VIEWS: 3 PAGES: 75 POSTED ON: 12/8/2011 Public Domain
PARTIAL DIFFERENTIAL EQUATIONS AND BOUND ARY VALUE PROBLEMS VIBRATING STRINGS AN D HEAT CONDUCTION THE VIBRATING STRING: Suppose that a flexible string is pulled taut on the x-axis and fastened at two points, which w e denote by x = 0 and x = L. The string is then drawn aside into a certain curve y = f(x) in the xy plane and then released from rest. Problem: To find the vertical displacement yx (,t) from the equilibrium position of the string at the point (distance x from the origin) at time t. We make certain assumptions: The subsequent motion of the string is entir ely transverse (i.e. along the xy plane). (Thu s the displacement of the string from the eq uilibrium position is a function y=y (x, t).) We look at a small piece of the string whic h in the equilibrium position has length x . If m is the (uniform) density of the string, the mass of the piece is m x. By Newton’s second law of motion, the transver se force F acting on the piece is given by ¶2 y F = m Dx 2 ¶t Since the string is flexible, the tension T at any point is directed along the tangent (see the figure) and has T sin as its y- compon ent. T sin T We now assume that the motion of the string i s due strictly to the tension T in it. (That is the gravitational force is negligible.) Hence F = (T sin ) (T tan ) (as ¶y is small) and so = (T tan ) = D (T ) ¶x Thus we get ¶y ¶ y 2 D (T ) = m Dx 2 ¶x ¶t Dividing by x and letting x 0, we get ¶ ¶y ¶ y 2 (T )=m 2 ¶x ¶x ¶t Or as T is constant (the same for all x), we get ¶ y2 ¶ y 2 T =m 2 ¶x 2 ¶t T Putting a =2 M, We get the one-dimensional wave equation: ¶ y ¶ y 2 2 a2 = 2 ¶x 2 ¶t which is usually written as ¶ y 2 1 ¶ y 2 = 2 2 ¶x 2 a ¶t It is proved in Physics that y(x,t) satisfies the ‘o ne-dimensional wave equation’ 2 y 2 y a 2 2 x 2 t where a is a constant (depending on the mass o f the string and the tension of the string). We th us want to solve the boundary value-problem y y 2 2 (1) a2 2 x 2 t subject to the boundary conditions y (0, t ) y ( L, t ) 0 for all t > 0 (2) and the initial conditions y 0 for t = 0 (3) t y(x,0) = f(x), 0 ≤ x ≤ L (4) We find the solution by the method of ‘sep aration of variables’. That is we let y(x,t) = u(x) v(t) (a function of x a function of t ) Substituting for y in the equation (1), we get a u x ) v (t ) u ( x ) v t ) 2 ( ( or u ( x ) 1 v ( t ) 2 (5) u ( x) a v (t ) where the ‘dashes’ denote derivative w.r.t. the c oncerned variable. The L.H.S of (5) is indepen dent of t and the R.H.S is independent of x and hence both are independent of both x and t and hence equals a pure consta nt, say, - . u x ) 1 v t ) ( ( Thus, we get u ( x ) a 2 v (t ) y (0, t ) y ( L , t ) 0 for all t > 0 implies u(0) = u(L) =0. So we have to solve for u u u 0 with boundary conditions u(0) = u(L) =0. This eigenvalue problem has nonttrivial soluti ons only when np 2 2 l = 2 (n =1,2,...) L Corresponding solutions are np x u n = sin ( n = 1, 2, ...) L and non-zero multiples of them. y 0 for t = 0 implies v 0 ) 0 ( t Now for each n, v satisfies the d.e. n p a 2 2 2 v ¢¢ + 2 v = 0 L with the initial condition v 0 ) 0 ( np at np at Thus v = c1 cos + c 2 sin L L np a np at np a np at v¢ = - c1 sin + c2 cos L L L L v0) 0 c2 0 ( Hence np a v = c1 cos t L We shall take c1=1 and denote the correspo nding solution (which depends on n ) by vn . Thus np a v n = cos t L Thus for each n =1,2,3,… np x np a t y n = u n v n = s in cos L L is a solution to the given problem satisfying th e first three conditions. So also any finite or in finite lc of them. But no finite lc of them wou ld satisfy the initial condition y(x,0) = f(x), 0 ≤ x ≤ L So we consider their infinite lc so as to satisfy t he above initial condition. Thus look at the infinite linear combination ¥ np x np a t y = å yn = å n =1 b n s in L cos L Now t = 0 implies ¥ np x f ( x ) = å bn sin , 0£ x£ L n =1 L Thus bn= Fourier sine coefficient of f(x) in [0, L] n x L 2 L f ( x ) sin L dx 0 Thus the solution to the wave equation satisf ying the initial and boundary conditions is ¥ np x np a t y( x,t) = ån =1 b n s in L cos L 0 ≤ x ≤ L, t 0 where L 2 np x bn = L ò 0 f ( x ) sin L d x , n = 1, 2, ... Example: Solve the wave equation when f ( x ) = x ( L - x ), 0 £ x £ L Thus 2 L np x bn = ò x ( L - x )sin dx L0 L L é æ np x ö æ np x ö æ np x öù 2ê ç cos L ÷ ç sin ÷ ç cos ÷ú = ê x( L - x) ç - - ( L - 2 x) ç - 2 L ÷ + (-2) ç 3 L ÷ú Lê np ÷ ç n p2 ÷ ç n p 3 ÷ú ç ÷ ç ÷ ç ÷ú ê ë è L ø è L ø 2 è L 3 øû 0 ì 8L2 4L 2 ï 3 3 n = 1,3,.. = 3 3 (1 - cos np ) = ín p np ï0 n = 2,4,.. î Thus the solution to the given problem is y(x,t) 8L 2 ¥ 1 (2n -1)p x (2n -1)p at = 3 p å (2n -1)3 sin L cos L n=1 0 ≤ x ≤ L, t 0 Example: Solve the wave equation when L x 0x f ( x) 2 L L x xL 2 Thus L nx nx 2 L 2 bn sin x dx L x ) sin ( dx L 0 L L L 2 éì æ np x ö np x ü L/2 êï 2 êï ç - cos L ÷ s in L ï ï = íx ç ÷+ ý L êï np n p 2 2 ï ç ÷ êï è L ø ï0 ëî þ 2 L ì æ np x ö np x ü L ù ï ï ç - co s L ÷ sin L ï ï ú + í(L - x) ç ÷- ý ú ï np n 2p 2 ï ú ç ÷ ï î è L ø L 2 ïL/2 ú þ û 2 é L 2 np L 2 np = ê - 2 np c o s 2 + n 2p 2 s in L ë 2 L 2 np x L 2 np ù + co s + 2 sin ú 2 np L n p 2 2 û ì 0 n = 2, 4, 6, ... 4L np ï = 2 2 sin = í 4L / n p 2 2 n = 1, 5, 9, ... n p 2 ï - 4 L / n 2p 2 n = 3, 7 ,1 1, ... î Hence the solution to the wave equatio n is y( x,t) 4L ¥ 1 (2n -1)p x (2n -1)p at = 2 å(-1) n+1 sin cos p n=1 (2n -1) 2 L L Example: Solve the wave equation when ì x, 0£ x £p /4 ï f ( x ) = í p / 4, p / 4 £ x £ 3p / 4 ï p - x, 3p / 4 £ x £ p î p Thus 2 bn = p ò f ( x) sin nx dx 0 2é p /4 3p / 4 p = ê ò x sin nx dx + ò sin nx dx pë0 p /4 4 p + ò (p - x)sin nx dx] p 3 /4 2 éì p /4 3p / 4 cos nx sin nx ü p ì cos nx ü = ê í x( - ) + 2 ý + í- ý p êî ë n n þ0 4î n þp / 4 cos nx sin nx ü ù p ì + í(p - x)(- )- 2 ý ú î n n þ3p / 4 ú û 2 np 3np = 2 (sin + sin ) pn 4 4 Thus the solution to the given problem is y( x,t) 2 ¥ 1 np 3np = å 2 (sin + sin )sin nx cos nat p n=1 n 4 4 D’Alembert’s solution of the wave equation: Consider the wave equation y 1 y 2 2 x 2 a t 2 2 We change the independent variables x, t to u, v by substituting u = x + at , v = x - at y y u y v y y x u x v x u v y 2 y y u x v x 2 x y u y v y u y v 2 2 2 2 2 2 u x vu x uv x v x y2 u y 2 2 2 2 2 u uv v (assuming that all mixed partial derivates are continuous). Similarly y 2 2 y u y 2 2 a 2 2 2 u 2 t 2 uv v Hence the wave equation becomes y 2 y 2 or y 2 2 2 0 u v u v uv Integrating partially w.r.t u, we get y = a constant independent of u v = a function of v = g(v) say Again integrating partially w.r.t v, we get y (v ) dv a g (where a is a constant v independent of v) = G (v) + F (u)(say) = F (x + at) + G (x - at) We have to chose F,G so as to satisfy the initi al and boundary conditions. This is extremel y difficult. Hence we have to adopt the Fouri er Series method for solving the wave equati on. ONE-DIMENSIONAL HEAT EQUATION The heat equation: We consider the flow of heat in a thin cylin drical rod of cross-sectional area A whose l ateral surface is perfectly insulated so that n o heat flows through it. The word ‘thin’ me ans that the temperature is uniform on any cross- section and thus is a function of the t ime t and its horizontal distance x from one end, w(x,t). We note the following physical principles: • Heat flows in the direction of decreasing te mperature, that is from hot regions to cold re gions • The rate at which heat flows across an area i s proportional to the area and to the rate of ch ange of temperature with respect to the dista nce in a direction perpendicular to the area. ( This proportionality factor is denoted by k an d called the thermal conductivity of the subst ance.) c) The quantity of heat gained or lost by a b ody when its temperature changes is propo rtional to the mass of the body and to the c hange of temperature. (This proportionalit y factor is denoted by c and called the spe cific heat of the substance.) Using the above three principles, we can deri ve the heat equation describing the temperatu re w(x,t) as 2 w 1 w 2 x 2 a t Here the constant a2 depends on three things: ,density of the rod, c, specific heat of the su bstance and k, thermal conductivity of the subs tance. k In fact a = 2 cr We assume the boundary conditions w (0, t ) w ( L , t ) 0 This means that the ends are maintained at zer o temperature at all times. We also assume that the rod has an initial temp erature distribution f(x) at time t = 0 i.e. w(x,0) = f(x) 0<x<L Again ‘separating’ the variables, we find that w=uv where u satisfies the eigenvalue problem u u 0 u(0) u(L) 0 (1) and that v satisfies first order d.e v a 2 v (2) Thus for nontrivial solutions, n 2p 2 ln = 2 ( n = 1, 2,...) L Corresponding eigenfunctions are np x u n = sin L n 2p 2 a 2 t - For each n=1,2,3,… vn = e L2 np 2 2 is a solution of (2) with ln = 2 a Hence for each n = 1,2,3,… n 2p 2 a 2 t np x - wn = u n vn = sin e L2 L is a solution of the heat equation. Again we assume that the infinite linear combin ion n 2p 2 a 2 t ¥ ¥ np x - w = å b n w n = å b n s in e L2 n =1 n =1 L is also a solution. Putting t = 0, we get f ( x) b sin nx n n 1 L Hence bn = nth Fourier sine coefficient of f(x) n x L 2 in [0, L] f ( x ) sin dx L 0 L Problem Solve the heat equation given that w (0, t ) = w ( L , t ) = 0; ì x 0 < x £ L/2 w ( x , 0) = f ( x ) = í îL - x L / 2 £ x £ L Solution The solution to the heat equation subject to the given boundary and initial conditions are: n 2p 2 a 2 t ¥ ¥ np x - w = å b n w n = å b n s in e L2 n =1 n =1 L where bn = nth Fourier sine coefficient of f(x) n x L 2 in [0, L] L f ( x ) sin L dx 0 L nx nx L 2 2 bn sin x dx L x ) sin ( dx L 0 L L L 2 Continued on the next slide éì æ np x ö np x ü L/2 êï 2 êï ç - cos L ÷ s in L ï ï = íx ç ÷+ ý L êï np n p 2 2 ï ç ÷ êï è L ø ï0 ëî þ 2 L ì æ np x ö np x ü L ù ï ï ç - co s L ÷ sin L ï ï ú + í(L - x) ç ÷- ý ú ï np n 2p 2 ï ú ç ÷ ï î è L ø L 2 ïL/2 ú þ û 2 é L 2 np L 2 np = ê - 2 np c o s 2 + n 2p 2 s in L ë 2 L 2 np L 2 np ù + co s + 2 sin ú 2 np 2 n p 2 2 û ì 0 n = 2, 4, 6, ... 4L np ï = 2 2 sin = í 4L / n p 2 2 n = 1, 5, 9, ... n p 2 ï - 4 L / n 2p 2 n = 3, 7 ,1 1, ... î The solution to the heat equation subject to the given boundary and initial conditions are: w( x , t ) = ( 2 n -1) 2 p 2 a 2t 4L ¥ (2 n - 1)p x - 2 å n +1 ( -1) sin e L2 p n =1 L Steady state temperature We assume that when t is large, w a function i ndependent of t, w(x) say. This is called the ste ady state temperature. And noting that in stead y state w 0 t w d w 2 2 we get 2 0 x 2 dx Therefore w = c1 x + c 2 Suppose that the ends are maintained at temper ature w1,w2 at all times. Hence x 0 w w1 c2 w1 1 x = L Þ w = w2 Þ c1 = ( w2 - w1 ) L Hence the steady state temperature is 1 w w1 ( w2 w1 ) x 0xL L Problem Find the solution of the one-dimensional heat e quation satisfying the boundary and initial cond itions w ( 0 , t ) 100 , w ( L , t ) 0 , for all t. px w ( x , 0) = 100 cos 0xL 2L Solution We write the temperature distribution as w( x, t ) = wS ( x) + wT ( x, t ) = steady state temperature + Transient tempe rature. We impose the boundary conditions wS(0 ) = 100, wS(L) = 0 on wS(x). Thus 1 x ws ( x) 100 (0 100) x 100 0 x L 1 L L We impose on wT(x,t) the boundary conditions wT(0,t) = wT(L,t) = 0 for all t and the initial condition px wT ( x, 0) = 100 cos - wS ( x) 0£ x£ L 2L That is : wT(x, t) satisfies the boundary conditio ns wT(0,t)=wT(L,t) = 0 for all t and the initial condition px x wT ( x, 0) = 100 cos - 100(1 - ) 0£ x£ L 2L L n 2p 2 a 2 t Thus ¥ np x - w T ( x , t ) = å b n sin e L2 n =1 L where 2 L px x np x bn = L ò [1 0 0 co s 2 L - 1 0 0 (1 - L )] sin L d x 0 L 2 px np x = ò 1 0 0 co s 2 L sin L d x L 0 L 2 x np x - ò 1 0 0 (1 - ) sin dx L 0 L L Now 2 L px np x ò 1 0 0 co s 2 L sin L d x L 0 ( 2 n 1)x ( 2 n 1)x L 100 L sin 2 L sin 0 L dx L 100 2L (2n 1)x 2L (2n 1)x (2n 1) cos 2L (2n 1) cos 2L L x0 200 cos( 2 n 1) 2 1 cos( 2 n 1) 2 1 ( 2 n 1) ( 2 n 1) 200 é 1 1 ù = ê (2n + 1) + (2n - 1) ú p ë û 800 n ( n 1,2,3,..) 4n 1 2 np x L And 2 x ò 1 0 0 (1 - L ) sin L d x L 0 L ì np x np x ü ï co s sin ï 200 ï x L )- 1 L ï = í (1 - )( - ý L ï L np L n 2p 2 ï ï î L L 2 ï x=0 þ 200 L 200 = = L np np Thus w( x, t ) n2p 2a2 x 200 æ 4n¥ 1 ö np x - 2 t = 100(1- ) + L åç 4n2 -1 - n ÷ sin L e p n=1 è ø L Problem The temperatures at the ends x = 0 and x = 100 of a rod 100 cm in length with insulated sides a re held at 0 and 100 degree centigrade respectiv ely, until steady-state conditions prevail. Then a t the instant t = 0, the temperatures of the two e nds are interchanged. Find the resultant tempera ture distribution of the rod as a function of x an d t. Solution We note that the steady state temperature (w hen the ends x = 0 and x = 100 of the rod 10 0 cm in length are held at 0 and 100 degree centigrade respectively) is f ( x ) = x , 0 < x < 100 Now, the temperatures of the two ends are i nterchanged. And we have to find the new temperature di stribution in the rod. Hence we have to solve the heat equation wi th the boundary and initial conditions w (0, t ) = 100, w (100, t ) = 0, for all t. w ( x , 0) = x , 0xL We write the temperature distribution as w( x, t ) = wS ( x) + wT ( x, t ) = (new) steady state temperature + Transient temperature. We impose the boundary conditions wS(0 ) = 100, wS(100) = 0 on wS(x). Thus 1 ws ( x) = 100 + (0 - 100) x = 100 - x 0 £ x £ 100 100 We impose on wT(x,t) the boundary conditions wT(0,t) = wT(100,t) = 0 for all t and the initial condition wT ( x,0) = x - wS ( x) 0 £ x £ 100 i.e. wT ( x,0) = 2 x - 100 0 £ x £ 100 That is : wT(x, t) satisfies the boundary conditio ns wT(0,t)=wT(100,t) = 0 for all t and the initial condition wT ( x,0) = 2 x - 100 0 £ x £ 100 n 2p 2 a 2 t Thus ¥ np x - w T ( x , t ) = å b n s in e 1002 n =1 100 100 where b n = 2 np x 100 ò0 [ 2 x - 1 0 0 ] sin 100 dx é æ np x ö 2 ê ç co s 1 0 0 ÷ = ê(2 x - 100) ç - ÷ 100 ê ç np / 1 0 0 ÷ ê ë è ø 100 æ np x öù ç sin ÷ú + 2 ç 2 2 100 2 ÷ú ç n p /100 ÷ú è øú0 û 200 400 = (1 - co s np ) = , n odd np np Hence the desired temperature distribution is: w( x, t ) = wS ( x) + wT ( x, t ) = 100 - x (2 n-1)2 p 2a2t 400 ¥ 1 (2n -1)p x - + å (2n -1) sin 100 e p n=1 1002 END OF WAVE AND HE AT EQUATIONS TWO-DIMENSIONAL H EAT EQUATION - LAPLACE’S EQUATIO N The two-dimensional heat equation: A thin rectangular plate has its boundary kep t at temperatures 0,0,0 and f(x) at all times. F ind the steady temperature distribution in the plate. Thus we have to solve the Laplace’s equation w w 2 2 2 0 x 2 y subject to the boundary conditions w ( 0, y ) w ( L , y ) 0 0 yM w( x, M ) = 0, w( x,0) = f ( x) 0 £ x £ L We again solve by the method of separation of variables. We assume w(x,y) = X(x) Y(y) Substituting and separating the variables we get X Y (a constant) X Y Thus X satisfies the boundary value problem X ¢¢ + l X = 0, X(0) = X(L) = 0 We know nontrivial solutions exist when n p2 2 l = ln = 2 ( n = 1, 2, 3, ...) L and the corresponding eigenfunction is nx X n sin L For each n, Y satisfies the BV problem Y n Y 0 , Y ( M ) 0 ny ny Now Y B sinh C cosh L L np M np M Y ( M ) = 0 Þ B sinh + C cosh =0 L L np M np M Taking B = - cosh , C = + sinh L L we find np ( M - y ) Y = sinh L Thus for each n, np x np ( M - y ) wn = bn sin sinh L L is a solution satisfying the first three boundary conditions. Hence ¥ np x np ( M - y ) w = å wn = å bn sin sinh n =1 L L is also a solution. Putting y = 0, we get nM nx f ( x) bn sinh sin n 1 L L nM Hence bn sinh L = nth Fourier sine coefficient of f(x) in [0, L] L 2 np x = ò f ( x ) sin dx ( n = 1, 2,3,...) L0 L Thus we have found out the steady state temp erature distribution. THE END BEST OF LUCK IN T HE COMPRE