ECE 3025 - The Wave Equation 1. Acoustic Waves We consider...

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Unformatted text preview: The Wave Equation 1. Acoustic Waves We consider a general conservation statement for a region U Ð R 3 containing a fluid which 3 3x is flowing through the domain U with velocity field V = VÝ3, t Þ . Let _ = _Ý3, t Þ denote the x 3 = FÝ3, t Þ denote the fluid flux at Ý3, t Þ. Then 3x (scalar) fluid density at Ý3, t Þ, and let F x x 3x FÝ3, t Þ = _Ý3, t Þ VÝ3, t Þ describes the direction and speed of the fluid flow at Ý3, t Þ. Proceeding x 3x x as we have in previous examples, we obtain the following equation asserting that the fluid mass is conserved during the flow x x 3x / t _Ý3, t Þ + div _Ý3, t ÞVÝ3, t Þ =0 for all Ý3, t Þ 5 U × Ý0, T Þ x This is another special case of the equation / t u + div F ? s = 0 we have seen before, this 3 3 time with u = _, F = _V, and s = 0. This is one equation for four unknowns, 3 _ and the 3 components of V. An additional equation is obtained from the assertion that momentum is conserved during the flow. This conservation statement, that the time rate of change of momentum equals the sum of the applied forces, can be expressed in terms of the state variables by the vector equation, d/dt X _Ý3, t Þ VÝ3, t Þ dx = ? X p 3 dSÝ3Þ, x 3x n x B /B where B denotes an arbitrary ball in U and p = pÝ3, t Þ denotes the scalar pressure field in the x fluid. Then by an integral identity that is related to the divergence theorem, X/B p 3 dSÝ3Þ = XB 4p dx, n x we arrive at d dt _Ý3, t Þ VÝ3, t Þ x 3x 3 = / t _Ý3, t Þ VÝ3, t Þ + V 6 4 _Ý3, t Þ VÝ3, t Þ x 3x x 3x = ?4p. This adds three equations to the system but also adds a new unknown, p, so the unknowns now consist of _, V 1 , V 2 , V 3 , and p. To complete the system we add the so called equation of state, a constitutive equation which asserts that p = f Ý_ Þ, where f denotes a fluid dependent function relating pressure to density. In one dimension, this system has the form 1 / t _Ýx, t Þ + / x Ý_Ýx, t ÞVÝx, t ÞÞ = 0, (1.1) / t Ý_VÞ + V/ x Ý_ V Þ + / x p = 0, p = f Ý_ Þ This is a system of nonlinear first order equations. The solution of this system is, in general, quite difficult, even in one dimension. Therefore we consider the simpler problem of modeling the propagation of acoustic waves in the fluid. Acoustic waves are small amplitude perturbations in the density field in a quiescent fluid. That is, VÝx, t Þ = 0 + fÝx, t Þ where _Ýx, t Þ = _ 0 Ý1 + dÝx, t ÞÞ |f | << 1 where _ 0 = const and |d | << 1 pÝx, t Þ = fÝ_ Þ = fÝ_ 0 Þ + f v Ý_ 0 ÞÝ_ ? _ 0 Þ = p 0 + _ 0 f v Ý_ 0 Þ dÝx, t Þ. These equations express that the unperturbed velocity and density fields are equal to zero and _ 0 = const, respectively, while the perturbations in these fields, f and d, are much less than 1 in magnitude. The perturbation in the pressure field is determined from the density perturbation and the equation of state. Substituting these expressions into the equations Ý1.1 Þ and neglecting any terms that involve products of perturbations, leads to / t dÝx, t Þ + / x fÝx, t Þ = 0 / t fÝx, t Þ + f v Ý_ 0 Þ / x dÝx, t Þ = 0. Then it is easy to show that both dÝx, t Þ and fÝx, t Þ satisfy the same second order equation, / tt uÝx, t Þ ? a 2 / xx uÝx, t Þ = 0, Ý1.2 Þ where f v Ý_ 0 Þ = a 2 in this case. Equation (1.2) is referred to as the wave equation due to the fact that its solutions exhibit wave-like behavior. 2. Electromagnetic Waves In a region U Ð R 3 with no charges present and no currents, the electric force field 3 3x 3 3x E = EÝ3, t Þ, and the magnetic force field, B = BÝ3, t Þ, satisfy Maxwell’s equations 2 3x iÞ 0 = div EÝ3, t Þ, 3x 3x iiÞ 0 = curl EÝ3, t Þ + 1 / t BÝ3, t Þ C 3x 3x iiiÞ 0 = curl BÝ3, t Þ ? 1 / t EÝ3, t Þ C 3x ivÞ 0 = divBÝ3, t Þ Apply the curl operator to ii) and recall that 3 3 3x curl curl EÝ3, t Þ = grad div E ? div 4 E , to obtain 3 3 grad div E ? div 4 E + 1 C 3 / t curl B = 0. Then it follows from i) and iii) that 3x ? 4 2 EÝ3, t Þ + and similarly, 1 C2 3x / tt EÝ3, t Þ = 3, 0 3x ? 4 2 BÝ3, t Þ + 1 C2 3x / tt BÝ3, t Þ = 3. 0 Evidently, each component of the electric and magnetic fields satisfy the 3-dimensional wave equation. As we have seen in the past, very different physical phenomena can be modelled by the same mathematical description. 3. Some Problems for the Wave Equation We can add various auxiliary conditions to the wave equation to try to get a well posed problem. (a) Pure Initial Value Problem (Cauchy Problem) x x x / tt uÝ3, t Þ ? 4 2 uÝ3, t Þ = FÝ3, t Þ 3 5 Rn, 0 < t < T x uÝ3, 0 Þ = fÝ3Þ, x x 35R x / t uÝ3, 0 Þ = gÝ3Þ, x x (3.1) 3 5 Rn. x (b) Initial-Boundary Value Problem x x x / tt uÝ3, t Þ ? 4 2 uÝ3, t Þ = FÝ3, t Þ n 3 5 Rn, 0 < t < T x uÝ3, 0 Þ = fÝ3Þ, x x 3 5 Rn x / t uÝ3, 0 Þ = gÝ3Þ, x x (3.2) 3 5 Rn. x BCßuÝ3, t Þà = hÝ3, t Þ, x x 3 5 /U, 0 < t < T x # 3 where BCßuÝ3, t Þà denotes one of the types of boundary conditions we have discussed. x (c) Dirichlet Problem for the Wave Equation x x / tt uÝ3, t Þ ? 4 2 uÝ3, t Þ = 0 3 5 U Ð Rn, 0 < t < T x uÝ3, 0 Þ = fÝ3Þ, x x 35UÐR x uÝ3, T Þ = gÝ3Þ, x x 3 5 U Ð Rn x uÝ3, t Þ = 0, x (3.3) 3 5 /U, 0 < t < T x n Problems (a) and (b) are examples of well posed problems for the wave equation, while (c) is not well posed. Problem 23 Consider the Dirichlet problem for the 1-dimensional wave equation, / tt uÝx, t Þ ? / xx uÝx, t Þ = 0 uÝ x , 0 Þ = 0 uÝx, T Þ = 0, uÝ0, t Þ = uÝL, t Þ = 0, 0 < x < L, 0 < t < T, 0 < x < L, 0 < x < L, 0 < t < T. Show that if T/L = m/n, for integers m and n, then the problem has infinitely many solutions u mn Ýx, t Þ = C sinÝn^x/L Þ sinÝm^t/T Þ , but if T/L is irrational, then the trivial solution is the only solution. This shows that the solution does not depend continuously on the data, which in this case is the shape (dimensions) of the domain, 0 < x < L, 0 < t < T . 4. The One Dimensional Wave Equation We will begin by considering the simplest case, the 1-dimensional wave equation. Recall that for arbitrary differentiable functions of one variable, F and G, Ý/ t ? a / x ÞFÝx + at Þ = 0, and Ý/ t + a / x ÞGÝx ? at Þ = 0. This implies Ý/ tt ? a 2 / xx ÞßFÝx + at Þ + GÝx ? at Þà = Ý/ t + a / x ÞÝ/ t ? a / x ÞßFÝx + at Þ + GÝx ? at Þà = 0 i.e., uÝx, t Þ = FÝx + at Þ + GÝx ? at Þ solves Ý/ tt ? a 2 / xx Þ uÝx, t Þ = 0 Now consider the initial value problem Ý3.1 Þ for n = 1, / tt uÝx, t Þ ? a 2 / xx uÝx, t Þ = 0 uÝx, 0 Þ = fÝxÞ, x 5 R, 0 < t < T, x 5 R, Ý4.1 Þ 4 / t uÝx, 0 Þ = gÝx Þ, Then x 5 R. uÝx, t Þ = FÝx + at Þ + GÝx ? at Þ uÝx, 0 Þ = FÝx Þ + GÝx Þ = fÝx Þ / t uÝx, 0 Þ = F v Ýx Þ Ýa Þ + G v Ýx Þ Ý?a Þ = gÝx Þ. FÝx Þ + GÝx Þ = fÝx Þ, a ßF v Ýx Þ ? G v Ýx Þ à = gÝx Þ or This leads to and 1 2 F Ýx Þ = x fÝx Þ + 1 X gÝs Þ ds, 0 2a Then uÝx, t Þ = = 1 2 x FÝx Þ ? GÝx Þ = 1 X gÝs Þ ds a0 GÝx Þ = 1 2 x fÝx Þ ? 1 X gÝs Þ ds.. 0 2a x+at x?at ßfÝx + at Þ + fÝx ? at Þà+ 1 X gÝs Þ ds ? 1 X gÝs Þ ds 0 0 2a 2a 1 2 x+at ßfÝx + at Þ + fÝx ? at Þà + 1 X gÝs Þ ds x?at 2a Ý4.2 Þ This is referred to as the D’ Alembert solution of the wave equation. Note that for a fixed Ýx 0 , t 0 Þ, the solution value uÝx 0 , t 0 Þ depends only on the data values in the interval DÝx 0 , t 0 Þ = ßx 0 ? at 0 , x 0 + at 0 à . This interval is referred to as the domain of dependence for the point Ýx 0 , t 0 Þ. Data values outside this interval have no influence on the value of u at the point Ýx 0 , t 0 Þ. 2 time axis x + t = 2, x ? t = ?2 -2 0 x axis 2 Domain of Dependence The values of the data at a point x 0 have an effect on the value of the solution at a point Ýx, t Þ, t > 0, only when the point Ýx, t Þ lies inside the wedge shaped region between the lines x + at = x 0 and x ? at = x 0 . This region is referred to as the domain of influence for the point x 0 . For a fixed time t 1 > 0, the data values at x 0 influence the solution values uÝx, t 1 Þ for all x in the interval ßx 0 ? at 1 , x 0 + at 1 à. At the later time t 2 > t 1 , the data values at x 0 influence the solution values uÝx, t 2 Þ for all x in the larger interval ßx 0 ? at 2 , x 0 + at 2 à. In the amount of time from t 1 to t 2 , the interval of influence expands by the amount aÝt 2 ? t 1 Þ, which is to say the domain of influence is expanding at the rate a. 5 1 time axis -2 x + t = ?1, x ? t = ?1 0 -1 x axis Domain of Influence If f 5 C 2 ÝR Þ, and g 5 C 1 ÝR Þ, then the D’Alembert solution, Ý4.2 Þ, solves the Cauchy problem Ý4.1 Þ. In addition, it is evident from Ý4.2 Þ that if u 1 , u 2 are solutions corresponding to data pairs áf 1 , g 1 â and áf 2 , g 2 â, then max | u 1 ? u 2 | ² max | f 1 ? f 2 | + t 0 max |g 1 ? g 2 | x x Ýx,t Þ Ý4.3 Þ where for each t 0 > 0, the solution max is taken over the interval áÝx, t Þ : x 1 ? at ² x ² x 2 + at, 0 ² t ² t 0 â the data max is taken over the interval ßx 1 , x 2 à. Note that the estimate (4.3) implies both the uniqueness of the solution and the continuous dependence on the data. 5. Energy Integrals Suppose uÝx, t Þ is a solution for Ý4.1 Þ for smooth data fÝx Þ, gÝx Þ and assume that the data vanishes for |x | > C where C denotes some positive but finite constant. Then for each finite positive t it follows from the discussion of the domain of influence, that uÝx, t Þ vanishes for |x | > C + at. Now, for t ³ 0, let EÝt Þ be given by E Ýt Þ = 1 2 XR ß / t uÝx, t Þ 2 + a 2 / x uÝx, t Þ 2 à dx Ý5.1 Þ Then 0 ² E Ý0 Þ = 1 2 XR ß gÝx Þ 2 + a 2 f v Ýx Þ 2 à dx = 1 2 C X ?C ß gÝx Þ 2 + a 2 f v Ýx Þ 2 à dx < K. Moreover, E v Ýt Þ = X ß / t uÝx, t Þ / tt uÝx, t Þ + a 2 / x uÝx, t Þ / tx uÝx, t Þà dx R But XR / x uÝx, t Þ / tx uÝx, t Þ dx = / x uÝx, t Þ / t uÝx, t Þ | K ? XR / xx uÝx, t Þ / t uÝx, t Þ dx ?K and since u vanishes for |x | > C + at, 6 E v Ýt Þ = X / t uÝx, t Þ ß / tt uÝx, t Þ ? a 2 / xx uÝx, t Þà dx = 0. R Then EÝt Þ is a non-negative constant. Physically, EÝt Þ is related to the total energy in the system. In particular, / t uÝx, t Þ 2 is proportional to the kinetic energy in the system, while / x uÝx, t Þ 2 is proportional to the potential energy stored in the system. As the system evolves in time,the energy simply changes from kinetic to potential and back again. This is because the equation in (4.1) contains no term which dissipates energy. Note that if (4.1) is modified as follows, / tt uÝx, t Þ ? a 2 / xx uÝx, t Þ + b 2 / t uÝx, t Þ = 0, for x 5 R, t > 0, then, in the same way, we find E v Ýt Þ = X / t uÝx, t Þ ß / tt uÝx, t Þ ? a 2 / xx uÝx, t Þà dx = R = X / t uÝx, t Þ ß ? b 2 / t uÝx, t Þà dx ² 0. R . Evidently, the additional term dissipates energy, causing E(t) to decrease as long as / t uÝx, t Þ is different from zero. Since E(t) is non-negative, E(t) will decrease steadily toward zero. The added term here has the physical interpretation of being related to friction. Adding other lower order terms to the wave equation does not produce the same dissipative effect. For example, if uÝx, t Þ solves the equation / tt uÝx, t Þ ? a 2 / xx uÝx, t Þ + b 2 uÝx, t Þ = 0, then for x 5 R, t > 0, E v Ýt Þ = X / t uÝx, t Þ ß / tt uÝx, t Þ ? a 2 / xx uÝx, t Þà dx = R = X / t uÝx, t Þ ß ? b 2 uÝx, t Þà dx = ? 1 b 2 2 R This implies E D Ýt Þ = E Ýt Þ + 1 2 d dt XR uÝx, t Þ 2 dx. b 2 X uÝx, t Þ 2 dx = constant. R Note that the energy integral provides an additional method for proving uniqueness of the solution to the Cauchy problem for the wave equation. To use the arguments presented here, however, we have to suppose that the data f and g have compact support. 6. The Inhomogeneous Wave Equation Consider the problem with homogeneous initial conditions but inhomogeneous equation, / tt vÝx, t Þ ? a 2 / xx vÝx, t Þ = FÝx, t Þ, x 5 R, t > 0, (6.1) vÝx, 0 Þ = / t vÝx, 0 Þ = 0, x 5 R, 7 where FÝx, t Þ is given. Then, for a fixed s 5 R, let wÝx, t; s Þ be given by wÝx, t; s Þ = 1 2a x+at X x?at FÝz, s Þ dz (6.2) It is easy to check that wÝx, t; s Þ solves (just note that wÝx, t; s Þ = PÝx + atÞ ? PÝx ? atÞ), / tt wÝx, t Þ ? a 2 / xx wÝx, t Þ = 0, wÝx, 0 Þ = 0, / t wÝx, 0 Þ = FÝx, s Þ, x 5 R, t > 0, x 5 R, and v Ýx , t Þ = t X 0 wÝx, t ? s; s Þ ds = 1 2a t x+aÝt?sÞ X 0 X x?aÝt?sÞ FÝz, s Þ dz ds. (6.3) To see that v(x,t) must be given by (6.3), differentiate (6.3) with respect to t, t / t vÝx, t Þ = wÝx, 0; t Þ + X / t wÝx, t ? s; s Þ ds 0 t t / tt vÝx, t Þ = / t wÝx, 0; t Þ + X / tt wÝx, t ? s; s Þ ds = FÝx, t Þ + X a 2 / xx wÝx, t ? s; s Þ ds 0 0 t = FÝx, t Þ + a 2 / xx X wÝx, t ? s; s Þ ds = FÝx, t Þ + a 2 / xx vÝx, t Þ. 0 It follows now that the solution of / tt uÝx, t Þ ? a 2 / xx uÝx, t Þ = FÝx, t Þ, x 5 R, t > 0 uÝx, 0 Þ = fÝx Þ, x 5 R, / t uÝx, 0 Þ = gÝx Þ, (6.4) x 5 R, is given by uÝx, t Þ = 1 ßfÝx + at Þ + fÝx ? at Þà + 1 2a 2 x+at X x?at gÝs Þ ds + 1 2a t x+aÝt?sÞ X 0 X x?aÝt?sÞ FÝz, s Þ dz ds (6.5) 7. A Fundamental Solution for the Wave Equation We can show that a fundamental solution for the 1-dimensional wave equation is given by uÝx, t Þ = 1 HÝat ? |x | Þ. 2a This means, that uÝx, t Þ solves L W uÝx, t Þ = Ý/ tt ? a 2 / xx ÞuÝx, t Þ = NÝx, t Þ, ? K < x < K, t > 0. 8 Here a > 0 and HÝ6Þ denotes the Heaviside step function that is zero when its argument is negative and is one when the argument is positive. Since the delta notation is only formal at this point, we restate the condition defining the fundamental solution as follows, K K X ?K X 0 dÝx, t ÞL W uÝx, t Þdtdx = dÝ0, 0 Þ Ý7.1 Þ for all smooth functions d which are zero for |x |, t sufficiently large. To show that u does, in fact, satisfy (7.1), we first use integration by parts to write K K K K X ?K X 0 dÝx, t ÞL W uÝx, t Þdtdx = X ?K X 0 uÝx, t Þ L W dÝx, t Þdtdx. Here we used the fact that dÝx, t Þ is zero for |x |, t sufficiently large to eliminate all the boundary terms in the integration by parts. Now we note that HÝat ? |x | Þ = x 1 if ?K < x < 0, t > ? a x 1 if 0 < x < K, t > a 0 if otherwise in order to write K K KK X ?K X 0 uÝx, t Þ L W dÝx, t Þdtdx = 1 X 0 X x Ý/ t ? a / x ÞÝ/ t + a / x ÞdÝx, t Þdtdx 2a a 0 K + 1 X X x Ý/ t + a / x ÞÝ/ t ? a / x ÞdÝx, t Þdtdx. 2a ?K ? a x x In the first of these integrals, let t = b + a so as t varies over a , K , b varies over ß0, K à x x and / t = / b .In the second integral, let t = b ? a so as t varies over ? a , K , b varies over ß0, K à and / t = / b . Then K K KK x X ?K X 0 uÝx, t Þ L W dÝx, t Þdtdx = 1 X 0 X 0 Ý/ b ? a / x ÞÝ/ b + a / x Þd x, b + a dbdx 2a 0 K x + 1 X X Ý/ b + a / x ÞÝ/ b ? a / x Þd x, b ? a dbdx. ?K 0 2a But so d d x, b + dx d d x, b ? dx x a x a = / x d + 1 / b d, a = / x d ? 1 / b d, a K K KK x X ?K X 0 uÝx, t Þ L W dÝx, t Þdtdx = 1 X 0 X 0 d Ý/ b ? a / x Þd x, b + a dxdb 2 dx 9 K0 d Ý/ b + a / x Þd x, b ? x dxdb, ?1 X X a 0 ?K dx 2 K K = ? 1 X Ý/ b ? a/ x ÞdÝ0, b Þ db ? 1 X Ý/ b + a / x ÞdÝ0, b Þdb, 20 20 K = ? X / b dÝ0, b Þ db = dÝ0, 0 Þ. 0 Using this fundamental solution, we can proceed as in section 9 of the previous chapter and construct Green’s functions for the wave operator. In particular, we can show that (6.5) can be interpreted in terms of the Green’s function for the 1-d wave equation. 8. The Wave Equation in R n In R n , the wave equation has the form x x / tt uÝ3, t Þ ? 4 2 uÝ3, t Þ = 0 for x 5 R n , t > 0. We will consider several simple solutions for this equation. Plane Wave Solutions cx Let 3 denote a fixed unit vector in R n . Then for each constant, C, 3 6 3 ? at = C, is the c equation of a plane in R n having 3 as its normal vector. As t varies, the plane moves in the c direction of the normal vector with speed equal to a. Now for F a smooth function of one variable, let uÝ3, t Þ = FÝ3 6 3 ? at Þ and note that x cx ¾ x x cx cx cc / tt uÝ3, t Þ ? 4 2 uÝ3, t Þ = F ”Ý3 6 3 ? at Þ a 2 ? a 2 F ”Ý3 6 3 ? at Þ 3 6 3 = 0. Then uÝ3, t Þ = FÝ3 6 3 ? at Þ is called a plane wave solution to the wave equation. If the wave x cx x x operator is changed to the related operator, / tt uÝ3, t Þ ? / f A /uÝ3, t Þ = 0, where A denotes a symmetric n by n matrix, then uÝ3, t Þ = FÝ3 6 3 ? at Þ solves the new equation if x cx x x cx cx cc / tt uÝ3, t Þ ? / f A /uÝ3, t Þ = F ”Ý3 6 3 ? at Þ a 2 ? F ”Ý3 6 3 ? at Þ 3f A 3 = 0; cc i.e., if a 2 ? 3f A 3 = 0. It is evident that plane wave solutions for this equation propagate with a different speed in every direction 3. c Note that uÝ3, t Þ = FÝ3 6 3 ? at Þ + GÝ3 6 3 + at Þ is not the general solution to the wave x cx cx equation in R n since it is clear that any unit vector 3 5 R n produces a solution to the PDE, c whereas in R 1 the only unit vector is c = 1. Spherical Wave Solutions In spherical coordinates, ¾ 10 / tt uÝ3, t Þ ? 4 2 uÝ3, t Þ = / tt uÝ3, t Þ ? r 1?n / r Ýr n?1 / r u Þ ? r ?2 C g u = 0 x x x and for spherically symmetric waves, u = uÝr, t Þ this reduces to / tt uÝ3, t Þ ? 4 2 uÝ3, t Þ = / tt uÝ3, t Þ ? r 1?n / r Ýr n?1 / r u Þ = 0. x x x For n = 3, this is, / tt uÝ3, t Þ ? r ?2 / r Ýr 2 / r u Þ = / tt uÝr, t Þ ? / rr u ? 2 / r u = 0. x r Let wÝr, t Þ = r uÝr, t Þ, so that / r w = r / r u + u, and / rr w = r / rr u + 2 / r u. Then if uÝr, t Þ is any spherically symmetric solution of the wave equation in R 3 , r / tt uÝr, t Þ ? r / rr u ? 2 / r u = / tt w ? / rr w = 0. But this implies wÝr, t Þ = FÝr ? t Þ + GÝr + t Þ for smooth functions F and G and then uÝr, t Þ = 1 FÝr ? t Þ + 1 GÝr + t Þ; r r i.e., uÝr, t Þ is the sum of a contracting spherical wave and an expanding spherical wave. For n = 2, the spherically symmetric wave equation has the form / tt uÝr, t Þ ? 4 2 uÝr, t Þ = / tt uÝr, t Þ ? r ?1 / r Ýr / r u Þ = / tt uÝr, t Þ ? / rr u ? 1 / r u = 0 r and there is no simple analogue of the trick that worked for n = 3. Solutions to the wave equation are very sensitive to changes in dimension. As another illustration of this fact, note that in 1-dimension, the solution of the Cauchy problem / tt uÝx, t Þ ? / xx uÝx, t Þ = 0, is given by uÝx, t Þ = X x+t 1 x?t 2 Then / t uÝx, t Þ = 1 2 and / tt uÝx, t Þ = 1 2 uÝx, 0 Þ = 0, / t uÝx, 0 Þ = gÝx Þ, gÝs Þ ds. ßgÝx + t Þ ? gÝx ? t Þà ßg v Ýx + t Þ + g v Ýx ? t Þà so that if g is continuous with a continuous derivative, then / tt uÝx, t Þ is continuous. On the other hand, in 3-dimensions the solution of / tt uÝr, t Þ ? 4 2 uÝr, t Þ = 0, 11 uÝr, 0 Þ = 0, is given by Ý1 ? r 2 Þ / t uÝr, 0 Þ = 3/2 if r 2 < 1 = gÝr Þ, if r 2 > 1 0 uÝr, t Þ = 1 FÝt + r Þ ? 1 FÝt ? r Þ, for F an even function. r r Note that if F is even, then uÝr, 0 Þ = 1 FÝr Þ ? 1 FÝ?r Þ = 0, r r and / t uÝr, 0 Þ = 1 F v Ýr Þ ? 1 F v Ý?r Þ = 2 F v Ýr Þ. r r r Note that since F is even, it follows that the derivative is odd, so F v Ýr Þ ? F v Ý?r Þ = 2F v Ýr Þ. Then the initial condition implies F v Ýr Þ = Ýr/2Þ gÝr Þ. In addition, lim r¸0 1 FÝt + r Þ ? 1 FÝt ? r Þ r r = 2F v Ýt Þ, which leads finally to the conclusion v lim r¸0 uÝr, t Þ = uÝ0, t Þ = 2F Ýt Þ = t gÝt Þ = tÝ1 ? t 2 Þ 3/2 if t 2 < 1 0 if t 2 > 1 . Evidently, even though u, / t u, and / tt u are all continuous at t = 0 for all r ³ 0, it is still the case that when t = 1 and r = 0 then u, and / t u, are continuous but / tt u is discontinuous. This singularity in the second derivative is a dimension dependent phenomenon referred to as ”focusing”. 9. Energy Integrals in R n Let U denote a bounded open set in R n and consider the following initial boundary value problem / tt uÝ3, t Þ ? 4 2 uÝ3, t Þ = 0 x x uÝx, 0 Þ = / t uÝx, 0 Þ = 0 uÝ x , t Þ = 0 Let E Ýt Þ = X 2 U x 5 U, t > 0, x 5 U, x 5 /U, t > 0. / t uÝ3, t Þ + | 4uÝ3, t Þ| 2 dx x x for t ³ 0. Then EÝ0 Þ = 0 and x x x x E v ÝtÞ = X 2ß/ t uÝ3, t Þ / tt uÝ3, t Þ + 4uÝ3, t Þ 6 4/ t uÝ3, t Þà dx. U But XU 4uÝ3, t Þ 6 4/ t uÝ3, t Þ dx = X/U / t uÝ3, t Þ 4uÝ3, t Þ 6 3 dS ? XU 4 2 uÝ3, t Þ / t uÝ3, t Þdx x x x x n x x 12 hence E v ÝtÞ = X 2/ t uÝ3, t Þ ß / tt uÝ3, t Þ ? 4 2 uÝ3, t Þ à dx. + X x x x U /U / t uÝ3, t Þ / N uÝ3, t Þ dS. x x Then it follows from the wave equation and the boundary condition that E v ÝtÞ = 0. This, in turn, implies that / t u and |4u | are both zero in Ū × ß0, KÞ, which is to say uÝx, t Þ is constant on the same set. But uÝx, 0 Þ = 0 for x 5 U, hence we conclude that the solution vanishes identically in Ū × ß0, KÞ. This is the essential part of the proof that the solution to the IBVP for the wave equation is unique. The energy integral EÝt Þ can also be used to show that solutions to the n-dimensional wave equation obey the principle of causality. Proposition 9.1 Suppose uÝ3, t Þ satisfies x x x / tt uÝ3, t Þ ? 4 2 uÝ3, t Þ = 0 x 5 R n , t > 0. Suppose, in addition, for some fixed, but arbitrary 30 5 R n , t 0 > 0, we have x uÝx, 0 Þ = / t uÝx, 0 Þ = 0 for xx | 3 ? 30 | ² t 0 . Then uÝ3.t Þ = 0 for all Ý3, t Þ in the light cone x x x xx Ý3, t Þ : 0 ² t ² t 0 , | 3 ? 30 | ² t 0 ? t . C= Remark: Let UÝt Þ denote the n-ball of radius t 0 ? t with center at x 0 ; i.e., UÝt Þ = 3 : | 3 ? 30 | ² t 0 ? t x xx Ð Rn. Then the proposition asserts that the value of uÝ30 , t 0 Þ depends only on the values of the x data inside UÝ0 Þ, and if the data is zero in this ball, then the solution is zero everywhere in the cone that is obtained by drawing all the backward characteristic lines through Ý30 , t 0 Þ. x 30 , t 0 Þ. This cone formed in this way is called the light cone at Ýx Proof- Let E Ýt Þ = X UÝt Þ / t uÝ3, t Þ 2 + | 4uÝ3, t Þ| 2 dx x x for 0 ² t ² t 0 Then .E v ÝtÞ = X UÝt Þ 2ß/ t uÝ3, t Þ / tt uÝ3, t Þ + 4uÝ3, t Þ 6 4/ t uÝ3, t Þà dx x x x x ?X 2 /UÝt Þ / t uÝ3, t Þ + | 4uÝ3, t Þ| 2 dSÝx Þ. x x Here we made use of the fact that dX fÝx Þ dx dr B r Ýx 0 Þ r = d XX fÝr, g Þ dg dr dr 0 /B r Ýx 0 Þ =X /B r Ýx 0 Þ fÝr, g Þ dg 13 x and B r Ý30 Þ = UÝt Þ = B t 0 ?t Ýx 0 Þ. Now XUÝt Þ 4uÝ3, t Þ 6 4/ t uÝ3, t Þ dx = X/UÝt Þ / t uÝ3, t Þ 4uÝ3, t Þ 6 3 dS ? XUÝt Þ 4 2 uÝ3, t Þ / t uÝ3, t Þdx x x x x n x x hence E v Ýt Þ = X UÝt Þ 2/ t uÝ3, t Þß / tt uÝ3, t Þ ? 4 2 uÝ3, t Þà dx x x x +X 2 /UÝt Þ 2/ t u / N u ? Ý/ t uÝ3, t Þ + | 4uÝ3, t Þ| 2 Þ dSÝx Þ. x x Notice that n x2 x | 2/ t u / N u | = 2 | / t u 4u 6 3 | ² 2 |/ t u | |4u | ² / t uÝ3, t Þ + | 4uÝ3, t Þ| 2 from which it follows that E v ÝtÞ ² 0. Since EÝ0Þ = 0, we conclude EÝt Þ is identically zero which implies that the solution is constant inside the light cone. Since the solution is zero on the ”floor” of the light cone, the solution must be zero throughout the light cone.n This is referred to as the ”causality” principle because it asserts that the effects from a cause cannot be felt at a point before the effects have had time to reach the point, travelling from the source to the point with the finite speed dictated by the governing partial differential equation. Another way to say this is that the value of the solution to the wave equation at a specified point Ý30 , t 0 Þ is not influenced by data values that lie outside the x domain of dependence for the point Ý30 , t 0 Þ. This domain of dependence is just the floor of x the light cone obtained by drawing all backward characteristics through Ý30 , t 0 Þ. x 10. Wave-Like Evolution We use the term diffusion-like evolution to describe the properties of the solutions to the heat equation. These properties include ¾ 1. 2. 3. 4. 5. instantaneous smoothing of the data infinite speed of propagation time irreversible existence of a Max-min principle monotone (non-oscillatory) evolution The wave equation also described the time evolution of a system but the properties of solutions to the wave equation are rather different from those of solutions to the heat equation. Solutions of the wave equation exhibit the following properties ¾ 1. 2. 3. no smoothing of the data finite speed of propagation time reversible 14 no Max-min principle oscillatory time behavior 4. 5. Points 1 and 2 are clearly illustrated by the D’Alembert solution (4.2) for the Cauchy problem (4.1). Consider the special case where g = 0 and f is given by f Ýx Þ = 1 ? | x| 0 if | x | ² 1 if | x| > 1 . Then f is a continuous but not differentiable triangular pulse centered at x = 0. In this case the solution uÝx, tÞ has exactly the same smoothness as the data since, uÝx, tÞ = ßfÝx + at Þ + fÝx ? at Þà/2. In addition, at positive integer values of t, t = N, the solution is seen to be composed of two triangular pulses, one centered at x = Na, and one centered at x = ?Na, and each of the pulses has half the amplitude of the original pulse. Evidently, the pulse at x = 0 splits into two halves and each half pulse has travelled a distance equal to Na in time t = N, which is to say the pulses travel with finite speed equal to a. Points 3,4, and 5 are illustrated by the solution for the following IBVP on a bounded interval. / tt uÝx, t Þ = a 2 / xx uÝx, t Þ, uÝx, 0 Þ = fÝx Þ, / t uÝx, 0 Þ = 0, uÝ0, t Þ = uÝL, t Þ = 0, 0 < x < L, 0 < t, 0 < x < L, 0 < t. It will be shown later that the solution for this problem can be written K uÝx, t Þ = > n=1 f n cosÝn^at/L Þ sinÝn^x/L Þ where f n denote the Fourier coefficients for fÝx Þ; i.e., L f n = 2 X fÝx Þ sinÝn^x/L Þ dx, n = 1, 2, ... L0 It is known from the theory of Fourier series that the series K > n=1 f n sinÝn^x/L Þ converges (in some sense) to *Ýx Þ, the odd, L-periodic extension of fÝx Þ. Then, since f cosÝn^at/L Þ sinÝn^x/L Þ = 1 2 ßsinÝn^Ýx + at Þ/L Þ + sinÝn^Ýx ? at Þ/L Þà, we see that 15 K uÝx, t Þ = 1 > n=1 f n ßsinÝn^Ýx + at Þ/L Þ + sinÝn^Ýx ? at Þ/L Þà 2 f = 1 *Ýx + at Þ + *Ýx ? at Þ . f 2 Clearly this solution is oscillatory in time and, since positive time is interchangeable with negative time, the solution is time reversible. The energy, EÝt Þ, has been shown to be constant in time so the solution does not die out with increasing time like the solution to the heat equation. Finally, if fÝx Þ is a triangular pulse of height one, centered at x = L/2, then plotting uÝx, t Þ versus x for various times shows that for t = L/a, we see uÝx, L/a Þ is a triangular pulse centered at x = L/2, but with height equal to minus one. Clearly there can be no max-min principle for this equation. The behavior exhibited by this solution of the IBVP problem in 1-dimension is typical of solutions to the wave equation in other settings. Collectively we refer to this behavior as wave-like evolution, as opposed to diffusion-like evolution, which is the behavior exhibited by the solutions to the heat equation. The heat equation and wave equation are prototypes for the classes of parabolic and hyperbolic partial differential equations, respectively. The Laplace equation is the prototype for the class of elliptic partial differential equations. We will now explain the meaning of this terminology. 11. Classification The Laplace, heat and wave operators in n-dimensions can be written in the following way / /x x 4 2 uÝ3Þ = 3f I n 3uÝ3Þ where 3f = ß/ 1 , ..., / n à / x x x / /x / t uÝ3, t Þ ? 4 2 uÝ3, t Þ = / t uÝ3, t Þ ? 3f I n 3uÝ3, t Þ 23 x x x / tt uÝ3, t Þ ? 4 uÝx, t Þ = / tt uÝ3, t Þ ? / f I n /uÝ3, t Þ where I n denotes the n by n identity matrix. If we do not distinguish between the space and time variables but think of these as operators in n variables (thus the time variable becomes x n in the heat and wave operators) the second order terms in each of these operators are all /x of the form 3f J n 3uÝ3Þ where the n by n matrix J n is as follows, / Laplace J n = I n , heat J n = I n?1 0 0 0 , wave J n = I n?1 0 0 ?1 Now consider the operator defined by an arbitrary symmetric n by n matrix, A. x LßuÝ3Þà= / f A n /uÝ3Þ x Recall that since A is symmetric, it has n real (but not necessarily distinct) eigenvalues. Then we say that the operator L is: 16 elliptic if the eigenvalues of A are all of one sign parabolic if zero is an eigenvalue for A of multiplicity one, and the remaining n ? 1 eigenvalues have one sign hyperbolic if A has n ? 1 eigenvalues of one sign and one eigenvalue with the opposite sign Note that when n = 2 this classification is exhaustive in the sense that every matrix A falls into one of these classes. If n is greater than 2 then there are matrices which are in none of the 3 classes and, consequently, there are partial differential operators that are neither elliptic, parabolic nor hyperbolic. On the other hand, many of the partial differential equations that occur in applications are one of these three types. The reason for these names is that when n = 2, the locus of points satisfying 3f Ax = C, is x3 an ellipse, parabola or hyperbola, respectively, when the matrix A has the properties assigned to each of the names. 17 ...
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This note was uploaded on 08/30/2011 for the course ECE 3041 taught by Professor Brewer during the Spring '08 term at Georgia Institute of Technology.

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