ T ∥ ∞ , 1 4 ⁢ ∥ ζ u ⁢ 1 n + 1 − ζ u ⁢ 1 n Δ ⁢ t ∥ 2 + ε 2 ⁢ Δ ⁢ t ⁢ ∥ ∇ ζ u ⁢ 1 n + 1 − ζ u ⁢ 1 n Δ ⁢ t ∥ 2 + ε 3 ⁢ Δ ⁢ t ⁢ ∥ ∇ ζ T ⁢ 1 n + 1 − ζ T ⁢ 1 n Δ ⁢ t ∥ 2 ,

(A.8) T 6 = − Δ ⁢ t ⁢ c ̄ ⁢ ( u 1 , h n + 1 − u 1 , h n Δ ⁢ t , e T ⁢ 1 n + 1 , ζ T ⁢ 1 n + 1 − ζ T ⁢ 1 n Δ ⁢ t ) = − Δ ⁢ t ⁢ c ̄ ⁢ ( u n + 1 − u n Δ ⁢ t , e T ⁢ 1 n + 1 , ζ T ⁢ 1 n + 1 − ζ T ⁢ 1 n Δ ⁢ t ) + Δ ⁢ t ⁢ Δ ⁢ t ⁢ c ̄ ⁢ ( S u ⁢ 1 n + 1 , e T ⁢ 1 n + 1 , ζ T ⁢ 1 n + 1 − ζ T ⁢ 1 n Δ ⁢ t ) ≤ C ⁢ Δ ⁢ t ⁢ ∥ ∇ e T ⁢ 1 n + 1 ∥ 2 ⁢ ∥ ∇ η u ⁢ 1 n + 1 − η u ⁢ 1 n Δ ⁢ t ∥ 2 + C ⁢ Δ ⁢ t ⁢ ∥ ∇ e T ⁢ 1 n + 1 ∥ 4 ⁢ ∥ ζ u ⁢ 1 n + 1 − ζ u ⁢ 1 n Δ ⁢ t ∥ 2 + C ⁢ Δ ⁢ t ⁢ ∥ ∇ e T ⁢ 1 n + 1 ∥ 2 ⁢ ∥ ∇ u t ⁢ ( θ n + 1 ) ∥ 2 + ε 2 ⁢ Δ ⁢ t ⁢ ∥ ∇ ζ u ⁢ 1 n + 1 − ζ u ⁢ 1 n Δ ⁢ t ∥ 2 + ε 3 ⁢ Δ ⁢ t ⁢ ∥ ∇ ζ T ⁢ 1 n + 1 − ζ T ⁢ 1 n Δ ⁢ t ∥ 2 ,

(A.9) T 7 = − Δ ⁢ t ⁢ c ̄ ⁢ ( e u ⁢ 1 n , T n + 1 − T n Δ ⁢ t , ζ T ⁢ 1 n + 1 − ζ T ⁢ 1 n Δ ⁢ t ) ≤ C ⁢ Δ ⁢ t ⁢ ∥ ∇ e u ⁢ 1 n ∥ 2 ⁢ ∥ ∇ T t ⁢ ( θ n + 1 ) ∥ 2 + ε 3 ⁢ Δ ⁢ t ⁢ ∥ ∇ ζ T ⁢ 1 n + 1 − ζ T ⁢ 1 n Δ ⁢ t ∥ 2 ,

(A.10) T 8 = − Δ ⁢ t ⁢ c ̄ ⁢ ( u 1 , h n , η T ⁢ 1 n + 1 − η T ⁢ 1 n Δ ⁢ t , ζ T ⁢ 1 n + 1 − ζ T ⁢ 1 n Δ ⁢ t ) ≤ C ⁢ Δ ⁢ t ⁢ ∥ ∇ u 1 , h n ∥ 2 ⁢ ∥ ∇ η T ⁢ 1 n + 1 − η T ⁢ 1 n Δ ⁢ t ∥ 2 + ε 3 ⁢ Δ ⁢ t ⁢ ∥ ∇ ζ T ⁢ 1 n + 1 − ζ T ⁢ 1 n Δ ⁢ t ∥ 2 ,

(A.11) T 9 = ( η u ⁢ 1 n + 1 − 2 ⁢ η u ⁢ 1 n + η u ⁢ 1 n − 1 Δ ⁢ t , ζ u ⁢ 1 n + 1 − ζ u ⁢ 1 n Δ ⁢ t ) ≤ C ⁢ Δ ⁢ t ⁢ ∥ η u ⁢ 1 , t ⁢ t ⁢ ( θ n + 1 ) ∥ − 1 2 + ε 2 ⁢ Δ ⁢ t ⁢ ∥ ∇ ζ u ⁢ 1 n + 1 − ζ u ⁢ 1 n Δ ⁢ t ∥ 2 ,

(A.12) T 10 = ( η T ⁢ 1 n + 1 − 2 ⁢ η T ⁢ 1 n + η T ⁢ 1 n − 1 Δ ⁢ t , ζ T ⁢ 1 n + 1 − ζ T ⁢ 1 n Δ ⁢ t ) ≤ C ⁢ Δ ⁢ t ⁢ ∥ η T ⁢ 1 , t ⁢ t ⁢ ( θ 1 n + 1 ) ∥ − 1 2 + ε 3 ⁢ Δ ⁢ t ⁢ ∥ ∇ ζ T ⁢ 1 n + 1 − ζ T ⁢ 1 n Δ ⁢ t ∥ 2 ,

(A.13) T 11 = ( ( u n + 1 − u n Δ ⁢ t − u t ⁢ ( t n + 1 ) ) − ( u n − u n − 1 Δ ⁢ t − u t ⁢ ( t n ) ) , ζ u ⁢ 1 n + 1 − ζ u ⁢ 1 n Δ ⁢ t ) = Δ ⁢ t 2 ⁢ ( u t ⁢ t ⁢ t ⁢ ( θ n + 1 ) , ζ u ⁢ 1 n + 1 − ζ u ⁢ 1 n Δ ⁢ t ) ≤ C ⁢ Δ ⁢ t 3 ⁢ ∥ u t ⁢ t ⁢ t ⁢ ( θ 1 n + 1 ) ∥ 2 + ε 2 ⁢ Δ ⁢ t ⁢ ∥ ∇ ζ u ⁢ 1 n + 1 − ζ u ⁢ 1 n Δ ⁢ t ∥ 2 ,

where θ 1 n + 1 ∈ ( t n − 1 , t n + 1 ) , and

(A.14) T 12 = ( ( T n + 1 − T n Δ ⁢ t − T t ⁢ ( t n + 1 ) ) − ( T n − T n − 1 Δ ⁢ t − T t ⁢ ( t n ) ) , ζ T ⁢ 1 n + 1 − ζ T ⁢ 1 n Δ ⁢ t ) ≤ C ⁢ Δ ⁢ t 3 ⁢ ∥ T t ⁢ t ⁢ t ⁢ ( θ n + 1 ) ∥ 2 + ε 3 ⁢ Δ ⁢ t ⁢ ∥ ∇ ζ T ⁢ 1 n + 1 − ζ T ⁢ 1 n Δ ⁢ t ∥ 2 ,

(A.15) T 13 = Pr ⁡ Pa ⁡ Δ ⁢ t ⁢ ( ξ ⁢ S T ⁢ 1 n + 1 , ζ u ⁢ 1 n + 1 − ζ u ⁢ 1 n Δ ⁢ t ) ≤ C ⁢ Δ ⁢ t ⁢ ∥ ζ T ⁢ 1 n + 1 − ζ T ⁢ 1 n Δ ⁢ t ∥ 2 + C ⁢ Δ ⁢ t ⁢ ∥ ∇ η T ⁢ 1 n + 1 − η T ⁢ 1 n Δ ⁢ t ∥ 2 + ε 2 ⁢ Δ ⁢ t ⁢ ∥ ∇ ζ u ⁢ 1 n + 1 − ζ u ⁢ 1 n Δ ⁢ t ∥ 2 .

Setting ε 2 = 1 18 , ε 3 = 1 12 and applying (A.3)–(A.15) to (A.2), and multiplying it by 2 ⁢ Δ ⁢ t , we can obtain

In order to use Lemma 2.2, we need Δ ⁢ t ⁢ ( ∥ ∇ e u ⁢ 1 n + 1 ∥ 4 + ∥ ∇ e T ⁢ 1 n + 1 ∥ 4 ) ≤ C uniformly for all 𝑛. It can be deduced from Theorem 3.4 that

On the other hand, using the inverse inequality and Theorem 3.4, we have

Thus, if h 4 ⁢ k ≤ Δ ⁢ t , we use (A.17) to bound Δ ⁢ t ⁢ ( ∥ ∇ e u ⁢ 1 n + 1 ∥ 4 + ∥ ∇ e T ⁢ 1 n + 1 ∥ 4 ) ; otherwise, we use (A.18). In any case, we have Δ ⁢ t ⁢ ( ∥ ∇ e u ⁢ 1 n + 1 ∥ 4 + ∥ ∇ e T ⁢ 1 n + 1 ∥ 4 ) ≤ C . Furthermore, since

are not defined for n = 0 , we can only sum (A.16) from n = 1 to n = l , which impels us to bound the terms ∥ S u ⁢ 1 1 ∥ 2 , ∥ S T ⁢ 1 1 ∥ 2 , ∥ ∇ S u ⁢ 1 1 ∥ 2 and ∥ ∇ S T ⁢ 1 1 ∥ 2 . We bound these terms as follows. At time t 1 , subtracting (3.2) from (3.1), we have

Setting u 1 , h 0 = R h ⁢ u 0 = u 0 , T 1 , h 0 = Q h ⁢ T 0 = T 0 and p 1 , h 0 = π h ⁢ p 0 = p 0 , we have

Setting v h = 2 ⁢ Δ ⁢ t ⁢ ζ u ⁢ 1 1 , q h = 2 ⁢ Δ ⁢ t ⁢ ζ p ⁢ 1 1 , ψ h = 2 ⁢ Δ ⁢ t ⁢ ζ T ⁢ 1 1 , adding the above three equations together, and using (2.5), (2.6), we can obtain

We bound the terms of the right side in (A.19) as follows. For all ε 4 , ε 5 > 0 ,

Similarly, we bound the remaining terms by using the Cauchy–Schwarz inequality as follows:

Setting ε 4 = Pr 4 , ε 5 = k 2 , and combining the above estimates with (A.19), we have

Since Δ ⁢ t is sufficiently small, we can easily obtain

We can draw the following conclusion from the trigonometric inequality:

Hence

We sum (A.16) from n = 1 to n = l and use Lemma 2.2 and Theorem 3.4 to obtain

Finally, using the triangle inequality and combining (A.20), we can complete the proof.