Chapter 2 PARTIAL DIFFERENTIAL EQUATIONS OF SECOND ORDER

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Draft PDE Lecture NotesKhanday M.A.Chapter 2PARTIAL DIFFERENTIAL EQUATIONS OF SECOND ORDERINTRODUCTION: An equation is said to be of order two, if it involves at least oneof the differential coefficients r ( 2z / 2x), s ( 2z / x y), t ( 2z / 2y), butnow of higher order; the quantities p and q may also enter into the equation. Thus thegeneral form of a second order Partial differential equation is𝑓( đ‘„, 𝑩, 𝑧, 𝑝, 𝑞, 𝑟, 𝑠, 𝑡) 0.(1)The most general linear partial differential equation of order two in two independentvariables x and y with variable coefficients is of the form𝑅𝑟 𝑆𝑠 𝑇𝑡 𝑃𝑝 𝑄𝑞 𝑍𝑧 đč. . . (2)where 𝑅, 𝑆, 𝑇, 𝑃, 𝑄, 𝑍, đč are functions of đ‘„ and 𝑩 only and not all 𝑅, 𝑆, 𝑇 are zero.Ex.1: Solve 𝑟 6đ‘„. 2𝑧Sol. The given equation can be written as đ‘„ 2 6đ‘„ đ‘§đ‘„ đ‘„ 3đ‘„ 2 1 (𝑩)Integrating (1) w. r. t.(1).(2)where 1 (𝑩) is an arbitrary function of 𝑩.Integrating (2) w. r. t. we getđ‘„ 𝑧 đ‘„ 3 đ‘„ 1 (𝑩) 2 (𝑩)where 2(y) is an arbitrary function of y.Ex.2. 𝑎𝑟 đ‘„đ‘Š 2𝑧1Sol: Given equation can be written as đ‘„ 2 𝑎 đ‘„đ‘Š.(1)Integrating (1) w. r. t., đ‘„, we get 𝑧 đ‘„ đ‘Šđ‘„2𝑎2 1(y).(2)where 1(y) is an arbitrary function of yIntegrating (2) w. r. t., x,z 𝑩3𝑎6 x 1(y) 2(y)Department of Mathematics, University of Kashmir, Srinagar-1900061

Draft PDE Lecture NotesKhanday M.A.𝑩z 2𝑎 x 1(y) 2(y)orwhere 2(y) is an arbitrary function of y.Ex.3: Solve r 2y2Sol:Try yourself.Ex. 4. Solve 𝑡 sin(đ‘„đ‘Š) 2𝑧Sol. Given equation can be written as 𝑩 2 sin(đ‘„đ‘Š).(1)Integrating (1) w. r. t., 𝑧 𝑩 1đ‘„ycos(đ‘„đ‘Š) 1 (đ‘„). . . (2)Integrating (2) w. r. t., y𝑧 1đ‘„2sin đ‘„đ‘Š 𝑩 1 đ‘„ 2 đ‘„which is the required solution, 1, 2 being arbitrary functions.Exercises:đ‘„đ‘Šđ‘  1 2𝑧Sol: We know that 𝑠 đ‘„ 𝑩Thereforeor 2đ‘§đ‘„đ‘Š đ‘„ 𝑩 1 2𝑧 đ‘„ 𝑩1 đ‘„đ‘ŠIntegrating w.r.t., y we have 𝑧 1 log 𝑩 𝑓 đ‘„ đ‘„ đ‘„Again integrating w.r.t., x we get𝑧 log đ‘„ log 𝑩 0r𝑓 đ‘„ đ‘‘đ‘„ đč 𝑩𝑧 log đ‘„ log 𝑩 𝑔 đ‘„ đč 𝑩Department of Mathematics, University of Kashmir, Srinagar-1900062

Draft PDE Lecture NotesKhanday M.A.Exercises:2đ‘„ 2𝑩 𝑠Sol: The given equation can be written as 2𝑧 2đ‘„ 2𝑩 đ‘„ 𝑩Integrating w.r.t., 𝑩, we have 𝑧 𝑩 2 2đ‘„đ‘Š 𝑓 đ‘„ đ‘„Integrating w.r.t., đ‘„, we have𝑧 𝑩2đ‘„ đ‘„2 𝑩 𝑓 đ‘„ đ‘‘đ‘„ đč 𝑩 𝑧 𝑩2đ‘„ đ‘„2𝑩 𝑔 đ‘„ đč 𝑩Exercises:đ‘„đ‘Ÿ 𝑝 9đ‘„ 2 𝑩 3Sol: The given equation can be written asđ‘„ 2𝑧 𝑝 9đ‘„ 2 𝑩 3 đ‘„ 2 đ‘„ 𝑝 𝑝 𝑝 9đ‘„ 2 𝑩 3 đ‘„đ‘ đ‘„ đ‘„ 9đ‘„đ‘Š 3 (1)which is linear first order differential equation in 𝑝 I. F. is 𝑒 log đ‘„ đ‘„Multiplying (1) by đ‘„ we getđ‘„ 𝑝 𝑝 9đ‘„ 2 𝑩 3 đ‘„ đ‘„ đ‘đ‘„ 9 đ‘đ‘„ 9đ‘„ 2 𝑩 3 đ‘‘đ‘„đ‘„3𝑩3 𝑓 𝑩3 đ‘đ‘„ 3đ‘„ 3 𝑩 3 𝑓 𝑩Department of Mathematics, University of Kashmir, Srinagar-1900063

Draft PDE Lecture NotesKhanday M.A.3đ‘„ 3 𝑩 3 𝑓 𝑩 𝑝 đ‘„ 𝑧𝑓 𝑩 3đ‘„ 2 𝑩 3 đ‘„đ‘„ Integrating with respect to đ‘„ we get𝑧 đ‘„ 3 𝑩 3 𝑓 𝑩 log đ‘„ đč 𝑩Exercises:𝑩𝑡 𝑞 đ‘„đ‘ŠSol: Please try yourself.Exercises:𝑡 đ‘„đ‘ž đ‘„ 2Sol: Please try yourself.Exercises:𝑟 2𝑩 2Sol: The given equation can be written as 2𝑧 2𝑩 2 đ‘„ 2 𝑝 2đ‘„ 2 đ‘„Integrating with respect to đ‘„ we get𝑝 2𝑩 2 đ‘„ 𝑓 𝑩 𝑧 2𝑩 2 đ‘„ 𝑓 𝑩 đ‘„Integrating we get𝑧 𝑩2đ‘„2 𝑓 𝑩 đ‘‘đ‘„ đč 𝑩 𝑧 𝑩 2 đ‘„ 2 đ‘„đ‘“ 𝑩 đč 𝑩Exercises:𝑡 sin đ‘„đ‘ŠSol: Please try yourself.Exercises:log 𝑠 đ‘„ 𝑩Sol: The given equation can be written asDepartment of Mathematics, University of Kashmir, Srinagar-1900064

Draft PDE Lecture NotesKhanday M.A.log 𝑞 đ‘„ 𝑩 đ‘„ 𝑞 𝑒 đ‘„ 𝑩 đ‘„ 𝑞 𝑒 đ‘„ 𝑒𝑩 đ‘„Integrating w.r.t. đ‘„ we get𝑞 𝑒 đ‘„ 𝑒𝑩 𝑓 𝑩 𝑧 𝑒 đ‘„ 𝑒𝑩 𝑓 𝑩 𝑩Integrating w.r.t., 𝑩, we get𝑧 𝑒 đ‘„ 𝑒𝑩 𝑓 𝑩 𝑑𝑩 đč đ‘„đ‘§ 𝑒 đ‘„ 𝑒𝑩 𝑔 𝑩 đč đ‘„orđ‘„Exercises:𝑠 𝑡 𝑩 2Sol: Please try yourself.Exercises:𝑡 𝑠 𝑞 0Sol: The given equation can be written as 𝑞 𝑝 𝑧 0 𝑩 𝑩 𝑩Integrating with respect to 𝑩, we get𝑞 𝑝 𝑧 𝑓 đ‘„ 𝑝 𝑞 𝑓 đ‘„ 𝑧It is of the form𝑃𝑝 𝑄𝑞 𝑅Its auxiliary system isđ‘‘đ‘„1 𝑑𝑩1 đ‘“đ‘‘đ‘§đ‘„ 𝑧 (1)From first two fractions of (1) we getDepartment of Mathematics, University of Kashmir, Srinagar-1900065

Draft PDE Lecture NotesKhanday M.A.đ‘‘đ‘„ 𝑑𝑩 đ‘„ 𝑩 𝑎From first and third fractions of (1) we getđ‘‘đ‘„đ‘‘đ‘§ 1𝑓 đ‘„ 𝑧 𝑓 đ‘„ 𝑧 đ‘‘đ‘„ 𝑑𝑧 𝑑𝑧 𝑓 đ‘„ đ‘§đ‘‘đ‘„ 𝑑𝑧 𝑧 𝑓 đ‘„đ‘‘đ‘„It is first order linear differential equation in 𝑧Its integrating factor isTherefore𝑧𝑒 đ‘„ đ‘’đ‘‘đ‘„ đ‘’đ‘„đ‘“ đ‘„ 𝑒 đ‘„ đ‘‘đ‘„ 𝑧𝑒 đ‘„ 𝑓 đ‘„ 𝑒 đ‘„ đ‘‘đ‘„ 𝑓 𝑩Exercise:𝑡 𝑠 𝑞 1Sol: Please try yourself.Exercise: Find the surface passing through the parabolas,𝑩 2 4đ‘Žđ‘„,𝑩 2 4đ‘Žđ‘„,and𝑧 0𝑧 1and satisfying the equationđ‘„đ‘Ÿ 2𝑝 0.Sol: The given second order partial differential equation isđ‘„đ‘Ÿ 2𝑝 0 𝑝2 (1) đ‘„ đ‘„ 𝑝 0It is first order linear differential equation in 𝑝.Its integrating factor is𝑒2đ‘‘đ‘„đ‘„2 𝑒 2 log đ‘„ 𝑒 log đ‘„ đ‘„ 2Department of Mathematics, University of Kashmir, Srinagar-1900066

Draft PDE Lecture NotesKhanday M.A.From (1) we getđ‘„2 𝑝 đ‘„2 𝑝 0 đ‘‘đ‘„0 đ‘‘đ‘„ 𝑓(𝑩) 𝑝 𝑓(𝑩)đ‘„2Integrating w. r. t. đ‘„ we have1 (2)𝑧 đ‘„đ‘“ 𝑩 đč 𝑩𝑩2Using the given condition 𝑧 0, đ‘„ 4𝑎 , in equation (2), we have0 orđč 𝑩 4𝑎𝑓 𝑩 đč 𝑩𝑩24𝑎𝑓 𝑩 (3)𝑩2Also for 𝑧 1, and đ‘„ 𝑩 24𝑎, we have from (2) we have1 4𝑎𝑓 𝑩 đč 𝑩𝑩2Using (3) we getor1 4𝑎𝑓 𝑩𝑩2 4𝑎𝑓 𝑩𝑩2 1 8𝑎𝑓 𝑩𝑩2 𝑓 𝑩 𝑩28𝑎Substituting 𝑓 𝑩 , in (3)4𝑎 𝑩 2đč 𝑩 2𝑩 8𝑎 đč 𝑩 12Department of Mathematics, University of Kashmir, Srinagar-1900067

Draft PDE Lecture NotesKhanday M.A.Therefore from (1) we get𝑧 𝑩 2 1 8đ‘Žđ‘„ 2Which is the required surface passing through the parabolas.Exercise: Find the surface satisfying 𝑡 6đ‘„ 3 𝑩 containing the two lines𝑩 0 𝑧and𝑩 1 𝑧Sol: The given 2nd order PDE is𝑡 6đ‘„ 3 𝑩 𝑞 6đ‘„ 3 𝑩 𝑩Integrating w. r. t., 𝑩, we have𝑞 6đ‘„ 3 𝑩 2 𝑓 đ‘„2 𝑧 3đ‘„ 3 𝑩 2 𝑓 đ‘„ 𝑩Integrating w. r. t., 𝑩,𝑧 3đ‘„ 3 𝑩 3 𝑩𝑓 đ‘„ đč đ‘„3 𝑧 đ‘„ 3 𝑩 3 𝑩𝑓 đ‘„ đč đ‘„ (1)Using given conditions 𝑩 0 𝑧, in (1), we have0 0 0 đč đ‘„ (2) đč đ‘„ 0Also using 𝑩 1 𝑧in equation (1) we get,1 đ‘„3 𝑓 đ‘„ đč đ‘„Using (2), we get1 đ‘„3 𝑓 đ‘„ 0𝑓 đ‘„ 1 đ‘„3 (3)Using (2) and (3) in (1) we getDepartment of Mathematics, University of Kashmir, Srinagar-1900068

Draft PDE Lecture NotesKhanday M.A.𝑧 đ‘„3 𝑩3 𝑩 1 đ‘„3Which is the required surface containing the two lines.Exercise: Find the surface satisfying 𝑟 𝑠 0, and touching the elliptic paraboloid𝑧 4đ‘„ 2 𝑩 2 along the surface of plane 𝑩 2đ‘„ 1. 𝑝 𝑞Sol: From the given equation we have đ‘„ đ‘„ 0.Integrating with respect to đ‘„, we have𝑝 𝑞 𝑓 𝑩Now, the auxiliary system isđ‘‘đ‘„1 𝑑𝑩1𝑑𝑧 𝑓 (1)𝑩Taking first two fractions we getđ‘‘đ‘„ 𝑑𝑩 11Integrating we getđ‘„ 𝑩 𝑎 (2)đ‘„ 𝑩 𝑎Also from 2nd and 3rd fractions of (1), we get𝑑𝑩𝑑𝑧 1𝑓 𝑩 𝑑𝑧 𝑓 𝑩 𝑑𝑩 𝑧 𝜑 𝑩 𝑏𝑧 𝜑 𝑩 đč 𝑎or 𝑧 𝜑 𝑩 đč đ‘„ 𝑩 (3)From (3), we get 𝑧𝑝 đ‘„ đč â€Č đ‘„ 𝑩 𝑧q 𝑩 𝜑 â€Č 𝑩 đč â€Č đ‘„ 𝑩Department of Mathematics, University of Kashmir, Srinagar-190006 (4) (5)9

Draft PDE Lecture NotesKhanday M.A.Since 𝑧 4đ‘„ 2 𝑩 2 𝑧 (6) 𝑝 đ‘„ 8đ‘„& 𝑧 (7)q 𝑩 2𝑩From (4) and (6)đč â€Č đ‘„ 𝑩 8đ‘„ (8)From (5) and (7)𝜑 â€Č 𝑩 đč â€Č đ‘„ 𝑩 2𝑩 (9)Adding (8) and (9) we get𝜑 â€Č 𝑩 8đ‘„ 2𝑩 8𝑩 1 2𝑩2 6𝑩 4Integrating w. r. t., 𝑩, we get𝜑 𝑩 3𝑩 2 4𝑩 𝑏 (10)Also, from (8) đč â€Č đ‘„ 𝑩 8đ‘„ 8 𝑩 đ‘„ 1 8 đ‘„ 𝑩 1Integrating w. r. t., đ‘„ 𝑩 we get đč đ‘„ 𝑩 4 đ‘„ 𝑩2 8 đ‘„ 𝑩 𝑐 (11)Substituting (10) and (11) in (3) we get𝑧 3𝑩 2 4𝑩 𝑏 4 đ‘„ 𝑩2 8 đ‘„ 𝑩 𝑐 4đ‘„ 2 𝑩 2 4𝑩 8đ‘„ 8đ‘„đ‘Š 𝑑From the given condition,4đ‘„ 2 2đ‘„ 12 4đ‘„ 2 2đ‘„ 1 8đ‘„ 2 2 2đ‘„ 122 4 2đ‘„ 1 8đ‘„ 8đ‘„ 2đ‘„ 1 𝑑 4 2đ‘„ 1 8đ‘„ 8đ‘„ 2đ‘„ 1 𝑑 8đ‘„ 2 8đ‘„ 2 2 8đ‘„ 8đ‘„ 4 8đ‘„ 16đ‘„ 2 8đ‘„ 𝑑Department of Mathematics, University of Kashmir, Srinagar-19000610

Draft PDE Lecture NotesKhanday M.A. 𝑑 2𝑧 4đ‘„ 2 𝑩 2 4𝑩 8đ‘„ 8đ‘„đ‘Š 2Thereforewhich is required surface.Exercise: Show that the surface satisfying 𝑟 6đ‘„ 2 and touching 𝑧 đ‘„ 3 𝑩 3along its section by the plane đ‘„ 𝑩 1 0 is 𝑧 đ‘„ 3 𝑩 3 đ‘„ 𝑩 1 2 .Sol: Try yourself.Partial differential equations with constant coefficients:We know that the general form of a linear partial differential equation 𝑛 𝑧 𝑛 𝑧 𝑛 𝑧 𝑛 𝑧𝐮𝑛 đ‘„ 𝑛 𝐮𝑛 1 đ‘„ 𝑛 1 𝑩 𝐮𝑛 2 đ‘„ 𝑛 2 𝑩 2 𝐮1 𝑩 𝑛 𝑓 đ‘„, 𝑩 (1)Where the coefficients 𝐮𝑛 , 𝐮𝑛 1 , 𝐮𝑛 2 , . . . , 𝐮1 are constants or functionsof đ‘„ and 𝑩. If 𝐮𝑛 , 𝐮𝑛 1 , 𝐮𝑛 2 , . . . , 𝐮1 are all constants, then (1) is calleda linear partial differential equation with constant coefficients.We denote đ‘„and by đ· 𝑜𝑟 đ·đ‘„ 𝑩and đ·â€Č 𝑜𝑟 đ·đ‘Šrespectively.Therefore (1) can be written as𝐮𝑛 đ·đ‘› 𝐮𝑛 1 đ·đ‘› 1 đ·â€Č 𝐮𝑛 2 đ·đ‘› 2 đ·â€Č2 𝐮1 đ·â€Č𝑛 𝑧 𝑓 đ‘„, 𝑩 (2)𝜑 đ·, đ·â€Č 𝑧 𝑓 đ‘„, 𝑩orThe complementary function of (2) is given by𝐮𝑛 đ·đ‘› 𝐮𝑛 1 đ·đ‘› 1 đ·â€Č 𝐮𝑛 2 đ·đ‘› 2 đ·â€Č2 𝐮1 đ·â€Č𝑛 𝑧 0or (3)𝜑 đ·, đ·â€Č 𝑧 0Let 𝑧 đč 𝑩 đ‘šđ‘„be the part of the solution đ‘§đ·đ‘§ đ‘„ 𝑚đč â€Č 𝑩 đ‘šđ‘„ 2đ‘§đ·2 𝑧 đ‘„ 2 𝑚2 đč â€Čâ€Č 𝑩 đ‘šđ‘„.Department of Mathematics, University of Kashmir, Srinagar-19000611

Draft PDE Lecture NotesKhanday M.A. 𝑛 đ‘§đ·đ‘› 𝑧 đ‘„ 𝑛 𝑚𝑛 đč 𝑛 𝑩 đ‘šđ‘„And đ‘§đ·â€Č 𝑧 𝑩 đč â€Č 𝑩 đ‘šđ‘„ 2đ‘§đ·â€Č2 𝑧 𝑩 2 đč â€Čâ€Č 𝑩 đ‘šđ‘„. . .đ·â€Č𝑛 𝑧 𝑛 𝑧 𝑩 𝑛. đč 𝑛 𝑩 đ‘šđ‘„Substitute these values in (3), we get𝐮𝑛 𝑚𝑛 𝐮𝑛 1 𝑚𝑛 1 𝐮𝑛 2 𝑚𝑛 2 . . . 𝐮1 đč𝑛𝑩 đ‘šđ‘„ 0which is true if â€Č𝑚â€Č is a root of the equationIf 𝑚1 , 𝑚2 ,𝑚𝑛 , are distinct roots, then complementary functions is𝑧 𝜑1 𝑩 𝑚1 đ‘„ 𝜑2 𝑩 𝑚2 đ‘„ . . . 𝜑𝑛 𝑩 𝑚𝑛 đ‘„where 𝜑1 , 𝜑2 , . . .,𝜑𝑛 are arbitrary functions. 𝜑 đ·, đ·â€Č 𝑧 0we replace đ· by m and đ·â€Č by 1 to get the auxiliary equation from which we getroots.Linear partial differential equations with constant coefficientsHomogenous and Non homogenous linear equations with constant coefficients: Apartial differential equation in which the dependent variable and its derivatives appearonly in the first degree and are not multiplied together, their coefficients beingconstants or functions of x and y, is known as a linear partial differential equation.The general form of such an equation is 𝑛 𝑧 𝑛 𝑧 𝑛 𝑧 𝑛 𝑧 𝑛 1 𝑧 𝑛 1 𝑧𝐮0 đ‘„ 𝑛 𝐮1 𝑩 đ‘„ 𝑛 1 𝐮1 𝑩 2 đ‘„ 𝑛 2 . . . 𝐮𝑛 𝑩 𝑛 đ”0 đ‘„ 𝑛 1 đ”1 𝑩 đ‘„ 𝑛 2 𝑛 1 𝑧 𝑛 1 𝑧 𝑧 đ‘§đ”1 𝑩 2 đ‘„ 𝑛 3 . . . đ”đ‘› 𝑩 𝑛 1 𝑀0 đ‘„ 𝑀1 𝑩 𝑁0 𝑧 𝑓(đ‘„, 𝑩)Department of Mathematics, University of Kashmir, Srinagar-190006.(1)12

Draft PDE Lecture NotesKhanday M.A.where the coefficients A0, A1, . . . An, B0, B1, . . . , Bn-1, M0, M1 and N0 are allconstants, then (1) is called a linear partial differential equation with constantcoefficients. For convenience đ‘„ and 𝑩 will be denoted by D and đ·Êč respectively.2.1. Short method for finding the P.I. in certain cases of F(D,DÊč) z f(x, y)2.1.a. Short method I. when f(x, y) is of the form ( ax by )Ex.1. Solve (D2 3DDÊč 2DÊč2)z x ySol. The Auxiliary equation of the given equation ism2 3m 2 0 giving m -1,-2therefore C.F. 1(y-x) 2(y-2x), 1, 2 being arbitrary functions1P.I. D2 3DDÊč 2DÊč2(x y)Now 1𝑣𝑑𝑣𝑑𝑣, where v x y12 3.1.1 2.12v221 𝑣3dv 661 36 (x y)3Hence the required general solution is z C.F. P.I.1or z 1(y-x) 2(y-2x) 36 (x y)3Ex. 2. Solve (2đ·2 5đ·đ·Êč 2đ·Êč2 ) 𝑧 24 (𝑩 – đ‘„ )Sol: Try yourselfEx.3. Solve (đ·2 3đ·đ·Êč 2đ·Êč2 )𝑧 2đ‘„ 3 𝑩Sol: Try yourself2.1.b.Short method II.When f(x,y) is of the form đ‘„ 𝑚 𝑩 𝑛 or a rational integralEx.1. Solve (D2 –a2DÊč2)z xSol. Here auxiliary equation is m2 –a2 0 so that m a, -aThereforeC.F. 1 𝑩 đ‘Žđ‘„ 2 𝑩 đ‘Žđ‘„ ,Department of Mathematics, University of Kashmir, Srinagar-190006.(1)13

Draft PDE Lecture NotesKhanday M.A. 1 , 2 being arbitrary functions.Now P.I. 1đ· 2 𝑎 2 đ·Êč21 đ·21 đ·2đ‘„ 1đ·21 𝑎21 𝑎 2đ·Êč2đ·2đ· Êč2đ·2đ‘„ 1đ‘„[1 a2 (DÊč2 /D2) .]x1 đ·2x x3.(2)6Hence the required solution is z C.F. P.I.Z 1(y a x ) 2 (y-a x) x36Exercise: 1.2𝑟 5𝑠 2𝑡 0Sol: It is a second order pole with constant coefficients, we have 2𝑧 2𝑧 2𝑧2 2 5 2 2 0 đ‘„ đ‘„ 𝑩 𝑩2đ·2 5đ·đ·â€Č 2đ·â€Č2 𝑧 0orNow the auxiliary equations is given by2𝑚2 5𝑚 2 01 𝑚 2 , -21Therefore the complementary function is 𝑧 𝜑1 𝑩 2 đ‘„ 𝜑2 𝑩 2đ‘„which is required solution.Exercise: 2.𝑟 𝑎2 𝑡Sol: Try Yourself 3𝑧(Ans: 𝑧 𝜑1 𝑩 đ‘Žđ‘„ 𝜑2 𝑩 đ‘Žđ‘„ ) 3𝑧 3𝑧Exercise: 3. đ‘„ 3 3 đ‘„ 2 𝑩 2 đ‘„ 2 𝑩 0Sol: It is a third order pole with constant coefficients, we haveDepartment of Mathematics, University of Kashmir, Srinagar-19000614

Draft PDE Lecture NotesKhanday M.A. 3𝑧 3𝑧 3𝑧 3 2 2 0 đ‘„ 3 đ‘„ 𝑩 đ‘„ 2 𝑩orđ·3 3đ·2 đ·â€Č 2đ·đ·â€Č2 𝑧 0Now the auxiliary equations is given by𝑚3 3𝑚2 2𝑚 0 𝑚 𝑚2 3𝑚 2 0 𝑚 𝑚 1 𝑚 2 0 𝑚 0, 1 , 2Therefore the complementary function is 𝑧 𝜑1 𝑩 𝜑2 𝑩 đ‘„ 𝜑3 𝑩 2đ‘„Which is required solution. 3𝑧 3𝑧 3𝑧 3𝑧Exercise: 4. đ‘„ 3 6 đ‘„ 2 𝑩 11 đ‘„ 2 𝑩 6 𝑩 3 0Sol: Try yourselfExercise: 5. 25𝑟 40𝑠 16𝑡 0Sol: Try yourselfExercise: 6. đ·4 đ·â€Č4 𝑧 0Sol: The auxiliary equation is given by𝑚4 1 0 𝑚2 1 𝑚2 1 0 𝑚 1 𝑚 1 𝑚 𝑖 𝑚 𝑖 0 𝑚 1, 1, 𝑖, 𝑖Therefore the complementary function is𝑧 𝜑1 𝑩 đ‘„ 𝜑2 𝑩 đ‘„ 𝜑3 𝑩 đ‘–đ‘„ 𝜑4 𝑩 đ‘–đ‘„which is required solution.Exercise: 7. đ·3 4đ·2 đ·â€Č 4đ·đ·â€Č2 𝑧 0Sol: Try yourself.Department of Mathematics, University of Kashmir, Srinagar-19000615

Draft PDE Lecture NotesKhanday M.A.Exercise: 8. đ·2 2đ·đ·â€Č đ·â€Č2 𝑧 12đ‘„đ‘ŠSol: The auxiliary equations corresponding to these linear system of equations isgiven by𝑚2 2𝑚 1 0 𝑚 1 𝑚 1 0 𝑚 1,1Therefore the complementary function is𝑧 𝑓1 𝑩 đ‘„ đ‘„đ‘“2 𝑩 đ‘„Also,P. I. 1đ· đ·â€Č 212đ‘„đ‘Š12đ·â€Č 2 1 đ·đ· 2đ‘„đ‘Š2 112đ·đ· 2 1 2 đ·đ·â€Čđ·â€Č12đ·đ· 2 1 2 đ·đ·â€Čđ·â€Č đ‘„đ‘Š2 đ‘„đ‘Š122đ‘„đ‘Š đ‘„đ·2đ·12 đ‘„ 2 𝑩 đ‘„ 3 đ· 23 12đ‘„3 𝑩 đ‘„4 612 2đ‘„ 3 𝑩 đ‘„ 4Therefore the complete solution isz C. F. P. I.i.e., 𝑧 𝑓1 𝑩 đ‘„ đ‘„đ‘“2 𝑩 đ‘„ 2đ‘„ 3 𝑩 đ‘„ 4Exercise: 9. 2đ·2 5đ·đ·â€Č 2đ·â€Č2 𝑧 24 đ‘„ 𝑩Sol: Try yourself.Exercise: 10. đ·3 đ·â€Č3 𝑧 đ‘„ 3 𝑩 3Department of Mathematics, University of Kashmir, Srinagar-19000616

Draft PDE Lecture NotesKhanday M.A.Sol: The auxiliary equations is𝑚3 1 0 𝑚 1 𝑚2 𝑚 1 0 𝑚 1,𝑧 𝜑1 𝑩 đ‘„ 𝜑2 𝑩 Complementary function is 1 𝑖 32P. I. 1 𝑖 3 1 𝑖 3,22 1 𝑖 32đ‘„ 𝜑3 𝑩 đ‘„ 1đ· 3 đ·â€Č 3đ‘„3 ïżœïżœ31đ·31đ·31 đ·â€Č 3đ·1 1 1đ·â€Č 3đ‘„3 𝑩3đ·đ·â€Č 3đ‘„3𝑩3 đ·â€Č 6 đ·đ·đ·â€Č 3đ‘„3 𝑩3 đ·đ·â€Č 2đ·3 . . .đ‘„3 𝑩3 đ·â€Č 6đ·đ‘„3 𝑩3đ‘„3𝑩3 . . .3đ‘„ 3 𝑩 2 0 . . .đ·â€Čđ‘„ 3 𝑩 3 đ· 3 6đ‘„ 3 𝑩1đ‘„ 3 𝑩 3 đ· 3 6đ‘„ 31đ‘„3 𝑩3 đ·2 61đ‘„44đ‘„5đ‘„ 3 𝑩 3 đ· 6 20đ‘„6đ‘„ 3 𝑩 3 201đ‘„4𝑩 3đ·241 đ‘„5𝑩 3đ·đ‘„3𝑩320đ‘„7 140đ‘„8 1120Department of Mathematics, University of Kashmir, Srinagar-19000617

Draft PDE Lecture Notesđ‘„6𝑩 3 Khanday M.A.đ‘„9 10080120The complete solution is C.F. P.I.𝑧 𝜑1 𝑩 đ‘„ 𝜑2 𝑩 1 𝑖 3 1 𝑖 3đ‘„6 𝑩3đ‘„9đ‘„ 𝜑3 𝑩 đ‘„ 22120 10080Exercise: Find the real function â€Č𝑣â€Č of đ‘„ and 𝑩 reducing to zero when 𝑩 0 andsatisfying 2𝑣 đ‘„ 2 2𝑣 𝑩 2 4𝜋 đ‘„ 2 𝑩 2Sol: We have to find the P. I. onlyP. I. 1đ· 2 đ·â€Č 2 4𝜋 đ‘„ 2 𝑩 21 4𝜋 đ‘„ 2 𝑩 2đ·â€Č 2đ· 2 1 2đ· 4đœ‹đ·2 4đœ‹đ·2 4đœ‹đ·2 4đœ‹đ·2 4đœ‹đ·2 4đœ‹đ·2 4đœ‹đ·2 4đœ‹đ·1 1 đ·â€Č 2 1đ·2đ·â€Č 2đ·2 đ‘„2 𝑩2đ·â€Č 22đ·2đ‘„2 𝑩2 đ·â€Č 2đ·2 . . .đ‘„2 𝑩2đ‘„2 𝑩2 0đ·â€Čđ‘„ 2 𝑩 2 đ· 2 2𝑩1đ‘„2 𝑩2 đ·2 21đ‘„ 2 𝑩 2 đ· 2đ‘„đ‘„2 𝑩2 đ‘„2đ‘„đ‘Š 2đ‘„2𝑩2 4𝜋 2đœ‹đ‘„ 2 𝑩 22Theorem: If 𝑱1 , 𝑱2 , 𝑱3 , . . . , 𝑱𝑛 are the solutions of the homogeneous linearPDE đč đ·, đ·â€Č 𝑧 0, then 𝑛𝑟 1 𝑐𝑟 𝑱𝑟 where 𝑐𝑟 â€Č𝑠 are arbitrary constants, is also asolution.Department of Mathematics, University of Kashmir, Srinagar-19000618

Draft PDE Lecture NotesKhanday M.A.Proof: Since ur , r 1, 2, 3, . . . , n are solutions of the PDEđč đ·, đ·â€Č 𝑧 0So, we have 𝑱𝑟 is one of the solutions i.e.,đč đ·, đ·â€Č 𝑱𝑟 0 ,𝑟 1, 2, . . . , 𝑛 đč đ·, đ·â€Č 𝑐𝑟 𝑱𝑟 𝑐𝑟 â€Č đč đ·, đ·â€Č 𝑱𝑟andđč đ·, đ·â€Č𝑱𝑟 đč đ·, đ·â€Č 𝑱𝑟 for any set of functions 𝑱𝑟 , we have𝑛đč đ·, đ·đ‘›â€Čđč đ·, đ·â€Č 𝑐𝑟 𝑱𝑟𝑐𝑟 𝑱𝑟 𝑟 1𝑟 1𝑛𝑐𝑟 đč đ·, đ·â€Č 𝑱𝑟 𝑟 1 0Therefore𝑛𝑟 1 𝑐𝑟 𝑱𝑟acts as a solution for the homogeneous system.Reducible and irreducible:If an operator đč đ·, đ·â€Č can be expressed as a product of linear factors, it is said to bereducible. If it can not be factorised, then it is said to be irreducible.Theorem: If đ›Œđ‘Ÿ đ· đ›œđ‘Ÿ đ·â€Č đ›Ÿđ‘Ÿ is a factor of đč đ·, đ·â€Č and 𝜑𝑟 𝜉 , then𝑱𝑟 đ‘’đ‘„đ‘. đ›Ÿđ‘Ÿ đ‘„đ›Œđ‘Ÿđœ‘đ‘Ÿ đ›œđ‘Ÿ đ‘„ đ›Œđ‘Ÿ 𝑩 for đ›Œđ‘Ÿ is a solution of the equation đč đ·, đ·â€Č 𝑧 0.Proof: The given equation isIn order to prove 𝑱𝑟 đ‘’đ‘„đ‘. đč đ·, đ·â€Č 𝑧 0 . . .đ›Ÿđ‘Ÿ đ‘„đ›Œđ‘Ÿis a solution of (1), we have to prove𝜑𝑟 đ›œđ‘Ÿ đ‘„ đ›Œđ‘Ÿ 𝑩 ;(1)đ›Œđ‘Ÿ 0 . . .(2)đč đ·, đ·â€Č 𝑱𝑟 0Diff. eq.(2) w.r.t. đ‘„ and 𝑩, we getđ·đ‘ąđ‘Ÿ đ›Ÿđ‘Ÿđ›Ÿđ‘Ÿ đ‘„đ‘ąđ‘Ÿ đ›œđ‘Ÿ đ‘’đ‘„đ‘. 𝜑â€Č đ›œđ‘Ÿ đ‘„ đ›Œđ‘Ÿ đ‘Šđ›Œđ‘Ÿđ›Œđ‘ŸDepartment of Mathematics, University of Kashmir, Srinagar-19000619

Draft PDE Lecture NotesKhanday M.A.Andđ·â€Č𝑱𝑟 đ›Œđ‘Ÿ đ‘’đ‘„đ‘. đ›Ÿđ‘Ÿ đ‘„đœ‘â€Č đ›œđ‘Ÿ đ‘„ đ›Œđ‘Ÿ đ‘Šđ›Œđ‘Ÿ đ›Œđ‘Ÿ đ· đ›œđ‘Ÿ đ·â€Č đ›Ÿđ‘Ÿ 𝑱𝑟 đ›Ÿđ‘Ÿ 𝑱𝑟 đ›Œđ‘Ÿ đ›œđ‘Ÿ đ‘’đ‘„đ‘. đ›Œđ‘Ÿ đ›œđ‘Ÿ đ‘’đ‘„đ‘. đ›Ÿđ‘Ÿ đ‘„đ›Œđ‘Ÿđœ‘ â€Č đ›œđ‘Ÿ đ‘„ đ›Œđ‘Ÿ 𝑩 đ›Ÿđ‘Ÿ 𝑱𝑟 0 . . .đ›Ÿđ‘Ÿ đ‘„ â€Č𝜑 đ›œđ‘Ÿ đ‘„ đ›Œđ‘Ÿ đ‘Šđ›Œđ‘Ÿ(3)Since đ›Œđ‘Ÿ đ· đ›œđ‘Ÿ đ· â€Č đ›Ÿđ‘Ÿ is a factor of đč đ·, đ·â€ČTherefore đč đ·, đ·â€Č 𝑧 𝑔 đ·, đ·â€Č đ›Œđ‘Ÿ đ· đ›œđ‘Ÿ đ·â€Č đ›Ÿđ‘Ÿ 𝑧,using (3), we getđč đ·, đ·â€Č 𝑱𝑟 0Therefore 𝑱𝑟 is a solution of đč đ·, đ·â€Č 𝑧 0Solution of Reducible Equations:Let đč đ·, đ·â€Č 𝑧 𝑓 đ‘„, 𝑩. . .(1)be a partial differential equation. Since (1) is reducible therefoređ‘›đ›Œđ‘Ÿ đ· đ›œđ‘Ÿ đ·â€Č đ›Ÿđ‘Ÿ 𝑧đč đ·, đ·â€Č 𝑧 𝑟 1If 𝑧 satisfies đ›Œđ‘Ÿ đ· đ›œđ‘Ÿ đ·â€Č đ›Ÿđ‘Ÿ 𝑧 0, 𝑟 0, 1, 2, . . . , 𝑛, then it gives uscomplementary function 𝑧 𝑧Now đ›Œđ‘Ÿ đ‘„ đ›œđ‘Ÿ 𝑩 đ›Ÿđ‘Ÿ 𝑧 0The subsidiary system isđ‘‘đ‘„ 𝑑𝑩 𝑑𝑧 đ›Œđ‘Ÿđ›œđ‘Ÿ đ›Ÿđ‘Ÿ 𝑧From the first two membersđ›œđ‘Ÿ đ‘„ đ›Œđ‘Ÿ 𝑩 𝑐𝑟From first and last members we getđ‘‘đ‘§đ›Ÿđ‘Ÿ đ‘‘đ‘„đ‘§đ›Œđ‘ŸDepartment of Mathematics, University of Kashmir, Srinagar-19000620

Draft PDE Lecture NotesKhanday M.A.đ›Ÿlog 𝑧 đ›Œđ‘Ÿ đ‘„ 𝐮𝑟Integrating we get𝑟 𝑧 log đ›œđ‘Ÿ exp đ›Ÿđ‘Ÿđ‘„đ›Œđ‘Ÿ 𝑧 𝜑 𝑟 exp 𝑒 𝐮𝑟 đ›œđ‘Ÿđ›Ÿđ‘Ÿđ‘„đ›Œđ‘Ÿ 𝑧 𝜑 đ›œđ‘Ÿ đ‘„ đ›Œđ‘Ÿ 𝑩 exp đ›Ÿđ‘Ÿđ‘„đ›Œđ‘ŸAlsođ‘‘đ‘§đ›Ÿđ‘Ÿ đ‘‘đ‘Šđ‘§đ›œđ‘Ÿ 𝑧 𝜑 đ›œđ‘Ÿ đ‘„ exp Example: Let đ›Œđ‘Ÿ đ· đ›œđ‘Ÿ đ·â€Č đ›Ÿđ‘Ÿ 𝑧1 0 wheređ›Ÿđ‘Ÿđ‘Šđ›Œđ‘Ÿđ‘§1 đ›Œđ‘Ÿ đ· đ›œđ‘Ÿ đ·â€Č đ›Ÿđ‘Ÿ 𝑧𝑧1 𝜑 đ›œđ‘Ÿ đ‘„ đ›Œđ‘Ÿ 𝑩 exp đ›Ÿđ‘Ÿđ‘„đ›Œđ‘Ÿ đ›Œđ‘Ÿ đ· đ›œđ‘Ÿ đ·â€Č đ›Ÿđ‘Ÿ 𝑧 𝜑𝑟 đ›œđ‘Ÿ đ‘„ đ›Œđ‘Ÿ 𝑩 exp đ›Œđ‘Ÿđ›Ÿđ‘Ÿđ‘„đ›Œđ‘Ÿ 𝑧 đ‘§đ›Ÿđ‘Ÿ đ›œđ‘Ÿ 𝜑𝑟 đ›œđ‘Ÿ đ‘„ đ›Œđ‘Ÿ 𝑩 exp đ‘„ đ›Ÿđ‘Ÿ 𝑧 đ‘„ đ‘Šđ›Œđ‘ŸAuxiliary system isđ‘‘đ‘„ 𝑑𝑩𝑑𝑧 đ›Œđ‘Ÿđ›œđ‘Ÿ 𝜑 đ›œ đ‘„ đ›Œ 𝑩 exp đ›Ÿđ‘Ÿ đ‘„ đ›Ÿ 𝑧𝑟 đ‘Ÿđ‘Ÿđ‘Ÿđ›Œđ‘ŸFrom first two we getđ‘‘đ‘„ 𝑑𝑩 đ›Œđ‘Ÿđ›œđ‘Ÿ đ›œđ‘Ÿ đ‘‘đ‘„ đ›Œđ‘Ÿ 𝑑𝑩 đ›œđ‘Ÿ đ‘„ đ›Œđ‘Ÿ 𝑑𝑩 𝑐𝑟From first and third we getDepartment of Mathematics, University of Kashmir, Srinagar-19000621

Draft PDE Lecture NotesKhanday M.A.đ‘‘đ‘„đ‘‘đ‘§ đ›Œđ‘Ÿ 𝜑 đ›œ đ‘„ đ›Œ 𝑩 exp đ›Ÿđ‘Ÿ đ‘„ đ›Ÿ 𝑧𝑟 đ‘Ÿđ‘Ÿđ‘Ÿđ›Œđ‘Ÿđ›Ÿđ‘Ÿđ‘‘đ‘§ 𝜑𝑟 đ›œđ‘Ÿ đ‘„ đ›Œđ‘Ÿ 𝑩 exp đ›Œđ‘Ÿ đ‘„ đ›Ÿđ‘Ÿ 𝑧 đ‘‘đ‘„đ›Œđ‘Ÿđ›Ÿđ‘Ÿđ‘‘đ‘§ đ›Ÿđ‘Ÿ 𝑧 𝜑𝑟 đ›œđ‘Ÿ đ‘„ đ›Œđ‘Ÿ 𝑩 exp đ›Œđ‘Ÿ đ‘„ đ‘‘đ‘„ đ›Œđ‘Ÿđ›Œđ‘Ÿđ›Ÿđ‘Ÿđ‘„Here I. F. is 𝑒 đ›Œ 𝑟therefore the above equation can be written asđ›Ÿđ‘Ÿ đ‘„đ‘‘ 𝑧𝑒 đ›Œ đ‘Ÿđ›Ÿđ‘Ÿ đ‘„ 𝑧𝑒 đ›Œ 𝑟 đ›Ÿđ‘Ÿ đ‘„ 𝑧𝑒 đ›Œ 𝑟 𝑒 1đ›Œđ‘Ÿđ›Ÿ đ‘„ đ‘Ÿđ›Œđ‘Ÿđœ‘đ‘Ÿ đ›œđ‘Ÿ đ‘„ đ›Œđ‘Ÿ đ‘Šđ›Œđ‘Ÿđœ‘đ‘Ÿ đ›œđ‘Ÿ đ‘„ đ›Œđ‘Ÿ 𝑩 đ‘‘đ‘„ đ›œđ‘Ÿđœ‘đ‘Ÿ đ›œđ‘Ÿ đ‘„ đ›Œđ‘Ÿ 𝑩 𝜓𝑟 đ›œđ‘Ÿ đ‘„ đ›Œđ‘Ÿ 𝑩Example: If 𝑧 𝑒 đ‘Žđ‘„ 𝑏𝑩Then đč đ·, đ·â€Č 𝑧 đč 𝑎, 𝑏 𝑒 đ‘Žđ‘„ 𝑏𝑩𝑧 acts as the solution of đč đ·, đ·â€Č 𝑧, where đč đ·, đ·â€Č 𝑧 is reducible if đč 𝑎, 𝑏 0. 3𝑧 3𝑧 3𝑧 3𝑧Exercise: đ‘„ 3 2 đ‘„ 2 𝑩 đ‘„ 𝑩 2 2 𝑩 3 𝑒 đ‘„ 𝑩Sol: The given differential equations can be written asđ·3 2đ·2 đ·â€Č đ·đ·â€Č2 2đ·â€Č3 𝑧 𝑒 đ‘„ 𝑩Auxiliary equations 𝑚3 2𝑚2 𝑚 2 0 𝑚 1 𝑚2 𝑚 2 0 𝑚 1, 1,2Therefore the C. F. is𝑧 𝑓1 𝑩 đ‘„ 𝑓1 𝑩 đ‘„ 𝑓1 𝑩 2đ‘„P. I. 1đ· 3 2đ· 2 đ· â€Č đ·đ·â€Č 2 2đ·â€Č 3𝑒 đ‘„ 𝑩Department of Mathematics, University of Kashmir, Srinagar-19000622

Draft PDE Lecture Notes Now letKhanday M.A.1đ·2đ· 2đ·â€Č đ· â€Č2đ· đ·â€Č1đ· 2 đ· â€Č2đ· 2đ·â€Č𝑒 đ‘„ 𝑩1đ· đ· â€Čđ· đ·â€Č đ· 2đ·â€Čđ· đ· â€Č1 1 1 211 2 đ· đ· â€Č𝑒 đ‘„ 𝑩𝑒 đ‘„ 𝑩𝑒 đ‘„ 𝑩𝑒 đ‘„ 𝑩1đ‘€ đ· đ·â€Č 𝑒 đ‘„ 𝑩 đ· đ·â€Č đ‘€ 𝑒 đ‘„ 𝑩The auxiliary equations aređ‘‘đ‘„ đ‘‘đ‘Šđ‘‘đ‘€ đ‘„ 𝑩1 1 𝑒From first two members we haveđ‘‘đ‘„ 𝑑𝑩 1 1 đ‘‘đ‘„ 𝑑𝑩 0 đ‘„ 𝑩 𝑐From first and third member we getđ‘‘đ‘„đ‘‘đ‘€ đ‘„ 𝑩1𝑒 đ‘‘đ‘„ đ‘‘đ‘€đ‘’đ‘ đ‘‘đ‘€ 𝑒 𝑐 đ‘‘đ‘„ đ‘€ 𝑒𝑐 đ‘„ đ‘€ đ‘„đ‘’ đ‘„ đ‘Šđ‘€1Therefore the particular integral 2 2 đ‘„đ‘’ đ‘„ 𝑩Hence the complete solution is 𝑧 đ¶. đč 𝑃. đŒDepartment of Mathematics, University of Kashmir, Srinagar-19000623

Draft PDE Lecture NotesKhanday M.A.1𝑧 𝑓1 𝑩 đ‘„ 𝑓1 𝑩 đ‘„ 𝑓1 𝑩 2đ‘„ đ‘„đ‘’ đ‘„ 𝑩2Laplace Equation 2 2 𝑧 𝑧 0 đ‘„ 2 𝑩 22Exercise: Find the solution of the equation 2 z e x cos yWhich tends to 0 as đ‘„ and cos 𝑩 for đ‘„ 0.Sol: The given pde is 2 z e x cos y 2𝑧 2𝑧 2 2 e x cos y đ‘„ 𝑩 đ·2 đ·â€Č2 𝑧 e x cos y. . .On comparing with đč đ·, đ·â€Č đ·2 đ·â€Č2(1)and𝑓 đ‘„, 𝑩 e x cos yLet 𝑧 𝑒 đ‘Žđ‘„ 𝑏𝑩 be the solution of (1) đ·2 đ·â€Č2 𝑧 𝑎2 𝑒 đ‘Žđ‘„ 𝑏𝑩 𝑏 2 𝑒 đ‘Žđ‘„ 𝑏𝑩𝑎2 𝑏 2 𝑒 đ‘Žđ‘„ 𝑏𝑩 Where 𝑎2 𝑏 2 0 đč 𝑎, 𝑏Therefore the complementary function is đ¶. đč. 𝑟 0 𝐮𝑟𝑒 đ‘Žđ‘„ 𝑏𝑩𝐮𝑟 â€Č𝑠 being the constants and 𝑎𝑟2 𝑏𝑟2 0AlsoP. I. 1đ· 2 đ·â€Č 2e x cos y1 cos 𝑩 đ· 2 1 e x đ‘„ cos 𝑩 2đ· e x 2 cos 𝑩 e x1đ‘„Therefore the complete solution is 𝑧 𝑟 0đ‘„đŽđ‘Ÿ 𝑒 đ‘Žđ‘„ 𝑏𝑩 cos 𝑩 e x2Department of Mathematics, University of Kashmir, Srinagar-19000624

Draft PDE Lecture NotesKhanday M.A.Using 𝑧 0 as đ‘„ , we write𝑎𝑟 𝜆𝑟 where 𝜆𝑟 0Since 𝑎𝑟2 𝑏𝑟2 0 𝑏𝑟 𝑖 𝑎𝑟2 𝑖𝜆𝑟 𝑧 𝑟 0 𝑟 0đ‘„đŽđ‘Ÿ 𝑒 𝜆 𝑟 đ‘„ 𝑒 𝑖𝜆 𝑟 đ‘„ cos 𝑩 e x2đ‘„đ”đ‘Ÿ 𝑒 𝜆 𝑟 đ‘„ cos 𝜆𝑟 𝑩 𝜖𝑟 cos 𝑩 e x2Using the boundary condition cos 𝑩 đ”đ‘Ÿ cos 𝜆𝑟 𝑩 𝜖𝑟𝑟 0Where đ”đ‘Ÿ 1 𝑎𝑛𝑑 𝜆𝑟 1 𝑓𝑜𝑟 𝑟 0 andđ”đ‘Ÿ 0 𝑎𝑛𝑑 𝜆𝑟 0 𝑓𝑜𝑟 𝑟 0đ‘„Therefore 𝑧 cos 𝑩 e x 2 cos 𝑩 e x is the required solution.Exercise: Show that the equation 2𝑩 𝑡 2 𝑩 2𝑩 2𝑘 𝑡 𝑐 2 đ‘„ 2đ¶đ‘Ÿ 𝑒 𝑘𝑡 cos đ›Œđ‘Ÿ đ‘„

Chapter 2 PARTIAL DIFFERENTIAL EQUATIONS OF SECOND ORDER INTRODUCTION: An equation is said to be of order two, if it involves at least one of the differential coefficients r (ĂČ 2z / ĂČ 2x), s (ĂČ 2z / ĂČ x ĂČ y), t (ĂČ 2z / ĂČ 2y), but now of higher order; the quantities p and q may also enter into the equation. Thus the

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