Partial Differential Equations lecture notes

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B Tech Mathematics III Lecture Note PARTIAL DIFFERENTIAL EQUATION A differential equation containing terms as partial derivatives is called a partial differential equation (PDE). The order of a PDE is the order of highest partial derivative. The dependent variable z depends on independent variables x and y. 2 2 2 zz z z z p = , q= , r= , s= , t= 2 2 x yxxyy For example: q + px = x + y is a PDE of order 1 2 s + t = x is a PDE of order 2 Formation of PDE by eliminating arbitrary constant: For f(x,y,z,a,b) = 0 differentiating w.r.to x,y partially and eliminating constants a,b we get a PDE 2 2 2 Example 1: From the equation x + y + z = 1 form a PDE by eliminating arbitrary constant. 2 2 2 Solution: z = 1 - x - y Differentiating w.r.to x,y partially respectively we get zz 2z 2x and 2z 2y xy z z p = = - x/z and q= = - y/z xy z = - x/p = -y/ q qx = py is required PDE Example 2 From the equation x/2 + y/3 + z/4 = 1 form a PDE by eliminating arbitrary constant. Solution: Differentiating w.r.to x,y partially respectively we get 1 1z 1 1z  0 and 0 2 4x 2 4y 1zz  0  4xy  B Tech Mathematics III Lecture Note z z p = = q = x y p = q is required PDE Formation of PDE by eliminating arbitrary function Let u= f(x,y,z), v= g(x,y,z) and ϕ(u,v) = 0 We shall eliminate ϕ and form a differential equation Example 3 From the equation z = f(3x-y)+ g(3x+y) form a PDE by eliminating arbitrary function. Solution: Differentiating w.r.to x,y partially respectively we get zz p 3f '(3x y) 3g'(3x y) and q f '(3x y)g'(3x y) xy 2 2  z z r 9f ''(3x y) 9g''(3x y) and t f ''(3x y)g''(3x y) 2 2 xy From above equations we get r = 9t which is the required PDE. 11.1 An equation involving atleast one partial derivatives of a function of 2 or more independent variable is called PDE. A PDE is linear if it is of first degree in the dependent variable and its partial derivatives. If each term of such an equation contains either dependent variable or one of its derivatives the equation is called homogeneous. Important Linear PDE of second order 2 U = c U (One dimensional Wave equation) tt xx 2 U = c U (One dimensional Heat equation) t xx U + U = 0 (Two dimensional Laplace equation) xx yy U + U + U = 0 (Three dimensional Laplace equation) xx yy zz U + U = f(x,y) (Two dimensional Poisson equation) xx yy PROBLEMS -t 1. Verify that U = e Sin 3x is a solution of heat equation. -t -t Solution: U = -e Sin 3x and U = -9e Sin 3x t xx 2 U = c U (One dimensional heat equation) .......... (1) t xx B Tech Mathematics III Lecture Note Putting the partial deivativers in equation (1) we get -t 2 -t -e Sin 3x = -9c e Sin 3x 2 Hence it is satisfied for c = 1/9 2 One dimensional heat equation is satisfied for c = 1/9. Hence U is a solution of heat equation. 2. Solve Uxy = -Uy p  p Solution: Put Uy = p then x p x p Integrating we get ln p = - x + ln c(y) –x  U/  y = p = e c(y) –x  U = e c(y) y –x Integrating we get U = e (y) ϕ(y) + D(x) where ϕ(y) = ∫c(y)  y 11.2 Modeling: One dimensional Wave equation We shall derive equation of small transverse vibration of an elastic string stretch to length L and then fixed at both ends. Assumptions. 1. The string is elastic and does not have resistance to bending. 2. The mass of the string per unit length is constant. 3. Tension caused by stretching the string before fixing it is too large. So we can neglect action of gravitational force on the string. 4. The string performs a small transverse motion in vertical plane. So every particle of the string moves vertically. Consider the forces acting on a small portion of the string. Tension is tangential to the curve of string at each point.Let T and T be tensions at end points. Since there is no motion in horizontal direction, horizontal 1 2 components of tension are T Cos α= T Cos β = T = Constant ..... ...... (1) 1 2 The vertical components of tension are - T Sin α and T Sin β of T and T 1 2 1 2 By Newton’s second law of motion, resultant force = mass x acceleration 2  u T Sin β - T Sin α = x 2 1 2 t 2 T Sin  T Sin x u 2 1 - = 2 T T Tt B Tech Mathematics III Lecture Note 2 T Sin  T Sin  x u 2 1 - = 2 T Cos  T Cos  Tt 2 1 2 x u tan  - tan  = 2 Tt ...... .................(2) As tan β = ( u/  x) = Slope of the curve of string at x x tan α = ( u/  x) = Slope of the curve of string at x+x x+Δx 2 x u Hence from equation (2) ( u/  x) - ( u/  x) = x+Δx x 2 Tt 2  u ( u/  x) - ( u/  x) /x = Dividing both sides by x x+Δx x 2 Tt Taking limit as x →0 we get 2  u xx = Lim →0 ( u/ x)x+Δx - ( u/ x)x / 2 Tt 2 u u  = 2  xx Tt  2 2  u u = 2 2 x Tt 2 2  u T u  2 2 tx 2 2  u u T 2 2 OR C whereC 2 2 tx which is One dimensional Wave equation 11.3 Solution of One dimensional Wave equation (separation of variable method) 2 One dimensional wave equation is u = c u .............................................(1) tt xx Boundary Condition u( 0, t) = 0, u(L,t) = 0 .............................................(2) Initial Condition u( x,0) = f(x) = initial deflection ..................(3) u (x, 0) = g(x) = initial velocity ..................(4) t Step I Let u(x,t) = F(x) A(t) Then u = F(x) Ӓ(t) and u = F″ (x) A(t) tt xx 2 Equation (1) becomes F(x) Ӓ(t) = C F″ (x) A(t) B Tech Mathematics III Lecture Note 2 Ӓ(t)/ C A(t) = F″ (x)/ F(x) L.H.S. involves function of t only and R.H.S. involves function of x only. Hence both expression must be equal to some constant k. 2 Ӓ(t)/ C A(t) = F″ (x)/ F(x) = k = constant F″ (x) - k F(x) = 0 -(6) 2 Ӓ(t) - C kA(t) = 0 .........................(7) Step II We have to find solutions of F and G of equations (6) and (7) so that u satisfies equation(2) . Hence u(0,t) = F(0) A(t)=0 and u(L,t) = F(L) A(t)=0 If A = 0 then u = 0 and we can not get a valid solution of deflection u. Let A is non zero then F(0) = 0 and F(L) = 0 .............................(8) Three cases may arise. Case I : K = 0 From eq (6) F” = 0 Integrating we get F = ax + b Using (8) we get a = 0, b = 0 Hence F = 0 and u =0 which is of no interest. 2 Case II : K = α (Positive) 2 From eq (6) F” - α F= 0 α x -αx Integrating we get F = ae + be Using (8) we get a = 0, b = 0 Hence F = 0 and u =0 which is of no interest. 2 Case III : K = -p (Positive) 2 From eq (6) F” + p F= 0 Integrating we get F = C Cos px + B Sin px Using (8) we get F(0) = C = 0, F(L) = B Sin pL = 0 Let B ≠ 0 then Sin pL = 0 Hence pL = nπ and p = nπ/L Putting B=1 we get F(x) = Sin nπx/L ........(9) So Fn (x) = Sin nπx/L where n=1,2,3, ... Thus we get infinitely many solutions satisfying equation (8). 2 2 2 Putting k = -p in equation (7) we get Ӓ(t) + p C A(t) = 0 2 2 2 2 Ӓ(t) + (C n π /L ) A(t) = 0 B Tech Mathematics III Lecture Note 2 OR Ӓ(t) + (λn) A(t) = 0 where λn = cn π/L General Solution An (t) = Bn Cos λn t + Bn Sin λn t ..................(10) Hence un (x,t) = (Bn Cos λn t + Bn Sin λn t) Sin nπx/L for n=1,2,3...... ...................(11) Are solutions of equation (1) satisfying boundary condition (2). These functions are called eigen functions and the values λn = cn π/L are calledeigen values or characteristic values of the vibrating string. Step III A single solution un(x,t) shall not satisfy initial Conditions (3) and (4). To get a solution that satisfies (3) and (4) we consider the series u(x,t) = Σ u (x,t) = Σ (Bn Cos λn t + Bn Sin λn t) Sin nπx/L..........(12) n From equations (12) and (3) we get u (x,0) = Σ (Bn Sin nπx/L) = f(x) .............................(13) Bn must be chosen so that u (x,0) must be a half range expansion of f(x) L 2 nx f(x) Sin dx i.e. Bn =  0 L L ..........................(14) Differentiating (12) w.r.to t and using (4) we get Σ (Bn λn Sin nπx/L) = g(x) For equation (12) to satisfy (4) the coefficient Bn should be chosen so that for t = 0, ut becomes Fourier Sine series of g(x) L 2 nx B  g(x) Sin dx n  0 Cn L PROBLEMS 1. Find the defection u(x,t) of the vibrating string of length L=π, ends fixed, C=1, with zero initial velocity and initial deflection x(π-x) Solution: Given length L=π, C=1, initial velocity g(x) = 0 .Hence Bn = 0 and λn = cn π/L = n The initial deflection f(x) = x(π-x) L 2 nx 2 2 Bn = f(x) Sin dx  (x - x ) Sin nxdx  0 0 L L  2 2 (x - x ) ( -2x) 2  - Cos nx Sin nx Cos nx  3 2  n n n  0 B Tech Mathematics III Lecture Note 4  1- Cos n 3 n B =8/ π, B =0, B =8/27π 1 2 3 The defection u(x,t) of the vibrating string u(x,t) = Σ (Bn Cos λn t + Bn Sin λn t) Sin nπx/L = Σ (Bn Cos n t ) Sin nx (as Bn = 0 and L = π ) = B Cos t Sin x + B Cos 2t Sin 2x + .............. 1 2 = (8/π) Cos t Sin x + (8/27π )Cos 3t Sin 3x + .............. 2. Using separation of variable solve the PDE U = U xy  Solution: Let U = F(x) G(Y) then U = F’ G and U = U / y = F’ G x xy x Where F’ =  F/  x and G = G/  y Putting these partial derivatives the given PDE becomes F’ G = F G By separation of variables we get F’ / F = G/G =k = Constant (Since L.H.S. is a function of x and R.H.S. is a function of y) F’ / F =k and G/G =k  F/ F = k x and  G/ G = y/k Integrating both sides of these equations we get ln F = kx + ln C and ln G = y/k + ln D kx y/k F = C e and G = D e kx + y/k U = F G = C D e 11.4 D ALEMBERT’S SOLUTION OF WAVE EQUATION 2 One dimensional wave equation is u = c u .............................................(1) tt xx We have to transform equation (1) by using new independent variables v = x + ct and z = x-ct u = u(x,t) will become a function of v and z. The partial derivatives are  v/ x = 1 =  z/ x,  v/ t = c and  z/ t = -c .....................(2) Using chain rule for function of several variables we get u = u v + u z = u + u x v x z x v z u = ( / x)(u + u ) xx v z B Tech Mathematics III Lecture Note vzvz uuuu uuuuu 2uu v v z z vv vz vz zz vv vz zz vxzxvxzx Hence u u 2uu .........(3) xx vv vz zz Similarly u = u v + u z = cu -c u t v t z t v z u = ( / t)(cu -c u ) = c( / t)u - c( / t) u tt v z v z vzvz 2 2 2 2  cucucucuc uc uc uc u v v z z vv vz vz zz vtztvtzt 2 uc (u 2uu ) ................................................(4) tt vv vz zz 2 2 Using (3) and (4) in equation (1) we get c (u 2uu ) c (u 2uu ) vv vz zz vv vz zz OR -2u = 2u Hence u = 0 vz vz vz u = c(v) v u= φ(v) + ψ(z) = φ(x+ ct) + ψ(x-ct) This is D Alemberts solution of wave equation where φ(v) = ∫c(v) v TYPES AND NORMAL FORM OF LINEAR PDE: An equation of the form A Uxx + 2B Uxy+ C Uyy = F(x,y,U,Ux,Uy) is said to be 2 elliptic if AC – B 0 2 2 parabolic if AC – B = 0 and hyperbolic if AC – B 0 For parabolic equations the transform v= x, z = ψ(x, y) is used to transform to normal form For hyperbolic equations the transform v=φ (x, y), z = ψ(x, y) is used to transform to normal form 2 Where φ = constant and ψ = constant are solutions of equation Ay’ – 2By’ + C = 0 PROBLEMS 2 1. Given f(x) = k(x – x ), L=1, k =0.01, g(x)= 0 Find the deflection of the string. 2 Solution: f(x) = k(x – x ) 2 2 f(x + ct) = k (x + ct) -(x + ct) and f(x - ct) = k (x - ct) -(x - ct) The deflection of the string is u(x,t) = f(x + ct) + f(x - ct) /2 2 2 = k x + ct -(x + ct) + x - ct -(x - ct) /2 2 2 2 = 0.01x – x - c t 2. Transform the PDE 4u - u = 0 to normal form and solve xx yy B Tech Mathematics III Lecture Note Solution : 4u - u = 0 ...................(1) xx yy 2 Here A=4, B = 0 and C= -1, hence AC – B = -4 0 Given equation is a hyperbolic type equation. 2 2 From the equation Ay’ – 2By’ + C = 0 we have 4y’ – 1 = 0 Solving we get x + 2y = c and x- 2y = c 1 2 We have to transform equation (1) by using new independent variables v = x + 2y and z = x-2y u = u(x,t) will become a function of v and z.  The partial derivatives are v/ x = 1 = z/ x, v/ y = 2 and z/ y = -2 .....................(2) Using chain rule for function of several variables we get u = u v + u z = u + u x v x z x v z u = ( / x)(u + u ) xx v z vzvz   u u u u uuuuu 2uu v v z z vv vz vz zz vv vz zz vxzxvxzx Hence u u 2uu .........(3) xx vv vz zz Similarly u = u v + u z = 2u -2 u y v y z y v z u = ( / y)(cu -c u ) = 2( / y)u - 2( / y) u yy v z v z vzvz  2u 2u 2u 2u 4u 4u 4u 4u v v z z vv vz vz zz vyzyvyzy u 4(u 2uu ) ................................................(4) yy vv vz zz Using (3) and (4) in equation (1) we get 4(u 2uu ) 4(u 2uu ) vv vz zz vv vz zz OR -2u = 2u Hence u = 0 vz vz vz u = c(v) v u= φ(v) + ψ(z) = φ(x+ 2y) + ψ(x-2y) This is D Alemberts solution of wave equation where φ(v) = ∫c(v) v 11.5 Solution of One dimensional Heat equation (separation of variable method) 2 One dimensional wave equation is u = c u .......................................(1) t xx Boundary Condition u( 0, t) = 0, u(L,t) = 0 .............................................(2) Initial Condition u( x,0) = f(x) = initial temperature ..................(3) Step I Let u(x,t) = F(x) G(t) ....................................................(4) Then u = F(x) G(t) and u = F″ (x) G(t) where F’ =  F/  x and G = G/  t t xx B Tech Mathematics III Lecture Note 2 Equation (1) becomes F(x) G(t) = C F″ (x) G(t) 2 G(t)/ C G(t) = F″ (x)/ F(x) ...................................................(5) L.H.S. involves function of t only and R.H.S. involves function of x only. Hence both expression must be equal to some constant k. 2 G(t)/ C G(t) = F″ (x)/ F(x) = k = constant F″ (x) - k F(x) = 0 -(6) 2 G(t) - C kG(t) = 0 .........................(7) Step II We have to find solutions of F and G of equations (6) and (7) so that u satisfies equation(2) . Hence u(0,t) = F(0) G(t)=0 and u(L,t) = F(L) G(t)=0 If G = 0 then u = 0 and we can not get a valid solution of deflection u. Let G is non zero then F(0) = 0 and F(L) = 0 .............................(8) Three cases may arise. Case I : K = 0 From eq (6) F” = 0 Integrating we get F = ax + b Using (8) we get a = 0, b = 0 Hence F = 0 and u =0 which is of no interest. 2 Case II : K = α (Positive) 2 From eq (6) F” - α F= 0 α x -αx Integrating we get F = ae + be Using (8) we get a = 0, b = 0 Hence F = 0 and u =0 which is of no interest. 2 Case III : K = -p (Positive) 2 From eq (6) F” + p F= 0 Integrating we get F = A Cos px + B Sin px Using (8) we get F(0) = A = 0, F(L) = B Sin pL = 0 Let B ≠ 0 then Sin pL = 0 Hence pL = nπ and p = nπ/L Putting B=1 we get F(x) = Sin nπx/L ........(9) So Fn (x) = Sin nπx/L where n=1,2,3, ... Thus we get infinitely many solutions satisfying equation (8). 2 2 2 Putting k = -p in equation (7) we get G(t) + p C A(t) = 0 B Tech Mathematics III Lecture Note 2 2 2 2 G(t) + (C n π /L ) G(t) = 0 2 OR G(t) + (λn) G(t) = 0 where λn = cn π/L –λn2 t General Solution Gn (t) = Bn e ..................(10) –λn2 t Hence un (x,t) = Bn Sin nπx/L e for n=1,2,3...... ...................(11) Are solutions of equation (1) satisfying boundary condition (2). Step III A single solution un(x,t) shall not satisfy initial Conditions (3) and (4). To get a solution that satisfies (3) and (4) we consider the series –λn2 t u(x,t) = Σ u (x,t) = Σ Bn Sin nπx/L e ..........(12) n From equations (12) and (3) we get u (x,0) = Σ (Bn Sin nπx/L) = f(x) .............................(13) Bn must be chosen so that u (x,0) must be a half range expansion of f(x) L 2 nx i.e. Bn = f(x) Sin dx  0 L L ..........................(14) PROBLEMS 1. Find the temperature u(x,t) in a bar of length L= 10 cm, c=1,constant cross section area, which is perfectly 0 insulated laterally and ends are kept at 0 C, the initial temperature is x(10-x) Solution: Given length L=10 λn = cn π/L = n π/10 The initial deflection f(x) = x(10-x) 10 2 nx 1 Sin nx 2 Bn = f(x) Sin dx  (10x - x ) dx  0 0 10 10 5 10  2  1 (10x - x ) (10-2x) 2  - Cos nx Sin nx Cos nx  3 2 5 n n n  0 400  1- Cos n 3 3 n 3 3 B =800/ π , B =0, B =800/27π 1 2 3 The temp u(x,t) of the bar –λn2 t u(x,t) = Σ Bn (Sin nπx/L) e -π2/100 4(-0.017π2t) = B (Sin πx/10) e + B (Sin 3πx/10) e +.................. 1 2 3 -0.017π2t 3 9(π2/100) = (800/ π )Sin πx/10 e + (800/27π ) Sin 3πx/10 e +.................. B Tech Mathematics III Lecture Note Insulated ends(Adiabatic Boundary Conditions) L 1 A f(x) dx 0  0 L L 2 nx An f(x) Cos dx  0 L L  nx 2 u(x,t) A A Cos e t  0 n n L n1 2. Find the temperature u(x,t) in a bar of length L=π , c=1, which is perfectly insulated laterally and also ends are insulated, the initial temperature is x Solution: Given length L=π λn = cn π/L = n The initial deflection f(x) = x L 1 1 A f(x)dx xdx / 2 0  0 0 L  L 2 nx 2 2 x Sin nx Cos nx  An f(x) Cos dx xCos nxdx  2  0 0 L L n n  0 2 Cosn1 2 n  2 nx  t n u(x,t) A A Cos e  0 n L n1 2 2 x 2x  t t 1 2  A A Cos e A Cos e .............. . 0 1 2 L L 9t  e Cos 3x t  / 2 4 / e Cos x............. .  9  B Tech Mathematics III Lecture Note UNIT-II 11.8 RECTANGULAR MEMBRANE 2 Two dimensional wave equation is u = c (u u ) ....................................(1) tt xx + yy Boundary Condition u( x,y, t) = 0 on the boundary of the membrane for all t ≥ 0 ....... .......(2) Initial Conditions: u( x,y,0) = f(x, y) = initial deflection ..................(3) ut (x, y,0) = g(x, y) = initial velocity ..................(4) Step I Let u(x,y,t) = F(x, y) A(t) Then u = F(x, y) Ӓ(t) and u = F A(t), u = F A(t) tt xx xx yy yy 2 Equation (1) becomes F(x,y) Ӓ(t) = C (F + F )A(t) xx yy 2 Ӓ(t)/ C A(t) = (F + F )/F .....................(5) xx yy L.H.S. involves function of t only and R.H.S. involves function of x only. Hence both expressions must be equal to some constant D. For D ≥ 0, as F= 0, hence u = o and we can not get solution. 2 For D 0 let D = - ν (negative) 2 2 Ӓ(t)/ C A(t) = (F + F )/F = - ν = constant xx yy 2 2 Ӓ(t) + ν C A(t) = 0 2 or Ӓ(t) + λ A(t) = 0 where λ= c ν .........................(6) 2 F + F + ν F = 0 .........................(7) xx yy In equation (7) two variables x and y are present and we want to separate them. Let F(x,y) = H(x)Q(y) ...........................................(8) 2 2 2  d H d Q d Q 2 2 Then from equation (7) QH HQH Q  2 2 2 dx dy dy  2 2  1 d H 1 d Q 2  Q  2 2 H dx Q dy  L.H.S. is a function of x only and R.H.S. is a function of y only. Hence the expressions on both sides equal to a 2 constant k. As negative value of constant leads to solution let the constant be –k then, 2 2  1 d H 1 d Q 2 2  Qk  2 2 H Q dx dy  2 d H 2 k H0 ................(9) 2 dx B Tech Mathematics III Lecture Note 2 d Q 2 2 2 2  p Q0where pk .......................(10) 2 dy Step II General solution of equations (9) and (10) are H(x) = A Cos kx + B Sin kx Q(y) = C Cos py + D Sin py where A,B,C and D are constants. From equations (5) and (2) we have F = HQ = 0 on the boundary. Hence x=0,x=a, y=0, y=b implies H(0) = 0, H(a) = 0, Q(0) = 0, Q(b) = 0 Now H(0) = 0 implies A = 0 H(a) = 0 implies B Sin ka = 0 Assume B ≠ 0 then Sin ka = 0 (Because if B= 0 then H = 0 and hence F =0) ka =mπ or k= mπ/a, m is integer Again Q(0) = 0 implies C = 0 Q(b) = 0 implies D Sin pb = 0 Assume D ≠ 0 then Sin pb = 0 (Because if D= 0 then Q = 0 and hence F =0) pb =nπ or p=nπ/b, n is integer Thus H (x) = Sin mπx/a , m= 1,2,...... m Q (y) = Sin nπy/b , n= 1,2,...... n F (x, y) = Sin mπx/a Sin nπy/b , m= 1,2,...... and n= 1,2,...... are solutions of equation (7) which are zero on the mn boundary of the membrane. 2 2 c k p λ=cν= 2 2 m n c m= 1,2,...... and n= 1,2,...... mn 2 2 a b The numbers λ are called eigen values or characteristic values. mn The general solution of (6) is Amn (t) = B Cos λ t+ B Sin λ t mn mn mn mn Hence umn(x,y,t) = (Bmn Cos λmn t+ BmnSin λmnt) Sin (mπx/a) Sin (nπy/b).......................(13) Step III B Tech Mathematics III Lecture Note We consider the series  mx ny u (x,y,t) u (x,y,t)B Cos tB Sin tSin Sin .................(14)  mn mn mn mn mn a b m1 n1 m1 n1 From equations (14) and (3) we get  mx ny u (x,y,0)BSin Sin f (x,y)  mn a b m1 n1 ................(15) This series is called a double Fourier series.  ny To find the Fourier coefficient B , we put K (y) = B Sin in equation (15) mn m  mn b n1  mx we get f (x,y) Km(y)Sin  a m1 a 2 mx The coefficient K (y) = f(x, y) Sin dx m  0 a a b 2 ny Hence B K (y) Sin dy mn m  0 b b b a 4 mx ny  f(x, y) Sin Sin dxdy  0 0 ab a b  u mx ny   BSin Sin g(x,y)  mn mn  t a b  m1 n1 t0 b a 4 mx ny B g(x,y) Sin Sin dxdy where m= 1,2,...... and n= 1,2,...... mn  0 0 ab a b mn PROBLEMS 1. Find the deflection u(x,y,t) of the square membrane a=1, b=1 and c=1 if the initial velocity is zero and 2 2 initial deflection is k(x-x )(y-y ) Solution: Given a=1, b=1 and c=1 . Hence we have 2 2 m n 2 2 c mn mn 2 2 a b The initial velocity g(x,y) is zero . Hence B = 0 mn 2 2 The initial deflection f(x,y) = k(x-x )(y-y ) B Tech Mathematics III Lecture Note b a 1 1 4 mx ny 2 2 B f(x, y) Sin Sin dxdy 4 k(x - x )(y - y ) Sin mxSin nydxdy mn  0 0 0 0 ab a b 1 1 2 2  4k(y- y )Sin ny (x - x )Sin mxdxdy ...................(1)  0 0 1 2 Now (y- y )Sin nydy  0 1 1 Cos ny Cos ny  2 =  (y- y ) (1- 2y) dy   0 n n  0 1  1 1 Sin ny Sin ny   00 (1- 2y) (-2) dy   0 n n n   0  1 1 2Cos ny 2   0 0 1 Cos n .................(2)  2 2 3 3 n n n  0 Putting this in equation (1) 1 2 2 Bmn 4k 1-Cos n(x - x )Sin mxdx 3 3  0 n 1 8k 2 1-Cos n (x - x )Sin mxdx 3 3 0 n 1 8k Cos mx Sin mx 2  2 1-Cos n - (x - x ) (1- 2x) Cos mx 3 3 2 2 3 3  m n m m  0 8k 2  1-Cos n1 Cos m 3 3 3 3 n m  16k 1-Cos n1 Cos m 3 3 6 m n  mx ny Deflectionu (x,y,t)B Cos tB Sin tSin Sin  mn mn mn mn a b m1 n1  B Cos tSinmxSinny  mn mn m1 n1 as Bmn = 0 and a=1, b=1  16k 2 2 1Cosn1CosmCos mn tSinmxSinny  3 3 6 m n m1 n1 2. Find the double Fourier series of f(x,y)= xy, 0xπ and 0yπ, B Tech Mathematics III Lecture Note Solution: Here a= π , b= π, f(x,y) = xy . Hence we have b a 4 mx ny 4 B f(x, y) Sin Sin dxdy xy Sin mx Sin nydxdy mn  2 0 0 0 0 ab a b   4 - y Sin ny  ...................(1)  Cos ny xSin mxdx 2 2  0 n  n  0  4 - 4   Cos n xSin mxdx Cos n xSin mxdx 2  0 0  n n   4 - x Sin mx 4   Cosn Cos mx CosnCosm  2 n m m mn  0 .................(2) The double Fourier series is  mx ny f (x,y)BSin SinBSinmxSinny  mn mn a b m1 n1 m1 n1  4   CosmCosn SinmxSinny   mn  m1 n1 11.9 LAPLACIAN IN POLAR COORDINATE 2 2  u u 2  u0 (Laplaceequation ) 2 2 xy ...............(1) To convert Laplace equation (1) into polar form we put x = r Cos θ ...............(2) and y = r Sin θ ...............(3) 2 2 2 . Squaring and adding equations (2) and (3) we have x +y = r ...............(4) -1 Dividing equations (3) by equation (2), tan θ = y/x , hence θ = tan (y/x) ...................(5) r 2x 2r x Differentiating equation (2) partially w.r.to x we get r 2y 2r y Differentiating equation (2) partially w.r.to y we get r xr y Hence r and r x y x ry r ...................(6) Again differentiating equation (6) partially w.r.to x and y respectively we get B Tech Mathematics III Lecture Note 2 2 2 2 2 2 r r xr r x(x /r ) r x x y x y  x r xx 2 2 3 3 3  xx r r r r r  2 2 2 2 2 2 r yr r r y(y /r ) r y x y y x y r yy  2 2 3 3 3 yy r r r r r  Differentiating equation (5) partially w.r.to x we get  1 y 11yy   y x 2 2 2 2 2 2 2 2  x 1y /xx x 1y /x x xy r  .............(7) Differentiating equation (5) partially w.r.to y we get  1 y 1 1 x x   y 2 2 2 2 2 2 2  y 1y /xy x 1y /x x xy r  ...............(8) Again differentiating equation (7) partially w.r.to x we get  1 2r 2 x 2xy  yyy xx 2 3 3 4  xxx r rx r r r  Again differentiating equation (8) partially w.r.to y we get  1 2r 2 y 2xy   x x x yy  2 3 3 4  yyy r ry r r r   Using chain rule for function of several variables we get uuru u u ru x r x x xrxx  u r uu ru u = xx r x x r x x xxx    u rr (u )u (u ) r x x r x x xxxx rr   u rr (u ) (u )u (u ) (u ) r xx x r r xx x  rxxrxx  2 y 2xy   urr u ( u )ur u ( u ) r x x r x r x x r x  3 r 4  r r   2 2 2 y x xy 2xy xy y  u u u u u u r r r r  r 3 2 3 4 3 4 r r r r r r B Tech Mathematics III Lecture Note 2 2 2 x 2xy y y 2xy u u u u u u xx r r  r r 2 3 4 3 4 r r r r r .....................(9) 2 2 2 y 2xy x x 2xy u u u u u u yy r r  r 2 r 3 4 3 4 r r r r r ..... ...............(10) Adding equations (9) and (10) we get 2 2 2 2 2 2 x y x y x y uu u u u xx yy rr  r 2 4 3 r r r 1 1 uuu u u xx yy r  r r 2 r r 1 1 Laplace equation in polar form is u u u 0 rr  r 2 r r PROBLEMS 1. Show that the only solutions of Laplace equation depending only on r is u = a ln r + b Solution: 1 1 Laplace equation in polar form is u u u 0 r  r r 2 r r As u depends only on r, u is a function of r only. Hence u =0 and u = 0 θ θθ 1 1 Hence u u 0 or u u rr r rr r r r Let u p then up /r r rr p ppr Hence or r r p r Integrating both sides we get ln p = - ln r + ln a u a ar Hence p oru r r r Integrating again both sides we get ln u = a ln r + b 2. Find the electrostatic potential (Steady state temperature distribution) in the disk r1 corresponding to 2 the boundary values 4cos θ Solution 2 The boundary value f(θ) = 4cos θ which is even function, -π θ π .Hence Bn = 0  1 1 1 1  2 A f( )d 4Cosd1 Cos2d 0.5Sin 2 2 0   2 2 B Tech Mathematics III Lecture Note  1 1 1 2 An f( ) Cos nd 4Cos Cos n d 21 Cos2 Cos n d     2 2  Cos n d Cos n Cos2 d     2 1   Sin nCos n 2 Cos n 2d    n  1 Sin n 2 Sin n 2   0 0 (except n 2)   n 2 n 2   1 For n 2, An A2 21 Cos2 Cos 2 d    2 2 2 1  2  Cos 2 d Cos 2 dSin 21 Cos 4d     2  Sin4  1  0 2   4   The electrostatic potential (Steady state temperature distribution) in the disk n   r 2 u (r, ) AA Cosn B Sin n 2 2r Cos2  0 n n n1 R  (Since R= radius of disk=1 and B = 0) n 11.10 CIRCULAR MEMBRANE 2 2 2 Two dimensional wave equation is u = c (u u ) c u tt xx + yy = Using Laplacian in polar form we have 1 1 2 Laplace equation in polar form is uu u u 0 r  r r 2 r r 2  u 1 1  2 c u u u rr  r 2 2 t r r  AS circular membrane is radially symmetric, u depends on r only and u does not depend on θ Hence u =0 and u = 0 θ θθ 2  u 1  2 Hencec u u ....................................(1) r r r 2 t r  Boundary Condition u( R, t) = 0 ....... .......(2)

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