Electronic Journal of Differential Equations, Vol. 2025 (2025), No. 04, pp. 1–17. ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu DOI: 10.58997/ejde.2025.04 EXACT CONTROLLABILITY FOR DEGENERATE AND SINGULAR WAVE EQUATIONS GUANG ZHANG, SHUGEN CHAI Abstract. In this article, we study exact controllability for degenerate and singular wave equations with a general coefficient. We estimate the observabil- ity inequality by the multiplier method and determine the observability time. We also deduce the exact controllability of the corresponding degenerate and singular control problem at a sufficiently large time, employing the Hilbert uniqueness method. 1. Introduction Control issues for non-degenerate parabolic and hyperbolic problems have been a mainstream topic over the past several years, and a lot of attention has led to numerous developments being pursued (see [10, 20, 22, 23, 25, 26, 28, 32]). Let us recall that exact controllability for the nondegenerate wave equation, which is characterized by the system of equations utt − uxx = 0, (t, x) ∈ (0, T )× (0, 1), u(t, 0) = 0, u(t, 1) = f(t), t ∈ (0, T ), u(0, x) = u0(x), x ∈ (0, 1), ut(0, x) = u1(x), x ∈ (0, 1), (1.1) where u is the state, f acts as a boundary control and is used to drive the solution to zero at a given time T . To be more precise, for given the initial data (u0, u1) in a suitable space, we look for a control f such that u(T, x) = ut(T, x) = 0, ∀x ∈ (0, 1). (1.2) Because of the finite speed of propagation of solutions to the wave equation, exact controllability can only be achieved at a sufficiently large time T (in the parabolic case, we have null controllability at any final time T ). As a general conclusion, we consider (1.1) to represent exact controllability if T > 2. The degenerate wave equations began to receive some attention within the past decade, and had developed rapidly [4, 5, 7, 29]. Different from the case non- degenerate equations, the main difficulty with degenerate wave equations is in- troducing a suitable function space to deal with degenerate terms, requiring the 2020 Mathematics Subject Classification. 35L80, 35L81, 35L05, 93B05. Key words and phrases. Degenerate wave equation; singular wave equation; controllability; boundary control. ©2025. This work is licensed under a CC BY 4.0 license. Submitted September 18, 2024. Published January 9, 2025. 1 2 G. ZHANG, S. CHAI EJDE-2025/04 development of new rules for analyzing observability and controllability. Gueye [19] considered the boundary control about the degenerate wave equation utt − (xαux)x = 0, (t, x) ∈ (0, T )× (0, 1). (1.3) We also refer to the work of Zhang and Gao [30, 31] for additional controllability results obtained through the use of a locally distributed control. Later, Alabau- Boussouira et al. [1] consider the equation utt − (aux)x = 0, (t, x) ∈ (0, T )× (0, 1), (1.4) where a is positive on (0, 1] and a(0) = 0. The degeneracy of (1.4) at x = 0 is measured by the parameter µa defined by µa := sup 0 0 on (0, 1], g(0) = 0 and sup 0 0 ∀x ∈ (0, 1], a(0) = b(0) = 0, Ka ∈ [0, 2) \ {1},Kb ∈ [0, 2] and Ka +Kb ≤ 2. (2.1) From Definition 1.1 and Assumption 2.1, it is easy to draw the following conse- quences. Remark 2.2. By integrating the inequality sg′(s) ≤ Kgg(s), ∀s ∈ (0, 1] 4 G. ZHANG, S. CHAI EJDE-2025/04 over [x, 1], we obtain g(x) ≥ g(1)xKg , ∀x ∈ [0, 1]. Hence, for all x ∈ [0, 1] we deduce that a(x) ≥ a(1)xKa , b(x) ≥ b(1)xKb . (2.2) Proposition 2.3 (Hardy-Poincare inequality). Under Assumption 2.1, there exists Ca,b > 0 such that∫ 1 0 u2 b dx ≤ Ca,b ∫ 1 0 a(u′) 2 dx, ∀u ∈ C∞ c (0, 1), (2.3) where Ca,b = 4 a(1)b(1)(1−Ka)2 . Proof. By Remark 2.2 and generalized Hardy inequality [11, chap. 5.3], (1−Ka) 2 4 ∫ 1 0 u2 x2−Ka dx ≤ ∫ 1 0 xKau2 x dx ∀u ∈ C∞ c (0, 1), we have∫ 1 0 u2 b dx ≤ 1 b(1) ∫ 1 0 u2 xKb dx ≤ 1 a(1)b(1) ∫ 1 0 a u2 xKa+Kb dx ≤ 1 a(1)b(1) ∫ 1 0 a u2 x2 dx ≤ 4 a(1)b(1)(1−Ka)2 ∫ 1 0 a(u′) 2 dx. (2.4) □ Assumption 2.4. The constant λ ∈ R satisfies λ < 1 Ca,b . (2.5) Under Assumptions 2.1 and 2.4, as in [1] and [3], we introduce the following spaces with related inner product V 1 a (0, 1) = {u ∈ L2(0, 1) ∩H1 loc(0, 1] : √ au′ ∈ L2(0, 1)}, ∥u∥2V 1 a (0,1) = ∫ 1 0 u2 + a(u′) 2 dx, ∀u ∈ V 1 a (0, 1), ⟨u, v⟩V 1 a (0,1) = ∫ 1 0 uv + au′v′ dx, ∀u, v ∈ V 1 a (0, 1), V 2 a (0, 1) = {u ∈ V 1 a (0, 1)|au′ ∈ H1(0, 1)}; and V 1 a,b(0, 1) = {u ∈ L2(0, 1) ∩H1 loc(0, 1]|a(u′) 2 − λ b u2 ∈ L1(0, 1)}, ∥u∥2V 1 a,b(0,1) = ∫ 1 0 u2 + a(u′) 2 − λ b u2 dx, ∀u ∈ V 1 a,b(0, 1), ⟨u, v⟩V 1 a,b(0,1) = ∫ 1 0 uv + au′v′ − λ b uv dx, ∀u, v ∈ V 1 a,b(0, 1), V 2 a,b(0, 1) = {u ∈ V 1 a,b(0, 1)|(au′)′ − λ b u ∈ L2(0, 1)}. Under the boundary conditions of (1.7), we introduce the space H1 a,b(0, 1). EJDE-2025/04 EXACT CONTROLLABILITY 5 (i) If Ka ∈ [0, 1), H1 a,b(0, 1) = {u ∈ V 1 a,b(0, 1)|u(0) = u(1) = 0}; (ii) If Ka ∈ (1, 2), H1 a,b(0, 1) = {u ∈ V 1 a,b(0, 1)|u(1) = 0}. Also, H−1 a,b (0, 1) denotes the conjugate space of H1 a,b(0, 1). We set H2 a,b(0, 1) = V 2 a,b(0, 1) ∩H1 a,b(0, 1). From Assumption 2.4, when λ > 0 there exists θ ∈ (0, 1) such that λ = 1 Ca,b − θ Ca,b . (2.6) Further, one can prove the next result. Lemma 2.5. Under Assumptions 2.1 and 2.4, we have∫ 1 0 a(u′) 2 dx ≤ 1 Cθ ∫ 1 0 a(u′) 2 − λ b u2 dx, (2.7) where Cθ = θ if λ ∈ ( 0, 1 Ca,b ) , Cθ = 1 if λ ≤ 0. Proof. (i) If λ ∈ ( 0, 1 Ca,b ) , then by (2.3) (2.6), we deduce that∫ 1 0 a(u′) 2 − λ b u2 dx ≥ ∫ 1 0 a(u′) 2 dx− (1− θ) ∫ 1 0 a(u′) 2 dx = θ ∫ 1 0 a(u′) 2 dx. (ii) If λ ≤ 0, obviously, ∫ 1 0 a(u′) 2 − λ b u2 dx ≥ ∫ 1 0 a(u′) 2 dx. □ Assumption 2.6. Under Assumptions 2.1 and 2.4, the function x → xKb b(x) is nondecreasing in a right neighborhood of x = 0. Remark 2.7. It is clear that, if Assumption 2.6 holds, then lim x→0+ xγ b(x) = 0, γ > Kb. (2.8) In particular, lim x→0+ x2 b(x) = 0. (2.9) Lemma 2.8. Under Assumption 2.6, for all u ∈ H1 a,b(0, 1), we have lim x→0+ x b(x) u2 = 0. (2.10) 6 G. ZHANG, S. CHAI EJDE-2025/04 Proof. If 0 ≤ Ka < 1, using that u(0) = 0, we have |u(x)| ≤ ∫ x 0 |u(ξ)|dξ ≤ √ x∥u′∥L2(0,1). Then x b(x) u2(x) ≤ x2 b(x) ∥u′∥L2(0,1). By equation (2.9), the lemma follows. If 1 < Ka < 2, then 0 ≤ Kb < 1, the conclusion follows directly from Remark 2.7. □ 3. Well-posedness First, we consider the degenerate/singular wave problem with Dirichlet/Neumann boundary conditions: utt − (a(x)ux)x − λ b(x) u = 0, (t, x) ∈ (0, T )× (0, 1), u(t, 1) = 0, t ∈ (0, T ), u(t, 0) = 0, Ka ∈ [0, 1), t ∈ (0, T ), lim x→0+ aux(t, x) = 0, Ka ∈ (1, 2), t ∈ (0, T ), u(0, x) = u0(x), ut(0, x) = u1(x), x ∈ (0, 1). (3.1) Let us recall the typical abstract setup of semigroup theory, which provides weak and classical solutions for the above system. Consider the Hilbert space H = H1 a,b(0, 1)× L2(0, 1) endowed with the inner product ⟨(u, v), (ũ, ṽ)⟩H = ∫ 1 0 vṽ + au′ũ′ − λ b uũ dx, ∀(u, v), (ũ, ṽ) ∈ H. By Assumption 2.4 and the Hardy-Poincare inequality (2.3), we have ⟨(u, v), (u, v)⟩H = ∫ 1 0 v2 + a(u′) 2 − λ b u2 dx ≥ 0, ∀(u, v) ∈ H. Thus, ⟨·, ·⟩H forms the scalar product. Arguing as for the classical wave equation, the unbounded operator A : D(A) ⊂ H → H is defined by A(u, v) = ( v, a(u′)′ + λ b u ) , ∀(u, v) ∈ D(A) (3.2) with D(A) = H2 a,b(0, 1)×H1 a,b(0, 1), if Ka ∈ [0, 1), or D(A) = {(u, v) ∈ H2 a,b(0, 1)×H1 a,b(0, 1) : aux(0) = 0}, provide Ka ∈ (1, 2). Proposition 3.1. Under Assumption 2.6, the operator A is maximally dissipative on H. EJDE-2025/04 EXACT CONTROLLABILITY 7 Proof. Let (u, v) ∈ D(A). Then ⟨A(u, v), (u, v)⟩H = ∫ 1 0 ( a(u′)′v + λ b uv + au′v′ − λ b uv ) dx = 0. Therefore, A is dissipative. It remains to be proved that the operator is maximally dissipative, which is equivalent to showing that I −A is surjective. Specifically, for any (g1, g2) ∈ H, we need to find (u, v) ∈ D(A) such that the problem v = u− g1, u− a(u′)′ − λ b u = g1 + g2. (3.3) So far that, we consider the bilinear form β : H1 a,b(0, 1) ×H1 a,b(0, 1) → R given by β(u, φ) = ∫ 1 0 ( uφ+ au′φ′ − λ b uφ ) dx, and the linear functional L : H1 a,b(0, 1) → R given by Lφ = ∫ 1 0 (g1 + g2)φdx. One can verify that β is a continuous and coercive bilinear functional on H. Also, L is a continuous linear functional. Consequently, by the Lax-Milgram theorem, there exist a unique solution u ∈ H1 a,b(0, 1) to the variational problem β(u, φ) = Lφ, ∀φ ∈ H1 a,b(0, 1). (3.4) Since C∞ c (0, 1) ⊂ H1 a,b(0, 1), we have∫ 1 0 ( uφ+ au′φ′ − λ b uφ ) dx = ∫ 1 0 (g1 + g2)φdx, ∀φ ∈ C∞ c (0, 1). (3.5) By duality, this implies that u− a(u′)′ − λ b u = g1 + g2 in the sense of distributions. Thus, u ∈ H2 a,b(0, 1) and u − a(u′)′ − λ b u = g1 + g2 almost everywhere in (0, 1). Setting v = u − g1, we conclude that (u, v) ∈ D(A) and the problem (3.3) is solved. □ Therefore A is the generator of a contraction semigroup in H, denoted by etA. For any U0 = (u0, u1) ∈ H, U(t) = etAU0 can be seen as the solution of the Cauchy problem U ′(t) = AU(t), U(0) = U0. (3.6) Hence, as in [1] or [3], we have the following conclusions. Corollary 3.2. Assume Assumption 2.6, for given (u0, u1) ∈ H1 a,b(0, 1)×L2(0, 1), there exist a unique mild solution u to the problem (3.1) satisfying u ∈ C1([0, T ];L2(0, 1)) ∩ C([0, T ];H1 a,b(0, 1)); If (u0, u1) ∈ D(A), then the solution u is classical, in the sense that u ∈ C2([0, T ];L2(0, 1)) ∩ C1([0, T ];H1 a,b(0, 1) ∩ C([0, T ];H2 a,b(0, 1)). 8 G. ZHANG, S. CHAI EJDE-2025/04 4. Energy estimate In this section, we establish an estimate of the energy and a direct inequality associated to the solution of the initial value problem, which will be used to prove a controllability in Section 6. Definition 4.1. Using Assumption 2.6, we define the generalized energy of a mild solution u of (3.1) as follows, Eu(t) = 1 2 ∫ 1 0 ( u2 t + au2 x − λ b u2 ) dx, ∀t ≥ 0. (4.1) Computations show that the conservation of the energy Eu remains valid in the degenerate and singular situation. Proposition 4.2. Under Assumption 2.6 and considering (u0, u1) ∈ H1 a,b(0, 1) × L2(0, 1), The energy Eu(t) of the mild solution u of (3.1) is constant in time, that is, Eu(t) = Eu(0), ∀t ≥ 0. (4.2) Proof. First, suppose that u is a classical solution. Then, multiplying the equation by ut and integrating over (0, 1), we obtain 0 = ∫ 1 0 ut(t, x) ( utt(t, x)− (a(x)ux(t, x))x − λ b(x) u(t, x) ) dx = ∫ 1 0 ( ut(t, x)utt(t, x)− a(x)ux(t, x)utx(t, x)− λ b(x) u(t, x)ut(t, x) ) dx︸ ︷︷ ︸ = d dtEu(t) − [a(x)ux(t, x)ut(t, x)] 1 0. (4.3) Using the boundary conditions and Alabau-Boussouira et al. [1, Proposition 2.5], we have that the boundary a(x)ux(t, x)ut(t, x) vanishes at x = 1 and x = 0. Now, let u be the mild solution associated with the initial data (u0, u1) ∈ H1 a,b(0, 1) × L2(0, 1). Consider a sequence {un 0 , u n 1}n∈N ⊂ D(A) = H2 a,b(0, 1) × H1 a,b(0, 1) that approximates (u0, u1), and let un be the classical solution of (3.1) associated to (un 0 , u n 1 ). u n satisfies (4.2) and un x is a Cauchy sequence in L2(0, 1). Therefore, we extend (4.2) to the mild solution. □ To facilitate the subsequent proof of controllability results, we prove the following direct inequality. Proposition 4.3. Under Assumption 2.6, if u is a classical solution of (3.1), then a(1) ∫ T 0 u2 x(t, 1)dt = ∫ Q ( u2 t + (a− xa′)u2 x + λ b− xb′ b2 ) dx dt + 2 [∫ 1 0 xuxutdx ]T 0 . (4.4) EJDE-2025/04 EXACT CONTROLLABILITY 9 As a consequence, if u is a mild solution, then ux(·, 1) ∈ L2(0, T ) for every T > 0 and a(1) ∫ T 0 u2 x(t, 1)dt ≤ 4max{ 1 a(1)Cθ , 1}Eu(0) + 2T Cθ (1 +Ka + Ca,b|λ|(1 +Kb))Eu(0). (4.5) Proof. Suppose first that (u0, u1) ∈ H2 a,b(0, 1) ×H1 a,b(0, 1), so that u is a classical solution of (3.1). Then, multiplying (3.1) by xux and integrating over Q, we obtain 0 = ∫ Q xux ( utt − (aux)x − λ b u ) dx dt = [∫ 1 0 xuxutdx ]T 0 − ∫ Q xuxtut dx dt − ∫ Q ( xa′u2 x + xauxuxx + λ b xuux ) dx dt = [∫ 1 0 xuxutdx ]T 0 − ∫ Q xa′u2 x dx dt − ∫ Q ( x (u2 t 2 ) x + xa (u2 x 2 ) x + λ 2 x(u2)x b ) dx dt. (4.6) Arguing as in the proof of Alabau-Boussouira et al. [1, Lemma 3.2],[ x u2 t 2 ]1 0 = 0, [xau2 x] 1 0 = a(1)u2 x(t, 1). From the boundary conditions and Lemma 2.8, we have[xu2 b ]1 0 = 0. Hence, ∫ Q x (u2 t 2 ) x dx dt = −1 2 ∫ Q u2 t dx dt, (4.7)∫ Q xa (u2 x 2 ) x dx dt = −1 2 ∫ Q (a+ xa′)u2 x dx dt+ 1 2 a(1) ∫ T 0 u2 x(t, 1)dt, (4.8) λ 2 ∫ Q x(u2)x b dx dt = −λ 2 ∫ Q b− xb′ b2 u2 dx dt . (4.9) Then (4.4) follows by inserting (4.7)–(4.9) into (4.6). Next, we estimate the term on the right side of equation (4.4) separately. According the Hölder inequality and Lemma 2.5, we have 2 ∫ 1 0 xuxutdx ≤ ∫ 1 0 ( x2u2 x + u2 t ) dx ≤ 1 a(1)Cθ ∫ 1 0 ( au2 x − λ b u2 ) dx+ ∫ 1 0 u2 t dx ≤ 2max { 1 a(1)Cθ , 1 } Eu(0). (4.10) 10 G. ZHANG, S. CHAI EJDE-2025/04 Moreover, Using the definition of Kg and Hardy’s inequality (2.3), we have∫ 1 0 (a+ xa′)u2 xdx ≤ (1 +Ka) ∫ 1 0 au2 x dx ≤ (1 +Ka) Cθ Eu(0), (4.11) and λ ∫ 1 0 b− xb′ b2 u2 dx ≤ ∫ 1 0 λ b ( 1− xb′ b ) u2dx ≤ ∫ 1 0 |λ| b (1 +Kb)u 2dx ≤ 2Ca,b|λ|(1 +Kb) Cθ Eu(0). (4.12) Hence, by (4.4) and the inequalities (4.10)–(4.12), we obtain (4.5). As before, to extend (4.5) to the mild solution associated with the initial data (u0, u1) ∈ H1 a,b(0, 1)×L2(0, 1), it suffices to approximate such data by (un 0 , u n 1 ) ∈ H2 a,b(0, 1)× H1 a,b(0, 1), and thanks to (4.5), we can show that the normal derivatives of the corresponding classical solutions give a Cauchy sequence in L2(0, 1). □ 5. Boundary observability Lemma 5.1. Under Assumption 2.6, for any mild solution u of (3.1) and every T ≥ 0, we have∫ Q ( a(x)u2 x − u2 t − λ b(x) u2 ) dx dt+ [∫ 1 0 uut dx ]T 0 = 0. (5.1) Proof. As before, suppose that u is the classical solution of (3.1). Multiplying (3.1) by u and integrating over the domain Q = (0, T )× (0, 1), we obtain 0 = ∫ 1 0 u ( utt − (a(x)ux)x − λ b(x) u ) dx = ∫ Q ( a(x)u2 x − u2 t − λ b(x) u2 ) dx dt+ [∫ 1 0 uut dx ]T 0 − ∫ T 0 [a(x)uux] 1 0 dt. (5.2) Thanks to the boundary conditions and Alabau-Boussouira et al. [12, Proposition 2.5], we have that auux also vanishes at x = 0 and at x = 1. The conclusion can be extended to mild solution by an approximation argument. □ Theorem 5.2. Under Assumption 2.6, let u be a mild solution of (3.1). Then, for every T ≥ 0, a(1) ∫ T 0 u2 x(t, 1)dt ≥ − ( 4max{ 1 a(1)θ , 1}+ 2Ka 1√ θa(1) ) Eu(0) + T{(2−Ka) + λCa,bθ (2−Ka −Kb)}Eu(0), (5.3) for λ ∈ (0, 1 Ca,b ), and a(1) ∫ T 0 u2 x(t, 1)dt ≥ − ( 4max{ 1 a(1) , 1}+ 2Ka 1√ a(1) ) Eu(0) + T{(2−Ka)− |λ|Ca,b (2−Ka −Kb)}Eu(0), (5.4) for λ ∈ (−∞, 0], where the constants Ca,b and θ are given in (2.3) and (2.6), respectively. EJDE-2025/04 EXACT CONTROLLABILITY 11 Proof. As usual, let us suppose that u is a classical solution of (3.1). Multiplying both sides of equation (5.1) by Ka 2 and summing the corresponding ones to both side of equation (4.4), we obtain a(1) ∫ T 0 u2 x(t, 1)dt = 2 [∫ 1 0 xuxutdx ]T 0 + Ka 2 [∫ 1 0 uut dx ]T 0 + ( 1− Ka 2 )∫ Q u2 t dx dt + ∫ Q [( 1 + Ka 2 ) a− xa′ ] u2 x dx dt+ ∫ Q λ b (b− xb′ b − Ka 2 ) u2 dx dt = ∫ Q ( 1− Ka 2 ) u2 t + [( 1 + Ka 2 ) a− xa′ ] u2 x + ( 1− Ka 2 )λ b u2 dx dt + 2 [∫ 1 0 xuxutdx ]T 0 + Ka 2 [∫ 1 0 uut dx ]T 0 + ∫ Q λ b ( 2− xb′ b −Ka ) u2 dx dt. (5.5) Using Remark 2.2 and the inequality xa′ ≤ Kaa, we have∫ Q ( 1− Ka 2 ) u2 t + [( 1 + Ka 2 ) a− xa′ ] u2 x + ( 1− Ka 2 )λ b u2 dx dt ≥ ( 1− Ka 2 )∫ Q ( u2 t + au2 x − λ b u2 ) dx dt ≥ (2−Ka)TEu(0). (5.6) By (4.10), we obtain 2 [∫ 1 0 xuxutdx ]T 0 ≤ 4max{ 1 a(1)Cθ , 1}Eu(0). (5.7) Furthermore, applying the Hardy inequality in its pure degenerate form (see [1]), we obtain ∫ 1 0 u2dx ≤ 4 a(1) ∫ 1 0 au2 xdx, ∀u ∈ C∞ c (0, 1), from which we can deduce that∫ 1 0 uut dx ≤ 1 2 ∫ 1 0 (√Cθa(1) 2 u2 + 2√ Cθa(1) u2 t ) dx ≤ 1 2 ∫ 1 0 ( 2 √ Cθ a(1) au2 x + 2√ Cθa(1) u2 t ) dx ≤ 1√ Cθa(1) ∫ 1 0 ( u2 t + au2 x − λ b u2 ) dx = 2√ Cθa(1) Eu(0). (5.8) Finally, if λ ∈ ( 0, 1 Ca,b ) , we have∫ Q λ b ( 2− xb′ b −Ka ) u2 dx dt ≥ ∫ Q λ b (2−Kb −Ka)u 2 dx dt ≥ λCa,bCθ (2−Ka −Kb)TEu(0); (5.9) 12 G. ZHANG, S. CHAI EJDE-2025/04 If λ ∈ (−∞, 0], then∫ Q λ b ( 2− xb′ b −Ka ) u2 dx dt ≤ ∫ Q |λ| b (2−Kb −Ka)u 2 dx dt ≤ |λ|Ca,bCθ (2−Kb −Ka)TEu(0). (5.10) Notice that Cθ = θ if λ ∈ (0, 1 Ca,b ); Cθ = 1 if λ ∈ (−∞, 0]. Therefore, (5.3) and (5.4) by substituting (5.6)-(5.10) into (5.5). □ We recall that (3.1) is said to be observable in time T ≥ 0 via the normal derivative at x = 1, if there exists a constant C > 0 such that for any (u0, u1) ∈ H1 a(0, 1)× L2(0, 1), the mild solution of (3.1) satisfies∫ T 0 u2 x(t, 1) dt ≥ CEu(0). (5.11) Moreover, any constant satisfying (5.11) is called an observability constant for (3.1) in time T ≥ 0. The supremum of all observability constants for (3.1) is denoted by CT , namely, CT := sup{C > 0, C satisfies (5.11)}. We said that (3.1) is observable if CT = inf (u0,u1) ̸=(0,0) ∫ T 0 u2 x(t, 1)dt Eu(0) > 0. (5.12) From the definition of observability, and Theorem 5.2, we have following corollary. Corollary 5.3. Under Assumption 2.6 and λ > 0, (3.1) is observable in time T , provided that T > Ta,b := C2 C1 . In this case, CT ≥ 1 a(1) (C1T − C2), where C1 = (2−Ka) + λCa,bθ (2−Ka −Kb) , C2 = ( 4max{ 1 a(1)θ , 1}+ 2Ka 1√ θa(1) ) . Corollary 5.4. Under Assumption 2.6, Ka−2 Ca,b(2−Kb−Ka) < λ ≤ 0, and λ ≤ 0 if Ka +Kb = 2. We have that (3.1) is observable in time T , provided that T > Ta,b := C4 C3 . Moreover, CT ≥ 1 a(1) (C3T − C4), where C3 = {(2−Ka)− |λ|Ca,b (2−Ka −Kb)}, C4 = ( 4max{ 1 a(1) , 1}+ 2Ka 1√ θa(1) ) . EJDE-2025/04 EXACT CONTROLLABILITY 13 6. Controllability In this section, we study the problem of exact controllability for (1.7). By its linearity and reversibility, it is straightforward to verify that exact controllability will hold as long as it is valid for any initial data (y0, y1) and a zero final state. Equivalently, given (y0, y1) ∈ H1 a,b(0, 1) × L2(0, 1), we seek a control function f ∈ L2(0, T ) such that the solution of (1.7) satisfies (y, y′)(T, ·) ≡ 0. Definition 6.1. Let f ∈ L2 loc(0, T ) and (y0, y1) ∈ L2(0, 1)×H−1 a,b (0, 1) be arbitrarily fixed. We say that y is a solution by transposition of (1.7) if y ∈ C1([0, T ];H−1 a,b (0, 1) ∩ C([0, T ];L2(0, 1)) and for any T > 0, ⟨yt(T ), w0 T ⟩H−1 a (0,1)×H1 a(0,1) − ∫ 1 0 y(T )w1 T dx = ⟨y1, w(0)⟩H−1 a,b(0,1)×H1 a,b(0,1) − ∫ 1 0 y(0)w′(0) dx+ a(1) ∫ T 0 f(t)wx(t, 1) dt (6.1) for all (w0 T , w 1 T ) ∈ H1 a,b(0, 1) × L2(0, 1), where w is the solution of the backward equation wtt − (a(x)wx)x − λ b(x) w = 0, (t, x) ∈ (0, T )× (0, 1), w(t, 1) = 0, t ∈ (0, T ), w(t, 0) = 0, Ka ∈ [0, 1), t ∈ (0, T ), a(x)wx(t, 0) = 0, Ka ∈ (1, 2), ]; t ∈ (0, T ), w(T, x) = w0 T (x), wt(T, x) = w1 T (x), x ∈ (0, 1). (6.2) By setting y(t, x) = w(T − t, x), we leverage the time reversibility of the wave equation to assert that the solution y maintains the same regularity as w for t ≤ 0. Consequently, the backward equation (6.2) admits a unique solution w ∈ C1([0, T ];L2(0, 1) ∩ C([0, T ];H1 a,b(0, 1)) which depends continuously on the initial data WT = (w0 T , w 1 T ) ∈ H. By Proposition 4.2, the energy Ew(t) of w is conserved through time, which implies that the direct inequality (4.5) and observabilities in- equality (5.3) and (5.4) remain valid for w. Therefore, there is a unique solution by transposition w ∈ C1([0, T ];H−1 a,b (0, 1)) ∩ C([0, T ];L2(0, 1)). Now, consider the bilinear form Λ : H×H → R defined as Λ(WT , W̃T ) = a(1) ∫ T 0 wx(t, 1)w̃x(t, 1)dt, (6.3) where wx, w̃x are the solution of (1.7) associated with the final dataWT := (w0 T , w 1 T ), W̃T := (w̃0 T , w̃ 1 T ), respectively. To prove that (1.7) is exactly controllable, the fol- lowing Lemma is key. Lemma 6.2. Under Assumption 2.6, the bilinear form Λ is continuous and coer- cive. 14 G. ZHANG, S. CHAI EJDE-2025/04 Proof. By the direct inequality and the result of energy conservation, |Λ(WT , W̃T )| ≤ a(1) ∫ T 0 |wx(t, 1)w̃x(t, 1)| dt ≤ ( a(1) ∫ T 0 w2 x(t, 1) dt )1/2( a(1) ∫ T 0 w̃2 x(t, 1) dt )1/2 ≤ CE1/2 w (T )E 1/2 w̃ (T ) ≤ C∥WT ∥H∥W̃T ∥H. (6.4) By the observability inequality, we have Λ(WT ,WT ) = a(1) ∫ T 0 w2 x(t, 1) dt ≥ CTEw(T ) = C∥WT ∥2H. (6.5) □ Theorem 6.3. Under Assumption 2.6, for all T > Ta,b and for all (y0, y1) ∈ L2(0, 1) ×H−1 a,b (0, 1), there exists a control f ∈ L2(0, T ) such that the solution (in the sense of transposition) satisfies y(T, x) = yt(T, x) = 0, ∀x ∈ (0, 1). Proof. We define the continuous linear map L(WT ) = ∫ 1 0 y0wt(0) dx−⟨y1, w(0)⟩H−1 a,b(0,1)×H1 a,b(0,1) , ∀WT ∈ H1 a,b(0, 1)×L2(0, 1). Thanks to Lemma 6.2 and the Lax-Milgram theorem, there exists a unique WT ∈ H1 a,b(0, 1)× L2(0, 1) such that Λ(WT , W̃T ) = L(W̃T ), ∀W̃T ∈ H1 a,b(0, 1)× L2(0, 1). (6.6) We set f = wx(t, 1) and denote by y the solution by transposition of (1.7). Then we have a(1) ∫ T 0 f(t)w̃x(t, 1) dt = a(1) ∫ T 0 wx(t, 1)w̃x(t, 1)dt = Λ(WT , W̃T ) = L(W̃T ) = ∫ 1 0 u0w̃t(0) dx− ⟨u1, w̃(0)⟩H−1 a,b(0,1)×H1 a,b(0,1) , (6.7) for all (w̃0 T , w̃ 1 T ) ∈ H1 a,b(0, 1)×L2(0, 1). On the other hand, by the definition of the solution by transposition, for all (w̃0 T , w̃ 1 T ) ∈ H1 a,b(0, 1)× L2(0, 1) we have a(1) ∫ T 0 f(t)w̃x(t, 1) dt = ∫ T 0 y(T )w̃1 T dt− ⟨yt(T ), w̃0 T ⟩H−1 a,b(0,1)×H1 a,b(0,1) + ⟨y1, w̃(0)⟩H−1 a (0,1)×H1 a,b(0,1) − ∫ 1 0 y(0)wt(0) dx. (6.8) By equations (6.7) and (6.8), we deduce that ⟨yt(T ), w̃0 T ⟩H−1 a,b(0,1)×H1 a,b(0,1) = ∫ T 0 y(T )w̃1 T dt, ∀(w̃0 T , w̃ 1 T ) ∈ H1 a,b(0, 1)× L2(0, 1). EJDE-2025/04 EXACT CONTROLLABILITY 15 Hence, we conclude that y(T, x) = yt(T, x) = 0, ∀x ∈ (0, 1). □ 7. Conclusion In this article, we have considered the controllability of degenerate and singu- lar wave equations, and obtained the exact controllability of the system under certain assumptions. We have adopted the coefficient settings from Reference [1], enhanced the equation system from Reference [3], and replaced the particular expo- nential form of the degenerate and singular coefficients presented in [3] with a more generalized form. Furthermore, we do not necessitate the relationship between the generalized coefficients to be as stringent as in [3]; it is only necessary that their sum falls within a certain range (Ka +Kb ≤ 2). This allows for the application of diverse technical approaches when addressing the singular term, leading to various trade-offs in parameter λ selection. From this perspective, article [3] can be viewed as a special case of this paper, which is also one of the novel aspects of this work. To more intuitively illustrate this point, we have provided an example other than xα as follows. Example. Let θ ∈ (0, 2) be given, we construct the functions a(x), b(x) as follows a(x) = { xθ ( 1 + sin2(lnxα) ) if x ∈ (0, 1] 0 if x = 0, where α ∈ (0, 1− θ/2); b(x) = { x2−θ ( 1 + sin2(lnxβ) ) if x ∈ (0, 1] 0 if x = 0, where β ∈ [θ/2 − 1,−α]. Then the functions a(x), b(x) satisfy Assumption 2.1. Indeed, a′(x) = θxθ−1 ( 1 + sin2(lnxα) ) + 2αxθ−1 sin(lnxα) cos(lnxα) so that Ka ≤ θ+ 2α < 2. It is also easy to discover that Ka is not always equal to 1. b′(x) = (2− θ)x1−θ ( 1 + sin2(lnxβ) ) + 2βx1−θ sin(lnxβ) cos(lnxβ) so that Kb ≤ 2− θ + 2β ≤ 2, and Ka +Kb ≤ 2 + 2α+ 2β ≤ 2. We summarize the comparison between this paper and existing literature as follows. Degenerate term Singular term In [3] xα, α ∈ [0, 2) \ {1} x2−α µ ≤ (1−α)2 4 Here Ka := sup0