where F denoting the force acting on V through V, ν is the unit outnormal on V. Newton s law says (assume the mass is 1) that

Transcription

1 Chapter 5 The Wave Equation In this chapter we investigate the wave equation 5.) u tt u = and the nonhomogeneous wave equation 5.) u tt u = fx, t) subject to appropriate initial and boundary conditions. unknown function u = ux, t) : [, ) R. Here x R n, t > ; the We shall discover that solutions to the wave equation behave quite differently from solutions of Laplaces equation or the heat equation. For example, these solutions are generally not C and exhibit the finite speed of propagation of given disturbances. 5.. Derivation of the wave equation The wave equation is a simplified model for a vibrating string n = ), membrane n = ), or elastic solid n = 3). In this physical interpretation ux, t) represents the displacement in some direction of the point at time t. Let V represent any smooth subregion of. The acceleration within V is then d dt udx = u tt dx, and the net force is V V V F νds, where F denoting the force acting on V through V, ν is the unit outnormal on V. Newton s law says assume the mass is ) that u tt = F νds. V This identity is true for any region, hence the divergence theorem tells that V u tt = div F.

2 5. The Wave Equation For elastic bodies, F is a function of Du, i.e., F = F Du). For small u and small Du, we use the linearization adu to approximate F Du), and so u tt a u =, when a =, the resulting equation is the wave equation. The physical interpretation strongly suggests it will be mathematically appropriate to specify two initial conditions, ux, ) and u t x, ). 5.. One-dimensional wave equations and d Alembert s formula This section is devoted to solving the Cauchy problem for one-dimensional wave equation: u tt u xx = in R, ), 5.3) ux, ) = gx), u t x, ) = hx), x R, where g, h are given functions. Note that u tt u xx = u t u x ) t + u t u x ) x = v t + v x where v = u t u x. So vx, ) = u t x, ) u x x, ) = hx) g x) := ax). From v t + v x =, we have vx, t) = ax t). Then u solves the nonhomogeneous transport equation Solve this problem for u to obtain ux, t) = gx + t) + = gx + t) + = gx + t) + u t u x = vx, t), t t vx + t s, s) ds x+t ux, ) = gx). ax + t s) ds = gx + t) + x t hy) g y)) dy, from which we deduce d Alembert s formula: 5.4) ux, t) = gx + t) + gx t) + x+t x t hy) dy. x+t x t ay) dy This formula defines a classical solution u C if and only if g C and h C. Remark 5.. i) When the initial data g and h are not smooth, we still use formula 5.4) to define a weak) solution to the Cauchy problem. If g C k and h C k, then u C k. Thus the wave equation does not have the smoothing effect like the heat equation has. ii) Any solution to the wave equation u tt = u xx has the form ux, t) = F x + t) + Gx t) for appropriate functions F and G. Usually, F x + t) is called a traveling wave to the left with speed ; Gx t) is called a traveling wave to the right with speed. iii) The value ux, t ) depends only on the values of initial data from x t to x +t ; in this sense, the interval [x t, x + t ] is called the domain of dependence for the point x, t ). The Cauchy data at x, ) influence the value of u in the set Ix, t ) = x, t) x t < x < x + t, t > },

3 5.. One-dimensional wave equations and d Alembert s formula 3 which is called the domain of influence for x, ). The domain of determinacy of [x, x ]; that is, the set of points x, t) at which the solution is completely determined by the values of initial data on [x, x ], is given by Dx, x ) = x, t) x + t x x t, t [, x ]} x. Lemma 5. Parallelogram property). Let be an open set in R R +. Then any solution u of the one-dimensional wave equation in satisfies 5.5) ua) + uc) = ub) + ud), where ABCD is any parallelogram contained in with the slope or, with A and C being two opposite points as shown in Figure 5.. t B C A D x Figure 5.. Parallelogram ABCD in Lemma 5.. We may use this property to solve certain initial and boundary problems. Example 5.. Solve the initial boundary value problem: u tt u xx =, x >, t >, ux, ) = gx), u t x, ) = hx), x >, u, t) = kt), t >, where g, h, k are given functions satisfying certain smoothness and compatibility conditions. Solution. See Figure 5..) If point E = x, t) is in the region I); that is, x t >, then ux, t) is given by 5.4): In particular, ux, t) = ux, x) = gx + t) + gx t) gx) + g) + + x x+t x t hy) dy. hy)dy.

4 4 5. The Wave Equation t Region II) A D B E Region I) C F G x Figure 5.. Mixed-value problem in x >, t >. If point A = x, t) is in the region II); that is, x < t, then we use the parallelogram ABCD as shown in the figure to obtain ux, t) = ub) + ud) uc), where ub) = u, t x) = kt x), ud) = u x + t, x + t gx + t) + g) ) = uc) = u t x, t x gt x) + g) ) = Hence, for x < t, ux, t) = kt x) + gx + t) gt x) x+t t x x+t t x hy) dy, hy) dy. hy) dy. Therefore the solution to this problem is given by gx + t) + gx t) + x+t 5.6) ux, t) = x t hy) dy t x), gx + t) gt x) kt x) + + x+t t x hy) dy x < t). Of course, some smoothness and compatibility conditions on g, h, k are needed in order for ux, t) to be a true solution of the problem. Derive such conditions as an exercise. Example 5.3. Solve the initial-boundary value problem u tt u xx =, x, π), t >, ux, ) = gx), u t x, ) = hx), x, π), u, t) = uπ, t) =, t >. Solution. See Figure 5.3.) We divide the strip, π), ) by line segments of slope ± starting first at, ) and π, ) and then at all the intersection points with the boundaries.

5 5.. One-dimensional wave equations and d Alembert s formula 5 t IV) A II) B III) E I) π x Figure 5.3. Mixed-value problem in < x < π, t >. We can solve u in the region I) by formula 5.4). parallelogram formula 5.5). In all other regions we use the Another way to solve this problem is the method of separation of variables. First try ux, t) = Xx)T t), then we should have which implies that where λ is a constant. From X x)t t) = T t)xx), X) = Xπ) =, T t) T t) = X x) Xx) = λ, X x) = λxx), X) = Xπ) =, we find that λ = j with all j =,,..., and X j x) = sinjx), T j t) = a j cosjt) + b j sinjt). To make sure that u satisfies the initial condition, we consider 5.7) ux, t) = [a j cosjt) + b j sinjt)] sinjx). To determine coefficients a j and b j, we use ux, ) = gx) = a j sinjx), u t x, ) = hx) = j= j= jb j sinjx). i.e., the a j and jb j are Fourier coefficients of functions gx) and hx). That is, a j = π π gx) sinjx)dx, b j = jπ π j= hx) sinjx)dx. Substitute these coefficients into 5.7) and we obtain a formal solution u in terms of trigonometric series; the issue of convergence will not be discussed here.

6 6 5. The Wave Equation 5.3. The method of spherical means Suppose u solves the Cauchy problem of n-dimensional wave equation u tt u =, x, t) R n R +, 5.8) ux, ) = gx), u t x, ) = hx), x R n. The idea is to reduce the problem to a problem of one-dimensional wave equation. This reduction requires the method of spherical means. For x R, r >, t >, define Note that Gx; r) = Ux; r, t) = uy, t) ds y = M u,t) x, r), Bx,r) gy) ds y = M g x, r), Hx; r) = Bx,r) Ux; r, t) = hy) ds y = M h x, r). Bx,r) nα n r n uy, t)ds y = ux + rξ, t)ds ξ. Bx,r) B,) So, for fixed x, the function Ux; r, t) extends as a function of r R and t R +. One can recover ux, t) from Ux; r, t) in terms of ux, t) = lim Ux; r, t). r The Euler-Poisson-Darboux equation. Let ux, t) be a smooth solution to 5.8). We show that, for each x R n, the function Ux; r, t) solves a PDE for r, t) R + R +. Theorem 5.4 Euler-Poisson-Darboux equation). Let u C m R n [, )) with m solve 5.8). Then, for each x R n, the function Ux; r, t) C m [, ) [, )) and solves the Cauchy problem of the Euler-Poisson-Darboux equation U tt U rr n 5.9) r U r = in, ), ), U = G, U t = H on, ) t = }. Proof.. Note that, for each x R n, the regularity of U on r, t) follows easily as Ux; r, t) = ux + rξ, t)ds ξ. B,). Using the formula above, we have U r = r n uy, t) dy = Bx,r) nα n r n u tt y, t) dy. Bx,r) Hence r n U r = r ) u tt y, t)ds y dρ nα n Bx,ρ) and so, differentiating with respect to r yields r n U r ) r = u tt y, t)ds y = r n U tt, nα n Bx,r) which expands into the Euler-Poisson-Darboux equation. The initial condition is satisfied easily; this completes the proof of the theorem.

7 5.3. The method of spherical means Solutions in R 3 and Kirchhoff s formula. For the most important case of threedimensional wave equations, we can easily see that the Euler-Poisson-Darboux equation implies ru) rr = ru rr + U r = r r U r ) r = ru tt = ru) tt. That is, for fixed x R 3, the function Ũr, t) = rux; r, t) solves the -D wave equation Ũ tt = Ũrr in r >, t >, with Ũ, t) =, Ũr, ) = rg := G, Ũr, ) = rh := H. This is the mixed-value problem studied in Example 5.; hence, by 5.6) we have We recover ux, t) by Ũr, t) = [ Gr + t) Gt r)] + t+r t r Hy) dy r t). Ũr, t) ux, t) = lim Ux; r, t) = lim r + r + r [ Gr + t) Gt r) = lim + ] r+t Hy) dy = r + r r G t) + Ht). Therefore, we have obtained the so-called Kirchhoff s formula for 3-D wave equation: ux, t) = ) t gy) ds y + t hy) ds y t 5.) = = 4πt Bx,t) Bx,t) r+t Bx,t) thy) + gy) + Dgy) y x)) ds y Bx,t) thy) + gy) + Dgy) y x)) ds y x R 3, t > ) Solutions in R by Hadamard s method of descent. Assume u C R [, )) solves problem 5.8) with n =. We would like to derive a formula of u in terms of g and h. The trick is to consider u as a solution to a 3-dimensional wave problem with one added dimension x 3 and then to use Kirchhoff s formula to find u. This is the well-known Hadamard s method of descent. Define ũ x, t) = ux, t) for x = x, x 3 ) R 3, t >, where x = x, x ) R. Then ũ solves ũ tt ũ =, x, t) R 3, ), ũ x, ) = g x), ũ t x, ) = h x), x R 3, where is the Laplacian in R 3, g x) = gx) and h x) = hx). Let x = x, ) R 3. Then, by Kirchhoff s formula, ux, t) = ũ x, t) = ) t gỹ) t B x,t) dsỹ + t hỹ) dsỹ, Bx,t) where B x, t) is the ball in R 3 centered at x of radius t. To evaluate this formula, we parametrize B x, t) by parameter z = z, z ) Bx, t) R as ỹ = z, ±γz)), γz) = t z x z Bx, t)).

8 8 5. The Wave Equation Note that Dγz) = x z)/ t z x and thus dsỹ = + Dγz) dz = t dz t z x ; hence, noting that B x, t) has the top and bottom parts with y 3 = ±γz), gỹ) B x,t) dsỹ = 4πt gỹ) B x,t) dsỹ = gz)t dz 4πt Bx,t) t z x = gz) dz πt Bx,t) t z x = t gz) dz t z x. Bx,t) Similarly, we obtain the formula for B x,t) hỹ) dsỹ; finally, we have obtained the so-called Poisson s formula for -D wave equation: ux, t) = t ) gz) dz + t hz) dz t Bx,t) t z x Bx,t) t z x 5.) = tgz) + t hz) + tdgz) z x) πt dz x R, t > ). Bx,t) t z x Remark 5.. i) There are some fundamental differences for the wave equation between the one dimension and the dimensions n =, 3. In both Kirchhoff s formula n = 3) and Poisson s formula n = ), the solution u depends on the derivative Dg of the initial data ux, ) = gx). For example, if g C m, h C m for some m then u is only C m and hence u t is only C m but u t x, ) = hx) C m ); therefore, there is loss of regularity for the wave equation when n =, 3 in fact for all n ). However, this does not happen when n =, in which u, u t are at least as smooth as g, h. ii) There are also some fundamental differences between the 3-D wave equation and the -D wave equation. In R, we need the information of initial data g, h in the whole disc Bx, t) to compute the value ux, t); that is, the domain of dependence for x, t) is the whole disc Bx, t), while in R 3 we only need the information of g, h on the sphere Bx, t) to compute the value ux, t); that is, the domain of dependence for x, t) is the sphere Bx, t), not the solid ball Bx, t). In R 3, a disturbance initiated at x propagates along the sharp wavefront Bx, t) and does not affect the value of u elsewhere; this is known as the strong Huygens s principle. In R, a disturbance initiated at x will affect the values ux, t) in the whole region x x t. In both cases n =, 3 in fact all cases n ), the domain of influence of the initial data grows with time t) at speed ; therefore, the wave equation has the finite speed of propagation. iii) To illustrate the differences between n = and n = 3 of the wave equation, imagine you are at position x in R n and there is a sharp initial disturbance at position x away from you at time t =. If n = 3 then you will only feel the disturbance e.g., hear a screaming) once, exactly at time t = x x ; however, if n =, you will feel the disturbance e.g., you are on a boat in a large lake and feel the wave) at all times t x x, although the effect on you will die out at t.

9 5.4. Solutions of the wave equation for general dimensions Solutions of the wave equation for general dimensions We now try to find the solution ux, t) of problem 5.8) in R n R + by solving its spherical mean Ux; r, t) from the Euler-Poisson-Darboux equation. We need the following useful result. Lemma 5.5. Let f : R R be C m+, where m. Then for k =,,..., m, i) ii) d dr ) r r d dr )k r k fr)) = r k d dr )k r k fr)) = j= where β k = k )!! = k )k 3) 3 and βk j d dr )k r k f r)); β k j r j+ dj f dr j r), are independent of f. Proof. Homework Solutions for the odd-dimensional wave equation. Assume that n = k + k ) and u C k+ R n [, )) is a solution to the problem 5.8) in R n R +. As above, let Ux; r, t) be the spherical mean of ux, t). Then for each x R n the function Ux; r, t) is C k+ in r, t) [, ) [, ) and solves the Euler-Poisson-Darboux equation. Let V r, t) = r r )k r k Ux; r, t)) r >, t ). Lemma 5.6. We have that V tt = V rr and V +, t) = for r >, t >. Proof. By part i) of Lemma 5.5, we have V rr = r ) r r )k r k U) = r = r r )k [ r rk U r ) r ] = r r )k r k U r ) r )k [r k U tt ] = V tt, where we have used the Euler-Poisson-Darboux equation r k U r ) r follows from part ii) of Lemma 5.5 that V +, t) =. Now that = r k U tt. Finally, it r )k r k Ux; r, )) = r r )k [r k M g x, r)] := Gr) V r, ) = r and V t r, ) = r r )k r k U t x; r, )) = r Hence, by 5.6), we have 5.) V r, t) = [ Gr + t) Gt r)] + By ii) of Lemma 5.5, V r, t) = r r )k [r k M h x, r)] := Hr). r+t t r Hy)dy k r )k r k Ux; r, t)) = βj k r j= j+ j r t). Ux; r, t) rj

10 5. The Wave Equation and hence ux, t) = Ux; + V r, t), t) = lim r + β k [ r = Gr + t) Gt r) β k lim + r + r r = β k [ G t) + Ht)]. r+t t r Hy)dy ] Therefore, we obtain the formula for C k+ -solution of k + )-dimensional wave equation: 5.3) ux, t) = [ β k t))] t t t )k t k M g x, + ) } t t )k t k M h x, t). Note that when n = 3 so k = ) this formula agrees with Kirchhoff s formula derived earlier. In fact, the formula 5.3) defines indeed a classical solution to problem 5.8) under some smoothness assumption on initial data. Theorem 5.7 Solution of wave equation in odd-dimensions). If n = k + 3, g C k+ R n ) and h C k+ R n ), then the function ux, t) defined by 5.3) belongs to C R n, )), solves the wave equation u tt = u in R n, ), and satisfies the Cauchy condition in the sense that, for each x R n, lim ux, t) = gx ), x x, t + lim u tx, t) = hx ). x x, t + Proof. We may separate the proof in two cases: a) g, and b) h. The proof in case a) is given in the text. Here we give a similar proof for case b) by assuming h.. The function ux, t) defined by 5.3) becomes ux, t) = [ β k ) ] t t t )k t k Gx; t), Gx; t) = M g x, t). By Lemma 5.5ii), ux, t) = β k k βj k j= as x, t) x, + ). Also from this formula, Note that G t x, t) = t n [ ] j + )t j j G t j + t j+ j+ G t j+ Gx, + ) = gx ) lim u tx, t) = x x,t + β k Hence u t x, t) as x, t) x, + ).. By Lemma 5.5i), 5.4) u t x, t) = β k t gy) dy = Bx,t) lim G tx, t). x x,t + ) kt k G t ), u tt x, t) = t β k nα n t k gy) dy. Bx,t) t t ) kt k G t ). t

11 5.4. Solutions of the wave equation for general dimensions Since we have Hence G t x, t) = u tt x, t) = β k On the other hand, and nα n t k gy) dy = Bx,t) t t nα n t k t t k G t ) t = t k gy) ds y. Bx,t) ) k ) t t tk G t ) t = β k ux, t) = β k t t Gx; t) = This proves u tt = u in R n, ). t t ) k ) t k Gx; t) t gy) ds y. Bx,t) ) gy) ds y dρ, Bx,ρ) ) k ) t k g ds. t Bx,t) Remark 5.3. i) In the above theorem, the solution ux, t) defined by 5.3) can be extended to t in the same way as M g x, t) and M h x, t). Then the extended function u C R n [, )) takes the initial data in the classical sense: ux, ) = gx), u t x, ) = hx). ii) Let n = k + 3. To compute ux, t) we need the information of g, Dg,, D k g and that of h, Dh,, D k h only on Bx, t), not on the whole ball Bx, t). Therefore, for the odd-dimensional wave equations, the domain of dependence for x, t) is also the sharp wavefront Bx, t); so one still has the strong Huygens principle. iii) If n =, in order for u to be C, we need g C and h C. However, if n = k + 3, in order for u to be C, we need g C k+ and h C k+. So, the solutions in general lose k-orders of smoothness from the initial data Solutions for the even-dimensional wave equation. Assume that n = k is even. Suppose u is a C k+ solution to the Cauchy problem 5.8). Again we use Hadamard s method of descent similarly as in the case n =. Let x = x, x n+ ) R n+, where x R n. Set ũ x, t) = ux, t), g x) = gx), h x) = hx). Then ũ is a C k+ solution to the wave equation in R n+, ) with initial data ũ x, ) = g x), ũ t x, ) = h x). Since n + = k + is odd, we use 5.3) to obtain ũ x, t) and then ux, t) = ũ x, t), where x = x, ) R n+. In this way, we obtain ux, t) = Note that k )!! t M g x, t) = [ t = t )k t k M g x; t))] + t t )k t k M h x; t) t n gy, y n+ )ds n + )α n+ yn+ + x y =t gy) t n n + )α n+ t y x dy, Bx,t) ) }.

12 5. The Wave Equation since ds = t dy on the surface t y x y n+ + y x = t. Similarly we have M h x, t) = t n n + )α n+ Bx,t) hy) t y x dy. Therefore, we have the following representation formula for even n: [ ux, t) = n + )!!α n+ t t t ) n gy) dy ] Bx,t) t y x 5.5) + t t ) n hy) dy }. t y x Bx,t) When n =, this reduces to Poisson s formula for -D wave equation obtained above. Theorem 5.8 Solution of wave equation in even-dimensions). If n = k, g C k+ R n ) and h C k+ R n ), then the function ux, t) defined by 5.5) belongs to C R n, )), solves the wave equation u tt = u in R n, ), and satisfies the Cauchy condition in the sense that, for each x R n, lim ux, t) = gx ), x x, t + lim u tx, t) = hx ). x x, t + Proof. This follows from the theorem in odd dimensions. Remark 5.4. i) In the theorem, note that gy) dy t y x = tn Bx,t) and that the function Gx, t) = B,) gx+tz) dz z B,) gx + tz) dz z is in C k+ R n R). Hence, the solution ux, t) defined by 5.5) can be extended to t and the extended function u C R n [, )) takes the initial data in the classical sense: ux, ) = gx), u t x, ) = hx). ii) Let n = k. To compute ux, t) we need the information of g, Dg,, D k g and that of h, Dh,, D k h in the solid ball Bx, t). Therefore, for the even-dimensional wave equations, the domain of dependence for x, t) is the whole solid ball Bx, t). iii) If n = k, in order for u to be C, we need g C k+ and h C k+. So, again, the solutions in general lose k-orders of smoothness from the initial data Solution of the wave equation from the heat equation*. We study another method of solving the odd-dimensional wave equations by the heat equation. Suppose u is a bounded, smooth solution to the Cauchy problem u tt u = in R n, ) 5.6) ux, ) = gx), u t x, ) = on x R n, where n is odd and g is smooth with nice decay at. We extend u to negative times by even extension of t and then define vx, t) = 4πt) / ux, s)e s 4t ds x R n, t > ).

13 5.4. Solutions of the wave equation for general dimensions 3 Then v is bounded, vx, t) = = = = 4πt) / 4πt) / 4πt) / 4πt) / ux, s)e s 4t ds u ss x, s)e s 4t ds u s x, s) s s e 4t ds t s ux, s) 4t )e s 4t ds t and s v t x, t) = 4πt) / ux, s) 4t )e s 4t ds. t Moreover lim vx, t) = gx) x t Rn ). + Therefore, v solves the Cauchy problem for the heat equation: v t v = in R n, ) vx, ) = gx) on x R n. As v is bounded, by uniqueness, we have vx, t) = 4πt) / gy)e y x 4t dy x R n, t > ). R n We have two formulas for vx, t) and take 4t = /λ in the two formulas to obtain ux, s)e λs ds = λ ) n e λ y x gy) dy π R 5.7) n = nα n λ ) n e λr r n Gx; r) dr π for all λ >, where Gx; r) = M g x, r) = gy) ds y. Bx,r) So far, we have not used the odd dimension assumption. We will solve for u from 5.7) when n = k + 3 is odd. Noticing that d r dr e λr ) = λe λr, we have λ n e λr r n Gx; r) dr = nα n π k k+ = )k k λ k e λr r k Gx; r) dr [ d r dr = [ k r r r ) ke λr )] r k Gx; r) dr ) kr k Gx; r))] e λr dr, where we integrated by parts k times be careful with the operator r write 5.7) as [ ux, r)e λr dr = r r ) kr k Gx; r))] e λr dr r d dr )k ). We can then

14 4 5. The Wave Equation for all λ >. If we think of r as τ, then this equation says that the Laplace transforms of two functions of τ are the same; therefore, the two functions of τ must be the same, which also implies the two functions of r are also the sam. So we obtain 5.8) ux, t) = nα n π k k+ t t t ) kt k Gx; t)) = nα n π k k+ t [ t t )k t k Gx; t) which, except for the constant, agrees with the formula 5.3) with h. In fact the constant here, using α n = πn/ Γ n+ ), Γ ) = π/ and Γs + ) = sγs) for all s >, nα n π k k+ = is also in agreement with the constant in 5.3). k + )π/ k+ Γk + + ) = k )!! = β k 5.5. Nonhomogeneous wave equations and Duhamel s principle We now turn to the initial value problem for nonhomogeneous wave equation u tt u = fx, t) in R n, ) 5.9) ux, ) =, u t x, ) = on x R n, where fx, t) is a given function. Motivated by Duhamel s principle used to solve the nonhomogeneous heat equations, for each s, let Ux, t; s) be the solution to the homogeneous Cauchy problem U tt x, t; s) Ux, t; s) = in R n s, ), 5.) Ux, s; s) =, U t x, s; s) = fx, s) on x R n. Define 5.) ux, t) = t Ux, t; s)ds Note that if v = vx, t; s) is the solution to the Cauchy problem v tt v = in R n, ), 5.) vx, ) =, v t x, ) = fx, s) on x R n, then Ux, t; s) = vx, t s; s) for all x R n, t s. Therefore, ux, t) = t Ux, t; s) ds = t vx, t s; s) ds x R n, t > ). Theorem 5.9. Assume n and f C [ n ]+ R n [, )). Let Ux, t; s) be the solution of 5.). Then the function u defined by 5.) is in C R n [, )) and a solution to 5.9). Proof.. The regularity of f guarantees a solution Ux, t; s) is given by 5.3) if n is odd or 5.5) if n is even. In either case, u C R n [, )).. A direct computation shows that u t x, t) = Ux, t; t) + u tt x, t) = U t x, t; t) + t t U t x, t; s)ds = t U tt x, t; s)ds = fx, t) + U t x, t; s)ds, t U tt x, t; s)ds, ) ],

15 5.5. Nonhomogeneous wave equations and Duhamel s principle 5 ux, t) = t Ux, t; s)ds. Hence u tt u = fx, t). Clearly ux, ) = u t x, ) =. Example 5.. Find a solution of the following problem u tt u xx = te x, x, t) R, ), ux, ) =, u t x, ) =. Solution. First, for each s solve v tt = v xx in R, ), vx, ) =, v t x, ) = se x. From d Alembert s formula, we have v = vx, t; s) = x+t x t se y dy = sex+t e x t ). Hence Ux, t; s) = vx, t s; s) = sex+t s e x+s t ) and so ux, t) = t Ux, t; s)ds = t se x+t s e x+s t ) ds = [ex+t te t e t + ) e x t te t e t + )] = tex + e x+t e x t ). Example 5.. Find a solution of the Cauchy problem u tt u = fx, t), x R 3, t >, ux, ) =, u t x, ) =, x R 3. Solution. By Kirchhoff s formula, the solution v of the Cauchy problem v tt v =, x R 3, t >, ux, ) =, u t x, ) = fx, s), x R 3 is given by Hence v = vx, t; s) = fy, s) ds y x R 3, t > ). 4πt Bx,t) t ux, t) = vx, t s; s) ds = 4π = t fy, t r) 4π r = 4π Bx,t) Bx,r) fy, t y x ) y x t ds y dr dy. Bx,t s) fy, s) t s ds yds Note that the domain of dependence on f) is the finite set y, t y x ) y Bx, t)}, which is the boundary of a solid cone in R 3 R + ; the integrand on the right is called a retarded potential.

16 6 5. The Wave Equation 5.6. Energy method and the uniqueness There are some subtle issues about the uniqueness of the Cauchy problem of wave equations. The formulas 5.3) for odd n or 5.5) for even n hold under more and more smoothness conditions of the initial data g, h as the dimension n gets larger and larger. For initial data not too smooth we cannot use such formulas to claim the uniqueness. Instead, we use certain quantities that behave nicely for the wave equation. One such quantity is the energy Domain of dependence. Let u C be a solution to the wave equation u tt = u in R n, ). Fix x R n, t >, and consider the backward wave cone with apex x, t ): Kx, t ) = x, t) t t, x x t t}. Theorem 5. Domain of dependence). Let u C R n [, )) solve the wave equation u tt = u in R n, ). If u = u t = on Bx, t ) t = }, then u within Kx, t ). Proof. Define the local energy et) = Then et) = Bx,t t) t t u t x, t) + Dux, t) ) dx t t ). Bx,ρ) u t x, t) + Dux, t) ) ds x dρ and so, by the divergence theorem, t e t ) t) = u t u tt x, t) + Du Du t x, t) ds x dρ 5.3) Bx,ρ) u t x, t) + Dux, t) ) ds x Bx,t t) = t u tt + Du Du t )dx Bx,t t)u = u t u tt u)dx + = Bx,t t) Bx,t t) Bx,t t) Bx,t t) u t x, t) + Dux, t) ) ds Bx,t t) u t u ν ds u t u ν u t x, t) Dux, t) ) ds because the last integrand is less than zero; in fact, u t x, t) + Dux, t) ) ds u u t ν u t Du u t u ν u t Du u t Du ). Now that e t) implies that et) e) = for all t t. Thus u t = Du =, and consequently u in Kx, t ). Theorem 5.3 Uniqueness of Cauchy problem for wave equation). Given any f, g, h the Cauchy problem u tt u = fx, t), x R n, t >, ux, ) = gx), u t x, ) = hx), x R n

17 5.6. Energy method and the uniqueness 7 can have at most one solution u in C R n [, )). Example 5.4. a) Show that there exists a constant K such that ux, t) K t U) < t < T whenever T > and u is a smooth solution to the Cauchy problem of 3-D wave equation u tt u =, x R 3, < t < T, ux, ) = gx), u t x, ) = hx), x R 3, where U) = R 3 g + h + Dg + Dh + D g ) dy. b) Let u be a smooth solution to the wave equation u tt = u in R 3, ) satisfying lim ux, t) + Dux, t) + u t x, t) + Du t x, t) + D ux, t) ) dx =. t t R 3 Show that u on R 3, ). Proof. The details are left as Homework. Part b) follows from a) by considering ũx, t) = ux, T t) on R n, T ) Energy method for mixed-value problem of wave equation. Let be a bounded smooth domain in R n and let T =, T ]. Let Γ T = T = T \ T. We are interested in the uniqueness of initial-boundary value problem mixed-value problem) u tt u = f in T, 5.4) u = g on Γ T, u t = h on t = }. Theorem 5.5 Uniqueness of mixed-value problem for wave equation). There exists at most one solution u C T ) of mixed-value problem 5.4). Proof. Let u, u be any two solutions; then v = u u is in C T ) and solves v tt v = in T, v = on Γ T, v t = on t = }. Define the energy et) = Then, by the divergence theorem e t) = v t v tt + Dv Dv t )dx = v t x, t) + Dvx, t) ) dx t T ). v v t v tt v)dx + v t ν ds = since v = on for all < t < T implies that v t = on for all < t < T. Therefore v from v = on Γ T.

18 8 5. The Wave Equation Other initial and boundary value problems. Uniqueness of wave equation can be used to find the solutions to some mixed-value problems. Since solution is unique, any solution found in special forms will be the unique solution. Example 5.6. Solve the Cauchy problem of the wave equation u tt u = in R 3, ), where hr) is a given function. ux, ) =, u t x, ) = h x ), Solution. In theory, we could use Kirchhoff s formula to find the solution; however, the computation would be too complicated. Instead, we can try to find a solution in the form of ux, t) = v x, t) by solving an equation for v, which becomes exactly the Euler-Poisson- Darboux equation that can be solved easily when n = 3; some condition on h is needed in order to have a classical solution. Details are left as an exercise. Example 5.7. Let λ R and λ. Solve u tt u + λu = in R n, ), ux, ) = gx), u t x, ) = hx). Solution. We use the idea of Hadamard s descent method. We first make u a solution to the wave equation in R n+, ) and recover u by this solution. If λ = µ > the equation is called the Klein-Gordon equation), let v x, t) = ux, t) cosµx n+ ), where x = x, x n+ ) R n+ and x R n. If λ = µ <, let v x, t) = ux, t)e µx n+. Then, in both cases, v x, t) solves the wave equation and can be solved by using the formula 5.3) or 5.5). Then we recover ux, t) in both cases by ux, t) = v x, t), where x = x, ) R n+. Example 5.8. Solve u tt x, t) ux, t) =, x = x, x n ) R n R +, t >, ux, ) = gx), u t x, ) = hx), x n >, ux,, t) =, x R n. Solution. We extend the functions g, h to odd functions g and h in x n ; e.g., gx, x n ) = gx, x n ) for all x n R and gx, x n ) = gx, x n ) when x n >. We then solve ũ tt ũ =, x R n, t >, ũx, ) = gx), ũ t x, ) = hx), x R n. Since V x, t) = ũx, x n, t) + ũx, x n, t) solves V tt V =, V x, ) =, V t x, ) =, the uniqueness result implies V, i.e., ũ is an odd function in x n. Hence u = ũ xn> is the solution to the original problem. Example 5.9. Let be a bounded domain. Solve u tt u = fx, t) in, ), ux, ) = gx), u t x, ) = hx), x, ux, t) =, x, t.

19 5.7. Finite speed of propagation for second-order linear hyperbolic equations 9 Solution. Use the method of separation variables and try to find the solution of the form u = u j x)t j t), j= where u j = λ j u j, u j =, T j t) λ j T t) = w j t), T j ) = a j, T j) = b j, By the elliptic theory, eigenfunctions u j x)} j= form an orthonormal basis of L ), and w j t), a j, b j are the Fourier coefficients of wx, t), gx) and hx) with respect to u j x), respectively. The question is whether the series gives indeed a true solution; we do not study such questions in this course Finite speed of propagation for second-order linear hyperbolic equations We study a class of special second-order linear partial differential equations of the form where L has a special form u tt + Lu = x R n, t > ), Lu = a ij x)d ij u, with smooth symmetric coefficients a ij x)) satisfying uniform ellipticity condition on R n. In this case, we say the operator tt + L is uniformly hyperbolic. Let x, t ) R n, ). Assume qx) is a continuous function on R n, positive and smooth in R n \ x } and qx ) =. Consider a curved backward cone and for each t t let C = x, t) R n, t ) qx) < t t} C t = x R n qx) < t t}. Assume C t = x R n qx) = t t} is a smooth surface for each t [, t ). In addition, we assume 5.5) a ij x)d i qx)d j qx) x R n \ x }). Lemma 5.. Let βx, t) be a smooth function and αt) = βx, t) dx < t < t ). C t Then α βx, t) t) = β t x, t) dx C t C t Dqx) ds x. Proof. This follows from the co-area formula. Theorem 5. Domain of dependence). Let u be a smooth solution to u tt + Lu = in R n [, ). If u = u t = on C, then u with the cone C.

20 5. The Wave Equation Proof. Define the local energy et) = C t Then the lemma above implies e t) = C t u t u tt + u t + ) a ij D i ud j u dx t t ). ) a ij D i ud j u t dx u t + C t := A B. ) a ij D i ud j u Dq ds Note that a ij D i ud j u t = D j a ij u t D i u) u t D j a ij D i u) no sum); hence, integrating by parts and by the equation u tt + Lu =, we have A = C t u t = u tt u t C t ) D j a ij D i u) dx + u t a ij D i u)ν j ds C t u t a ij D i u)ν j ds, D i ud j a ij dx + C t where ν = µ, ν,, ν n ) is the outer unit normal on C t. In fact, ν j = D jq Dq j =,,, n) on C t. Since v, w = n aij v i w j defines an inner product on v, w R n, by Cauchy-Schwartz s inequality, a ij D i u)ν j ) / a ij D i ud j u a ij ν i ν j ) /. Therefore A Cet) + u t C t Cet) + C t u t + ) / a ij D i ud j u a ij ν i ν j ) /ds ) ) /ds. a ij D i ud j u a ij ν i ν j However, since ν j = D j q)/ Dq, by 5.5), we have a ij ν i ν j ) / Dq on C t. Consequently, we derive that A Cet) + B and thus e t) Cet) < t < t ). Sine e) =, this gives et). Hence u within C.

21 5.7. Finite speed of propagation for second-order linear hyperbolic equations Energy method for mixed-value problems. Let be a bounded smooth domain in R n and let T =, T ]. Let Γ T = T = T \ T. We are interested in the uniqueness of initial-boundary value problem mixed-value problem) u tt + Lu + Bx, t) Du + cx, t)u = f in T, 5.6) u = g on Γ T, u t = h on t = }, where Lu is defined as above, B and c are bounded functions on T, and g, h are given functions. Theorem 5. Uniqueness of mixed-value problem). There exists at most one solution u C T ) of mixed-value problem 5.6). Proof. Let u, u be any two solutions; then v = u u is in C T ) and solves v tt + Lv + Bx, t) Dv + cx, t)v = in T, Define the energy et) = Then 5.7) e t) = v = on Γ T, v t = on t = }. [ v t x, t) + v x, t) + ] a ij x)v xi v xj dx t T ). v t v tt + vv t + ) a ij v xi v xj t dx. Since a ij v xi v xj t = a ij v xi v t ) xj a ij x j v xi v t a ij v xi x j v t and since v = on for all < t < T implies that v t = on for all < t < T, it follows by the divergence theorem that a ij x)v xi v xj t dx = a ij v xi v t ν j ds a ij x j v xi v t + a ij v xi x j v t ) dx = a ij x j v xi v t + a ij v xi x j v t ) dx. Therefore, using the equation v tt + Lv = B Dv cv, by 5.7), we obtain that ) e t) = v t v B Dv cv a ij x j v xi dx C vt + v + Dv ) dx C v t + v + a ij v xi v xj )dx, where the last inequality follows from the uniform positivity of matrix a ij ). Hence e t) Cet) on, T ). Note that et) and e) = ; this implies et) on [, T ], proving v on T.