Classification of partial differential equations of second order: In the fields of wave propagation, heat conduction, vibrations, elasticity, boundary layer theory, etc., second order partial differential equations are of particular interest. The general form of a second order P.D.E. in the function u of the two independent variables x, y is given by ( ) ( ) ( ) 0 y u , x u , y , x f y u y , x C y x u y , x B x u y , x A 2 2 2 2 2 = | | ¹ | \ | ∂ ∂ ∂ ∂ + ∂ ∂ + ∂ ∂ ∂ + ∂ ∂ . (i) This equation is linear in second order terms. PDE (i) is said to be “linear or quasi-linear” according as f is linear or non-linear. PDE (i) is classified as Elliptic, Parabolic or Hyperbolic. • If 0 AC 4 B 2 < − , then PDE (i) is classified as elliptic. • If 0 AC 4 B 2 = − , then PDE (i) is classified as parabolic. • If 0 AC 4 B 2 > − , then PDE (i) is classified as hyperbolic. 2 22 2 nd nd nd nd Topic Topic Topic Topic Applications of Applications of Applications of Applications of Partial Differential Equations Partial Differential Equations Partial Differential Equations Partial Differential Equations Wave equation Vibration of a stretching string, solution of wave equation, D’Alembert’s solution of wave equation Prepared by: Dr. Sunil NIT Hamirpur (HP) (Last updated on 25-09-2007) Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 2 Examples: Elliptic: ( ) 0 AC 4 B 2 < − : Laplace’s equation in two dimensions: 0 y u x u 2 2 2 2 = ∂ ∂ + ∂ ∂ . Poisson’s equation: ( ) y , x f y u x u 2 2 2 2 = ∂ ∂ + ∂ ∂ . Parabolic: ( ) 0 AC 4 B 2 = − : One dimensional heat-flow equation: t u x u a 2 2 2 ∂ ∂ = ∂ ∂ . Hyperbolic: ( ) 0 AC 4 B 2 > − : One-dimensional wave equation: 2 2 2 2 2 x u a t u ∂ ∂ = ∂ ∂ . Note: Laplace’s equation, one dimensional heat-flow equation and one-dimensional wave equation are homogeneous, whereas Poisson’s equation is non- homogeneous. Vibrations of a stretched string, One-dimensional wave equation: The classical one-dimensional wave equation, which is hyperbolic, arises in the study of transverse vibrations of an elastic (flexible) string or torsional oscillations or longitudinal vibrations of a rod. Vibrating string: Consider a uniform elastic string of length l stretched tightly between two points O and A, and displayed slightly its equilibrium position OA. Taking the end O as origin, OA as x-axis and perpendicular line through O as the y-axis, we shall find the displacement y as a function of the distance x and time t. 2 2 2 2 2 x y c t y ∂ ∂ = ∂ ∂ Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 3 We shall obtain the equation of motion for the string under the following assumptions: 1. The motion takes place entirely in the xy-plane and each particle of the string moves perpendicular to the equilibrium position OA of the string 2. The string is perfectly flexible and offers no resistance to bending. 3. The tension in the string is so large that the forces due to weight of the string can be neglected. 4. The displacement y and the slope x y ∂ ∂ are small, so that their higher powers can be neglected. Let m be the mass per unit length of the string. Consider the motion of an element PQ of length s δ . Since the string does not offer resistance to bending (by assumption), the tensions T 1 and T 2 at P and Q, respectively are tangential to the curve. Since there is no motion in the horizontal direction, we have T cos T cos T 2 1 = β = α (constant). (i) Mass of element PQ is s mδ . Then, by Newton’s second law of motion, the equation of motion in the vertical direction is α − β = ∂ ∂ δ sin T sin T t y s m 1 2 2 2 α α − β β = ∂ ∂ δ ⇒ cos T sin T cos T sin T t y T s m 1 1 2 2 2 2 [by using (i)] ( ) α − β δ = ∂ ∂ ⇒ tan tan s m T t y 2 2 ( ¸ ( ¸ | ¹ | \ | ∂ ∂ − | ¹ | \ | ∂ ∂ δ = ∂ ∂ ⇒ δ + x x x 2 2 x y x y s m T t y . [Since x s δ = δ to a first approximation, and α tan and β tan are the slopes of the curve of the string at x and x x δ + ] P x-axis y-axis O Q A T 1 T 2 β α x y δx x+δx δs Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 4 2 2 x x x 2 2 x y m T x x y x y m T t y ∂ ∂ = ( ( ( ( ¸ ( ¸ δ | ¹ | \ | ∂ ∂ − | ¹ | \ | ∂ ∂ = ∂ ∂ ⇒ δ + , as 0 x → δ 2 2 2 2 2 x y c t y ∂ ∂ = ∂ ∂ ⇒ , where m T c 2 = . Thus, . This is the partial differential equation giving the transverse vibrations of the string. It is also called the one-dimensional wave equation or vibrations of a stretched string and y y (x, y) is called displacement function. Boundary conditions: The boundary conditions, satisfied by the equation 2 2 2 2 2 x y c t y ∂ ∂ = ∂ ∂ are: ( ¸ ( = = = = l when x 0 y (ii). 0 when x 0 y ). i ( . This means that ( ) ( ) ) ` ¹ = = 0 t , y 0 t , 0 y l . These should be satisfied for every value of t. Initial conditions: If the string is made to vibrate by pulling it into a curve y =f(x) and then releasing it, the initial conditions are: (i). ( ) x f y = , when t = 0, (ii). 0 t y = ∂ ∂ when t = 0. Solution of the one-dimensional wave equation by separation of variables: The wave equation is 2 2 2 2 2 x y c t y ∂ ∂ = ∂ ∂ . (i) Assume that y is separable, therefore let ( ) ( ) t T x X y = , (ii) be a solution of (i). Then T X t y 2 2 ′ ′ = ∂ ∂ and T X x y 2 2 ′ ′ = ∂ ∂ . 2 2 2 2 2 x y c t y ∂ ∂ = ∂ ∂ Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 5 Substituting in (i), we have T X c T X 2 ′ ′ = ′ ′ . Separating the variables, we get T T . c 1 X X 2 ′ ′ = ′ ′ . (iii) Now the LHS of (iii) is a function of x only and RHS is a function of t only. Since x and t are independent variables, this equation can hold only when both sides reduce to a constant say, k. Then, equation (iii) leads to the ordinary linear differential equations 0 kX X = − ′ ′ and 0 T kc T 2 = − ′ ′ . (iv) Solving (iv), we get (i). When k is positive and = p 2 , (say) px 2 px 1 e c e c X − + = and cpt 4 cpt 3 e c e c T − + = . (ii). When k is negative and 2 p − = , (say) px sin c px cos c X 2 1 + = and cpt sin c cpt cos c T 4 3 + = . (iii). When k = 0, 2 1 c x c X + = and 4 3 c t c T + = . Thus, the various solutions of the wave equation (i) are: • ( )( ) cpt 4 cpt 3 px 2 px 1 e c e c e c e c y − − + + = , • ( )( ) cpt sin c cpt cos c px sin c px cos c y 4 3 2 1 + + = , • ( )( ) 4 3 2 1 c t c c x c y + + = . Of these three solutions, we have to choose that solution which is consistent with the physical nature of the problem. Since, we are dealing with a problem on vibrations, y most be a periodic function of x and t. Therefore, the solution must involve trigonometric terms. Accordingly, ( )( ) cpt sin c cpt cos c px sin c px cos c y 4 3 2 1 + + = , (v) is only suitable solution of the wave equation and it corresponds to 2 p k − = . Case 1: If boundary conditions are given: Now applying boundary conditions that y = 0 when x = 0 and y = 0 when l = x Therefore ( ) cpt sin c cpt cos c c 0 4 3 1 + = (vi) Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 6 and ( )( ) cpt sin c cpt cos c p sin c p cos c 0 4 3 2 1 + + = l l . (vii) From (vi), we have 0 c 1 = . Then, equation (vii) reduce to ( ) 0 cpt sin c cpt cos c p sin c 4 3 2 = + l , which is satisfied when l l l π = ⇒ π = ⇒ = n p n p 0 p sin , where n = 1, 2, 3,…… ∴ A solution of the wave equation satisfying the boundary condition is l l l x n sin ct n sin c ct n cos c c y 4 3 2 π | ¹ | \ | π + π = . On replacing c 2 c 3 by a n and c 2 c 4 by b n , we get l l l x n sin ct n sin b ct n cos a y n n π | ¹ | \ | π + π = . Adding up the solutions for different values of n, we get l l l x n sin ct n sin b ct n cos a y n n 1 n π | ¹ | \ | π + π = ∑ ∞ = , (viii) is also a solution. Case 2: If boundary conditions and initial conditions are given: Now applying the initial conditions. y = f(x) and 0 t y = ∂ ∂ , when t = 0, then from (viii), we have ( ) l x n sin a x f n 1 n π = ∑ ∞ = (ix) and l l x n sin b c n 0 n 1 n π π = ∑ ∞ = . (x) Since equation (ix) represents Fourier series for f(x), we have ( ) dx x n sin x f 2 a 0 n l l l π = ∫ . (xi) From (x), we get b n = 0, for all n. Hence, (viii) reduces to | ¹ | \ | π π = ∑ ∞ = l l x n sin ct n cos a y n 1 n , (xii) Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 7 where ( ) dx x n sin x f 2 a 0 n l l l π = ∫ and ( ) ( ) x f 0 , x y = . D’Alembert’s solution of the one-dimension wave equation: The wave equation is 2 2 2 2 2 x y c t y ∂ ∂ = ∂ ∂ . (i) Let us introduce new independent variables ct x v , ct x u − = + = , so that y becomes a function of u and v. Then v y u y x y ∂ ∂ + ∂ ∂ = ∂ ∂ and 2 2 2 2 2 2 2 v y v u y 2 u y v y u y v v y u y u v y u y x x y ∂ ∂ + ∂ ∂ ∂ + ∂ ∂ = | ¹ | \ | ∂ ∂ + ∂ ∂ ∂ ∂ + | ¹ | \ | ∂ ∂ + ∂ ∂ ∂ ∂ = | ¹ | \ | ∂ ∂ + ∂ ∂ ∂ ∂ = ∂ ∂ Similarly, | | ¹ | \ | ∂ ∂ + ∂ ∂ ∂ − ∂ ∂ = ∂ ∂ 2 2 2 2 2 2 2 2 v y v u y 2 u y c t y . Substituting in (i), we get 0 v u y 2 = ∂ ∂ ∂ . (ii) Integrating (ii) w.r.t. v, we get ) u ( f u y = ∂ ∂ , (iii) where f(u) is an arbitrary function of u. Since the integral is a function of u alone, we may denote it by ( ) u φ . Thus ( ) ( ) v u y ψ + φ = ( ) ( ) ct x ct x ) t , x ( y − ψ + + φ = ⇒ . (iv) This is the general solution of (i). Now to determine φ and ψ, suppose u(x, 0) = f(x) and ( ) 0 t 0 , x y = ∂ ∂ . Differentiating (iv) w.r.t. t, we get ( ) ( ) ct x c ct x c t y − ψ′ − + φ′ = ∂ ∂ At t = 0, ( ) ( ) x x ψ′ = φ′ (v) and ( ) ( ) ( ) ( ) x f x x 0 , x y = ψ + φ = (vi) (v) gives, ( ) ( ) k x x + ψ = φ ∴ (vi) becomes ( ) ( ) ( ) x f k x 2 0 , x y = + ψ = Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 8 ( ) ( ) [ ] k x f 2 1 x − = ψ ⇒ and ( ) ( ) [ ] k x f 2 1 x + = φ . Hence the solution of (iv) takes the form ( ) ( ) [ ] ( ) [ ] ( ) ( ) ct x f ct x f k ct x f 2 1 k ct x f 2 1 t , x y − + + = − − + + + = , (vii) which is D’Alembert’s solution of the wave equation (i). ********************** FINAL CONCLUSIONS 1. The solution of one-dimensional wave equation 2 2 2 2 2 x y c t y ∂ ∂ = ∂ ∂ satisfying the boundary conditions ( ) ( ) ) ` ¹ = = 0 t , y 0 t , 0 y l is given by ( ) l l l x n sin ct n sin b ct n cos a t , x y y n n 1 n π | ¹ | \ | π + π = = ∑ ∞ = . 2. The solution of one-dimensional wave equation 2 2 2 2 2 x y c t y ∂ ∂ = ∂ ∂ satisfying the boundary conditions ( ) ( ) ) ` ¹ = = 0 t , y 0 t , 0 y l and the initial conditions ( ) ( ) ¦ ) ¦ ` ¹ = | ¹ | \ | ∂ ∂ = = 0 t y x f 0 , x y 0 t is given by ( ) | ¹ | \ | π π = = ∑ ∞ = l l x n sin ct n cos a t , x y y n 1 n , where ( ) dx x n sin x f 2 a 0 n l l l π = ∫ and ( ) ( ) x f 0 , x y = . ***************************************** Now let us solve some problems related to one-dimensional wave equation: Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 9 Q.No.1.: A string is stretched and fastened to two points l apart. Motion is started by displacing the string in the form l x sin a y π = from which it is released at time t = 0. Show that the displacement of any point at a distance x from one end at time t is given by ( ) l l ct cos x sin a t , x y π π = . Sol.: The given is the problem of vibration of a stretched string, therefore we find the solution of 2 2 2 2 2 x y c t y ∂ ∂ = ∂ ∂ . (i) Boundary conditions: As the end points of the string are fixed, for all time, therefore the boundary conditions are ( ) ( ) 0 t , y t , 0 y = = l . (ii) Initial conditions: Since the initial transverse velocity of any point of the string is zero, therefore, the initial conditions are ( ) l x sin a 0 , x y π = and 0 t y = ∂ ∂ , when t = 0. (iii) Now we have to solve (i) subject to the boundary conditions (ii) and initial conditions (iii). Then ( ) l l x n sin ct n cos a t , x y n 1 n π π = ∑ ∞ = , where ( ) dx x n sin 0 , x y 2 a 0 n l l l π = ∫ dx x n sin x sin a 2 0 l l l l π | ¹ | \ | π = ∫ dx x n sin x sin a 2 0 l l l l π π = ∫ , which vanishes for all values of n except n = 1. dx x sin a 2 a 2 0 1 l l l π = ∴ ∫ dx x 2 cos 1 a 0 | ¹ | \ | π − = ∫ l l l a x 2 sin 2 x a 0 = ( ¸ ( ¸ π π − = l l l l . Hence, the required solution is ( ) l l ct cos x sin a t , x y π π = . Q.No.2.: The points of trisection of a string are pulled aside through the same distance on opposite sides of the position of equilibrium and string is released from rest. Derive an expression for the displacement of the string at subsequent time and show that the mid-point of the string always remains at rest. Sol.: Let B and C be the points of trisection of the string OA of length l , say. Initially the string is held in the form A C B O ′ ′ , where a C C B B = ′ = ′ (say) Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 10 The equation of B O ′ is x a 3 y l = . The equation of C B ′ ′ is | ¹ | \ | − − = − 3 x 3 a 2 a y l l i.e. ( ) x 2 a 3 y − = l l . The equation of A C′ is ( ) l l − = x a 3 y . Here the boundary conditions are ( ) ( ) 0 t , y t , 0 y = = l . And the initial conditions are ( ) ( ) ( ) x 3 2 , x a 3 3 2 x 3 1 , x 2 a 3 3 1 x 0 , x a 3 t , 0 y ¦ ¦ ¦ ¹ ¦ ¦ ¦ ´ ¦ ≤ ≤ − ≤ ≤ − ≤ ≤ = l l l l l l l l and 0 t y = ∂ ∂ when t = 0. We have ( ) l l x n sin ct n cos a t , x y n 1 n π π = ∑ ∞ = , where ( ) dx x n sin 0 , x y 2 a 0 n l l l π = ∫ ( ) ( ) ( ¸ ( ¸ π − + π − + π = ∫ ∫ ∫ dx x n sin x a 3 dx x n sin x 2 a 3 dx x n sin ax 3 2 3 / 2 3 / 2 3 / 3 / 0 l l l l l l l l l l l l l l ¸ | | ¹ | \ | π π − − | ¹ | \ | π π − = 3 / 0 2 2 2 2 x n sin n . 1 x n cos n x a 6 l l l l l l ( ) ( ) 3 / 2 3 / 2 2 2 x n sin n . 2 x n cos n x 2 l l l l l l l | | ¹ | \ | π π − − − | ¹ | \ | π π − − + x-axis y-axis O B C A a a | ¹ | \ | ′ a , 3 B l | ¹ | \ | − ′ a , 3 2 C l ( ) 0 , l Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 11 ( ) ( ( ( ¸ ( | | ¹ | \ | π π − − | ¹ | \ | π π − − + l l l l l l l 3 / 2 2 2 2 x n sin n . 1 x n cos n x \ | π π + π π − π π ¸ + | | ¹ | \ | π π + π π − = 3 n cos n 3 3 n 2 sin n 2 3 n 2 cos n 3 3 n sin n 3 n cos n 3 a 6 2 2 2 2 2 2 2 2 2 2 l l l l l l ( ( ¸ ( | | ¹ | \ | π π + π π − | | ¹ | π π + 3 n 2 sin n 3 n 2 cos n 3 3 n sin n 2 2 2 2 2 2 2 2 l l l ( ) [ ] n 2 2 2 2 2 2 1 1 3 n sin n a 18 3 n 2 sin 3 n sin n 3 . a 6 − + π π = | ¹ | \ | π − π π = l l ( ) ( ) ( ¸ ( ¸ π − − = π − − = π π − π π = | ¹ | \ | π − π = π 3 n sin 1 3 n sin 1 0 3 n sin n cos 3 n cos n sin 3 n n sin 3 2n sin Since n n 0 a n = ∴ , when n is odd. 3 n sin n a 36 2 2 π π = , when n is even. i.e. 3 m 2 sin m a 9 2 2 π π , taking n = 2m Hence, ( ) l l x m 2 sin ct m 2 cos 3 m 2 sin m 1 a 9 t , x y 2 1 m 2 π π π π = ∑ ∞ = . Also 0 m sin ct m 2 cos 3 m 2 sin m 1 a 9 t , 2 y 2 1 m 2 = π π π π = | ¹ | \ | ∑ ∞ = l l , since 0 m sin = π . ⇒The displacement of the mid-point of the string is zero for all values of t. Thus, the mid-point of the string is always at rest. Q.No.3.: A tightly stretched string with fixed end points x = 0 and l = x is initially at rest in its equilibrium position. If it is set vibrating by giving to each of its points a velocity ( ) x x − λ l , find the displacement of the string at any distance x from one end at any time t. Sol.: Here the boundary conditions are ( ) ( ) 0 t , y t , 0 y = = l . ( ) l l l x n sin ct n sin b ct n cos a t , x y n n 1 n π | ¹ | \ | π + π = ∑ ∞ = . (i) Since the string was at rest initially, y(x, 0) = 0. (Initial condition) Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 12 ∴ From (i), we have 0 a x n sin a 0 n n 1 n = ⇒ π = ∑ ∞ = l . Thus, ( ) l l x n sin ct n sin b t , x y n 0 n π π = ∑ ∞ = (ii) and l l l l l l x n sin ct n cos nb c x n sin ct n cos b c n t y n 1 n n 1 n π π π = π π π = ∂ ∂ ∑ ∑ ∞ = ∞ = But ( ) x x t y − λ = ∂ ∂ l when t = 0. (Initial condition) ( ) l l l l l x n sin nb c x n sin nb c x x n 1 n n 1 n π | ¹ | \ | π = π π = − λ ∴ ∑ ∑ ∞ = ∞ = , where ( ) dx x n sin x x 2 nb c 0 n l l l l l π − λ = π ∫ ( ) ( ) ( ) l l l l l l l l l l 0 3 3 3 2 2 2 x n cos n 2 x n sin n x 2 x n cos n x x 2 ( ( ¸ ( ¸ | | ¹ | \ | π π − + | | ¹ | \ | π π − − − | ¹ | \ | π π − − λ = ( ) ( ) [ ] n 3 3 2 3 3 2 1 1 n 4 n cos 1 n 4 − − π λ = π − π λ = l l ¦ ¹ ¦ ´ ¦ π λ = odd is n when , n 8 even is n when , 0 3 3 2 l = ( ) 3 3 2 1 m 2 8 − π λl , taking 1 m 2 n − = . ( ) 4 4 3 n 1 m 2 c 8 b − π λ = ⇒ l . ∴ From ( ) l l x n sin ct n sin b t , x y n 0 n π π = ∑ ∞ = , the required general solution is ( ) ( ) ( ) ( ) l l l x 1 m 2 sin ct 1 m 2 sin 1 m 2 1 c 8 t , x y 4 1 m 4 3 π − π − − π λ = ∑ ∞ = . Ans. Q.No.4.: A tightly stretched string of length l with fixed ends is initially in equilibrium position. It is set vibrating by giving each point a velocity l x sin v 3 0 π . Find the displacement y(x, t). Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 13 or A string of length l is initially at rest in equilibrium position and each of its points is given the velocity l x sin b t y 3 0 t π = | ¹ | \ | ∂ ∂ = . Find the displacement y(x, t). Sol.: The given is the problem of a vibrating stretched string. So consider one-dimensional wave equation 2 2 2 2 2 x y c t y ∂ ∂ = ∂ ∂ , (i) where ( ) ( ) ) ` ¹ = = 0 t , y 0 t , 0 y l . Boundary conditions (ii) and ( ) ¦ ) ¦ ` ¹ π = | ¹ | \ | ∂ ∂ = = l x sin b t y 0 0 , x y 3 0 t . Initial condition (iii) Now, we have to solve (i) subject to the boundary conditions (ii) and initial conditions (iii). Now the solution of (i) satisfying (ii) is given by ( ) l l l x n sin ct n sin b ct n cos a t , x y n n 1 n π | ¹ | \ | π + π = ∑ ∞ = . (iv) Now ( ) l l x n sin ct n cos a t , x y n 1 n π π = ∑ ∞ = [from (iii)] n 0 a n ∀ = ⇒ . Hence, (iv) reduces to ( ) l l x n sin ct n sin b t , x y n 1 n π π = ∑ ∞ = . (v) To find b n : Differentiating (v) w.r.t. t, partially, we have l l l x n sin ct n cos . c n . b t y n 1 n π π π = ∂ ∂ ∑ ∞ = l l l x sin b x n sin . c n . b t y 3 n 1 n 0 t π = π π = | ¹ | \ | ∂ ∂ ⇒ ∑ ∞ = = [from (iii)] ( ) l l l x n sin nb x sin c b n 1 n 3 π = π π ⇒ ∑ ∞ = , which is a half-range sine series in ( ) l , 0 , hence Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 14 dx x n sin x sin . c b 2 nb 3 0 n l l l l l π π π = ∫ dx x n sin 4 x 3 sin x sin 3 cn b 2 b 0 n l l l l π π − π π = ⇒ ∫ [ A sin 4 A sin 3 A 3 sin 3 − = Q where we take l x A π = ] dx x 3 sin x n sin x sin x n sin 3 cn 2 b b 0 n | ¹ | \ | π π − π π π = ⇒ ∫ l l l l l (vi) dx x 3 sin x n sin 2 cn 4 b dx x sin x n sin 2 cn 4 b 3 0 0 l l l l l l π π π − π π π = ∫ ∫ ( ) ( ) dx x 1 n cos x 1 n cos cn 4 b 3 0 | ¹ | \ | π + − π − π = ∫ l l l ( ) ( ) dx x 3 n cos x 3 n cos cn 4 b 0 | ¹ | \ | π + − π − π − ∫ l l l ( ) ( ) [ ] B A cos B A cos B sin A sin 2 + − − = Q ( ) ( ) ( ) ( ) l l l l l 0 n 1 n x 1 n sin 1 n x 1 n sin cn 4 b 3 b ( ( ( ( ¸ ( ¸ π + π + − π − π − π = ⇒ ( ) ( ) ( ) ( ) l l l l l 0 3 n x 3 n sin 3 n x 3 n sin cn 4 b ( ( ( ( ¸ ( ¸ π + π + − π − π − π − n 0∀ = except n = 1, n = 3. ( Z n 0 n sin ∈ ∀ = π Also n varies fro 1 to ∞ 3 , 0 n − ≠ ∴ . Now for n = 1, dx x 3 sin x sin x sin 3 c 2 b b 2 0 1 | ¹ | \ | π π − π π = ∫ l l l l dx sin x 3 sin 2 2 1 2 x 2 cos 1 3 c 2 b 0 ( ( ( ( ¸ ( ¸ | ¹ | \ | π π − | | | | ¹ | \ | π − π = ∫ l l l l ( ( ¸ ( ¸ | ¹ | \ | π + π − π − π = ∫ dx x 4 cos x 2 cos x 2 cos 3 3 c 4 b 0 l l l l ( ) ( ) [ ] B A cos B A cos B sin A sin 2 + − − = Q ( ( ( ( ¸ ( ¸ π π + π π − π π − π = l l l l l l 4 x 4 sin 2 x 2 sin 2 x 2 sin 3 x 3 c 4 b [ ] 0 3 c 4 b − π = l [ ] Z n 0 n sin = ∀ = π Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 15 c 4 b 3 π = l . Again from (vi), we have dx x 3 sin x sin x 3 sin 3 c 6 b b 2 0 3 | ¹ | \ | π − π π π = ∫ l l l l dx x 3 sin 2 x sin x 3 sin 2 3 c 12 b 2 0 ( ¸ ( ¸ π − | ¹ | \ | π π π = ∫ l l l l ( ( ¸ ( ¸ | ¹ | \ | π − − | ¹ | \ | π − π π = ∫ dx x 6 cos 1 x 4 cos x 2 cos 3 c 12 b 0 l l l l ( ) ( ) [ ] B A cos B A cos B sin A sin 2 + − − = Q l l l l l l l 0 6 x 6 sin x 4 x 4 sin 3 2 x 2 sin 3 c 12 b ( ( ( ( ¸ ( ¸ π π + − π π − π π π = c 12 b π − = l . [ ] Z n 0 n sin = ∀ = π Hence, from (v), the required solution is given by ( ) l l x n sin ct n sin b t , x y n 1 n π π = ∑ ∞ = ........ 0 0 x 3 sin ct 3 sin b x sin ct sin b 3 1 + + + π π + π π = l l l l l l l l l l x 3 sin ct 3 sin c 12 b x sin ct sin c 4 b 3 π π | ¹ | \ | π − + π π π = [ n 0 b n ∀ = ∴ except n = 1, 3] ( ¸ ( ¸ π π − π π π = l l l l l x 3 sin ct 3 sin x sin ct sin 9 c 12 b . Ans. Q.No.5.:Find the deflection y(x, t) of the vibrating string of length π and ends fixed, corresponding to zero initial velocity and initial deflection ( ) ( ) x 2 sin x sin k x f − = , given 1 c 2 = Sol.: We know that the partial differential equation of the vibrating string is giving by 2 2 2 2 2 2 2 x y x y c t y ∂ ∂ = ∂ ∂ = ∂ ∂ , π < < x 0 [ ] 1 c 2 = (i) Also the boundary conditions are ( ) ( ) ) ` ¹ = π = 0 t , , y 0 t , 0 y (ii) and the initial conditions are ( ) ¦ ) ¦ ` ¹ = | ¹ | \ | ∂ ∂ − = = 0 t y x 2 sin x sin k ) 0 , x ( y 0 t . (iii) Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 16 Now, we have to solve (i) subject to the boundary conditions (ii) and initial conditions (iii). The required solution of (i) satisfying (ii) and 0 t y 0 t = | ¹ | \ | ∂ ∂ = is given by ( ) π π π π = = ∑ ∞ = x n sin t n cos a t , x y y n 1 n nx sin nt cos a n 1 n ∑ ∞ = = . (iv) Now using ( ) ( ) x 2 sin x sin k 0 , x y − = , (iv) gives ( ) x 2 sin x sin k nx sin a n 1 n − = ∑ ∞ = , which is half-range Fourier sine series in ( ) π , 0 and hence, ( ) nxdx sin x 2 sin x sin k 2 a 0 n − π = ∫ π (v) ( ) ( ) [ ]dx nx sin x 2 sin 2 nx sin x sin 2 k 0 − π = ∫ π ( ) ( ) ( ) ( ) ( ) [ ] x 2 n cos x 2 n cos x 1 n cos x 1 n cos k 0 + + − − + − − π = ∫ π [ ] ) B A cos( ) B A cos( B sin A sin 2 − − − = Q ( ) ( ) ( ) ( ) π ( ¸ ( ¸ + + + − − − + + − − − π = 0 2 n x 2 n sin 2 n x 2 n sin 1 n x 1 n sin 1 n x 1 n sin k [ ] Z n 0 n sin ∈ ∀ = π Q n 0∀ = except n = 1, 2 (As n varies from 1 to ∞ 2 n − ≠ ∴ ) Also n = 1, (v) gives ( ) xdx sin x 2 sin x sin k 2 a 0 1 − π = ∫ π ( )dx x sin x 2 sin 2 x sin 2 k 2 0 − π = ∫ π ( )dx x 3 cos x cos x 2 cos 1 k 0 + − − π = ∫ π π ( ¸ ( ¸ + − − π = 0 3 x 3 sin x sin 2 x 2 sin x k [ ] k 0 0 0 k = + − − π π = . [ ] Z n 0 n sin ∈ ∀ = π Q For n = 2, (v) gives Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 17 ( ) xdx 2 sin x 2 sin x sin k 2 a 0 2 − π = ∫ π ( )dx x 2 sin 2 x 2 sin x sin 2 k 2 0 − π = ∫ π ( )dx x 4 cos 1 x 3 cos x cos k 0 + − − π = ∫ π [ ] A sin 2 1 A 2 cos 2 − = Q π ( ¸ ( ¸ + − − π = 0 4 x 4 sin x 3 x 3 sin x sin k [ ] k k = π − π = [ ] Z n 0 n sin ∈ ∀ = π Q Therefore, from (iv), we have ( ) nx sin nt cos a t , x y n 1 n ∑ ∞ = = = ......... x 2 sin t 2 cos a x sin t cos a 2 1 + + = x 2 sin t 2 cos k x sin t cos k − = [ ] 2 1, n except n 0 a n = ∀ = ( ) x 2 sin t 2 cos x sin t cos k y − = , is the required solution of (i). Q.No.6.: A tightly stretched flexible string has its ends fixed at x = 0 and l = x . At time t = 0, the string is given a shape defined by ( ) x x ) x ( f − µ = l , where µ is a constant, and then released. Find the displacement of any point x of the string at any time t > 0. Sol.: The given is the problem of vibration of a stretched string, therefore we find the solution of 2 2 2 2 2 x y c t y ∂ ∂ = ∂ ∂ . (i) Also the boundary conditions are ( ) ( ) ) ` ¹ = = 0 t , y 0 t , 0 y l (ii) and the initial conditions are ( ) ¦ ) ¦ ` ¹ = | ¹ | \ | ∂ ∂ − µ = = 0 t y x x ) 0 , x ( y 0 t l . (iii) Now we have to solve (i) subject to the boundary conditions (ii) and initial conditions (iii). Here the solution of (i) satisfying (ii) and (iii) is given by ( ) l l t n sin ct n cos a t , x y y n 1 n π π = = ∑ ∞ = , l < < x 0 , (iv) where ( ) dx x n sin x f 2 a n l l l l π = ∫ , ( ) ( ) ( ) x x x f 0 , x y − µ = = l . Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 18 Now ( ) dx x n sin x x 2 a II I 0 n l l l l π − µ = ∫ , Integrating by parts, we get ( ) ( ) ( ) l l l l l l l l l l 0 3 3 3 2 2 2 n n x n cos 2 n x n sin x 2 n x n cos . x x 2 a ( ( ( ( ( ¸ ( ¸ | | | | | ¹ | \ | π π − + | | | | | ¹ | \ | π π − − − | | | | ¹ | \ | π π − − µ = ( ( ¸ ( ¸ | | ¹ | \ | π − − | | ¹ | \ | π π − − µ = 3 3 3 3 3 3 n 2 0 n cos n 2 0 0 2 l l l (As n varies from 1 to ∞ 0 n ≠ ∴ ) ( ) π − π µ = n cos 1 n 2 . 2 3 3 3 l l ( ) ( ) 3 n 3 2 n 1 1 4 − − π µ = l ¦ ¹ ¦ ´ ¦ π µ = odd. is n , n 8 even is n , 0 3 3 2 l Hence, from (iv), the required solution is given by ( ) l l l x n sin ct n cos n 1 . 8 t , x y y 3 3 2 odd n π π π µ = = ∑ ∞ = ( ) ( ) ( ) l l l x 1 m 2 sin ct 1 m 2 cos 1 m 2 1 8 3 1 m 3 2 π − π − − π µ = ∑ ∞ = , Ans. where 1 m 2 n − = , odd. Q.No.7.: A string is stretched between the fixed points (0, 0) and ( ) 0 , l and released at rest from the initial deflection given by ( ) l l l l l l < < − = < < = x 2 when , x k 2 2 x 0 when , x k 2 ) x ( f . Find the deflection of the string at time t. Sol.: The given is the problem of a vibrating stretching string. Therefore, we consider one- dimensional wave equation 2 2 2 2 2 x y c t y ∂ ∂ = ∂ ∂ . (i) Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 19 Also given the boundary conditions are ( ) ( ) ) ` ¹ = π = 0 t , , y 0 t , 0 y (ii) and the initial conditions are ( ) ¦ ¦ ¦ ) ¦ ¦ ¦ ` ¹ ¦ ¦ ¹ ¦ ¦ ´ ¦ < < − < < = = | ¹ | \ | ∂ ∂ = l l l l l l x 2 , x 2k 2 x 0 , x 2k ) 0 , x ( y 0 t y 0 t . (iii) Now we have to solve (i) subject to the boundary conditions (ii) and initial conditions (iii). Here, the solution of (i) satisfying (ii) and (iii) is given by ( ) l l x n sin ct n cos a t , x y n 1 n π π = ∑ ∞ = , l < < x 0 , (iv) where ( ) dx x n sin x f 2 a 0 n l l l π = ∫ , y(x, 0) = f(x) ( ) ( ( ( ¸ ( ¸ π − + π = ⇒ ∫ ∫ II I 2 / II I 2 / 0 n dx x n sin x k 2 dx x n sin x k 2 2 a l l l l l l l l l Integrating by parts, we get ( ) 2 / 0 2 2 2 2 n n x n sin ) 1 ( n x k cos . x k 4 a l l l l l l ¦ ¦ ) ¦ ¦ ` ¹ ¦ ¦ ¹ ¦ ¦ ´ ¦ | | | | | ¹ | \ | π π − | | | | ¹ | \ | π π − = ⇒ ( ) 0 2 / 2 2 2 2 n x n sin ) 1 ( n x n cos . x k 4 l l l l l l l ¦ ¦ ) ¦ ¦ ` ¹ ¦ ¦ ¹ ¦ ¦ ´ ¦ | | | | | ¹ | \ | π π − − | | | | ¹ | \ | π π − − + ( ( ¸ ( ¸ | | ¹ | \ | π π + π π + + + | | ¹ | \ | π π + π π − = 2 n sin n 2 n cos n 2 0 0 2 n sin n 2 n cos n 2 k 4 2 2 2 2 2 2 2 2 2 l l l l l 2 n sin n k 8 2 n sin n 2 k 4 2 2 2 2 2 2 π π = ( ( ¸ ( ¸ π π = l l [ 0 n ≠ , as n varies from 1 to ∞] (v) Hence, from (iv) and (v), the required solution is given by l l x n sin ct n cos 2 n sin n 1 k 8 ) t , x ( y 2 1 n 2 π π π π = ∑ ∞ = . Q.No.8.: The ends of tightly stretched string of length l are fixed at x = 0 and l = x . The Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 20 string is at rest with the point x = b drawn aside through a small distance d and released at time t = 0. Show that ( ) ( ) l l l l l ct n cos b n sin b n sin n 1 b b d 2 t , x y 2 1 n 2 2 π π π − π = ∑ ∞ = . Sol.: The given is the problem of a vibrating stretched string of length l . So consider 2 2 2 2 2 x y c t y ∂ ∂ = ∂ ∂ . (i) Also the boundary conditions are ( ) ( ) ) ` ¹ = = 0 t , y 0 t , 0 y l (ii) and the initial conditions are ) ` ¹ = | ¹ | \ | ∂ ∂ = 0 t y 0 t . (iii) Now we have to solve (i) subject to the boundary conditions (ii) and initial conditions (iii). To find y(x, 0), equation of OA is given by ( ) x b d 0 x 0 b 0 d 0 y = − − − = − . Also, equation of AB is given by ( ) l l − − − = − x b 0 d 0 y ( ) l l − − = x b d . Hence, ( ) ¦ ¦ ¹ ¦ ¦ ´ ¦ ≤ ≤ − ≤ ≤ = l l l x b , x - b d b x 0 , x b d ) 0 , x ( y (iv) Therefore, the solution of (i), satisfying (ii), (iii) and (iv) is given by y-axis x-axis A(b, d) O(0, 0) d b B L Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 21 l l x n sin ct n cos a ) t , x ( y n 1 n π π = ∑ ∞ = , l < < x 0 , where ( ) dx x n sin x f 2 a 0 n l l l π = ∫ , and y(x, 0) = f(x). ( ) ( ( ( ¸ ( ¸ π − − + π = ⇒ ∫ ∫ II I b II I b 0 n dx x n sin x b d dx x n sin x b d 2 a l l l l l l Integrating by parts, we get ( ) b 0 2 2 2 n n x n sin ) 1 ( n x n cos . x b d 2 a ¦ ¦ ) ¦ ¦ ` ¹ ¦ ¦ ¹ ¦ ¦ ´ ¦ | | | | | ¹ | \ | π π − | | | | ¹ | \ | π π − = ⇒ l l l l l ( ) ( ) l l l l l l l l 0 2 2 2 n x n sin ) 1 ( n x n cos . x b d 2 ¦ ¦ ) ¦ ¦ ` ¹ ¦ ¦ ¹ ¦ ¦ ´ ¦ | | | | | ¹ | \ | π π − | | | | ¹ | \ | π π − − − + | | ¹ | \ | − π π − π π − = 0 b n sin n b n cos n b b d 2 2 2 2 l l l l l ( ) ( ) | | ¹ | \ | π π + π π − + + − + l l l l l l l b n sin n b n cos n b 0 0 b d 2 2 2 2 ( ) l l l l l l l b n sin b n d 2 b n cos n d 2 b n sin bn d 2 b n cos n d 2 2 2 2 2 π − π + π π + π π − π π − = ( ) l l l l l l l b n sin b b b b n d 2 b n sin b 1 b 1 n d 2 2 2 2 2 π | | ¹ | \ | − − − π − = π | ¹ | \ | − − π − = ( ) ( ) l l l l l l b n sin n 1 . b b d 2 b n sin n b b d 2 2 2 2 2 2 2 π − π = π π − = . (v) Therefore, from (iv) and (v), the required solution is given by ( ) l l x n sin ct n cos a t , x y n 1 n π π = ∑ ∞ = ( ) l l l l l x n sin ct n cos b n sin n 1 b b d 2 2 1 n 2 2 π π π − π = ∑ ∞ = . Q.No.9.: A tightly stretched string l and fixed at the both ends is plucked at 3 x l = and assumes initially the shape of a triangle of height h. Find the displacement y(x, t) after the string is released from rest. Sol.: The given is the problem of a one dimensional wave equation Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 22 2 2 2 2 2 x y c t y ∂ ∂ = ∂ ∂ . (i) Also the boundary conditions are ( ) ( ) ) ` ¹ = = 0 t , y 0 t , 0 y l (ii) and the initial conditions are ) ` ¹ = | ¹ | \ | ∂ ∂ = 0 t y 0 t . (iii) Now we have to solve (i) subject to the boundary conditions (ii) and initial conditions (iii). We find y(x, 0). Now equation of OB is given by ( ) 0 x 0 3 0 h 0 y − − − = − l l hx 3 y = ⇒ , 3 x 0 l ≤ ≤ Also equation of AB is given by ( ) l l l − − − = − x 3 0 h 0 y ( ) l l − − = ⇒ x 2 h 3 y , l l ≤ ≤ x 3 Hence, ( ) ¦ ¦ ¹ ¦ ¦ ´ ¦ ≤ ≤ − ≤ ≤ = l l l l l l x b 3 , x 2 3h - 3 x 0 , x h 3 ) 0 , x ( y (iv) Now, the solution of (i) satisfying (ii) and (iii) is given by ( ) l l x n sin ct n cos a t , x y y n 1 n π π = = ∑ ∞ = , l < < x 0 (v) 3 / l y-axis x-axis B( ) h , 3 / l O(0, 0) h A L 3 / 2l ( ) 0 , l Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 23 where ( ) dx x n sin x f 2 a 0 n l l l π = ∫ , y(x, 0) = f(x) Now ( ) ( ( ( ¸ ( ¸ π − − + π = ∫ ∫ dx x n sin x 2 h 3 dx x n sin hx 3 2 a II I 3 / II I 3 / 0 n l l l l l l l l l Integrating by parts, we get ( ) 3 / 0 2 2 2 2 n n x n sin 1 n x n cos x h 6 a l l l l l l ( ( ( ( ( ¸ ( ¸ π | ¹ | \ | π − − π | ¹ | \ | π − = ( ) ( ) l l l l l l l l 3 / 2 2 2 2 n x n sin 1 n x n cos x h 3 ( ( ( ( ( ¸ ( ¸ π | ¹ | \ | π − − | | | | ¹ | \ | π π − − − ( ( ¸ ( ¸ π π + π π − = 3 n sin . n 3 n cos . n . 3 h 6 2 2 2 2 l l l l ( ( ¸ ( ¸ π π − π π − − 3 n sin n 3 n cos n 3 2 0 h 3 2 2 2 2 2 l l l 3 n sin n h 3 3 n cos n h 2 3 n sin n h 6 3 n cos n h 2 2 2 2 2 π π + π π + π π + π π − = 3 n sin n h 9 2 2 π π = , 0 n ≠ . (vi) Hence, from (v) and (vi), the required solution is given by l l x n sin ct n cos 3 n sin n 1 h 9 ) t , x ( y y 2 1 n 2 π π π π = = ∑ ∞ = . Ans. Q.No.10.: A tightly stretched string with fixed end points x = 0 and l = x is initially in a position given by | ¹ | \ | π = l x sin y y 3 0 . If it is released from rest from this position, find the displacement y(x, t). Sol.: The given is the problem of a one dimensional wave equation 2 2 2 2 2 x y c t y ∂ ∂ = ∂ ∂ . (i) Also, the boundary conditions are ( ) ( ) ) ` ¹ = = 0 t , y 0 t , 0 y l (ii) and the initial conditions are ) ` ¹ = | ¹ | \ | ∂ ∂ = 0 t y 0 t . (iii) Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 24 Now we have to solve (i) subject to the boundary conditions (ii) and initial conditions (iii). We know that the solution of (i), satisfying (ii) and (iii) is given by ( ) l l x n sin ct n cos a t , x y n 1 n π π = ∑ ∞ = , l < < x 0 , (*) where ( ) dx x n sin x f 2 a 0 n l l l π = ∫ , y(x, 0) = ) x ( f x sin y 3 0 = π l Now dx x n sin x sin y 2 a 3 0 0 n l l l l π π = ∫ dx x n sin 4 x 3 sin sin 3 y 2 0 0 l l l l l π π − π = ∫ ( ¸ ( ¸ π = − = l Q x A take we where A sin 4 A sin 3 A 3 sin 3 dx x n sin x 3 sin x n sin x sin 3 2 y a 0 0 n | ¹ | \ | π π − π π = ⇒ ∫ l l l l l l (iv) dx x 3 sin x n sin 2 4 y dx x sin x n sin 2 4 y 3 0 0 0 0 l l l l l l l l π π − π π = ∫ ∫ ( ) ( ) dx x 1 n cos x 1 n cos 4 y 3 0 0 ( ¸ ( ¸ π + − π − = ∫ l l l l ( ) ( ) dx x 3 n cos x 3 n cos 4 y 0 0 ( ¸ ( ¸ π + − π − − ∫ l l l l [ ] ) B A cos( ) B A cos( B sin a sin 2 + − − = Q ( ) ( ) ( ) ( ) l l l l l l 0 0 1 n x 1 n sin 1 n x 1 n sin 4 y 3 ( ( ( ( ¸ ( ¸ π + π + − π − π − = ( ) ( ) ( ) ( ) l l l l l l 0 0 3 n x 3 n sin 3 n x 3 n sin 4 y ( ( ( ( ¸ ( ¸ π + π + − π − π − − n 0 a n ∀ = ⇒ except n = 1, n = 3. Also n varies from 1 to ∞ 3 n − ≠ ∴ Now from (iv), we have dx x sin x 3 sin x sin x sin 3 2 y a 0 0 1 | ¹ | \ | π π − π π = ∫ l l l l l l dx x sin x 3 sin 2 2 1 . 2 y dx x sin 2 y 3 0 0 2 0 0 l l l l l l l π π − π = ∫ ∫ dx x 4 cos x 2 cos 4 y dx 2 x 2 cos 1 2 y 3 0 0 0 0 | ¹ | \ | π − π − | ¹ | \ | π − = ∫ ∫ l l l l l l l Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 25 [ ] ) B A cos( ) B A cos( B sin a sin 2 + − − = Q l l l l l l l l l l 0 0 0 0 4 x 4 sin 2 x 2 sin 4 y 2 x 2 sin x 4 y 3 π π − π π − π π − = ( ) 4 y 3 4 y 3 0 0 = = l l [ ] Z n 0 n sin ∈ ∀ = π Q Again from (iv), we have dx x 3 sin x 3 sin x 3 sin x sin 3 2 y a 0 0 3 | ¹ | \ | π π − π π = ∫ l l l l l l = dx x 3 sin 2 y dx x sin x 3 sin 2 4 y 3 2 0 0 0 0 l l l l l l l π − π π ∫ ∫ dx x 6 cos 1 4 y dx x 4 cos x 2 cos 4 y 3 0 0 0 0 | ¹ | \ | π − − | ¹ | \ | π − π = ∫ ∫ l l l l l l l ( ¸ ( ¸ π = − = l Q x 3 A take we where A sin 2 1 A 2 cos 2 l l l l l l l l l l 0 0 0 0 6 x 6 sin x 4 y 4 x 4 sin 2 x 2 sin 4 y 3 π π − − π π − π π = Hence from (*), we get ( ) l l x n sin ct n cos a t , x y y n 1 n π π = = ∑ ∞ = .......... 0 0 x 3 sin ct 3 cos a x sin ct cos a 3 1 + + + π π + π π = l l l l l l l l x 3 sin ct 3 cos 4 y x sin ct cos 4 y 3 0 0 π π − π π = [Since n 0 a n ∀ = except n = 1, 3] ( ¸ ( ¸ π π − π π = l l l l x 3 sin ct cos x sin ct cos 3 4 y 0 . Thus, | ¹ | \ | π π − π π = l l l l ct 3 cos x 3 sin ct cos x sin 3 4 y ) t , x ( y 0 . Ans. Q.No.11.: Solve the boundary value problem 2 2 2 2 x y 4 t y ∂ ∂ = ∂ ∂ , y(0, t) = y(5, t) = 0, y(x, 0) = 0, ( ) x f t y 0 t = | ¹ | \ | ∂ ∂ = , Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 26 where (i) x 5 sin 2 x 2 sin 3 ) x ( f π − π = , (ii) x sin 5 ) x ( f π = . Sol.: (i). We know that the solution of one dimensional wave equation 2 2 2 2 x y 4 t y ∂ ∂ = ∂ ∂ , (i) satisfying the boundary conditions y(0, t) = y(5, t) = 0 (ii) is given by 5 x n sin ct n sin b ct n cos a ) t , x ( y y n n 1 n π | ¹ | \ | π + π = = ∑ ∞ = l l , 0 < x < 5, where c 2 = 4 and 5 = l . Hence, we have 5 x n sin 5 t n sin b 5 t n cos a ) t , x ( y y n n 1 n π | ¹ | \ | π + π = = ∑ ∞ = (iii) To find a n : Use y(x, 0) = 0, (iii) gives ( ) 0 5 x n sin 0 0 cos a ) 0 , x ( y n 1 n = π + = ∑ ∞ = (given) n 0 a n ∀ = ⇒ . Hence (iii) reduces to 5 x n sin 5 t n 2 sin . b ) t , x ( y y n 1 n π π = = ∑ ∞ = (iv) To find b n : Differentiate (iv) w.r.t. t partially, we have 5 x n sin . 5 t n 2 cos . 5 n 2 . b t y n 1 n π π π = ∂ ∂ ∑ ∞ = x 5 sin 2 x 2 sin 3 5 x n sin b 5 n 2 t y n 1 n 0 t π − π = π π = | ¹ | \ | ∂ ∂ ⇒ ∑ ∞ = = , (given) which is a Fourier half-range sine series in (0, 5), and hence, for 0 < x < 5, we have ( ) dx 5 x n sin x 5 sin 2 x 2 sin 3 5 2 b 5 n 2 5 0 n π π − π = π ∫ dx 5 x n sin x 5 sin 2 5 x n sin x 2 sin 3 n 1 b 5 0 n | ¹ | \ | π π − π π π = ⇒ ∫ (v) Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 27 xdx 5 sin 5 x n sin 2 n 1 xdx 2 sin 5 x n sin 2 n 2 3 5 0 5 0 π π π − π π π = ∫ ∫ ( ) [ ] ) B A cos( B A cos B sin A sin 2 + − − = Q 5 0 5 0 dx x 2 5 n cos x 2 5 n cos n 2 3 ( ( ¸ ( ¸ ) ` ¹ ¹ ´ ¦ | ¹ | \ | π + π − | ¹ | \ | π − π π = ∫ 5 0 5 0 dx x 5 5 n cos x 5 5 n cos n 1 ( ( ¸ ( ¸ ) ` ¹ ¹ ´ ¦ | ¹ | \ | π + π − | ¹ | \ | π − π π − ∫ n 0∀ = except 0 2 5 n = π − π or n = 10 [As n varies from 1 to ∞, 25 n − ≠ ∴ ] and 25 n 0 5 5 n = ⇒ = π − π . [ ] Z n 0 n sin ∈ ∀ = π Now ( )dx x 2 sin x 5 sin 2 x 2 sin x 2 sin 3 10 1 b 5 0 10 π π − π π π = ∫ ( )dx x 7 cos x 3 cos dx 10 1 xdx 2 sin 2 2 1 . 10 3 5 0 2 5 0 π − π π − π π = ∫ ∫ ( ) ( )dx x 7 cos x 3 cos 10 1 dx x 4 cos 1 20 3 5 0 5 0 π − π π − π − π = ∫ ∫ [ ] x 2 A Take , A sin 2 1 A 2 cos 2 π = − = Q 5 0 5 0 7 x 7 sin 3 x 3 sin 10 1 4 x 4 sin x 20 3 ( ¸ ( ¸ π − π π − ( ¸ ( ¸ π π − π = ( ) π = π = 4 3 5 20 3 [ ] Z n 0 n sin ∈ ∀ = π Q . Also ( )dx x 5 sin x 5 sin 2 x 5 sin x 2 sin 3 25 1 b 5 0 25 π π − π π π = ∫ [Using (v)] xdx 5 sin dx 25 1 xdx 5 sin x 2 sin 2 2 1 . 25 3 2 5 0 5 0 π π − π π π = ∫ ∫ ( ) ( )dx x 10 cos 1 25 1 dx x 7 cos x 3 cos 50 3 5 0 5 0 π − π − π − π π = ∫ ∫ Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 28 ( ) [ ] B A cos ) B A cos( B sin A sin 2 + − − = Q 5 0 5 0 10 x 10 sin x 25 1 7 x 7 sin 3 x 3 sin 50 3 ( ¸ ( ¸ π − π − ( ¸ ( ¸ π π − π π π = π − = 5 1 [ ] Z n 0 n sin ∈ ∀ = π Therefore, from (iv), we have ( ) 5 x n sin 5 t n 2 sin b t , x y y n 1 n π π = = ∑ ∞ = ....... 0 0 x 5 sin t 10 sin b x 2 sin t 4 sin b 25 10 + + π π + π π = [ ] 25 10, n except n 0 b n = ∀ = Q x 5 sin t 10 sin 5 1 x 2 sin t 4 sin 4 3 y π π π − π π π = ⇒ . Ans. (ii). Proceed similarly up to equation (iv), we have 5 x n sin 5 t n 2 sin . b ) t , x ( y y n 1 n π π = = ∑ ∞ = (iv) To find b n : Differentiate (iv) w.r.t. t, partially, we have x 5 n sin . t 5 n 2 cos . 5 n 2 . b t y n 1 n π π π = ∂ ∂ ∑ ∞ = x sin 3 x 5 n sin . 5 n 2 b t y n 1 n 0 t π = π π = | ¹ | \ | ∂ ∂ ⇒ ∑ ∞ = = , (given) which is a Fourier half-range sine series for 0 < x < 5, we have ( ) xdx 5 n sin . x sin 5 5 2 b 5 n 2 5 0 n π π = π ∫ dx xdx 5 n sin . x sin 5 n 1 b 5 0 n | ¹ | \ | π π π = ⇒ ∫ (v) xdx sin x 5 n sin 2 n 2 5 5 0 π π π = ∫ dx nx 1 5 n cos nx 1 5 n cos n 2 5 5 0 ) ` ¹ ¹ ´ ¦ | ¹ | \ | + π − | ¹ | \ | − π π = ∫ ( ) [ ] ) B A cos( B A cos B sin A sin 2 + − − = Q Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 29 5 0 1 5 n x 1 5 n sin 1 5 n x 1 5 n sin n 2 5 ( ( ( ( ¸ ( ¸ π | ¹ | \ | + π | ¹ | \ | + − π | ¹ | \ | − π | ¹ | \ | − π = n 0∀ = except 5 n 0 1 5 n = ⇒ = π | ¹ | \ | − [ ] Z n 0 n sin ∈ ∀ = π Q 0 1 5 n ≠ π | ¹ | \ | + as n varies from 1 to ∞] Put n = 5, using (v), we have dx n sin x sin 5 5 1 b 5 0 5 π π π = ∫ dx 2 x 2 cos 1 1 xdx sin n 1 5 0 2 5 0 π − π = π = ∫ ∫ [ ] x sin 2 1 x 2 cos 2 − = Q π = π = π π − π = 2 5 5 2 1 2 x 2 sin x 2 1 5 0 [ ] Z n 0 n sin ∈ ∀ = π Q Hence, from (iv) the required solution is given by x 5 n sin t 5 n 2 sin . b ) t , x ( y y n 1 n π π = = ∑ ∞ = .......... 0 x sin t 2 sin b ...... 0 0 n + + π π + + + = [ n 0 b n ∀ = Q except n = 5] x sin t 2 sin 2 5 π π π = . Ans. Q.No.12.: Solve completely the equation, 2 2 2 2 2 x y c t y ∂ ∂ = ∂ ∂ , representing the vibrations of a string of length l , fixed at both ends, given that y(0, t) = 0; ( ) 0 t , y = l ; (Boundary conditions) ( ) ) x ( f 0 , x y = , ( ) l < < = ∂ ∂ x 0 , 0 0 , x y t . (Initial conditions) Sol.: One-dimensional wave equation is 2 2 2 2 2 x y c t y ∂ ∂ = ∂ ∂ , (given) (i) satisfying ( ) ( ) t , y 0 t , 0 y l = = , (ii) Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 30 ( ) l < < ¦ ) ¦ ` ¹ = | ¹ | \ | ∂ ∂ = = x 0 0 t y ) x ( f 0 , x y 0 t . (iii) We now apply the method of separation of variables to solve (i). Here y is dependent variable and x and t are independent variables, so we get XT y = (iv) to be the required solution of (i) Now T X t y 2 2 ′ ′ = ∂ ∂ , T X x y 2 2 ′ ′ = ∂ ∂ , where 2 2 dt T d T = ′ ′ and 2 2 dx X d X = ′ ′ . Substituting these values in (i), we get T X c T X 2 ′ ′ = ′ ′ . Separating the variables, we get T T c 1 X X 2 ′ ′ = ′ ′ . (v) As LHS of (v) is a function of x only and RHS is a function of t only, thus (v) can hold only when each side is equal to some constant, say k k T T c 1 X X 2 = ′ ′ = ′ ′ ∴ k X X = ′ ′ ⇒ k dx X d X 1 2 2 = ⇒ 0 kX dx X d 2 2 = − ⇒ ( ) 0 X k D 2 = − ⇒ , | ¹ | \ | = dx d D (vi) and k T T c 1 2 = ′ ′ k dt T d T c 1 2 2 2 = ⇒ ( ) 0 T kc D 2 2 = − ⇒ . (vii) Case I: When k = positive, say p 2 , then (vi) gives ( ) 0 X p D 2 2 = − . Its auxiliary equation is 0 p m 2 2 = − p m ± = ⇒ . ∴CF px 2 px 1 e c e c − + = . Also PI = 0. PI CF X + = ∴ px 2 px 1 e c e c X − + = ⇒ . (vii) gives ( ) 0 T kc D 2 2 = − . Its auxiliary equation is 0 c p m 2 2 2 = − pc m ± = ⇒ . ∴CF pct 4 pct 3 e c e c − + = . Also PI = 0. PI CF T + = ∴ pcct 4 pct 3 e c e c T − + = ⇒ Case 2: When 2 p k − = , then (vi) gives ( ) 0 X p D 2 2 = + . Its auxiliary equation is (vii) gives ( ) 0 T c p D 2 2 2 = + . Its auxiliary equation is Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 31 0 p m 2 2 = + pi m ± = ⇒ . ∴CF px sin c px cos c 2 1 + = . Also PI = 0. PI CF X + = ∴ px sin c px cos c X 2 1 + = ⇒ . 0 c p m 2 2 2 = + pci m ± = ⇒ . ∴CF pct sin c pct cos c 4 3 + = . Also PI = 0. PI CF T + = ∴ pct sin c pct cos c T 4 3 + = ⇒ . Case 3: When 0 k = , then (vi) gives 0 X D 2 = . Its auxiliary equation is 0 m 2 = 0 , 0 m = ⇒ . ∴CF ( ) x c c e x c c 2 1 x 0 2 1 + = + = . Also PI = 0. PI CF X + = ∴ x c c X 2 1 + = ⇒ . (vii) gives 0 T D 2 = . Its auxiliary equation is 0 m 2 = 0 , 0 m = ⇒ . ∴CF ( ) t c c e t c c 4 3 t 0 4 3 + = + = . Also PI = 0. PI CF T + = ∴ t c c T 4 3 + = ⇒ . Hence, from (iv), the various possible solutions of (i), are given by Y = XT ( )( ) pct 4 pct 3 px 2 px 1 e c e c e c e c y − − + + = ⇒ (viii) ( )( ) pct sin c pct cos c px sin c px cos c y 4 3 2 1 + + = ⇒ (ix) ( )( ) t c c x c c y 4 3 2 1 + + = ⇒ (x) Now,, out of these solutions, we must choose that solution which is consistent with the physical nature of the given problem. As we are dealing with the problem of a wave equation and since waves are periodic in nature therefore y must be a periodic function i.e., y must involve trigonometric identities. Hence from (ix), ) t , x ( y y = ( )( ) pct sin c pct cos c px sin c px cos c 4 3 2 1 + + = (xi) is the required solution where 2 p k − = Using the given condition (2), we have from (11) ( ) pct sin c pct cos c c 0 ) t , 0 ( y 4 3 1 + = = 0 c 1 = ⇒ . Also ( ) pct sin c pct cos c p sin c 0 ) t , ( y 4 3 2 + = = l l 0 p sin = l [ , 0 c 2 ≠ otherwise (xi) reduces to y = 0, which is not possible] π = ⇒ n pl l π = ⇒ n p , n = 0, ... ,......... 2 , 1 ± ± Hence (xi) reduces to Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 32 ( ) | ¹ | \ | π + π π = t c n sin c t c n cos c x n sin c t , x y 4 3 2 l l l x n sin t c n sin c c t c n cos c c 4 2 3 2 l l l π | ¹ | \ | π + π Replacing c 2 c 3 , by a n , c 2 c 4 by b n , and adding all solutions, we have ( ) x n sin t c n sin b t c n cos a t , x y n n 1 n l l l π | ¹ | \ | π + π = ∑ ∞ = . (xii) Further, we find a n and b n , by using (iii) Now, from (xii) x n sin . c n . t c n cos b t c n sin a t y n n 1 n l l l l π π ( ¸ ( ¸ π + | ¹ | \ | π − = ∂ ∂ ∑ ∞ = 0 x n sin . c n . b t y n 1 n 0 t = π π = | ¹ | \ | ∂ ∂ ⇒ ∑ ∞ = = l l n 0 b n ∀ = ⇒ Hence (xii) reduces to ( ) x n sin t c n cos a t , x y n 1 n l l π π = ∑ ∞ = (xiii) Finally, ( ) x n sin a ) x ( f 0 , x y n 1 n l π = ∑ ∞ = , Which is a Fourier half-range sine series for l < < x 0 , and hence xdx n sin ) x ( f 2 a 0 n l l l π = ∫ . (xiv) Hence from (xiii), the required solution of (i) is given by ( ) x n sin t c n cos a t , x y n 1 n l l π π = ∑ ∞ = where xdx n sin ) x ( f 2 a 0 n l l l π = ∫ . *** *** *** *** *** *** *** *** *** Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 33 Home Assignments Q.No.1.: Find the solution of the wave equation 2 2 2 2 2 x y c t y ∂ ∂ = ∂ ∂ , corresponding to the triangular initial deflection ( ) x k 2 x f l = when 2 x 0 l < < . ( ) x k 2 − = l l when l l < < x 2 , and initial velocity zero. Ans.: | | ¹ | \ | + π π − π π π = ..... ct 3 cos x 3 sin 3 1 ct cos x sin k 8 y 2 2 l l l l Q.No.2.: A tightly stretched string of length l has its ends fastened at x = 0, l = x . The mid point of the string is then taken to height h and then released from rest in that position. Find the lateral displacement of a point of the string at time t from the instant of release. Ans.: | | ¹ | \ | + π π − π π π = ..... ct 3 cos x 3 sin 3 1 ct cos x sin h 8 y 2 2 l l l l Q.No.3.: A tightly stretched string with fixed end points at x = 0 and x = 1, is initially in a position given by ( ) ¦ ¦ ¹ ¦ ¦ ´ ¦ ≤ ≤ − ≤ ≤ = 1 x 2 1 , x 1 2 1 x 0 x, x f It is released from this position with velocity a, perpendicular to the x-axis, show that the displacement u(x, t) at any point x of the string any time t > 0, given by ( ) ( ) [ ] ( ) ( ) ¦ ¦ ¹ ¦ ¦ ´ ¦ − ( ¸ ( ¸ π − π − π − ¸ π = ∑ ∞ = 2 1 n 2 3 n 4 4 at 3 n 4 cos x 3 n 4 sin 2 4 t , x u Applications of Partial Differential Equations: Wave Equation Prepared by: Dr. Sunil, NIT Hamirpur (HP) 34 ( ) [ ] ( ) ( ) ( ( ( ( ¸ ( ¦ ¦ ) ¦ ¦ ` ¹ − ( ¸ ( ¸ π + π − π − − 2 1 n 4 4 at 1 n 4 cos x 1 n 4 sin . Ans.: Q.No.4.: A taut string of length 20 cms. fastened at both ends is displaced from its position of equilibrium, by imparting to each of its points an initial velocity given by v = x in 10 x 0 ≤ ≤ x 20 v − = in 20 x 10 ≤ ≤ , x being the distance from one end . Determine the displacement at any subsequent time. Ans.: | | ¹ | \ | + π π − π π π = ..... 20 at 3 sin 20 x 3 sin 3 1 20 at sin 20 x sin a 1600 y 3 3 Q.No.5.: Show that the wave equation 2 2 2 2 2 x u c t u ∂ ∂ = ∂ ∂ , under the conditions u(0, t), 0 ) t , ( u = l for all t, u(x, t) = f(x), ( ) x g t u 0 t = | ¹ | \ | ∂ ∂ = has the solution of the form ( ) ( ) x n sin t sin C t cos B t , x u n n n n 1 n l π λ + λ = ∑ ∞ = , where dx x n sin ) x ( f 2 B 1 0 n l l π = ∫ , dx x n sin ) x ( g cn 2 C 1 0 n l π π = ∫ . Ans.: Q.No.6.: Using D’Alembert’s method, find the deflection of a vibrating string of unit length having fixed ends, with initial velocity zero and initial deflection: (i) ( ) ( ) 3 x x a x f − = , (ii) ( ) x sin a x f 2 π = . Ans.: (i). ( ) 2 2 2 t c 3 x 1 ax ) t , x ( y − − = , (ii). ( ) ct 2 cos x 2 cos 1 2 a ) t , x ( y π π − = . ************************************************ **************************************** ****************************