Two problems in parabolic PDEs
A DISSERTATION
SUBMITTED TO THE FACULTY OF THE GRADUATE SCHOOL OF THE UNIVERSITY OF MINNESOTA
BY
Lu Li
Under the supervision of Vladim´ır ˇSver´ak
IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF
Doctor Of Philosophy
c
Lu Li 2011
Acknowledgements
There are many people that have earned my gratitude for their contribution to my time in graduate school.
I owe my deepest gratitude to my adivisor Vladim´ır ˇSver´ak for his supervision, encouragement, and support. It would have been impossible to write this thesis without his constant help and guidance. I have learned from him innumerable lessons and insights on all aspects of the workings of academic research.
Special thanks goes to Professor Pol´aˇcik, who has been very supportive and helped numerously in all stages of my graduate study.
I also take this chance to thank Professor Escauriaza for helpful advices on the project in the first half of this thesis, for many inspiring discussions during my visit to UPV/EHU in beautiful Bilbao.
This work is partially supported by Doctoral Dissertation Fellowship of Univer-sity of Minnesota. Some of the work appearing in this thesis was carried out when
I thank my husband Weiyi Zhang for standing by my side during the past few years. Last but not least, I would like to express my gratitude to my parents for their everlasting love and support.
Abstract
In this thesis, we will study two problems in parabolic Partial Differential Equa-tions.
In the Chapter 2 we study the problem of backward uniqueness of the linear heat equation. We briefly recall some results about backward uniqueness under the assumption of Dirichlet boundary condition and results of boundary controllability in control theory. It was shown in [9, 25] that a bounded solution of the heat equation in a half-space which becomes zero at some time must be identically zero, even though no assumptions are made on the boundary values of the solutions. In a recent example, Luis Escauriaza showed that this statement fails if the half-space is replaced by cones with opening angle smaller than 90◦. The main result of the chapter is that the result remains true for cones with opening angle larger than 110◦.
In Chapter 3 we study the profiles of the singularity of the 1-D complex Burgers equation. The Cauchy problem for the 1-D real-valued viscous Burgers equation ut + uux = uxx is globally well-posed [13]. For complex-valued solutions finite
time blow-up is possible from smooth compactly supported initial data, see [23]. It is also proved in [23] that the singularities for the complex-valued solutions are
more detail. In particular, we classify the possible blow-up rates and blow-up profiles. It turns out that all singularities are of type II and that the blow-up profiles are regular steady state solutions of the equation.
Contents
Acknowledgements i
Abstract iii
List of Figures vii
1 Introduction 1
1.1 Backward uniqueness of the heat equation . . . 1 1.2 Isolated singularities of the 1-D complex Burgers equation . . . . 4
2 Backward Uniqueness of the Heat Equation 7
2.1 Background and relative results . . . 8 2.1.1 Backward uniqueness with the Dirichlet boundary condition 9 2.1.2 Bounded domains are controllable . . . 14 2.1.3 Two Backward uniqueness results via Fourier Transform . 23
2.2 Backward uniqueness and Carleman inequalities . . . 34
2.2.1 Backward uniqueness . . . 36
2.3 Proof of Lemma 2.2 . . . 45
2.4 Proof of the Carleman inequality . . . 49
3 Isolated singularities of the 1D complex viscous Burgers equation 57 3.1 Structure of the isolated singularities . . . 57
3.2 Representation in caloric polynomials and outline of the proof . . 60
3.2.1 Caloric polynomials . . . 60
3.2.2 Proof of Theorem 3.1.1 . . . 62
3.3 Proof of the lemmas . . . 65
References 71
List of Figures
2.1 Abstract linear control system . . . 16 2.2 Abstract linear observed system . . . 16
Chapter 1
Introduction
1.1
Backward uniqueness of the heat equation
In Chapter 2, we study the problem of backward uniqueness of the heat equation. Consider an open set Ω ⊂ Rn. Let u be a bounded solution of the equation
ut− ∆u + b(x, t) · ∇u + c(x, t)u = 0 in Ω × (0, T ), (1.1)
where the coefficients b = (b1, . . . , bn), c are measurable and bounded. We say
that Ω has the backward uniqueness property if the following statement holds: (BU) If a bounded u : Ω × (0, T ) → R satisfies (1.1) and u( · , T ) = 0, then u ≡ 0 in Ω × (0, T ).
It is important to emphasize that no assumptions are made about u at the
parabolic boundary (∂Ω × (0, T )) ∪ (Ω × {0}) in the statement (BU). With bound-ary condition, the backward uniqueness of the heat equation is standard. We will discuss this case later in section 2.1.1.
In fact, we can think about the problem in terms of the control theory: we are given some initial data u0: Ω → R, and we wish to find a suitable boundary
con-dition g on the lateral boundary ∂Ω × (0, T ) so that when we solve equation (1.1) with u0 as the initial condition and g as the boundary condition, the solution will
become exactly zero at time t = T . In this interpretation condition (BU) means that we can never achieve the exact boundary control of any non-trivial solution. By classical results we know that in bounded domains we can achieve exact control, which is discussed in section 2.1.2 and therefore any domain satisfying (BU) has to be unbounded. Classical backward uniqueness results for parabolic equations imply that Ω = Rn satisfies (BU). It turns out that the half-space Ω = Rn
+ also satisfies (BU), although this is harder to prove, see [9]. In general,
the smaller the domain, the harder it will be to show that it satisfies (BU). It is immediate that if Ω1 ⊂ Ω2 and Ω1 satisfies (BU), then also Ω2 satisfies (BU).
While the control theory for PDEs seems to be the most natural background for (BU), the problem also appeared in regularity theory of parabolic equations, such as the Navier-Stokes equations, harmonic map heat flows, or semi-linear heat equations, see [9, 22, 29]. The specific unbounded domains which arise in this
connection are complements of closed balls (for interior regularity), half-spaces (for boundary regularity at C1 boundaries), or cones (for boundary regularity in Lipschitz domains).
In Chapter 2 we consider the question for cones with opening angle θ. In suitable coordinates
Oθ = {x = (x1, x0), x0 ∈ Rn−1, x1 > |x| cos(θ/2)}.
Luis Escauriaza [5] recently showed that - surprisingly - (BU) fails when θ < π/2. We will briefly recall his counterexample in section 2.1.4. Since we also know that (BU) is true for θ = π, it is easy to see that there exists a borderline angle θ0 ∈ [π/2, π] such that (BU) is true for θ > θ0, and (BU) fails for θ < θ0. The
borderline case θ = θ0 might perhaps present an extra difficulty.
The main result of Chapter 2 is that the cones Oθ satisfy (BU) for
θ > 2 arccos(1/√3) ∼ 109.52◦.
The proof of the main result is in section 2.2. It is tempting to conjecture that θ0 = π/2. This is supported by the fact that θ = π/2 is the borderline case for the
construction of Escauriaza, as can be seen from the Phragm´en-Lind¨olef principle. For the classical heat equation, corresponding to the case b = 0 and c = 0 in (1.1), and θ = π (the half-space), the statement (BU) can be proved by a relatively simple application of Fourier transformation and some classical complex analysis
results, see [25]. We were not able to find such simple proof of the case b = 0, c = 0 when θ < π. However, with additional condition that the solution is also 0 at the initial time, we can show the uniqueness using Fourier transformation in a similar way. The results are included in section 2.1.3.
1.2
Isolated singularities of the 1-D complex
Burg-ers equation
A second problem we consider in the thesis in Chapter 3 is about the singularities of the Burgers equation. In a recent paper [23] Pol´aˇcik and ˇSver´ak study the occurrence of the singularities for the complex 1D viscous Burgers equation. They prove that there exist well-behaved initial data for which the solutions blow up in finite time. ( In an earlier paper [17] D. Li and Sinai show that singularities are possible for the complex-valued Navier-Stokes equation. For other results regarding singularities of complex equations, see [3] [18] [19] [26] and [27].) Using the well known Cole-Hopf transformation, Pol´aˇcik and ˇSver´ak reduce the problem to the study of the zeros of solutions of the 1D complex heat equation and show that the singularities are isolated if they are not presented in the initial data, see Theorem 3.3 of [23]. This chapter can be considered as a continuation of [23] aiming at the description of these isolated singularities and in particular at the
study of their profiles. The Burgers equation [2]
ut+ uux = uxx , (1.2)
named after the Dutch physicist Johannes Martinus Burgers (1895-1981), is some-times used for modeling certain aspects of equations of fluid mechanics. It can be reduced to the standard heat equation by the Cole-Hopf transformation [4] [13]. Let v(x, t) be a positive function satisfying the heat equation vt= vxx, then
u = −2vx
v , (1.3)
called the Cole-Hopf transformation of v, solves the Burgers equation. Conversely, let u be a (sufficiently regular) solution to the Burgers equation. Let
v(x, t) = exp −1 2 Z x 0 u(ξ, t) d ξ .
It is easy to check by direct calculation that v satisfies ((vt− vxx)/v)x = 0, and
thus there exists a function C(t) such that vt− vxx = C(t)v. Then the function
˜
v(x, t) = v(x, t) exp R −C(t) dt satisfies the heat equation ˜vt = ˜vxx and u =
−2vx/v = −2˜vx/˜v.
Singularities of u are related to zeros of v. One can see that for real and “sufficiently regular” initial data, equation 1.2 has a unique smooth global solution (in some natural classes of functions) given by u = −2vx/v, see [13]. By contrast,
for complex-valued Burgers equation blow-up is possible from regular, compactly supported initial data (complex-valued) [23]. In addition, the singularities are isolated if they are not present in the initial data (as quoted in the following theorem).
Theorem 3.3 of [23] Let v be a bounded complex-valued solution of the heat equation in R × (0, ∞). Assume v has no zeros in some neighborhood of R × {0}. Then all zeros of v in R × (0, ∞) are isolated.
In Chapter 3 we consider the structure of an isolated singularity via the re-scaling procedure, which is well-known to be very useful in studying singularities of PDEs. In particular, we classify the possible blow-up rates and blow-up profiles. It turns out that all singularities are of type II and that the blow-up profiles are regular steady state solutions of the equation.
Chapter 2
Backward Uniqueness of the Heat
Equation
In this chapter we study the problem of backward uniqueness of the heat equation. Let us recall some settings mentioned in the introduction chapter. Consider an open set Ω ⊂ Rn. Let u be a bounded solution of the equation
ut− ∆u + b(x, t) · ∇u + c(x, t)u = 0 in Ω × (0, T ), (2.1)
where the coefficients b = (b1, . . . , bn), c are measurable and bounded. We say
that Ω has the backward uniqueness property if the following statement holds: (BU) If a bounded u : Ω × (0, T ) → R satisfies (2.1) and u( · , T ) = 0, then u ≡ 0 in Ω × (0, T ).
We emphasize that no assumptions are made about u at the parabolic bound-ary (∂Ω × (0, T )) ∪ (Ω × {0}) in the statement (BU).
In Section 2.1 we includes some relative results on the topic of backward uniqueness of the heat equation. The main result of this chapter is in Section 2.2
2.1
Background and relative results
In this section we first briefly recall the standard results about backward unique-ness under the assumption of Dirichlet boundary condition and some classical results of boundary controllability in control theory. We then include the elegant proof of [25] on backward uniqueness without boundary condition, in the domain of half space. In other words, (BU) holds with the equation ut− ∆u = 0 in half
space. The same technique is applied to prove a result with a π/2 cone, with some additional condition. Lastly, we include the surprising example of Escauriaza – that (BU) fails for cones with angle smaller than 90◦.
2.1.1
Backward uniqueness with the Dirichlet boundary
condition
Let us first consider the simplest case of the standard heat equation with the Dirichlet boundary condition. Let Ω ∈ Rn be a bounded domain and u is a sufficient regular solution to
ut− ∆u = 0 Ω × (0, T ), (2.2)
u(x, t) = 0 ∂Ω × (0, T ). (2.3)
The Laplacian operator −∆ with the Dirichlet boundary condition (2.3) is self-adjoint in the space L2(Ω), with domain H2(Ω) ∩ H01(Ω). Let uk(x) be the
or-thonormal system of eigenfunctions of −∆ corresponding to eigenvalues λk, where
λk> 0. Then u(x, t) = ∞ X k=1 cke−λktuk(x).
If u(·, T ) = 0, then all the coefficients ck are 0, hence u(x, t) ≡ 0 in Ω × (0, T ).
Another way to see the backward uniqueness of the heat equation with the Dirichlet boundary condition is from the concept of “log-convexity”. Let f (t) = ||u(·, t)||2
convex function. ˙ f (t) = Z 2uut = Z 2u∆u = −2 Z |∇u|2, ¨ f (t) = −4 Z ∇u∇ut = 4 Z ∆uut= 4 Z |∆u|2, ¨ F (t) = ¨ f f − ˙f2 f2 = 4R |∆u|2R u2− 4(R u∆u)2 f2 ≥ 0. We then have F (t) ≤ F (0)(T − t)/T + F (T )t/T, (2.4) and in turn, f (t) ≤ f (0)T −tT f (T ) t T. (2.5)
If u(x, T ) = 0, that is f (T ) = 0, then it follows that f (t) = 0, t ∈ [0, T ].
The concept of “log-convexity” was introduced by Agmon and Nirenberg in [1] in 1967. It works for more general equations with variable coefficients. In the rest part of this section, we mostly follow the presentation of the book [15], pp 41-45. For the following equation with bounded coefficients
ut− ∆u = b(x, t) · ∇u + c(x, t)u Ω × (0, T ), (2.6)
u(x, t) = 0 ∂Ω × (0, T ), (2.7)
one can derive a similar differential inequality, provided that b(x, t), c(x, t) have enough regularity.
¨
with some positive constant C depending on b(x, t), c(x, t). The differential in-equality (2.8) follows from the following lemma and the Schauder estimates for the heat equation.
Let A = A(t) be a linear operator in the Hilbert space H with domain D(t). By || · ||, (, ) we denote the norm and the scalar product in H. Consider the equation
ut− A(t)u = 0 on (0, T ). (2.9)
We assume that A = A+ + A− where A+, A− are the symmetric and
skew-symmetric parts of A. Moreover, they satisfy the following conditions.
||A−u||2 ≤ α(||A+u||||u|| + ||u||2) (2.10)
and
∂t(A+u, u) ≥ 2(A+u, ut) − α(||A+u||||u|| + ||u||2) (2.11)
for some constant α.
Lemma 2.1.1. Let u(t) ∈ D(t), u ∈ C1([0, T ]; H) be a solution of the
equa-tion (2.9). Let f (t) = ||u(t)||22, F (t) = log f (t). Then ¨F (t) + C| ˙F (t)| + C ≥ 0, on (0, T ). The constant C depends on α.
Proof. By straightforward calculations,
˙
F (t) = 2(u, ut) ||u||2 =
2(u, A+u)
¨ F (t) = 2∂t(u, A+u) ||u||2 − 4(u, A + u)2 ||u||4 ≥ 4(A+u, ut) ||u||2 −
2α(||A+u||||u|| + ||u||2)
||u||2 − 4(u, A + u)2 ||u||4 = 4||A+u|| 2 ||u||2 + 4(A+u, A−u) ||u||2 −
2α(||A+u||||u|| + ||u||2)
||u||2 −
4(u, A+u)2
||u||4 . (2.12)
We set (A+u, u) = ||A+u||||u||θ and represent A+u as βu + u⊥, where u⊥ is
orthogonal to u and β = (A+u, u)/||u||2. Then ||u⊥||2 = ||A+u||2(1 − θ2). We thus
have that
4(A+u, A−u) = 4(u⊥, A−u) ≥ −4||u⊥||||A−u||
≥ −2(||u⊥||2+ ||A
−u||2) ≥ −2||A+u||2(1 − θ2) − 2α(||A+u||||u|| + ||u||2). (2.13)
Combining (2.12) (2.13), it follows that ¨
F ≥ 2||A+u||
2(1 − θ2)
||u||2 −
4α(||A+u||||u|| + ||u||2)
||u||2 = 2σ
2(1−θ2)−4ασ −4α, (2.14)
where we set σ = ||A+u||/||u||.
Observe that | ˙F | = 2σ|θ|. If θ2 < 1 4, then ¨ F ≥ 3 2σ 2− 4ασ − 4α ≥ −3 4α 2− 4α. (2.15) If θ2 ≥ 1 4, then | ˙F | ≥ σ and ¨ F ≥ −4ασ − 4α, (2.16)
so that ¨F + 4α| ˙F | + 4α ≥ 0. In both cases we have the required inequality with C = 34α2+ 4α.
Our operator is A = ∆ + b(x, t) · ∇ + c(x, t) and the space H = L2(Ω). The
domain of D(t) is H2(Ω)∩H01(Ω). A+= ∆+(−12div b+c), A− = b(x, t)·∇+12div b.
By Schauder estimates, the operator A satisfies the conditions (2.10) (2.11), if b(x, t) has bounded derivatives up to order 2, and c(x, t) has bounded derivatives. The backward uniqueness of the problem (2.6) (2.7) follows from the next lemma. If u(x, T ) = 0, Lemma 2.1.2 implies that u(x, t) ≡ 0, Ω × [0, T ].
Lemma 2.1.2. From the inequality (2.8) we have that
f (t) ≤ C1f (0)1−λ(t)f (T )λ(t), t ∈ [0, T ], λ(t) ∈ [0, 1]. (2.17)
Proof. Let L be a solution to the differential equality ¨L + C| ˙L| + C = 0 on (0, T ), coincide with F at the end points 0, T . The existence of L can be proven by using a priori estimates on L, ˙L that follow from the observation that L satisfies the linear differential equation with the coefficient Csign ˙L. It follows that ¨F − ¨L + C| ˙F − ˙L| ≥ 0. Then F −L ≤ 0 by maximum principles, and it suffices to bound L. From the equation, L(t) has no local minimum on (0, T ), so ˙L > 0 on (0, τ ) and
˙
L < 0 on (τ, T ) for some τ in (0, T ). The cases τ = 0, T are follow along a similar proof. Therefore, L satisfies the linear differential equation ¨L + CτL + C = 0,˙
where Cτ is C on (0, τ ) and −C on (τ, T ).
We separate L into two parts, L = v + w, where v is the solution to the homogeneous equation ¨v + Cτ˙v = 0 coinciding with F at the end points, and
w = L − v solves the inhomogeneous equation with zero boundary data.
Since ˙v solves a linear first-order homogeneous ODE, it does not change its sign. Let F (0) ≤ F (T ), then ˙v ≥ 0. Consider the solution V to the ODE:
¨
V + C ˙V = 0 with the same boundary data. We have
¨
v − ¨V + C( ˙v − ˙V ) = (C − Cτ) ˙v ≥ 0.
By maximum principles, v ≤ V . We solve the equation for V ,
V (t) = F (0)e
−Ct− e−CT
1 − e−CT + F (T )
1 − e−Ct 1 − e−CT.
When F (0) ≥ F (T ), we obtain a similar bound with C replaced by −C.
To bound w we solve the equations: ¨w+C ˙w+C = 0 on (0, τ ) and ¨w−C ˙w+C = 0 on (τ, T ) and match the solution at τ . It is easy to see that w is bounded by some constant C2(T ). Replace F (t) by log f (t) we have the required result, with
λ = 1−e1−e−CT−Ct in the case F (0) ≤ F (T ) and C1 = eC2T. (We have a similar λ in the
case F (0) > F (T ).)
2.1.2
Bounded domains are controllable
By classical results in control theory we know that bounded domains are exact controllable. In other words, for bounded domains, (BU) cannot be satisfied. However, the proof of the controllability is highly non-trivial, even in 1-dim case.
In this section, we include the proof of the null-controllability of the 1-dim heat equation through the boundary. If u(x, t) satisfies the standard heat equation ut− ∆u = 0 with a reasonable initial condition u0, then we can find a boundary
condition h such that u(x, T ) = 0. The proof is in two steps. First, we show the equivalence of controllability and observability in the abstract framework of functional analysis. Second, we show the observability by a Carleman inequality.
Let us consider the following boundary control problem.
(I.1) ut− ∆u = 0 Ω × (0, T ), u(0, t) = 0 (0, T ), u(L, t) = h(t) (0, T ), u(x, 0) = u0(x) (0, L),
where u0 ∈ L2(0, L). We look for a control h(t) ∈ L2(0, T ) so that u(x, T ) = 0.
Controllability and Observability
In this part we follow the definitions in the 1978 survey paper of D. Russell, see [24]. Let X, Y , Z be Hilbert spaces and let S : X → Z be a bounded linear operator and C : D(C) ⊆ Y → Z a closed linear operator with domain D(C) dense in Y . Then {X, Y, Z, S, C} together constitute an abstract control system.
Definition 2.1.3. The abstract linear control system {X, Y, Z, S, C} is control-lable if R(C) ⊇ R(S). We denote by R(C), R(S) the ranges of the operators C
Y ⊇ D(C) −→ ZC
X
6
S
Figure 2.1: Abstract linear control system and S respectively.
The adjoint operators S∗ : Z → X and C∗ : D(C∗) ⊆ Z → Y may be defined by (S∗z, x)X = (z, Sx)Z, x ∈ X, z ∈ Z (2.18) (C∗z, y)Y = (z, Cy)Z, y ∈ D(C), z ∈ D(C∗). (2.19) Y C ∗ ←− D(C∗) ⊆ Z X? S∗
Figure 2.2: Abstract linear observed system
Definition 2.1.4. The abstract linear observed system { X, Y, Z, S∗, C∗} is ob-servable if there is a positive number K such that
The duality of controllability and observability is given by the following theo-rem.
Theorem 2.1.5. The abstract linear control system { X, Y, Z, S, C} is controllable if and only if the abstract linear observed system { X, Y, Z, S∗, C∗} is observable.
Proof. The proof is standard in functional analysis. We include it here for com-pleteness of the thesis.
Controllability =⇒ observability. Assuming R(C) ⊇ R(S), we need to show that there exists K > 0 such that ||C∗z||Y ≥ K||S∗z||X for z ∈ D(C∗). Otherwise,
there is a sequence zn ∈ D(C∗) such that C∗zn → 0, and ||S∗zn|| → ∞. For any
y ∈ D(C), (C∗zn, y)Y → 0, thus (zn, Cy)Z → 0. Because R(C) ⊇ R(S), for
any x ∈ X, we have (zn, Sx)Z = 0, thus (S∗zn, x)X = 0. Uniform Boundedness
Theorem implies that S∗zn is a bounded sequence. This is a contradiction.
Observability =⇒ Controllability. Define a linear functional for x ∈ X on R(C∗) ⊂ Y as the following.
Tx(C∗z) = (x, S∗z)X.
The operator Tx is well defined because that kerC∗ ⊆ kerS∗. In addition, Tx is
also bounded. |Tx(C∗z)| ≤ ||x||X||S∗z||X ≤ 1 K||x||X||C ∗ z||Y.
Extend Tx to a linear functional on Y . By Riesz representation theorem, there
exists some y ∈ Y such that
(y, C∗z)Y = (x, S∗z)X = (Sx, z)Z.
This y is in D(C∗∗). As C is a closed operator with domain dense in Y , we have that D(C∗∗) = D(C). Thus for each x ∈ X, we found a y ∈ Y such that (Cy, z)Z = (Sx, z)Z for any z ∈ Z. That is, R(C) ⊇ R(S).
Most controllability results for linear systems are proved by a method of estab-lishing the corresponding observability property of the dual system. We put the control problem of the 1-dim heat equation in terms of the abstract framework. X = L2(0, L), Y = L2(0, L), Z = L2(0, L), Su0 = u(x, T ) where u is the solution
of the following system with initial data u0.
(I.2) ut− uxx = 0 (0, L) × (0, T ), u(0, t) = u(L, t) = 0 (0, T ), u(x, 0) = u0(x) (0, L),
following system. (I.3) ut− uxx = 0 (0, L) × (0, T ), u(0, t) = 0 (0, T ), u(L, t) = h(t) (0, T ), u(x, 0) = 0 (0, L),
We calculate the adjoint operators of S and C respectively. Let u(x, t), v(x, t) be solutions to the system (I.2) with initial data u(x, 0) and v(x, 0), respectively. We have that ∂ ∂t Z L 0 u(x, t)v(x, T − t)dx = Z T 0 utv − uvtdx = Z ∆uv − u∆v
= ux(0, t)v(0, T −t)−ux(L, t)v(L, T −t)−u(0, t)vx(0, T −t)+u(L, t)vx(L, T −t) = 0,
(2.20) which implies that
Z L 0 u(x, T )v(x, 0) = Z L 0 u(x, 0)v(x, T ).
Thus S∗ = S. Using the same calculation with u(x, t) be a solution to (I.3) and v(x, t) be a solution to (I.2), we have
Z L 0 u(x, T )v(x, 0)dx − Z L 0 u(x, 0)v(x, T )dx = Z T 0 u(0, t)vx(0, T − t) − u(L, t)vx(L, T − t) dx = −h(t)vx(L, T − t). (2.21)
Hence C∗v0 = −vx(L, T − t), where v(x, t) is the solution to (I.2) with initial data
v0.
The controllability problem of the 1-dim heat equation is now reduced to the proof of the inequality ||C∗v0|| ≥ K||S∗v0||, with some constant K, that is,
Z T 0 vx(L, t)2dt ≥ K Z L 0 v(x, T )2dx, (2.22)
with v(x, t) being the solution to the equation (I.2) with initial data v0 ∈ L2(0, L).
proof of the observability
We apply a Carleman type estimate to show the observability (2.22) of the 1-dim heat equation. It is a modification of the estimate in the paper [14] of O. Y. Imanuvilov, (see Lemma 1.1). We prove a simplified version that is enough for our needs.
Lemma 2.1.6. Let ϕ, ψ be functions of the following form.
ϕ(x, t) = e
−x
t2(T − t)2, ϕ(x, t) =˜
e−x− eL
t2(T − t)2.
Then for any solution u of (I.2) the following estimate holds. Z T 0 Z L 0 s3ϕ3u2e2s ˜ϕdxdt ≤ c0 Z T 0 sϕ(L, t)ux(L, t)2e2s ˜ϕdt, , ∀s ≥ s0 (2.23)
Lv ≡ es ˜ϕ(∂tu − ∆u) = −∆v − s2|∇ ˜ϕ|2v − s∂tϕv + ∂˜ tv + 2s∇ ˜ϕ∇v + s∆ ˜ϕv. (2.24)
We decompose L into two parts L = S + A, where
Sv = −∆v − s2|∇ ˜ϕ|2v − s∂tϕv,˜ Av = ∂tv + 2s∇ ˜ϕ∇v + s∆ ˜ϕv. (2.25)
We use the following relation to show the inequality (2.23). Z |Lv|2dxdt = Z |Sv|2dxdt + Z |Av|2dxdt + 2 Z (Sv)(Av)dxdt, (2.26)
By simple calculations we have that
2 Z (Sv)(Av)dxdt = Z 4s ˜ϕ,klv,kv,ldxdt + Z 4s3ϕ˜,klϕ˜,kϕ˜,l− 2s2∂t|∇ ˜ϕ|2− s∆2ϕ + s∂˜ 2tϕ |v|˜ 2 dxdt + Z T 0 (−2sv2x(0, t) ˜ϕx(0, t) + 2svx2(L, t) ˜ϕx(L, t))dt. (2.27)
There is some positive constant s0 such that when s > s0 we have that
2 Z (Sv)(Av)dxdt ≥ Z s3ϕ3v2dxdt + Z T 0 (2sϕ(0, t)vx2(0, t) − 2sϕ(L, t)vx2(L, t))dt ≥ Z s3ϕ3v2dxdt − Z T 0 2sϕ(L, T )v2x(L, t)dt. (2.28)
Since Lv = 0 by the heat equation, we finally have that Z T 0 Z L 0 s3ϕ3u2e2s ˜ϕdxdt ≤ c0 Z T 0 sϕ(L, t)ux(L, t)2e2s ˜ϕdt. (2.29)
Now we are ready to proof the observability (2.22). It is easy to see that there exist two positive constant c1 and c2 that
ϕ(L, t)e2s ˜ϕ ≤ c1, t ∈ (0, T ), (2.30) and ϕ3(x, t)e2s ˜ϕ ≥ c2, (x, t) ∈ (0, L) × ( T 3, 2T 3 ). (2.31)
The Laplacian ∆ generates a strongly continuous semigroup et∆ for t ∈ (0, ∞) on
L2(0, L). There are positive constants M and µ ||et∆|| ≤ M eµt, t ≥ 0. Hence, for t ∈ (T3,2T3 ) Z L 0 u2(x, T )dx ≤ M2e2µ(T −t) Z L 0 u2(x, t)dx. By integration over (T3,2T3 ), T 3 Z L 0 u2(x, T )dx ≤ M2e4µT /3 Z 2T /3 T /3 Z L 0 u2(x, t)dxdt. (2.32) Combining (2.30) (2.31) (2.32), we have that
Z T 0 ux(L, t)2dt ≥ K Z L 0 u(x, T )2dx.
This is the end of the proof. 2
We have thus proved the null controllability of the system (I.1). For Neumann boundary control and inner control problems in 1-dim and higher, one can use the same strategy by proving the observability by suitable Carleman inequalities.
2.1.3
Two Backward uniqueness results via Fourier
Trans-form
Theorem 2.1.7 and the proof is taken from [25]. We introduce additional notation:
Rn+= {x ∈ R
n||x = (x
1, x2, ..., xn), xn > 0},
QT = Rn× (0, T ),
where T is a positive fixed number.
Theorem 2.1.7. Let u : QT → R be a bounded smooth function satisfying the
heat equation ∂tu = ∆u in QT. Assume that there exists a non-empty open set
Ω ⊂ Rn+ such that lim t→T −0 Z Ω |u(x, t)|dx = 0. Then u ≡ 0 in QT.
Proof. Using the known regularity theory for the heat equation and the fact that smooth solutions to the heat equation are analytic in spatial variables, we see that one can extend u by zero to the set Q = Rn× (0, ∞), and the extension,
also denoted by u, is smooth, satisfies the heat equation in Q, and vanishes for t ≥ T . Also, replacing u(x, t) by u(x1, x2, ..., xn−1, xn+ yn, t + s) for small yn> 0
continuous in the closure ¯Q of Q. Making these simplifying assumptions, we will now prove the theorem in several steps.
Step 1. Reduction to the case n = 1. The obvious idea here is to use the Fourier transformation along x0 = (x1, x2, ..., xn−1). For each t > 0 and xn ≥ 0,
we define a distribution ˜u(·, xn, t) on Rn−1 by
< ˜u(·, xn, t), φ(·). = Z n−1 R dx0u(x0, xn, t) Z Rn−1 dξ0et|ξ|2−ix0·ξ0φ(ξ0) Here, φ ∈ C0∞(Rn−1) and ξ = (ξ
1, ξ2, ...ξn−1). Under suitable assumptions,
˜
u(·, xn, t) is a function and we have
˜
u(ξ0, xn, t) = et|ξ
0|2Z
Rn−1
u(x0, xn, t)e−ix
0·ξ0
dx0.
A simple calculation shows that, for each fixed φ ∈ C0∞(Rn−1), the function
˜
u(xn, t) =< ˜u(·, xn, t), φ(·) >
is bounded in R+ × (0, +∞), satisfies the heat equation in R+ × (0, +∞), and
vanishes for t > T . We now see that it is enough to prove the case n = 1.
In what follows, we use the notation Q = R+× (0, +∞), and (x, t) stands for
points of Q.
Step 2. Reduction to the case |u(x, t)| ≤ Ce−αx. This can be achieved by the following change of variables:
This function v is, of course, defined in a domain different from QT , but we can
obviously achieve by a suitable shift that the domain of v contains a domain of the form Q in which the theorem is violated, if v does not identically vanish. Moreover, v has the required decay as x → ∞.
Step 3. Proof in the case n = 1 and |u(x, t)| < Ce−αx. We extend u to all R × R by requiring that the extension be an even function of x vanishing for t ∈ (−∞, 0) ∪ (T, +∞). The extended function has a discontinuity in t at t = 0, but it is smooth in t (for a fixed x) when t ∈ (0, ∞).
Let a(x) = u(x, 0) and let
g(t) = lim x→0+02 ∂u ∂x(x, t). Clearly, ∂u ∂t(x, t) − ∂2u ∂t (x, t) = −δ(x)g(t) + δ(t)a(x), x ∈ R, , t ∈ R, (2.33) where δ denotes the Dirac distribution.
Denoting by ˆg and ˆa the Fourier transformations of g and a, respectively, we note that (2.33) implies
ˆ
g(iξ2) = ˆa(ξ), ξ ∈ R. (2.34)
We now look at the function ˆg in more detail. We have
g(τ ) = Z R g(t)e−itτdt = Z T 0 g(t)e−itτdt (2.35)
Hence, ˆg is defined for each τ ∈ C and is holomorphic in C. Moreover, standard calculations together with (2.34) and the decay property of a imply that
ˆ
g(τ ) = O( 1
|τ |) as τ → ∞ in K, (2.36)
where K = {τ ∈ C || Re τ = 0 or Im τ = 0}, i.e., K is the union of the real and imaginary axes. In fact, (2.36) along the real axis and along the negative part of the imaginary axis follows from (2.35) and integration by parts. (2.36) along the positive part of imaginary axis follows from (2.34) and the fact that
ˆ
a(ξ) = O( 1
|ξ|2) as ξ → ∞ and ξ ∈ R.
Step 4. The last step in the proof is simple lemma about holomorphic func-tions.
Lemma 2.1.8. Let K ⊂ C be the union of the real and imaginary axes. Let f : C → C be a holomorphic function satisfying two conditions:
|f (z)| ≤ Aea|z|,
z ∈ C for some positive constants a and A, and
f (τ ) = O 1 |τ |
, as τ → ∞ in K.
Lemma 4.2 can be easily obtained from the Phragm´en-Lindel¨of theorem for an angle (for details, see Lemma 2.1.11, or [21], Theorem 7.5).
The next proposition says that if a solution u to the heat equation is 0 on both ends of the time interval [a, b] of the first quadrant, then it must be identically 0 in between.
Theorem 2.1.9. Let O be the first quadrant of R2. The bounded function u(x, y, t) is defined on O ×(0, T ) and satisfies the backward heat equation and the conditions
ut+ 4u = 0, O × (0, T ) (2.37)
u(x, y, 0) = u(x, y, T ) ≡ 0 (x, y) ∈ O. (2.38)
Then u ≡ 0 in O × [0, T ].
Proof. Replacing (x, y, t) by (x + 1, y + 1, t), we can assume that all derivatives of u are well-defined.
Step 1. Reduction to the case |u(x, y, t)| < ce−c(x2+y2). We take a series of change of variables and still denote the function by u. Replacing t by −t, u satisfies ut+ ∆u = 0 in O × (−T, 0). Shifting the time interval by T0 > T , we
have that u satisfies
ut+ 4u = 0, O × (−T + T0, T0)
We apply the Appell transform of u, v(x, y, t) = 1 4πte −x2+y24t v x t, y t, 1 t . (2.39)
The transformed function satisfies the heat equation in O × (−T1
0, −
1
−T +T0) and
has the desired decay.
Step 2. Reduction to the equation ∆v − v = 0. Now suppose that u(x, y, t) satisfies the heat equation in O × (0, T ), with the decay
|u(x, y, t)| < c e−c(x2+y2) (2.40) for some c > 0. In addition u(x, y, 0) = u(x, y, T ) = 0.
We apply the Laplace transform to u.
v(x, y, s) = Z ∞
0
e−stu(x, y, t)dt. (2.41)
The function v is analytic in s ∈ C because u is compact supported in t. By direct calculation, v satisfies the equation
4v − sv = 0, (x, y) ∈ O, (2.42)
and the decay |v| < ce−c(x2+y2)
for all s with Re s > 0. Of course c depend on s. If we can prove that v(x, y, 1) = 0 in O, then v(x, y, s) = 0, for all s ∈ R and s > 0. By analyticity of v(x, y, s) in s, we have that v(x, y, s) ≡ 0 in O × C. Hence u(x, y, t) ≡ 0 in O × (0, T ).
Step 3. Proof of uniqueness of the equation ∆v − v = 0. Suppose that smooth function v(x, y) satisfies the equation
4v − v = 0, (x, y) ∈ O. (2.43)
In addition for some constant c > 0, |v| < c e−c(x2+y2)
. We want to show that v ≡ 0.
For simplicity in writing we assume c = 1. We extend v(x, y) to the whole plane R2 by v(x, y) = v(|x|, |y|), for (x, y) ∈ R2\ O. (2.44) Let f (x) = −2 lim y→0+ ∂v
∂y(x, y) g(y) = −2 limx→0+
∂v
∂x(x, y) (2.45)
The extended function v(x, y) satisfies in distribution sense the equation
4v − v = f (x)δ(y) + g(y)δ(x), (x, y) ∈ O. (2.46)
Apply Fourier transform to the equation we have
−(ξ2+ η2+ 1)˜v(ξ, η) = ˜f (ξ) + ˜g(η). (2.47)
We now look at the function ˜f in more detail. We have
˜ f (ξ) =
Z
R
The decay of f (x) implies that
| ˜f (ξ)| ≤ Z
e−x2ex Im ξdx ≤ Z
e−(x−Im ξ/2)2+(Im ξ)2/4dx < c e(Im ξ)2/4 (2.49) for some absolute constant c. In the same way we see that ˜f (ξ) is analytic in ξ ∈ C. Moreover, it follows from (2.48) and integration by parts that the function
˜
f (ξ) = O( 1
|ξ|2), as ξ → ∞, for ξ ∈ R.
Similarly, the function ˜g(η) is analytic and
˜
g(η) = O( 1
|η|2), as η → ∞, for η ∈ R. (2.50)
From (2.47), we have that
˜
f (ξ) = O( 1
|ξ|2), as ξ → ∞, for ξ ∈ K, (2.51)
where K = {ξ ∈ C, Re ξ = 0, or Im ξ = 0}, and
| ˜f (ξ)| < c e(Im ξ)2/4. (2.52) Step 4. The last step in the proof is a lemma about holomorphic functions. Lemma 2.1.10. Let K ⊂ C be the union of the real and imaginary axes. Let h : C → C be a holomorphic function satisfying two conditions:
|h(z)| ≤ c e(Im ξ)2/4
for some positive constants c, and h(τ ) = O 1 |τ | , as τ → ∞ in K. Then h ≡ 0.
If the rate of increasing of h is bounded by e|z|α for α < 2, then u is bounded directly from Phragm´en-Lindel¨of theorem. We have α = 2. However, the bound in solely in Im z. By a simple trick we can also apply Phragm´en-Lindel¨of theorem in this case. We notice that for a fixed σ > 0, if θ greater than and close to 0, the function
eiσz2f (z),˜ z = x + iy
is bounded by some C (independent of σ or θ ) on the boundary of the angle
Gθ = {z ∈ C, θ < arg z < π/2},
and is bounded by C eC|z|2 in the interior of the angle Gθ. Apply the Phragm´
en-Lindel¨of theorem, see Lemma 2.1.11, we have that |eiσz2f (z)| ≤ C on G˜
θ. Passing
θ → 0 and then σ → 0, it follows that |h(z)| < C in the first quadrant. In the same way h is bounded in other quadrants. By Liouville theorem, h is a constant, and this constant has to be zero due to the decay on the axes.
of an angle of απ radians, (0 < α ≤ 2), with boundary Γ, and let f (z) be ana-lytic on G, continuous up to the boundary. Suppose f (z) satisfies the following conditions: 1. f (z) ≤ C < ∞ on Γ. 2. lim inf r→∞ ln M (r) r1/α ≤ 0, (2.53) where M (r) = sup |z|=r,z∈G |f (z)|. Then |f (z)| ≤ C. (2.54)
2.1.4
Counterexample to BU in cones of angle less than
π/2
Recall that Oθ is defined as a cone with opening angle θ.
Oθ = {x = (x1, x0), x0 ∈ Rn−1, x1 > |x| cos(θ/2)}. (2.55)
Luis Escauriaza [5] showed that (BU) fails when θ < π/2. We briefly recall the counterexample in this section. Let us denote by Γ the standard heat kernel,
i. e. Γ(x, t) = (4πt)−n/2exp (−|x|2/4t), and recall Appell transformation
u(x, t) = Γ(x, t)v(y, s), y = x
t, s = 1
t . (2.56)
This transformation takes the solutions u(x, t) of the heat equation into the solutions v(y, s) of the backward heat equation
vs+ ∆v = 0 . (2.57)
By taking u(x, t) = h(x) for a suitable harmonic function h in Oθ, we can get
a counterexample to the backward heat equation form of (BU) for θ < π/2. In dimension 2 we can use the real part of the holomorphic function z → exp(−Azα) (for suitable A > 0 and a parameter α > 2) to obtain an explicit formula:
v(y, s) = Re 1 s exp(−A (y1+ iy2)α sα + |y|2 4s ) . (2.58)
The function v(y1, y2, s) is bounded in a sector that | arctanyy2
1| < π/(2α), away
from the origin. We can shift it to v(y1+1, y2, s). The resulted function is bounded
in a sector with angle π/α satisfying the backward heat equation vs+ 4v = 0,
and v(·, ·, 0) = 0. We remark that π/2 is the borderline of the construction above. The key element of the construction is the existence of a harmonic function in cones with the right decay rates, e−|z|α for some α > 2. For cones with angle smaller than π/2, one can simply take the real part of e−|z|α as above. However,
for cones with angle greater than or equal to π/2, such a harmonic function does not exist.
We note that it is enough to construct a counterexample in dimension n = 2. The higher-dimensional example can then be constructed by simply consider-ing the two-dimensional function as a function of n variables, independent of x3, . . . , xn.
2.2
Backward uniqueness and Carleman
inequal-ities
In this section we prove the main result of this chapter. Let us recall the some settings and the results first. Consider an open set Ω ⊂ Rn. Let u be a bounded
solution of the equation
ut− ∆u + b(x, t) · ∇u + c(x, t)u = 0 in Ω × (0, T ), (2.59)
where the coefficients b = (b1, . . . , bn), c are measurable and bounded. Recall that
Ω has the backward uniqueness property if the following statement holds:
(BU) If a bounded u : Ω × (0, T ) → R satisfies (2.59) and u( · , T ) = 0, then u ≡ 0 in Ω × (0, T ).
From the results in [9], (BU) holds for θ ≥ π. On the other hand, the con-struction of Escauriaza shows that (BU) fails for θ < π/2. It is easy to see that there exists some critical angle θ0 such that (BU) holds for θ > θ0 and (BU) fails
for θ0.
Theorem 2.2.1. The cones Oθ satisfy (BU) for
θ > 2 arccos(1/√3) ∼ 109.52◦.
In other words, the critical angle θ0 introduced above satisfies
θ0 ≤ 2 arccos(1/
√ 3).
Our proof of Theorem (2.2.1) relies on two Carleman-type inequalities, along lines similar to [9]. The first inequality, Proposition 2.2.2, is taken from [9] and is applied in the same way to obtain fast decay rates for the solutions, see Lemma 2.2.3. We note that Carleman inequalities of this type can be found already in [6, 10, 7, 8].
The second inequality, Proposition 2.2.4, is the main new tool used in our proof. The heuristics behind this inequality is somewhat similar to the heuristics behind the second Carleman-type inequality in [9] (Proposition 6.2). However, the proof of Proposition 2.2.4 requires a new idea, as for θ < π certain critical terms appearing in the proofs of the inequalities lose convexity.
In addition to determining the critical angle, another interesting open problem is to optimize the assumptions on the coefficient b(x, t) and c(x, t), in the spirit of [16]. For example, it is conceivable that the result remains true for b ∈ Ln+2x,t and c ∈ L(n+2)/2x,t , but it might be a difficult problem to decide whether this is the case.
In what follows we will work with the inequality
|ut− ∆u| ≤ c1(|∇u| + |u|) (2.60)
rather than (2.59). It is not hard to see that when assuming the boundedness of b and c, the two formulations are equivalent.
2.2.1
Backward uniqueness
Without loss of generality we assume T = 1 and work with the backward form of (2.60). Recall that
Oθ = {x = (x1, x0), x0 ∈ Rn−1, x1 > |x| cos(θ/2)}. (2.61)
Suppose that u(x, t) is a solution to the backward heat equation for some θ > 2 arccos(1/√3).
|ut+ ∆u| ≤ c1(|∇u| + |u|) in Oθ×(0, 1), (2.62)
In addition,
|u| < M in Oθ×(0, 1). (2.64)
Then u ≡ 0.
To prove the above statement we firstly need the following Carleman inequality from [9] (Proposition 6.1), by which we obtain a decay result for solutions of backward heat equation.
Proposition 2.2.2 ([9]). For any function u ∈ C0∞(Rn × (0, 2); Rn) and any
positive number a, Z Rn×(0,2) h−2a(t)e−|x|24t a t|u| 2+ |∇u|2dxdt ≤ c0 Z Rn×(0,2) h−2a(t)e−|x|24t |∂ tu + ∆u|2dxdt, (2.65)
where c0 is a positive absolute constant and h(t) = te
1−t 3 .
Lemma 2.2.3 below immediately implies exponential decay of the solution u satisfying (2.62)-(2.64). The proof is by using the Carleman inequality in Propo-sition 2.2.2. The decay of u enables us to apply the Carleman inequality in Proposition 2.2.4 and reach the conclusion in Theorem 2.2.1.
Lemma 2.2.3. Let BR denote the ball with radius R in Rn. Assume that R >
constants c1 and M .
|ut+ 4u| ≤ c1(|∇u| + |u|) in BR× (0, T ), (2.66)
u(x, 0) = 0 in BR, (2.67)
|u| < M in BR× (0, T ). (2.68)
Then there exists some constants β, γ, such that for t ∈ (0, γ),
u(0, t) ≤ c2
min{1, T }M e
−βR2
t , (2.69)
where β is a small enough absolute constant, c2 depends on c1, γ depends on c1
and T .
Notice that the constants β, γ and c2 do not depend on R. It follows
immedi-ately the exponential decay of the bounded solution to the equation (2.60) with respect to the distance to the lateral boundary. We will give the proof of the lemma in the next section.
The following Carleman inequality in Proposition 2.2.2 is a key tool used in our proof of the backward uniqueness in cones. We define the set
Qθ = (Oθ∩{x1 > 1}) × (0, 1).
The purpose of “cutting off the corner” is to avoid singularities at the origin. Proposition 2.2.4. Let φ(x, t) = aΛ(t)ϕ(x)+t2, where Λ(t) = 1 − t
tα/2 , and ϕ(x) =
xα
1 − εαrα, where r = |x|, and ε = cos(θ/2). For any ε ∈ (0, 1/
√
θ ∈ (2 arccos(1/√3), π), there exists some α = α(ε) ∈ (1, 2) such that the following inequality holds for u ∈ C0∞(Qθ) and a > a0 for some constant a0.
Z Qθ e2φ(x,t)a (Λ(t) + ϕ(x)) u2+ |∇u|2 dxdt ≤ 4 Z Qθ e2φ(x,t)|∂tu + ∆u|2dxdt. (2.70)
We apply this Carleman inequality to prove the main result of the paper in the remaining part of this section. The proof of Proposition 2.2.4 is postponed to Section 2.4.
For x ∈ Oθ we denote by dθ(x) the distance between x and the boundary of
Oθ, explicitly given by
dθ(x) = x1sin(θ/2) − |x0| cos(θ/2). (2.71)
Let Oθ+2 = {x ∈ Oθ | dθ(x) > 2}. With any other number c, the set Oθ+c is
defined in the same way.
The next lemma is a consequence of the decay result from Lemma 2.2.3 and Proposition 2.2.4. It implies Theorem 2.2.1 immediately.
Lemma 2.2.5. Assume that for some θ ∈ (2 arccos(1/√3), π) a function u satis-fies (2.62) – (2.64), then there is a number γ1(c1) such that
u(x, t) ≡ 0 (2.72)
Proof. Lemma 2.2.3 implies that
|u(x, t)| ≤ c2M e−β
d2θ(x)
t (2.73)
for all (x, t) ∈ Oθ+2×(0, γ). By local gradient estimates for the heat equation [20]
we can assume that
|u(x, t)| + |∇u(x, t)| ≤ c3M e−
βd2 θ(x)
2t (2.74)
for all (x, t) ∈ Oθ+3×(0, γ/2].
By scaling we define a function v by
v(y, s) = u(λy, λs2− γ1) (2.75)
for (y, s) ∈ Oθ×(0, 1) with λ =
√
2γ1. This function satisfies the relations
|∂sv + ∆v| ≤ c1λ(|∇v| + |v|) in Oθ×(0, 1), (2.76) v(y, s) = 0 in Oθ×(0, 1/2), (2.77) and |v(y, s)| + |∇v(y, s)| ≤ c3M e −βλ2d2θ(y) 2(λ2s−γ1) ≤ c3M e−β d2 θ(y) 2s (2.78)
for 1/2 < s < 1 and y ∈ Oθ+3/λ = {y ∈ Oθ | dθ(y) > 3/λ}.
To apply Proposition 2.2.4, we need certain decay of |v(y, s)| when |y| is large. Notice that the preferred decay can be obtained by considering a cone
with slightly small opening. Proposition 2.2.4 holds for angles in the interval (2 arccos(1/√3), π). We thus consider the median of θ and 2 arccos(1/√3),
δ = θ + 2 arccos(1/ √
3)
2 .
In the smaller cone Oδ = {x ∈ Rn, x1 > |x| cos(δ/2)} we have the estimate
dθ(y) ≥ |y| sin θ−δ2 . It follows (2.78) that
|v(y, s)| + |∇v(y, s)| ≤ c3M e−β
0 |y|2
s (2.79)
for 1/2 < s < 1 and y ∈ Oδ∩ Oθ+3/λ, with the constant β0 = β sin2(θ−δ2 )/2. We
can further have
|v(y, s)| + |∇v(y, s)| ≤ c03M e−β0 |y|
2
s (2.80)
for 1/2 < s < 1 and y ∈ Oδ∩{y1 > 3/λ} for some other constant c03.
Next we work on the smaller cone Oδwith opening δ, where we have
exponen-tial decay (2.80) and the following properties inherited from Oθ.
|∂sv + ∆v| ≤ c1λ(|∇v| + |v|) in Oδ×(0, 1), (2.81)
v(y, s) = 0 in Oδ×(0, 1/2). (2.82)
Proposition 2.2.4 requires support condition for the Carleman inequality. For that purpose, let us fix two smooth cut-off functions such that
ψ1(y1) = 0, y1 < 3/λ + 2, 1, y1 > 3/λ + 3,
ψ2(τ ) = 0, τ < −3/4, 1, τ > −1/2. We set (for the definition of φ, see Proposition 2.2.4)
φB(y, s) = 1 aφ(y, s) − B = (1 − s) yα 1 − εα|y|α sα/2 + s2 a − B, where ε = cos(δ/2), B = 2 aφ(yλ, 1 2), with yλ = (3/λ + 3, 0, · · · , 0) and η(y, s) = ψ1(y1)ψ2( φB
B ), w(y, s) = η(y, s)v(y, s).
The function w is not compact supported in Qδ = (Oδ∩{x1 > 1}) × (0, 1).
How-ever, it follows from (2.78) and the special structure of the weight function φ in Proposition 2.2.4 that, with w replacing u in (2.70), integrals on both sides converge. If we multiply w by an additional cut-off function ξ such that
ξ(x) = 1, τ < R 0, τ > 2R
and |∇ξ| < c/R, |∇2ξ| < c/R2, then apply Proposition 2.2.4 to the compact
supported function wξ, and let R → ∞, we finally obtain
Z Qδ e2aφBa (Λ(s) + ϕ) w2+ |∇w|2 dyds ≤ 4 Z Qδ e2aφB|∂ sw + ∆w|2dyds. (2.83)
From (2.81) we have
|∂sw + 4w| ≤ c1λ(|∇w| + |w|)
+c4(|∇v| + |v|)(|∂sη + |∇η| + |∆η|). (2.84)
We can see that for a large enough a(Λ(s) + ϕ(y)) > 1 in the support of w, where φB/B ≥ −3/4 because of the definition of ψ2. In addition, we take γ1(c1) small
enough such that 16c21λ2 < 1/2. We then have
I ≡ Z Qδ e2aφB w2+ |∇w|2 dyds (2.85) ≤ 32c24 Z Qδ e2aφB(|v|2+ |∇v|2)(|∂ sη| + |∇η| + |∆η|)2dyds. (2.86)
To estimate the right hand side, we need to look into details of derivatives of η. In view of the definitions of ψ1 and ψ2, the support of derivatives of η(y, s) is
the closure of the set
{y1 > 3/λ + 2, −3B/4 < φB < −B/2}
∪ {3/λ + 2 < y1 < 3/λ + 3, φB > −B/2}.
However, the second set has empty intersection with Oδ×(1/2, 1), where the
func-tion v is nonzero. Hence the support of the term (|∇v| + |v|)(|∂sη + |∇η| + |∆η|)
is closure of the set
of which we denote by χ(y, s) the characteristic function.
Next we estimate the derivatives of η(y, s) in the support of the term (|∇v| + |v|)(|∂sη + |∇η| + |∆η|). Recall that φ(y, s) = aΛ(s)ϕ(y) + s2, where Λ(s) =
1 − s sα/2
and ϕ(y) = yα
1 − εα|y|α. The function Λ(s) and the derivative Λ
0(s) are bounded
for s ∈ (1/2, 1). The function ϕ and its derivatives up to the second order are bounded by (const. |y|α). The cut-off functions ψ
1and ψ2and derivatives up to the
second order are bounded by some absolute constant. The value of B is bounded from below regardless of the value of the parameter a. Thus
(|∂sη| + |∇η| + |∆η|)2 < c5|y|2α. (2.87)
We now estimate (2.86) by using (2.79) and (2.87). We see that
I ≤ c6M e−Ba Z Qδ |y|2αe−2β0 |y| 2 s χ(y, s)dyds
for some constant c6. The last integral is bounded. Passing to the limit as a → ∞
we see that v(y, s) = 0 where φB(y, s) > 0 and 1/2 < s < 1. Using the property
of unique continuation across the spatial boundaries (see Theorem 4.1 in [9]), we show that v(y, s) = 0 if y ∈ Oθ and 0 < s < 1. This proves the lemma.
2.3
Proof of Lemma 2.2
The proof of Lemma 2.2.3 is based on the Carleman inequality in [9], which we quoted in Proposition 2.2.2.
Proof of Lemma 2.2.3. The proof is similar to the one in [9]. We still include the proof here for the convenience of the reader.
In what follows, we always assume that the function u is extended by zero to negative values of t.
The assumption that R > 2 results in no loss, since the conclusion is only useful when R is large. According to the local gradient estimates of the heat equation [20], in the smaller cylinder (x, t) ∈ BR−1× (0, T /2), we can assume that
|u| + |∇u(x, t)| ≤ c7
min{1, T }M (2.88)
with some absolute constant c7.
We fix t ∈ (0, min{1, T }/12) and introduce a new function v by the usual parabolic scaling:
v(y, s) = u(λy, λ2s − t/2).
The function v is well defined on the set Qρ = Bρ× (0, 2), where ρ = (R − 1)/λ
and λ =√3t ∈ (0, min{1,√T }/2). We have the following relations for v.
|v(y, s)| + |∇v(y, s)| < c7
min{1, T }M (2.90)
for all (y, s) ∈ Qρ,
v(y, s) = 0 (2.91)
for (y, s) ∈ Bρ× (0, 1/6].
By the assumption that R > 2 and λ < 1/2, we have ρ > 2. In order to apply Proposition 2.2.2, we take two smooth cut-off functions in Qρ:
ψρ(y) = 0, |y| > ρ − 1/2, 1, |y| < ρ − 1, ψt(s) = 0, 7/4 < s < 2, 1, 0 < s < 3/2.
By assumption, these functions take values in [0, 1] and are such that |∇kψ
ρ| < Ck,
k = 1, 2, and |∂sψt| < C0. We set η(y, s) = ψρ(y)ψt(s) and
w(y, s) = η(y, s)v(y, s). (2.92)
It follows from (2.89) that
|∂sw + ∆w| ≤ c1λ(|∇w| + |w|) + c8χ(|∇v| + |v|), (2.93)
where c8is a positive constant depending only on c1 and Ck, k = 0, 1, 2, χ(y, s) = 1
χ(y, s) = 0 for (y, s) /∈ ω. The set ω is where the cut-off function η is not constantly 1 in Qρ. Obviously, the function w is compactly supported on R2× (0, 2). The
inequality (2.65) also holds for scale valued functions. Therefore we may apply Proposition 2.2.2 and obtain
Z
Qρ
h−2a(s)e−|y|24s
a s|w| 2+ |∇w|2dyds ≤ c0 Z Qρ
h−2a(s)e−|y|24s |∂sw + ∆w|2dyds. (2.94)
Taking a > 2, and applying (2.93) we finally obtain that
I ≡ Z
Qρ
h−2a(s)e−|y|24s (|w|2+ |∇w|2)dyds ≤ 4c0(c2
1λ 2 I + c28I1), (2.95) where I1 = Z Qρ
χ(y, s)h−2a(s)e−|y|24s (|∇v|2+ |v|2)dyds.
Taking a sufficiently small value for γ = γ(c1) such that in the range λ ∈ (0, γ), we
can assume that the inequality 4c0c21λ2 ≤ 1/2 holds, and therefore (2.95) implies
that
I ≤ 8c0c28I1. (2.96)
Notice that near the origin {y = 0, s = 0}, where the parametric function h−2a(s)e−|y|24s is not integrable, the characteristic function χ is 0. By (2.90) we
have I1 ≤ c2 7M2 min{1, T2} Z 2 3/2 Z |y|<ρ−1
h−2a(s)e−|y|24s dyds
+ Z 2
0
Z
ρ−1<|y|<ρ
h−2a(s)e−|y|24s dyds
(2.97) ≤ c9M 2 min{1, T2} h−2a(3/2) + Z 2 0 h−2a(s)e−(ρ−1)24s ds , (2.98)
where c9 is an absolute constant.
Using (2.98) we obtain the estimate
D ≡ Z B1 Z 1 1/2 |w|2dyds = Z B1 Z 1 1/2 |v|2dyds ≤ c10 Z Qρ
h−2a(s)e−|y|24s (|w|2+ |∇w|2)dyds
≤ c11M 2 min{1, T2} h−2a(3/2) + Z 2 0 h−2a(s)e−16sρ2 ds = c11M 2 min{1, T2}e −βρ2 h−2a(3/2)eβρ2 + Z 2 0 h−2a(s)eβρ2−16sρ2 ds .
We take β < 1/64 and then let
a = βρ2/(2 log h(3/2)).
This choice of a leads to the estimate
D ≤ c11e−βρ 2 1 + Z 2 0 g(s)ds ,
where g(s) = h−2a(s)e−32sρ2 . By simple calculation we have that
g0(s) = h−2a(s)e−32sρ2 − βρ 2 log h(3/2) 1 s − 1 3 + ρ 2 32s2 .
One can readily verify that g(2) < 1 and g0(s) ≥ 0 for any s ∈ (0, 2) if β < 1 64log h(3/2). Therefore, D ≤ 3 c11M 2 min{1, T2}e −βρ2 = 3 c11M 2 min{1, T2}e −β12tR2 .
On the other hand, the regularity theory implies that
|u(0, t)|2 = |v(0, 1/2)|2 ≤ c 12D.
Finally we obtain that
|u(0, t)| ≤ c2
min{1, T }M e
−βR224t.
2.4
Proof of the Carleman inequality
One of the difficulties in the proof of the Carleman inequality in Proposition 2.2.4 is that – by comparison with the case θ = π – some loss of the convexity of the weight ϕ cannot be avoided. Therefore we have to investigate in more detail some of the terms which can be neglected when θ ≥ π.
Proof of Proposition 2.2.4. Let u be an arbitrary function in C0∞(Qθ) and v = eφu.
Then
We decompose L into symmetric and skew-symmetric parts L = S + A, where Sv = ∆v + |∇φ|2v − ∂tφv (2.100) and Av = ∂tv − 2∇φ∇v − ∆φv. (2.101)
The right hand side of the inequality (2.70) is Z |Lv|2dxdt = Z |Sv|2dxdt + Z |Av|2dxdt + Z ([S, A]v)vdxdt, (2.102)
where [S, A] = SA − AS is the commutator of S and A. By simple calculations we have that ([S, A]v, v) = Z 4φ,klv,kv,ldxdt (2.103) + Z 2∇φ∇|∇φ|2− ∆2φ + ∂2 tφ − 2∂t|∇φ|2 |v|2dxdt.(2.104)
The Hessian of the function φ = aΛ(t)(xα
1 − εαrα) is not positive-definite. To
compensate the term R 4φ,klv,kv,ldxdt in (2.103) we introduce a function F (x, t)
to be determined. (Sv, F v) = Z ∆vF v + (|∇φ|2− ∂tφ)F v2dx = Z −F |∇v|2+ (1 24F + |∇φ| 2F − ∂ tφF )v2dxdt.
Cauchy-Schwartz inequality implies that (Sv, Sv) ≥ −(Sv, F v) −1 4 Z F2v2dxdt ≥ Z F |∇v|2− (1 24F + |∇φ| 2F − ∂ tφF + 1 4F 2)v2dxdt. (2.105)
Combining (2.103) and (2.105) we have
([S, A]v, v) + (Sv, Sv) ≥ Z 4φ,klv,kv,l+ F |∇v|2dxdt (2.106) + Z 2∇φ∇|∇φ|2− ∆2φ + ∂2 tφ − 2∂t|∇φ|2 v2dxdt (2.107) + Z −(1 24F + |∇φ| 2F − ∂ tφF + 1 4F 2)v2dxdt. (2.108)
By calculation the Hessian of ϕ is
D2ϕ(x) = α(α − 1) xα−21 0 · · · 0 0 0 · · · 0 .. . ... . .. ... 0 0 · · · 0 − α εαrα−2E n (2.109) +α(2 − α) εαrα−4xTx, (2.110)
where En reprensents the n dimensional identity matrix, x = (x1, . . . , xn) is the
row vector and xT denotes the transpose of x. It is easy to see that
D2ϕ(x) + α εαrα−2En≥ 0.
We thus let f (x) = α εαrα−2 and
With this choice of F (x, t), the right hand side of line (2.106) is positive and Z
4φ,klv,kv,l+ F |∇v|2dxdt ≥
Z
|∇v|2dxdt.
Grouping the remaining terms according to the orders of the parameter a, with A3 denoting the terms with a3 and etc., we have that
([S, A]v, v) + (Sv, Sv) ≥ Z |∇v|2dxdt + Z (A3+ A2+ A1+ A0)v2dxdt, (2.112) where A3 = 2∇φ∇|∇φ|2− 4a3Λ3f |∇ϕ|2, (2.113) A2 = −4a2ΛΛ0|∇ϕ|2+ 4a2ΛΛ0ϕf − 4a2Λ2(t)f2− a2Λ2|∇ϕ|2, (2.114)
A1 = −aΛ∆2ϕ + aΛ00ϕ − 2aΛ(t)∆f + aΛ0ϕ − 2aΛ(t)f + 8atΛf, (2.115)
A0 = 7/4 + 2t. (2.116)
We analyze A3 first. With a3 as a coefficient A3 must be non-negative in the
set Qθ. By letting x1/r → ε, we see easily that ε < 1/
√
3 is a necessary condition for A3 ≥ 0. Next we show that under the condition ε < 1/
√
3, we indeed have that A3 ≥ 0. Denoting by ∇ϕT the transpose of the row vector ∇ϕ, we notice
that
It is easy to see that D2ϕ(x) − α εαrα−2 ≥ α(α − 1) xα−21 0 · · · 0 0 0 · · · 0 .. . ... . .. ... 0 0 · · · 0 − 2α εαrα−2En. (2.118)
By the fact that xα−21 > rα−2, we have
D2ϕ(x) − α εαrα−2 ≥ αrα−2 α − 1 − 2 εα 0 0 −2 εαE n−1 , (2.119)
where En−1 is the n − 1 dimensional identity matrix.
The first derivatives of φ are as follows.
ϕ,1(x, t) = αxα−11 − α εαrα−2x1, ϕ,k(x, t) = −α εαrα−2xk, k = 2, . . . , n.
We notice that ϕ,1≥ α(1 − εα)xα−11 . Thus
A3 ≥ 4a3Λ3(t)α3rα−2(α − 1 − 2 εα)(1 − εα)2x2α−21 − 2 ε 3α
r2α−4|x0|2 .
Taking into account that x1/r > ε and |x0|2/r2 < 1 − ε2,
A3 ≥ 4a3Λ3(t)α3r3α−4ε2α−2(α − 1 − 2 εα)(1 − εα)2− 2 εα+2(1 − ε2) .
Let us denote the quantity in the bracket above by m(α, ε).
A3 is non-negative if m(α, ε) is. In the set (α, ε) ∈ (1, 2) × (0, 1/
√
3), the function m(α, ε) is monotone increasing with respect to α and is monotone decreasing with respect to ε. Notice that
m(2, 1/√3) = 0.
Hence for any ε smaller than and close to 1/√3, we can choose a corresponding α < 2 such that m(α, ε) ≥ 0, and in turn A3 ≥ 0. We fix this α in the rest of the
proof.
For the estimate of A2 we notice that ∇v = (∇φ)v + eφ∇u, as v = eφu. To
bound |eφ∇u|, we want to apply the inequality |eφ∇u|2/2 ≤ |∇v|2+ |∇φ|2v2. We
thus look at the inequality (2.112) in the following way.
([S, A]v, v) + (Sv, Sv) ≥ Z (|∇v|2+ |∇φ|2v2)dxdt + Z [A3+ (A2 − |∇φ|2) + A1+ A0]v2dxdt. Next we estimate A2− |∇φ|2. A2− |∇φ|2 = −4a2Λ(t)Λ0(t) 1 + Λ(t) 2Λ0(t) |∇ϕ|2− ϕf + Λ(t) Λ0(t)f 2 . (2.120) Λ(t) = t1−tα/2 and Λ 0(t) = −α/2+(1−α/2)t tα/2+1 . |Λ(t)/Λ 0(t)| < 1 2α. A2− |∇φ|2 ≥ −4a2Λ(t)Λ0(t) 1 − 1 4α |∇ϕ|2− ϕf − 1 2αf 2 . (2.121) Notice that |∇ϕ|2 ≥ |ϕ ,1|2 ≥ α2(1 − εα)2x2α−21 and ϕf ≤ α εα(1 − εα)x 2α−2 1 . For
Taking into account that r ≥ x1 > 1, A2− |∇φ|2 ≥ −4a2Λ(t)Λ0(t) 3 8x 2α−2 1 − 1 4r 2α−4 ≥ −a 2 2Λ(t)Λ 0 (t)x2α−21 . (2.122)
Finally, we estimate A1. Recall that
A1 = −aΛ∆2ϕ + aΛ00ϕ − 2aΛ(t)∆f + aΛ0ϕ − 2aΛ(t)f + 8atΛf.
A simple observation is that
A1 ≥ a(Λ00ϕ + aΛ0)ϕ − aΛ(∆2ϕ + 2∆f + 2f ), (2.123)
and Λ00(t) + Λ0(t) > α − 1 > 0. The terms in the second parenthesis are of homo-geneity less than or equal to α − 2, thus under the control of A2. Consequently,
there exists some constant a0 depending on ϕ such that for a > a0,
A2+ A1 ≥ −
a2 4Λ(t)Λ
0
(t)x2α−21 + a(α − 1)ϕ. (2.124)
In addition |Λ0(t)| ≥ 1 and x1 > 1. We thus have
([S, A]v, v) + (Sv, Sv) ≥ Z (|∇v|2+ |∇φ|2v2)dxdt + Z a2 4Λ(t)v 2 dxdt + Z a(α − 1)ϕv2dxdt In turn Z Qθ e2aΛ(t)ϕ(x)+2t2 a 2 4 Λ(t) + a(α − 1)ϕ u2+ 1 2|∇u| 2 dxdt ≤ Z Qθ e2aΛ(t)ϕ(x)+2t2|∂tu + ∆u|2dxdt. (2.125)
To simplify, we can assume α > 3/2, and a > 2, it follows that Z Qθ e2aΛ(t)ϕ(x)+2t2a (Λ + ϕ) u2+ |∇u|2 dxdt ≤ 4 Z Qθ e2aΛ(t)ϕ(x)+2t2|∂tu + ∆u|2dxdt. (2.126)
Chapter 3
Isolated singularities of the 1D
complex viscous Burgers equation
3.1
Structure of the isolated singularities
In this chapter, we consider the structure of the isolated singularities of the Burgers equation
ut+ uux = uxx , (3.1)
via the re-scaling procedure, which is well-known to be very useful in studying singularities of PDEs. We will present the settings and the results first and give the proofs in the sections following.
With initial data u0 ∈ L1(R), we define the solution u by means of the
Cole-Hopf transformation, that is, u = −2vx/v, where v is the bounded solution to
the heat equation with initial data v0(x) = exp −12
Rx
0 u0(ξ) dξ. The function
v is analytic in R × (0, ∞) and the zero set is isolated. Thus the function u is analytic and solves the Burgers equation in R × (0, ∞) excluding the zero set of v. By shifting the origin and re-scaling the function we can assume that (0, 0) is an isolated singularity in a neighborhood Q = (−1, 1) × (−1, 1) and
G(t) = sup
|x|<1
|u(x, t)|
is attained at some x in (−1, 1) for any fixed t in the interval (−1, 0). Let
M (t) = sup
|x|<1,s<t
|u(x, s)|.
There exists a sequence tj → 0− (that is, tj < 0, tj → 0) such that G(tj) = M (tj).
Let xj be such that M (tj) = |u(xj, tj)|. Of course xj need not be unique. The
Burgers equation has a natural scaling invariance
u(x, t) → λu(λx, λ2t).
We consider the re-scaled functions
wj(y, s) = 1 M (tj) u(xj + y M (tj) , tj + s M (tj)2 ).
They are equilibria of equation (1) in the (y, s) variables defined in the cylinders Qj = {(y, s) | (xj + M (ty
j), tj +
s
subset of R × (−∞, 0] if j is large enough. We are interested in the limiting process as tj → 0−. What is the rate of blow-up? Does the limit of wj exist? If
so, in which sense? What is the limit function? We answer these questions in the following theorem.
Theorem 3.1.1. With the notations introduced above, we have
(i) there exists a constant C0 > 0 such that |M (t)| > C0/|t| for t ∈ (−1, 0),
(ii) there is a T0 > 0 such that when −T0 < t < 0, M (t) = G(t),
(iii) for a suitable choice of x(t) (it may not be unique) such that M (t) = |u(x(t), t)|, the re-scaled functions
w(t)(y, s) := 1 M (t)u x(t) + y M (t), t + s M2(t)
converge as t → 0− uniformly in any bounded subset of R × (−∞, 0]. Moreover
lim
t→0−w
(t)(y, s) = −2
y ± 2i, (3.2)
which are steady state solutions of the Burgers equation.
Singularities for which some scale-invariant quantity is bounded are often called type I singularities. The most common definition of type I singularity for equations with the scaling invariance u(x, t) → λu(λx, λ2t) uses the quantity sup√−t |u(x, t)|. Singularities which are not of type I are called type II. In this terminology the blow-up of the solution u to the Burgers equation is of type II.
The proof of Theorem 3.1.1 uses analyticity and explicit calculation. Namely, the function v in the Cole-Hopf transformation u = −2vx/v is analytic and in
turn has a representation in caloric polynomial series.
3.2
Representation in caloric polynomials and
outline of the proof
3.2.1
Caloric polynomials
Let us first recall the definition of caloric polynomials. A function f (x, t) is said to be parabolically m-homogeneous if f (λx, λ2t) = λmf (x, t). The m-th caloric polynomial Pm(x, t) is a parabolically m-homogeneous polynomial satisfying the
heat equation. It is unique, modulo a multiplicative factor, and can be given by:
Pm(x, t) = [m/2] X k=0 m! k!(m − 2k)!x m−2ktk.
For m even, m = 2k, the polynomial Pm is of the form
Pm(x, t) = (x2+ a1t) · · · (x2+ akt), 0 < a1 < · · · < ak.
For m odd, m = 2k + 1, Pm is of the form
The aj,s might be different. The caloric polynomials are simply the heat extension
of the polynomials xm.
Poisson transform, see [30]:
P[φ] = Z ∞
−∞
Γ(x − y, t)φ(y)dy. (3.3)
The Poisson transform of xm is the heat polynomial of degree m.
P[xm] = P m(x, t) = m! [m/2] X k=0 xm−2k (m − 2k)! tk k!. (3.4)
To verify, change variable and using integration by parts as follows.
P[xm] = √1 4πt Z ∞ −∞ e−(x−y)2/4tymdy (3.5) = √1 4πt( √ 4t)m+1 Z ∞ −∞ e−z2(z + √x 4t) mdz (3.6) = √1 π [m/2] X k=0 m 2k 4kxm−2ktk Z ∞ −∞ e−z2z2kdz (3.7) = Pm(x, t). (3.8)
In the last equation, we use integration by parts to get that Z ∞ −∞ e−z2z2kdz = 2k − 1 2 Z ∞ −∞ e−z2z2k−2dz, (3.9) and the fact that
Z ∞
−∞
A solution v(x, t) to the heat equation can be written as an absolutely convergent caloric polynomial series
v(x, t) =
∞
X
k=0
αkPk(x − x0, t − t0)
in a strip |t − t0| < σ (with the coefficients αk = ∂xkv(x0, t0)/k!) if and only if it
has the property
v(x, t) = Z ∞ −∞ 1 p4π(t − s)e −4(t−s)|x−y| v(y, s)dy
for every pair t, s such that t0−σ < s < t < t0+σ, see Definition 8.1 and Theorem
11.1 in [28].
3.2.2
Proof of Theorem 3.1.1
In our setting, the bounded solution v of the heat equation satisfies the above conditions and has an isolated zero at (0, 0). We thus have the following series starting from a positive integer m.
v(x, t) =
∞
X
k=m
αkPk(x, t), αk ∈ C.
The series converges absolutely in a strip R × (−σ, σ), where σ is the length from the initial time to t = 0 (recall that we have shifted the origin). We can assume αm = 1. If not, we replace v(x, t) by v(x, t)/αk, which does not affect
