**2. Background**

Let x¨, ¨y be the standard Euclidean scalar product in R*<sup>n</sup>*. For *x* P R*<sup>n</sup>*, denote }*x*}"x*<sup>x</sup>*, *x*y<sup>1</sup>{2.

For *α* in R*<sup>n</sup>*zt0u, we write *rα* for the reflection with respect to the hyperplane *α*<sup>K</sup> orthogonal to *α* defined by

$$r\_{\mathfrak{a}}(\mathfrak{x}) := \mathfrak{x} - 2 \frac{\langle \mathfrak{a}, \mathfrak{x} \rangle}{\langle \mathfrak{a}, \mathfrak{a} \rangle} \mathfrak{a}, \quad \mathfrak{x} \in \mathbb{R}^n.$$

A finite set R Ă R*n*zt0u is called a root system if *<sup>r</sup>α*pRq Ă R for every *α* P R. The finite group *G* Ă *O*p*n*q generated by the reflections t*<sup>r</sup>α* : *α* P Ru is called the finite Coxeter group associated with R. A multiplicity function for *G* is a function *k* : R Ñ Rě0 which is constant on *G*-orbits.

For 1 ď *j* ď *n*, the Dunkl operator is defined in [17] by

$$T\_{\hat{\jmath}}f(\mathbf{x}) = \hat{c}\_{\hat{\jmath}}f(\mathbf{x}) + \frac{1}{2} \sum\_{\mathfrak{a} \in \mathcal{R}} k(\mathfrak{a}) \frac{f(\mathbf{x}) - f(r\_{\mathfrak{a}}(\mathbf{x}))}{\langle \mathfrak{a}, \mathfrak{x} \rangle} \langle \mathfrak{a}, \mathfrak{e}\_{\hat{\jmath}} \rangle, \quad \mathfrak{x} \in \mathbb{R}^n.$$

where B*j* is the standard directional derivative and t*<sup>e</sup>*1, ... ,*en*u is the canonical orthonormal basis in R*<sup>n</sup>*. The most important property of these operators is that they commute. The operators *Tj* and B*j* are intertwined by the following Laplace type operator

$$V\_k f(\mathbf{x}) = \int\_{\mathbb{R}^n} f(y) d\mu\_{\mathbf{x}}^k(y),\tag{1}$$

where *μk x* is a unique compactly supported probability measure with suppp*μ<sup>k</sup> x*qĂt*y* P R*n* : }*y*}ď}*x*}u (see [17,18]).

The Dunkl Laplacian, which is akin to the Euclidean Laplace operator Δ, is defined by Δ*k* :" *T*<sup>2</sup> 1`¨¨¨` *T*<sup>2</sup> *n* and is given explicitly, for suitable function *f* , by

$$
\Delta\_k f(\mathbf{x}) = \Delta f(\mathbf{x}) + \sum\_{a \in \mathcal{R}} k(a) \left( \frac{\langle \nabla f(\mathbf{x}), a \rangle}{\langle a, \mathbf{x} \rangle} - \frac{\|a\|^2}{2} \frac{f(\mathbf{x}) - f(\mathbf{r}\_a(\mathbf{x}))}{\langle a, \mathbf{x} \rangle^2} \right), \quad \mathbf{x} \in \mathbb{R}^n,\tag{2}
$$

where ∇ is the gradient. It is worth mentioning that if *k*p*α*q " 0 for all *α* P R, then Δ*k* reduces to the Euclidean Laplacian Δ. We refer the reader to [19] for the theory of Dunkl's operators. This theory, which started with the seminal paper [17], was developed extensively afterwards and continues to receive considerable attention (see, e.g., [20–32]).

Next we will introduce some definitions and results for the generalized Fourier transform; for details we refer to [14]. For *a* ą 0, let

$$\Delta\_{k,a} := \|\mathbf{x}\|^{2-a} \Delta\_k - \|\mathbf{x}\|^a \Delta\_k$$

where }*x*}*<sup>a</sup>* on the right-hand side of the formula stands for the multiplication operator by }*x*}*<sup>a</sup>*. The operator <sup>Δ</sup>*k*,*<sup>a</sup>* is symmetric on the Hilbert space *<sup>L</sup>*<sup>2</sup>pR*<sup>n</sup>*, *<sup>ϑ</sup>k*,*<sup>a</sup>*q consisting of square integrable functions against the measure *<sup>ϑ</sup>k*,*<sup>a</sup>*p*x*q*dx* :" }*x*}*<sup>a</sup>*´<sup>2</sup> ś*α*P<sup>R</sup> |x*<sup>α</sup>*, *<sup>x</sup>*y|*<sup>k</sup>*p*α*q*dx*.

The p*k*, *a*q-generalized Fourier transform F*k*,*<sup>a</sup>* was defined in [14] to be

$$\mathcal{F}\_{k,a} := \mathbf{e}^{i\frac{\pi}{2}\left(\frac{n+2\langle k\rangle+s-2}{a}\right)} \exp\left(i\frac{\pi}{2a}\Delta\_{k,a}\right)\rho$$

where x*k*y :" 12 ř*α*P<sup>R</sup> *k*p*α*q. We pin down that F*k*,*<sup>a</sup>* is a unitary operator from *<sup>L</sup>*<sup>2</sup>pR*<sup>n</sup>*, *<sup>ϑ</sup>k*,*<sup>a</sup>*q onto itself and the inversion formula is given as

$$\mathcal{F}\_{k, \frac{1}{r}}^{-1} = \mathcal{F}\_{k, \frac{1}{r}} \qquad \left(\mathcal{F}\_{k, \frac{2}{2r+1}}^{-1} f\right)(\mathbf{x}) = \left(\mathcal{F}\_{k, \frac{2}{2r+1}} f\right)(-\mathbf{x}) ,\tag{3}$$

where *r* is any nonnegative integer. The transform F*k*,*<sup>a</sup>* reduces to the Euclidean Fourier transform if *k* " 0 and *a* " 2; to the Kobayashi–Mano Hankel transform [33] if *k* " 0 and *a* " 1; to the Dunkl transform [34] if *k* ą 0 and *a* " 2. In this paper we consider the case *k* ą 0 and *a* " 1. For more details, we refer the reader to ([14] Sections 4 and 5) (see also [35–41]).

Let us collect the main properties of the p*k*, 1q-transform F*k*,<sup>1</sup> :" F*k*. In ([14] Theorem 4.23), the authors proved that for *n* ` 2x*k*y ą 1, there exists a kernel *Bk*p*<sup>x</sup>*, *y*q such that for every *f* P *<sup>L</sup>*<sup>2</sup>pR*<sup>n</sup>*, *<sup>ϑ</sup>k*,<sup>1</sup>q,

$$\mathcal{F}\_k f(\mathbf{x}) = c\_k^{-1} \int\_{\mathbb{R}^n} f(y) B\_k(\mathbf{x}, y) \vartheta\_{k, 1}(y) dy, \quad \mathbf{x} \in \mathbb{R}^n, \mathbf{y}$$

where, for *x* " *rθ*<sup>1</sup> and *y* " *<sup>t</sup>θ*<sup>2</sup>, the kernel *Bk* is given by

$$B\_k(\mathfrak{x}, \mathfrak{y}) = V\_k\left(\widetilde{J}\_{\frac{n-3}{2} + \langle k \rangle}\left(\sqrt{2rt(1 + \langle \theta', \cdot \rangle)}\right)\right)(\theta'').$$

Here *Vk* is the Dunkl intertwining operator (1) and *Jν*p*z*q is the normalized Bessel function. Above,

r

$$c\_k := \Gamma(n + 2\langle k \rangle - 1) \int\_{S^{n-1}} \prod\_{\alpha \in \mathcal{R}} |\langle \alpha, \eta \rangle|^{k(\alpha)} d\sigma(\eta).$$

It is noteworthy mentioning that

$$
\mathcal{F}\_k \circ \|\mathbf{x}\| = -\|\mathbf{x}\| \Delta\_k \mathcal{F}\_{k'} \qquad \mathcal{F}\_k \circ (-\|\mathbf{x}\| \Delta\_k) = -\|\mathbf{x}\| \circ \mathcal{F}\_k. \tag{4}
$$

Recently, in [42] the authors defined a translation operator *τx*, for *x* P R*<sup>n</sup>*, on the space *L*1 X *<sup>L</sup>*<sup>8</sup>pR*<sup>n</sup>*, *<sup>ϑ</sup>k*,<sup>1</sup>q by

$$\mathcal{F}\_k(\pi\_\mathfrak{x}f)(\mathfrak{f}) = B(\mathfrak{x}, \mathfrak{f})\mathcal{F}\_k(f)(\mathfrak{f}), \qquad \mathfrak{f} \in \mathbb{R}^n.$$

Here are some basic properties of the translation operator:


By means of the translation operator, a convolution - on the Schwartz space SpR*n*q was defined by

$$f \circledast g(\mathfrak{x}) = c\_k^{-1} \int\_{\mathbb{R}^n} f(y) \pi\_{\mathfrak{x}} g(y) \theta\_{k,1}(y) dy, \quad \mathfrak{x} \in \mathbb{R}^n.$$

In particular, *f* - *g* " *g* - *f* and F*k*p*f* - *g*q " F*k f* ¨ F*kg* (see [42] for more details).

Next we turn our attention to the convolution of distributions (see [42,43]). Denote by S<sup>1</sup>pR*n*q the dual of the Schwartz space SpR*<sup>n</sup>*q. If *T* P S<sup>1</sup>pR*<sup>n</sup>*q, then F*k*p*T*q is defined by

$$
\langle \mathcal{F}\_k(T), \mathfrak{q} \rangle := \langle T, \mathcal{F}\_k(\mathfrak{q}) \rangle, \qquad \forall \mathfrak{q} \in \mathcal{S}(\mathbb{R}^n).
$$

It is worth mentioning that SpR*n*q is stable by F*k* (see [36]). The convolution *T* - *f* of *T* P S<sup>1</sup>pR*n*q and *f* P SpR*n*q is defined in [42] by

$$T \circledast f(\mathbf{x}) = \langle T, \pi\_{\mathbf{x}} f \rangle.$$

In particular, a result analogous to the Euclidean convolution shows that *T* - *f* P S<sup>1</sup>pR*n*q X *C*<sup>8</sup>pR*n*q and F*k*p*<sup>T</sup>* - *f*q " F*kT* ¨ F*k f* .

### **3. The Deformed Wave Equation and Huygens' Principle**

For *n* ` 2x*k*y ´ 1 ą 0, where x*k*y " 12 ř*α*P<sup>R</sup> *k*p*α*q, we consider the following Cauchy problem for the wave equation

$$\begin{aligned} \mathbf{2} \|\mathbf{x}\| \Delta\_k^\mathbf{x} \mu\_k(\mathbf{x}, t) &= \hat{c}\_{tt} \mu\_k(\mathbf{x}, t), \quad (\mathbf{x}, t) \in \mathbb{R}^M \times \mathbb{R}, \\\\ \mu\_k(\mathbf{x}, 0) &= f(\mathbf{x}), \quad \hat{c}\_t \mu\_k(\mathbf{x}, 0) = \mathbf{g}(\mathbf{x}), \end{aligned} \tag{5}$$

where the functions *f* and *g* belong to the space DpR*n*q of smooth functions with compact support. Here the superscript in Δ*xk* indicates the relevant variable, while the subscript *t* indicates differentiation in the *t*-variable. Next, we will prove the following statements:

pSq Let *uk*p*<sup>x</sup>*, *t*q, *x* P R*n* and *t* P R, satisfy <sup>2</sup>}*x*}Δ*xkuk*p*<sup>x</sup>*, *<sup>t</sup>*q´B*ttuk*p*<sup>x</sup>*, *t*q " 0, then *uk* does not satisfy Huygen's principle. In other words, the solution *uk* is expressed as a sum of --convolution of *f* and *g* with distributions that are not supported entirely on the set tp*<sup>x</sup>*, *t*q P R*n* ˆ R : }*x*} " 12 *<sup>t</sup>*<sup>2</sup>u.

For *t* P R, denote by *Pk*,*<sup>t</sup>* the 2 ˆ 2 matrix of tempered distributions on R*n*

$$P\_{k,t} = \begin{pmatrix} P\_{k,t}^{11} & P\_{k,t}^{12} \\ P\_{k,t}^{21} & P\_{k,t}^{22} \end{pmatrix} := \begin{pmatrix} \mathcal{F}\_k(\cos(t\sqrt{2\|\cdot\|})) & \mathcal{F}\_k(\sin(t\sqrt{2\|\cdot\|})/\sqrt{2\|\cdot\|}) \\ \mathcal{F}\_k(-\sqrt{2\|\cdot\|}\sin(t\sqrt{2\|\cdot\|})) & \mathcal{F}\_k(\cos(t\sqrt{2\|\cdot\|})) \end{pmatrix}. \tag{6}$$

Set *Uk*p*<sup>x</sup>*, 0q :" ˜*f*p*x*q *g*p*x*q¸, where the initial data p*f* , *g*q P DpR*n*q ˆ DpR*<sup>n</sup>*q. Thus, we may define the vectorcolumn*Uk*p*<sup>x</sup>*,*t*qby

 

$$\begin{split} \mathcal{U}\_{k}(\mathbf{x},t) &:= \quad \left\{ P\_{k,t} \otimes \mathcal{U}\_{k}(\cdot, 0) \right\}(\mathbf{x}) \\ &= \quad \left\{ \begin{pmatrix} P\_{k,t}^{11} & P\_{k,t}^{12} \\ P\_{k,t}^{21} & P\_{k,t}^{22} \end{pmatrix} \otimes \begin{pmatrix} f \\ \mathcal{S} \end{pmatrix} \right\}(\mathbf{x}). \end{split} \tag{7}$$

By applying the Fourier transform F*k* to (7), in the *x*-variable, we ge<sup>t</sup>

$$\mathcal{F}\_k(\mathcal{U}\_k(\cdot, t))(\xi) = e^{t\mathcal{U}\_k} \mathcal{F}\_k(\mathcal{U}\_k(\cdot, 0))(\xi), \tag{8}$$

where

$$\mathbb{A} := \begin{pmatrix} 0 & 1 \\ -2\|\underline{\mathcal{J}}\| & 0 \end{pmatrix}. \tag{9}$$

Above we have used the fact that F ´1 *k* " F*k* (see (3)). That is F*k*p*Uk*p¨, *t*qqp*ξ*q is a solution to the following ordinary differential equation

$$
\partial\_t \mathcal{F}\_k(\mathcal{U}\_k(\cdot, t))(\xi) = \mathbb{A} \mathcal{F}\_k(\mathcal{U}\_k(\cdot, t))(\xi) = \begin{pmatrix} 0 & 1 \\ -2\|\xi\| & 0 \end{pmatrix} \mathcal{F}\_k(\mathcal{U}\_k(\cdot, t))(\xi). \tag{10}
$$

Now, recall from (4) that ´}*ξ*}F*k*p*f*qp*ξ*q " F*k*p}*x*}Δ*k f*qp*ξ*q, and using the injectivity of the Fourier transform F*k*, we deduce that

$$
\hat{c}\_l \mathcal{U}\_k(\mathbf{x}, t) = \begin{pmatrix} 0 & 1 \\ 2 \|\mathbf{x}\| \Delta\_k & 0 \end{pmatrix} \mathcal{U}\_k(\mathbf{x}, t). \tag{11}
$$

Hence, if we write *Uk*p*<sup>x</sup>*, *t*q " ˜*uk*p*x*, *t*q *vk*p*<sup>x</sup>*, *t*q¸ , then *uk*p*<sup>x</sup>*, *t*q satisfies the following equation

$$
\hat{\sigma}\_{\text{H}} u\_k(\mathbf{x}, t) = \mathcal{Z} \|\mathbf{x}\| \Delta\_k u\_k(\mathbf{x}, t) \dots
$$

Moreover, since *f* , *g* P DpR*<sup>n</sup>*q, it follows from (7) and the properties of the --convolution that *uk*p¨, *t*q P *C*<sup>8</sup>pR*n*q for all *t* P R.

Furthermore, *uk*p*<sup>x</sup>*, *t*q Ñ *f*p*x*q as *t* Ñ 0. Indeed, if *δ* denotes the Dirac functional, then, as *t* Ñ 0, <sup>F</sup>*k*pcosp*t*a2} ¨ }qq Ñ *δ* in S<sup>1</sup>pR*n*q and thus in <sup>D</sup><sup>1</sup>pR*<sup>n</sup>*q. On the other hand <sup>F</sup>*k*psinp*t*a2} ¨ }q{a2} ¨ }q Ñ 0 as *t* Ñ 0. Using the continuity of the convolution -, we deduce that

$$
\mu\_k(\mathfrak{x}, t) \to (\delta \otimes f)(\mathfrak{x}) = f(\mathfrak{x}) \qquad \text{as} \quad t \to 0.
$$

Similarly, one can prove that pB*tuk*qp*<sup>x</sup>*, *t*q Ñ *g*p*x*q as *t* Ñ 0.

We mention that the solution *uk* constructed above is unique. To prove this claim, we need the lemma below. Let

$$\mathcal{E}\_{k}[\boldsymbol{u}\_{k}](t) = \int\_{\mathbb{R}^{n}} \left\{ |\hat{c}\_{t} \mathcal{F}\_{k}(\boldsymbol{u}\_{k}(\cdot, t))(\boldsymbol{\xi})|^{2} + 2 \|\boldsymbol{\xi}\| |\mathcal{F}\_{k}(\boldsymbol{u}\_{k}(\cdot, t))(\boldsymbol{\xi})|^{2} \right\} \theta\_{k, 1}(\boldsymbol{\xi}) d\boldsymbol{\xi}. \tag{12}$$

**Lemma 1.** *Assume that n* ` 2x*k*y ´ 1 ą 0 *and that the initial data f* , *g* P <sup>D</sup>pR*<sup>N</sup>*q. *Then the total energy* E*k*r*uk*s *is independent of t*.

**Proof.** Since

$$\mathcal{F}\_k(u\_k(\cdot, t))(\xi) = \cos(t\sqrt{2\|\xi\|})\mathcal{F}\_k f(\xi) + \frac{\sin(t\sqrt{2\|\xi\|})}{\sqrt{2\|\xi\|}}\mathcal{F}\_k g(\xi), \qquad \text{for all } t \in \mathbb{R}\_+$$

we deduce that

$$\begin{split} |\mathcal{F}\_{k}(\boldsymbol{u}\_{k}(\cdot,t))(\boldsymbol{\xi})|^{2} &= \quad \cos^{2}(t\sqrt{2|\boldsymbol{\xi}|})|\mathcal{F}\_{k}f(\boldsymbol{\xi})|^{2} + \frac{\sin^{2}(t\sqrt{2|\boldsymbol{\xi}|})}{2|\boldsymbol{\xi}|}|\mathcal{F}\_{k}\mathcal{F}(\boldsymbol{\xi})|^{2} \\ &+ \sqrt{2} \frac{\cos(t\sqrt{|\boldsymbol{\xi}|})\sin(t\sqrt{|\boldsymbol{\xi}|})}{\sqrt{|\boldsymbol{\xi}|}} \text{Re}\left(\mathcal{F}\_{k}f(\boldsymbol{\xi})\overline{\mathcal{F}\_{k}\mathcal{F}(\boldsymbol{\xi})}\right) . \end{split}$$

and

$$\begin{split} \left| \left| \partial\_{t} \mathcal{F}\_{k} (\boldsymbol{u}\_{k} (\cdot, t)) (\boldsymbol{\xi} ) \right|^{2} \right| &= \left| \cos^{2} (t \sqrt{2 \| \boldsymbol{\xi} \|}) | \mathcal{F}\_{k} \mathcal{g} (\boldsymbol{\xi} ) |^{2} + 2 | \boldsymbol{\xi} | \sin^{2} (t \sqrt{2 \| \boldsymbol{\xi} \|}) | \mathcal{F}\_{k} \boldsymbol{f} (\boldsymbol{\xi} ) |^{2} \\ &\quad - 2 \sqrt{2 \| \boldsymbol{\xi} \|} \cos (t \sqrt{2 \| \boldsymbol{\xi} \|}) \sin (t \sqrt{2 \| \boldsymbol{\xi} \|}) \text{Re} \left( \mathcal{F}\_{k} \boldsymbol{f} (\boldsymbol{\xi} ) \overline{\mathcal{F}\_{k} \boldsymbol{g} (\boldsymbol{\xi} )} \right). \end{split}$$

.

Thus we have

$$\mathcal{E}\_k[\mu\_k](t) = \int\_{\mathbb{R}^n} \left\{ 2|\xi| |\mathcal{F}\_k f(\xi)|^2 + |\mathcal{F}\_k \mathcal{g}(\xi)|^2 \right\} \theta\_{k,1}(\xi) d\xi. \tag{13}$$

Hence, we established the lemma.

Now let us go back to the uniqueness of the solution *uk*. Assume that *u*p1<sup>q</sup> *k* and *u*p2<sup>q</sup> *k* are two solutions of the wave equation with the same initial data, then *u*p1<sup>q</sup> *k* ´ *u*p2<sup>q</sup> *k* is a solution of the wave equation with zero initial data. Therefore, by (13), we have <sup>E</sup>*k*r*u*p1<sup>q</sup> *k* ´ *u*p2<sup>q</sup> *k* sp*t*q " 0. Hence, (12) implies <sup>B</sup>*t*F*k*pp*u*p1<sup>q</sup> *k* ´ *u*p2<sup>q</sup> *k* qp¨, *t*qqp*ξ*q " 0 for every *t* P R. That is, the function *t* ÞÑ <sup>F</sup>*k*pp*u*p1<sup>q</sup> *k* ´ *u*p2<sup>q</sup> *k* qp¨, *t*qqp*ξ*q is a constant, which implies <sup>F</sup>*k*pp*u*p1<sup>q</sup> *k* ´ *u*p2<sup>q</sup> *k* qp¨, *t*qqp*ξ*q " <sup>F</sup>*k*pp*u*p1<sup>q</sup> *k* ´ *u*p2<sup>q</sup> *k* qp¨, 0qqp*ξ*q " 0. Using the injectivity of the Fourier transform F*k*, we deduce that p*u*p1<sup>q</sup> *k* ´ *u*p2<sup>q</sup> *k* qp*<sup>x</sup>*, *t*q " 0 for all *x* P R*n* and *t* ą 0. This proves that the solutions of the wave equation are uniquely determined by the initial Cauchy data.

The following theorem collects all the above facts and discussions.

**Theorem 1.** *The solution to the Cauchy problem* (5) *is given uniquely by*

$$
\mu\_k(\mathfrak{x}, t) = (P\_{k,t}^{11} \circledast f)(\mathfrak{x}) + (P\_{k,t}^{12} \circledast \mathfrak{g})(\mathfrak{x})\_{\mathsf{x}},
$$

*where, for a fixed t*, *P*11*k*,*tand P*12*k*,*tare the tempered distributions on* R*n given by*

$$P\_{k,t}^{11} = \mathcal{F}\_k\left(\cos(t\sqrt{2\|\cdot\|})\right), \qquad P\_{k,t}^{12} = \mathcal{F}\_k\left(\sin(t\sqrt{2\|\cdot\|})/\sqrt{2\|\cdot\|}\right).$$

*The distributions Pij k*,*twill be called the propagators.*

We shall now prove the statement (S). To do so, we will assume that the propagators *P*11*k*,*t* and *P*12*k*,*t* are supported entirely on the set C " tp*<sup>x</sup>*, *t*q P R*n* ˆ R : }*x*} " 12 *t*2u and we will show that this assumption cannot hold. Our approach uses the representation theory of the Lie algebra slp2, Rq, following [43,44].

Assume that the propagators *P*11*k*,*t* and *P*12*k*,*t* are supported entirely on the set C " tp*<sup>x</sup>*, *t*q P R*n* ˆ R : }*x*} " 12*<sup>t</sup>*<sup>2</sup>u.

We start by investigating certain properties of the wave equation, which are reflected in properties of the propagators. To see this, we define the 2 ˆ 2 matrix *Pk* " ˜*P*11*k <sup>P</sup>*12*k <sup>P</sup>*21*k <sup>P</sup>*22*k* ¸ of entrywise distributions onR*n*`1,where

$$P\_k^{ij}(\psi\_1 \otimes \psi\_2) := \int\_{\mathbb{R}} P\_{k,t}^{ij}(\psi\_1) \psi\_2(t) dt, \qquad i, j = 1, 2, 3$$

for *ψ*1 P SpR*n*q and *ψ*2 P SpRq. Here we used the fact that SpR*n*`<sup>1</sup>q » SpR*n*qb<sup>p</sup>SpRq is the unique topological tensor product of SpR*n*q and SpRq as nuclear spaces. From the constructive proof of Theorem 1, it follows that

$$2\|\mathbf{x}\|\Delta\_k P\_k^{i\bar{j}} = \partial\_{tt} P\_k^{i\bar{j}} \qquad i,\mathbf{j} = 1,2... $$

Next, we will investigate the dilations of the propagators under a dilation operator. This will inform us on the degree of the "homogeneity" of the distributions *Pij k* , with *i*, *j* " 1, 2. For *λ* ą 0 and a function *ψ* on R*n*`1, let

$$S^\chi\_\lambda \psi(\mathbf{x}, t) := \psi(\lambda^2 \mathbf{x}, t), \qquad S^t\_\lambda \psi(\mathbf{x}, t) := \psi(\mathbf{x}, \lambda t),$$

where the superscript denotes the relevant variable. Set *Sλ* :" *Sxλ* ˝ *<sup>S</sup><sup>t</sup>λ*. By duality, the operators *Sxλ*, *<sup>S</sup><sup>t</sup>λ*, and *Sλ* act on distributions in the standard way.

We begin by looking to the properties of *Pij k*,*t* under the dilation *Sλ*. Observe that if *uk*p*<sup>x</sup>*, *t*q is a solution to (5) with initial data p*f*p*x*q, *g*p*x*qq, then *Sλuk*p*<sup>x</sup>*, *t*q solves the wave equation with initial data p*Sxλ f*p*x*q, *<sup>λ</sup>Sxλg*p*x*qq. Thus

$$S\_{\lambda} \wr L\_{k}(\mathbf{x}, t) = P\_{k, t} \oplus \begin{bmatrix} S\_{\lambda}^{x} f \\ \lambda S\_{\lambda}^{x} g \end{bmatrix}. \tag{14}$$

On the other hand

$$\begin{aligned} S\_{\lambda} \mathcal{U}\_{k}(\mathbf{x},t) &= \left[ \begin{array}{c} S\_{\lambda} u\_{k}(\mathbf{x},t) \\ \end{array} \right] = \left[ \begin{array}{c} u\_{k}(\lambda^{2} \mathbf{x}, \lambda t) \\ \lambda \{\hat{c}\_{t} u\_{k}(\lambda^{2} \mathbf{x}, \lambda t) \end{array} \right] \\ &= \left[ \begin{array}{c} u\_{k} \\ \lambda \hat{c}\_{t} u\_{k} \end{array} \right] (\lambda^{2} \mathbf{x}, \lambda t) \\ &= \left[ \begin{array}{c} 1 & 0 \\ 0 & \lambda \end{array} \right] \left[ \begin{array}{c} u\_{k} \\ \end{array} \right] (\lambda^{2} \mathbf{x}, \lambda t) \\ &= \left[ \begin{array}{c} 1 & 0 \\ 0 & \lambda \end{array} \right] \left\{ P\_{k, \lambda t} \otimes \begin{bmatrix} f \\ g \\ \end{bmatrix} \right\} (\lambda^{2} \mathbf{x}) \\ &= \left[ \begin{array}{c} 1 & 0 \\ 0 & \lambda \end{array} \right] S\_{\lambda}^{x} \left\{ P\_{k, \lambda t} \otimes \begin{bmatrix} f \\ g \\ \end{bmatrix} \right\} (\lambda^{2} \mathbf{x}). \end{aligned}$$

Using the fact that *Sxλ* preserves the convolution of a distribution with a function, a fact that can be easily checked using the properties of the translation operator, we ge<sup>t</sup>

$$\begin{aligned} \label{eq:SDAL-1} S\_{\lambda} \boldsymbol{L} \mathbf{f}\_{k} (\mathbf{x}, t) &= \begin{bmatrix} 1 & 0 \\ 0 & \lambda \end{bmatrix} \left\{ S\_{\lambda}^{\mathbf{x}} \boldsymbol{P}\_{k, \lambda t} \oplus \begin{bmatrix} S\_{\lambda}^{\mathbf{x}} f \\ S\_{\lambda}^{\mathbf{x}} \boldsymbol{g} \end{bmatrix} \right\} (\mathbf{x}) \\ &= \begin{bmatrix} 1 & 0 \\ 0 & \lambda \end{bmatrix} \left\{ S\_{\lambda}^{\mathbf{x}} \boldsymbol{P}\_{k, \lambda t} \oplus \begin{bmatrix} 1 & 0 \\ 0 & \lambda^{-1} \end{bmatrix} \begin{bmatrix} S\_{\lambda}^{\mathbf{x}} f \\ \lambda S\_{\lambda}^{\mathbf{x}} \boldsymbol{g} \end{bmatrix} \right\} (\mathbf{x}). \end{aligned} \tag{15}$$

Comparing (14) with (15) gives *SxλPij k*,*λt* " *<sup>λ</sup>j*´*iPij k*,*t*, for *i*, *j* " 1, 2. Now we can obtain the dilation properties of *Pij k* as follows: For *ψ*1 P SpR*n*q and *ψ*2 P SpRq, we have

$$\begin{split} S\_{\lambda}(P\_{k}^{ij})(\psi\_{1}\otimes\psi\_{2}) &= \ &P\_{k}^{ij}(S\_{\lambda-1}^{x}(\psi\_{1})\otimes S\_{\lambda-1}^{t}(\psi\_{2})) \\ &= &\int\_{\mathbb{R}}P\_{k,t}^{ij}(S\_{\lambda-1}^{x}(\psi\_{1}))S\_{\lambda-1}^{t}(\psi\_{2})(t)dt \\ &= &\lambda\int\_{\mathbb{R}}P\_{k,\lambda t}^{ij}(S\_{\lambda-1}^{x}(\psi\_{1}))\psi\_{2}(t)dt \\ &= &\lambda\int\_{\mathbb{R}}S\_{\lambda}^{x}(P\_{k,\lambda t}^{ij}(\psi\_{1}))\psi\_{2}(t)dt \\ &= &\lambda^{1+j-i}\int\_{\mathbb{R}}P\_{k,t}^{ij}(\psi\_{1})\psi\_{2}(t)dt \\ &= &\lambda^{1+j-i}P\_{k}^{ij}(\psi\_{1}\otimes\psi\_{2}). \end{split}$$

We summarize the above computations.

**Proposition 1.** *For n* ` 2x*k*y ´ 1 ą 0*, we have*

(1) *The distribution Pij ksatisfies the deformed wave equation, i.e.,*

$$(\|\mathbf{x}\|\Delta\_k - \frac{1}{2}\hat{c}\_{\text{It}})P\_k^{ij} = 0, \qquad i, j = 1, 2. \tag{16}$$

(2) *For λ* ą 0,

$$S\_{\lambda}P\_k^{i\dot{j}} = \lambda^{1+j-i}P\_{k\,\,\,\prime}^{i\dot{j}} \qquad i,\,\dot{j} = 1,2...$$

Next we shall describe the structure of a representation of the Lie algebra slp2, Rq on <sup>S</sup>pR*<sup>n</sup>*`<sup>1</sup>q. This structure, together with Proposition 1, will allow us to prove that the Assumption 3 does not hold true.

We take a basis for the Lie algebra slp2, Rq as

$$\mathbf{e} := \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \qquad \mathbf{f} := \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \qquad \mathbf{h} := \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$

The triple t**<sup>e</sup>**,**f**, **h**u satisfies the commutation relations

$$[\mathbf{e}, \mathbf{f}] = \mathbf{h}, \qquad [\mathbf{h}, \mathbf{e}] = 2\mathbf{e}, \qquad [\mathbf{h}, \mathbf{f}] = -2\mathbf{f}\_{\prime}$$

where r*A*, *B*s :" *AB* ´ *BA*.

Choose *x*1, *x*2,..., *xn* as the usual system of coordinates on R*<sup>n</sup>*. Let

$$\mathbb{E}\_{n,1} := i(\|\mathbf{x}\| - \frac{1}{2}t^2), \quad \mathbb{F}\_{n,1} := i(\|\mathbf{x}\| \Delta\_k - \frac{1}{2}\hat{c}\mathbb{I}\_\ell), \quad \mathbb{H}\_{n,1} := n + 2\langle k \rangle - \frac{1}{2} + 2\sum\_{\ell=1}^n \mathbf{x}\_\ell \hat{c}\_\ell + t\hat{c}\_\ell.$$

Using ([14] Theorem 3.2), the following commutation relations hold

$$\left[\mathbb{E}\_{n,1}, \mathbb{F}\_{n,1}\right] = \mathbb{H}\_{n,1}, \quad \left[\mathbb{H}\_{N,1}, \mathbb{E}\_{N,1}\right] = 2\mathbb{E}\_{N,1}, \quad \left[\mathbb{H}\_{n,1}, \mathbb{F}\_{n,1}\right] = -2\mathbb{F}\_{n,1}.\tag{17}$$

These are the commutation relations of a standard basis of the Lie algebra slp2, Rq. Equation (17) gives rise to a representation *ωk* of the Lie algebra slp2, Rq on the Schwartz space SpR*n*`<sup>1</sup>q by setting

$$
\omega\_k(\mathbf{h}) = \mathbb{H}\_{n,1\prime} \qquad \omega\_k(\mathbf{e}) = \mathbb{E}\_{n,1\prime} \qquad \omega\_{k,a}(\mathbf{f}) = \mathbb{F}\_{n,1} \tag{18}
$$

An analogue of the representation *ωk* has been intensively studied in [14].

Recall that the Huygens' principle is equivalent to the fact that the propagators *<sup>P</sup>*11*k* and *<sup>P</sup>*12*k* are supported on the set C " tp*<sup>x</sup>*, *t*q P R*n* ˆ R | }*x*} ´ 12 *t*2 " 0u. Since C is the locus of zeros of }*x*} ´ 12 *t*2, then, *Pij k*is supported on C if and only if

$$\mathbb{E}\_{n,1}^m \cdot P\_k^{jj} = 0 \tag{19}$$

for some positive integer *m* (see, for instance, ([44] p. 173)). In the light of Proposition 1(1) together with the dilatation property of *Pij k* , which implies that *Pij k* is an eigen-distribution for H*<sup>n</sup>*,1, Equation (19) amounts to saying the distribution *Pij k* generates a finite-dimensional representation *<sup>ω</sup>*˚*k* for slp2, Rq on <sup>S</sup><sup>1</sup>pR*<sup>n</sup>*`<sup>1</sup>q. Thus, the qualitative part of Huygens' principle holds.

**Theorem 2.** *Huygens' principle holds if and only if Pij k is supported on the set* C*, if and only if Pij k generates a finite-dimensional representation <sup>ω</sup>*˚*k for* slp2, Rq *on* <sup>S</sup><sup>1</sup>pR*<sup>n</sup>*`<sup>1</sup>q.

**Theorem 3.** *Huygens' principle cannot hold when*

$$n - 1 + 2\langle k \rangle \notin \mathbb{Z}\_{\cdot}$$

**Proof.** In ([14] Theorem 3.21) the authors proved that the spectrum of the element **k** :" *i*p**f** ´ **e**q acting on S<sup>1</sup>pR*n*`<sup>1</sup>q via the dual representation *<sup>ω</sup>*˚*k* is *n* ´ 1 ` 2x*k*y ` 2Z, whereas, it is well known, the spectrum of **k** in finite-dimensional representations of slp2, Rq is contained in Z.

The above theorem leaves the possibility that the wave equation may satisfy Huygens' principle when *n* ´ 1 ` 2x*k*y P Z.

Now, using Proposition 1(2), we ge<sup>t</sup>

$$\left\{ 2\sum\_{\ell=1}^{n} \mathbf{x}\_{\ell} \hat{\boldsymbol{\mathcal{O}}}\_{\ell} + t \hat{\boldsymbol{\mathcal{O}}}\_{\ell} \right\} P\_k^{ij} = (1 + j - i) P\_k^{ij}.$$

Therefore

$$\mathbb{H}\_{\mathfrak{n},1} P\_k^{ij} = -\left(n + 2\langle k \rangle - \frac{1}{2} + i - j - 1\right) P\_{k'}^{ij}, \qquad i, j = 1, 2.1$$

That is *Pij k* is an eigendistributation for H*<sup>n</sup>*,<sup>1</sup> with eigenvalue ´p*n* ` 2x*k*y ´ 12 ` *i* ´ *j* ´ <sup>1</sup>q. Keeping in mind the fact that F*<sup>n</sup>*,<sup>1</sup> ¨ *Pij k* " 0, and in the light of Theorem 3, clearly each distribution *Pij k* cannot generate a finite-dimensional *<sup>ω</sup>*˚*k* for slp2, Rq on <sup>S</sup><sup>1</sup>pR*<sup>n</sup>*`<sup>1</sup>q; otherwise *n* ` 2x*k*y ´ 12 ` *i* ´ *j* ´ 1 P Z which is impossible in view of Theorem 3. That is our Assumption 3 does not hold true.

**Theorem 4.** *The solution uk*p*<sup>x</sup>*, *t*q *to the Cauchy problem* (5) *does not satisfy the Huygens' principle.*

**Author Contributions:** Investigation, S.a.-B., M.a.-K., A.a.-M., F.a.-S.; methodology, S.B.S.; supervision, S.B.S.; writing–original draft preparation, S.B.S.; writing–review and editing, S.B.S. All authors have read and agreed to the published version of the manuscript.

**Funding:** This research was funded by United Arab Emirates University (UAEU), SURE Plus 2019 gran<sup>t</sup> # G00003109.

**Acknowledgments:** The authors would like to thankfully acknowledge the financial support awarded by UAEU through the SURE Plus 2019 gran<sup>t</sup> # G00003109. We thank the referees for their valuable comments.

**Conflicts of Interest:** The authors declare no conflict of interest.
