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- 1. MA 201, Mathematics III, July-November 2012,Part II: Partial Diﬀerential Equations and Integral Transforms October 9, 2012 () MA 201, PDE October 9, 2012 1 / 48
- 2. Classiﬁcation of PDEs () MA 201, PDE October 9, 2012 2 / 48
- 3. Classiﬁcation of PDEsAn ODE is classiﬁed according toits order and whether it is linear or nonlinear. () MA 201, PDE October 9, 2012 2 / 48
- 4. Classiﬁcation of PDEsAn ODE is classiﬁed according toits order and whether it is linear or nonlinear.PDE models are more diﬃcult to classify () MA 201, PDE October 9, 2012 2 / 48
- 5. Classiﬁcation of PDEsAn ODE is classiﬁed according toits order and whether it is linear or nonlinear.PDE models are more diﬃcult to classifyOrder and linear arenot the only issues. () MA 201, PDE October 9, 2012 2 / 48
- 6. Classiﬁcation of PDEsAn ODE is classiﬁed according toits order and whether it is linear or nonlinear.PDE models are more diﬃcult to classifyOrder and linear arenot the only issues.In Factthe structure of the PDE dictates what type of boundary and initial conditions are tobe imposed. () MA 201, PDE October 9, 2012 2 / 48
- 7. Classiﬁcation of PDEs () MA 201, PDE October 9, 2012 3 / 48
- 8. Classiﬁcation of PDEsWhat is the structure of a PDE? () MA 201, PDE October 9, 2012 3 / 48
- 9. Classiﬁcation of PDEsWhat is the structure of a PDE?Let us consider the following second-order equations: ut − αuxx = 0 utt − c2 uxx = 0 uxx + uyy = 0. () MA 201, PDE October 9, 2012 3 / 48
- 10. Classiﬁcation of PDEsWhat is the structure of a PDE?Let us consider the following second-order equations: ut − αuxx = 0 utt − c2 uxx = 0 uxx + uyy = 0.The ﬁrst two equations areevolution equations that describe how a process evolves in time. () MA 201, PDE October 9, 2012 3 / 48
- 11. Classiﬁcation of PDEsWhat is the structure of a PDE?Let us consider the following second-order equations: ut − αuxx = 0 utt − c2 uxx = 0 uxx + uyy = 0.The ﬁrst two equations areevolution equations that describe how a process evolves in time.We expect thateach would require initial data that give the state of the system at time t = 0. () MA 201, PDE October 9, 2012 3 / 48
- 12. Classiﬁcation of PDEs () MA 201, PDE October 9, 2012 4 / 48
- 13. Classiﬁcation of PDEsThe third one isan equilibrium equation where time is not an issue () MA 201, PDE October 9, 2012 4 / 48
- 14. Classiﬁcation of PDEsThe third one isan equilibrium equation where time is not an issuebut we expect that boundary conditions are appropriate () MA 201, PDE October 9, 2012 4 / 48
- 15. Classiﬁcation of PDEsThe third one isan equilibrium equation where time is not an issuebut we expect that boundary conditions are appropriateMoreover we have a feeling thatthe solutions of these will behave diﬀerently. () MA 201, PDE October 9, 2012 4 / 48
- 16. Classiﬁcation of PDEsThe third one isan equilibrium equation where time is not an issuebut we expect that boundary conditions are appropriateMoreover we have a feeling thatthe solutions of these will behave diﬀerently.We observe • Diﬀusion tends to smear out signals • Waves tend to propagate with some coherence • Solutions to Laplace’s equation are steady-state and do not vary in time at all. () MA 201, PDE October 9, 2012 4 / 48
- 17. Classiﬁcation of PDEsThe third one isan equilibrium equation where time is not an issuebut we expect that boundary conditions are appropriateMoreover we have a feeling thatthe solutions of these will behave diﬀerently.We observe • Diﬀusion tends to smear out signals • Waves tend to propagate with some coherence • Solutions to Laplace’s equation are steady-state and do not vary in time at all.If physical classiﬁcations are allowed,we would call these equations diﬀusive, wave-like and static. () MA 201, PDE October 9, 2012 4 / 48
- 18. Classiﬁcation of Second order PDEs () MA 201, PDE October 9, 2012 5 / 48
- 19. Classiﬁcation of Second order PDEsGeneral second-order partial diﬀerential equation in two independent variablesx, t is of the form Auxx + Buxt + Cutt + Dux + Eut + F (x, t)u = G(x, t),where A, B, C are assumed to be constant. () MA 201, PDE October 9, 2012 5 / 48
- 20. Classiﬁcation of Second order PDEsGeneral second-order partial diﬀerential equation in two independent variablesx, t is of the form Auxx + Buxt + Cutt + Dux + Eut + F (x, t)u = G(x, t),where A, B, C are assumed to be constant.which may conveniently be written as Auxx + Buxt + Cutt + f (x, t, u, ux , ut ) = 0, (1) () MA 201, PDE October 9, 2012 5 / 48
- 21. Classiﬁcation of Second order PDEsGeneral second-order partial diﬀerential equation in two independent variablesx, t is of the form Auxx + Buxt + Cutt + Dux + Eut + F (x, t)u = G(x, t),where A, B, C are assumed to be constant.which may conveniently be written as Auxx + Buxt + Cutt + f (x, t, u, ux , ut ) = 0, (1)Note • Second-order derivatives appear linearly • The expression Lu ≡ Auxx + Buxt + Cutt is called the Principal part of the equation. • Classiﬁcation of such PDEs are based on this principal part. • If the problem involves only spatial variables, replace t by y, say. () MA 201, PDE October 9, 2012 5 / 48
- 22. Classiﬁcation of Second order PDEs (Contd.) () MA 201, PDE October 9, 2012 6 / 48
- 23. Classiﬁcation of Second order PDEs (Contd.)The classiﬁcation is basedon the sign of the quantity D ≡ B 2 − 4AC, called the discriminant. () MA 201, PDE October 9, 2012 6 / 48
- 24. Classiﬁcation of Second order PDEs (Contd.)The classiﬁcation is basedon the sign of the quantity D ≡ B 2 − 4AC, called the discriminant.We say • Equation (1) is hyperbolic if D > 0. • Equation (1) is parabolic if D = 0. • Equation (1) is elliptic if D < 0. () MA 201, PDE October 9, 2012 6 / 48
- 25. Classiﬁcation of Second order PDEs (Contd.)The classiﬁcation is basedon the sign of the quantity D ≡ B 2 − 4AC, called the discriminant.We say • Equation (1) is hyperbolic if D > 0. • Equation (1) is parabolic if D = 0. • Equation (1) is elliptic if D < 0.Under this classiﬁcation • Wave equation is hyperbolic. • Diﬀusion or heat equation is parabolic. • Laplace’s equation is elliptic. () MA 201, PDE October 9, 2012 6 / 48
- 26. Classiﬁcation of Second order PDEs (Contd.) () MA 201, PDE October 9, 2012 7 / 48
- 27. Classiﬁcation of Second order PDEs (Contd.)The idea behind this terminology: () MA 201, PDE October 9, 2012 7 / 48
- 28. Classiﬁcation of Second order PDEs (Contd.)The idea behind this terminology:Familiar classiﬁcation of plane curves () MA 201, PDE October 9, 2012 7 / 48
- 29. Classiﬁcation of Second order PDEs (Contd.)The idea behind this terminology:Familiar classiﬁcation of plane curvesFor exampleAx2 + Ct2 = 1, where A, C > 0, B = 0. Hence D < 0. It gives us a graph which is anellipse in the xt-plane. () MA 201, PDE October 9, 2012 7 / 48
- 30. Classiﬁcation of Second order PDEs (Contd.)The idea behind this terminology:Familiar classiﬁcation of plane curvesFor exampleAx2 + Ct2 = 1, where A, C > 0, B = 0. Hence D < 0. It gives us a graph which is anellipse in the xt-plane.SimilarlyAx2 − Ct2 = 1, where D > 0. It gives us a graph which is hyperbola.Similarlyx2 − 4at = 0, where D = 0. It gives us a graph which is parabola. () MA 201, PDE October 9, 2012 7 / 48
- 31. Classiﬁcation of Second order PDEs (Contd.) () MA 201, PDE October 9, 2012 8 / 48
- 32. Classiﬁcation of Second order PDEs (Contd.)If A, B, C are not constants, but are functions of x and t,the discriminant depends on x and t () MA 201, PDE October 9, 2012 8 / 48
- 33. Classiﬁcation of Second order PDEs (Contd.)If A, B, C are not constants, but are functions of x and t,the discriminant depends on x and tSame classiﬁcations apply to these cases, butthe sign of D can change, depending upon the domain. () MA 201, PDE October 9, 2012 8 / 48
- 34. Classiﬁcation of Second order PDEs (Contd.)If A, B, C are not constants, but are functions of x and t,the discriminant depends on x and tSame classiﬁcations apply to these cases, butthe sign of D can change, depending upon the domain.That isthe classiﬁcation may change when values of the variables change. () MA 201, PDE October 9, 2012 8 / 48
- 35. Classiﬁcation of Second order PDEs (Contd.)If A, B, C are not constants, but are functions of x and t,the discriminant depends on x and tSame classiﬁcations apply to these cases, butthe sign of D can change, depending upon the domain.That isthe classiﬁcation may change when values of the variables change.The principal part Lu in the equation can be simpliﬁed for each of the three typesby introducing a new set of variablesThis idea of introducing new variables, common in many diﬀerential equations,renders the given diﬀerential equation to a much simpliﬁed form: enabling integrationwithout much diﬃculty. () MA 201, PDE October 9, 2012 8 / 48
- 36. Method of Characteristics () MA 201, PDE October 9, 2012 9 / 48
- 37. Method of CharacteristicsAssume A, B, C to be constants and seek a linear transformation: () MA 201, PDE October 9, 2012 9 / 48
- 38. Method of CharacteristicsAssume A, B, C to be constants and seek a linear transformation: ξ = ax + bt, η = cx + dtthat simpliﬁes Lu. () MA 201, PDE October 9, 2012 9 / 48
- 39. Method of CharacteristicsAssume A, B, C to be constants and seek a linear transformation: ξ = ax + bt, η = cx + dtthat simpliﬁes Lu.Hereξ and η are new independent variables and a, b, c, d are to be determined for diﬀerentcases. () MA 201, PDE October 9, 2012 9 / 48
- 40. Method of CharacteristicsAssume A, B, C to be constants and seek a linear transformation: ξ = ax + bt, η = cx + dtthat simpliﬁes Lu.Hereξ and η are new independent variables and a, b, c, d are to be determined for diﬀerentcases.Assume that ad − bc = 0 () MA 201, PDE October 9, 2012 9 / 48
- 41. Method of CharacteristicsAssume A, B, C to be constants and seek a linear transformation: ξ = ax + bt, η = cx + dtthat simpliﬁes Lu.Hereξ and η are new independent variables and a, b, c, d are to be determined for diﬀerentcases.Assume that ad − bc = 0so thatthe transformation is invertible, that is, () MA 201, PDE October 9, 2012 9 / 48
- 42. Method of CharacteristicsAssume A, B, C to be constants and seek a linear transformation: ξ = ax + bt, η = cx + dtthat simpliﬁes Lu.Hereξ and η are new independent variables and a, b, c, d are to be determined for diﬀerentcases.Assume that ad − bc = 0so thatthe transformation is invertible, that is,x and t can be solved in terms of ξ and η. () MA 201, PDE October 9, 2012 9 / 48
- 43. Method of Characteristics (Contd.) Figure: Characteristic Plane (ξ, η) () MA 201, PDE October 9, 2012 10 / 48
- 44. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 11 / 48
- 45. Method of Characteristics (Contd.)The dependent variable u in the new variables will be denoted by () MA 201, PDE October 9, 2012 11 / 48
- 46. Method of Characteristics (Contd.)The dependent variable u in the new variables will be denoted by U = U (ξ, η) () MA 201, PDE October 9, 2012 11 / 48
- 47. Method of Characteristics (Contd.)The dependent variable u in the new variables will be denoted by U = U (ξ, η)that is u(x, t) = U (ax + bt, cx + dt) () MA 201, PDE October 9, 2012 11 / 48
- 48. Method of Characteristics (Contd.)The dependent variable u in the new variables will be denoted by U = U (ξ, η)that is u(x, t) = U (ax + bt, cx + dt)By chain rule () MA 201, PDE October 9, 2012 11 / 48
- 49. Method of Characteristics (Contd.)The dependent variable u in the new variables will be denoted by U = U (ξ, η)that is u(x, t) = U (ax + bt, cx + dt)By chain rule ux = Uξ ξx + Uη ηx = aUξ + cUη , ut = Uξ ξt + Uη ηt = bUξ + dUη . () MA 201, PDE October 9, 2012 11 / 48
- 50. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 12 / 48
- 51. Method of Characteristics (Contd.)The second-order partial derivatives can be obtained by another application ofthe chain rule () MA 201, PDE October 9, 2012 12 / 48
- 52. Method of Characteristics (Contd.)The second-order partial derivatives can be obtained by another application ofthe chain ruleThey can be found as () MA 201, PDE October 9, 2012 12 / 48
- 53. Method of Characteristics (Contd.)The second-order partial derivatives can be obtained by another application ofthe chain ruleThey can be found as uxx = a2 Uξξ + 2acUξη + c2 Uηη , utt = b2 Uξξ + 2bdUξη + d2 Uηη , uxt = abUξξ + (ad + bc)Uξη + cdUηη () MA 201, PDE October 9, 2012 12 / 48
- 54. Method of Characteristics (Contd.)The second-order partial derivatives can be obtained by another application ofthe chain ruleThey can be found as uxx = a2 Uξξ + 2acUξη + c2 Uηη , utt = b2 Uξξ + 2bdUξη + d2 Uηη , uxt = abUξξ + (ad + bc)Uξη + cdUηηSubstituting these into the principal part Auxx + Buxt + Cutt = (Aa2 + Bab + Cb2 )Uξξ +(2acA + B(ad + bc) + 2Cbd)Uξη +(Ac2 + Bcd + Cd2 )Uηη . (2) () MA 201, PDE October 9, 2012 12 / 48
- 55. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 13 / 48
- 56. Method of Characteristics (Contd.)We can select a, b, c, d so thatsome of the second partial derivatives in the new variables disappear. () MA 201, PDE October 9, 2012 13 / 48
- 57. Method of Characteristics (Contd.)We can select a, b, c, d so thatsome of the second partial derivatives in the new variables disappear.This process must be handled diﬀerentlydepending on the sign of the discriminant D. () MA 201, PDE October 9, 2012 13 / 48
- 58. Method of Characteristics (Contd.)We can select a, b, c, d so thatsome of the second partial derivatives in the new variables disappear.This process must be handled diﬀerentlydepending on the sign of the discriminant D.KEEP IN MIND THAT WHATEVER TRANSFORMATION IS USED, THETYPE OF THE PDE DOES NOT GET CHANGEDThe transformation only allows us to handle a simpliﬁed PDE (in new variables) () MA 201, PDE October 9, 2012 13 / 48
- 59. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 14 / 48
- 60. Method of Characteristics (Contd.)Hyperbolic Case, D > 0:Let us choose a = c = 1 () MA 201, PDE October 9, 2012 14 / 48
- 61. Method of Characteristics (Contd.)Hyperbolic Case, D > 0:Let us choose a = c = 1In that casethe coeﬃcients of Uξξ and Uηη , which have the same form () MA 201, PDE October 9, 2012 14 / 48
- 62. Method of Characteristics (Contd.)Hyperbolic Case, D > 0:Let us choose a = c = 1In that casethe coeﬃcients of Uξξ and Uηη , which have the same formbecome quadratic expressions in b and d, respectively. () MA 201, PDE October 9, 2012 14 / 48
- 63. Method of Characteristics (Contd.)Hyperbolic Case, D > 0:Let us choose a = c = 1In that casethe coeﬃcients of Uξξ and Uηη , which have the same formbecome quadratic expressions in b and d, respectively.We can force these coeﬃcients to vanish by choosing √ √ −B + D −B − D b= , d= (3) 2C 2C () MA 201, PDE October 9, 2012 14 / 48
- 64. Method of Characteristics (Contd.)Hyperbolic Case, D > 0:Let us choose a = c = 1In that casethe coeﬃcients of Uξξ and Uηη , which have the same formbecome quadratic expressions in b and d, respectively.We can force these coeﬃcients to vanish by choosing √ √ −B + D −B − D b= , d= (3) 2C 2Cby simply using the quadratic formula. () MA 201, PDE October 9, 2012 14 / 48
- 65. Method of Characteristics (Contd.)Hyperbolic Case, D > 0:Let us choose a = c = 1In that casethe coeﬃcients of Uξξ and Uηη , which have the same formbecome quadratic expressions in b and d, respectively.We can force these coeﬃcients to vanish by choosing √ √ −B + D −B − D b= , d= (3) 2C 2Cby simply using the quadratic formula.The remaining coeﬃcient, that of Uξη is nonzero. () MA 201, PDE October 9, 2012 14 / 48
- 66. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 15 / 48
- 67. Method of Characteristics (Contd.)Consequentlywe ﬁnd that () MA 201, PDE October 9, 2012 15 / 48
- 68. Method of Characteristics (Contd.)Consequentlywe ﬁnd thatthe transformation √ √ −B + D −B − D ξ =x+ t, η = x + t (4) 2C 2C () MA 201, PDE October 9, 2012 15 / 48
- 69. Method of Characteristics (Contd.)Consequentlywe ﬁnd thatthe transformation √ √ −B + D −B − D ξ =x+ t, η = x + t (4) 2C 2Ctransforms the PDE(1) into a simpler equation of the form () MA 201, PDE October 9, 2012 15 / 48
- 70. Method of Characteristics (Contd.)Consequentlywe ﬁnd thatthe transformation √ √ −B + D −B − D ξ =x+ t, η = x + t (4) 2C 2Ctransforms the PDE(1) into a simpler equation of the form Uξη + G(ξ, η, U, Uξ , Uη ) = 0 (5) () MA 201, PDE October 9, 2012 15 / 48
- 71. Method of Characteristics (Contd.)Consequentlywe ﬁnd thatthe transformation √ √ −B + D −B − D ξ =x+ t, η = x + t (4) 2C 2Ctransforms the PDE(1) into a simpler equation of the form Uξη + G(ξ, η, U, Uξ , Uη ) = 0 (5)where only the mixed derivative appears () MA 201, PDE October 9, 2012 15 / 48
- 72. Method of Characteristics (Contd.)Consequentlywe ﬁnd thatthe transformation √ √ −B + D −B − D ξ =x+ t, η = x + t (4) 2C 2Ctransforms the PDE(1) into a simpler equation of the form Uξη + G(ξ, η, U, Uξ , Uη ) = 0 (5)where only the mixed derivative appearsThus there is a signiﬁcant simpliﬁcation over (1) where all the second derivativesoccur. () MA 201, PDE October 9, 2012 15 / 48
- 73. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 16 / 48
- 74. Method of Characteristics (Contd.)The new transformed equation is calledthe canonical form of a hyperbolic equation () MA 201, PDE October 9, 2012 16 / 48
- 75. Method of Characteristics (Contd.)The new transformed equation is calledthe canonical form of a hyperbolic equationand the coordinates ξ and η deﬁned by (4) are called thecharacteristic coordinates and ξ(x, y) = c1 and η(x, y) = c2 are called characteristiccurves. () MA 201, PDE October 9, 2012 16 / 48
- 76. Method of Characteristics (Contd.)The new transformed equation is calledthe canonical form of a hyperbolic equationand the coordinates ξ and η deﬁned by (4) are called thecharacteristic coordinates and ξ(x, y) = c1 and η(x, y) = c2 are called characteristiccurves.Note that if C = 0,(3) is not valid () MA 201, PDE October 9, 2012 16 / 48
- 77. Method of Characteristics (Contd.)The new transformed equation is calledthe canonical form of a hyperbolic equationand the coordinates ξ and η deﬁned by (4) are called thecharacteristic coordinates and ξ(x, y) = c1 and η(x, y) = c2 are called characteristiccurves.Note that if C = 0,(3) is not validIn this case choose b = d = 1 () MA 201, PDE October 9, 2012 16 / 48
- 78. Method of Characteristics (Contd.)The new transformed equation is calledthe canonical form of a hyperbolic equationand the coordinates ξ and η deﬁned by (4) are called thecharacteristic coordinates and ξ(x, y) = c1 and η(x, y) = c2 are called characteristiccurves.Note that if C = 0,(3) is not validIn this case choose b = d = 1Then we have Auxx + Buxt = (Aa2 + Ba)Uξξ + (2acA + B(a + c))Uξη + (Ac2 + Bc)Uηη () MA 201, PDE October 9, 2012 16 / 48
- 79. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 17 / 48
- 80. Method of Characteristics (Contd.)In order that the coeﬃcients of Uξξ and Uηη vanish: −B −B a = 0, or a = ; c = 0, or c = A A () MA 201, PDE October 9, 2012 17 / 48
- 81. Method of Characteristics (Contd.)In order that the coeﬃcients of Uξξ and Uηη vanish: −B −B a = 0, or a = ; c = 0, or c = A ANow observe that if we take a = 0 and c = 0 at the same time, we getthe same characteristics ξ = t, η = t () MA 201, PDE October 9, 2012 17 / 48
- 82. Method of Characteristics (Contd.)In order that the coeﬃcients of Uξξ and Uηη vanish: −B −B a = 0, or a = ; c = 0, or c = A ANow observe that if we take a = 0 and c = 0 at the same time, we getthe same characteristics ξ = t, η = t −B −BSimilarly if we take a = A and c = A at the same time, we getthe same characteristics again: −B −B ξ= x + t, η = x+t A A () MA 201, PDE October 9, 2012 17 / 48
- 83. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 18 / 48
- 84. Method of Characteristics (Contd.)But we know that ξ and η cannot have the same expressions () MA 201, PDE October 9, 2012 18 / 48
- 85. Method of Characteristics (Contd.)But we know that ξ and η cannot have the same expressionsHence we consider those values of a and c at the same time so as to gettwo diﬀerent expressions for ξ and η. () MA 201, PDE October 9, 2012 18 / 48
- 86. Method of Characteristics (Contd.)But we know that ξ and η cannot have the same expressionsHence we consider those values of a and c at the same time so as to gettwo diﬀerent expressions for ξ and η.Any one of the following pairs can be the characteristics: B ξ=− x + t, η = t A B ξ = t, η = − x + t A () MA 201, PDE October 9, 2012 18 / 48
- 87. Method of Characteristics (Contd.)But we know that ξ and η cannot have the same expressionsHence we consider those values of a and c at the same time so as to gettwo diﬀerent expressions for ξ and η.Any one of the following pairs can be the characteristics: B ξ=− x + t, η = t A B ξ = t, η = − x + t AEither of them will reduce the given equation again to the simpliﬁed one: Uξη + G(ξ, η, U, Uξ , Uη ) = 0 () MA 201, PDE October 9, 2012 18 / 48
- 88. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 19 / 48
- 89. Method of Characteristics (Contd.)Parabolic Case, D = 0 () MA 201, PDE October 9, 2012 19 / 48
- 90. Method of Characteristics (Contd.)Parabolic Case, D = 0In this case, equations (3) give b = d. () MA 201, PDE October 9, 2012 19 / 48
- 91. Method of Characteristics (Contd.)Parabolic Case, D = 0In this case, equations (3) give b = d.The resulting transformation ξ = x + bt, η = x + btis not invertible. () MA 201, PDE October 9, 2012 19 / 48
- 92. Method of Characteristics (Contd.)Parabolic Case, D = 0In this case, equations (3) give b = d.The resulting transformation ξ = x + bt, η = x + btis not invertible.Observe that if we choose a = c = 1, d = −B/(2C), and b = 0, () MA 201, PDE October 9, 2012 19 / 48
- 93. Method of Characteristics (Contd.)Parabolic Case, D = 0In this case, equations (3) give b = d.The resulting transformation ξ = x + bt, η = x + btis not invertible.Observe that if we choose a = c = 1, d = −B/(2C), and b = 0,then the coeﬃcients ofUηη and Uξη in (2) vanish. () MA 201, PDE October 9, 2012 19 / 48
- 94. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 20 / 48
- 95. Method of Characteristics (Contd.)Hence () MA 201, PDE October 9, 2012 20 / 48
- 96. Method of Characteristics (Contd.)Hencethe transformation B ξ = x, η = x − t 2C () MA 201, PDE October 9, 2012 20 / 48
- 97. Method of Characteristics (Contd.)Hencethe transformation B ξ = x, η = x − t 2Ctransforms equation (1) into Uξξ + H(ξ, η, U, Uξ , Uη ) = 0 (6)where only one double derivative appears and () MA 201, PDE October 9, 2012 20 / 48
- 98. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 21 / 48
- 99. Method of Characteristics (Contd.)Elliptic Case, D < 0: () MA 201, PDE October 9, 2012 21 / 48
- 100. Method of Characteristics (Contd.)Elliptic Case, D < 0:Now b and d in (3) arecomplex conjugate numbers () MA 201, PDE October 9, 2012 21 / 48
- 101. Method of Characteristics (Contd.)Elliptic Case, D < 0:Now b and d in (3) arecomplex conjugate numbersthat is d = b. () MA 201, PDE October 9, 2012 21 / 48
- 102. Method of Characteristics (Contd.)Elliptic Case, D < 0:Now b and d in (3) arecomplex conjugate numbersthat is d = b.Selecting a = c = 1, we obtain a complex transformation: ξ = x + bt, η = x + bt () MA 201, PDE October 9, 2012 21 / 48
- 103. Method of Characteristics (Contd.)Elliptic Case, D < 0:Now b and d in (3) arecomplex conjugate numbersthat is d = b.Selecting a = c = 1, we obtain a complex transformation: ξ = x + bt, η = x + btA real transformation can be found by taking real variables: 1 1 α= (ξ + η), β = (ξ − η). 2 2i () MA 201, PDE October 9, 2012 21 / 48
- 104. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 22 / 48
- 105. Method of Characteristics (Contd.)This transforms (1) into Uαα + Uββ + K(α, β, U, Uα , Uβ ) = 0. (7)where both double partial derivatives are present and the mixed derivative is missing. () MA 201, PDE October 9, 2012 22 / 48
- 106. Method of Characteristics (Contd.)This transforms (1) into Uαα + Uββ + K(α, β, U, Uα , Uβ ) = 0. (7)where both double partial derivatives are present and the mixed derivative is missing.Equation (8) is the canonical form of elliptic equations. () MA 201, PDE October 9, 2012 22 / 48
- 107. Method of Characteristics (Contd.)This transforms (1) into Uαα + Uββ + K(α, β, U, Uα , Uβ ) = 0. (7)where both double partial derivatives are present and the mixed derivative is missing.Equation (8) is the canonical form of elliptic equations.We recognize the combination of second partial derivatives asthe Laplacian operator. () MA 201, PDE October 9, 2012 22 / 48
- 108. Method of Characteristics (Contd.)This transforms (1) into Uαα + Uββ + K(α, β, U, Uα , Uβ ) = 0. (7)where both double partial derivatives are present and the mixed derivative is missing.Equation (8) is the canonical form of elliptic equations.We recognize the combination of second partial derivatives asthe Laplacian operator.Selecting a = c = 1, we obtain a complex transformation: ξ = x + bt, η = x + bt () MA 201, PDE October 9, 2012 22 / 48
- 109. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 23 / 48
- 110. Method of Characteristics (Contd.)Generallycharacteristic coordinates are useful only for hyperbolic equations. () MA 201, PDE October 9, 2012 23 / 48
- 111. Method of Characteristics (Contd.)Generallycharacteristic coordinates are useful only for hyperbolic equations.They do not play any important role in elliptic and parabolic equations. () MA 201, PDE October 9, 2012 23 / 48
- 112. Method of Characteristics (Contd.)Generallycharacteristic coordinates are useful only for hyperbolic equations.They do not play any important role in elliptic and parabolic equations.Note thatﬁrst order euqations, like a reaction-advection equations, are classiﬁed as hyperbolicbecause () MA 201, PDE October 9, 2012 23 / 48
- 113. Method of Characteristics (Contd.)Generallycharacteristic coordinates are useful only for hyperbolic equations.They do not play any important role in elliptic and parabolic equations.Note thatﬁrst order euqations, like a reaction-advection equations, are classiﬁed as hyperbolicbecauseThey propagate signals like wave equations. () MA 201, PDE October 9, 2012 23 / 48
- 114. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 24 / 48
- 115. Method of Characteristics (Contd.)Example:Find the characteristics of the following equation and reduce it to the appropriatestandard form and then obtain the general solution: 3uxx + 10uxy + 3uyy = 0. () MA 201, PDE October 9, 2012 24 / 48
- 116. Method of Characteristics (Contd.)Example:Find the characteristics of the following equation and reduce it to the appropriatestandard form and then obtain the general solution: 3uxx + 10uxy + 3uyy = 0.Solution:Here A=3,B=10,C=3 and we have B 2 − 4CA = 100 − 36 = 64 > 0 () MA 201, PDE October 9, 2012 24 / 48
- 117. Method of Characteristics (Contd.)Example:Find the characteristics of the following equation and reduce it to the appropriatestandard form and then obtain the general solution: 3uxx + 10uxy + 3uyy = 0.Solution:Here A=3,B=10,C=3 and we have B 2 − 4CA = 100 − 36 = 64 > 0Hence the equation is of hyperbolic type. () MA 201, PDE October 9, 2012 24 / 48
- 118. Method of Characteristics (Contd.)Example:Find the characteristics of the following equation and reduce it to the appropriatestandard form and then obtain the general solution: 3uxx + 10uxy + 3uyy = 0.Solution:Here A=3,B=10,C=3 and we have B 2 − 4CA = 100 − 36 = 64 > 0Hence the equation is of hyperbolic type.The characteristics are given by √ √ −B + D −B − D ξ =x+ y = x − (1/3)y, η = x + y = x − 3y 2C 2C () MA 201, PDE October 9, 2012 24 / 48
- 119. Method of Characteristics (Contd.)Example:Find the characteristics of the following equation and reduce it to the appropriatestandard form and then obtain the general solution: 3uxx + 10uxy + 3uyy = 0.Solution:Here A=3,B=10,C=3 and we have B 2 − 4CA = 100 − 36 = 64 > 0Hence the equation is of hyperbolic type.The characteristics are given by √ √ −B + D −B − D ξ =x+ y = x − (1/3)y, η = x + y = x − 3y 2C 2CHere we have ξx = 1, ηx = 1, ξy = −1/3, ηy = −3. () MA 201, PDE October 9, 2012 24 / 48
- 120. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 25 / 48
- 121. Method of Characteristics (Contd.)Therefore, under this transformation ux = uξ ξx + uη ηx = uξ + uη , uy = uξ ξy + uη ηy = (1/3)uξ − 3uη . () MA 201, PDE October 9, 2012 25 / 48
- 122. Method of Characteristics (Contd.)Therefore, under this transformation ux = uξ ξx + uη ηx = uξ + uη , uy = uξ ξy + uη ηy = (1/3)uξ − 3uη .expressions for the second order derivatives are uxx = uξξ ξx + uξη ηx + uηξ ξx + uηη ηx = uξξ + 2uξη + uηη , uxy = uξξ ξy + uξη ηy + uηξ ξy + uηη ηy = −(1/3)uξξ − (10/3)uξη − 3uηη , uyy = (1/3)[uξξ ξy + uξη ηy ] − 3[uηξ ξy + uηη ηy ] = (1/9)uξξ + 2uξη + 9uηη . () MA 201, PDE October 9, 2012 25 / 48
- 123. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 26 / 48
- 124. Method of Characteristics (Contd.)Substituting these values in the given equation () MA 201, PDE October 9, 2012 26 / 48
- 125. Method of Characteristics (Contd.)Substituting these values in the given equation uξη = 0. () MA 201, PDE October 9, 2012 26 / 48
- 126. Method of Characteristics (Contd.)Substituting these values in the given equation uξη = 0.Now integrating with respect to ξ partially uη = F (η). () MA 201, PDE October 9, 2012 26 / 48
- 127. Method of Characteristics (Contd.)Substituting these values in the given equation uξη = 0.Now integrating with respect to ξ partially uη = F (η).Again integrating with respect to η partially u = F (η) + G(ξ). () MA 201, PDE October 9, 2012 26 / 48
- 128. Method of Characteristics (Contd.)Substituting these values in the given equation uξη = 0.Now integrating with respect to ξ partially uη = F (η).Again integrating with respect to η partially u = F (η) + G(ξ).Hence the solution of the original equation is u(x, y) = F (x − 3y) + G(3x − y),where F and G are arbitrary functions. () MA 201, PDE October 9, 2012 26 / 48
- 129. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 27 / 48
- 130. Method of Characteristics (Contd.)Example:Find the characteristics of the following equation and reduce it to the appropriatestandard form and then obtain the general solution: uxx + 4uxy = 0. () MA 201, PDE October 9, 2012 27 / 48
- 131. Method of Characteristics (Contd.)Example:Find the characteristics of the following equation and reduce it to the appropriatestandard form and then obtain the general solution: uxx + 4uxy = 0.Solution:Here A=1,B=4,C=0 and we have B 2 − 4CA = 16 > 0 () MA 201, PDE October 9, 2012 27 / 48
- 132. Method of Characteristics (Contd.)Example:Find the characteristics of the following equation and reduce it to the appropriatestandard form and then obtain the general solution: uxx + 4uxy = 0.Solution:Here A=1,B=4,C=0 and we have B 2 − 4CA = 16 > 0Hence the equation is of hyperbolic type. () MA 201, PDE October 9, 2012 27 / 48
- 133. Method of Characteristics (Contd.)Example:Find the characteristics of the following equation and reduce it to the appropriatestandard form and then obtain the general solution: uxx + 4uxy = 0.Solution:Here A=1,B=4,C=0 and we have B 2 − 4CA = 16 > 0Hence the equation is of hyperbolic type.The characteristics are given by −B ξ= x + y = −4x + y, η = y A () MA 201, PDE October 9, 2012 27 / 48
- 134. Method of Characteristics (Contd.)Example:Find the characteristics of the following equation and reduce it to the appropriatestandard form and then obtain the general solution: uxx + 4uxy = 0.Solution:Here A=1,B=4,C=0 and we have B 2 − 4CA = 16 > 0Hence the equation is of hyperbolic type.The characteristics are given by −B ξ= x + y = −4x + y, η = y AHere we have ξx = −4, ηx = 0, ξy = 1, ηy = 1. () MA 201, PDE October 9, 2012 27 / 48
- 135. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 28 / 48
- 136. Method of Characteristics (Contd.)Therefore, under this transformation ux = uξ ξx + uη ηx = −4uξ , uy = uξ ξy + uη ηy = uξ + uη . () MA 201, PDE October 9, 2012 28 / 48
- 137. Method of Characteristics (Contd.)Therefore, under this transformation ux = uξ ξx + uη ηx = −4uξ , uy = uξ ξy + uη ηy = uξ + uη .expressions for the second order derivatives are uxx = uξξ ξx + uξη ηx + uηξ ξx + uηη ηx = 16uξξ , uxy = uξξ ξy + uξη ηy + uηξ ξy + uηη ηy = −4(uξξ + uξη ). () MA 201, PDE October 9, 2012 28 / 48
- 138. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 29 / 48
- 139. Method of Characteristics (Contd.)Substituting these values in the given equation () MA 201, PDE October 9, 2012 29 / 48
- 140. Method of Characteristics (Contd.)Substituting these values in the given equation uξη = 0. () MA 201, PDE October 9, 2012 29 / 48
- 141. Method of Characteristics (Contd.)Substituting these values in the given equation uξη = 0.Now integrating with respect to ξ partially uη = F (η). () MA 201, PDE October 9, 2012 29 / 48
- 142. Method of Characteristics (Contd.)Substituting these values in the given equation uξη = 0.Now integrating with respect to ξ partially uη = F (η).Again integrating with respect to η partially u = F (η) + G(ξ). () MA 201, PDE October 9, 2012 29 / 48
- 143. Method of Characteristics (Contd.)Substituting these values in the given equation uξη = 0.Now integrating with respect to ξ partially uη = F (η).Again integrating with respect to η partially u = F (η) + G(ξ).Hence the solution of the original equation is u(x, y) = F (y) + G(y − 4x),where F and G are arbitrary functions. () MA 201, PDE October 9, 2012 29 / 48
- 144. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 30 / 48
- 145. Method of Characteristics (Contd.)Example:Find the characteristics of the following equation and reduce it to the appropriatestandard form and then obtain the general solution: uxx + 4uxy + 4uyy = 0. () MA 201, PDE October 9, 2012 30 / 48
- 146. Method of Characteristics (Contd.)Example:Find the characteristics of the following equation and reduce it to the appropriatestandard form and then obtain the general solution: uxx + 4uxy + 4uyy = 0.Solution: () MA 201, PDE October 9, 2012 30 / 48
- 147. Method of Characteristics (Contd.)Example:Find the characteristics of the following equation and reduce it to the appropriatestandard form and then obtain the general solution: uxx + 4uxy + 4uyy = 0.Solution:B 2 − 4CA = 0Hence the equation is parabolic. () MA 201, PDE October 9, 2012 30 / 48
- 148. Method of Characteristics (Contd.)Example:Find the characteristics of the following equation and reduce it to the appropriatestandard form and then obtain the general solution: uxx + 4uxy + 4uyy = 0.Solution:B 2 − 4CA = 0Hence the equation is parabolic.The characteristics are given by B ξ = x, η = x − y = x − (1/2)y 2C () MA 201, PDE October 9, 2012 30 / 48
- 149. Method of Characteristics (Contd.)Example:Find the characteristics of the following equation and reduce it to the appropriatestandard form and then obtain the general solution: uxx + 4uxy + 4uyy = 0.Solution:B 2 − 4CA = 0Hence the equation is parabolic.The characteristics are given by B ξ = x, η = x − y = x − (1/2)y 2CWe have ξx = 1, ηx = 1, ξy = 0, ηy = −(1/2) () MA 201, PDE October 9, 2012 30 / 48
- 150. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 31 / 48
- 151. Method of Characteristics (Contd.)Under this transformation ux = uξ ξx + uη ηx = uξ + uη , uy = uξ ξy + uη ηy = −(1/2)uη . () MA 201, PDE October 9, 2012 31 / 48
- 152. Method of Characteristics (Contd.)Under this transformation ux = uξ ξx + uη ηx = uξ + uη , uy = uξ ξy + uη ηy = −(1/2)uη .The expressions for the second order derivatives: uxx = uξξ + 2uξη + uηη , uxy = −(1/2)uξη − (1/2)uηη , uyy = (1/4)uηη . () MA 201, PDE October 9, 2012 31 / 48
- 153. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 32 / 48
- 154. Method of Characteristics (Contd.)Substituting these values in the given equation: uξξ = 0. () MA 201, PDE October 9, 2012 32 / 48
- 155. Method of Characteristics (Contd.)Substituting these values in the given equation: uξξ = 0.Integrating with respect to ξ partially: uξ = f (η). () MA 201, PDE October 9, 2012 32 / 48
- 156. Method of Characteristics (Contd.)Substituting these values in the given equation: uξξ = 0.Integrating with respect to ξ partially: uξ = f (η).Again integrating with respect to ξ partially: u = ξf (η) + g(η). () MA 201, PDE October 9, 2012 32 / 48
- 157. Method of Characteristics (Contd.)Substituting these values in the given equation: uξξ = 0.Integrating with respect to ξ partially: uξ = f (η).Again integrating with respect to ξ partially: u = ξf (η) + g(η).Hence the solution of the original equation is u(x, y) = xf (x − (1/2)y) + g(x − (1/2)y),where f and g are arbitrary functions. () MA 201, PDE October 9, 2012 32 / 48
- 158. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 33 / 48
- 159. Method of Characteristics (Contd.)Please note thatmethod of characteristics is not found suitable for elliptic equations since () MA 201, PDE October 9, 2012 33 / 48
- 160. Method of Characteristics (Contd.)Please note thatmethod of characteristics is not found suitable for elliptic equations sinceeven after using the transformation, that is,the characteristics (in new variables), the equation gets reduced to Laplace’s equationform only. () MA 201, PDE October 9, 2012 33 / 48
- 161. Method of Characteristics (Contd.)Please note thatmethod of characteristics is not found suitable for elliptic equations sinceeven after using the transformation, that is,the characteristics (in new variables), the equation gets reduced to Laplace’s equationform only.In other words:the given equation gets reduced marginally only with two double derivatives stillremaining. () MA 201, PDE October 9, 2012 33 / 48
- 162. Method of Characteristics (Contd.)Please note thatmethod of characteristics is not found suitable for elliptic equations sinceeven after using the transformation, that is,the characteristics (in new variables), the equation gets reduced to Laplace’s equationform only.In other words:the given equation gets reduced marginally only with two double derivatives stillremaining.For this reasonwe will not apply method of characteristics to elliptic equations. () MA 201, PDE October 9, 2012 33 / 48
- 163. Method of Characteristics (Contd.)Please note thatmethod of characteristics is not found suitable for elliptic equations sinceeven after using the transformation, that is,the characteristics (in new variables), the equation gets reduced to Laplace’s equationform only.In other words:the given equation gets reduced marginally only with two double derivatives stillremaining.For this reasonwe will not apply method of characteristics to elliptic equations.Instead let us considera non-homogenous equation and apply method of characteristics. () MA 201, PDE October 9, 2012 33 / 48
- 164. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 34 / 48
- 165. Method of Characteristics (Contd.)Example:Find the characteristics of the following equation and reduce it to the appropriatestandard form and then obtain the general solution: uxx − 4uxy + 4uyy = cos(2x + y). () MA 201, PDE October 9, 2012 34 / 48
- 166. Method of Characteristics (Contd.)Example:Find the characteristics of the following equation and reduce it to the appropriatestandard form and then obtain the general solution: uxx − 4uxy + 4uyy = cos(2x + y).Solution:The given equation is a parabolic one. () MA 201, PDE October 9, 2012 34 / 48
- 167. Method of Characteristics (Contd.)Example:Find the characteristics of the following equation and reduce it to the appropriatestandard form and then obtain the general solution: uxx − 4uxy + 4uyy = cos(2x + y).Solution:The given equation is a parabolic one.The characteristics are: B ξ = x, η = x − y = x + (1/2)y 2C () MA 201, PDE October 9, 2012 34 / 48
- 168. Method of Characteristics (Contd.)Example:Find the characteristics of the following equation and reduce it to the appropriatestandard form and then obtain the general solution: uxx − 4uxy + 4uyy = cos(2x + y).Solution:The given equation is a parabolic one.The characteristics are: B ξ = x, η = x − y = x + (1/2)y 2CThe canonical form will be uξξ = cos η.. () MA 201, PDE October 9, 2012 34 / 48
- 169. Method of Characteristics (Contd.) () MA 201, PDE October 9, 2012 35 / 48
- 170. Method of Characteristics (Contd.)Integrating partially with respect to ξ: uξ = ξ cos η + f (η). () MA 201, PDE October 9, 2012 35 / 48
- 171. Method of Characteristics (Contd.)Integrating partially with respect to ξ: uξ = ξ cos η + f (η).Again integrating w.r.t. ξ ξ2 u(ξ, η) = cos η + ξf (η) + g(η). 2 () MA 201, PDE October 9, 2012 35 / 48
- 172. Method of Characteristics (Contd.)Integrating partially with respect to ξ: uξ = ξ cos η + f (η).Again integrating w.r.t. ξ ξ2 u(ξ, η) = cos η + ξf (η) + g(η). 2The solution: x2 u(x, y) = xf (y + 2x) + g(y + 2x) + cos(y + 2x), 2where f and g are arbitrary functions. () MA 201, PDE October 9, 2012 35 / 48
- 173. The D’Alembert solution of the wave equation () MA 201, PDE October 9, 2012 36 / 48
- 174. The D’Alembert solution of the wave equationMethod of characteristics is very usefulfor hyperbolic equations. () MA 201, PDE October 9, 2012 36 / 48
- 175. The D’Alembert solution of the wave equationMethod of characteristics is very usefulfor hyperbolic equations.Please note • Two families of characteristics of hyperbolic equations, being real and distinct, are of considerable practical value. • In one-dimensional progressive wave propagation, consideration of characteristics can give us a good deal of information about the propagation of the wave fronts. • This solution of one-dimensional wave equation, known as D’Alembert’s solution, was discovered by a French mathematician named Jean Le Rond D’Alembert. () MA 201, PDE October 9, 2012 36 / 48
- 176. The D’Alembert solution of the wave equation (Contd.) () MA 201, PDE October 9, 2012 37 / 48
- 177. The D’Alembert solution of the wave equation (Contd.)Consider the one-dimensional wave equation: utt = c2 uxx . (8) () MA 201, PDE October 9, 2012 37 / 48
- 178. The D’Alembert solution of the wave equation (Contd.)Consider the one-dimensional wave equation: utt = c2 uxx . (8)Here B 2 − 4AC = c2 > 0. () MA 201, PDE October 9, 2012 37 / 48
- 179. The D’Alembert solution of the wave equation (Contd.)Consider the one-dimensional wave equation: utt = c2 uxx . (8)Here B 2 − 4AC = c2 > 0.The characteristics are given by ξ = x − ct, η = x + ct. () MA 201, PDE October 9, 2012 37 / 48
- 180. The D’Alembert solution of the wave equation (Contd.)Consider the one-dimensional wave equation: utt = c2 uxx . (8)Here B 2 − 4AC = c2 > 0.The characteristics are given by ξ = x − ct, η = x + ct.Under this transformation ux = uξ ξx + uη ηx = uξ + uη , ut = uξ ξt + uη ηt = −cuξ + cuη , uxx = (uξξ ξx + uξη ηx ) + (uηξ ξx + uηη ηx ) = uξξ + 2uξη + uηη , utt = −c(uξξ ξt + uξη ηt ) + c(uηξ ξt + uηη ηt ), = c2 (uξξ − 2uξη + uηη ). () MA 201, PDE October 9, 2012 37 / 48
- 181. The D’Alembert solution of the wave equation (Contd.) () MA 201, PDE October 9, 2012 38 / 48
- 182. The D’Alembert solution of the wave equation (Contd.)Substituting these into the given equation uξη = 0. (9) () MA 201, PDE October 9, 2012 38 / 48
- 183. The D’Alembert solution of the wave equation (Contd.)Substituting these into the given equation uξη = 0. (9)Integrating partially with respect to η: uξ = F (ξ). () MA 201, PDE October 9, 2012 38 / 48
- 184. The D’Alembert solution of the wave equation (Contd.)Substituting these into the given equation uξη = 0. (9)Integrating partially with respect to η: uξ = F (ξ).Integrating partially w.r.t. ξ u = F (ξ) + G(η). () MA 201, PDE October 9, 2012 38 / 48
- 185. The D’Alembert solution of the wave equation (Contd.)Substituting these into the given equation uξη = 0. (9)Integrating partially with respect to η: uξ = F (ξ).Integrating partially w.r.t. ξ u = F (ξ) + G(η).The solution in physical variables: u(x, t) = F (x − ct) + G(x + ct) (10)where F and G are arbitrary functions. () MA 201, PDE October 9, 2012 38 / 48
- 186. The D’Alembert solution of the wave equation (Contd.) () MA 201, PDE October 9, 2012 39 / 48
- 187. The D’Alembert solution of the wave equation (Contd.)The physical interpretation of these functions is quite interesting. () MA 201, PDE October 9, 2012 39 / 48
- 188. The D’Alembert solution of the wave equation (Contd.)The physical interpretation of these functions is quite interesting.The functions F and G represent two progressive waves travelling in oppositedirections with the speed c. () MA 201, PDE October 9, 2012 39 / 48
- 189. The D’Alembert solution of the wave equation (Contd.)The physical interpretation of these functions is quite interesting.The functions F and G represent two progressive waves travelling in oppositedirections with the speed c.To see this let us ﬁrst consider the solution u = F (x − ct). () MA 201, PDE October 9, 2012 39 / 48
- 190. The D’Alembert solution of the wave equation (Contd.)The physical interpretation of these functions is quite interesting.The functions F and G represent two progressive waves travelling in oppositedirections with the speed c.To see this let us ﬁrst consider the solution u = F (x − ct).At t = 0, it deﬁnes the curve u = F (x), and after time t = t1 , it deﬁnes thecurve u = F (x − ct1 ). () MA 201, PDE October 9, 2012 39 / 48
- 191. The D’Alembert solution of the wave equation (Contd.)The physical interpretation of these functions is quite interesting.The functions F and G represent two progressive waves travelling in oppositedirections with the speed c.To see this let us ﬁrst consider the solution u = F (x − ct).At t = 0, it deﬁnes the curve u = F (x), and after time t = t1 , it deﬁnes thecurve u = F (x − ct1 ).But these curves are identical except that the latter is translated to the right adistance equal to ct1 . () MA 201, PDE October 9, 2012 39 / 48
- 192. Method of Characteristics (Contd.) Figure: A Progressive Wave () MA 201, PDE October 9, 2012 40 / 48
- 193. The D’Alembert solution of the wave equation (Contd.) () MA 201, PDE October 9, 2012 41 / 48
- 194. The D’Alembert solution of the wave equation (Contd.)Thus the entire conﬁguration moves along the positive direction of x-axis adistance of ct1 in time t1 . () MA 201, PDE October 9, 2012 41 / 48
- 195. The D’Alembert solution of the wave equation (Contd.)Thus the entire conﬁguration moves along the positive direction of x-axis adistance of ct1 in time t1 .The velocity with which the wave is propagated is, therefore, ct1 v= =c t1 () MA 201, PDE October 9, 2012 41 / 48
- 196. The D’Alembert solution of the wave equation (Contd.)Thus the entire conﬁguration moves along the positive direction of x-axis adistance of ct1 in time t1 .The velocity with which the wave is propagated is, therefore, ct1 v= =c t1Similarly the function G(x + ct) deﬁnes a wave progressing in the negativedirection of x-axis with constant velocity c. () MA 201, PDE October 9, 2012 41 / 48
- 197. The D’Alembert solution of the wave equation (Contd.)Thus the entire conﬁguration moves along the positive direction of x-axis adistance of ct1 in time t1 .The velocity with which the wave is propagated is, therefore, ct1 v= =c t1Similarly the function G(x + ct) deﬁnes a wave progressing in the negativedirection of x-axis with constant velocity c.The total solution is, therefore, the algebraic sum of these two travelling waves. () MA 201, PDE October 9, 2012 41 / 48
- 198. The D’Alembert solution of the wave equation (Contd.)Thus the entire conﬁguration moves along the positive direction of x-axis adistance of ct1 in time t1 .The velocity with which the wave is propagated is, therefore, ct1 v= =c t1Similarly the function G(x + ct) deﬁnes a wave progressing in the negativedirection of x-axis with constant velocity c.The total solution is, therefore, the algebraic sum of these two travelling waves.Solution (11) is a very convenient representation for progressive waveswhich travel large distances through a uniform medium. () MA 201, PDE October 9, 2012 41 / 48
- 199. The D’Alembert solution of the wave equation (Contd.) () MA 201, PDE October 9, 2012 42 / 48
- 200. The D’Alembert solution of the wave equation (Contd.)Let us consider the following two initial conditions for a uniform medium−∞ < x < ∞. () MA 201, PDE October 9, 2012 42 / 48
- 201. The D’Alembert solution of the wave equation (Contd.)Let us consider the following two initial conditions for a uniform medium−∞ < x < ∞. Displacement: u(x, 0) = φ(x), (11) Velocity: ut (x, 0) = ψ(x), (12) () MA 201, PDE October 9, 2012 42 / 48
- 202. The D’Alembert solution of the wave equation (Contd.)Let us consider the following two initial conditions for a uniform medium−∞ < x < ∞. Displacement: u(x, 0) = φ(x), (11) Velocity: ut (x, 0) = ψ(x), (12)that iswe consider the vibration of a thin string of inﬁnite length with some initialdisplacement and initial velocity. () MA 201, PDE October 9, 2012 42 / 48
- 203. The D’Alembert solution of the wave equation (Contd.)Let us consider the following two initial conditions for a uniform medium−∞ < x < ∞. Displacement: u(x, 0) = φ(x), (11) Velocity: ut (x, 0) = ψ(x), (12)that iswe consider the vibration of a thin string of inﬁnite length with some initialdisplacement and initial velocity.From the solution (11) we ﬁnd that F (x) + G(x) = φ(x), (13) −cF (x) + c G (x) = ψ(x), (14)for all values of x. () MA 201, PDE October 9, 2012 42 / 48
- 204. The D’Alembert solution of the wave equation (Contd.) () MA 201, PDE October 9, 2012 43 / 48
- 205. The D’Alembert solution of the wave equation (Contd.)Integrating the second equation with respect to x x 1 −F (x) + G(x) = ψ(s)ds. (15) c x0 () MA 201, PDE October 9, 2012 43 / 48
- 206. The D’Alembert solution of the wave equation (Contd.)Integrating the second equation with respect to x x 1 −F (x) + G(x) = ψ(s)ds. (15) c x0From (14) and (16) x 1 1 F (x) = φ(x) − ψ(s)ds , (16) 2 c x0 x 1 1 G(x) = φ(x) + ψ(s)ds , (17) 2 c x0where A is an integration constant and s is a dummy variable. () MA 201, PDE October 9, 2012 43 / 48
- 207. The D’Alembert solution of the wave equation (Contd.)Integrating the second equation with respect to x x 1 −F (x) + G(x) = ψ(s)ds. (15) c x0From (14) and (16) x 1 1 F (x) = φ(x) − ψ(s)ds , (16) 2 c x0 x 1 1 G(x) = φ(x) + ψ(s)ds , (17) 2 c x0where A is an integration constant and s is a dummy variable.Substituting these expressions into (11) x+ct 1 1 u= [φ(x − ct) + φ(x + ct)] + ψ(s)ds. (18) 2 2c x−ct () MA 201, PDE October 9, 2012 43 / 48
- 208. The D’Alembert solution of the wave equation (Contd.)Integrating the second equation with respect to x x 1 −F (x) + G(x) = ψ(s)ds. (15) c x0From (14) and (16) x 1 1 F (x) = φ(x) − ψ(s)ds , (16) 2 c x0 x 1 1 G(x) = φ(x) + ψ(s)ds , (17) 2 c x0where A is an integration constant and s is a dummy variable.Substituting these expressions into (11) x+ct 1 1 u= [φ(x − ct) + φ(x + ct)] + ψ(s)ds. (18) 2 2c x−ctThis is D’Alembert’s solution of wave equation. () MA 201, PDE October 9, 2012 43 / 48
- 209. The D’Alembert solution of the wave equation (Contd.) () MA 201, PDE October 9, 2012 44 / 48
- 210. The D’Alembert solution of the wave equation (Contd.)Thus for a given initial displacement and velocity in the vertical directionthe wave equation is completely solved and this solution is usually called theprogressive wave solution. () MA 201, PDE October 9, 2012 44 / 48
- 211. The D’Alembert solution of the wave equation (Contd.)Thus for a given initial displacement and velocity in the vertical directionthe wave equation is completely solved and this solution is usually called theprogressive wave solution.It is to be noted thatthe use of string problem to demonstrate the solution of the wave problem is a matterof convenience. However, any variables satisfying the wave equation possess themathematical properties developed for the string. () MA 201, PDE October 9, 2012 44 / 48
- 212. The D’Alembert solution of the wave equation (Contd.) () MA 201, PDE October 9, 2012 45 / 48
- 213. The D’Alembert solution of the wave equation (Contd.)It is clear thatthe wave equation can be handled very easily by introducing the characteristic variables(ξ, η). () MA 201, PDE October 9, 2012 45 / 48
- 214. The D’Alembert solution of the wave equation (Contd.)It is clear thatthe wave equation can be handled very easily by introducing the characteristic variables(ξ, η).The relationship between the physical plane and the characteristic plane for thisparticular example can be demonstrated graphically. () MA 201, PDE October 9, 2012 45 / 48
- 215. The D’Alembert solution of the wave equation (Contd.)It is clear thatthe wave equation can be handled very easily by introducing the characteristic variables(ξ, η).The relationship between the physical plane and the characteristic plane for thisparticular example can be demonstrated graphically.Equation (19) represents the solution as the sum of two progressive waves; one goingto the right and the other to the left. () MA 201, PDE October 9, 2012 45 / 48
- 216. The D’Alembert solution of the wave equation (Contd.)It is clear thatthe wave equation can be handled very easily by introducing the characteristic variables(ξ, η).The relationship between the physical plane and the characteristic plane for thisparticular example can be demonstrated graphically.Equation (19) represents the solution as the sum of two progressive waves; one goingto the right and the other to the left.The wave celerity is c. () MA 201, PDE October 9, 2012 45 / 48
- 217. The D’Alembert solution of the wave equation (Contd.) () MA 201, PDE October 9, 2012 46 / 48
- 218. The D’Alembert solution of the wave equation (Contd.)For each of the two progressive waves, we can also follow the wave motion byobserving that in the xt-plane 1 F (x − ct) is constant along each line x − ct = constant 2 1and similarly 2 G(x + ct) is constant along each line x + ct = constant. () MA 201, PDE October 9, 2012 46 / 48
- 219. The D’Alembert solution of the wave equation (Contd.)For each of the two progressive waves, we can also follow the wave motion byobserving that in the xt-plane 1 F (x − ct) is constant along each line x − ct = constant 2 1and similarly 2 G(x + ct) is constant along each line x + ct = constant.Thus there are two families of parallel lines called the characteristics along which thewaves are propagated. () MA 201, PDE October 9, 2012 46 / 48
- 220. Method of Characteristics (Contd.) Figure: Relationship between characteristic plane and physical plane () MA 201, PDE October 9, 2012 47 / 48
- 221. The D’Alembert solution of the wave equation (Contd.) () MA 201, PDE October 9, 2012 48 / 48
- 222. The D’Alembert solution of the wave equation (Contd.)Furthermorealong x-axis the values of u(x, 0) and ut (x, 0) are given as initial conditions ofdisplacement and velocity () MA 201, PDE October 9, 2012 48 / 48
- 223. The D’Alembert solution of the wave equation (Contd.)Furthermorealong x-axis the values of u(x, 0) and ut (x, 0) are given as initial conditions ofdisplacement and velocityand they just suﬃce to determine the constant values of F and G along the individualcharacteristic. () MA 201, PDE October 9, 2012 48 / 48
- 224. The D’Alembert solution of the wave equation (Contd.)Furthermorealong x-axis the values of u(x, 0) and ut (x, 0) are given as initial conditions ofdisplacement and velocityand they just suﬃce to determine the constant values of F and G along the individualcharacteristic.The characteristics, therefore, represent the paths in xt-plane along which disturbancesin the medium propagated. () MA 201, PDE October 9, 2012 48 / 48
- 225. The D’Alembert solution of the wave equation (Contd.)Furthermorealong x-axis the values of u(x, 0) and ut (x, 0) are given as initial conditions ofdisplacement and velocityand they just suﬃce to determine the constant values of F and G along the individualcharacteristic.The characteristics, therefore, represent the paths in xt-plane along which disturbancesin the medium propagated.Finally, since the solution of wave equation is u = F (x − ct) + G(x + ct) () MA 201, PDE October 9, 2012 48 / 48
- 226. The D’Alembert solution of the wave equation (Contd.)Furthermorealong x-axis the values of u(x, 0) and ut (x, 0) are given as initial conditions ofdisplacement and velocityand they just suﬃce to determine the constant values of F and G along the individualcharacteristic.The characteristics, therefore, represent the paths in xt-plane along which disturbancesin the medium propagated.Finally, since the solution of wave equation is u = F (x − ct) + G(x + ct)the value of u at any point in the xt-plane is the sum of the values of F and G on therespective characteristics which pass through that point. () MA 201, PDE October 9, 2012 48 / 48

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