The wave equation and energy conservation

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1 The wave equation an energy conservation Peter Haggstrom May 1, 17 1 Problem 1, Chapter 3 of Fourier Analysis: An Introuction by Elias Stein an Rami Shakarchi Problem 1 in Chapter 3, page 9, of Elias Stein an Rami Shakarchi s textbook Fourier Series: An Introuction [1] contains the following problem which exhibits some funamental techniques for physics an analysis stuents. Here is the problem. Consier a vibrating string whose isplacement u(x, t) at time t satisfies the wave equation: 1 u c u, c τ ρ (1) ρ is the constant ensity an τ is the coefficient of tension of the string. The string is subject to the initial conitions: u(x, ) f(x) an (x, ) g(x) () where it is assume that f C 1 an g is continuous. The total energy for the string is efine as: E(t) ρ L () τ x + L () x (3) 1

2 (there is a typo in the textbook where T appears instea of τ). The length of the string is L. The first term in (3) correspons to the kinetic energy of the string (in analogy with 1 mv, the kinetic energy of a particle of mass m an velocity v), an the secon term correspons to the potential energy. Show that the total energy of the string is conserve, in the sense that E(t) is constant. Therefore: E(t) E() ρ L g(x) x + τ L f (x) x (4) Detaile solution The crux of this problem is the efinition of total energy in (3) which is merely given, not erive. The kinetic energy component is simply the ifferential velocitymass integrate over the length of the string. If the string is isplace a small amount so that is small then the ifferential arc length s 1 + ( x ) (this is just Pythagoras theorem s x + u ). Because approximate s 1 + ( so that the kinetic energy is ρ L is small we can x ) by x an so the mass is simply ρ x an velocity ( ) x. The potential energy is a bit trickier. If we assume that the potential energy is proportional to a ifferential increase in the length of the string compare with its rest length of L we can get an expression for the potential energy as follows. The factor of proportionality is the tension τ. The change in length is (using Taylor s theorem): L 1 + ( ) x L L ( () ) x L 1 L () x (5) Accoringly the potential energy is τ L ( ) x. There is a more inirect way of arriving at the expression for total energy E(t). Because this involves ifferentiating uner the integral sign, a general theorem is as follows (see [] pages 68-9):

3 .1 Theorem Let g : R R R be a continuous function such that g(x,t) exists an is continuous. Suppose g(x, t) x an g(x,t) x an exist for each t an that g(x,t) x as R uniformly in t on every interval [a, b]. Then x >R t g(x, t) x is ifferentiable with: g(x, t) x g(x,t) x Let us leave for the moment whether we can actually o the ifferentiation insie the integral (I will come to the subtleties of that shortly) an simply formally ifferentiate (4) as follows: E (t) ρ L ρ L u x+τ u [ x+τ L L L u x ρ ρ c L u x ρ L L u x τ L u x+τ u x τ L L u x (6) Note that in equating u x with u x we are implicitly assuming continuity of the secon erivative. In the step involving integration by parts we are assuming initial conitions on the erivatives ie (, t) (L, t) for all t. In other wors the ens are fixe so that there is no movement an hence no velocity. Equation (6) shows that E(t) is a constant so that E(t) E() ρ L g(x) x + τ L f (x) x where () has been use. We can erive equation (3) in a more general context by starting with the kinetic energy ie: u x u x KE ρ () x (7) To get convergence of the integral we have to assume that the integran vanishes outsie of some large interval x R. To see whether the KE is conserve in time we ifferentiate with respect to time an we get: 3

4 t KE ρ u x ρ u x ρ u c x τ The final integral in (8) oes not look like it woul be zero but if we integrate by parts we come up with something useful: Now τ u x τ (8) u x (9) shoul vanish because the spee of propogation is finite an the erivatives ought to approach zero in the limit. So we are left with: u x t KE u x t( 1 τ ( ) ) x (1) If we efine the potential energy to be: P E 1 τ ( ) x (11) Then what we have is this: t KE t P E or (KE + P E) (1) t So (1) says that the total energy efne by the sum of (7) an (11) ( which is just (3) ) is conserve. Now let s go right back to basics. If we start with the classic plucke string set up (see [3] pages 39-4) we get the following situation. The string is stretche between (, ) an (, ) an the mi-point (1, ) is raise to a height h above the x axis. Thus the initial position of the string is efine by: f(x) { hx when x 1 hx + h when 1 x (13) The Fourier sine series of (13) has coefficients b n 8h sin nπ. The formal solution π n to the wave equation (obtaine by separation of variables) u a u for < x < c an t > is (see [3] pages 39-4) : 4

5 u(x, t) Thus we get the formal solution: n1 b n sin nπx c cos nπat c (14) u(x, t) 8h π n1 1 n sin nπ nπx sin cosnπat (15) which can be shown to be represente as: u(x, t) 8h π m1 ( 1) m+1 (m 1)πx sin (m 1) cos (m 1)πat (16) Neither (15) or (16) is actually twice ifferentiable with repect to x or t. To see this look at (15) for instance an note that the amping factor 8h 1 sin nπ will π n be neutralise by two ifferentiations of either the sin nπx factor or the cos nπat factor so you will get a pure non-ampe oscillatory infinite sum ie something that will not converge. More etaile inforrmation concerning ifferentiating uner the integral sign can be foun in [5]. Clearly, then, further conitions are neee to ensure existence an uniqueness of a solution to the wave equation. Churchill provies the proof in Chapter 1 of [3]. He generalizes the problem as follows: u(x, t) a u(x, t) + φ(x, t), < x < c, t > (17) u(, t) p(t), u(c, t) q(t) t (18) u(x, ) f(x), (x, ) g(x) x c (19) It is now assume that u is of class C in the region efine by x c, t. Thus u an its first an secon erivatives incluing the mixe secon partial erivatives are assume to be continuous in this region R. Now suppose there are two C solutions u 1 (x, t) an u (x, t) in the region. Hence the ifference z u 1 u is of class C in the region an it will satisfy the homogeneous problem: 5

6 z(x, t) a z(x, t), < x < c, t > () z(, t), z(c, t), t (1) z(x, ), z (x, ), x c () We have to show that z throughout the region so that u 1 u. Churchill states ( [3], page 4) that the integran of the integral: I(t) 1 c satisfies conitions such that we can write: I (t) c (z x + 1 a z t ) x (3) (z x z xt + 1 a z t z tt ) x. (4) Equation ( (4) ought to look familiar. When c is substitue for a, I(t) 1 ) c ( τ τ z x + ρ z t ) x which is 1 E(t) (taking z(x, t) u(x, t) an c L). τ Now because z tt a z xx, the integran in (4) becomes: Hence I (t) c using (). z x z xt + 1 a z t z tt z x z xt + z t z tt (z x z t ) (5) [ ] xc (z x z t ) x z x (x, t) z t (x, t) z x (c, t) z t (c, t) z x (, t) z t (, t) x (6) Equation (6) shows that I(t) is constant. Now z(x, ) z x (x, ) an z t (x, ) so I() an hence I(t). We have a continuous non-negative integran which is zero at the en-points, hence it must vanish. This is just a special case of a stanar result in the calculus of variations (see [4] page 9). We can conclue therefore that z x (x, t) z t (x, t) for x c an t. Thus z must be constant ie z(x, t) because z(x, ). 6

7 What this shows that the original bounary value problem set out in (17)-(19) cannot have more than one C solution in R. Churchill makes the comment ( [4]. p.41) that the requirement of continuity of erivatives of u is severe. Solutions of many simple problems in the wave equation have iscontinuities in their erivatives. Churchill shows ( [3[, pages ) how to resolve the problem of the nonconvergence of the ifferentiate series commente upon above. 3 Differentiating uner the integral sign As can be seen from the Theorem, we nee continuity of g (x, t) where g(x, t) is ( ) ( an ) (see (3) ). From physical consierations the energy is finite so that we shoul have existence of L g(x, t) x. As for the integral g (x, t) x, it has to exist for each t. In our case there is a finite spatial interval of integration [, L] so that given the continuity of the erivatives on this interval there ought to be no blow ups so we can assume this conition is satisfie for all t. The final conition in the theorem requires g (x, t) x as R an x >R since we are ealing with a finite spatial interval this shoul also be satisfie. Hence, ifferentiating uner the integral sign is kosher - certainly goo enough to reprouce the erivation for an exam! However, as Churchill pointe out above, it is not at all improbable to see iscontinuities in the erivatives in even simple wave equation problems so that the analysis becomes much more complex. 4 Energy conservation for the -imensional wave equation The analysis given above for energy conservation of the wave equation in 1 imension is essentially a bit of a foothol on a much longer an aruous trip to the summit of Mount Everest. In general one woul want to establish energy conservation in imensions an this is a much more ifficult process than for one imension. Textbooks on partial ifferential equations (PDEs) usually approach the -imensional proof in the context of so-calle energy conservation methos. There is a variety of approaches some of which omit vast tracts of calculus. In my view one of the best approaches is that of Elias Stein an Rami Shakarchi [1, Chapter 6 ] because they start off with the 1- imensional case an then gently show that the solution to a certain wave equation problem (the Cauchy problem) gives rise to energy conservation. Then they leave as an exercise (with hints) the 7

8 general case. In what follows I retrace what Stein an Shakarchi have one as a necessarily preliminary an then give the proof of the general case in terms of the problem they pose (see Problem 3, [ 1], pages ) The wave equation in imensions is: u + + u 1 This equation can be rewritten using the Laplacian operator: where c u (7) u 1 c u (8). In what follows we take c 1. The Cauchy problem is to fin a solution to (8 ) subject to these initial conitions: } u(x, ) f(x) (x,) g(x) (9) Both f, g S( ) i.e. they inhabit Schwartz space. Recall that Schwartz space consists of all inefinitely ifferentiable functions on such that: ( ) β sup x α f(x) < (3) x for all inices α, β. Hence f, an all its erivatives are rapily ecreasing. See [5.1] for more on Schwartz space. Given that ultimately we want to prove energy conservation in imensions we nee pretty strong assumptions on the nature of the unerlying functions so that we can ifferentiate uner the integral sign an that the integrals at issue actually converge. 5 The Cauchy Problem Stein an Shakarchi start with the Cauchy problem as follows ( [1], pages ). Suppose u solves Cauchy problem escribe by (8)-(9). We take Fourier transforms of (8) in the space variable as follows by first recalling that: 8

9 ( (x, t) ) F πiξ k û(ξ, t) (31) k û(ξ, t) is the Fourier transform of u(x, t) with respect to the space variable. Hence: Hence: ( u(x, t) ) F (πiξ k ) û(ξ, t) 4πξk û(ξ, t) (3) k ( ) F u(x, t) 4π ξ û(ξ, t) (33) The Fourier transform of the RHS of (8) (with c1) is: To see this note that: ( u(x, t) ) F û(ξ, t) (34) F(u(x, t)) u(x, t) e πix ξ x where ξ (35) (x, t) F(u(x, t)) e πix ξ x where ξ (36) R Note that x ξ i1 x i ξ i. Then (8) becomes the following Fourier-transforme equation: 4π ξ û(ξ, t) û(ξ, t) (37) For each ξ, (37) is a manageable orinary ifferential equation whose solution is: û(ξ, t) A(ξ) cos(π ξ t) + B(ξ) sin(π ξ t) (38) 9

10 Now we have to fin A(ξ) an B(ξ) for each ξ an to o this we have to use the initial conitions. We start by taking the Fourier transforms of the initial conitions in (9): F(u(x, )) F(f(x)) ie û(ξ, ) ˆf(ξ) A(ξ) (39) (x, ) F( ) F(g(x)) û(ξ, ) ie ĝ(ξ) (4) Differentiating (38) we thus have from (4) that: ĝ(ξ) π ξ B(ξ) B(ξ) ĝ(ξ) π ξ (41) Thus we have: û(ξ, t) ˆf(ξ) cos(π ξ t) + ĝ(ξ) π ξ Our solution to the Cauchy problem is this: u(x, t) sin(π ξ t) (4) [ ˆ f(ξ) cos(π ξ t) + ĝ(ξ) π ξ sin(π ξ t) ] e πix ξ ξ (43) Because f, g S( ) we can ifferentiate uner the integral sign an so u is at least C. In fact since both f, g S( ) their Fourier transforms also inhabit Schwartz space ( see [1] page 184 ). So performing the spatial erivative we have for each j: (x, t) j [ f(ξ) ˆ cos(π ξ t) + ĝ(ξ) ] j π ξ sin(π ξ t) e πix ξ ξ [ πiξ j f(ξ) ˆ cos(π ξ t) + ĝ(ξ) ] π ξ sin(π ξ t) e πix ξ ξ (44) 1

11 Therefore: u(x, t) j [ πiξ j f(ξ) ˆ cos(π ξ t) + ĝ(ξ) ] R j π ξ sin(π ξ t) e πix ξ ξ [ 4π ξj f(ξ) ˆ cos(π ξ t) + ĝ(ξ) ] π ξ sin(π ξ t) e πix ξ ξ (45) Thus the Laplacian is: u(x, t) u(x, t) j 4π ξj 4π ξ [ ˆ f(ξ) cos(π ξ t) + ĝ(ξ) π ξ sin(π ξ t) ] e πix ξ ξ [ ˆ f(ξ) cos(π ξ t) + ĝ(ξ) π ξ sin(π ξ t) ] e πix ξ ξ (46) Now we fin u(x,t) : (x, t) [ f(ξ) ˆ cos(π ξ t) + ĝ(ξ) ] π ξ sin(π ξ t) e πix ξ ξ [ ] π ξ f(ξ) ˆ sin(π ξ t) + ĝ(ξ) cos(π ξ t) e πix ξ ξ (47) u(x, t) [ ] π ξ f(ξ) ˆ sin(π ξ t) + ĝ(ξ) cos(π ξ t) e πix ξ ξ R [ ] 4π ξ f(ξ) ˆ cos(π ξ t) π ξ ĝ(ξ) sin(π ξ t) e πix ξ ξ R [ 4π ξ f(ξ) ˆ cos(π ξ t) + ĝ(ξ) ] R π ξ sin(π ξ t) e πix ξ ξ This emonstrates that: (48) 11

12 u(x, t) 4π ξ u(x, t) (49) When t we have: u(x, ) ˆf(ξ) e πix ξ ξ f(x) using the Fourier Inversion Theorem (5) Similarly: (x, ) ĝ(ξ) e πix ξ ξ g(x) using the Fourier Inversion Theorem (51) Stein an Shakarchi note that since the existence of a solution to the Cauchy problem for the wave equation has been establishe it is reasonable to woner about uniqueness. Are there other solutions other than that given above? The answer is no an to prove this energy methos are employe ( see Problem 3, [1], pages 13-14). I will come to that problem shortly but first the authors exten the result of time conservation of total energy of a vibrating string in 1 imension to higher imensions. They efine energy as follows: where u(x, t) E(t) 1. The authors then pose this as a theorem: Theorem x 1 u(x, t) x (5) 1

13 If u is a solution to the wave equation given by (8) - (9), then E(t) is conserve in the sense that E(t) E() t R. Note here that time can be both positive an negative. Unlike the heat equation where time reversal is not possible, the wave equation oes amit time reversal. As a preliminary step in the proof the authors show that if a, b C an α R then: a cos α + b sin α + a sin α + b cos α a + b (53) This can be pove by expaning the LHS of (53) or noting that e 1 ( cos α sin α) an e ( ) ( sin α cos α are a pair of orthonormal vectors so that with Z a ) b C we have: Z Z e 1 + Z e (54) where represents the inner prouct in C. With these preliminairies the authors use Plancherel s formula which states that: ˆf(ξ) ξ f(x) x (55) so: x ( ĝ(ξ) ) ˆf(ξ) cos(π ξ t) + R R π ξ sin(π ξ t) ξ π ξ f(ξ) ˆ sin(π ξ t) + ĝ(ξ) cos(π ξ t) ξ (56) The next relationship is this: x j π ξ f(ξ) ˆ cos(π ξ t) + ĝ(ξ) sin(π ξ t) ξ (57) To see this we nee to apply Plancherel s theorem an the relationship between a erivative an its Fourier transform ( see [1], page 181 ). Thus we have: 13

14 {( ) α } ( α F u(x, t) πiξ) û(ξ, t) (58) In the present context we have α 1 so that for each spatial coorinate: Hence we have: {( ) } ) F u(x, t) (πiξ j û(ξ, t) (59) j { ( ) F u(x, t) } πiξ j û(ξ, t) πξ j û(ξ, t) (6) j Now, by Plancherel we have (using (6) ): x j πξ j û(ξ, t) ξ ĝ(ξ) πξ j ˆf(ξ) cos(π ξ t) + π ξ sin(π ξ t) ξ ( ) ĝ(ξ) π ξ ˆf(ξ) cos(π ξ t) + π ξ sin(π ξ t) ξ π ξ ˆf(ξ) cos(π ξ t) + ĝ(ξ) sin(π ξ t) ξ (61) Now using (53) with a π ξ ˆf(ξ), b ĝ(ξ) an α π ξ t we have: E(t) x 1 (4π ξ ˆf(ξ) ) + ĝ(ξ) ξ (6) Since (6) is inepenent of t an because of (5)-(51), E(t) E() for all t thus proving the claim. 14

15 6 Proving uniqueness of the solution to the Cauchy problem using the energy metho Stein an Shakarchi evelop the proof of the uniqueness of the solution to the Cauchy problem using the energy metho in Problem 3 of Chapter 6 ( [1], pages 13-14) as follows. They initially observe that the solution to the wave equation given by (43) epens only on the initial ata on the base of the backwar light cone. Note that (43) is a highly inirect solution. In this context the authors say ( [1], page 196) that the solution to at a point (x,t) epens only on the ata at the base of the backwar light cone originating at (x,t). In fact when > 1 is o, only the ata in an immeiate neighborhoo of the bounary of the base will affect u(x, t). (x,t) x-space Backwar light cone originating at (x,t) They then go on to ask if this property is share by any solution of the wave equation with the implication of an affirmative answer being that the solution is unique. They efine B(x, r ) to be the close ball in the hyperplane T centre at x an of raius r. The backwar light cone with base B(x, r ) is then efine to be: L B(x,r ) {(x, t) R : x x r t, t r } (63) More information on the concept of a light cone an the ifferences between the heat an wave equations in terms of propagation spees can be foun in ( [7], pages ). Thus Problem 3 becomes the proof of the following theorem: Theorem Suppose that u(x, t) is a C function on the close upper half-plane {(x, t) : x, t } that solves the wave equation u u. If u(x, ) (x, ) for all x B(x, r ), then u(x, t) for all (x, t) L B(x,r ). 15

16 If the initial ata of the Cauchy problem for the wave equation vanishes on a ball B, then any solution of the problem vanishes in the backwar light cone with base B. The steps in the proof are as follows. It is assume that u is real an for t r let B t (x, r ) {x : x x r t} an u(x, t) The energy integral is given by: 1.. E(t) 1 1 B t(x,r ) B t(x,r ) u(x, t) x {( ) + ( j ) } x (64) Clearly, from (64) E(t). Also E() because of the following. First, from ( ) the initial conitions u(x, ) for all x B(x, r ) so in the ball B (x, r ). Secon, we have: u(x, ) u(x, ) j j j u(x, ) j }{{} this is not inentically zero for all j since c 1 (65) Hence since the LHS of (65) is zero for all x B (x, r ) we must have that j u(x, ) for all j. Thus ( ) j an so E(). The authors invite you to prove the following: E (t) B t(x,r ) { u + j u } x 1 u(x, t) σ(γ) j B t(x,r ) (66) 16

17 Note that σ(γ) enotes the surface element on the sphere S 1 using spherical coorinates. This is a version of the Reynols Transport Equation which, in the context of flui ynamics, generalises ifferentiation uner the integral sign (Leibniz s Rule) with a moving bounary. Harley Flaners has given a physicist s proof of the general formula ([6]) which I essentially reprouce below in the Appenix. There are very few proofs, let alone goo ones, of this formula in unergrauate mathematical textbooks. There is a eep cross-over with the theory of partial erivatives in the context of harmonic functions an hence there is a substantial overhea in getting one s min aroun the etail. One can fin proofs in physics textbooks such as [8] or partial ifferential equation textbooks such as [7] an [1]. It is common to fin short proofs in partial ifferentiation textbooks (especially those pitche at grauate level) which assume a whole eifice of knowlege about harmonic functions. In fact in his article over 4 years ago Flaners sai this ( [6.1], page 616): We shall iscuss generalizations of the Leibniz rule to more than one imension. Such generalizations seem to be common knowlege among physicists, some ifferential geometers, an applie mathematicians who work in continuum mechanics,, but are virtually unhear of among most mathematicians. I cannot fin a single mention of such formulas in the current avance calculus an several variable texts, except for Loomis an Sternberg. Not much has change since Rather than go into what Flaners i here, I have put the etail in the Appenix which you shoul rea if you want to unerstan where the Reynols Transport Equation comes from, at least as a mathematical exercise (he was not trying to give the sort of proof one fins in flui mechanics textbooks). There is a lot of etail (I have expane on Flaners treatment) because there is no short way to unerstan where the formula comes from. The Reynols Transport Equation (RTE) can be researche at a high level in Wikipeia: org/wiki/leibniz_integral_rule The RTE can be expresse as follows: F ( x, t) V t D t D t F ( x, t) V + F ( x, t) v b ˆn S (67) D t Here F ( x, t) is a scalar function an D t an D t enote time varying connecte regions an its bounary which moves at velocity v b an ˆn is the unit normal component of the surface element. 17

18 Once you know the RTE it is easy to prove (66). From (64): F ( x, t) 1 {( ) + ( j ) } (68) Hence the first integral on the RHS of (67) is: B t(x,r ) F ( x, t) V B t(x,r ) B t(x,r ) B t(x,r ) 1 {( ) + {( ) u + {( ) u + j j ( j ) } V u } V j u } x j (69) Note that the authors use x to mean V x 1 x... x. To work out the secon integral on the RHS of (67) we have to remember that the velocity of the bounary, v b is the velocity of the base of the light cone which we have normalise to c 1 so, v b 1. The next thing to note is that the unit normal has a negative sign (see the iagram of the light cone). Thus recalling (64) we have: B t(x,r ) F ( x, t) v b ˆn S 1 Putting (69)-(7) together we o get (66) ie: B t(x,r ) u(x, t) σ(γ) (7) E (t) B t(x,r ) { u + j The next step in the proof is to note that: u } x 1 u(x, t) σ(γ) j B t(x,r ) (71) [ ] j j }{{} Φ i j j u j + u j (7) 18

19 Using (71) an the ivergence theorem an the fact that u solves the wave equation we have to show that: E (t) B t(x,r ) j ν j σ(γ) 1 u(x, t) σ(γ) (73) B t(x,r ) where v j is the j th coorinate of the outwar normal to B t (x, r ). At this point recall that the Divergence Theorem says this for 3-space an more generally for imensions with appropriate moifications: D F V D F ˆn S (74) where ˆn is the outwar unit normal to the surface. Using (7) in (71) we have: { E (t) B t(x,r ) { B t(x,r ) B t(x,r ) u + { Φj u } } x 1 j j } x 1 u + Φ u { u } u } {{} Φ x 1 B t(x,r ) Φ ˆn σ(γ) B t(x,r ) } {{ } Divergence Theorem B t(x,r ) x + 1 B t(x,r ) B t(x,r ) j v j σ(γ) 1 B t(x,r ) B t(x,r ) Φ x 1 u(x, t) σ(γ) u(x, t) σ(γ) B t(x,r ) B t(x,r ) u(x, t) σ(γ) u(x, t) σ(γ) B t(x,r ) u(x, t) σ(γ) u(x, t) σ(γ) To get a boun on v j j where v j is the j th coorinate of the unit outwar normal to B t (x, r ) we note that: 19 (75)

20 Letting x a k an y b k we then have: xy 1 x + 1 y x, y R (76) a k b k 1 a k + 1 b k (77) (77) is a species of a Höler inequality ( see [9], pages ). Hence: j }{{} a j v j }{{} b j ( ) 1 + j ( ) 1 ( + j ( ) v j ) {( ) ( ) } + j v j }{{} 1 (78) 1 u(x, t) Thus going back to (75) we have: E (t) 1 B t(x,r ) j v j σ(γ) 1 B t(x,r ) u(x, t) σ(γ) 1 B t(x,r ) B t(x,r ) u(x, t) σ(γ) u(x, t) σ(γ) (79) Thus we have that E (t) an from (64)-(65) we also know that E(t) for all t an E(). From the efinition of the erivative we have for h > :

21 E( + h) E() E () lim h + h E(h) lim h + h (8) But E(h) so we have a contraiction an so E(t) for all t. Alternatively the negative erivative implies that the energy is a ecreasing function of increasing time but E(t) an E() so energy must always be zero. Also because u(x, ) (x,) for all x B(x, r ) we have that u since the terms in the expression for E(t) in (64) must be zero. 7 Appenix Flaners starts out by noting that if: Φ(u, v, t) v where u g(t) an v h(t) the chain rule gives: u F (x, t) x (81) Φ[g(t), h(t), t] t ( ) Φ ġ + Φ v ḣ }{{} interval variation term + Φ (8) The brackete terms measure changes ue to the variation in the interval of integration [g(t), h(t)] while the remaining term measure changes ues to the variation of the integran. Assuming sufficient smoothness to justify interchange of integration an ifferentation operators we have: Φ v v F (x, t) F (x, t) x x (83) u u To eal with the multiimensional case the problem is with the moving omain rather than the varying integran but (8) shows us how to separate the two components. So Flaners starts with this: 1

22 t h(t) g(t) F (x) x (84) The omain of integration is an interval C t [g(t), h(t)] which is moving in time but what happens in the interior of the omain is not clear. What Flaners oes is to imagine that the interval C t is a worm crawling along the x-axis so that as it stretches an shrinks each point of its boy can only shrink so much, so >. Thus for each t, the map u x(u, t) is smooth one-one with smooth (continuously ifferentiable ) inverse. He then writes: φ t (u) x(u, t) φ t : [a, b] [φ t (a), φ t (b)] [g(t), h(t)] C t (85) Thus by changing the variable in an integral we have: h(t) g(t) F (x) x φt(b) φ t(a) F (x) x b a F [x(u, t)] u (86) The significance of this is that moving omain is now fixe but the integran is now time-varying. Flaners now ifferentiates uner the integral sign: t h(t) g(t) a b F (x) x F [x(u, t)] t a u b { F [x(u, t)] } u a b { F [x(u, t)] + F [x(u, t)] x } u (87) Flaners goes on to explain that the focus is now on the moving omain whose instantaneous velocity is v v(u, t) which he treats as function of x an t via the transformation (u, t) (x, t). Thus when t is fixe: x v ( ) v v (88)

23 Hence (with t fixe throughout): t h(t) g(t) F (x) x b a Φt(b) Φ t(a) Φt(b) Φ t(a) Φt(b) Φ t(a) h(x) g(x) { F [x(u, t)] { F (x) v { F (x) v + F (x) v [vf (x)] x [vf (x)] x + F [x(u, t)] x } u v + F (x) } x } u (89) Flaners notes that the erivative has been expresse as an integral over a moving omain an the integran epens on the velocity v at each point of the omain. However, the integran is an exact erivative so the answer epens only on the bounary values. At the bounary points g(t) an h(t) the respective velocities are ġ(t) an ḣ(t) so that: t h(t) g(t) h(t) F (x) x [ g(t) vf (x) [vf (x)] x ] h(t) g(t) F [h(t)] ḣ(t) F [g(t)] ġ(t) (9) Flaners concees that this may be a silly approach because of the introuction of the unnecessary quantity v an by sie-stepping the use of the funamental theorem initially only to use it in the en anyway. However, the ultimate goal was the reuction to a fixe omain. Next Flaners imagines a moving omain D t (see Figure 1) in the x y plane. Flaners procees by separating the bounary variation from the integran variation. 3

24 He fixes t t an uses the chain rule as before (see (8) ). Thus we have: F (x, y, t) x y F (x, y, t ) x y + t D t t D t tt tt D t F x y tt (91) To work out the LHS of (91) Flaners uses what he calls a physicist s argument. All this means is that the argument oesn t have epsilons an eltas in it! The logic of the erivation is common to both isciplines. It is how I remember such proofs an if you are pushe you ot all the i s an cross all the t s. He starts by consiering two successive omains as shown in Figure. If we let v v(x, y, t) enote the velocity at a bounary point (x, y) of D t an let n enote the outwar unit normal in the following ifference everything in the overlap of D t an D t+t cancels so that only a thin bounary strip makes a contribution: F (x, y, t) x y D t+t F (x, y, t) x y D t (9) 4

25 It can be seen from Figure that the relevant contribution is given by: F (x, y, t)(vt) (n s) (93) In time t the bounary which has velocity v moves a istance v n t an the istance along the bounary is s. Thus the prouct is an elemental area weighte by F (x, y, t). Higher orer terms are ignore an in a mathematician s proof you woul use regularity of the function an region to show that the higher orer terms on t matter. Thus Flaners arrives at this approximation: ( ) 1 F (x, y, t) x y F (x, y, t) x y F (x, y, t) v n s (94) t D t+t D t D t where enotes the bounary as usual. To get v n s Flaners rotates the unit tangent ( x, y y ) backwars through a right angle to obtain n (, x ). Thus s s s s we have: Thus: v n s (u, v) (y, x) u y v x (95) F (x, y, t) x y F (x, y, t)(u y v x) (96) t D t D t 5

26 Equation (96) screams out for the application of Green s Theorem in the context of (91) an (96): F F (x, y, t) x y F (x, y, t)(u y v x) + t D t D t D t [ iv(f v) + F ] x y D t x y (97) Recall that: (1) ivφ A Φ n s an D D () iv(f v) (F u) + (F v) gra F v + F iv v. y Flaners then moves on to a 3-imensional formula. He consiers a flui flowing through a region of space whose position is x x(u, t) at time t of a particle of flui originally at point u. The velocity is v v(x, t) at present time t of a particle now at position x. Now the omain D t moves with the flow an we suppose that we are given a function F (x, t) on the region of flow. Flaners then presents a mathematician s proof of a formula foun in [8, pages 48-43] which was the subject of a physicist s proof. The formula is: F F (x, t) x y z F v S + x y z t D t D t D t [ iv (F v) + F ] x y z D t (98) Here S is an elemental vectorial area on the close surface D t with: S (y z, z x, x y) n S (99) where n is the outwar unit normal an S is the element of area. The form of (99) may not be immeiately obvious. Flaners was an expert in ifferential forms an the form of (99) reflects this backgroun. Inee, in [6., page 43] he efines the vectorial area element as (σ 1 σ ) e 3 as a vector irecte along the normal e 3 with magnitue σ 1 σ. This arises from the cross prouct of x x (σ 1 e 1 +σ e ) (σ 1 e 1 +σ e ) (σ 1 σ )e 3. You have to pay attention ot 6

27 the orering of terms in exterior algebra eg σ σ 1 (e e 1 ) ( σ 1 σ )( e 1 e ) (σ 1 σ ) e 3. Another way of eriving (99) is to think of elementary Cartesian rectangles in each plane: S x y z n x S y z x n y (1) S z x y n z The S in (99) is interprete to reflect the relevant factors in (1). Flaners mathematician s (as istinct from a physicist s ) proof of (98) begins by efining the initial position as u (u 1, u, u 3 ), the moving point is x (x 1, x, x 3 ) an the velocity is ẋ (v 1, v, v 3 ) (ẋ 1, ẋ, ẋ 3 ) where the ot represents. There is a omain C in u-space an for each t it evolves (which Flaners escribes as an imbeing Φ t : C D t of C into x - space. The mapping (u, t) Φ t (u) is assume twice continuously ifferentiable an he writes Φ t u) x(u, t). For each fixe t he writes the Jacobian matrix of Φ t as: [ ] i j [ (11) The Jacobian matrix is non-singular everywhere an its inverse is. The eterminant of the Jacobian matrix is. Flaners next relies upon Jacobi s formula for the erivative of a eterminant. Thus if A A(t) is a non-singular matrix function then: (1) is prove in [5.] j i A A trace(ȧa 1 ) (1) The next step is to apply (1) to the Jacobian matrix noting that, because of the assume continuity of the erivatives: ( i ) ( i ) ẋ i j j vi (13) j j 7 ]

28 Hence we have: {( trace {[ ) ( ) 1 } trace i,j v i j v i j j i v i i i iv v ] [ j k ]} (14) So using (1) we have: ( iv v) (15) t Flaners now sets: f(t) F (x, t) x 1 x x 3 D t (16) Then the change of variables rule gives: f(t) There is now a fixe omain an he ifferentiates: C F [x(u, t), t] u 1 u u 3 (17) f(t) t { } [ F C i vi + F ] + F (x, t) ( iv v) u 1 u u 3 i { } ( gra F ) v + F iv v + F x 1 x x 3 D t { } iv (F v) + F x 1 x x 3 D t (18) 8

29 since ( gra F ) v+f iv v iv (F v). This mechanically emonstrate by: ( gra F ) v i v i F i (19) F iv v i F vi i (11) iv (F v) i F vi i 1 3 F v 1 F v ( gra F ) v + F iv v (111) F v 3 Thus after some work oes (18) oes equal (98). 8 Now for the physicist s proof If we go to [8, pages 48-43] we can fin a quite concise physicist type proof which runs like this. Sokolnikoff an Reheffer start with a volume integral: T u(p, t) τ (11) where u(p, t) is a continuous scalar function in a simply connecte region T. Because the position of points P in T can in principle vary with time, the region of integration will change with t. When the region is fixe so that T is inepenent of t an u(p, t) an are continuous in T for all relevant t you get the stanar Leibniz result: u(p, t) τ t T T (P, t) τ (113) But when the omain of integration varies with time the RHS of (113) shoul inclue another term to account for that change of region. To fin this term consier the surface S bouning T at a certain time t. We suppose that S is such that the ivergence theorem is applicable an consier an element S of S with surface area σ. In a small time interval (t, t + t) the points P of S sweep out a region of space whose volume is: 9

30 τ (v n) t σ (114) where v is the velocity of P an n is the exterior unit normal to the small surface element S. Now we sum the proucts of u(p, t) an the elemental areas over the whole surface S as follows: I t S u(p, t) (v n) σ (115) Now we ivie by t an pass to the limit an we get the formal result: I t u(p, t) (v n) σ (116) S This then is the term that shoul be ae to the RHS of (113) ue to the motion of the region. Thus: u(p, t) τ t T T (P, t) τ + S u(p, t) (v n) σ (117) (There is a typo in equation 18-5 of [8], page 49 which is correcte in (117) ) Clearly because u is a scalar function we have u(v n) (u v n) an so we can write: Thus we get: S u(v n) σ T iv (u v) τ (118) [ (P, t) ] u(p, t) τ + iv (u v) τ (119) t T T This is what Flaners arrive in equation (98). As you can see it is a much shorter proof which actually contains the motivation for the more mathematical treatment of Flaners. 3

31 9 The style of proof one fins in PDE textbooks In finishing this article I come finally to the style of proof one funs in PDE textbooks. Because these textbooks are usually irecte at grauate level or senior unergrauate level they assume a egree of familiarity with harmonic theory. For instance in [7], a grauate PDE textbook, the author exhibits the following 5 line proof. Let u be a solution of the wave equation u tt (x, t) u(x, t) for x, t >. The energy norm of u is efine as follows: It is claime that: E(t) 1 { u t (x, t) + } u x i(x, t) x (1) i1 E { } t u t u tt + u x i u x i t x i1 { } u t (u tt u) + (u t u x i) x i x if u(x, t) for sufficiently large x which may epen on t. i1 (11) Clearly u t (u tt u) x because u is a solution of the wave equation but why oes i1 (u tu x i) x i x? The author has use a fact about harmonic functions that if they are zero on a bounary of a volume then they are zero everywhere in that volume. Hence the erivatives in the sum are zero an the result follows. 1 References [1] Elias M Stein & Rami Shakarchi, Fourier Analysis: An Introuction, Princeton University Press, 3 [] T. W. Körner, Fourier Analysis, Cambrige University Press,

32 [3] Ruel V Churchill, Fourier Series an Bounary Value Problems, Secon Eition, McGraw-Hill International Stuent Eition, [4] I M Gelfan an S V Fomin, Calculus of Variations, Dover, 1963 [5.1] Peter Haggstrom, Basic Fourier Integrals, com/basic%fourier%integrals.pf [5.] Peter Haggstrom, Jacobi s formula for the erivative of a eterminant, http: // erivative%of%a%eterminant.pf [6.1] Harley Flaners, Differentiatiuon uner the integral sign, American Mathematical Monthly, 8 (1973), No.6, pages , [6.] Harley Flaners, Differential Forms with Applications to the Physical Sciences, Dover Publications, [7] Jürgen Jost, Partial Differential Equations, Springer, [8] I. S. Sokolnikoff an R.M. Reheffer, Mathematics of Physics an Moern Engineering, Secon Eition, McGraw-Hill [9] J. Michael Steele The Cauchy-Schwarz Master Class, Cambrige University Press, 4. [1] Michael E. Taylor, Partial Differential Equations Part 1, Springer, History Create 5 November 13 8 December 15 - fixe some typos an ae comment about resolution of 3

33 non-convergence of erive series - Churchill [3], pages May 17 - major extension with etaile proof of energy conservation in imensions. In the process there is a etaile examination of ifferentiation uner the integral sign in more than one imension ( effectively how the Reynols Transport Equation generalises Leibniz s rule). 33