Math 342 Partial Differential Equations «Viktor Grigoryan
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1 Math 342 Partial Differential Equations «Viktor Grigoryan 6 Wave equation: solution In this lecture we will solve the wave equation on the entire real line x R. This correspons to a string of infinite length. Although physically unrealistic, as we will see later, when consiering the ynamics of a portion of the string away from the enpoints, the bounary conitions have no effect for some (nonzero) finite time. Due to this, one can neglect the enpoints, an make the assumption that the string is infinite. The wave equation, which escribes the ynamics of the amplitue u(x, t) of the point at position x on the string at time t, has the following form u tt = c 2 u xx, or u tt c 2 u xx =. (1) As we saw in the last lecture, the wave equation has the secon canonical form for hyperbolic equations. One can then rewrite this equation in the first canonical form, which is u ξη =. (2) This is achive by passing to the characteristic variables ξ = x + ct, η = x ct. (3) To see that (2) is equivalent to (1), let us compute the partial erivatives of u with respect to x an t in the new variables using the chain rule. u t = cu ξ cu η, u x = u ξ + u η. We can ifferentiate the above first orer partial erivatives with respect to t, respectively x using the chain rule again, to get u tt = c 2 u ξξ 2c 2 u ξη + c 2 u ηη, u xx = u ξξ + 2u ξη + u ηη. Substituting these expressions into the left han sie of equation (1), we see that u tt c 2 u xx = c 2 u ξξ 2c 2 u ξη + c 2 u ηη c 2 (u ξξ + 2u ξη + u ηη ) = 4c 2 u ξη =, which is equivalent to (2). Equation (2) can be treate as a pair of successive ODEs. Integrating first with respect to the variable η, an then with respect to ξ, we arrive at the solution u(ξ, η) = f(ξ) + g(η). Recalling the efinition of the characteristic variables (3), we can switch back to the original variables (x, t), an obtain the general solution u(x, t) = f(x + ct) + g(x ct). (4) Another way to solve equation (1) is to realize that the secon orer linear operator of the wave equation factors into two first orer operators L = 2 t c 2 2 x = ( t c x )( t + c x ). Hence, the wave equation can be thought of as a pair of transport (avection) equations ( t c x )v =, (5) ( t + c x )u = v. (6) 1
2 It is no coincience, of course, that x + ct = constant, (7) an x ct = constant, (8) are the characteristic lines for the transport equations (5), an (6) respectively, hence our choice of the characteristic coorinates (3). We also see that for each point in the xt plane there are two istinct characteristic lines, each belonging to one of the two families (7) an (8), that pass through the point. This is illustrate in Figure 1 below x 6.1 Initial value problem Figure 1: Characteristic lines for the wave equation with c =.6. Along with the wave equation (1), we next consier some initial conitions, to single out a particular physical solution from the general solution (4). The equation is of secon orer in time t, so values must be specifie both for the initial ispalcement u(x, ), an the initial velocity u t (x, ). We stuy the following initial value problem, or IVP utt c 2 u xx = for < x < +, (9) u(x, ) = φ(x), u t (x, ) = ψ(x), where φ an ψ are arbitrary functions of single variable, an together are calle the initial ata of the IVP. The solution to this problem is easily foun from the general solution (4). All we nee to o is fin f an g from the initial conitions of the IVP (9). To check the first initial conition, set t = in (4), u(x, ) = φ(x) = f(x) + g(x). (1) To check the secon initial conition, we ifferentiate (4) with respect to t, an set t = u t (x, ) = ψ(x) = cf (x) cg (x). (11) Equations (1) an (11) can be treate as a system of two equations for f an g. To solve this system, we first integrate both sies of (11) from to x to get ri of the erivatives on f an g (alternatively we coul ifferentiate (1) instea), an rewrite the equations as f(x) + g(x) = φ(x), f(x) g(x) = 1 c ˆ x ψ(s) s + f() g(). We can solve this system by aing the equations to eliminate g, sn subtracting them to eliminate f. This leas to the solution f(x) = 1 2 φ(x) + 1 2c g(x) = 1 2 φ(x) 1 2c ˆ x ˆ x ψ(s) s + 1 [f() g()], 2 ψ(s) s 1 [f() g()]. 2 2
3 Finally, substituting these expressions for f an g back into the solution (4), we obtain the solution of the IVP (9) u(x, t) = 1 2 φ(x + ct) + 1 2c ψ(s) s φ(x ct) 1 2c ˆ x ct ψ(s) s. Combining the integrals using aitivity, an the fact that flipping the integration limits changes the sign of the integral, we arrive at the following form for the solution u(x, t) = 1 2 [φ(x + ct) + φ(x ct)] + 1 2c x ct ψ(s) s. (12) This is Alambert s solution for the IVP (9). It implies that as long as φ is twice continuously ifferentiable (φ C 2 ), an ψ is continuously ifferentiable (ψ C 1 ), (12) gives a solution to the IVP. We will also consier examples with iscontinuous initial ata, which after plugging into (12) prouce weak solutions. This notion will be mae precise in later lectures. Example 6.1. Solve the initial value problem (9) with the initial ata φ(x), ψ(x) = sin x. Substituting φ an ψ into Alambert s formula, we obtain the solution u(x, t) = 1 2c x ct sin s s = 1 ( cos(x + ct) + cos(x ct)). 2c Using the trigonometric ientities for the cosine of a sum an ifference of two angles, we can simplify the above to get u(x, t) = 1 sin x sin(ct). c You shoul verify that this inee solves the wave equation an satisfies the given initial conitions. 6.2 The Box wave When solving the transport equation, we saw that the initial values simply travel to the right, when the spee is positive (they propagate along the characteristics (8)); or to the left (they propagate along the characteristics (7)), when the spee is negative. Since the wave equation is mae up of two of these type of equations, we expect similar behavior for the solutions of the IVP (9) as well. To see this, let us consier the following example with simplifie initial ata. Example 6.2. Fin the solution of IVP (9), with the initial ata φ(x) = ψ(x). h, x a,, x > a, (13) This ata correspons to an initial isturbance of the string centere at x = of height h, an zero initial velocity. Notice that φ(x) is not continuous, let alone twice ifferentiable, though one can still substitute it into Alambert s solution (12) an get a function u, which will be a weak solution of the wave equation. Since ψ(x), only the first term in (12) is nonzero. We compute this term using the particular φ(x) in (13). First notice that h, x + ct a, φ(x + ct) = (14), x + ct > a, an h, x ct a, φ(x ct) =, x ct > a. 3 (15)
4 Hence, the solution u(x, t) = 1 [φ(x + ct) + φ(x ct)] 2 is piecewise efine in 4 ifferent regions in the xt half-plane (we consier only positive time t ), which correspon to pairings of the intervals in the expressions (14) an (15). These regions are I : x + ct a, x ct a}, u(x, t) = h II : x + ct a, x ct > a}, u(x, t) = h 2 (16) III : x + ct > a, x ct a}, u(x, t) = h 2 IV : x + ct > a, x ct > a}, u(x, t) = The regions are epicte in Figure 2. Notice that x + ct a is equivalent to Similarly for the other inequalities. a x + ct a, or 1 c (x + a) t 1 (x a). c IV IV II I III IV a a Figure 2: Regions where u has ifferent values. Using the values for the solution in (16), we can raw the graph of u at ifferent times, some of which are epicte in Figures 3-6. We see that the initial box-like isturbance centere at x = splits into two isturbances of half the size, which travel in opposite irections with spee c. The graphs hint that the initial isturbance will not be felt at a point x on the string (for x > a) before the time t = 1 x a. We will shortly see that this is a general property for the wave equation. In this particular c case a box-like isturbance appears at the time t = 1 x a, an lasts exactly t = 2a units of time, after which c c it completely moves along. In general, the initial velocity may slow own the spee, an subsequently make the isturbance last longer, however, the spee cannot excee c. t t a c h a a Figure 3: The solution at t =. Figure 4: The solution for < t < a/c. 4
5 t a c t a c h 2 2a 2a Figure 5: The solution at t = a/c. Figure 6: The solution for t > a/c. 6.3 Causality The value of the solution to the IVP (9) at a point (x, t ) can be foun from Alambert s formula (12) u(x, t ) = 1 2 [φ(x + ct ) + φ(x ct )] + 1 2c ˆ x +ct x ct ψ(s) s. (17) We can see that this value epens on the values of φ at only two points on the x axis, x + ct, an x ct, an the values of ψ only on the interval [x ct, x + ct ]. For this reason the interval [x ct, x + ct ] is calle interval of epenence for the point (x, t ). The triangular region with vertices at x ct an x + ct on the x axis an the vertex (x, t ) is calle the omain of epenence, or the past history of the point (x, t ), as epicte in Figure 7. The sies of this triangle are segments of characteristic lines passing through the point (x, t ). Thus, we see that the initial ata travels along the characteristics to give the values at later times. In the previous example of the box wave, we saw that the jump iscontinuities travel along the characteristic lines as well. omain of influence x, t past history x c t, t t x c t, t x c t x c t x Figure 7: Domain of epenence for (x, t ). Figure 8: Domain of influence of x. An inverse notion to the omain of epenence is the notion of omain of influence of the point x on the x axis. This is the region in the xt plane consisting of all the points, whose omain of epenence contains the point x. The region has an upsie-own triangular shape, with the sies being the characteristic lines emanating from the point x, as shown in Figure 8. This also means that the value of the initial ata at the point x effects the values of the solution u at all the points in the omain of influence. Notice that at a fixe time t, only the points satisfying x ct x x + ct are influence by the point x on the x axis. Example 6.3 (The hammer blow). Analyze the solution to the IVP (9) with the following initial ata φ(x), h, x a, ψ(x) =, x > a. (18) From Alambert s formula (12), we obtain the following solution u(x, t) = 1 2c 5 x ct ψ(s) s. (19)
6 Similar to the previous example of the box wave, there are several regions in the xt plane, in which u has ifferent values. Inee, since the initial velocity ψ(x) is nonzero only in the interval [ a, a], the integral in (19) must be compute ifferently accoring to how the intervals [ a, a] an [x ct, x + ct] intersect. This correspons to the cases when ψ is zero on the entire integration interval in (19), on a part of it, or is nonzero on the entire integration interval. These ifferent cases are: I : x ct < x + ct < a < a}, u(x, t) = II : x ct < a < x + ct < a}, u(t, x) = 1 2c a ˆ a h s = h x + ct + a 2c III : x ct < a < a < x + ct}, u(t, x) = 1 2c IV : a < x ct < x + ct < a}, u(t, x) = 1 2c V : a < x ct < a < x + ct}, u(t, x) = 1 2c VI : a < a < x ct < x + ct}, u(x, t) = a h s = h a c x ct ˆ a x ct h s = ht a (x ct) h s = h 2c (2) The regions are epicte in Figure 9 below. Notice that to fin the value of the solution at a point (x, t ), one simply nees to trace the point back to the x axis along the characteristic lines, an etermine how the interval of epenence intersects the segment [ a, a]. x, t III II V I IV x c t a a x c t VI Figure 9: Regions where u has ifferent values. 6.4 Conclusion The wave equation, being a constant coefficient secon orer PDE of hyperbolic type, posseses two families of characteristic lines, which correspon to constant values of respective characteristic variables. Using these variables the equation can be treate as a pair of successive ODEs, integrating which leas to the general solution. This general solution was use to arrive at Alambert s solution (12) for the IVP on the whole line. Unfortunatelly this simple erivation relies on having two families of characteristics an oes not work for the heat an Laplace s equations. Exploring a few examples of initial ata, we establishe causality between the initial ata an the values of the solution at later times. In particular, we saw that the initial values travel with spees boune by the wave spee c, an that the iscountinuities of the initial ata travel along the characteristic lines. 6
7 7 The energy metho 7.1 Energy for the wave equation Let us consier an infinite string with constant linear ensity ρ an tension magnitue T. The wave equation escribing the vibrations of the string is then ρu tt = Tu xx, < x <. (21) Since this equation escribes the mechanical motion of a vibrating string, we can compute the kinetic energy associate with the motion of the string. Recall that the kinetic energy is 1 2 mv2. In this case the string is infinite, an the spee iffers for ifferent points on the string. However, we can still compute the energy of small pieces of the string, a them together, an pass to a limit in which the lengths of the pieces go to zero. This will result in the following integral KE = 1 2 ρu 2 t x. We will assume that the initial ata vanishes outsie of a large interval x R, so that the above integral is convergent ue to the finite spee of propagation. We woul like to see if the kinetic energy KE is conserve in time. For this, we ifferentiate the above integral with respect to time to see whether it is zero, as is expecte for a constant function, or whether it is ifferent from zero. t KE = 1 2 ρ 2u t u tt x = ρu t u tt x. Using the wave equation (21), we can replace the ρu tt by Tu xx, obtaining t KE = T u t u xx x. The last quantity oes not seem to be zero in general, thus the next best thing we can hope for, is to convert the last integral into a full erivative in time. In that case the ifference of the kinetic energy an some other quantity will be conserve. To see this, we perform an integration by parts in the last integral t KE = Tu tu x Tu xt u x x. Due to the finite spee of propagation, the enpoint terms vanish. The last integral is a full erivative, thus we have ˆ t KE = Tu xt u x x = ( ˆ 1 ) Tu 2 x x. t 2 Defining PE = 1 2 T u 2 x x, we see that t KE = t PE, or (KE + PE) =. t The quantity E = KE + PE is then conserve, which is the total energy of the string unergoing vibrations. Notice that PE plays the role of the potential energy of a stretche string, an the conservation of energy implies conversion of the kinetic energy into the potential energy an back without a loss. Another way to see that the energy E = 1 2 (ρu 2 t + Tu 2 x) x (22) 7
8 is conserve, is to multiply equation (21) by u t an integrate with respect to x over the real line. = ρu tt u t x Tu xx u t x. The first integral above is a full erivative in time. Integrating by parts in the secon term, an realizing that the subsequent integral is a full erivative as well, while the bounary terms vanish, we obtain the ientity ( ˆ 1 ) ρu 2 t + Tu 2 x x =, t 2 which is exactly the conservation of total energy. The conservation of energy provies a straightforwar way of showing that the solution to an IVP associate with the linear equation is unique. We emonstrate this for the wave equation next, while a similar proceure will be applie to establish uniqueness of solutions for the heat IVP in the next section. Example 7.1. Show that the initial value problem utt c 2 u xx = f(x, t) for < x < +, u(x, ) = φ(x), u t (x, ) = ψ(x), (23) has a unique solution. Arguing from the inverse, let as assume that the IVP (23) has two istinct solutions, u an v. But then their ifference w = u v will solve the homogeneous wave equation, an will have the initial ata w(x, ) = u(x, ) v(x, ) = φ(x) φ(x), w t (x, ) = u t (x, ) v t (x, ) = ψ(x) ψ(x). Hence the energy associate with the solution w at time t = is E[w]() = 1 2 [(w t (x, )) 2 + c 2 (w x (x, )) 2 ] x = This iffers from the energy efine above by a constant factor of 1/ρ (recall that T/ρ = c 2 ), an is thus still a conserve quantity. It will subsequently be zero at any later time as well. Thus, E[w](t) = 1 2 [(w t (x, t)) 2 + c 2 (w x (x, t)) 2 ] x =, t. But since the integran in the expression of the energy is nonnegative, the only way the integral can be zero, is if the integran is uniformly zero. That is, w(t, x) = (w t (x, t), w x (x, t)) =, x, t. This implies that w is constant for all values of x an t, but since w(x, ), the constant value must be zero. Thus, u(x, t) v(x, t) = w(x, t), which is in contraiction with our initial assumption that u an v are ifferent. This implies that the solution to the IVP (23) is unique. The proceure use in the last example, calle the energy metho, is quite general, an works for other linear evolution equations possessing a conserve (or ecaying) positive efinite energy. The heat equation, consiere next, is one such case. 8
9 7.2 Energy for the heat equation We next consier the (inhomogeneous) heat equation with some auxiliary conitions, an use the energy metho to show that the solution satisfying those conitions must be unique. Consier the following mixe initial-bounary value problem, which is calle the Dirichlet problem for the heat equation ut ku xx = f(x, t) for x l, t > u(x, ) = φ(x), (24) u(, t) = g(t), u(l, t) = h(t), for given functions f, φ, g, h. Example 7.2. Show that there is at most one solution to the Dirichlet problem (24). Just as in the case of the wave equation, we argue from the inverse by assuming that there are two functions, u, an v, that both solve the inhomogeneous heat equation an satisfy the initial an Dirichlet bounary conitions of (24). Then their ifference, w = u v, satisfies the homogeneous heat equation with zero initial-bounary conitions, i.e. wt kw xx = for x l, t > w(x, ) =, (25) u(, t) =, u(l, t) =, Now efine the following energy E[w](t) = 1 2 ˆ l [w(x, t)] 2 x, (26) which is always positive, an ecreasing, if w solves the heat equation. Inee, ifferentiating the energy with respect to time, an using the heat equation we get t E = ˆ l ww t x = k Integrating by parts in the last integral gives t E = kww x l ˆ l ˆ l ww xx x. w 2 x x, since the bounary terms vanish ue to the bounary conitions in (25), an the integran in the last term is nonnegative. Due to the initial conition in (25), the energy at time t = is zero. But then using the fact that the energy is a nonnegative ecreasing quantity, we get E[w](t) E[w]() =. Hence, ˆ 1 l [w(x, t)] 2 x =, for all t, 2 which implies that the nonnegative continuous integran must be ientically zero over the integration interval, i.e w, for all x [, l], t >. Hence also u 1 u 2, which finishes the proof of uniqueness. The energy (26) arises if one multiples the heat equation by w an integrates in x over the interval [, l]. Then realizing that the first term will be the time erivative of the energy, an performing the same integration by parts on the secon term as above, we can reprove that this energy is ecreasing. Notice that all of the above arguments hol for the case of the infinite interval < x < as well. In this case one ignores the effect of the infinitely far enpoints an consiers an IVP. 9
10 7.3 Conclusion Using the energy motivate by the vibrating string moel behin the wave equation, we erive a conserve quantity, which correspons to the total energy of motion for the infinite string. This positive efinite quantity was then use to prove uniqueness of solution to the wave IVP via the energy metho, which essentially asserts that zero initial total energy preclues any (nonzero) ynamics. A similar approach was use to prove uniqueness for the heat IBVP, concluing that zero initial heat implies steay zero temperatures at later times. 1
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