4.3 Numerical simulations and experiments
4.3.3 Numerical spectrum
4.3. NUMERICAL SIMULATIONS AND EXPERIMENTS 131
0 20 40 60
-2 -1 0 1 2
g) frequency ω (blue) and speed c(red)
Figure 4.4: Frequency ω (blue) and velocity c (red) in the numerical simulation of the freezing method in (QCGL) with parameters from (4.28).
We apply this algorithm to (QCGL) with the parameter set (4.28) from the previous Section 4.3.1 and initial value u0(x) = (12tanh(1000x) + 12,0)⊤. The results are shown in Figure 4.3 and Figure 4.4. The spatial step size is chosen to be ∆x = 0.1 and the domain of computation is Ω = [−50,50]. For the time discetization we use ∆t = 0.1 and the implicit equations are solved with Newton’s methods using a tolerance of 10−5. We see that the numerical solution of the freezing method converge to the profile of the TOF as Theorem 4.10 guarantees. In addition, the variable ν converges to the frequency and velocity of the TOF which are numerically given by ω ≈ −1.5821 and c ≈ 1.29 respectively. Note that the numerical value of the frequency coincides with the a-priori calculated value in (4.30). In particular, the freezing method is a powerful tool to compute the profile of a TOF as well as its frequency and velocity. Moreover, it enables us to pursue the TOF for arbitrary times without increasing the computational domain and therefore the numerical effort.
the case of (QCGL) with (4.28) the dispersion set consists of σdisp,˜+ µ(L) =
s∈C:
s=−ν2+i(c+ 2˜µ)ν+ ˜µ2−c˜µ+g1′(|v∞|2)|v∞|2± |g′1(|v∞|2)|v∞|2| and
σdisp,˜− µ(L) =
s∈C:s=−ν2 +i(c+ 2˜µ)ν+ ˜µ2+c˜µ+µ1±iω .
It is easy to see that the dispersion set describes four parabolas in the complex plane opened to the left. Since we have c > 0, µ1 < 0 and g′1(|v∞|2) = β1 + 2γ1|v∞|2 = 1−2|v∞|2 <0, Assumption 4 is satisfied. In particular, in Figure 4.5 the parabolas are shown in the cases µ˜ = 0 and µ˜ = 0.05. In the latter case, we see that the dispersion curves are included in the strict left half-plane.
-2 -1.5 -1 -0.5 0
-3 -2 -1 0 1 2 3
a) µ˜= 0
-2 -1.5 -1 -0.5 0
-3 -2 -1 0 1 2 3
b)µ˜ = 0.05
Figure 4.5: The dispersion sets σdisp,˜+ µ(L) (blue) and σ−disp,˜µ(L) (red) in (QCGL) with (4.28).
In the example (4.28) the dispersion curves are given by four parabolas, since the imaginary part of the diffusion coefficient vanishes. From (1.12) and (1.13) we see that the dispersion curves can be much more complicated if there is a non-vanishing imaginary part of the diffusion coefficient, i.e. α2 6= 0. As an example for this case we use the parameters (4.28) but set α2 = −103. The dispersion curves in this case are shown in Figure 4.6, again for the exponential growth rates µ˜= 0 and µ˜= 0.05. Also in this case the essential spectrum is included in the strict left half-plane if the exponential growth rate is chosen to be µ˜= 0.05. This strongly depends on the magnitude and sign of the imaginary part of the diffusion coefficient. It might happen that the curveσdisp,µ(L)forms a dovetail due to the fourth order terms, see Figure 4.7. Then Assumption 4 is violated.
But as Figure 4.7 shows there also might be an exponential growth rate such that the
4.3. NUMERICAL SIMULATIONS AND EXPERIMENTS 133
-2 -1.5 -1 -0.5 0
-3 -2 -1 0 1 2 3
a) µ˜= 0
-2 -1.5 -1 -0.5 0
-3 -2 -1 0 1 2 3
b)µ˜ = 0.05
Figure 4.6: The dispersion sets σdisp,˜+ µ(L) (blue) and σ−disp,˜µ(L) (red) in (QCGL) with (4.28) but α= 1−103i.
dispersion set is still included in the left half-plane. We expect that our stability results also apply in this case. However, one has to be careful using the exponential growth rates µ.
-2 -1.5 -1 -0.5 0
-3 -2 -1 0 1 2 3
a) µ˜= 0
-2 -1.5 -1 -0.5 0
-3 -2 -1 0 1 2 3
b)µ˜ = 0.05
Figure 4.7: The dispersion sets σdisp,˜+ µ(L) (blue) and σ−disp,˜µ(L) (red) in (QCGL) with (4.28) but α= 1 + 103 i.
Now let us consider the point spectrum of the linearized operator L and verify As-sumption 3. In Section 3.3 we have shown that for the essential spectrum it holds σess(L) =σess(L) for the operators L ∈ C[Xη] and L ∈ C[L2η]. As we will see, a similar relation holds true for the point spectrum. For this purpose, recall L defined by (0.26)
reading as
L :Yη →Xη, u
ρ
7→
Auxx+cux+Sωu+Df(v⋆)u SωρDf(v∞)ρ
and L from (0.12) reading as
L:Hη2 →L2η, u7→Auxx+cux+Sωu+Df(v⋆)u.
It follows that ifs∈Cis an eigenvalue ofL with eigenfunctionu0 ∈Hη2 thens is also an eigenvalue of L with eigenfunction u0 = (u0,0)⊤ ∈Yη, i.e. σpt(L)⊂σpt(L). Conversely, let s ∈ C be an eigenvalue of L with eigenfunction u0 = (u0, ρ0)⊤. If ρ0 = 0 then s is also an eigenvalue of L with eigenfunction u0. Now assume ρ0 6= 0. Then we have (sI −Sω−Df(v∞))ρ0 = 0 and thus either s = 2g1′(|v∞|2)|v∞|2 or s = 0. But 0∈σ(L) and we obtain
σpt(L)⊂σpt(L)∪ {2g′1(|v∞|2)|v∞|2}.
The possible eigenvalue 2g′1(|v∞|2)|v∞|2 is of no interest since it is included in the strict left half-plane by Assumption 2. So neglecting this additional eigenvalue it is sufficient to compute the spectrum of the operatorLinstead of the spectrum of the operatorL to verify Assumption 4. Moreover, we have
σ(L)⊂σ(L)∪ {2g′1(|v∞|2)|v∞|2}.
Now if s∈σpt(L)for L∈ C[L2η]with eigenfunction u0 ∈Yη, thenu0 ∈Y and s∈σpt(L) for L ∈ C[L2]. In particular, the point spectrum does not move by taking exponential weights into account. More precisely, if sbelongs to the point spectrum of LonL2 then s belongs to the point spectrum of L on L2η, unless the essential spectrum has moved to encompass s, cf. [36, Sec. 3.1.1.2]. Thus to verify that the point spectrum of L, respectively L, is included in the strict left half-plane it is sufficient to compute the spectrum on L2 instead of L2η.
Finally, let us compute the point spectrum of L, respectively L, numerically in the case of (QCGL) with (4.28). In order to do so, we use a finite difference approximation of L on the truncated domainΩ = [−1000,1000]with spatial step size∆x= 0.1and periodic boundary conditions. The numerical results are shown in Figure 4.8 and Figure 4.9.
We see that there are no eigenvalues in the right half-plane or on the imaginary axis expect for the zero eigenvalue. Moreover, there are even no eigenvalues in left half-plane. Therefore we expect the point spectrum of L on X to be empty. The isolated eigenvalues between the dispersion curves belong to the essential spectrum since in these regions the operator sI− L is not Fredholm of index 0. This can be seen by computing the corresponding Morse indices, cf. [36] and Figure 3.3. In particular, the dispersion curves from the numerical spectrum do not fit exactly to the dispersion set calculated in
4.3. NUMERICAL SIMULATIONS AND EXPERIMENTS 135
-1.5 -1 -0.5 0 0.5
-3 -2 -1 0 1 2 3
Figure 4.8: Numerical spectrum of the linearized operator.
-1.5 -1 -0.5 0 0.5
-3 -2 -1 0 1 2 3
Figure 4.9: Numerical spectrum of the linearized operator with dispersion set (red) and zero eigenvalue (green).
(1.12) and (1.13). The reason for this is that we approximate the operatorLby an finite difference approximation on a large, but bounded, domain. As a result the parabolas from the dispersion set are approximated by ellipses depending on the size of the domain
of computation. This can be seen by considering the whole spectrum of the linearized operator on a truncated domain, see Figure 4.10.
-400 -300 -200 -100 0
-15 -10 -5 0 5 10 15
Figure 4.10: Whole numerical spectrum of the linearized operator on the truncated domain with periodic boundary conditions.
Chapter 5
Stability in polynomially weighted spaces
In Chapter 3 we proved a nonlinear stability result for TOFs when the perturbation u0
of the initial data u(0) =v⋆+u0 converges exponentially fast to some limit r∞ at +∞ and to zero at −∞. A natural question arises whether the assumption on the initial perturbation can be weakened from exponential to polynomial decay. In this chapter we prove a nonlinear stability result for polynomially decaying initial perturbations, see Theorem 1.13. In this case we have to assume r∞ = 0, i.e. u0 ∈ Hη1 with a polynomial weight function η.
Throughout the chapter we setη=ηpoly with the polynomial weight function from (0.29) reading as
ηpoly(x) = (x2+ 1)12.
and consider the weighted spacesL2k,Hkℓ from (0.30). We assume the existence of a TOF with profile v⋆ and speeds (ω, c) and consider the perturbed co-moving equation from (0.11)
ut=Auxx+cux+Sωu+f(u), u(0) =v⋆+u0
with an initial perturbation u0 ∈ Hk2 for some k ∈ N. Since u0 → 0 at ±∞ we expect that the limit at+∞ of the solutionu of (0.11) stays constant in the time evolution, cf.
(0.18). In particular, a TOF with frequency ω can be seen as a traveling wave solution of the co-rotated equation
ut=Auxx+Sωu+f(u).
For that reason we seek for solutions of (0.11) in the affine linear spaces, cf. (0.31) Mk = ¯v+L2k, Mkℓ = ¯v+Hkℓ, v¯=v∞v.ˆ
137
5.1 Polynomially weighted Sobolev spaces
Before investigating the nonlinear stability we collect some properties concerning the weighted spaces L2k, Hkℓ from (0.30). The function η = ηpoly ∈ C∞(R,R) from (0.29) is a function of linear growth, i.e. η(x) ∼ |x|. More precisely, for |x| > 1, k ∈ N0 the following estimates hold true
|x|k≤η(x)k ≤2k2|x|k. Furthermore, we note the first and second derivative
ηx(x) =x(x2+ 1)−12, ηxx(x) = (x2+ 1)−32.
Then |ηx(x)|,|ηxx(x)| ≤1,x∈R and the function spaces L2k, Hkℓ are Hilbert spaces with the inner products
(u, v)L2
k = (ηku, ηkv)L2, (u, v)Hℓ
k = (ηku, ηkv)Hℓ.
Moreover, Hkℓ is dense in L2k and since ηk(x) ≤ ηℓ(x) for all x ∈ R as long as k ≤ ℓ, it follows immediately thatL2ℓ ⊂L2k and the inclusion is dense as well. As in Chapter 3 we consider the multiplication operator mku=ηku for u∈ Hkℓ, k ∈N0, ℓ = 1,2. Similarly to Lemma 3.1 we have that mk defines a continuous isomorphism from Hkℓ to Hℓ. Lemma 5.1. Let k ∈ N0 and mku = ηku define the multiplication operator associates with ηk. Then
i) mk :L2k →L2 is an isometric isomorphism.
ii) mk :Hkℓ →Hℓ, ℓ= 1,2 is a continuous isomorphism.
Remark 5.2. It also holds true that mk : Hkℓ → Hℓ is a continuous isomorphism for arbitrary ℓ ∈ N. However, the proof is more involved and we are only interested in the cases ℓ= 0,1,2 as in Lemma 5.1.
Proof. We show that mk : Hkℓ → Hℓ, ℓ = 1,2 is continuous. Then the claim follows as in the proof of Lemma 3.1. First let u∈Hk1. Then
k(ηku)xkL2 =kkηk−1ηxu+ηkuxkL2 ≤kkηk−1ukL2 +kηkuxkL2 ≤(k+ 1)kukHk1. Thus,
kηkuk2H1 =kuk2L2k +k∂(ηku)k2L2 ≤(k2+ 2k+ 2)kuk2Hk1. For u∈Hk2 we have
k(ηu)xxkL2 =kk(k−1)ηk−2η2xu+kηk−1ηxxu+ 2kηk−1ηxux+ηkuxxkL2
≤(k(k−1) +k)kukL2k+ 2kkuxkL2k+kuxxkL2k ≤(k+ 1)2kukHk2.
5.1. POLYNOMIALLY WEIGHTED SOBOLEV SPACES 139 To show resolvent estimates of the linearized operator later on we need a integration by parts formula inL2k which is slightly different from the standard integration by parts formula in L2.
Lemma 5.3. Suppose u, v ∈ Hk1(R,Rn). Then there holds the following integration by parts formula
−(u, vx)L2k = (ux, v)L2k+ 2k(η−1ηxu, v)L2k. Moreover, if u, v ∈Hk2(R,Rn), then there holds
(u, vxx)L2
k = (uxx, v)L2
k + 4k(η−1ηxux, v)L2
k+ (4k2−2k)(η−2ηx2u, v)L2
k+ 2k(η−1ηxxu, v)L2
k. Proof. We write (·,·)L2k = (·,·). The assertion is a direct consequence of the standard integration by parts formula in L2, since
−(u, vx) =− Z
R
η2k(x)u(x)⊤vx(x)dx = Z
R
∂x(η2k(x)u(x))⊤v(x)dx
= Z
R
η2k(x)ux(x)⊤v(x)dx+ 2k Z
R
η2k−1(x)ηx(x)u(x)⊤v(x)dx
= (ux, v) + 2k(η−1ηxu, v).
The second formula follows by applying integration by parts in L2η twice. We obtain (u, vxx) =−(ux, vx)−2k(η−1ηxu, vx)
= (uxx, v) + 2k(η−1ηxux, v) + 2k[(2k−1)(η−2ηx2u, v) + (η−1ηxxu, v) + (η−1ηxux, v)]
= (uxx, v) + 4k(η−1ηxux, v) + (4k2−2k)(η−2ηx2u, v) + 2k(η−1ηxxu, v).
In Lemma 3.3 we have proven that the space C0∞(R,Rn) of smooth function with compact support are dense in Hη1(R,Rn) if η is an exponential weight function as in (0.24). However, the proof of Lemma 3.3 is independent of the choice of the weight function η and we conclude that C0∞(R,Rn) is dense in Hk1(R,Rn). Since we look for traveling waves in the space L2k we have to collect some smoothness properties of the shift u 7→ u(· −τ), τ ∈ R on the polynomially weighted spaces. As in the exponential case, cf. Lemma 3.4, it turns out that the shift is continuous onL2k and locally Lipschitz continuous on Hk1.
Lemma 5.4. Supposek ∈N0. i) Ifu∈L2k and τ ∈R, then
ku(· −τ)kL2k ≤CτkkukL2k, Cτ := 1 +|τ|.
ii) If u∈Hk1 and τ ∈R, then
ku(· −τ)−ukL2k ≤Cτk|τ|kuxkL2k. iii) If u∈L2k, then
ku(· −τ)−ukL2k →0 as τ →0.
Further, the estimate in ii) holds true if u is replaced by vˆ from (0.19) or v⋆ from Assumption 2.
Proof. The case k = 0 is the usual case inL2. Thus, let k ≥1. We only show i) then ii) and iii) follow exactly as in the proof of Lemma 3.4.
First note for all x, τ ∈R η(x+τ)2
η(x)2 = x2+ 2τ x+τ2+ 1
x2+ 1 ≤1 + 2|τ| |x|
x2+ 1 + τ2
x2+ 1 ≤1 + 2|τ|+τ2 = (1 +|τ|)2. Now we obtain
ku(· −τ)k2L2k = Z
R
η2k(x+τ)|u(x)|2dx ≤(1 +|τ|)2k Z
R
η2k(x)|u(x)|2dx=Cτ2kkuk2L2k.