Wave Equations: Explicit Formulas In this lecture we derive the representation formulas for the wave equation in the whole space:

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1 Math 57 Fall 009 Lecture 7 Sep. 8, 009) Wave Equations: Explicit Formulas In this lecture we derive the representation formulas for the wave equation in the whole space: u u t t u = 0, R n 0, ) ; u x, 0) = g x), u t x, 0) = h x). ) It turns out that the properties of the solutions depend on the dimension. More specifically, there are three cases: n =, n > even; n > odd. We will discuss in detail the three representative cases: n =,, 3 the order is actually n =, 3,, for reasons that will be clear soon).. n =. We consider the D wave equation u t t u x x = 0, R 0, ) ; u x, 0) = g x), u t x, 0) = h x). ) This equation can be solved via the following change of variables: and to make things clearer we set ũ ξ, η) = u x, t). With this change of variable we compute Therefore ξ = ; η =, 3) u t = ũ ξ ξ t + ũ η η t = ũ ξ ũ η 4) u t t = ũ ξξ ũ ξη + ũ ηη 5) u x = ũ ξ + ũ η 6) u x x = ũ ξξ + ũ ξη + ũ ηη 7) u t t u x x = 0 ũ ξη = 0 ũ ξ, η) = φ ξ) + ψ η) φ ) + ψ ). 8) Now using the initial values we have which yields This is d Alembert s formula. φ x) + ψ x) = g x) ; φ x) ψ x) = h x) 9) [ g ) + g ) ] + Theorem. Assume g C R), h C R), define u by Then i. u C R [ 0, ) ) ; ii. u t t u x x = 0 in R 0, ) ; [ g ) + g ) ] + iii. u takes the correct boundary values: x, t) x 0, 0) t > 0 x, t) x 0, 0) t > 0 h y) dy. 0) h y) dy. ) g x 0 ) ; ) u t x, t) = h x 0 ). 3) Proof. The proof is by direct calculation and is left as an exercise. Remark. It is easy to generalize the above theorem to the case g C k, h C k u C k. 4)

2 But in general u cannot be smoother in contrast to the heat equation and the Laplace equation). For example, consider the case h = g, then g ). It is clear that u cannot have better regularity than g. Remark 3. One can show that the formula remains true even for distributional solutions of the D wave equation. φ ) + ψ ) 5). Spherical means and Euler-Poisson-Darboux equation. The case n is much more complicated. The idea is to reduce the wave equation to a D equation which can be solved explicitly. The reduction is fulfilled through introducing the following auxiliary functions. Let u = u x, t). We define at each x R n, U x; r, t) u w, t) ds w, 6) G x; r) g w) ds w, 7) H x; r) h w) ds w. 8) Note that when u is continuous, we can recover u from U by taking r 0. It turns out that U x; r, t) as a function of r and t satisfies a D equation. Lemma 4. Euler-Poisson-Darboux equation) Fix x R n. Let u x, t) C m, m solves the wave equation. Then U x; r, t) u w, t) ds w 9) belongs to C m R + [ 0, ) ), and satisfies U t t U r r n r Remark 5. Notice that r r n r Proof. Recall that U r x; r, t) = r n B r x) U r = 0 R + 0, ) ; U r, 0) = G r), U t r, 0) = H r). 0) B r x) r is just in R n with radial symmetry. B r x) y u y, t) dy = n α n) r n y u y, t) dy ) This shows U C, and we can define U r x; 0, t) = 0 since its it as r 0 is 0. Differentiating w. r. t r again, [ U r r x; r, t) = d ] dr n α n) r n y u y, t) dy = n y u + y u. ) n B r This shows U C and also U r r x; 0, t) can be defined. As y u = B r B B x ) y u) x + r z) dz, 3) It is clear that the regularity of the LHS is the same as the regularity of y u. Similar argument shows that the same holds for the term y u. 4)

3 Therefore U r r as the same regularity as y u, which shows U C m when u C m. We further have U t t x; r, t) = u t t = y u 5) using the equation. 3. n = 3, Kirchhoff s formula. Let U, G, H be the spherical means. We set B r x) Some calculation yields Ũ = r U, G = r G, H = r H. 6) Ũ t t Ũ r r = 0 R + 0, ) ; Ũ = G, Ũ t = H. 7) Remark 6. Note that here we used the fact that n = 3. Thus we need to solve the wave equation in the first quadrant. Example 7. Consider the wave equation in the first quadrant: Let u t t u x x = 0, x > 0, t > 0; u x, 0) = g x), u t x, 0) = h x), u = 0 for x = 0, t > 0. 8) ũ x, t) = { u x, t) x > 0 u x, t) x < 0 9) and define similarly g, h. Then it is clear that ũ solves the wave equation with initial values g, h. Thus we have ũ x, t) = [ g ) + g ) ] + h y) dy. 30) Therefore the solution to the original problem is [ g ) + g ) ] + [ g ) g t x) ] + t+ x t x h y) dy x t 0 h y) dy t x 0 Now for our purpose, we only need the case t r remember that finally we will let r 0 and recover u from U). In this case Ũ x; r, t) = [ ] G r + t) G t r) + t+ r H y) dy. 3) We have Further computation yields Ũ x; r, t) r 0 r = G t) + H t) = t which is Kirchhoff s formula. t B t x ) B t x ) g w) ds w ) 4. n=, Method of descent and Poisson s formula. + t t r B t x) 3 ) h w) ds w. 33) [ t h w) + g w) + g w) w x) ] ds w 34)

4 It is not possible to simplify as we did in the n = 3 case. Instead, we use the so-called method of descent, which treats the solution u x, t) of the D wave equation as a solution to the 3D equation. We set and define ḡ, h similarly. Using the Kirchhoff s formula we have ū x, t) = B t x ) ū x, x, x 3, t) u x, x, t). 35) B t x ) where x = x, x 3 ) and B t x ) is the ball in R 3. From definitions of the variaous bar-ed functions, we have 4 π t where w = ) y, ± t y. B t x ) t h w ) + ḡ w ) + x ḡ w ) w x ) ds w. 36) t h y) + g y) + y g y) y x) ds w 37) Finally, let D t x) denote the ball in R centered at x with radius t, we have t h y) + g y) + y g y) y x) 4 π t ) D t x ) y x dy / t This is the Poisson s formula. = D t D t x ) t g y) + t h y) + t g y x) t y x ) / dy. 38) Remark 8. Huygens Principle) We notice that the behavior of the solutions for the D and 3D wave equations are drastically different. In D, u x, t) depends on initial data in the whole ball D t x) while in 3D it only depends on the data on the boundary of the ball B t x). Or equivalently, in 3D the effect of a vibration is only felt at the front of its propagation while in D it is felt forever after the front passed. This is the so-called Huygens principle. Remark 9. For general n, we define Ũ r, t) = ) k r r r k U x; r, t) ) 39) and define G, H accordingly. Some calculation yields the solution ) ) n 3 ) ) n 3 t n g ds + t n γ n t t t t t B t x ) B t x ) h ds ) 40) for n odd, where γ n = 3 ) t t t γ n ) n n ). Then the method of descent yields t n B t for n even, where γ n = 4 n ) n. See Evans pp for details. B t ) g y) dy t y x ) + / t t ) n t n B t Remark 0. Nonhomogeneous problem) For the nonhomogeneous problem B t ) h y) dy t y x ). / u = f, u = 0, u t = 0, 4 ). If u x, t) also depends on data in the whole ball B t x) in 3D, we would not be able to clearly hear anything!

5 we use the Duhamel s principle, obtaining where u x, t; s) solves In particular, we have n = : n = 3: 0 t u x, t; s) ds 4) u t t u = 0, u x, s; s) = 0, u t x, s; s) = f, s). 43) 4 π 0 t x s B t x ) x + s Here the integrand is called the retarded potential. f y, t s) dy ds. 44) f y, t y x ) y x dy. 45)