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A Appendix
To prove the statement of Theorem 3.1, we first note a simplification of [33, Th. 4.1] which can be easily verified by considering block Toeplitz matrices for the special case of scalar block entries.
Theorem A.1
Let A ∈ ℝM×M, M ∈ ℕ, be a Toeplitz matrix generated by some function f ∈ L∞((−π, π), ℂ), that is,
(A.1)
akl=12π∫−ππf(x)e−i(k+l)xdx, k,l=1,…,M
Then the spectral norm of A is bounded from above by
(A.2)
∥ A ∥ 2 ⩽ sup x ∈ Q | f ( x ) | .
Based on this result, Theorem 3.1 can be shown by some algebraic manipulations.
Proof of Theorem 3.1. First of all, let us consider the special case f2 = 0. Then the matrix F−1E coincides with f −1 1 E, which is well defined because f1 does not vanish by assumption of the theorem. Therefore, we have
(F−1E)⊤F−1E=f1−2(e12+e22e1e2e1e2⋱⋱⋱e12+e22e1e2e1e2e12).
Obviously, this matrix is positive semidefinite while the 12maximal eigenvalue is bounded from above by e+ e+2|e1e2| according to the Gershgorin circle theorem 22[14]. Then the statement of the theorem directly follows by distinguishing between e1e2 ⩾ 0 and e1e2 < 0.
If f2 ≠ 0, we first note that the inverse of F reads
F−1=(f1−1(−f1−1f2)f1−1f1−1⋮⋱⋱(−f1−1f2)M−1f1−1⋯−(f1−1f2)f1−1f1−1)
and, hence,
F − 1 E = f 1 − 1 − f 1 − 1 f 2 f 1 − 1 f 1 − 1 ⋮ ⋱ ⋱ − f 1 − 1 f 2 M − 1 f 1 − 1 ⋯ − f 1 − 1 f 2 f 1 − 1 f 1 − 1 e 1 e 2 e 1 ⋱ ⋱ e 2 e 1 = f 1 − 1 e 1 f 1 − 1 ( e 2 − f 2 f 1 − 1 e 1 ) f 1 − 1 e 1 ⋮ ⋱ ⋱ − f 1 − 1 f 2 M − 2 ( e 2 − f 2 f 1 − 1 e 1 ) ⋯ − f 1 − 1 ( e 2 − f 2 f 1 − 1 e 1 f 1 − 1 e 1 = s 1 s 2 s 1 ⋮ ⋱ ⋱ s 2 M − 2 s 3 ⋯ s 3 s 1
where
s1=f1−1e1,s2=−f1−1f2∈(−1,1),s3=f1−1(e2−f2f1−1e1).
Therefore, the matrix F−1E is a Toeplitz matrix and generated by
f ( x ) := s 1 + s 3 ∑ l = 1 ∞ s 2 l − 1 e i l x = s 1 + s 3 e i x ∑ l = 0 ∞ s 2 e i x l = s 1 + s 3 e i x 1 − s 2 e i x − 1 = s 1 + s 3 e − i x − s 2 − 1 .
Function f reaches its absolute extremum at x = 0 or x = π due to the fact that s1, s2, and s3 are real values.
Thus, the spectral norm of F−1E is bounded from above by
‖ F−1E ‖2⩽max(|f(0)|,|f(π)|)=max(| e1−e2 || f1−f2 |,| e1+e2 || f1+f2 |)
which can be easily verified by
|f(0)|=| s1+s3(1−s2)−1 |=| e1f1+e2−f2f1−1e1f1(1+f1−1f2) |=| e1f1+e2f1−f2e1f1(f1+f2) |=| e1+e2f1+f2 ||f(π)|=| s1−s3(1+s2)−1 |=| e1f1−e2−f2f1−1e1f1(1−f1−1f2) |=| e1f1−e2f1−f2e1f1(f1−f2) |=| e1−e2f1−f2 |.
This completes the proof of Theorem 3.1.
Proof of Theorem 5.3. First of all, we note that
2mii−(1−2ϑ)δtdii(2+ζ)mii−2(1−2ϑ)δtdii−23=6mii−3(1−2ϑ)δtdii−(4+2ζ)mii+4(1−2ϑ)δtdii3((2+ζ)mii−2(1−2ϑ)δtdii)=2(1−ζ)mii+(1−2ϑ)δtdii3((2+ζ)mii−2(1−2ϑ)δtdii).
Therefore, the value of ω0 is equal to 2/3 if and only if
(A.3)
2(1−ζ)mii+(1−2ϑ)δtdii⩽0
and, particularly, ϑ ⩾ 1/2 is mandatory. If condition (A.3) is satisfied and d(ℓ) > dii, the first argument of the maximum in (5.5b) is bounded from above by 1/3 because
ω 0 2 m ( ℓ ) − ( 1 − 2 ϑ ) δ t d ( ℓ ) 2 m i i − ( 1 − 2 ϑ ) δ t d i i − 1 − 1 3 = 2 3 2 m ( ℓ ) − ( 1 − 2 ϑ ) δ t d ( ℓ ) 2 m i i − ( 1 − 2 ϑ ) δ t d i i − 1 − 1 3 = 2 2 m ( ℓ ) − ( 1 − 2 ϑ ) δ t d ( ℓ ) − 4 2 m i i − ( 1 − 2 ϑ ) δ t d i i 3 2 m i i − ( 1 − 2 ϑ ) δ t d i i = 4 m ( ℓ ) − 2 m i i ⏞ ⩽ 0 by ( 2.8 ) + 2 ( 1 − 2 ϑ ) ⏞ ⩽ 0 δ t 2 d i i − d ( ℓ ) ⏞ ⩾ 0 by ( 2.7 ) 3 2 m i i − ( 1 − 2 ϑ ) δ t d i i ⩽ 0 1 − ω 0 2 m ( ℓ ) − ( 1 − 2 ϑ ) δ t d ( ℓ ) 2 m i i − ( 1 − 2 ϑ ) δ t d i i − 1 3 = 1 − 2 3 2 m ( ℓ ) − ( 1 − 2 ϑ ) δ t d ( ℓ ) 2 m i i − ( 1 − 2 ϑ ) δ t d i i − 1 3 = 2 2 m i i − ( 1 − 2 ϑ ) δ t d i i − 2 2 m ( ℓ ) − ( 1 − 2 ϑ ) δ t d ( ℓ ) 3 2 m i i − ( 1 − 2 ϑ ) δ t d i i = 4 m i i − m ( ℓ ) − 2 ( 1 − 2 ϑ ) δ t d i i − d ( ℓ ) 3 2 m i i − ( 1 − 2 ϑ ) δ t d i i
which is nonpositive for finite differences due to the fact that mii = m(ℓ) = 1, ϑ ⩾ 1/2, and d(ℓ) > dii while we have
4(mii−m(l))−2(1−2ϑ)δt(dii−d(l))=2(2sl2−1)mii−2(1−2ϑ)δt(dii−d(l)) =2dii−1(d(l)−dii︸⩾0(mii+(1−2ϑ)δtdii)︸⩽0 by (A.3) ⩽0
in the context of linear finite elements. On the other hand, the second argument is not greater than 1/3 due to the fact that
| 1−23dii−1d(l) |=1−23dii−1d(l)⩽1−23=13 if d(l)⩽32dii| 1−23dii−1d(l) |=23dii−1d(l)−1⩽43−1=13 if d(l)⩾32dii
by virtue of (2.7). Let us now assume that
(A.4)
2(1−ζ)mii+(1−2ϑ)δtdii>0.
Then the relaxation parameter ω0 attains the second argument of the maximum in (5.9) and, hence, the first expression in the definition of B(ℓ) is bounded from above by
(2−ζ)mii(2+ζ)mii−2(1−2ϑ)δtdii.
Indeed, we have
1−ω02m(l)−(1−2ϑ)δtd(l)2mii−(1−2ϑ)δtdii=(2+ζ)mii−2(1−2ϑ)δtdii−(2m(l)−(1−2ϑ)δtd(l))(2+ζ)mii−2(1−2ϑ)δtdii =(2−ζ)mii+2ζmii−2m(l)+4(1−sl2)(1−ζ)mii︷=0!(2+ζ)mii−2(1−2ϑ)δtdii−2(1−sl2)︷⩾0>0 by (A.4) (2(1−ζ)mii+(1−2ϑ)δtdii)(2+ζ)mii−2(1−2ϑ)δtdii ⩽(2−ζ)mii(2+ζ)mii−2(1−2ϑ)δtdii
because
2ζmii−2m(l)+4(1−sl2)(1−ζ)mii={ 2−2=0,ζ=1mii(1−3+2sl2+2(1−sl2))=0,ζ=12
and, on the other hand,
ω02m(l)−(1−2ϑ)δtd(l)2mii−(1−2ϑ)δtdii−1=−(2+ζ)mii+2(1−2ϑ)δtdii+(2m(l)−(1−2ϑ)δtd(l))(2+ζ)mii−2(1−2ϑ)δtdii =(2−ζ)mii−4mii+2(1−2ϑ)δtdii+(2m(l)−(1−2ϑ)δtd(l))(2+ζ)mii−2(1−2ϑ)δtdii ⩽(2−ζ)mii(2+ζ)mii−2(1−2ϑ)δtdii
due to the fact that
4mii−2(1−2ϑ)δtdii−(2m(l)−(1−2ϑ)δtd(l)) ={ 2(ζmii−(1−2ϑ)δtdii)︸⩾0 by (2.10)+2((2−ζ)mii−m(l))︸⩾0 by (2.8)+(1−2ϑ)δtd(l)︸⩾0⩾0, ϑ⩽122(2mii−m(l))︸⩾0 by (2.8)−(1−2ϑ)︸⩽0δt(2dii−d(l))︸⩾0 by (2.7) ⩾0,ϑ>12.
Finally, the second argument of the maximum in (5.5b) satisfies
1 − ω 0 d ( ℓ ) d i i = 1 − 2 m i i d ( ℓ ) − ( 1 − 2 ϑ ) δ t d i i d ( ℓ ) d i i ( 2 + ζ ) m i i − 2 ( 1 − 2 ϑ ) δ t d i i = ( 2 + ζ ) m i i d i i − 2 ( 1 − 2 ϑ ) δ t d i i 2 − 2 m i i d ( ℓ ) + ( 1 − 2 ϑ ) δ t d i i d ( ℓ ) d i i ( 2 + ζ ) m i i − 2 ( 1 − 2 ϑ ) δ t d i i = ( 2 − ζ ) m i i d i i + 2 ( 1 − ζ ) m i i + ( 1 − 2 ϑ ) δ t d i i ⏞ ⩾ 0 by (A.4) d ( ℓ ) − 2 d i i ⏞ ⩽ 0 + ( 4 − 2 ζ ) ⏞ ⩾ 0 m i i d i i − d ( ℓ ) ⏞ ⩽ 0 d i i ( 2 + ζ ) m i i − 2 ( 1 − 2 ϑ ) δ t d i i ⩽ ( 2 − ζ ) m i i ( 2 + ζ ) m i i − 2 ( 1 − 2 ϑ ) δ t d i i
ω0d(l)dii−1=2miid(l)−(1−2ϑ)δtdiid(l)−(2+ζ)miidii+2(1−2ϑ)δtdii2dii((2+ζ)mii−2(1−2ϑ)δtdii) =(2−ζ)miidii+(d(l)−2dii)︷⩽0 by (2.7)⩾0 by (2.10) (2mii−(1−2ϑ)δtdii)dii((2+ζ)mii−2(1−2ϑ)δtdii) ⩽(2−ζ)mii(2+ζ)mii−2(1−2ϑ)δtdii
which proves the validity of inequality (5.10).
To prove identity (5.11), we have to show the validity of
| 1−ω2m(l)−(1−2ϑ)δtd(l)2mii−(1−2ϑ)δtdii |⩽1−ωdii−1d(l)
whenever d(ℓ) ⩽ dii and ω ∈ (0, 1]. For this purpose, we first note that
(1−ωdii−1d(l))−(1−ω2m(l)−(1−2ϑ)δtd(l)2mii−(1−2ϑ)δtdii)=ω(2m(l)−(1−2ϑ)δtd(l)2mii−(1−2ϑ)δtdii−dii−1d(l)) =ω2diim(l)−(1−2ϑ)δtdiid(l)−2miid(l)+(1−2ϑ)δtdiid(l)dii(2mii−(1−2ϑ)δtdii) =ω2(diim(l)−miid(l))dii(2mii−(1−2ϑ)δtdii)⩾0
due to the fact that
m(l)dii−miid(l)=dii−d(l)⩾0 if ζ=1m(l)dii−miid(l)=mii(32−sl2)dii−mii2diisl2=32miidii(1−2sl2)=32mii(dii−d(l))⩾0 if ζ=12 ,
On the other hand, we have
1 − ω d i i − 1 d ( ℓ ) − ω 2 m ( ℓ ) − ( 1 − 2 ϑ ) δ t d ( ℓ ) 2 m i i − ( 1 − 2 ϑ ) δ t d i i − 1 = 2 − ω 2 m i i d ( ℓ ) − ( 1 − 2 ϑ ) δ t d i i d ( ℓ ) + 2 d i i m ( ℓ ) − ( 1 − 2 ϑ ) δ t d i i d ( ℓ ) d i i 2 m i i − ( 1 − 2 ϑ ) δ t d i i = 2 2 m i i d i i − ( 1 − 2 ϑ ) δ t d i i 2 − ω m i i d ( ℓ ) + d i i m ( ℓ ) + ω ( 1 − 2 ϑ ) δ t d i i d ( ℓ ) d i i 2 m i i − ( 1 − 2 ϑ ) δ t d i i = 2 ( 2 − ζ ) m i i d i i − ω ( 1 − ζ ) m i i d ( ℓ ) + d i i m ( ℓ ) d i i 2 m i i − ( 1 − 2 ϑ ) δ t d i i + ζ m i i − ( 1 − 2 ϑ ) δ t d i i ⏞ ⩾ 0 by ( 2.10 ) d i i − ω d ( ℓ ) ⏞ ⩾ d i i − d ( ℓ ) ⩾ 0 d i i 2 m i i − ( 1 − 2 ϑ ) δ t d i i
which is nonnegative either for ζ = mii = m(ℓ) = 1 in case of finite differences or according to
(2−ζ)miidii−ω((1−ζ)miid(l)+diim(l))=miidii(32−ω(sl2+32−sl2))=32miidii(1−ω)⩾0
for linear finite elements and ζ = 1/2.
Although the statement of Theorem 6.1 is true for finite element and finite difference discretizations, we prove the result by considering both spatial discretization techniques individually.
Proof of Theorem 6.1 for finite differences. To prove that spr(J(Cor,ℓ)(J(Jac,ℓ))ν) is smaller than 1 for all ℓ = 1, . . . , ̄N, we directly estimate the eigenvalues λ± ∈ ℂ of J(Cor,ℓ)(J(Jac,ℓ))ν which are the roots of the characteristic polynomial p : ℂ → ℂ
p ( λ ) = det J ( Cor , ℓ ) J ( Jac , ℓ ) v − λ I = a ¯ ( ℓ ) − 2 det a ¯ ( ℓ ) − c ℓ 4 a ( ℓ ) j ( Jac , ℓ ) v − a ¯ ( ℓ ) λ s ℓ 2 c ℓ 2 a ( N + 1 − ℓ ) j ( Jac , N + 1 − ℓ ) v s ℓ 2 c ℓ 2 a ( ℓ ) j ( Jac , ℓ ) v a ¯ ( ℓ ) − s ℓ 4 a ( N + 1 − ℓ ) j ( Jac , N + 1 − ℓ ) v − a ¯ ( ℓ ) λ = a ¯ ( ℓ ) − 2 a ¯ ( ℓ ) − c ℓ 4 a ( ℓ ) j ( Jac , ℓ ) v − a ¯ ( ℓ ) λ a ¯ ( ℓ ) − s ℓ 4 a ( N + 1 − ℓ ) j ( Jac , N + 1 − ℓ ) v − a ¯ ( ℓ ) λ − s ℓ 4 c ℓ 4 a ( ℓ ) a ( N + 1 − ℓ ) j ( Jac , ℓ ) v j ( Jac , N + 1 − ℓ ) v = a ¯ ( ℓ ) − 2 a ¯ ( ℓ ) λ 2 − a ¯ ( ℓ ) λ a ¯ ( ℓ ) − c ℓ 4 a ( ℓ ) j ( Jac , ℓ ) v + a ¯ ( ℓ ) − s ℓ 4 a ( N + 1 − ℓ ) j ( Jac , N + 1 − ℓ ) v + a ¯ ( ℓ ) 2 − a ¯ ( ℓ ) c ℓ 4 a ( ℓ ) + s ℓ 4 a ( N + 1 − ℓ ) j ( Jac , ℓ ) v j ( Jac , N + 1 − ℓ ) v = a ¯ ( ℓ ) − 2 a ¯ ( ℓ ) λ 2 − a ¯ ( ℓ ) λ s ℓ 2 a ¯ ( ℓ ) + c ℓ 2 j ( Jac , ℓ ) v + c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 j ( Jac , N + 1 − ℓ ) v + 2 s ℓ 2 c ℓ 2 a ¯ ( ℓ ) j ( Jac , ℓ ) v j ( Jac , N + 1 − ℓ ) v .
Here, the last identity is valid because
(A.5a)
a¯(l)−cl4a(l)=1+2ϑδtdiisl2cl2−cl4−2ϑδtdiisl2cl4=sl2+cl2−cl4+2ϑδtsl4cl2=sl2(a¯(l)+cl2)
(A.5b)
a¯(l)−sl4a(N+1−l)=1+2ϑδtdiisl2cl2−sl4−2ϑδtdiisl4cl2=sl2+cl2−sl4+2ϑδtsl2cl4=cl2(a¯(l)+sl2).
Therefore, the eigenvalues λ± satisfy
(A.6)
(a¯(l)λ±−12(sl2(a¯(l)+cl2)(j(Jac,l))v+cl2(a¯(l)+sl2)(j(Jac,N+1−l))v))2 =14(sl2(a¯(l)+cl2)(j(Jac,l))v+cl2(a¯(l)+sl2)(j(Jac,N+1−l))v)2−2sl2cl2a¯(l)(j(Jac,l))v(j(Jac,N+1−l))v.
Let us now consider the special case (j(Jac,ℓ))ν(j(Jac,N+1−ℓ))ν ⩽ 0. Then the right hand side of (A.6) is obviously nonnegative and can be estimated by
(A.7)
0 ⩽ 1 4 s ℓ 2 a ¯ ( ℓ ) + c ℓ 2 j ( J a c , ℓ ) v + c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 j ( J a c , N + 1 − ℓ ) v 2 − 2 s ℓ 2 c ℓ 2 a ¯ ( ℓ ) j ( J a c , ℓ ) v j ( J a c , N + 1 − ℓ ) v = 1 4 s ℓ 2 a ¯ ( ℓ ) + c ℓ 2 j ( J a c , ℓ ) v 2 + 1 4 c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 j ( J a c , N + 1 − ℓ ) v 2 + s ℓ 2 c ℓ 2 j ( J a c , ℓ ) v j ( J a c , N + 1 − ℓ ) v ⏟ ⩽ 0 1 2 a ¯ ( ℓ ) + c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 − 2 a ¯ ( ℓ ) ⏟ > − 1 2 a ¯ ( ℓ ) + c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 ⩽ 1 4 s ℓ 2 a ¯ ( ℓ ) + c ℓ 2 j ( J a c , ℓ ) v − c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 j ( J a c , N + 1 − ℓ ) v 2
because
(A.8)
(a¯(l)+cl2)(a¯(l)+sl2)−2a¯(l)=(a¯(l))2−a¯(l)+sl2cl2=a¯(l) (¯(l)−1 )︸=ϑδta¯(l)+sl2cl2>0.
Thus, both eigenvalues are real and satisfy
a ¯ ( ℓ ) λ ± ⩽ 1 2 s ℓ 2 a ¯ ( ℓ ) + c ℓ 2 j ( J a c , ℓ ) v + c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 j ( J a c , N + 1 − ℓ ) v + 1 2 s ℓ 2 a ¯ ( ℓ ) + c ℓ 2 j ( J a c , ℓ ) v − c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 j ( J a c , N + 1 − ℓ ) v = 1 2 s ℓ 2 a ¯ ( ℓ ) + c ℓ 2 j ( J a c , ℓ ) v − c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 j ( J a c , N + 1 − ℓ ) v ∣ + 1 2 s ℓ 2 a ¯ ( ℓ ) + c ℓ 2 j ( J a c , ℓ ) v + c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 j ( J a c , N + 1 − ℓ ) v = max s ℓ 2 a ¯ ( ℓ ) + c ℓ 2 j ( J a c , ℓ ) v , c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 j ( J a c , N + 1 − ℓ ) v ⩽ a ¯ ( ℓ ) max j ( J a c , ℓ ) v , j ( J a c , N + 1 − ℓ ) v < a ¯ ( ℓ )
due to the reverse triangle inequality, Theorem 5.1, (A.5) and the fact that (j(Jac,ℓ))ν(j(Jac,N+1−ℓ))ν ⩽ 0.
On the other hand, for (j(Jac,ℓ))ν(j(Jac,N+1−ℓ))ν > 0, we can assume that
jmax:=max((j(Jac,l))v,(j(Jac,N+1−l))v)>0.
Otherwise, consider −(J(Jac,ℓ))ν instead of (J(Jac,ℓ))ν. Then estimate (A.8) can be exploited as in (A.7) to prove
(A.9)
14(sl2(a¯(l)+cl2)(j(Jac,l))v+cl2(a¯(l)+sl2)(j(Jac,N+1−l))v)2−2sl2cl2a¯(l)(j(Jac,l))v(j(Jac,N+1−l))v >14(sl2(a¯(l)+cl2)(j(Jac,l))v−cl2(a¯(l)+sl2)(j(Jac,N+1−l))v)2⩾0
and, hence, both eigenvalues are real and positive because
a¯(l)λ±>12(sl2(a¯(l)+cl2)(j(Jac,l))v+cl2(a¯(l)+sl2)(j(Jac,N+1−l))v) −12| sl2(a¯(l)+cl2)(j(Jac,l))v−cl2(a¯(l)+sl2)(j(Jac,N+1−l))v | =min(sl2(a¯(l)+cl2)(j(Jac,l))v,cl2(a¯(l)+sl2)(j(Jac,N+1−l))v)⩾0
by (A.6) and (A.9). Furthermore, the maximal eigenvalue λ+ satisfies
a ¯ ( ℓ ) λ + = 1 2 s ℓ 2 a ¯ ( ℓ ) + c ℓ 2 j ( J a c , ℓ ) v + c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 j ( J a c , N + 1 − ℓ ) v + 1 4 s ℓ 2 a ¯ ( ℓ ) + c ℓ 2 j ( J a c , ℓ ) v + c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 j ( J a c , N + 1 − ℓ ) v 2 − 2 s ℓ 2 c ℓ 2 a ¯ ( ℓ ) j ( J a c , ℓ ) v j ( J a c , N + 1 − ℓ ) ) v ⩽ 1 2 s ℓ 2 a ¯ ( ℓ ) + c ℓ 2 j max + c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 j max + 1 4 s ℓ 2 a ¯ ( ℓ ) + c ℓ 2 j max + c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 j max 2 − 2 s ℓ 2 c ℓ 2 a ¯ ( ℓ ) j max 2 = 1 2 a ¯ ( ℓ ) + 2 s ℓ 2 c ℓ 2 j max + 1 4 a ¯ ( ℓ ) + 2 s ℓ 2 c ℓ 2 2 − 2 s ℓ 2 c ℓ 2 a ¯ ( ℓ ) j max = 1 2 a ¯ ( ℓ ) + 2 s ℓ 2 c ℓ 2 j max + 1 2 a ¯ ( ℓ ) − 2 s ℓ 2 c ℓ 2 j max = max a ¯ ( ℓ ) , 2 s ℓ 2 c ℓ 2 j max < a ¯ ( ℓ )
by Theorem 5.1 and due to the fact that
2sl2cl2⩽1/2⩽1⩽a¯(l)
because λ+ grows monotonically with 22respect to
(j(Jac ,l))vand(j(Jac,N+1−l))v.
Indeed, for instance, we have
a ¯ ( ℓ ) ∂ λ + ∂ j ( Jac , N + 1 − ℓ ) v = 1 2 c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 + 1 2 1 2 s ℓ 2 a ¯ ( ℓ ) + c ℓ 2 j ( J a c , ℓ ) v + c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 j ( J a c , N + 1 − ℓ ) v c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 − 2 s ℓ 2 c ℓ 2 a ¯ ( ℓ ) j ( J a c , ℓ ) v 1 4 s ℓ 2 a ¯ ( ℓ ) + c ℓ 2 j ( J a c , ℓ ) v + c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 j ( J a c , N + 1 − ℓ ) v 2 − 2 s ℓ 2 c ℓ 2 a ¯ ( ℓ ) j ( J a c , ℓ ) v j ( J a c , N + 1 − ℓ ) v ⩾ 1 4 c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 s ℓ 2 a ¯ ( ℓ ) + c ℓ 2 j ( J a c , ℓ ) v − c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 j ( J a c , N + 1 − ℓ ) v 1 4 s ℓ 2 a ¯ ( ℓ ) + c ℓ 2 j ( J a c , ℓ ) v + c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 j ( J a c , N + 1 − ℓ ) v 2 − 2 s ℓ 2 c ℓ 2 a ¯ ( ℓ ) j ( J a c , ℓ ) v j ( J a c , N + 1 − ℓ ) v + 1 2 1 2 s ℓ 2 a ¯ ( ℓ ) + c ℓ 2 j ( J a c , ℓ ) v + c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 j ( J a c , N + 1 − ℓ ) v c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 − 2 s ℓ 2 c ℓ 2 a ¯ ( ℓ ) j ( J a c , ℓ ) v 1 4 s ℓ 2 a ¯ ( ℓ ) + c ℓ 2 j ( J a c , ℓ ) v + c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 j ( J a c , N + 1 − ℓ ) v 2 − 2 s ℓ 2 c ℓ 2 a ¯ ( ℓ ) j ( J a c , ℓ ) v j ( J a c , N + 1 − ℓ ) v
by (A.9), which is nonnegative because
1 2 c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 s ℓ 2 a ¯ ( ℓ ) + c ℓ 2 j ( J a c , ℓ ) v − c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 j ( J a c , N + 1 − ℓ ) v + 1 2 s ℓ 2 a ¯ ( ℓ ) + c ℓ 2 j ( J a c , ℓ ) v + c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 j ( J a c , N + 1 − ℓ ) v c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 − 2 s ℓ 2 c ℓ 2 a ¯ ( ℓ ) j ( J a c , ℓ ) v = c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 max s ℓ 2 a ¯ ( ℓ ) + c ℓ 2 j ( J a c , ℓ ) v , c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 j ( J a c , N + 1 − ℓ ) v − 2 s ℓ 2 c ℓ 2 a ¯ ( ℓ ) j ( J a c , ℓ ) v ⩾ c ℓ 2 a ¯ ( ℓ ) + s ℓ 2 s ℓ 2 a ¯ ( ℓ ) + c ℓ 2 j ( J a c , ℓ ) v − 2 s ℓ 2 c ℓ 2 a ¯ ( ℓ ) j ( J a c , ℓ ) v = s ℓ 2 c ℓ 2 j ( J a c , ℓ ) v a ¯ ( ℓ ) + s ℓ 2 a ¯ ( ℓ ) + c ℓ 2 − 2 a ¯ ( ℓ ) ⩾ 0
according to (A.8). This proves the statement of Theorem 6.1 for finite differences by exploiting (6.11).
Proof of Theorem 6.1 for finite elements. For finite elements, we first note that J(Cor,ℓ) is singular because
a ¯ ( ℓ ) 2 det J ( Cor, ℓ ) = a ¯ ( ℓ ) 2 det 1 − ζ − 1 c ℓ 4 a ¯ ( ℓ ) − 1 a ( ℓ ) ζ − 1 s ℓ 2 c ℓ 2 a ¯ ( ℓ ) − 1 a ( N + 1 − ℓ ) ζ − 1 s ℓ 2 c ℓ 2 a ¯ ( ℓ ) − 1 a ( ℓ ) 1 − ζ − 1 s ℓ 4 a ¯ ( ℓ ) − 1 a ( N + 1 − ℓ ) = a ¯ ( ℓ ) − 2 c ℓ 4 a ( ℓ ) a ¯ ( ℓ ) − 2 s ℓ 4 a ( N + 1 − ℓ ) − 4 s ℓ 4 c ℓ 4 a ( ℓ ) a ( N + 1 − ℓ ) = a ¯ ( ℓ ) a ¯ ( ℓ ) − 2 s ℓ 4 a ( N + 1 − ℓ ) − 2 c ℓ 4 a ( ℓ ) = 0
according to
(A.11)
2sl4a(N+1−l)+2cl4a(l)=mii(3sl4−2sl4cl2+3cl4−2sl2cl4)+9δtdii(4sl4cl2+4sl2cl4) =mii(3−8sl2cl2)+ϑδtdii(4sl2cl2)=m¯(l)+ϑδtd¯(l)=a¯(l).
Therefore, the matrix J(Cor,ℓ)(J(Jac,ℓ))ν1+ν2 has a vanishing eigenvalue, too, and its spectral radius coincides with the absolute value of the trace, that is,
(A.12)
spr J ( C o r , ℓ ) J J a c , ℓ ) v = tr J C O r , ℓ ) J J a c , ℓ ) v = 1 − 2 c ℓ 4 a ¯ ( ℓ ) − 1 a ( ℓ ) j ( J a c , ℓ ) v + 1 − 2 s ℓ 4 a ¯ ( ℓ ) − 1 a ( N + 1 − ℓ ) j ( J a c , N + 1 − ℓ ) v = 2 a ¯ ( ℓ ) − 1 s ℓ 4 a ( N + 1 − ℓ ) j ( J a c , ℓ ) v + c ℓ 4 a ( ℓ ) j ( J a c , N + 1 − ℓ ) v ⩽ 2 a ¯ ( ℓ ) − 1 s ℓ 4 a ( N + 1 − ℓ ) j ( J a c , ℓ ) v + c ℓ 4 a ( ℓ ) j ( J a c , N + 1 − ℓ ) v < 2 a ¯ ( ℓ ) − 1 s ℓ 4 a ( N + 1 − ℓ ) + c ℓ 4 a ( ℓ ) = 1
by virtue of (5.3) and (A.11). This proves the statement of Theorem 6.1 for finite elements by exploiting (6.11).
Proof of Lemma 6.1. To prove the inequalities, we first note that
(A.13)
2m¯(l)−(1−2ϑ)δtd¯(l)⩾32mii>0, l=1,…,N¯
because
(A.14)
2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) = m i i 6 − 16 s ℓ 2 c ℓ 2 − 2 s ℓ 2 c ℓ 2 + 2 s ℓ 2 c ℓ 2 m i i − 2 ( 1 − 2 ϑ ) δ t d i i ⏟ ⩾ 0 by ( 2.10 ) ⩾ 6 m i i 1 − 3 s ℓ 2 c ℓ 2 ⩾ 6 m i i 1 − 3 4 ⩾ 3 2 m i i if ζ = 1 2
(A.15)
2m¯(l)−(1−2ϑ)δtd¯(l)=2(1−sl2cl2)+2sl2cl2(1−(1−2ϑ)δtdii)︸⩾0 by (2.10) ⩾2(1−sl2cl2)⩾2(1−14)⩾32 if ζ=1
due to the fact that
sl2cl2∈(0,1/4].
We now find upper bounds 22for the spectral norm of the submatrices by using Theorem 3.1, where different values for e1 and e2 are considered while
f1=m¯(l)+ϑδtd¯(l), f2=−m¯(l)+(1−ϑ)δtd¯(l).
Indeed, the requirement |f2|< |f1| made in this theorem is valid because
f 2 2 − f 1 2 = − δ t d ¯ ( ℓ ) 2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) ⏟ > 0 by (A .13) < 0
which can be shown as in (5.8).
– Then, according to Theorem 43.1, the spectral norm of IK
−ζ−1cl4(S¯K(l))−1 SK(l)
is bounded from above by
I K − ζ − 1 c ℓ 4 S ¯ K ( ℓ ) − 1 S K ( ℓ ) 2 max 1 − ζ − 1 c ℓ 4 2 m ( ℓ ) − ( 1 − 2 ϑ ) δ t d ( ℓ ) 2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) , d ¯ ( ℓ ) − ζ − 1 c ℓ 4 d ( ℓ ) d ¯ ( ℓ ) .
using the quantities
e1=(m¯(l)+ϑδtd¯(l))−ζ−1cl4(m(l)+ϑδtd(l))e2=(−m¯(l)+(1−ϑ)δtd¯(l))−ζ−1cl4(−m(l)+(1−ϑ)δtd(l)).
This bound can be simplified by exploiting the identities
d¯(l)−ζ−1cl4d(l)d¯(l)=d¯(l)−cl2d¯(l)d¯(l)=1−cl2=sl2
due to (6.5) and
(A.16)
1 − ζ − 1 c ℓ 4 2 m ( ℓ ) − ( 1 − 2 ϑ ) δ t d ( ℓ ) 2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) = 1 − c ℓ 4 + ζ − 1 c ℓ 4 2 ζ m ¯ ( ℓ ) − m ( ℓ ) − ( 1 − 2 ϑ ) δ t ζ d ¯ ( ℓ ) − d ( ℓ ) 2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) = 1 − c ℓ 4 + ζ − 1 c ℓ 4 2 ζ m ¯ ( ℓ ) − m ( ℓ ) + 2 ( 1 − 2 ϑ ) δ t s ℓ 4 d i i 2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ )
where the numerator of the last fraction is nonpositive if ϑ ⩾ 1/2 by (6.4) or due to the fact that
(A.17)
2ζm¯(l)−2m(l)+2ζsl4mii−2sl4(ζmii−(1−2ϑ)δtdii)︷⩾0 by (2.10)⩽mii(3−8sl2cl2−3+2sl2+sl4) =3miiso2(3so2−2)⩽0
for finite elements because
sl2⩽1/2 for all l=1,…,N¯.
For ζ = 1 (and ϑ < 1/2), we observe
1−ζ−1cl42m(l)−(1−2ϑ)δtd(l)2m¯(l)−(1−2ϑ)δtd¯(l)=1−cl4+cl42sl4−2sl4(1−(1−2ϑ)δtdii)2−2sl2cl2+2sl2cl2(1−(1−2ϑ)δtdii) ⩽1−cl4+cl42sl42(1−sl2cl2)=1−cl41−sl2(cl2+sl2)1−sl2cl2=1−cl61−sl2cl2⩽1−cl6.
On the other hand, we have
1 − ζ − 1 c ℓ 4 2 m ( ℓ ) − ( 1 − 2 ϑ ) δ t d ( ℓ ) 2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) = 1 − c ℓ 2 2 c ℓ 2 − ( 1 − 2 ϑ ) δ t c ℓ 2 d ( ℓ ) 2 − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) ⩾ 1 − c ℓ 2 2 − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) 2 − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) = 1 − c ℓ 2 ⩾ 0 if ζ = 1 1 − ζ − 1 c ℓ 4 2 m ( ℓ ) − ( 1 − 2 ϑ ) δ t d ( ℓ ) 2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) = 2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) − 4 c ℓ 4 m ( ℓ ) + 2 c ℓ 4 ( 1 − 2 ϑ ) δ t d ( ℓ ) 2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) = 2 m ¯ ( ℓ ) − 2 c ℓ 4 m ( ℓ ) − ( 1 − 2 ϑ ) δ t d i i 4 s ℓ 2 c ℓ 2 − 4 c ℓ 4 s ℓ 2 2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) = 2 m ¯ ( ℓ ) − 2 c ℓ 4 m ( ℓ ) − s ℓ 4 c ℓ 2 m i i + 2 s ℓ 4 c ℓ 2 ⩾ 0 by (2.10) m i i − 2 ( 1 − 2 ϑ ) δ t d i i 2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) ⩾ 2 m i i 3 − 8 s ℓ 2 c ℓ 2 − 3 c ℓ 4 + 2 s ℓ 2 c ℓ 4 − s ℓ 4 c ℓ 2 2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) = 6 m i i s ℓ 6 2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) ⩾ 0 if ζ = 1 2
according to (A.13).
– To estimate the spectral norm of
IK−ζ−1sl4(S¯K(l))−1 SK(N+1−l),
we can proceed similarly by replacing m(ℓ) and d(ℓ) by m(N+1−ℓ) and d(N+1−ℓ), respectively, while 22sℓhas to be substituted by cℓand vice versa. However, the numerator occurring in (A.16) does not have to be nonpositive for ζ = 1/2 and ϑ ⩾ 1/2 while the last inequality of (A.17) is not valid any more either. However, using the same ideas, we derive
(A.18)
1−ζ−1sl42m(N+1−l)−(1−2ϑ)δtd(N+1−l)2m¯(l)−(1−2ϑ)δtd¯(l)⩽1−sl4+sl46miicl2(3cl2−2)2m¯(l)−(1−2ϑ)δtd¯(l)
(A.19)
⩽1−sl4+sl46miicl2(3cl2−2)6mii(1−3sl2cl2)=1−sl41−3cl2(sl2+cl2)+2cl21−3sl2cl2⩽1−sl6
for
ζ=1/2,ϑ∈[0,1], and cl2⩾2/3,
because estimate (A.18) can be shown as in (A.16) and (A.17) while the first inequality 2in (A.19) is valid due to (A.14). For
ζ=1/2,ϑ∈[0,1], and cl2<2/3,
the same inequality can be easily verified because
1−ζ−1sl42m(N+1−l)−(1−2ϑ)δtd(N+1−l)2m¯(l)−(1−2ϑ)δtd¯(l) ⩽ 1−sl4+sl46miicl2(3cl2−2)2m¯(l)−(1−2ϑ)δtd¯(l) ⩽ 1−sl4 ⩽ 1−sl6.
– Invoking Theorem 3.1 using e1 = m(ℓ) + ϑδtd(ℓ) and e2 = −m(ℓ) + (1 − ϑ)δtd(ℓ), an upper bound of
‖ ζ−1sl2cl2(S¯K(l))−1 SK(l) ‖2
is given by
(A.20)
‖ ζ−1sl2cl2(S¯K(l))−1 SK(l) ‖2⩽ζ−1sl2cl2max(| 2m(l)−(1−2ϑ)δtd(l)2m¯(l)−(1−2ϑ)δtd¯(l) |,| d(l)d¯(l) |)
where
ζ−1sl2cl2d(l)d¯(l)=ζ−1sl2cl22sl2dii2ζ−1sl2cl2dii=sl2.
Furthermore, it is not necessary to take the absolute value of the first argument in the definition of the maximum in (A.20) because
ζ − 1 s ℓ 2 c ℓ 2 2 m ( ℓ ) − ( 1 − 2 ϑ ) δ t d ( ℓ ) 2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) ⩾ 0
by (2.13) and (A.13). On the other hand, the expression is bounded from above by
se2
for a finite difference approximation, that is, ζ = 1, because
ζ − 1 s ℓ 2 c ℓ 2 2 m ( ℓ ) − ( 1 − 2 ϑ ) δ t d ( ℓ ) 2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) − s ℓ 2 = s ℓ 2 2 c ℓ 2 − ( 1 − 2 ϑ ) δ t c ℓ 2 d ( ℓ ) − 2 + ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) 2 − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) = s ℓ 2 − 2 s ℓ 2 2 − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) ⩽ 0
while, in case of finite elements, an upper bound is given by
43Sl2
due to
ζ − 1 s ℓ 2 c ℓ 2 2 m ( ℓ ) − ( 1 − 2 ϑ ) δ t d ( ℓ ) 2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) − s ℓ 2 c ℓ 4 1 − 3 s ℓ 2 c ℓ 2 = 4 1 − 3 s ℓ 2 c ℓ 2 s ℓ 2 c ℓ 2 m ( ℓ ) − 2 s ℓ 2 c ℓ 4 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t 2 1 − 3 s ℓ 2 c ℓ 2 s ℓ 2 c ℓ 2 d ( ℓ ) − s ℓ 2 c ℓ 4 d ¯ ( ℓ ) 1 − 3 s ℓ 2 c ℓ 2 2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) = s ℓ 2 c ℓ 2 m i i 6 − 4 s ℓ 2 − 18 s ℓ 2 c ℓ 2 + 12 s ℓ 4 c ℓ 2 − 6 c ℓ 2 + 16 s ℓ 2 c ℓ 4 − 2 m i i s ℓ 2 − 3 s ℓ 4 c ℓ 2 − s ℓ 2 c ℓ 4 1 − 3 s ℓ 2 c ℓ 2 2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) + s ℓ 2 c ℓ 2 s ℓ 2 1 − 3 s ℓ 2 c ℓ 2 − c ℓ 4 ⩽ 0 by ( A .21 ) s ℓ 2 − 3 s ℓ 4 c ℓ 2 − s ℓ 2 c ℓ 4 ⩾ 0 by ( 2.10 ) 2 m i i − 4 ( 1 − 2 ϑ ) δ t d i i 1 − 3 s ℓ 2 c ℓ 2 2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) ⩽ s ℓ 2 c ℓ 2 m i i 6 − 18 s ℓ 2 c ℓ 2 − 6 s ℓ 2 + c ℓ 2 + 18 s ℓ 4 c ℓ 2 + 18 s ℓ 2 c ℓ 4 1 − 3 s ℓ 2 c ℓ 2 2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) = 0
by virtue of the fact that
(A.21)
c ℓ 4 1 − 3 s ℓ 2 c ℓ 2 = 3 1 − s ℓ 2 2 3 1 − 3 s ℓ 2 c ℓ 2 = 3 − 6 s ℓ 2 + 3 s ℓ 4 3 1 − 3 s ℓ 2 c ℓ 2 = 4 − 12 s ℓ 2 1 − s ℓ 2 ⏞ = c ℓ 2 − 1 − 3 s ℓ 2 2 ⏞ ⩾ 0 3 1 − 3 s ℓ 2 c ℓ 2 ⩽ 4 1 − 3 s ℓ 2 c ℓ 2 3 1 − 3 s ℓ 2 c ℓ 2 = 4 3 c ℓ 4 1 − 3 s ℓ 2 c ℓ 2 = 1 + c ℓ 4 − 1 + 3 s ℓ 2 c ℓ 2 1 − 3 s ℓ 2 c ℓ 2 = 1 + − 1 + c ℓ 2 c ℓ 2 + 3 s ℓ 2 1 − 3 s ℓ 2 c ℓ 2 = 1 + − 1 + 1 − s ℓ 2 1 + 2 s ℓ 2 1 − 3 s ℓ 2 c ℓ 2 = 1 + s ℓ 2 1 − 2 s ℓ 2 ⏞ ⩾ 0 1 − 3 s ℓ 2 c ℓ 2 ⩾ 1
because sℓ⩽ 1/2 for all ℓ = 1, . . . , ̄N.
– Finally, inequality (6.18) can be derived similarly by invoking
ζ − 1 s ℓ 2 c ℓ 2 d ( N + 1 − ℓ ) d ¯ ( ℓ ) = ζ − 1 s ℓ 2 c ℓ 2 2 c ℓ 2 d i i 2 ζ − 1 s ℓ 2 c ℓ 2 d i i = c ℓ 2 0 ⩽ ζ − 1 s ℓ 2 c ℓ 2 2 m ( N + 1 − ℓ ) − ( 1 − 2 ϑ ) δ t d ( N + 1 − ℓ ) 2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) = c ℓ 2 2 ζ − 1 s ℓ 2 m ( N + 1 − ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) 2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) ⩽ c ℓ 2 2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) 2 m ¯ ( ℓ ) − ( 1 − 2 ϑ ) δ t d ¯ ( ℓ ) ⩽ c ℓ 2
because
sl2⩽1/2<1 for ζ=1 and
ζ − 1 s ℓ 2 m ( N + 1 − ℓ ) − m ¯ ( ℓ ) = m i i 3 s ℓ 2 − 2 s ℓ 2 c ℓ 2 − 3 + 8 s ℓ 2 c ℓ 2 = m i i 6 s ℓ 2 c ℓ 2 − 3 c ℓ 2 = 3 c ℓ 2 m i i 2 s ℓ 2 − 1 ⩽ 0
in case of a finite element approximation.
