Wave equation II: Qualitative Properties of solutions

Chapter 5 Wave equation II: Qualitative Properties of solutions In this hapter, we disuss some of the important qualitative properties of solutions to wave equation. Solutions of wave equation in one spae dimension have a speial property alled parallelogram identity, whih an be used to find solutions of ertain initialboundary value problems for wave equation and is disussed in Setion 5.. In Setion 5.2, we introdue two important onepts (whih are dual to eah other), namely domain of dependene and domain of influene, whih are exlusive to hyperboli equations in any spae dimension. We ame aross these onepts for a first order quasilinear PDE in Setion 2.2. Setion 5.3 disusses a ausality priniple whih is equivalent to onepts of domains of dependene and influene. Another equivalent formulation of domains of dependene and influene is in terms of finite speed of propagation whih is disussed in Setion 5.4. Huygens priniple is studied in Setion 5.6. We study energy of a solution to wave equation in Setion 5.5. Conept of a weak solution is defined in Setion 5.7, and propagation of onfined disturbanes is studied in Setion 5.8 in all spae dimensions. Propagation of singularities in solutions of wave equation is studied in Setion 5.9. Deay of solutions of wave eqution in two and three spae dimensions is studied in Setion 5.. 5. Parallelogram identity Solutions of one dimensional wave equation enjoy a speial property alled Parallelogram identity. Definition 5. (Charateristi Parallelogram). A parallelogram in the x t-plane is said to be a harateristi parallelogram if the sides of the parallelogram are along the harateristis. Theorem 5.2 (Parallelogram identity). Let P, Q, R, S be the verties of a harateristi parallelogram PQRS with P R and QS as its diagonals. Let u be a funtion having the form u(x, t) = F (x t) + G(x + t). (5.) Then the values of u at the verties satisfy the relation u(p) + u(r) = u(q) + u(s). (5.2) 7

8 Chapter 5. Wave equation II R(ξ r + s,τ + r +s ) S(ξ r,τ + r ) Q(ξ + s,τ + s ). P(ξ,τ) Figure 5.. haraterisit parallelogram. Proof: Let us first draw a diagram of a harateristi parallelogram and identify the oordinates of its verties. Without loss of generality, let us assume that the side PQ lies along the family of harateristis: x t = onstant, and that verties P, Q, R, S are desribed in anti-lokwise manner. Let us fix the oordinates of P to be P(ξ,τ). Thus the point Q lies on the line x t = ξ τ, and hene the oordinates of Q are of the form Q(ξ + s,τ + s ) for some s >. Sine S must lie on a line belonging to the family of harateristis: x + t = onstant, it must lie on the line x + t = ξ + τ. Thus the oordinates of S are of the form S(ξ r,τ + r ) for some r >. Now the oordinates of R are fixed and are given by R(ξ r + s,τ + r +s ). Thus we have Note that P(ξ,τ), Q ξ + s,τ + s, R ξ r + s,τ + r + s, S ξ r,τ + r. (5.3) u(p) + u(r) = F (ξ τ) + G(ξ + τ) + F (ξ r + s τ s r ) + G(ξ r + s + τ + s + r ) and = F (ξ τ) + F (ξ τ 2r ) + G(ξ + τ) + G(ξ + τ + 2s), (5.4) u(q) + u(s) = F (ξ + s τ s) + G(ξ + s + τ + s) + F (ξ r τ r ) + G(ξ r + τ + r ) = F (ξ τ) + F (ξ τ 2r ) + G(ξ + τ) + G(ξ + τ + 2s). (5.5) Equations (5.4) and (5.5) together imply the required equality (5.2). Remark 5.3. Reall from (4.4) that any C 2 solution of the wave equation is of the form (5.), and hene parallelogram law is satisfied by any lassial solution of the wave equation. Converse of this result holds, and is the ontent of the next result. Theorem 5.4. Let u : 3 be a thrie ontinuously differentiable funtion satisfying the equality u(p) + u(r) = u(q) + u(s) (5.6) for every harateristi parallelogram PQRS with P R and QS as its diagonals. Then u solves the wave equation u t t 2 u xx =. IIT Bombay

5.2. Domain of dependene, Domain of influene 9 Proof. Let the verties of the parallelogram be given by P(ξ,τ), Q ξ + s,τ + s, R ξ r + s,τ + r + s, S ξ r,τ + r, (5.7) where r > and s >. The idea of the proof is to write Taylor expansion up to seond order along with remainder term around the point P, for u(q), u(r), u(s), substitute in the equation (5.6), and then pass to the limit as r and s. Proof is left to the reader as an exerise. 5.2 Domain of dependene, Domain of influene Let u be a solution of the Cauhy problem (4.2) for homogeneous wave equation. We are interested in knowing answers to the following two questions regarding the nature of the relation between solution at a point and Cauhy data. Question Let (x, t ) d (, ). How muh of the Cauhy data plays a role in determining the value of u(x, t )? Question 2 Let x d. What are all the points in spae-time (x, t) d (, ) suh that x plays a role in determining the value of u(x, t) through Cauhy data? Answers to the Questions and 2, are known as Domain of dependene and Domain of influene respetively. However the preise domains of dependene and influene need to be omputed for eah value of the dimension d. We also remark that the above questions are irrelevant for Cauhy problem (4.) for non-homogeneous wave equation, as the answers will be trivial, in the presene of soure terms, as an be seen from the formulae of solutions. In the rest of this setion, we will determine the domains of dependene and influene, using the formulae of solutions that were determined in Setion 4.. 5.2. Case of one dimensional wave equation Domain of Dependene The d Alembert formula for the solution of Cauhy problem for homogeneous wave equation is u(x, t ) = φ(x t ) + φ(x + t ) + x + t ψ(s) d s. 2 2 x t From the above formula, it is evident that in order to ompute the solution at a point (x, t ), the values of φ are needed at just two points x t and x + t, and the values of ψ are needed in the interval [x t, x + t ]. Thus the domain of dependene is the interval [x t, x + t ]. See Figure 5.2 for an illustration. In other words, if we onsider two sets of Cauhy data (φ,ψ) and (φ,ψ ) suh that φ(x) φ (x) and ψ(x) ψ (x) on the interval [x t, x + t ], then solutions of both the Cauhy problems oinide at the point (x, t ). In fat, both the solutions oinide in the whole of haraterisit triangle, a triangle with verties at (x, t ), (x t,), and (x + t,). In partiular, hanging the Cauhy data outside the interval [x t, x + t ] has no effet on the solution at the point (x, t ). That is the effet of hange in initial data is not felt at the point x for all times t t. Thus we may say that the solution at (x, t ) has a domain of dependene given by the interval [x t, x + t ]. September 7, 25

2 Chapter 5. Wave equation II t (x, t ) x t = x t x + t = x + t. x x t x + t Figure 5.2. Wave equation in one spae dimension: Domain of dependene for solution at (x, t ). Domain of Influene Let [a, b] be an interval on the x-axis on whih Cauhy data is presribed. We are interested in the domain of influene of the set of points in the interval, and it should be union of domains of influene of eah of the points in [a, b]. Domain of influene of a point x on the x-axis is preisely the set of all points (x, t) (, ) whose domain of dependene ontains the point (x,). Sine domain of dependene of solution at (x, t) is the interval [x t, x + t], the domain of influene of x is {(x, t) (, ) : x t x x + t}. (5.8) Thus domain of influene of the interval [a, b] turns out to be the set of all those points (x, t) suh that the domain of dependene of the solution at (x, t) has a non-empty intersetion with [a, b]. That is the domain of of influene of the interval [a, b] is given by {(x, t) (, ) : x t b, a x + t}. (5.9) Domain of influene of a point (x,) is illustrated in Figure 5.3, and that of an interval [a, b] is illustrated in Figure 5.4 respetively. 5.2.2 Case of two dimensional wave equation The solution of Cauhy problem for two dimensional wave equation is given by u(x, x 2, t) = t 2π + 2π D((x,x 2 ), t) D((x,x 2 ), t) φ(y, y 2 ) 2 t 2 (x y ) 2 (x 2 y 2 ) d y 2 d y 2 ψ(y, y 2 ) 2 t 2 (x y ) 2 (x 2 y 2 ) 2 d y d y 2, where D((x, x 2 ), t) denotes the open disk with enter at (x, x 2 ) having radius t. IIT Bombay

5.2. Domain of dependene, Domain of influene 2 x + t = x Domain of influene of (x,) x t = x t. x x Figure 5.3. Wave equation in one spae dimension: Domain of influene of the point (x,). Domain of influene of the interval [a, b] x + t = a x t = b t a. x b Figure 5.4. Wave equation in one spae dimension: Domain of influene of the interval [a, b]. In this ase, domain of dependene for the solution at (x, x 2 ) is learly the open disk D((x, x 2 ), t) having its enter at (x, x 2 ), and having radius t. Domain of influene of a point (y, y 2 ) is given by {(x, x 2, t) : (x, x 2 ) (y, y 2 ) < t}, (5.) whih is the olletion of all those points x whih an be reahed upto time t from y. 5.2.3 Case of three dimensional wave equation The solution of Cauhy problem for three dimensional wave equation is given by u(x, t) = φ(y) dσ + ψ(y) dσ. (5.) t 4π 2 t S(x, t) 4π 2 t S(x, t) In this ase, domain of dependene for the solution at (x, t) is the sphere S(x, t). Domain of influene of a point y is given by {(x, t) : x y = t}, (5.2) September 7, 25

22 Chapter 5. Wave equation II whih is the olletion of all those points x whih an be reahed exatly at time t from y. 5.3 Causality priniple Causality stands for ause and effet. What are the reasons (in the past) that are responsible for the urrent state? What will be the future events for whih the urrent state is responsible for? These questions were answered in Setion 5.2 using the expliit formulae for solutions of Cauhy problem. In this setion, we attempt to answer these questions without using the formulae. This is a typial illustration of an a priori analysis where onlusions are drawn on the solution without the knowledge of its existene. In this disussion we swith-off the nonhomogeneous term in the wave equation, and onsider effets of the Cauhy data. One an also keep the nonhomogeneous term and study these questions; whih is left as an exerise to the reader. Theorem 5.5 (Causality Priniple). Let u : d (, ) be a lassial solution of the Cauhy problem for a homogeneous wave equation, i.e., u is a solution of u u t t 2 u x x + u x2 x 2 + + u xd x d =, x d, t >. (5.3a) u(x,) = φ(x), x d, (5.3b) u t (x,) = ψ(x), x d. (5.3) Let (x, t ) d (, ). The value of u(x, t ) depends only on the values of φ and ψ in the losure of the ball B(x ; t ) with enter at x d, and having a radius of t, lying in d {}. Proof. The theorem follows immediately from the formulae for the solution of (5.3) derived in Setion 4., namely the equations (4.7) for d =, the equation (4.45) for d = 3, and the equation (4.56) for d = 2. However we would like to give a proof of this result without using the expliit formulae. In the next two subsetions we will prove this theorem diretly, for d = and then for a general d respetively. 5.3. Proof of ausality priniple for d = Proof. [of Causality priniple] Consider a point (x, t ) (, ). Consider the harateristi triangle, whih is a triangle formed by the harateristis x t = x t, x + t = x + t, and the x-axis. Fix a T suh that < T < t, and draw the line t = T. Consider the trapezium, denoted by F, formed by the lines x t = x t, x + t = x + t, t = T, t =. Multiplying the wave equation u t t 2 u x x = with u t, we see that = u t t 2 u xx IIT Bombay

5.3. Causality priniple 23 = 2 u2 t + 2 2 (u t u x ) x t = ( x, t ). 2 2 2 u t u x, u2t + (5.4) Integrating the equation (5.4) on the trapezium F, we get ( x, t ). 2 2 2 u t u x, u2t + d xd t =. F Using integration by parts formula in the last equation, we get = ( x, t ). 2 2 2 u t u x, u2t + F = 2 2 2 u t u x, u2t + F.n dσ, where n is the unit outward normal to the boundary of F. The boundary of the trapezium F onsists of four lines: They are (i) The base of the trapezium, denoted by B, given by the equation t =. The outward unit normal to B is given by n = (, ). (ii) A part of the harateristi, denoted by K, given by the equation x + t = x + t. The outward unit normal to K is given by n = (, ). + 2 (iii) The upper part of the trapezium, denoted by T, given by the equation t = T. The outward unit normal to T is given by n = (,). (iv) A part of the harateristi, denoted by K 2, given by the equation x t = x t. The outward unit normal to K 2 is given by n = (, ). + 2 Thus we have = F Thus we get 2 2 2 u t u x, u2t + = 2 2 2 u t u x, u2t + F.n dσ = B Observe that 2 u t u x + K + 2 and 2 u t u x + K 2 + 2.n dσ = 2 2 2 u t u x, u2t + B K T K 2.n dσ. (5.5) 2 u2 t + 3 2 u2 t + 3 + T 2 u2 t + 2 2 u2 t + 2 dσ + dσ + 2 u t u x + K + 2 2 u t u x + K 2 + 2 2 u2 t + 3 2 u2 t + 3 dσ 2 (u t u x ) 2 dσ, (5.7) + 2 K dσ = 2 (u t + u x ) 2 dσ. (5.8) + 2 K 2 dσ. dσ (5.6) September 7, 25

24 Chapter 5. Wave equation II We now onlude from (5.6), in view of (5.7) and (5.8), that T 2 u2 t + 2 dσ B 2 u2 t + 2 dσ. (5.9) The inequality (5.9) is known as domain of dependene inequality. Let u and v be solutions of the Cauhy problem (5.3) (for d = ). Let us denote by w the differene of u and v, i.e., w := u v. Note that w solves the wave equation (due to linearity of the wave equation), and the Cauhy data satisfied by the funtioin w are zero funtions, as both u and v are solutions of the same Cauhy problem. Thus w(x,) w t (x,) on B. As a onsequene of the domain of dependene inequality, we get 2 w2 t + 2 2 w2 x dσ. (5.2) T As a result, w t (x,t ) = w x (x,t ) =. Sine T < t is arbitrary, we onlude that w t (x, t) = w x (x, t) = (5.2) for every (x, t) belonging to the harateristi triangle. This implies that w is a onstant funtion, and sine w = on B, it follows that w is the zero funtion inside the harateristi triangle. In partiular, u(x, t ) = v(x, t ). This finishes proof of the theorem. Remark 5.6 (Consequenes of Domain of dependene inequality). The above disussion shows that u(x, t ) depends solely on the values of the Cauhy data on the base of the harateristi triangle. In other words, the Cauhy data φ,ψ at a spatial point x an influene the solution only in the region enlosed by the two harateristis starting from (x,) whih is given by {(x, t) (, ) : x t x x + t}. (5.22) 5.3.2 Proof of ausality priniple for general d Charateristi one a.k.a. Light one Definition 5.7 (Charateristi one, Light one).. The harateristi one (also alled light one) at a point (x, t ) d (, ) is defined as the set (x, t) d (, ) : x x = t t, (5.23) where x x is the eulidean distane in d between x and x. 2. The harateristi one at a point (x, t ) d (, ) together with its interior is alled the solid light one. That is, solid light one is defined as the set (x, t) d (, ) : x x t t, (5.24) IIT Bombay

5.3. Causality priniple 25 t F or war d C one F ut u r e C one (x, t ) 2 or 3 Bakwar d C one Pas t C one.. Figure 5.5. Wave equation in d (d = 2,3): Bakward and Forward (solid) light ones. Remark 5.8 (On light one). (i) Note that the definition of a haraterisit one (given in Definition 5.7) is onsistent with the definition of a haraterisit hypersurfae as introdued in Setion 3.5 of Chapter 3. From (3.7), reall that the analyti haraterization for a hypersurfae given by Γ : φ(x, t) = to be a haraterisit surfae is φ.(a φ) =, (5.25) where A is the diagonal matrix diag( 2, 2,, 2,) for the wave equation in d spae dimensions. Thus the equation (5.25) takes the form φ 2 t 2 (φ x x + φ x2 x 2 + + φ xd x d ) =. (5.26) Note that φ(x, t) = 2 (t t ) 2 x x 2 is a solution of the eqution (5.26), and φ(x, t) = is nothing but the haraterisit one through the point (x, t ). (ii) Charateristi one is alled light one, as it isthe union of all light rays that emanate from (x, t ) whih travel at the speed, i.e., d x =. In other words, {(x, t) : 2 (t t ) 2 = x x 2 } = d t v d, v = {(x, t) : x = x + v(t t )}. (iii) Eah t-ross-setion of the (solid) light one is a (resp. solid) sphere. That is, for eah fixed T, the intersetion of (solid) light one with the hyperplane t = T is a d- dimensional (resp. solid) sphere lying in the hyperplane t = T. When the t-setions September 7, 25

26 Chapter 5. Wave equation II are projeted in the spae d, they are spheres with enter x, and radius (t t ). Clearly as t, the spheres are expanding. (iv) Like in the ase of d =, we an see the solid light one as a union of the past and future half-ones. As in the ase of d =, we are going to integrate ertain quantities on the solid light one and then we would perform integration by parts, whih requires the knowledge of the outward unit normal to the light one, whih is the boundary of the solid light one. Let us now ompute the outward unit normal at points on the light one (5.23). Note that the light one an be written as a level surfae L(x, t) x x 2 2 (t t ) 2 =. (5.27) Normal to a level surfae is given in terms of the gradient of the funtion defining the level surfae. Thus the unit normal vetors are given by n = ± L L (5.28) Thus we have n = ± (x x, 2 (t t ) x x 2 + 4 (t t ) 2 (5.29) Sine the point (x, t ) lies on the light one (5.23), we have 4 (t t ) 2 = x x 2. This simplifies the expression (5.29) to x x n = ± + 2 x x, t t. (5.3) t t Proof. [of Causality priniple for general d] Multiplying the homogeneous wave equation (5.3a) with u t, and re-arranging the terms, yields d i= x i 2 u t u xi + t 2 u2 t + 2 2 u 2 = (5.3) Let T < t, and F denote the frustum of the solid light one, ontained between the hyperplanes t = and t = T. Integrating the equation (5.3) on the frustum F, we get F,,, x x n t. 2 u t u x,, 2 2 2 u t u xd, u2t + 2 u 2 d xd t =. Using integration by parts formula in the last equation, we get 2 u t u x,, 2 2 2 u t u xd, u2t + 2 u 2.n dσ =, (5.32) F where n is the unit outward normal to the boundary of F. The boundary of the Frustum F onsists of three parts: They are (i) The base of the frustum, denoted by B, given by the equation t =. The outward unit normal to B is given by n = (,,, ). IIT Bombay

5.3. Causality priniple 27 (ii) Mantle, denoted by K, is the lateral part of the Frustum, whih is part of the light one. The unit normals to K was already omputed earlier, see (5.3), and we need to hoose whih one is the outward unit normal. The outward unit normal has a positive t-omponent, and is thus given by n = + 2 x x x x, t t t t. (5.33) Sine t < t on the mantle K, the outward normal beomes x x n = + 2 x x,. (5.34) (iii) The upper part of the frustum, denoted by T, given by the equation t = T. The outward unit normal to T is given by n = (,,,). The equation (5.32) beomes 2 u t u x,, 2 2 2 u t u xd, u2t + 2 u 2.n dσ =, (5.35) B T K Let us ompute the integral on B, whih is given by 2 u t u x,, 2 2 2 u t u xd, u2t + B 2 u 2.n dσ = B 2 u2 t + 2 2 u 2 dσ. (5.36) Let us ompute the integral on T, whih is given by 2 u t u x,, 2 2 2 u t u xd, u2t + T 2 u 2.n dσ = T 2 u2 t + 2 2 u 2 dσ. (5.37) Let us ompute the integral on K, whih is given by K 2 u t u x,, 2 2 2 u t u xd, u2t + + 2 K 2 u 2.n dσ = 2 u t u x,, 2 2 2 u t u xd, u2t + 2 u 2 x x. x x, dσ. We prove that the integral on K is non-negative by showing that the integrand is a non-negative funtion. The integrand is 2 u t u x,, 2 2 2 x x u t u xd, u2t + 2 u 2. x x,, (5.38) whih is equal to 2 u2 t + 2 x x 2 u 2 u t u. x x, (5.39) whih an be easily seen to be equal to x x 2 u 2 t u. + 2 x x 2 u x x u. x x 2, (5.4) x x x x September 7, 25

28 Chapter 5. Wave equation II whih is learly non-negative. Using the information about the integrals on T,B,K that we have, and the equation(5.35), we get the inequality T 2 u2 t + 2 2 u 2 dσ B 2 u2 t + 2 2 u 2 dσ. (5.4) The inequality (5.4) is also known as domain of dependene inequality. Let u and v be solutions of the Cauhy problem (5.3). Let us denote by w the differene of u and v, i.e., w := u v. Note that w solves the wave equation (due to linearity of the wave equation), and the Cauhy data satisfied by the funtioin w are zero funtions, as both u and v are solutions of the same Cauhy problem. Thus w(x,) w t (x,) on B. As a onsequene of the domain of dependene inequality, we get 2 w2 t + 2 2 w 2 dσ. (5.42) T As a result, w t (x,t ) = w(x,t ) =. Sine T < t is arbitrary, we onlude that w t (x, t) = w(x, t) = (5.43) for every (x, t) belonging to the solid one. This implies that u is a onstant funtion, and sine w = on B, it follows that w is the zero funtion inside the solid one. In partiular, u(x, t ) = v(x, t ). This finishes the proof of the theorem. Remark 5.9 (Consequenes of Domain of dependene inequality). (i) The solid bakward one is alled the past history of the vertex (x, t ). (ii) In other words, the Cauhy data φ,ψ at a spatial point x an influene the solution only in the future one with vertex at (x,), whih is the forward solid light one emanating from (x,). See Figure 5.5 for a pitorial presentation of past and future ones loated at a point (x, t ). 5.4 Finite speed of propagation Finite speed of propagation is a ommon feature for wave equation in all dimensions. We are going to study the speed of propagation of the Cauhy data. Hene we swith-off the nonhomogeneous term from the Wave equation. We illustrate with examples. Example 5. (d = ). Let us onsider initial data φ and ψ that are zero outside of the interval (,). Thus u(2,) = as u(2,) = φ(2) =. Let us now study the behaviour of u(2, t) for t >. For t > suh that 2 t >, i.e., t <, u(2, t) =. That is the information (Cauhy data) at t = has not reahed the point x = 2 till t =, from whih time the information will be reeived at x = 2. Thus it took a time of t = to travel a distane of, and thus the speed is. We illustrate this with = and the interval (,) is replaed with ( 3, 2) in Figure. 5.6, and u(, t) remains zero till t = 2. Example 5. (d = 2,3). Let us onsider the ase of d = 2,3. Suppose that the Cauhy data φ and ψ is supported in the ball of radius R with enter at origin B(, R) d. IIT Bombay

5.5. Conservation of energy 29 4 t 3 2 4 3 Support of φ,ψ 2. x 2 3 4 Figure 5.6. Wave equation in one spae dimension: finite speed of propagation. We know that u(x, t) depends on the values of initial data on B(x, t) B(, R). If this intersetion is empty, then u(x, t) =. In other words, for eah fixed t > the support of the funtion x u(x, t) is ontained in y B(,R) B(y, t), whih is nothing but B(, R+ t). Thus for eah fixed t the support of the solution u(., t) is a ompat set if the Cauhy data is ompatly supported. In other words the support spreads with finite speed. Let y / B(, R). Then not only u(y,) = but also u(y, t) = for t < y R. This is referred to as finite speed of propagation. 5.5 Conservation of energy In this setion we will prove that the energy assoiated to the wave equation defined by E(t) := 2 u2 t + 2 2 u 2 d x (5.44) d is a onstant funtion. In other words, energy is onserved. Theorem 5.2. Let the Cauhy data φ and ψ be ompatly supported funtions defined on d. Let u be the solution of the Cauhy problem for the homogeneous wave equation. Then d d t 2 u2 t + 2 2 u 2 d x =. (5.45) d Proof. As in the proof of ausality priniple, on multiplying the homogeneous wave equation (5.3a) with u t, and re-arranging the terms, we get d i= 2 u x t u xi + i t 2 u2 t + 2 2 u 2 = (5.46) September 7, 25

3 Chapter 5. Wave equation II Integrating the last equality over d, we get d i= d 2 u x t u xi d x + i t 2 u2 t + 2 2 u 2 d x =. (5.47) d The first term in (5.47) is equal to zero under the ondition that for eah fixed t, the funtion x u(x, t) is identially equal to zero for suffiiently large values of x i.e., the funtion u(., t) is of ompat support for eah fixed t >. This is always the ase if the initial data φ,ψ are ompatly supported funtions. In suh a ase, the equation will redue to d t 2 u2 t + 2 2 u 2 d x =. (5.48) The last equation (5.48) is nothing but the desired equation (5.45). Thus the funtion E(t) is a onstant funtion. In other wrods, the energy is onserved. 5.6 Huygens priniple Huygens priniple is onerned with the propagation of information in spae-time. More preisely, Huygens priniple is onerned with solutions of the Cauhy problem for homogeneous wave equation in whih the initial data is assumed to be ompatly supported (see Remark 5.7 for an explanation on this restrition), and how this support propagates with time. There are two forms of Huygens priniple in the literature, known as weak, and strong forms of Huygens priniple respetively. Several authors present the Strong form as the Huygens priniple. Huygens observed that if a wave is sharply loalized at some time, then it will ontinue to be so for all later times, when the wave propagation is governed by homogeneous wave equation in spae whose dimension is an odd number greater than or equal to 3. This observation does not hold when the spae has dimension one or any even number. This observation is formulated as strong form of Huygens priniple. We present Huygens priniple in the two dual points-of-view, namely, in terms of domains of dependene, and influene. Definition 5.3 (Strong form of Huygens priniple). (i) The solution of Cauhy problem at the point (x, t) depends only on the values of the initial data on the sphere x x = t. (ii) The values of initial data at a point x influenes the solution of the wave equation at the points (x, t) belonging to the sphere x x = t. Remark 5.4 (Solutions to 3-d wave equation obey strong Huygens priniple). (i) We derived the Poisson-Kirhhoff formulae (4.4)-(4.45) for solution to the Cauhy problem in three spae dimensions. Note that these forumale give an expression for the solution at the point (x, t) 3 (, ) in terms of integrals on the sphere S(x, t). This is the statement (i) of the strong form of Huygens priniple. IIT Bombay

5.6. Huygens priniple 3 (ii) Imagine that initial data is onentrated at the point (x,) 3 {}. This data will affet the solution at all those points (x, t) suh that (x,) belongs to the sphere S(x, t), whih is preisely the set (x, t) 3 (, ) : x x = t. This is the statement (ii) of the strong form of Huygens priniple. (iii) For three dimensional wave equation, onsider the initial data φ and ψ that is supported inside a ball B of radius ε entered at a point x. Now Poisson-Kirhhoff formulae for three dimensional wave propagation suggests that the solution at the point (x, t) depends on the initial data only on the intersetion of B and the domain of dependene S(x, t) = {y : y x = t}, whih is a sphere entered at x having radius t. Note that this sphere is expanding as t inreases. As long as this intersetion is empty, the solution will be zero. Sine the sphere is expanding with t, there will be a time instant t e at whih the intersetion beomes non-empty. In fat t e = x x ε. Also a time t f after whih the intersetion will beome empty. Thus the solution beomes zero again after time t f. In fat t f = x x + ε. That is, an initial disturbane onfined to a small ball of radius ε, gives rise to an expanding spherial wave having a leading and a trailing edge having support in an annular region of width 2ε at eah time instant. This phenomenon is referred to as sharp signal propagation. (iv) Let us interpret this priniple in terms of a physially relevant example. Assuming that wave equation models sound waves that propagate in our three dimensional world, we an easily see that the strong form of Huygens priniple holds in three dimensions. For example the sound waves generated by a speaker will reah a listener after sometime depending on the distane from the speaker, and of ourse at sound speed. In fat, the listener hears at the time instant t + d, the sounds produed by the speaker at the time instant t, where d is the distane between the speaker and the listener. In other words, the listener hears only silene, and then suddenly some speeh is heard for a ertain duration of time, and then suddenly one again silene (whih happens when the speaker pauses his speeh). This is the strong Huygens priniple. (v) Mathematially speaking, we say that sound waves propagate exatly at the speed. Note however that we hear ehoes, and observe reverberation phenomenon in enlosed spaes like aves, but this does not ontradit the Poisson-Kirhhoff formulae whih were derived earlier. This only means that wave equation is not suited to this situation. We have the following weak form of Huygens priniple whih holds in all even spae dimensions. Definition 5.5 (Weak form of Huygens priniple). (i) The solution of Cauhy problem at the point (x, t) depends only on the values of the initial data in the ball x x t. (ii) The values of initial data at a point x influenes the solution of the wave equation at the points (x, t) belonging to the ball x x t. September 7, 25

32 Chapter 5. Wave equation II Remark 5.6 (Solutions to 2-d wave equation obey weak Huygens priniple). (i) We derived the Poisson-Kirhhoff formula (4.56) for solution to the Cauhy problem in two spae dimensions, whih gives an expression for the solution at the point (x, t) 2 (, ) in terms of integrals on the losed ball B(x, t). This is the statement (i) of the weak form of Huygens priniple. (ii) Imagine that initial data is onentrated at the point (x,) 2 {}. This data will affet the solution at all those points (x, t) suh that (x,) belongs to the ball B(x, t), whih is preisely the set (x, t) 2 (, ) : x x t. This is the statement (ii) of the weak form of Huygens priniple. (iii) For two dimensional wave equation, onsider the initial data φ and ψ that is supported inside a ball B of radius ε entered at a point x. Now Poisson-Kirhhoff formula for two dimensional wave propagation suggests that the solution at the point (x, t) depends on the initial data only on the intersetion of B and the domain of dependene B(x, t) = {y : y x t}, whih is a ball entered at x having radius t. Note that here also there will be a time instant t e at whih the intersetion beomes non-empty; however there is no time t f after whih the intersetion will beome empty. That is, one an initial disturbane reahes a point x at a time instant t e, the effet will stay on forever (unless the initial data have zero averages). This is the ase for one-dimensional equation as well. However, the solution of a Cauhy problem to two-dimensional wave equation deays i.e., at eah fixed x, the value of u(x, t) tends to zero as t (see Setion 5.). This phenomenon, namely that of a slowly deaying trailing edge, is known as diffusion of waves. When d = 3, there is no diffusion of waves as we saw that the solution beomes zero after a time instant t f, and then remains zero for t > t f. In the ase of one-dimensional wave equation, there is no deay in the solution, and in fat the solution is eventually onstant (when φ, ψ is of ompat support with non-zero average). (iv) Let us interpret this priniple in terms of an example. Assuming that wave equation models waves that are generated when a stone is thrown into a still pond of water. First we observe irular waves propagating from the point where stone touhed the water surfae. Seondly we see that these irular waves propagate forever. We also observe that new irles are formed within the expanding irular waves, whih is a result of waves propagating at all speeds less than or equal to. (v) Let us now be thankful that we do not live in a one-dimensional or in a two-dimensional world. The reason is that if the propagation of sound waves is governed by onedimensional (or two-dimensional) wave equation, then by d Alembert formula (respetively, Poisson-Kirhhoff formula (4.56)) we find that sound propagates at all speeds less than or equal to, resulting in ehoes, and the phenomenon of reverberation (take φ for instane). The following remark explains the relevane of the ontext in whih Huygens priniples are stated. IIT Bombay

5.7. Generalized solutions, Weak solutions 33 Support of φ, ψ. S t S t2 S t3 S t4 Figure 5.7. Signal propagation in d - Illustration of Huygens Priniple: Here t < t 2 < t 3 < t 4 are time instants. For eah i =,2,3,4, S ti denotes the set S ti = {y d : y x = t i }. Remark 5.7 (On Huygens priniples and sharp signal propagation). (i) We explain the reasons behind onsidering ompatly supported initial data in the disussion of Huygens priniples. As we saw Huygens priniples are onerned with propagation of supports of initial data with time, and thus onsidering arbitrary initial data is not meaningful. In fat, we should ideally onsider initial data that are supported at a single point like a Dira δ funtion whih is supported at a single point. To avoid tehnialities that arise by onsidering data whih is supported at a single point, we onsider ompatly supported initial data (whih an be taken to be a small interval and thus resembling point-support data). (ii) Also note that onsidering non-homogeneous wave equation is not meaningful (in the ontext of Huygens priniples) as the domain of dependene for the solution at a point (x, t) d (, ) is the bakward solid one region with vertex at (x, t) for all d =,2,3. Thus there are no speial dimensions where there will be sharp signal propagation. (iii) Huygens priniples are better understood from the domain of influene point of view. (iv) The minimum number of spae dimensions that allows sharp signal propagation (initial data is of ompat support implies the solution has ompat support at eah time instant) is three. Huygens priniple is illustrated in Figure 5.7. 5.7 Generalized solutions, Weak solutions In this setion, we motivate the onept of a generalized solution (also known as weak solution) to homogeneous wave equation, and to Cauhy problem for the homogeneous wave equation. This notion an be generalized to nonhomogeneous equations, and to problems posed on domains with boundaries. There are at least two reasons behind the neessity to introdue a generalized notion of solution. They are (i) Notion of lassial solution that we dealt with so far neessitates the Cauhy data to September 7, 25

34 Chapter 5. Wave equation II be suffiiently smooth (as required by the existene theorems), failing whih there is no solution to the Cauhy problem. In the ase of wave equation in one spae dimension, whih models vibrations of a string where the unknown funtion is the displaement, the initial displaement ould be that of a string whih is raised to a ertain height in some part of the string. This results in a φ whih is a pieewise onstant funtion, and is very far from being a twie differentiable funtion as required by the existene theorem. Thus there is a need to generalize the onept of a solution. (ii) We derived formulae whih represent lassial solutions, for example (4.4), (4.7), (4.4), (4.45), (4.53), (4.56), when Cauhy data is suffiiently smooth. We also showed that solutions of one-dimensional wave equation satisfy the parallelogram identity. However while dealing with Cauhy data that do not satisfy these smotthness requirements, these formulae no longer represent a lassial solution. Of ourse, it is not lear if the Cauhy problem admits a lassial solution in suh ases. However everything is not lost and we an retrieve something from them. For example, the formulae may give a lassial solution in a restrited (x, t) domain. We may use the formulae to study how the lak of smoothness in the Cauhy data propagates with time (see Setion 5.9). Also, these formulae might be the atual solutions of the Cauhy problems when Cauhy data is not smooth, with a new interpretation of solutions. Any new onept of a weak solution (generalized solution) is meaningful only when the lassial solutions still fulfill the requirements for solutions in the generalized sense. We derive the notion of weak solution based on (4.4) for one-dimensional wave equation. We an also disuss generalized solutions for all Cauhy problems for wave equation in higher dimensions onsidered so far. For further disussion of weak solutions, the reader is advised to refer to advaned texts on partial differential equations. Reall that the general solution of the homogeneous wave equation in one spae dimension is given by (4.4), whih is given by u(x, t) = F (x t) + G(x + t). Let us onsider F C 2 () and G for simpliity in presentation. Let ζ : be a twie ontinuously differentiable funtion with ompat support. Sine F (x t) solves the homogeneous wave equation, we have 2 t 2 2 2 x 2 F (x t) ζ (x, t) d x d t =. (5.49) Using integration by parts formula, the equation (5.49) transforms into 2 t 2 2 ζ (x, t) F (x t) d x d t =. (5.5) 2 x 2 Note that the equation (5.5) is meaningful even when the funtion F is only ontinuous. This motivates the following notion of a weak solution. Definition 5.8. A ontinuous funtion u : is said to be a weak solution of the homogeneous wave equation if for every twie ontinuously differentiable funtion ζ : with ompat support the following equality holds 2 u(x, t) t 2 2 2 x 2 ζ (x, t) d x d t =. (5.5) IIT Bombay

5.8. Propagation of onfined disturbanes 35 Suppose u is at least twie ontinuously differentiable funtion whih is a weak solution of the homogeneous wave equation, then we expet that u is a lassial solution as well. Indeed, applying integration by parts formula four times in the equation (5.5) results in 2 ζ (x, t) t 2 2 u(x, t) d x d t =. (5.52) 2 x 2 Sine (5.52) holds for every twie differentiable funtion ζ with ompat support, it follows that 2 t 2 2 u(x, t) =. (5.53) 2 x 2 In other words, u is a lassial solution. Remark 5.9. In this remark, we desribe how to arrive at the onept of a weak solution in general. (i) Multiply the given equation with a suffiiently differentiable funtion having ompat support (alled test funtion). Perform integration by parts till all the derivatives on the unknown are transferred to the test funtion. The resulting equation gives rise to a weak formulation, and this is meaninful for unknown funtions that are only ontinuous. This justifies the word weak solution. (ii) However note that if a weak solution u is suffiiently differentiable, then by performing integration by parts till all derivatives are shifted to the weak solution, we get that u is indeed a lassial solution sine the test funtions are in plenty. (iii) Using d Alembert formula also we an generalize the notion of a lassial solution, by requiring that φ to be ontinuous and ψ to be pieewise ontinuous. We may also onsider any ontinuous funtion satisfing the parallelogram identity as a weak solution of the one-dimensional wave equation. (iv) Similarly we an use Poisson-Kirhhoff formulae to arrive at the onepts of weak solutions for higher dimensional wave equations. We will not disuss further on suh weak formulations. 5.8 Propagation of onfined disturbanes In this setion, we study propagation of solutions of the Cauhy problem u t t 2 u x x + u x2 x 2 + + u xd x d =, x d, t >. (5.54a) u(x,) = φ(x), x d, (5.54b) u t (x,) = ψ(x), x d, (5.54) when the initial disturbanes φ and ψ are onfined i.e., φ and ψ are ompatly supported funtions. Even though there is abundane of funtions φ and ψ whih satisfy onditions that guarantee the existene of lassial solution to (5.54), we hose pieewise onstant initial data to make omputations simpler. We have to bear in mind that the orresponding solutions given by formulae like d Alembert and Poisson-Kirhhoff will not be lassial solutions. In suh a ase they an be interpreted as weak solutions (see Setion 5.7). Considering suh initial onditions will also help us in analysing propagation of singularities (see Setion 5.9). September 7, 25

36 Chapter 5. Wave equation II Example 5.2 (d =, ψ ). Assume that u(x,) = φ has a ompat support, say φ(x) = for x [,]. Assume that ψ. Now d Alembert formula gives the solution of the IVP as φ(x t) + φ(x + t) u(x, t) =. 2 Fix a t. Then if x [ t, + t φ(x t ) = ], otherwise if x [ t, t φ(x + t ) = ], otherwise Note that φ take only two values and. Therefore u(x, t ) is non-zero for all those x for whih x [ t, + t ] or x [ t, t ]. Observe that for t large, the intervlas are disjoint. Hene the support of u(x, t ) will be [ t, t ] [ t, + t ]. That is the support of the initial disturbane propagates with time. Example 5.2 (d =, φ ). Assume that φ and u t (x,) = ψ has a ompat support, say [,]. At the time instant t = t, d Alembert formula gives the solution u(x, t ) of the IVP as u(x, t ) = x+ t ψ(s) d s. 2 x t Now fix x = x. Note that as t beomes large, x t < and also x + t >. Thus for large enough t (t t ), we have u(x, t) = 2 ψ(s) d s. Note that RHS is a onstant, and is non-zero if ψ(s) d s. This shows that we are in for a big trouble if sound waves propagate aording to the one-dimensional wave equation. Example 5.22 (d = 3, φ ). [3] Let us onsider the Cauhy problem for wave equation in three spae dimensions (with = ) where the Cauhy data is given by if x, φ, ψ(x) = if x >. Sine φ, Poisson-Kirhhoff formula (4.44) redues to u(x, t) = ψ(y) dσ. (5.55) 4πt S(x,t) Sine the integral in (5.55) is effetively on the part of the sphere S(x, t) lying inside the ball B(, ), whenever the sphere lies ompletely outside the unit ball (whih happens for IIT Bombay

5.9. Propagation of singularities 37 t > + x or t < x ) the solution u(x, t) =. On the other hand, whenever the sphere lies ompletely inside the unit ball (whih happens for t < x ), we have u(x, t) = t. For t suh that x < t < x + or t > x, we get u(x, t) = 4πt S(x,t) B(,) dσ. (5.56) Sine the surfae area of the part of sphere S(x, t) B(,) (alled spherial ap) is equal to πt (t x ) 2 x Thus the solution is given by t if t x, u(x, t) = ( (t x ) 2 ) 4 x if x t x +, if t x or t > + x. (5.57) Consider an x with x = 2. From the expression (5.57), we get u(x, t) = for t < and also for t > 3. The funtion t u(x, t) is inreasing in the interval [,2], and is dereasing in the interval [2,3]. This behaviour is very different from that of d = as illustrated by Example 5.2, where the solution beomes non-zero after some time, and remains onstant thereafter. Example 5.23 (d = 3, ψ ). [3] Let us onsider the Cauhy problem for wave equation in three spae dimensions (with = ) where the Cauhy data is given by if x, ψ, φ(x) = if x >. Sine ψ, Poisson-Kirhhoff formula (4.44) redues to u(x, t) = φ(y) dσ. (5.58) t 4πt S(x,t) Thus the solution is given by differentiating w.r.t. t of the RHS in (5.57) if t x, x t u(x, t) = 2 x if x t x +, if t x or t > + x. (5.59) Consider an x with x = 2. From the expression (5.59), we get u(x, t) = for t <. The funtion t u(x, t) is dereasing in the interval [,3] (from 4 at t = to at t = 3), 4 and then beomes zero for t > 3. 5.9 Propagation of singularities 5.9. Singularities travel along harateristis for d = Assume that for a fixed time t, the solution u is a smooth funtion exept at one point (x, t ). As u is given by (4.4), this means that either F is not smooth at x + t or G is September 7, 25

38 Chapter 5. Wave equation II not smooth at x t. Now, observe that there are two harateristis (one eah of the two families) passing through (x, t ), given by x t = x t, x + t = x + t If F is not smooth at x + t, then u will not be smooth at all the points lying on the line x + t = x + t. If G is not smooth at x t, then u will not be smooth at all the points lying on the line x t = x t. This shows that the singularities of solutions of the wave equation are travelling only along harateristis. Also reord that at (x, t) for whih φ is C 2 at x t and x + t, and ψ is C in the interval [x t, x + t], d Alembert formula represents a lassial solution at (x, t). Thus singularities in the Cauhy data are transported along harateristis with finite speed. Singularity propagation takes plae without hanging the nature of singularity. This is a typial behaviour of solutions to hyperboli partial differential equations. 5.9.2 Propagation of singularities in higher dimensions In this subsetion we will present an example in three spae dimensions. The example is Example 5.23. Example 5.24. Let us onsider the homogeneous wave equation in three spae dimensions. Let the Cauhy data be given by funtions that are radial. In other words onsider the following Cauhy problem 2 u(r, t) = t 2 2 r 2 + 2 r u(r, t), r r, t >, (5.6a) u(r,) = f (r ), r, (5.6b) u (r,) = g(r ), r, (5.6) t where f, g are extended as ontinuously differentiable even funtions to r. This requires that f () = g () =. Defining v(r, t) := r u(r, t), we observe that v satisfies the following Cauhy problem for the one-dimensional wave equation: Thus u(r, t) is given by 2 t v(r, t) = 2 v (r, t), 2 r 2 r, t >, (5.6a) v(r,) = r f (r ), r, (5.6b) v (r,) = r g(r ), t r. (5.6) u(r, t) = [(r t)f (r t) + (r + t)f (r + t)] + 2r 2r r +t Let us now onsider the speial ase where g(r ), and f (r ) is given by if r, u(r,) = if r >. In this speial ase the solution of Cauhy problem is given by r t s g(s) d s. (5.62) u(r, t) = [(r t)f (r t) + (r + t)f (r + t)]. (5.63) 2r IIT Bombay

5.. Deay of solutions 39 Note that the Cauhy data f (.) has a disontinuity on the sphere r =. Let us now ompute lim r u(r, t). Using L Hospital s rule, and the fat that the funtion f is even, we get u(, t) = lim r u(r, t) = f (t) + t f (t). (5.64) Sine f is disontinuous at r =, we expet trouble for u(, t) as t approahes as f () is not meaningful. The singulairty whih was initially onfined to a two-dimensional surfae (whih is the unit sphere) gets onentrated at a point (the origin) at time t =. This is alled fousing of singularities, and it is also said that austis are formed. 5. Deay of solutions No deay for d = In this setion we study the deay properties of solutions to the Cauhy problems for homogeneous wave equation when the initial data φ and ψ are ompatly supported funtions, in one, two, and three spae dimensions. Note that properties of solutions for large times (i.e., as t ) are dominated by terms involving ψ when ompared to those with ϕ, whih follows from d Alembert formula (d = ), Poisson-Kirhhoff formulae (d = 2,3). Thus we may assume that φ. We annot expet deay of solutions to one dimensional wave equation unless ψ satisfies ψ(y) d y =, as shown in Example 5.2. Deay for d = 3 Theorem 5.25. Let ψ be a ompatly supported funtion having support in B(, R) 3. Then for t >, the following estimates hold (i) sup u(x, t) R2 x 3 2 t sup ψ(y). (5.65) y 3 (ii) sup u(x, t) x 4π 3 t ψ(y) d y. 3 (5.66) Proof. Proof of (i): Sine φ, Poisson-Kirhhoff formula (4.44) redues to u(x, t) = ψ(y) dσ. (5.67) 4π 2 t S(x, t) Note that the integral in (5.67) is over the sphere S(x, t), and the funtion ψ is supported in the ball B(, R). Thus effetively, the integral in (5.67) is over the part of the sphere S(x, t) that lies inside the ball B(, R), and let A denote its surfae area. September 7, 25

4 Chapter 5. Wave equation II From (5.67), we get u(x, t) A 4π 2 t sup y 3 ψ(y). (5.68) Note that A will be maximum if the sphere S(x, t) lies ompletely inside the ball B(, R). In suh a ase, A will be less than or equal to the surfae area of the ball B(, R) whih is 4πR 2. Thus we get the estimate u(x, t) R2 2 t sup y 3 ψ(y). (5.69) Sine RHS of (5.69) does not depend on x 3, the deay estimate (5.65) follows. Proof of (ii): Sine φ, Poisson-Kirhhoff formula (4.4) redues to By fundamental theorem of alulus, we write ψ(x + tν) = u(x, t) = t ψ(x + tν) dω. (5.7) 4π ν = Integrating the last relation w.r.t. ν yields ψ(x + tν) dω = ν = t ψ(x + ρν) dρ = ρ ν = t t ψ(x + ρν).ν dρ. (5.7) ψ(x + ρν).ν dρ dω. (5.72) Thus the expression for u given by (5.7) beomes u(x, t) = t ψ(x + ρν).ν dρ dω. (5.73) 4π From the last equation, we get u(x, t) 4π t ν = t t ψ(x + ρν) dω dρ. (5.74) ν = Sine ρ t, we get t 2 ρ2. Thus t ρ2. Also note that 2 2 t ρ 2 ψ(x + ρν) dω dρ = ψ(y) d y ψ(y) d y. (5.75) t ν = 3 \B(x, t) 3 Thus the estimate (5.74) gives rise to the following estimate u(x, t) ψ(y) d y. (5.76) 4π 2 t 3 Sine RHS of (5.74) does not depend on x 3, the deay estimate (5.66) follows. Note that the radius of the support of the initial data features expliitly in the first estimate (5.65), while it does not feature in the seond estimate (5.66). IIT Bombay

5.. Deay of solutions 4 Deay for d = 2 We present two kinds of deay results in the ase of two dimensional wave equation. The first result is onerned with the deay of the funtion t u(x, t) for a fixed x 2. The seond result is onerned with deay estimate whih is uniform in x 2. Theorem 5.26 (Deay at a fixed x 2 ). Let ψ be a ompatly supported funtion having support in the dis D(, R) 2. Then for eah x 2, there exists a onstant K = K(x) suh that for all t >, u(x, t) K t sup ψ(y). (5.77) y 2 Proof. Sine φ, Poisson-Kirhhoff formula (4.54) redues to u(x, t) = ψ(y) d y. (5.78) 2π D(x, t) 2 t 2 x y 2 Sine ψ outside the dis D(, R), Thus we have u(x, t) = ψ(y) d y. (5.79) 2π D(x, t) D(,R) 2 t 2 x y 2 u(x, t) 2π sup d y ψ(y) y 2 D(x, t) D(,R) 2 t 2 x y. (5.8) 2 Let us estimate the integral I := D(x, t) D(,R) d y 2 t 2 x y 2 (5.8) Note that y D(x, t) D(, R) implies that x y t and x y x + R. For reasons that will be lear later on, we obtain estimate on I in two steps: first we onsider times suh that < t < x +2R, and then we onsider the ase t > x +2R. Case (i): Let t be suh that < t < x + 2R. In this ase, after swithing to polar oordinates at x we have I := D(x, t) D(,R) t 2π d y t 2 t 2 x y 2π r d r 2 2 t 2 r 2 r d r 2 t 2 r 2π 2 t t 2π t 2 t 2 d u = 2π u 2 t 2 t 2 r d r r 2 2 t 2 2π t ( x + 2R)2. (5.82) September 7, 25

42 Chapter 5. Wave equation II Case (ii): Let t be suh that t > x +2R. In this ase, after swithing to polar oordinates at x we have I := D(x, t) D(,R) d y x +R 2 t 2 x y 2π r d r 2 2 t 2 r 2 x +R r d r 2π 2 t 2 r 2π 2 t 2π t π t x +R x + 2R x +R r d r R(2 x + 3R) r d r r 2 2 t 2 x + 2R R(2 x + 3R) ( x + R) 2. (5.83) Combining (5.82) and (5.83), the estimate (5.77) follows. Theorem 5.27 (Uniform deay estimate). Let ψ be a ompatly supported funtion having support in the dis D(, R) 2. Then there exists a onstant K and t > suh that for t > t, the following estimate holds sup x 2 u(x, t) K t ψ(y) d y + 2 ψ(y) d y 2. (5.84) Proof. Sine φ, Poisson-Kirhhoff formula (4.54) redues to Equation (5.85) may be re-written as u(x, t) = ψ(y) d y. (5.85) 2π D(x, t) 2 t 2 x y 2 u(x, t) = 2π = 2π = 2π D(, t) t t ψ(x + z) 2 t 2 z 2 d z ν = ψ(x + ρν) 2 t 2 ρ 2 ρ dρdω ρ 2 t 2 ρ 2 ν = ψ(x + ρν)dω dρ (5.86) We write the integral on the interval [, t] as sum of two terms, by splitting the domain of integration into [, t ε] and [ t ε, t] (ε > to be hosen later), whih we denote by I and I 2 respetively. Let us first estimate the integral I. Reall that I is given by I = t ε ρ 2π 2 t 2 ρ 2 ν = ψ(x + ρν)dω dρ, (5.87) IIT Bombay