Vibrating Strings and Heat Flow

Transcription

1 Vibrating Strings and Heat Flow Consider an infinite vibrating string Assume that the -ais is the equilibrium position of the string and that the tension in the string at rest in equilibrium is τ Let u(, t) denote the displacement at at time t Then the wave equation (in one space dimension) governs the motion snap shot at time t: u u(, t) = displacement Derivation (for small displacements) We make the following simplifying assumptions: the displacement of the string from equilibrium (and its slope) are small; each point on the string moves only in the vertical direction; the tension force T(, t) in the string (ie, the (vector) force which the part of the string to the right of eerts on the part to the left of, at time t) is tangential to the string and has magnitude proportional to the local stretching factor 1 + u T(,t) τu τ Since u = in equilibrium, the constant of proportionality is the equilibrium tension τ Thus the magnitude of T(, t) is τ 1 + u (, t), and the vertical component of T(, t) is τu Now consider the part of the string between and + The vertical component of Newton s second law (force = mass acceleration) applied to this part of the string is force }} τu ( +, t) τu (, t) = mass }} ρ accel }} u tt (, t), T(,t) T(+,t) where ρ is the density (mass per unit length; assumed constant) Dividing by and taking the limit as, we obtain τu = ρu tt Normalizing units so that ρ = τ, we obtain the wave equation (in one space dimension): u tt = u 99 +

2 1 Vibrating Strings and Heat Flow Solutions of u tt = u Change variables Let y = + t, z = t, so = 1(y + z), t = 1 (y z) Then y = ( y ) + ( y t) t = 1 ( + t ), z = ( z ) + ( z t) t = 1 ( t ), so u y = 1 (u + u t ) and u yz = 1 ( t ) 1 (u + u t ) = 1 4 (u u tt ) In the new coordinates, the wave equation becomes simply u yz = Thus u y is independent of z, ie, u y = ṽ(y) Integrating in y for each fied z, we get u = v(y) + w(z) (where v(y) = ṽ(y)dy) So any solution of the wave equation u tt = u is of the form ( ) u(, t) = v( + t) + w( t) Physically, this is a superposition of left-going and right-going waves: v() w() v( + t) (as t increases) w( t) Observation The derivation above shows that any C function of and t satisfying the wave equation is of the form ( ) Conversely, if v and w are C functions of one variable, it is easily checked that u(, t) = v( + t) + w( t) is a C solution of the wave equation But if v and w are only continuous, v( + t) + w( t) still makes sense; in what sense is this a solution of u tt = u? We will see later in the course that the equation holds in the sense of distributions Initial-Value Problem (IVP) (or the Cauchy Problem) If we think of the wave operator as an ordinary differential operator in time acting on functions of t taking values in functions of (overlooking considerations arising from the fact that is itself a differential operator), we should be able to determine u(, t) for R and t if we are given initial values u(, ) and u t (, ) for R (we need u and u t at t = since the equation is second-order in t) D Alembert s Formula (for the Cauchy Problem for u tt = u ) Consider the IVP: DE u tt = u ( R, t ) u(, ) = f() ( R) IC u t (, ) = g() ( R)

3 Vibrating Strings and Heat Flow 11 (To obtain a C solution u(, t), it will suffice for f C (R), g C 1 (R)) We will separately analyze the cases g and f, and then use superposition Case 1 g u(, ) = f() IC u t (, ) = ( R) We have u(, t) = v( + t) + w( t) for some v, w C (R) By the IC, v() + w() = u(, ) = f() v () w () = u t (, ) =, so v and w differ by a constant One solution is v() = w() = 1 f() Any other solution v() = 1 is f() + c w() = 1 for some constant c So the solution in Case 1 is f() c u(, t) = 1f( + t) + 1 f( t) Remark For a solution u(, t) of u tt = u, v and w are uniquely determined up to a constant This is because if v 1 ( + t) + w 1 ( t) = v ( + t) + w ( t), then v 1 ( + t) v ( + t) = w ( t) w 1 ( t) is independent of both y = +t and z = t, and is thus a constant Case f u(, ) = IC u t (, ) = g() ( R) Again, u(, t) = v( + t) + w( t) for some v, w C (R) By the IC, So v = 1 v() + w() = v () w () = g() Thus w = v v = 1g g and the solution in Case is u(, t) = 1 +t t g(s)ds Adding Cases 1 and, the solution of the IVP with IC d Alembert s formula: u(, ) = f() u t (, ) = g() is given by u(, t) = 1 f( + t) + 1 f( t) + 1 +t t g(s)ds

4 1 Vibrating Strings and Heat Flow Remarks t t +t (1) d Alembert s formula gives an eplicit demonstration of the finite domain of dependence of the solution of this IVP on the initial data (a general property of hyperbolic PDEs): for a fied R and fied t >, u(, t) depends only on f( + t), f( t), and g(s) : t s + t} () d Alembert s formula also provides a solution for negative t as u(, ) = f() well: u tt = u ( R, t ), ( final u t (, ) = g() conditions); like ODEs, hyperbolic PDEs in general can be advanced either in the +t direction or the t direction Initial-Boundary Value Problem (IBVP) Consider now a finite string ( π) fied at both ends, so u(, t) = u(π, t) Suppose the initial displacement is u(, ) = f() ( π) (where f() = f(π) = ), and for simplicity suppose the initial velocity is u t (, ) = ( π) This models a plucked violin string (moved to position u(, ) = f() at time t =, and then released with initial velocity u t (, ) = ) We obtain an IBVP with both initial conditions (IC) and boundary conditions (BC): DE u tt = u ( π, t ) u(, ) = f() IC ( π) u t (, ) = u(, t) = BC (t ) u(π, t) = t π We will solve this IBVP in two ways: 1 by d Alembert s formula, and by Fourier series Solution 1 (d Alembert) Find functions v, w defined on R so that u(, t) = v( + t) + w( t) satisfies the IC and BC The BC u(, t) = for t gives = v(t) + w( t) for t, or w(t) = v( t) for t [Note that to define u(, t) in the region π, t, we only need to give v(s) for s and w(s) for s π To simplify our calculations, we will find v and w defined on all of R, so that u(, t) satisfies the BC for t too] So we ask w(t) = v( t)( t R) Net, the BC u(π, t) = (now t R) gives = v(π+t)+w(π t), so v(π + t) = w(π t) = v(t π), ie, v(t + π) = v(t)( t R) So v is π-periodic, and thus w(t) = v( t) is also π-periodic The IC u t (, ) = ( π) gives

5 Vibrating Strings and Heat Flow 13 = v () w () = v () v ( ) for π, ie, v ( ) = v () for π Since v is π-periodic, we conclude that v is an even function on R We may assume v() = (if not, replace v by v(s) v() and replace w by w(s) + v()) Then v( ) = v (s)ds = v ( s)ds = v (s)ds = v()( R), so v is an odd function on R; moreover w = v since w(t) = v( t) Finally, the IC u(, ) = f() ( π) gives f() = v() + w() = v(), ie, v() = 1 f() for π This completes the determination of v: it is the π-periodic, odd function on R which agrees with 1 f on [, π] So d Alembert s solution can be summarized as follows: define f() = f() for π, f() = f( ) for π (the odd etension of f from [, π] to [ π, π]), and then etend f to be π-periodic on R [Note: if f() = f(π) = and f C 1 [, π], then f C 1 (R); if in addition f C [, π] and f () = f (π) =, then f C (R)] We obtain d Alembert s formula for the solution of this IBVP: u(, t) = 1 ( f( + t) + f( t) ) (remember, this is the special case where u t (, ) = ( π)) Solution (Fourier series) We use separation of variables We want to find simple harmonics of the string, that is, solutions of the form u(, t) = v()w(t) (often called fundamental modes) The v and w here are not the same v and w as above Using to mean d for v, and also d for w, the DE u d dt tt = u becomes v()w (t) = v ()w(t), or (wherever v()w(t) ) w (t) w(t) = v () v() The LHS is independent of and the RHS is independent of t, so both sides are equal to a constant; call it λ We end up with ODEs for v and w: v () + λv() = ( π) spatial ODE w (t) + λw(t) = (t ) temporal ODE Applying the BC to the spatial ODE, we get v() = v(π) =, leading to the following eigenvalue problem : determine for which (in this case real) values of λ there eists a non-trivial (ie, not ) solution v() of the boundary-value problem (BVP): DE v + λv = π BC v() = v(π) =

6 14 Vibrating Strings and Heat Flow Case (i) λ < The general solution of v + λv = is c 1 cosh( λ) + c sinh( λ) v() = c 1 =, and then v(π) = c = No nontrivial solutions Case (ii) λ = The general solution of v = is v() = c 1 + c v() = c 1 =, and then v(π) = c = No nontrivial solutions Case (iii) λ > The general solution of v + λv = is v() = c 1 cos( λ) + c sin( λ) v() = c 1 = Then v(π) = (and c so v is nontrivial) sin( λπ) = λ 1,, 3, } λ = n for n 1,, 3, } These are the eigenvalues of this eigenvalue problem The corresponding eigenfunctions are sin( λ) = sin(n) Applying the homogeneous IC u t (, ) = to the temporal ODE, we get w () = For λ = n, the general solution of w + λw = is c 1 cosnt + c sin nt The IC w () = implies c =, so w(t) = c 1 cos nt Thus the fundamental modes for this problem are u n (, t) = cos(nt) sin(n) n 1,, 3, } Linear combinations of these are also solutions of the DE, the BC, and the one IC u t (, ) = To satisfy the IC u(, ) = f() for π, we represent f() in a Fourier sine series: f() = A n sin(n) Then (provided this series converges appropriately), u(, t) = A n cos(nt) sin(n) satisfies the DE, the BC, and both IC (See Problem 3 on Problem Set 7 for details) Heat Flow Consider heat flow in a thin rod with insulated lateral surface + Assume that the temperature u(, t) is a function only of horizontal position and time t By Newton s law of cooling, the amount of heat flowing from left to right across the point in time t is κ u (, t) t (proportional to the gradient of temperature), where the constant of proportionality κ is called the heat conductivity of the rod So the net heat flowing into the part of rod between and + in the time interval from t to t + t is κ u ( +, t) t κ u(, t) t The net heat flowing into this part of the rod in this time interval can also be epressed as mass specific heat }}}} ρ c u }} u t t,

7 Vibrating Strings and Heat Flow 15 where ρ is the density (mass per unit length) of the rod, and c is the specific heat of the rod (the amount of heat needed to raise a unit mass by 1 unit of temperature) Equating these two epressions, dividing by t and, and taking the limit as, we obtain κu = ρcu t Normalizing units so that ρc = κ, we obtain the heat equation (in one space dimension): u t = u Fourier considered circular rods of length π, leading to the following IBVP with periodic BC: π IBVP: DE u t = u π, t IC u(, ) = f() π u(, t) = u(π, t) periodic BC t u (, t) = u (π, t) t π We can view u as defined on T [, ) (where T = S 1 ), or as a π-periodic function of R with t ) As with the wave equation, we separate variables and look for solutions of the form u(, t) = v()w(t) The DE u t = u becomes v()w (t) = v ()w(t), or (wherever v()w(t) ) w = v v ; both sides are equal to a constant; call it λ The spatial ODE is and the temporal ODE is w v () + λv() = w (t) + λw(t) =, (t )

8 16 Vibrating Strings and Heat Flow In this case our eigenvalue problem has periodic boundary conditions: v + λv =, ( π) v() = v(π), v () = v (π) Case (i) λ < The only solution is v Case (ii) λ = There is one linearly independent solution: v 1 Case (iii) λ > We must have λ = n for n 1,, 3, }, now with two linearly independent solutions: cos(nt) and sin(nt) (see Problem 1 on Problem Set 7 for details) For λ = n (with n, 1,, }), there is one linearly independent solution of w + λw = : w = e λt Thus the fundamental modes for this problem are: and for n 1,, 3, }: u 1 u(, t) = e nt cos nt, and u(, t) = e nt sin nt To satisfy the IC u(, ) = f() for π, we represent f() in a Fourier series: f() = a + (a n cos n + b n sin n) Then (provided this series converges appropriately) u(, t) = a + e nt (a n cosn + b n sin n) satisfies the DE, the periodic BC, and the IC Remark This form of the Fourier series of f (viewed as its π-periodic etension) is equivalent to the comple eponential form For n 1, cosn = 1 f() = n= c n e in ( e in + e in) and sin n = 1 i span the same two-dimensional subspace (over C) as ( e in e in) e in = cos n + i sin n and e in = cosn i sin n The coefficients are related as follows: c = a, and for n 1, c n = 1 (a n ib n ), c n = 1 (a n + ib n ),

9 Vibrating Strings and Heat Flow 17 In the inner product on L (S 1 ), the set a n = c n + c n, b n = i(c n c n ) g, f = 1 π g()f()d π π 1, cosn, sin n : n 1} is a complete orthonormal set, giving us the following formulae for a n and b n : and for n 1, Caution Many books will write in which case a = 1, f = 1 π a n = cosn, f = 1 π b n = sin n, f = 1 π π π π f()d, f() cosnd, f() sin nd f() = a + (a n cosn + b n sin n), a = 1, f = 1 π π f()d = 1 π π f() cos()d The solution u(, t) epressed in terms of comple eponentials is u(, t) = ξ Z f(ξ)e ξt e iξ where f(ξ) = e iξ, f = 1 π f()e iξ d Note that if f C 1 (T) (or even f is continuous π and piecewise C 1 on T, meaning f has only a finite number of jump discontinuities), then f l 1 (Z) Thus this series for u(, t) converges absolutely and uniformly for T and t, and u(, ) = f(); moreover, for t >, this is a C solution of u t = u This is a consequence of the rapid decay of e ξt as ξ for t > But for t <, we do not epect this series to converge unless f(ξ) etremely fast as ξ These properties are common for parabolic equations: the solution is smooth for t >, but we cannot go backwards in time Remark As for the wave equation, we can also solve IBVP of the form DE u t = u ( π, t ) IC u(, ) = f() ( π) BC u(, t) =, u(π, t) = (t ) (or with BC u (, t) =, u (π, t) =, etc)