Unit 3 Part 1 Applications of Partial Differential Equations PDF

UNIT-III APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 3.0 Introduction. 3.1 3.1 Classification of p.d.e. 3.1 Exercises 3.5 3.2 Method of seperation of variables...

UNIT-III APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS 3.0 Introduction. 3.1 3.1 Classification of p.d.e. 3.1 Exercises 3.5 3.2 Method of seperation of variables 3.5 Exercises 3.8 3.3 Fourier seriessolutions of One Dimensional Wave Equation. 3.9 Exercises 3.45, 3.61, 3.69 3.4 One Dimensional Equation of Heat conduction. 3.70 Exercises 3.85, 3.109 3.5 Steady State Solution of Two Dimensional 3.113 equation of Heat conduction (Cartesian Co-ordinates only) 3.116 Exercises 3.156 (x)  UNIT-I|| PROBLEMS Classification of p.d.e 1. Problems on 3.2 Problems on Method of separation of variables 3.5 2 with Zero 3. Problems on Vibrating String Initial Velocity. 3.16 4. Problems on Vibrating String withNon-zero Initial Velocity. 3.47 5. Problems on Vibrating String with Initial Velocity and Initial Displacement Given. 3.64 6. Problems with Zero Boundary Values (Temperature or Temperature Gradients), 3.77 7 Problems on Steady State Conditions and Zero Boundary Conditions. 3.85 8 Problems on Steady State Conditions and Non-zero Boundary Conditions. 3.93 9. Problems on Thermally Insulated Ends. 3.101 10. Problems on Finite Plates. Type 1. 3.120 Type 2. 3.132 Type 3. 3.139 11. Problems on Infinite Plates. 3.142 12. PART-A QUESTIONS AND 3.161 ANSWERS. (xi)  Partial Differential Equations 3.1 ahations of UNIT - III PPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS Cassification of PDE Method of separation of variables Solutions of one dimensional wave equation One dimensional equation of heat conduction - Steady state soution of twwo dimensional equation of heat conduction (excludinginsulated edges). 0 INTRODUCTION PartialIdifferential equations arise in connection with several physical which the functions involved depend on engineering problems in MOOr more independent variables such as time and co-ordinates in of equations are Pt In practical problems, the following types zerallyused. 9Wave equation OOne dimensional heat flow ü) Two dimensional + = 0 steady-state heat flow ) Radio equations DIFFERENTIAL EQUATIONS CLASSI La FICATION OF PARTIAL Second order p.d.e. in the function u of the two independent vcables x,y be of the form +B ou ou = 0... (1) +C ôx ay 3.2 Engineering Mathematics Applications of Partial Equation (1) is classified as elliptic, parabolic depending on or hyperbol, Sample 3.1.2 Classify Differential Equations + uyy = (uy + 3.3 B² - 4AC < 0(clliptic equation) Solution Given + Uyy (u,) B- 4AC = 0 (parabolic equation) (4,y +(uy)' [A.U A/M 2003] Second order p.d.e in the B²- 4AC > 0 (hyperbolic equation) function u of the form + B (1,y) au Example |A(4,y) +C (x,y) au Qu 1, Elliptic type Here, A=1, B= 0, =0 C=1 ou au B²- 44C (1) = 0(Laplace equation in two dimension) = 0 - 4(1) (1) = -4 < 0 The given equation is eliptic equation. (i) + = fo,y) (Poisson's equation) Example 3.1.3 Classify Ox y + 2. Parabolic type Ou au Solution : Given : Wxy = ly + uy [A.U. NND 2002] +xy Second order p.d.e in the (one dimensional heat flow equation) function u of the form au = 2.+ B(, 'u +C(«,y) au 3. Hyperbolic type (one dimensional wave equation) 9)ax ay du Ox' oy ou =0 Problems on classification of partial differential equations Here, A=0, B= 1, C=0 B 4AC = Example 3.1.1 Find the nature of the partial differential equation * The 1>0 given equation is hyperbolic equation. 4 Lyx t4 uxy t Uyy +2u, - u, = 0 [AU, Oct. 1996]JA.UCBT ND 2010 Exampl e 3.1.4 Classify olution : Given : uy -y Bo-yuy =2y'u [AU, May 1996] Solution : Given : 4uyx + 4 uyy + uyy t 2 u, - uy = 0 Second order p.d.e in the function u of the form |Second order p.d.e. in theu- 2y° uy 0 function u of the form A(, y) a'u + B (x, y) a'u a'u + B(, y) a'u au Ôu ax oy +C(1, y) e, A 1. B àx ay +C(*,y) +f =0 Here, A = 4, B = 4, C= 1 = 0. C = B y 4AC = 0 - 4(1)(y) = 4y B3- 4 AC = 16 - 4(4) (1) = 0 =0, we get B2 I y>0 or 4AC= 0 [Parabolic equation] i. The given equation is parabolic equaion. y 0 [Hyperbolic equation] 3.4 Example 3.1.5 Classify the p.d.e. (1 + x) Ug- 4x Engineering Mathem Aplications of Partial Differential [AU, March EXERCISE:3.1 Equations Solution : Given: (1+x)" uxx - 4x uxy t uyy Cassifythe following p.d.e 3.5 Second order p.d.e. in the function u of the form au o'u 1. dx ay Ou A (K,y) + B (3, y) + C (*,y) a'u +f du du Ox y 0 2 +4 au +2 [Ans. Parabolic) Oxôy +2 Here, A = (1+x), B = -4x, C= 1 x = 0 B'- 4AC = 162- 4(1 + x) = 16x - 4(1 3 3ux t+ 2ugy + 5Uyy + x uy 0 [Ans. byperbolic] +*+ 2:1 (Ans. elliptic] = 16x- 4- 48¢ = 12x - 8x4 4. Myr Lyy 2y uy [A.U May 1996] [A.UMay 2004] [Ans. hyperbolic] 12--n 12 ( 1) +) [Ans. x >0 > 12 METHOD OF eliptic; x =0* parabolic ; x hyperbolic) ADOwerful method of VARIABLES If x=1 (or) x= we get B 4 AC= 0 [Parabolic equaic known as separation of variables or obtaining particular solutions of a p.d.e. is 1 Let z be the product method. If - y(, ) (Agx +A0) (A11+A) solutions of (6)& () depends on The nature of the he value eguation say k= p Thusthe various possible solutions of the wave equations are be positive Case (i) Let k (Ajeex + Aze P) (Ag eP + -pX = 0 () > T" - (6) > X" (8) we get two ordinary diff. equations. = (A_ CoS px +A6 sin px) (Aq cos cpt + Ag sin Now, cpt). (9) are The auxiliary equations (Agx + Aj0) (Aj+ Ajp) (10) = 0 m² -p' = 0 Out of the three mathematically possible solutions derived, we the solution which is m =p, -p m = cp, -cp have to choose consistent with the physical nature dfthe problem and the given boundary conditions. In the case of A, eP T = Aq ePt + Ay e So, X = Aj eP + ibration of an elastic string, y(4, t) representing the displacement of thestring at any point x, y must be periodic in t. Hence solution (9), y = XT (2) > which consists of periodic functions in t is the suitable solution of the problems on vibration of strings. The arbitrary constants in the suitable y - (A,eP + Age P) Ag eP +Ay e-gi) solution are found out by using the boundary conditions of the problem. -p Case (ii) Let k be negative say k= In problems, we directly assume that (9) is the suitable solution 0 dvibration of string problems. (6) & (7) > X" + p X = 0 and T'" + cpT = The suitable solution (9) which is periodic in is incidentally The auxiliary equations are periodic in also. m+p = 0 m²+p = 0 Example 3.32 : Derive D' Alembert's solution of the wave equation: m = + pi m = t cpi Solution : Consider the one dimensional wave equation So, X = (A_ COS px + A6 sin px), T = (A, cos cpt + Ag Sihg... (1) (2) > y= XT Let y (,) = (A_ cos px t A_ sin px) (A, cos cpt + Ag sin cp!) D = and D' = 3.12 Engineering Mathemal Apolcatlonsof Partial Dif erential Equations D²-eD) y = 0 : State the Hence, (1) can be written as m'-= 0 Erample3.3.3 mensional wave equation, assumptions made in the derivation of 3.13 one The auxiliary equation is equation is y (A.U NID 2016 m = ±c Loion : The wave R-13) solution of the wave equation is (, ), the.. The general Thisgives tension T. In transverse vibration of ( -ct) y= f + ci) +g (2) 1 Constant AsSumptions, deriving this equation, we amake string stretched to the following arbitrary functions where f and g are The motion takes place entirely in one 1. plane i.e., xy plane. () and (r, 0) = 0 Suppose, initially y(, 0) = 2 We consider only transverse viisbrations, the horizontal From (2), y(r, 0) = fe) + 8) = () ofthe particles of the string negligible. displacement The tension T is constant at all times and i.e., () = f) + g() (3) deflected string. at all points of the From (2), (,) = c ð(c+ c) -8- c)) is considered to be so large compared with the at string and hence the force of gravity is negigible. weicht of the ,0) = 0 > cf ()-g' ()] = 0 S The effect of friction is negligible. i.e., f () -g'(*) = 0 6 The string is perfectly tlexable, 1.e., it can transmit tension but not bending or shearing forces. Integrating, we get fr) - g() = k (4) 1. The slope of the From (3) & (4), we get deflection curve at all points and at all instants is so small that sin a can be P)-k replaced by , where ca is the fr) = 2 and g() 2 inclination of the tangent to the deflection curve. Bxample 3.3.4 : Derive one.. (2) can be written as dimensional wave equation : Solution : with its end Consider tightly stretched elastic string of length a y(,0) = ;l(+ c) +k< +;-c) - points fixed. Let the string be released from rest and allowed to a any point vibrate. x The problem is to determine the deflection y(,t) and at any time t > 0. 1 For deriving the p.d.e. we make the following assumption. dimensic This solution is called D' Alemberts solution of the One See Example 3.3.3. wave equation.  Partial Agolcatlonsof Differential Equations 3.14 the differential equation Fngine rng Matheng The acceleration of the element in they direcion is 3.15 To obtain Newton's second Hence, by law of motion, Q(x+ Ax,x y+ Ay) P(«y) mAS = T Ay w where mis the mass per unit length of the string. the length of the element PO. TË T A m AS ar origin 0 and take X-axis Take one end of the string as limit as AS> 0, we get string. along Taking We assume tuat the motion takes place entirely in the ay T dy m dS... (1) Consider the forces acting on a small portion PQ of the sua where P is (x, y) and Q is ( + Ax,y + Ay) By assumption, the string does not offer resistance to dy represents the curvature of the curve at P. of the string is beending z: Now, dS shearing hence the tension at each point the curve of the string. tangentid Let T, and T, be the tensions at the points P and ). då dS 2 13/2 Let yand y + Ay be the angles made by the tangents at Pzd Qrespectively with the X axis. Let m be the mass per unit length of the string Since, there is no motion in the horizontal portion of the sti the tension must be constant. 1s very small and is negligible].:. TË cos ý = T, cos (p + A) which is a constant. Since ý is small, cos y and cos (p + Ay) are approximately oq ng this in (1), we get oy T ßy to 1. max Thus, T=T; = T, a constant. The vertical Component of the force acting on the element PQ where a = T m T sin ( + Ay) - T sin y = T (y + Ay) - Ty [: sin y = y where is Saall Wave This partial differential equation is known as the one dimensional cquation. It is a of second order. = TAY homogeneous  Agolications of Partial Differential Equations Engineering Math condition (iv), we get 3.17 3.16 vibrating strino i. } C, sin n12nX = f() p.d.e. ofa y(x, 0) = n=1 Example 33.5a : The is a? where Cn = b, 2 f) sin n nx Tension strim Substitute, C in (2) we get the general solution.... (3) Solution : = mass per unit length of the Example 33.5b : Explain why (instead of a) is used in the Erample33.6: Astring is stretched and fastened to r =0 and started by displacing thetwostringpoints r=lapart. Motion is string theformy=k(Ux-x) rom which it is released at timet = 0. intoFind of the vibrating at Tension the displacement of any point on the string at a distance of xfrom Solution : = T unit length of the time t. [AU, April 2001, A/M 2003] [A.U. m mass per string end at equation is N/D 2007] by a? (and not Solution : The wave [A.UA/M 2008, Tvli N/D 2010] As T is positive, it is denoted by a). [AU.T CH N/D 2011][AU N/D 2013, AM = 2015 R-13] [AUND 2015 R-08| with zero Problem on vibrating string initial velocity. JAUAM 2019 R-17]| JAUN/D 2018 R-17] Fromthe given problem, we get the JA.UA/M 2019 R-08] initial velocity : initial conditions. Type 1. Vibrating string with zero (ollowing boundary & The boundary and initial conditions of the deflection v(r 0y(0,1) = 0 for all t > 0 B.C] ()y (0,) = 0 [boundary () y (,) = 0 condition] [condi boundantium ) yl,) = 0for all t > 0[B.C] A (ü)0) = 0 [initial (iv) y (z, 0) =f) inital = 0, 00 mditions (OR) (0,!) = 0, () y >0 y(l,) = 0, for all t 1 (21 +) x sin (27 + 1)na () 0 X" -p X= § sOLUTION OF HEAT EQUATION The auxiliary equations are 0;T' -p a T =0 IA.U|AU, March 19: NID 2019, Rn m'-p' = 0, m -p' The heat equation is m = ±p m = pa.. (1) ge cet X = Aj eP + Age P, T Thermal conductivity = A PPat (density) (specific heat) Here, u is a function of x and t. (P)(C) u(z, t) = (Aj e + Aze P) (Ag ePay Case (ü) Let k be negative say k= -p So, assume that solution of (1) is of the form Then (3) & (4), we get u = XT (2) X" + p X = 0; T' + X is a function of x alone and T is a function of ' alone. The A.E are m +p = apT =0 = XT 0; m+a'p = 0 m = pi ßx = X'T We get X m = -ap? = A4 cos px + As Sin px = X"T T = A, e-apt u (x, t) ßu = XT' =(Aq cospx + As sinpx) (Agea Py Case (iü) Let k = 0 (1) > XT' = axT  Applcationsof Partial Engineering Matheai, Dif erential Equations 3.76 (( being the length of theonc 3.77 Then (3) & (4) > X" = 0 and T' = 0 The initial condition is dimensional rod) (iü) w(, 0) =f), 00 B, Sin n NX (0) uT (l, ) = uT(, ) = n =1 e. (IV) xta-c (ii) uT(, 0) = Now, applying conditions (iii)) in (IV), we get solution is Now, the suitable n NX sin ut, 0) =n=1 B, = eapt Xta-c (V) B sin px)... (A cos px + 4T (I,) = (), we get To finnd Bn we expand f(x) in a half range Fourier sine series Applying condition (i) in Ae-ap't = 0 f(*) =n=1 bn sin n NX úT(0, ) = (VI) e-a'p't 0 [ it is defincd for all r Here, A = 0) where bn = 720 s sin dx we get Substitute, A = 0 in eqn. (), From (V) and (VI), we get B, = b, uT ( ,) = Bsin px e-ap't... (II) Applying condition (ii) in (I), we get e-a p t 0 B, -1.*.-a ) uT (!, ) = B sin pl ,-ap't 0 [ it is defincd for all Here, Sin B0 [:If B= 0 alrcady A = 0 then we get a trivial solution sin pl = 0 cos n I + (a - sin pl = Sin n n [ sin n n =0] 2 ((a - c) - (b- d) (-1)"] p! = n JL DUDSlitute, the valuc of B, in IV, we get = p 2 an'i n Jt 8 |(a-c)-(b- d) (-1" in (|I), we get sin Substitute, p = 3.98 Engineering Malhemat dications of Partial Differential Equations 3.99 Hence, u (, )) = u, (r) + uT (,) = r+ 3 40 + uf (i, !).. (C) (a-c) - (b- d) (-1)a sin boundary and initial conditions arc -[ ends A and B of a rod 30 cms long (i) u(0, ) =40, for all >0 () u (30, ) = 60, for allI0 Brample 3.4c(2) : The at 80° have and the other C until temperature kept at 20° C (i) u(, 0) = 2x + 20, 0 Aee'p': = 0 The solution may be u (, ) =u, () + uT (x, ).. (B) # 0 it is Steady state 1 Steady state 2 defined for all t) A= 0 A A B Substitute, A =0 in equation (1), we get x=0 x=0 x=30 2 u=20 u=80 u=40 u(, ) = + 40 + B sin pxe-apt (2) |The steady state solution is The steady state solution is + Applying condition () in equation (2), we get U= + W= X=0 40 = b 60 = 20 + 40 + B sin (30 )e-apt x=0 X=30 ’ 80 = 30a + 20 x= 30 > 60 = 30 + Bsin(30 p)-apt = 0 a= 2 0 [it is defined for all ] B # 0[ suppose B = 0, already A =0 then we get a = 2x + 20 2x+ 40 sin(30 p) trivial solution] r+40 =0 i.e., u (z, 0) = 2r + 20 Ms () = 3 Sin (30 p) = sin n JI  opllcationsof Partial Differential Equations 3.100 Engineering Mathema 3.101 n I 30 (20) [1 + (-1)) 30p = nn incquation (2), we get -40 [1+ (-1)"] nn Substitute, p 30 -ant 0 if n is odd n tI 900 r t 40 + Bsin (3) -80 if n is even 30 u (r, ) = general solution is 2 -80 -an The most + 40 + sin (nnx 900 n II -ant n=even 30 e + B, sin 30 900 u (r, 1) = +40 n=1 -ant 2 x+40- 80 1 (ii) in equation (4), we get 900 Applying condition = n=even n 30 n JtX 2x + 20 = x+40 +n=1 B, sin 30 80 -4a n2 t - t 40- 1 Sin (2n IX 900 4 n=12n 30 sin n NX = 2r-x+ 20 - 40 =-20 =f 30 n=l t 40 1 n JLX 40 - sin 225 To find B, : n=1 15 av Fouricr sine series in the intenl Expand f() in a Half-range 14d. Thermally insulated ends n nX 2 sin Problems based on Thermally insulated ends 0 Eample 3.4.d(1) : Explain the term "Thermally insulated End's". 30 2 sin Solution : If an end of heat conducting body is means that no heat passes through that section. thermally insulated, - 30 30 -coS n NX -Sin 30 Mathematically, the temperature gradient is zero at that point. 30 2 2 i.e., = 0 30 30 |Lample 3.4.d(2) : Express the boundary conditions in respect of Sulated ends of a bar of length a and also the initial COS nI t 30 6) 900 istribution. Sdlution n: The temperature 0) boundary and initial conditions are (-1" = 0 for all t 30 X= 0  Engineering Malhemal pollcations of Partial Differential Equations 3.103 3.102 -a'p'i# = 0 for all t HerG, e 0 it is defined for all t (ü) p 0 x= a.:.(B = 0 for 0