Hyperbolic PDEs. Chapter 6

Transcription

1 Chapter 6 Hyperbolic PDEs In this chapter we will prove existence, uniqueness, and continuous dependence of solutions to hyperbolic PDEs in a variety of domains. To get a feel for what we might expect, we will start with the one-dimensional wave equation. It has the form u tt = c 2 u xx, u(x, ) = f(x), u t (x, ) = g(x). The wave equation follows from applying Newton s laws of motion to an elastic string or by combining Maxwell s equations in the context of electricity and magnetism. In the case of a string the term u tt is the acceleration, c is the speed of a wave propagating on the string. The PDE indicates the acceleration is proportional to the concavity of the string: the acceleration is down if the string is concave down, and up if the string is concave up. As with any PDE we have four major concerns we like to address: (1) Does a solution exist? (2) Is the solution unique? (3) Does solution depend continuously on the data? (4) What are the properties of the solutions? If the answer to the first three is yes, we say the PDE is s well-posed. Otherwise it is called ill Posed. In investigating the properties of the solutions we may like to know how the smoothness of the data affects the smoothness of the solution. Or we may like to verify the behavior of the solutions reflects the physics the PDE is meant to depict. 69

2 7 6. Hyperbolic PDEs 6.1. Bounded Domains In this section we apply our spectral method to the wave equation on bounded domains. Recall we represent the solution u(x, t) as span of eigenfunctions. The eigenvalue problem is motivated by the boundary conditions. Consider the wave equation with Dirichlet boundary conditions. That is, (6.1) u tt = c 2 u xx < x < L, t > with boundary conditions and initial conditions We represent u as u(, t) = = u(l, t) t >, u(x, ) = f(x), u t (x, ) = g(x) < x < L. u(x, t) = C n (t)ϕ n (x). To find appropriate basis functions, ϕ n we solve the eigenvalue problem (6.2) n=1 ϕ xx = λ 2 ϕ ϕ() = = ϕ(l), Advantages of solving this problem are that the ϕ satisfy the boundary conditions and ϕ xx is easy to compute. The general solution to (6.2) is ϕ(x) = C 1 cos(λx) + C 2 sin(λx) To satisfy the boundary conditions we need Thus C 1 =. At the other boundary ϕ() = = C 1. ϕ(l) = = C 2 sin(λl). If we want a nontrivial solution we need to require λl = nπ with n N. Thus the eigenvalues are λ n = nπ L and the eigenfunctions are ϕ n (x) = sin( nπ L x). We attempt to write the solution to the wave equation as (6.3) u(x, t) = C n (t) sin(λ n x). n= By attempting to express the solution using (6.3) we see the boundary condition are satisfied and the u xx derivative is easy to compute-provided we can move the derivatives inside the infinite sum. By putting (6.3) into the wave equation (6.1), we can turn the PDE into an infinite number of ODEs. Indeed, assuming derivatives and infinite sums commute, we find u tt = n= C n (t) sin(λ nx), u xx = λ 2 n C n(t) sin(λ n x). n=

3 6.1. Bounded Domains 71 Thus to solve the wave equation we need C n(t) + c 2 λ 2 nc n (t) sin(λ n x) =. n= The orthogonality of the eigenfunctions implies for all n N. Solving we find C n (t) + c2 λ 2 n C n(t) = C n (t) = a n cos(cλ n t) + b n sin(cλ n t) Thus our possible solution takes the form (6.4) u(x, t) = a n cos(cλ n t) + b n sin(cλ n t) sin(λ n x). n= If derivatives and the sum commute, we see u(x, t) solves the wave equation and the boundary conditions. We need only worry about the initial conditions. Somehow this will determine the a n and b n. If we put t = in (6.4), we find (6.5) u(x, ) = f(x) = a n sin(λ n x) n= This is a Fourier series discussed in Chapter 4. Fourier coefficient for n N. a n = 2 L L f(x) sin( nπ x) dx L The orthogonality implies the We can find the b n in a similar way. We need to require u t (x, ) = g(x) for < x < l. So we formally take the time derivative of (6.4) (by switching sum and derivative) and find (6.6) u t (x, ) = g(x) = b n cλ n sin(λ n x). Again we multiply by sin( mπ L for all n N. is n= x) and integrate both sides. We find b n = 2 cnπ L To Summarize: The formal solution to (6.7) u(x, t) = g(x) sin( nπ x) dx L u tt = c 2 u xx < x < L, t > u(, t) = = u(l, t) t >, u(x, ) = f(x), u t (x, ) = g(x) < x < L. n= a n cos(c nπ L t) + b n sin(c nπ L t) sin( nπ L x),

4 72 6. Hyperbolic PDEs where a n = 2 L L f(x) sin( nπ L x) dx, b n = 2 cnπ L g(x) sin( nπ x) dx. L Theorem 6.1. The one-dimensional wave equation with homogeneous Dirichlet boundary conditions has solution given by (6.7) provided f C 4 [, L], g C 3 [, L], f() = f(l) = f () = f (L) = and g() = g(l) =. Proof. We just have to justify our steps leading to (6.7). To satisfy (6.5) and (6.6), according to Dirichlet s Theorem, Theorem 4.13, we need f and g continuously differentiable on < x < L. To justify moving two derivatives past the infinite sum in (6.7) we need a n Cn 4 and b n Cn 3 for all n N and some constant C. To find conditions on f and g to make this happen we follow the ideas in Example 4.15 in Chapter 4. We have L a n = 2 L = 2 L f(x) sin( nπ x) dx L L nπ f(x) cos(nπ L x) L + L nπ L f (x) cos( nπ L x) dx The boundary term must vanish, so we require f() = f(l)= (Dirichlet s theorem already required this). Integrating by parts again a n = 2 L 2 L n 2 π 2 f (x) sin( nπ L x) L L + L2 n 2 π 2 f (x) sin( nπ L x) dx. The boundary term vanishes this time, but we get the added requirement that f C 2 [, L]. We can guess the pattern now. The next time we integrate by parts we will need to require f () = f (L) = and f C 3 [, L]. The next and last integration by parts will give us a n Cn 4 provided f C 4 [, l]. The requirement on b n is less severe (why?). We have,. b n = = 2 cnπ 2 cnπ L g(x) sin( nπ x) dx L L nπ g(x) cos(nπ L x) L + L nπ l g (x) cos( nπ L x) dx. The boundary term must vanish, so we require g() = g(l) =. The next integration by parts only requires g C 2 [, L] and no further condition on the boundary. This will give us b n cn 3. On more integration by parts gives us what we need, b n Cn 4 and requires g C 3 [, L] (the boundary term does not need to vanish). We may now apply Theorem D. The Weierstrass-M test and the smoothness of the basis functions ensures all three conditions are met in Theorem D. Moreover, our calculations show we can apply Theorem D again. This justifies our formal solution.

5 6.2. d Alembert s Solution d Alembert s Solution In this section we study the wave equation on infinite domains. We recall from Chapter 3 that the change of variables ξ = x + ct, η = x ct changes the wave equation to u ξη =. If we integrate and assume u ξη is continuous, we find = η ξ u xy dx dy = u(ξ, η) u(ξ, ) u(, η) + u(, ). This implies that the general solution, that is, all solutions, has the form u(ξ, η) = A(ξ) + B(η), for arbitrary smooth functions A and B. Translating back, the general solution to the wave equation is u(x, t) = A(x ct) + B(x + ct). Applying the initial conditions, we find Integrating the last equation gives Solving for A and B gives x f(x) = A(x) + B(x) g(x) = ca (x) cb (x). g = ca(x) cb(x) + c 1. A(x) = 1 2 f(x) + 1 2c x x g c 1 B(x) = 1 2 f(x) 1 g + c 1. 2c Putting this back in our equation for u give the solution (6.8) u(x, t) = 1 2 (f(x + ct) + f(x ct)) + 1 2c Equation (6.8) is called d Alembert s solution. x+ct Now were are in a position to show the wave equation on the real line is wellposed. The existence is confirmed by (6.8). The uniqueness follows from the way the solution is constructed. The requirement that the solution depends continuously on the data is somewhat vague. It could mean small changes in the data, f and g, imply small changes in the solution. More specifically, if f 1, f 2 C 2 (R) and g 1, g 2 C 1 (R) are such that f 1 (x) f 2 (x) < ǫ and g 1 (x) g 2 (x) < ǫ for all x R, and u 1, u 2 are the corresponding solutions, then (6.8) shows u 1 (x, t) u 2 (x, t) ǫ + tǫ. Thus for finite t the solution are close together and so depend continuously on the data. x ct g.

6 74 6. Hyperbolic PDEs The solution hints at properties of solutions to the wave equation (hyperbolic PDEs). First we see that the solution is no more smooth than the initial data f and it gains one derivative in g. If f = and g has ten derivatives, the u has eleven derivatives. This is not surprising since the initial conditions have the form u t (x, ) = g(x). We also notice the solution only depends on interval [x ct, x+ct]. Only information in the interval [x ct, x+ ct] affects the solution at x, t. Moreover, the solution goes forward and backward in time. Example 6.2. Consider the wave equation u tt = u xx, < x <, t >, u(x, ) = f(x), u(x, x) = g(x). u(, t) = h(t). Starting with the observation that A(x ± t) formally satisfies the PDE for an arbitrary smooth function A, derive d Alembert s solution. Give conditions on f, g and h so that u is twice differentiable. Solution. The solution has the form u(x, t) = A(x + t) + B(x t) for some functions A and B. To find them we apply the data. We find f(x) = A(x) + B(x) g(x) = A(2x) + B() h(t) = A(t) + B( t). This last equation will tell us how to define B for negative input. In particular, we have B( x) = h(x) A(x) Eliminating A between the first two equations, we find A(x) = g(x/2) B(). Back substituting into the first equation yields Thus, for x > t, B(x) = f(x) g(x/2) + B(). u(x, t) = g( x + t t ) + f(x t) g(x 2 2 ). If x < t, we use B( x) = h(x) A(x) instead of the previous B. We find u(x, t) = g( x + t x ) + h(t x) g(t 2 2 ). The candidate for the solution is g( x+t x t u(x, t) = 2 ) + f(x t) g( 2 ) x > t g( x+t t x 2 ) + h(t x) g( 2 ) x < t. We need the solution to have two derivatives. For continuity we require g(x) + f() g() = g(x) + h() g() or f() = h(). We also need u(x, x) = g(x). This requires u(x, x) = g(x) = g(x) + f() g() = g(x) + h() g(). Thus we need f() = g() = h().

7 6.3. Higher-Dimensional Wave Equation 75 For the partial with respect to x to match, we require 1 2 g (x) + f () 1 2 g () = 1 2 g (x) h () g () or f () + h () = g (). Checking the time derivatives match up gives the same conditions. Moving on to second derivatives, we find they all require f () = h (). These conditions could have be predicted by examining the original PDE (do this!). Summarizing, the PDE has a unique solution provided f, g, h C 2 (R + ) and f() = g() = h(), f () + h () = g (), and f () = h () Higher-Dimensional Wave Equation In this section we give some results, without proof, for the wave equation in higher dimensions and on infinite domains. In R N the wave equation is with initial conditions u tt = c 2 2 u x R n, t >, u(x, ) = f(x), u t (x, ) = g(x), x R n. Once again c is a given constant (the wave speed), f and g are given functions. Theorem 6.3. If f C 3 (R 2 ) and g C 2 (R 2 ), then the wave equations has a C 2 solution given by u(x, y, t) = 1 f(x + ctx, y + cty ) dx dy 2t 4π t 1 x 2 y 2 + t 4π x 2 +y 2 <1 x 2 +y 2 <1 g(x + ctx, y + cty ) dx dy 1 x 2 y 2 We note that there is a potential loss of regularity in higher dimensions. Also, similar to the one-dimensional case, the solutions depends on all of the data in the circle (x x) 2 + (y y) 2 ct. In three dimensions the situation is quite different. Theorem 6.4. If f C 3 (R 3 ) and g C 2 (R 3 ), then the wave equations has a C 2 solution given by u(x, t) = 1 t f(x + ctx ) ds x + t g(x + ctx ) ds x. 4π t 4π x =1 x =1 This last equation is called Kirchhoff s Formula. In the three-dimensional case the solution only depends information on a sphere. That is, to determine the solution at time t, we need only integrate on a sphere of radius ct. Once the information on the sphere passes the observation point, it no longer affects the solution. This is in contrast to the two-dimensional case where the solution at an observation point depends on the data everywhere in a circle of radius ct. In three.

8 76 6. Hyperbolic PDEs dimensions solutions travel in sharp fronts. In two dimensions (and in one) the solution is more like throwing a rock into a pond; there are ripples and the solution rings. It turns out, after computing the solution to the wave equation in n dimensions that a pattern emerges. In even dimensions the solution behaves as in the two-dimensional case - it depends on the data inside a hyper sphere. While in odd dimensions (excluding one dimension) the solution travels in sharp fronts - the solution depends only on the data on a hyper sphere. Crisp sounds in odd dimensions and ringing in even. This property of the wave equation is called Huygens s Principle General Two-Dimensional Hyperbolic Equations In this section we try to solve the general hyperbolic equation of the form Lu = u xy + a(x, y)u x + b(x, y)u y + c(x, y)u = F (x, y) with data given on a non characteristic curve Γ. Specifically we assume Γ = {(x, φ(x)) x I}, where I is an interval in φ is a smooth function. Moreover, u(x, φ(x)) = f(x), u x (x, φ(x)) = g(x), u y (x, φ(x)) = h(x), with f, g, and h given functions. We will construct a Green s function for this problem - or at least find conditions necessary to find the Green s functions. The procedure for finding the Green s function is the same as in the last chapter. We first need to find the adjoint operator. To do this we need a domain on which to integrate. As usual, let x, y be a given, fixed, observation point. Let D be region depicted in Figure 1. We start with (Lu, v) and integrate by parts to obtain (u, L v). Integration by parts in multi dimensions is the divergence theorem. We recall the fundamental theorem of calculus in two dimensions. Figure 1. The region D in the x y plane is bounded by Γ and Γ. The point P is at (x, y).

9 6.4. General Two-Dimensional Hyperbolic Equations 77 Theorem 6.5. (Green s Theorem) Suppose C is a piecewise smooth simple closed curve in a region D. If P and Q are continuously differentiable in D and R is the closed region bounded by C, then (Q x P y )dx dy = P dx + Q dy. C R To apply the theorem we need to make Lu v look like ( ) x + ( ) y + ul v. That is, the divergence of some quantity plus a part involving no derivatives on u. Recall we think of (x, y) as a fixed observation point. Only primed variables change. Here is how we compute the adjoint. Note and Averaging these two Similarly, If we set, then vu x y = (vu x ) y v y u x = (vu x ) y (v y u) x + v x y u vu x y = (vu y ) x v x u y = (vu y ) x (v x u) y + v x y u. vlu ul v = vu x y = 1 2 (vu y v y u) x (vu x v x u) y + uv x y. au x v = (auv) x (av) x u, C bu y v = (buv) y (bv) y u, L v = v x y (av) x (bv) y + cv, (vu y v y u) + auv + x 2 (vu x v x u) + buv. y Set T (x, y) to be the triangular region AP B. Since the solution only depends on information in this region, we require the Green s function to vanish outside of T (x, y). We will need to solve (6.9) L v = δ(x x )δ(y y ). Integrating inside the closed region D bounded by Γ and Γ we get (assuming (6.9)) A A (6.1) u(x, t) = D v(x, y, x, y )F (x, y ) dx dy + N 1 dx N 2 dy B B where and N 1 = 1 2 (vu x v x u) + buv N 2 = 1 2 (vu y v y u) + auv. No line integrals appear on Γ since v is zero outside of T (x, y). The linear integral is from B to A since in Green s theorem the line integral is counter clockwise. Equation (6.9) is of course formal. It really requires that certain derivatives of v have jump discontinuities. Based on our one-dimensional experience in the previous

10 78 6. Hyperbolic PDEs chapter, we find them by integrating (6.9). If we integrate in the y direction from y = y ǫ to y = y + ǫ across AP. We get v + x v x b(v+ v ) =, where the + and denote the limiting values as AP is approached from the outside and inside of T (x, y) respectively. Similarly integrating (6.9) in the x direction across BP we get v + y + v y a(v+ v ) =, Since v is zero outside of T (x, y), the terms are all zero, and so v x bv = v y av = on AP, on BP, where we have dropped the + notation since the Green s function is understood to be nonzero in T (x, y). Integrating (6.9) over a rectangle containing P. That is, over x = x ± ǫ and y = y ± ǫ. Taking the limit as ǫ, we find (6.11) v(x, y, x, y) = 1. Next we need to compute the limits of the line integrals in (6.1). The first is A B N 2 dy = lim A A B B A B N 2 dy, where A and B are outside T (x, y) as in Figure 1. The only trouble terms in N 1 and N 2 are terms involving derivatives of v. That is, uv x and uv y. We note that v y is defined on BP and in fact at B. It is discontinuous across BP, but the derivative is in the direction of BP -we are not differentiating across the line BP. This means lim A A B B A B uv y dy = To deal with the latter term we note A A + uv y dy = A + B uv y dy + A lim A A A A + [(uv) y vu y ] dy = (uv) A + A A + vu y dy. A + uv y dy. In the limit as A and A + approaches A the second term vanishes while the first is (uv) A. The subscript means uv evaluated at A. Thus lim A A B B A similar calculation shows lim A A B B B A B N 2 dy = 1 2 (uv) A A N 1 dx = 1 2 (uv) B + A B A B N 2 dy. N 1 dx.

11 6.4. General Two-Dimensional Hyperbolic Equations 79 where and Finally, returning to (6.1), we find u(x, t) = v(x, y, x, y )F (x, y ) dx dy T (x,y) 1 2 ((uv) A + (uv) B ) + A B N 1 = 1 2 (vu x v x u) + buv N 1 dx N 2 = 1 2 (vu y v y u) + auv. A B N 2 dy, Summarizing the conditions on v, L v = in T (x, y). v x bv = on AP or for x x and y = y, v y av = on BP or for x = x and y y, v(x, y, x, y) = 1. That is, v evaluated at P is minus one. These conditions can be met if we require L v = x x y y (6.12) v(x, y, x, y) = exp[ x x b(s, y)ds] x x. v(x, y, x, y ) = exp[ y y a(x, s)ds] y y Equation (6.12) is called a Goursat Problem (pronounced goo sa). Example 6.6. Solve, for y > x u xy yu x = f(x, y), u(x, x) = x, u x (x, x) = x 2, u y (x, x) = x 2 1. Solution. By the way, the PDE comes from writing the operator in canonical form. Lu = u tt u xx 2(t x)(u x + u t ) In this problem the curve Γ has slope 1 so Figure 1 has to be altered. In particular, (6.11) becomes v(x, y, x, y) = 1. The sign changes on the term ((uv) A + (uv) B ) as well. Here a = y, b = c =. Thus (6.12) becomes v x y + (y v) x = x x y y (6.13) v(x, y, x, y) = 1 x x v(x, y, x, y ) = e (y2 y 2 )/2 y y. Integrating L v = with respect to x gives v y + y v = c 1 (y ),

12 8 6. Hyperbolic PDEs where c 1 is a generic function of y. again. We find Using integrating factors we can integrate (6.14) v(x, y, x, y ) = c 1 (y ) + c 2 (x )e y 2 /2, where c 2 is a generic function of x. This is v in the interior of T (x, y). The Green s function v is continuous on T (x, y) and on its boundary, so it must match the other conditions in (6.13). Thus e (y2 y 2 )/2 = v(x, y, x, y ) = c 1 (y ) + c 2 (x)e y 2 /2 1 = v(x, y, x, y) = c 1 (y) + c 2 (x )e y2 /2. If we put y = y in the first of these an compare the two, we conclude c 2 (x ) = c 2 (x). Then the first implies e (y2 y 2 )/2 = v(x, y, x, y ) = c 1 (y ) + c 2 (x )e y 2 /2. Comparing this with (6.14), we conclude the Green s function is v(x, y, x, y ) = e (y2 y 2 )/2. Note that the Green s function is independent of the curve on which the data is given and it is independent of the forcing term f(x, y). The solution is now fun to compute. The point P = (x, y). Therefore, A = ( y, y) and B = (x, x). Thus 1 2 ((uv) A + (uv) B ) = 1 ( y)e (y2 y 2 )/2 + xe (y2 x 2 )/2. 2 We must evaluate N 1 on Γ and express all prime variables in terms of x since the integration is with respect to x. Thus N 1 = 1 2 (vu x v x u) + buv = 1 e (y2 x 2 )/2 x Similarly N 2 must be expressed in terms of y. Thus, for example, u on Γ is u( y, y ) = y. We get The solution is (6.15) N 2 = 1 2 (vu y v y u) + auv = 1 e (y2 y 2 )/2 (y 2 1) y e (y2 y 2 )/2 y + y 2 e (y2 y 2 )/2 2 = y 2 1 e (y2 y 2 )/2. 2 u(x, y) = + y x x x y e (y2 y 2)/2 f(x, y ) dx dy + 1 y 2 x y x 2 e (y2 x 2 )/2 dx y xe (y2 x 2 )/2 y e (y2 y 2 )/2 dy.

13 6.4. General Two-Dimensional Hyperbolic Equations 81 We should check the solution. We need only apply Leibniz s Formula. By writing the first integral as an iterated integral, we get y x y x 2 y 2 )/2 f(x, y ) dx dy = 2 y 2 )/2 f(x, y ) dx dy x where y e (y F (x, y, y ) = x = y e (y x y x y e (y F (x, y, y )dy, 2 y 2 )/2 f(x, y ) dx. If we call the first term in the solution, (6.15), w(x, y). That is, Then w(x, y) = y We see from above F (x, y, x) = and Thus It follows x F (x, y, y )dy. y w x = F (x, y, x) + F x (x, y, y )dy. x F x (x, y, y ) = e (y2 y 2 )/2 f(x, y ). w x = The above calculations show y x e (y2 y 2 )/2 f(x, y )dy. y w xy = f(x, y) + ye (y2 y 2 )/2 f(x, y )dy. x yw x = y x ye (y2 y 2 )/2 f(x, y )dy and we see w xy yw x = f(x, y). This is the particular solution. The remaining part of the solution must solve u xy yu x = (the homogeneous solution) and satisfy the Cauchy data. Putting y = x in the solution (6.15), we immediately see u(x, x) = x as required. Taking the x derivative of the homogeneous solution (by applying Leibniz s Formula), we get u x (x, y) = w x (x, y) + 1 e (y2 x 2 )/2 x 2 e (y2 x 2 )/ x2 e (y2 x 2 )/2 + x 2 1 e (y2 x 2 )/2. 2 Since w x (x, x) =, we see u x (x, x) = x 2 as required. We leave the verification of the last condition, u y (x, x) = x 2 1, as an exercise. One can check, almost by inspection, that u xy yu x = f.

14 82 6. Hyperbolic PDEs 6.5. Wave Equation on Semi-Bounded Domains In this section we consider the wave equation with boundary conditions. Here is a simple model of a wind instrument (trumpet, saxophone...). u tt = c 2 u xx + F (x, t), x >, t >, (6.16) u(, t) = h(t), t >, u(x, ) = f(x), u t (x, ) = g(x), x >. Here, h, f, and g are given functions. To find the solution we need to compute the adjoint. We have, for T > t and L > x, and by integrating by parts, T L (u t t c2 u x x )v dx dt = L c 2 T + (u t v uv t t =T t = dx T L To remove the unknown boundary terms we require v(x, t, x, T ) =, v t (x, t, x, T ) =, (u x v uv x x =L x = dt u(v t t c2 v x x ) dx dt and v(x, t,, t ) =, v(x, t, L, t ) =, v x (x, t, L, t ) =. These are of course satisfied if we require the Green s function to vanish for large x and t > t and v(x, t,, t ) =. Thus the Green s function must satisfy L v = v t t c2 v x x = δ(x x )δ(t t ), v(x, t, x, t ) = for large x, t > t, and v(x, t,, t ) =. Now imagine we were interested in constructing the Green s function for the adjoint problem (L v = f with initial and boundary conditions). That is, we reverse the problem. Denote the Green s function for the adjoint problem by G. Then G satisfies LG = δ(x x )δ(t t ) with G (x, t,, t ) =, G (x, t, x, ) =, G t (x, t, x, ) =, and (LG, G) = (G, L G). The two Green s functions are related. Here is a general result. Theorem 6.7. Suppose G and G satisfy LG = δ(x x ) and L G = δ(x x ). Moreover, suppose (LG, G) = (G, L G) (G and G satisfy appropriate boundary conditions). Then G(x, x ) = G (x, x). Proof. The proof is formal (assumes properties of delta functions are valid). Set u = G (x 1, x ), and v = G(x 2, x ). Then (Lu, v) = δ(x 1 x )G(x 2, x ) dx = G(x 2, x 1 ). Similarly, (Lu, v) = (u, L G) = G (x 1, x )δ(x 2 x ) dx = G (x 1, x 2 ). The result follows.

15 6.5. Wave Equation on Semi-Bounded Domains 83 Before proceeding we recall the Laplace transform of f is defined to be f(s) := L(f(t)) = e st f(t) dt. Table 1 gives the Laplace transforms of some relevant functions. f(t) = L 1 (f(s)) f(s) = L(f(t)) s s > 2. e at 1 s a s > a 3. H(t c)f(t c) e cs L(f(t)) 4. δ(t c) e cs 5. y (t) sl(y) y() 6. y (t) s 2 L(y) sy() y () Table 1. Laplace and inverse Laplace transforms for common functions. We need a theorem about when a function has a Laplace transform. Theorem 6.8. Suppose f : [, ) R satisfies (a) is Riemann integrable on [a, b] for all a b <. (b) There exists c R such that for all a >, lim b (c) for any b >, lim a b a f(t) dt exists. b a e ct f(t) dt exists. Then the Laplace transform of f exists for s > c and f(s) tends to zero as s. Moreover, if a function f(s) does not converge to zero as s, then f(s) is not the Laplace transform of a function satisfying the above conditions. Proof. See Sneddon (1972). We construct G for the problem outlined above. Theorem 6.9. Let Lu = u tt c 2 u xx. The solution to LG = δ(x x )δ(t t ) with boundary and initial conditions is G (x, t,, t ) =, G (x, t, x, ) =, G t (x, t, x, ) =. G (x, t, x, t ) = 1 2c H(t t 1 c x x ) 1 2c H(t t 1 c x + x ), where H(x), the Heaviside function, is zero for x < and one for x.

16 84 6. Hyperbolic PDEs Proof. To solve LG = δ(x x )δ(t t ) we take the Laplace transform with respect to t. Setting G (x, t, x, s) = L(G (x, t, x, t )) we find, using Table 1, (6.17) s 2 G c 2 G xx = δ(x x )e st, G (x, t,, s) =. This is an ODE and the general solution is G = c 1 e sx /c + c 2 e sx /c. The usual jump discontinuity (see Chapter 3) is found by integrating (6.17) across x = x. The result is x + s 2 x G dx c 2 G x (x, t, x+, t ) G x (x, t, x, t ) = e st. As x + and x approach x, the integral vanishes and the jump condition is G x (x, t, x+, t ) G x (x, t, x, t ) = e st c 2. Theorem 6.8 requires G to vanish as s. Thus we require (1) G (x, t,, s) =, (2) lim s G (x, t, x, s) =, (3) G (x, t, x +, s) = G (x, t, x, s), (4) G x (x, t, x+, t ) G x (x, t, x, t ) = e st c 2. Properties (1) and (2) and the general solution for G suggest G (x, t, x c 1 e s(x x)/c e s(x +x)/c x < x, s) = c 2 e sx /c x > x. Applying (3) and (4) we find Thus c 1 = e st 2cs, G (x, t, x, s) = 1 2cs 1 2cs c 2 = e st 2cs e sx/c e sx/c. e s(t 1 c (x x) e s(t+ 1 c (x+x ) e s(t 1 c (x x ) e s(t+ 1 c (x+x ) x < x x > x. Using Table 1, we find 1 G (x, t, x, t ) = 2c H(t t + 1 c (x x)) H(t t 1 c (x + x)) x < x 1 H(t t + 1 c (x x )) H(t t 1 c (x + x)) x > x. 2c We may write the more succinctly as G (x, t, x, t ) = 1 H(t t 1 2c c x x ) H(t t 1 c x + x ).

17 6.5. Wave Equation on Semi-Bounded Domains 85 Theorem 6.1. Consider the wave equation on the half line u tt = c 2 u xx + F (x, t), x >, t >, u(, t) = h(t), t >, u(x, ) = f(x), u t (x, ) = g(x), x >. Suppose f, h C 2 (, ), g C 1 (, ), and F C([, ) [, )). In addition, suppose the compatibility conditions h() = f(), h () = g(), and h () = c 2 f () + F (, ) are satisfied. Then the solution is given by u(x, t) = where F 2 = F 1 (x, t) (f(x + ct) + f(x ct)) + 1 2c F 2 (x, t) (f(x + ct) f(ct x)) + h(t x/c) + 1 2c F 1 = x/c+t x+c(t t ) t x/c x+c(t t ) + t x+c(t t ) x c(t t ) t F (x, t )dx dt + F (x, t )dx dt. F (x, t )dx dt. t x/c x+ct x ct g(s) ds x x+c(t t ) x c(t t ) Moreover, u is twice continuously differentiable in each variable. > ct. x+ct ct x g(s) ds x < ct, F (x, t )dx dt Proof. We have already constructed the Green s function for this problem. From the previous theorem and Theorem 6.7 the Green s function is G(x, t, x, t ) = 1 H(t t 1 2c c x x ) H(t t 1 c x + x ). Returning to the calculation below (6.16), the solution is u(x, t) = G(x, t, x, t )F (x, t ) + G(x, t, x, )g(x )dx G t (x, t, x, )f(x )dx + c 2 G x (x, t,, t )h(t )dt. Note the integrals are not improper since G vanishes for t > t and for large x. To compute the term in the solution involving f, we only need compute G t at t =. Since the derivative of the Heaviside function is the delta function, we have G t (x, t, x, ) = 1 2c δ(t 1 c x x ) + 1 2c δ(t 1 c x + x ). The first delta function is nonzero when t = 1 c x x, or x = x ct and x = x+ct. The second delta function is nonzero when t = 1 c x + x, or only when x = ct x. Thus, for x > ct, G t (x, t, x, )f(x )dx = 1 (f(x + ct) + f(x ct)). 2

18 86 6. Hyperbolic PDEs When x < ct G t (x, t, x, )f(x )dx = 1 (f(x + ct) f(ct x)). 2 We note that the 1/c canceled because δ(x/c)dx = c. Similarly, to compute the term in the solution involving h, we need only compute G x at x =. We have G x (x, t,, t ) = 1 2c 2 δ(t t x c ) + 1 2c 2 δ(t t x c ). Thus c 2 G x (x, t,, t )h(t )dt = x > ct h(t x/c) x < ct. To compute the term involving g, we need only find where the Green s is nonzero when t =. The first term in the Green s function is nonzero when t x x /c > or x x < ct. This requires x ct < x < x + ct. The second term requires t x + x /c > or (ct + x) < x < ct x. Since c and t are positive, the second term is nonzero when < x < ct x. Note that the second term is always zero when x > ct. So, when x < ct G(x, t, x, )g(x )dx = 1 2c x+ct g(x ) dx 1 2c ct x g(x ) dx, and so G(x, t, x, )g(x )dx = 1 x+ct 2c x ct g(x ) dx 1 x+ct 2c ct x g(x ) dx x > ct x < ct. Finally, the most difficult term is the one involving F. The First term in the Green s function is nonzero when t t x x /c > or x c(t t ) < x x < x + c(t t ). This is a triangular region in the x, t plane whose vertices are the points (x, t), (x ct, ), and (x + ct, ). The second term in the Green s function is nonzero when t t x + x /c >. This too is a triangular region in the x, t plane whose vertices are ( (x + ct), ), (ct x, ), and ( x, t). Of course we are only interested in the region of the triangles which are in the right quadrant (x >, t > ). Thus, for x > ct, the triangle associated with the first term is completely in the first quadrant, while the second triangle is disjoint from the first quadrant (a sketch of the triangles is necessary to see this). For x > ct we get G(x, t, x, t )F (x, t )dx dt = t x+c(t t ) x c(t t ) F (x, t )dx dt.

19 6.6. Uniqueness 87 For x < ct each triangle has a portion in the first quadrant. Integrating over the portion in the first quadrant, we get G(x, t, x, t )F (x, t )dx dt = x/c+t x+c(t t ) t + + x+c(t t ) t x/c x c(t t ) t x/c x+c(t t ) Putting these together we obtain the solution stated in the theorem. F (x, t )dx dt F (x, t )dx dt F (x, t )dx dt. To establish the compatibility conditions we simply check the solution. The two parts of the solution and its derivatives should match when x = ct. This is the source of the compatibility conditions. The details are left to the reader Uniqueness In this section we show a variety of hyperbolic PDEs have unique solutions. The proof of uniqueness is independent of the existence of solutions. We start with the PDE (6.18) u tt = c 2 u xx + a(x, t)u x + b(x, t)u y + c(x, t)u F (x, t), < x <, t >, u(x, ) = f(x) u t (x, ) = g(x). We suppose a smooth solution exists for < x < and < t < T, some T >. By making the usual change of variables, ξ = x + ct, η = x ct we can change the PDE into the form u ξη = ã(ξ, η)u ξ + b(ξ, η)u η + c(ξ, η)u F (ξ, η), u(ξ, ξ) = f(ξ), u ξ (ξ, ξ) = g(ξ), u η (ξ, ξ) = h(ξ). The exact form of the transformed known variables is not important. Theorem Suppose (6.18) has a twice continuously differentiable solutions on < x < and < t < T. Moreover, suppose a, b, c are continuously differentiable. Then the solution is unique on < x < and < t < T. Proof. Suppose u 1 and u 2 solve (6.18) with the given initial data. Set w = u 1 u 2. Then w solves w ξη = ã(ξ, η)w ξ + b(ξ, η)w η + c(ξ, η)w, w(ξ, ξ) =, w ξ (ξ, ξ) =, w η (ξ, ξ) =. Let ξ, η be arbitrary. Let T (ξ, η ) be the triangle in the ξ, η plane enclosed by the points (ξ, ξ ), (η, η ) and (ξ, η ). Since T (ξ, η ) is a closed bounded region, w, a, b, c all have maximums. Set M := max { w, ã, b, c, ã ξ, ã η, b ξ, b η }. T (ξ,η )

20 88 6. Hyperbolic PDEs Now let (ξ, η) T (ξ, η ). We will integrate over the triangle T (ξ, η) formed by the points (ξ, ξ), (η, η) and (ξ, η). Note T (ξ, η) T (ξ, η ). We have, ξ ξ w ξ η dξ dη = w ξ η dξ dη = w(ξ, η). It follows Thus (6.19) We have ξ ξ η T (ξ,η) w(ξ, η) = w(ξ, η) = + ξ ξ η η ξ ξ η η ξ ξ η ξ ξ + (ãw) ξ dξ dη η η η ξ ξ η ξ Similarly, we have ξ ξ ( bw) η dξ dη η η = η η η ãw ξ + bw η + cw dξ dη (ãw) ξ + ( bw) η + cw dξ dη η (ã ξ w + b η w) dξ dη. (ãw) ξ + ( bw) η + cw dξ dη η (ã ξ w + b η w) dξ dη. (ãw ξ η dη ξ ξ η η ξ We bound the remaining three terms in (6.19) as follows: and ξ ξ η η ξ ξ η cw dξ dη ξ ξ η η 2M 2 dη = 2M 2 ξ η. ( bw) η dη dξ 2M 2 ξ η. η M 2 dξ dη = M η (ã ξ w + b η w) dξ dη 2M Returning to (6.19) we get, for all (ξ, η) T (ξ, η ), (6.2) 2 (ξ η)2 2 2 (ξ η)2. 2 w(ξ, η) 4M 2 2 ξ η 2 ξ η + 3M 2 4M 2 ξ η + ξ η 2 2 We started by assuming that w is bounded by M on T (ξ, η ), and found a better bound on w. If we repeat the above process, we will obtain yet another bound for.

21 6.6. Uniqueness 89 w. So we use the new bound (6.2) and start over. We have ξ ξ ξ (ãw) ξ dξ dη ã(ξ, η )w(ξ, η ) ã(η, η )w(η, η ) dη η η η ξ 4M 2 2M ξ η + ξ η 2 dη η 2! ξ η = 4M 3 2 ξ η ! 3! Similarly, we can estimate the integral of cw, ã ξ, or b η w by ξ ξ ξ ξ M w dξ dη 4M ξ 3 η + ξ η 2 dξ dη η η η η 2! ξ η 4M 3 3 ξ η ! 4! We will have two terms like the first estimate and three like this last one. Combining we get, for all (ξ, η) T (ξ, η ), ξ η w(ξ, η) 4 4M 3 2 ξ η 3 ξ η M 3 3 ξ η ! 3! 3! 4! ξ η M ! ξ η 3 3! + ξ η 4. 4! We may continue as many times as we like. One more time makes the pattern clear. We have ξ η w(ξ, η) M 4 3 ξ η 4 ξ η 5 ξ η ! 4! 5! 6! We can assume that 8M 1 without loss of generality. Then, for any n 3, we have x w(ξ, η) M 3 n + x n+1 n! (n + 1)! + + x 2n, (2n)! where x = 8M ξ η. To show the right side goes to zero as n, and hence w(ξ, η) = for any ξ, η, we consider the sequence of the form n x k S n (x) = k!. The sequence converges, by the ratio test, for any x R. Call the limit S(x). Then x n n! k=1 + x n+1 (n + 1)! + + x 2n (2n)! S(x) S n 1(x), which goes to zero as n. We conclude w = or u 1 = u 2 and the solution is unique.

22 9 6. Hyperbolic PDEs Uniqueness on Bounded Domains. Next we turn to the wave equation on a bounded domain. Consider the PDE u tt = c 2 u xx + F (x, t), < x < L, t >, (6.21) u(, t) = l(t), u(l, t) = r(t) t >, u(x, ) = f(x), u t (x, ) = g(x) < x < L. Here c is a constant, F, l, r, f, and g are given functions. Before proceeding we recall a theorem from advanced calculus whose proof is in Chapter 1 (Theorem 1.34). Theorem Suppose f(x, t) and tf(x, t) are continuous on [a, b] [c, d]. Then for all t [c, d], d dt b a f(x, t) dx = b a f(x, t) dx. t Example Suppose (6.21) has a solution in C 2 ([, L] [, )). Then the solution is unique. Proof. Suppose (6.21) has two solutions u 1 and u 2. Set w = u 1 u 2. then w solves w tt = c 2 w xx, < x < L, t >, w(, t) = = w(l, t) t >, w(x, ) =, w t (x, ) = < x < L. We multiply Equation (6.21) by w t and integrate over x to obtain L L w tt w t dx c 2 w xx w t dx =. We rewrite the first term as L 1 2 t (w t) 2 dx. The second term we integrate by parts to find L w xx w t dx = w x (L, t)w t (L, t) w x (, t)w t (, t) Thanks to boundary and initial conditions vanishing, we get L w xx w t dx = L w x w xt dx = L L 1 2 t (w x) 2 dx. Putting the two together and applying Theorem 6.12, we obtain d L ((w t ) 2 + c 2 (w x ) 2 ) dx =. dt We set domain-averaged energy to be E(t) = L ((w t ) 2 + c 2 (w x ) 2 ) dx. w x w xt dx.

23 6.7. Uniqueness in Higher Dimensions 91 We have the energy is conserved since E (t) =. Moreover, integrating E (t) = we find E(t) = E(), or L ((w t (x, t)) 2 + c 2 (w x (x, t)) 2 ) dx = L ((w t (x, )) 2 + c 2 (w x (x, )) 2 ) dx =. Since w t and w x are continuous, we conclude w t = w x = on [, L] [, ). We need to show w = everywhere. To do this we first show w is constant. Let (x, t ) and (x 1, t 1 ) be any two points in [, L] [, ), and let x(s), t(s) and any smooth curve connecting the two points. More specifically, suppose (x(), t()) = (x, t ) and (x(1), t(1)) = (x 1, t 1 ). Then by the fundamental theorem of calculus and the chain rule w(x 1, t 1 ) w(x, t ) = = 1 1 =. dw(x(s), t(s)) ds ds (w x (x(s), t(s)), w t (x(s), t(s))) (x (s), t (s)) ds Thus w(x 1, t 1 ) = w(x, t ) for all points in [, L] [, ) and so w is constant. Since w is zero at x = and t =, we conclude w is identically zero. That is, u 1 = u Uniqueness in Higher Dimensions Consider the wave equation on an open connected set Ω R 3. We suppose the boundary of Ω is sufficiently smooth so that the divergence theorem applies (see Theorem 1.31 in Chapter 1). We assume the solutions are in C 2 ( Ω, [, )). u tt = c 2 2 u x Ω, t >, with Boundary conditions u =, t >, x Ω and with initial conditions u(x, ) = f(x), u t (x, ) = g(x), x Ω. Once again c is a given constant (the wave speed), f and g are given functions. To prove uniqueness we proceed as above. We suppose there are two smooth solutions and set w = u 1 u 2. As previously, we multiply by w t. This time we will have to integrate w t 2 w by parts. Note that (w t w) = w t w + w t 2 w. Now we apply the divergence theorem and obtain w t w n dσ = w t w dx + w t 2 w dx. Ω Ω Ω

24 92 6. Hyperbolic PDEs Thus Ω w t 2 w dx = Ω w t w n dσ d 1 dt 2 Ω Applying this to the wave equation we conclude d (wt ) 2 + c 2 w 2 dx =. dt Ω Again we set the energy, E(t) (in this case), to be (wt ) 2 + c 2 w 2 dx =. w 2 dx. Ω Then E (t) =, and so E(t) = E(). As before E() = and hence we conclude w t = and w = on Ω [, ). The same argument at the end of the previous example shows w is constant on this domain. Since w is zero on the boundary, w = everywhere on Ω [, ). That is, u 1 = u 2.

25 6.8. Homework Homework Exercise 6.1. Consider the wave equation u tt = c 2 u xx, < x <, t >, u(x, ) = f(x), u t (x, ) = g(x). (a) Starting with the observation that A(x ± ct) formally satisfies the PDE for an arbitrary smooth function A, derive d Alembert s solution. Give conditions on f and g so that u is twice differentiable. (b) If u i, i = 1, 2 are the solutions corresponding to the initial data f i and g i, respectively. Suppose that f 1 (x) f 2 (x) ǫ and g 1 (x) g 2 (x) ǫ. Show that for each T > there is a constant C(T ) >, depending only on T, such that u 1 (x, t) u 2 (x, t) C(T )ǫ for all x and t with t T. Exercise 6.2. Find in closed form (similar to d Alembert s formula) the solution u(x, t) of the one-dimensional semi-infinite wave equation u tt = c 2 u xx, x, t >, u(, t) =, t, u(x, ) = f(x), u t (x, ) = g(x) x >. Find conditions on f and g so that the solution has two continuous derivatives, even on the characteristic x = ct. What happens when a wave traveling towards x = hits the boundary at x =? Exercise Find in closed form the solution u(x, t) of wave equation with nonhomogeneous boundary condition u tt = c 2 u xx, x, t >, u(, t) = α(t), t, u(x, ) = f(x), u t (x, ) = g(x) x >. Find conditions on f, g, and α so that the solution is C 2. Exercise 6.4. Solve for x, t R the PDE u tt u xx = sin(x 2t) u(x, ) = = u t (x, ). Hint: try ū = A sin(x 2t), for some value of A, as a particular solution. Exercise 6.5. Find a closed-form solution to the one-dimensional, semi-infinite wave equation u tt = u xx, x, t >, u t (x, ) = f(x), x, u(x, x) = g(x), x, u(, t) = h(t), t. Derive conditions on f, g and h so that the formal solution is twice continuously differentiable.

26 94 6. Hyperbolic PDEs Exercise 6.6. Consider the damped wave equation u tt = u t + c 2 u xx, < x < π, t >, u(, t) = u(π, t) =, t >, u(x, ) = f(x), u t (x, ) = < x < π. Using an eigenfunction expansion, find the formal solution. Find conditions on f(x) so that the solution is twice differentiable in both variables. With these assumptions on f(x) prove that u(x, t) as t (You will need to apply Theorem 1.48 in Chapter 1). Exercise 6.7. Find the formal solution to u tt = γ 2 u + c 2 u xx, < x < π, t >, u(, t) = u(π, t) =, t >, u(x, ) = f(x), < x < π, u t (x, ) = g(x), < x < π. Exercise 6.8. Let f(x, y) be a continuous function on R 2. Solve u xy u x = f(x, y), u(x, x 3 ) = sin x, u x (x, x 3 ) = cos x, u y (x, x 3 ) = in the region above the curve Γ : y = x 3. Check the solution. Exercise 6.9. Solve u xy + xu x = f(x, y), u(x, x) = g(x), u x (x, x) = h(x), u y (x, x) =, < x <. We also impose the compatibility condition g (x) = h(x) (See Equation (3.2) in Chapter 3). (Hint: the Green s function is G(x, y, x, y ) = exp[x (y y)]). Check the solution. Exercise 6.1. Find the Green s function for Lu = u xy + xu x + yu y + xyu. Exercise Consider the damped wave equation on a bounded domain u tt = u t + u x + u xx, < x < π, t >, u(, t) = u(π, t) =, t >, u(x, ) = f(x), u t (x, ) =, < x < π. Write the solution, symbolically, to the PDE in terms of a Green s function. That is, find the properties a Green s function must satisfy to solve the problem (not the actual Green s function). Do this by finding the formal adjoint and the adjoint boundary conditions. Exercise Find the solution, for x, t R, by constructing a Green s function, to u tt = u xx + F (x, t), Check your answer. u(x, ) = f(x), u t (x, ) = g(x).

27 6.8. Homework 95 Exercise Solve, by constructing a Green s function, the wave equation on the half line u tt = c 2 u xx, x >, t >, u(x, ) =, u t (x, ) =, x >. u x (, t) = h(t), t >. State carefully the assumptions you make on h(t). Exercise Establish a theorem like Theorem 6.1 for the wave equation on a bounded domain. That is, solve u tt = c 2 u xx + F (x, t), < x < L, t >, u(, t) = h l (t), u(l, t) = h r (t), t >, u(x, ) = f(x), u t (x, ) = g(x), x >. by constructing a Green s function. Find conditions on the data for the solution you find to be twice continuously differentiable. Exercise Consider the damped wave equation u tt = u t + u xx, < x < π, t >, u(, t) = u(π, t) =, t >, u(x, ) = f(x), u t (x, ) = < x < π. Prove the solution is unique. Note that energy is not conserved here. Instead show E (t). Exercise Prove the solution to wave equation with dispersion on a normal domain Ω R 3 is unique. Specifically, show u tt = γ 2 u + 2 u + F (x, t), x Ω, t >, u Ω =, t >, u(x, ) = f(x), x Ω, u t (x, ) = g(x), x Ω has a unique solution in the class of functions in C 2 (Ω, [, )). Exercise Consider a flexible beam with clamped ends at x = and x = 1. Small wave motion in the beam satisfies u tt + γ 2 u xxxx =, < x < 1, t >, u(x, ) = f(x), u t (x, ) = g(x), < x < 1, u(, t) = = u(1, t), u x (, t) = = u x (1, t), t >. Show, using energy methods, the solution is unique. Argue carefully. Exercise Let Ω be a bounded, normal domain in R 3. Assume a solution exists to the following PDE in some space. Prove the solution is unique, and State the conditions you need on the solutions in order to conclude uniqueness: u tt = 2 u, x Ω, t >, (u + u n) =, t >, Ω u(x, ) = f(x), u t (x, ) = g(x), x Ω.

28 96 6. Hyperbolic PDEs Exercise Using the techniques in Section 6.5 find the to the following PDE by constructing a Green s function: u t + u x = F (x, t), and u(x, ) = f(x) with x R and t >.

29 6.9. Solutions Solutions 6.2 Find in closed form (similar to d Alembert s formula) the solution u(x, t) of the one-dimensional semi-infinite wave equation u tt = c 2 u xx, x, t >, u(, t) =, t, u(x, ) = f(x), u t (x, ) = g(x) x >. Find conditions on f and g so that the solution has two continuous derivatives, even on the characteristic x = ct. What happens when a wave traveling towards x = hits the boundary at x =? Solution. The general solution is u(x, t) = A(x + ct) + B(x ct). Applying the initial data we find f(x) = A(x) + B(x) g(x) = ca (x) cb (x). Thus A(x) = 1 2 f(x) + 1 x 2c g c 1 B(x) = 1 2 f(x) 1 x 2c g + c. 1 This gives us a solution when x ct. Indeed, u(x, t) = 1 2 (f(x + ct) + f(x ct)) + 1 2c x+ct x ct To determine the solution for x < ct we apply the boundary condition. The condition u(, t) = implies A(ct) + B( ct) = for t. That is, B( x) = A(x). Thus the solution for x < ct can be found. It is u(x, t) = A(x + ct) A(ct x). We have u(x, t) = 1 2 (f(x + ct) + f(x ct)) + 1 x+ct 2c x ct g x > ct u(x, t) = 1 2 (f(x + ct) f(ct x)) + 1 x+ct 2c ct x g x < ct. To find the conditions on f and g we see immediately that we need f C 2 (R + ) and g C 1 (R + ) to get two continuous derivatives of u. We also need the solution to match at x = ct. This requires f() = f(), or f() =. To get u x to match at x = ct requires g() =. To get u xx to match at x = ct requires f () =. The other derivatives do not give new requirements. To see what happens when a wave hits the wall at x =, suppose g =. Consider the wave equation on the entire line. If we require the initial data to be odd, then d Alembert s solution, (6.8), on the whole agrees with the above solution on the positive axis. Imagine a localized positive initial data on the positive x axis. Then its odd extension has a negative localized counterpart on the negative x axis. The parts of the solution moving toward the origin pass one another with the solution to our problem begin the sum of the two. This implies as the wave hits the origin it flips, reflects, and move out to +. g.