Partial Differential Equations

Transcription

1 Partial Differential Equations Lecture Notes Dr. Q. M. Zaigham Zia Assistant Professor Department of Mathematics COMSATS Institute of Information Technology Islamabad, Pakistan

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3 Contents 1 Lecture Motivation Basic Definitions Lecture Motivation Initial conditions Boundary conditions Solution of a Boundary Value Problem Linear Boundary Conditions Classification of Boundary Conditions Dirichlet Conditions Neumann Conditions Mixed Boundary Conditions or Robin s Boundary Conditions Superposition Principle Formation of Partial Differential Equation Lecture First Order Partial Differential Equations First Order Partial Differential Equation Solution of First Order Partial Differential Equation Lecture Solution of First Order Partial Differential Equations Method of Characteristics Examples Integral Surfaces Passing Through a Given Curve Parametric Equations Methodology Examples iii

4 iv CONTENTS 5 Lecture Compatible System of First Order Differential Equations Example Non-linear Partial Differential Equations Charpit s Method Examples Lecture Non-linear Partial Differential Equations Charpit s Method Examples Special Types of First Order Non-Linear Partial Differential Equations Equations involving p and q only Separable Equations Clairaut s Form Lecture Second Order Partial Differential Equation Definition Classification of Second Order Partial Differential Equations Examples Reduction of Second Order Partial Differential Equation into Canonical Form Lecture Reduction of Second Order Partial Differential Equations in Canonical Form Result Canonical Form for Hyperbolic Equation Examples Lecture Reduction of Second Order Partial Differential Equations in Canonical Form Result Canonical Form for Parabolic Equation Examples Lecture Reduction of Second Order Partial Differential Equations in Canonical Form Result

5 CONTENTS v 10.2 Canonical Form for Elliptic Equation Example Lecture Reduction of Second Order Partial Differential Equations in Canonical Form Result Mixed Type of Partial Differential Equation Example Lecture Mathematical Modeling Objective Background Approximations and Idealizations Modeling Model Validation Compounding Review of Physics Laws The Continuity Equation Lecture The Heat (Or Diffusion) Equation Initial Conditions Boundary Conditions Initial Boundary Value Problem Examples Solution

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7 1 Lecture Motivation Partial differential equations describe the behavior of many engineering phenomena: Heat flow and distribution Electric fields and potentials Diffusion of chemicals in air or water Electromagnetism and quantum mechanics Wave propagation Fluid Flow (air or liquid) - Air around wings, helicopter blade, atmosphere - Water in pipes or porous media - Material transport and diffusion in air or water - Weather: large system of coupled partial differential equations for momentum, pressure, moisture, heat, Vibration Mechanics of solids - Stress-strain in material, machine part, structure 1

8 2 Chapter 1 Lecture 01 Course Contents Introduction of partial differential equations Partial Differential equations of first order Linear and non-linear partial differential equations Applications of first order partial differential equations Partial Differential equations of second order Mathematical Modeling of heat, Laplace and wave equations Classification of second order partial differential equations Boundary and Initial value problems Reduction to canonical form and the solution of second order partial differential equations Sturm-Liouville system Technique of Separation of Variable for the solution of partial differential equations Laplace, Fourier and Hankel transforms for the solution of partial differential equations and their application to boundary value problems. Recommended Books Richard Haberman, Elementary partial differential equations, Prentice-Hall, INC., Englewood Cliffs, New Jersy K. Sankara Rao, Introduction to partial differential equations, Prentice-Hall of India New Delhi. M. Humi, W. B. Miller, Boundary value problems and partial differential equations, PWS-KENT publishing company, Boston. T. Myint-U, L. Debnath, Linear partial differential equations for scientists and engineers, Fourth Edition, Birkhauser, Berlin Grading Scheme 1.2 Basic Definitions Differential Equations Definition 1.1. A differential equation is an equation that relates the derivatives of a function depending on one or more variables.

9 1.2 Basic Definitions 3 For example d 2 u dx 2 + du dx = cos x is a differential equation involving an unknown function u(x) depending on one variable and 2 u x u 2 y = u t is a differential equation involving an unknown function u(t, x, y) depending on three variables. Partial Differential Equation (PDE) Definition 1.2. A partial differential equation (PDE) is an equation that contains, in addition to the dependent and independent variables, one or more partial derivatives of the dependent variable. Suppose that our unknown function is u and it depends on the two independent variables then the general form of the PDE is F (x, y,, u, u x, u y, u xx, u xy, u yy, ) = 0 Here subscripts denotes the partial derivatives, for example u x = u x, u y = u y, u xx = 2 u x 2, u xy = 2 u x y, u yy = 2 u y 2 Order of Partial Differential Equation Definition 1.3. The order of a partial differential equation is the order of the highest ordered partial derivative appearing in the equation Examples: In the following examples, our unknown function is u and it depends on three variables t, x and y. u xx + 2xu xy + u yy = e y ; Order is Two u xxy + xu yy + 8u = 7y; Order is Three u t 6uu x + u xxx = 0; Order is Three u t + uu x = u xx ; Order is Two u xxx + xu xy + yu 2 = x + y; Order is Three u x + u y = 0; Order is One

10 4 Chapter 1 Lecture 01 Degree of Partial Differential Equation Definition 1.4. The degree of a partial differential equation is the degree of the highest order partial derivative occurring in the equation. Examples: In the following examples, our unknown function is u and it depends on two variables t, x and y. u xx + 2xu xy + u yy = e y ; Degree is One u xxy + xu 2 yy + 8u = 7y; Degree is One u t 6uu x + u 3 xxx = 0; Degree is Three u t + uu 3 x = u xx ; Degree is One u 2 xxx + xu 3 xy + yu 2 = x + y; Degree is Two u x + u y = 0; Degree is One Linear Operator A linear operator L by definition satisfies L(c 1 u 1 + c 2 u 2 ) = c 1 L(u 1 ) + c 2 L(u 2 ) (1.2.1) eq:01 for any two functions u 1 and u 2, where c 1 and c 2 are arbitrary constants. / t and 2 / x 2 are the examples of linear operators since these two satisfy equation (4.2.1): t (c u 1 1u 1 + c 2 u 2 ) = c 1 t + c u 2 2 t 2 x 2 (c 2 u 1 1u 1 + c 2 u 2 ) = c 1 x 2 + c 2 u 2 2 x 2 Note: Any linear combination of linear operators is a linear operator. However ( / t) 2 is not a linear operator since it does not satisfy equation (4.2.1). To prove the non-linearity of the operator ( / t) 2, we do the following calculations. If our L = ( / t) 2 then Lu = ( u/ t) 2 u derivative }{{} u t Square }{{} ( ) u 2 t

11 1.2 Basic Definitions 5 Now ( ) (c1 u 1 + c 2 u 2 ) 2 = t = = c 2 1 ( c 1 u 1 ( u 1 c 1 t ( u1 t c 1 ( u2 t ) 2 t + c u 2 2 t ) 2 ( ) u 2 2 u 1 + c 2 + 2c 1 c 2 t t ) 2 ( ) 2 + c 2 u2 u c 1 c 2 t t ) 2 ( ) 2 u2 + c 2 t u 2 t u 2 t Definition 1.5. A partial differential equation is said to be a linear if i) it is linear in the unknown function and ii) all the derivatives of the unknown functions with constant coefficients or the coefficients depends on the independent variables. Examples: Suppose u is our unknown function which depends on three independent variables t, x and y then the following are linear partial differential equations Laplace s equations: u = 0, where = u xx + u yy Helmholtz s equation: = λu First-order linear transport equation: u t + cu x = 0 Heat or diffusion equation: u t u = 0 Schrodinger s equation: iu t + u = 0 Quasi-Linear Partial Differential Equation Definition 1.6. A partial differential equation is said to be a quasi-linear if the derivatives of unknown function are linear, while the coefficients depends on the independent variables as well as the dependent variables. Examples: Suppose u is our unknown function which depends on two independent variables x and y then the following are quasi-linear partial differential equations uu x + u y = 0 yu xx + 2xyuu yy + u = 1 u xxy + xuu yy + 8u = 7y

12 6 Chapter 1 Lecture 01 Semi-Linear Partial Differential Equation Definition 1.7. A partial differential equation is said to be a semi-linear if the derivatives of unknown function are linear and the coefficients of derivatives depends on the independent variables only. Examples: Suppose u is our unknown function which depends on two independent variables x and y then the following are semi-linear partial differential equations u x + u xy = u 2 yu xx + (2x + y)u yy + u 3 = 1 u xxy + xu yy + 8u = 7xyu 2 Non-Linear Partial Differential Equation Definition 1.8. A partial differential equation which is not linear is called nonlinear partial differential equation. Examples: Suppose u is our unknown function which depends on three independent variables t, x and y then the following are non-linear partial differential equations: Non linear Poisson equation: u = f(u) Burgers equation: u t + uu x = 0 Korteweg-deVries equation(kdv): u + uu x + u xxx = 0 Reaction-diffusion equation: u t u = f(u) Homogeneous Partial Differential Equation Definition 1.9. A partial differential equation is said to be a homogeneous partial differential equation if its all terms contain the unknown functions or its derivatives. The simplest way to test whether an equation is homogeneous is to substitute the function u identically equal to zero. If u = 0 satisfies the equation then the equation is called homogeneous. Examples: u xx + xu xy + yu 2 = 0 is a homogeneous equation u xx + 2xu xy + 5u = 0 is a homogeneous equation 3u x + uu y = f(x, y) is a non-homogeneous equation

13 1.2 Basic Definitions 7 Solution of Partial Differential Equation Definition Functions u = u(x, y, ) which satisfy the following partial differential equation F (x, y,, u, u, u x, u y,, u xx, u xy, u yy, ) = 0 identically in a suitable domain D of the n dimensional space R n in the independent variables x, y,, if exist are called solutions. Example: Functions u(x, y) = (x + y) 3 and u(x, y) = sin(x y) are solutions of the differential equation u xx u yy = 0. Principle of Superposition Definition If u 1 and u 2 are two solutions of a linear homogeneous partial differential equation then an arbitrary linear combination of them c 1 u 1 + c 2 u 2 also satisfies the same linear homogeneous differential equation. Examples: Consider a one dimensional heat equation 1 u k t = 2 u x 2. (1.2.2)?eq:2? If u 1 and u 2 are two solutions of equations (4.1.1) then they must satisfy equation (4.1.1), that is, 1 u 1 k t = 2 u 1 x 2, 1 u 2 k t = 2 u 2 x 2. According to the principle of superposition, c 1 u 1 + c 2 u 2 is again the solution of equation (4.1.1). t (c 1 1u 1 + c 2 u 2 ) = c 1 k L.H.S. = 1 k 2 u 1 = c 1 x 2 + c 2 u 2 2 x 2 u 1 t + c 1 2 k u 2 t = 2 x 2 (c 1u 1 + c 2 u 2 ) = R. H. S. Note: Here the superposition principle is stated for two solutions only, it is true for any finite linear combinations of solutions. Furthermore, under proper restrictions, it is also true for infinite number of solutions. If u i, i = 1, 2, are solutions of a homogeneous linear partial differential equations, then n w = c i u i is also a solution. i=1

14 8 Chapter 1 Lecture 01 Theorem If u 1 and u 2 are solutions of a linear inhomogeneous equation then u 1 u 2 is a solution of the corresponding homogeneous equation. Consider the general form of a non-homogeneous differential equation au + n b i (u) xi + = f(x) i=1 If u 1 and u 2 are two solutions of above equation then au 1 + n b i (u 1 ) xi + = f(x), au 2 + i=1 The process of subtraction gives us, a(u 1 u 2 ) + n b i (u 2 ) xi + = f(x) i=1 n b i (u 1 u 2 ) xi + = 0 i=1

15 2 Lecture Motivation The number of independent solutions for a partial differential equations are infinite. The boundaries of the regions of the independent variables over which we desire to solve the differential equations are not discrete points (as in one dimensional case) but are continuous curves or surfaces. Thus, the complete formulation of a physical system in terms of partial differential equations requires careful attention - not only the equations that govern the system, but - also to the correct formulation of the boundary conditions. Furthermore, most differential equations that we encounter in applications are mathematical expressions of physical laws (for example the heat equation is the expression of the law of energy conservation). Therefore, in order to obtain a unique solution, we must specify the initial conditions in addition to the boundary conditions. 2.2 Initial conditions Definition 2.1. Conditions at an initial time t = t 0 from which a given set of mathematical equations or physical system evolves are known as initial conditions. A system with initial conditions specified is known as the initial value problem. 9

16 10 Chapter 2 Lecture 02 Alternatively, if the conditions relates one value of the independent variable such as u(x 0 ) = A and u x (x 0 ) = B. Then these conditions are called Initial Conditions and x 0 is called the initial point. Example:1 As a simple example, we suppose that our unknown function u is dependent on one variable x. Then the following problem is known as initial value problem u xx + u x 2u = 0, u(0) = 3, u x (0) = 7. Example:2 Now we suppose that our unknown function u is dependent on two variable t, x. Then the following problem is known as initial value problem u xx + u t 2u = 0, u(0, x) = 3x, u t (0, x) = sin x. 2.3 Boundary conditions Definition 2.2. The set of conditions specified for the behavior of the solution to a set of differential equations or partial differential equations at the boundary of its domain are known as Boundary Conditions. A system with boundary conditions is known as the boundary value problem. Alternatively, the problem of finding the solution of a differential equation such that all the associated conditions relate to two different values of the independent variable is called a boundary value problem. Example: If u(x, t) is the displacement of a vibrating string and its ends are fixed at x = a and x = b, then the conditions are boundary conditions. u(a, t) = 0, and u(b, t) = Solution of a Boundary Value Problem By a solution to a boundary value problem on an open region D, we mean a function u that satisfies the differential equation on D and it is continuous on D D, and satisfies the specified boundary conditions on D. 2.5 Linear Boundary Conditions Definition 2.3. The boundary conditions are linear if they express a linear relationship between u and its partial derivatives (up to appropriate order) on D. (In other words, a boundary condition is linear if it is expressed as a linear equation between u and its derivatives on D. )

17 2.6 Classification of Boundary Conditions 11 Figure 2.1: Laterally insulated Wire 2.6 Classification of Boundary Conditions Dirichlet Conditions Definition 2.4. The boundary conditions specify the values of the unknown function u on the boundary. This type of boundary condition is called the Dirichlet condition Neumann Conditions Definition 2.5. The boundary conditions specify the derivatives of the unknown function u in the direction normal to the boundary, which is written as u/ n. This type of boundary condition is called the Neumann condition. Remark: The normal derivative on the boundary u/ n is defined as ( u u n = grad u n =,, u ) n, x 1 x n where n is the outward normal to D Mixed Boundary Conditions or Robin s Boundary Conditions Definition 2.6. The boundary conditions specify a linear relationship between u and its normal derivative on the boundary. These are referred to as mixed boundary conditions or Robin s boundary conditions. The general form of such a boundary condition is [ αu + β n] u = f(x) D, α, β are constants. D Example Consider the problem of heat conduction in a laterally insulated thin wire. u(x, t) is the temperature in the wire, the constant k is the diffusivity, which indicates the rate of diffusion of the heat along the wire, and the length of the wire is L. The initial condition is u(x, 0) = F (x), where F (x) is the initial temperature distribution in the wire. The three major types of boundary conditions are as follows: (1) Immerse the wire in melting ice (0 C) at each end point (See Figure) and let u(x, t) be measured in C. u(0, t) = u(l, t) = 0, for t > 0.

18 12 Chapter 2 Lecture 02 Figure 2.2: Laterally Insulated Wire with Ice on the Ends These are Dirichlet or fixed boundary conditions. Alternatively, prescribe the temperature of each end-point to be p(t) and q(t), respectively: u(0, t) = p(t), u(l, t) = q(t), for t > 0. These also are Dirichlet, or fixed, boundary conditions. 2) Insulate each end-point; thus, the wire is totally insulated (as in Figure): u x (o, t) = u x (L, t) = 0, for t > 0. These are Neumann, or free, boundary conditions. Figure 2.3: Totally Insulated Wire Alternatively, prescribe the flow of heat at each end-point to be p(t) and q(t), respectively: u x (0, t) = p(t) K, and u x(l, t) = q(t), for t > 0. K K > 0 is the thermal conductivity. These also are Neumann, or free, boundary conditions. 3) Each end-point is exposed and radiates heat into the surrounding medium which has a temperature of T (t): βu x (0, t) = α[u(0, t) T (t)], and δu x (L, t) = γ[u x (L, t) T (t)], for t > 0, which simplify to: αu(0, t) βu x (0, t) = αt (t), and γu(l, t) + δu x (L, t) = γt (t), for t > 0, where α, β, γ are positive constants. These are Robin, or mixed, boundary conditions. If the surrounding medium has a temperature of 0 C (i.e., T (t) = 0), and u(x, t) is measured in C, then we have αu(0, t) βu x (0, t) = 0, and γu(l, t) + δu x (L, t) = 0, for t > 0.

19 2.7 Superposition Principle 13 Figure 2.4: laterally Insulated Wire with Exposed Ends 2.7 Superposition Principle Superposition Principle for Linear Boundary Conditions Theorem 2.7. If u 1 and u 2 are the solutions of a linear homogeneous partial differential equation with linear boundary conditions [ αu 1 (x) + β u ] 1(x) = f(x) n D D [ αu 2 (x) + β u ] 2(x) = g(x) n D where α, β are constants, then w = u 1 + u 2 is a solution of the partial differential equation that satisfies the boundary conditions [ αw(x) + β w(x) ] = (f(x) + g(x)) n D D Note: The above result is particularly useful in applications in which the boundary conditions are complex. Example Consider the Laplace equation D 2 u x u y 2 = 0 in rectangle with the following linear boundary conditions u(x, 0) = f 1 (x), u(x, b) = f 2 (x), We split the problem in two parts: 2 u 1 x u 1 y 2 = 0 u 1 (x, 0) = f 1 (x) u 1 (x, b) = f 2 (x) u 1 (0, y) = 0 u 1 (a, y) = 0 u(0, y) = g 1 (y) u(a, y) = g 2 (y) 2 u 2 x u 2 y 2 = 0 u 2 (x, 0) = 0 u 2 (x, b) = 0 u 2 (0, y) = g 1 (y) u 2 (a, y) = g 2 (y)

20 14 Chapter 2 Lecture 02 Figure 2.5: Boundary Conditions on a Plate Obviously, if we solve u 1, u 2, then u 1 +u 2 is a solution of Laplace equation, which satisfies all the boundary conditions. Example Note: Neumann boundary conditions usually do not specify the unique solution of a boundary value problem. Consider the solution with the Neumann boundary conditions 2 u 2 x u 2 y 2 = 0 u y (x, 0) = f 1(x), u x (0, y) = g 1(x), u y (x, b) = f 2(x) u x = g 2(x). It is obvious that if u is the solution of this boundary value problem, then w = u + c (c is constant) is also a solution of the above boundary value problem. Thus Neumann boundary conditions determine the solution of this boundary value problem up to a constant. 2.8 Formation of Partial Differential Equation Suppose u, v are two given functions of x, y and z. Let F be an arbitrary function of u and v of the form F (u, v), or F (u(x, y, z), v(x, y, z)). A differential equation can be formulated by eliminating the arbitrary function F. Taking partial derivatives of F (u, v) with respect to x and y and taking z is a

21 2.8 Formation of Partial Differential Equation 15 function of x, y, we obtain [ F u u x + u [ F u u y + u z p ] z p ] + F v + F v [ v x + v [ v y + v z p ] z p ] = 0 = 0 Eliminating F/ u and F/ v from above equations, we obtain and where P = P p + Qq = R, (u, v) (u, v) (u, v), Q =, R = (y, z) (z, x) (x, y), p = z x, q = z y. Definition 2.8. The first order partial differential equation P p + Qq = R, is called Lagrange s partial differential equation of first order. Problem Sheet Form the partial differential equation by eliminating the arbitrary function from 1. z = f(x + it) + g(x it), where i = 1 2. f(x + y + z, x 2 + y 2 + z 2 ) = 0 3. z = xy + f(x 2 + y 2 ) 4. z = f(xy/z) 5. z = ax + by + ab Solution: 1. 2 z x z t 2 = (z y)p + (x z)q = y x 3. yp xq = y 2 x 2 4. px = qy 5. z = px + qy + pq

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23 3 Lecture First Order Partial Differential Equations Definition 3.1. An equation written in the form f(x, y, z, p, q) = 0, is called first order partial differential equation, where p = z x, q = z y ; x and y are independent variables and z is dependent variable. Example Form the partial differential equation from the following equation x 2 + y 2 + (z c) 2 = a 2 with a and c are arbitrary constants. Note: The equation in above example represents the set of all spheres whose center lie along the z-axis. Example Form the partial differential equation from the following equation x 2 + y 2 = (z c) 2 tan 2 α with c and α are arbitrary constants. Note: The equation in above example represents the set of all right circular cones whose axis coincides with the line OZ. Example Form the partial differential equation from the following equation x 2 + y 2 + (z c) 2 = a 2 with a and c are arbitrary constants. Solution: The obtained differential equation is yp xq = 0. 17

24 18 Chapter 3 Lecture 03 Example Form the partial differential equation from the following equation x 2 + y 2 = (z c) 2 tan 2 α with c and α are arbitrary constants. Solution: The obtained differential equation is Surfaces of Revolution yp xq = 0. Now a question arises that why these two different geometrical entities have the same partial differential equation. The spheres and cones are basically surfaces of revolution which have the line OZ as axes of symmetry. Definition 3.2. A surface of revolution is a surface in Euclidean space created by rotating a curve around a straight line in its plane (the axis). All surfaces of revolution with the property that they have the line OZ as axes of symmetry are characterized by an equation of the form z = f(x 2 + y 2 ) where the function f is arbitrary. The corresponding differential equation is yp xq = First Order Partial Differential Equation Definition 3.3. (Linear Partial Differential Equation) An equation of the form P p + Qq = R where P and R are given functions of x, y and Q is a function of x, y and z is known as first order linear partial differential equation. Definition 3.4. (Quasi-Linear Partial Differential Equation) An equation of the form P p + Qq = R where P and R are given functions of x, y and z (which do not involve p or q) is known as first order quasi-linear partial differential equation. Note: The above equation is referred to as Langrange s Equation.

25 3.3 Solution of First Order Partial Differential Equation Solution of First Order Partial Differential Equation Theorem 3.5. The general solution of linear partial differential equation P p + Qq = R can be written in the form F (u, v) = 0, where F is an arbitrary function, and u(x, y, z) = C 1 and v(x, y, z) = C 2 form a solution of the equation dx P (x, y, z) = dy Q(x, y, z) = dz R(x, y, z) Example Find the solution of following linear partial differential equation y 2 p xyq = x(z 2y) (y + 2x)p (x + yz)q = x 2 y 2

26 20 Chapter 3 Lecture 03

27 4 Lecture 04 Review Formation of First Order Partial Differential Equations Solution of First Order Partial Differential Equations 4.1 Solution of First Order Partial Differential Equations Method of Characteristics Solution of Partial Differential Equation of First Order Theorem 4.1. The general solution of first order quasi-linear partial differential equation P p + Qq = R (4.1.1) eq:02 can be written in the form F (u, v) = 0, where F is an arbitrary function, and u(x, y, z) = C 1 and v(x, y, z) = C 2 form a solution of the equation dx P (x, y, z) = dy Q(x, y, z) = dz R(x, y, z). The curves defined by u(x, y, z) = C 1 and v(x, y, z) = C 2 are called the families of characteristics curves of equation (4.1.1) Examples Example Find the solution of following partial differential equation x y z α β γ 1 = 0 z x z y 21

28 22 Chapter 4 Lecture 04 where z = z(z, y). Solution: The solution of the above partial differential equation is. F (x 2 + y 2 + z 2, αx + βy + yz) = 0 Example Find the general integrals of the following partial differential equations pz qz = z 2 + (x + y) 2 Solution: The solution of the above partial differential equations is F (x + y, 2x ln z 2 + (x + y) 2 ) = 0 Example Find the general integrals of the following partial differential equations (x 2 yz)p + (y 2 zx)q = z 2 xy. Solution: The solution of the above partial differential equations is ( x y F y z, y z ) z x Theorem 4.1. If u i (x 1, x 2,, x n, z) = c i (i = 1, 2,, n) are independent solutions of the equations dx 1 P 1 = dx 2 P 2 = = dx n P n = dz R then the relations F (u 1, u 2,, u n ) = 0, in which the function F is arbitrary, is a general solution of the linear partial differential equation P 1 z x 1 + P 2 z x P n z x n = R. Example If u is a function of x, y and z which satisfies the partial differential equation (y z) u + (z x) u + (x y) u x y z = 0. Show that u contains x, y and z only in combinations x + y + z and x 2 + y 2 + z 2. Solution: Auxiliary Equations are dx y z = This implies, (using multipliers) dy z x = dz x y = du 0 du = 0 dx + dy + dz = 0 xdx + ydy + zdz = 0

29 4.2 Integral Surfaces Passing Through a Given Curve 23 Integrating above equations we obtain u = C 1, x + y + z = C 2, x 2 + y 2 + z 2 = C 3 Where C 1, C 2, C 3 are arbitrary constants. Thus the general solution can be written in terms of an arbitrary function F in the form u = F (x + y + z, x 2 + y 2 + z 2 ) 4.2 Integral Surfaces Passing Through a Given Curve Parametric Equations Definition 4.2. Parametric equations are a set of equations that express a set of quantities as explicit functions of a number of independent variables, known as parameters. Examples: The equation of a circle in Cartesian coordinates is x 2 + y 2 = r 2, while its one set of parametric equations are x = r cos θ y = r sin θ Equation for a parabola y = x 2 can be parametrized by using a free parameter t. The One set of parametric equation is x = t, y = t 2. while the second set is x = t, y = t Methodology Integral Surfaces Passing Through a Given Curve Suppose for a given partial differential equations we obtained two integral curves described by u(x, y, z) = C 1, v(x, y, z) = C 2. (4.2.1) eq:01 Then the solution of the partial differential equation can be written in the form F (u, v) = 0.

30 24 Chapter 4 Lecture 04 Now we wish to determine an integral surface, containing a given curve Γ described by the parametric equation of the form x = x(t), y = y(t), z = z(t), where t is a parameter. Then our particular solution must be like u(x(t), y(t), z(t)) = C 1, v(x(t), y(t), z(t)) = C 2. Eliminating the parameter t from above equations to obtain a relation of the type F (C 1, C 2 ) = 0. Substituting the values of C 1 and C 2 from equation (4.2.1) leads to the integral surface passing through the given curve Γ Examples Example Find the integral surface of the partial differential equation x(y 2 + z)p y(x 2 + z)q = (x 2 y 2 )z containing the straight line x + y = 0, z = 1. Solution: The required integral surface is x 2 + y 2 2z + 2xyz 2 = 0. Example Find the integral surface of the partial differential equation xp + yq = z with the data x 2 + y 2 + z 2 = 4 and x + y + z = 2. Solution: The required integral surface is xy + xz + yz = 0

31 5 Lecture 05 Review Solution of First Order Partial Differential Equations Method of Characteristics Examples Integral Surfaces Passing Through a Given Curve Parametric Equations Methodology Examples 5.1 Compatible System of First Order Differential Equations Definition 5.1. Two first order partial differential equations are said to be compatible if they have a common solution. Necessary and Sufficient Condition for Compatible Differential Equations Consider two partial differential equations f(x, y, z, p, q) = 0, g(x, y, z, p, q) = 0. Then the necessary and sufficient condition for compatible differential equations is to satisfy the following equation (f, g) g) (f, g) (f, g) + p (f, + + q (x, p) (z, p) (y, q) (z, q) = 0. 25

32 26 Chapter 5 Lecture Example Example Show that the following partial differential equations are compatible xp yq = x, x 2 p + q = xz and, hence, find their solution. Example Show that the following partial differential equations are compatible Home Work xp = yq, and z(xp + yq) = 2xy. 5.2 Non-linear Partial Differential Equations Charpit s Method Consider a non-linear first order partial differential equation of the form f(x, y, z, p, q) = 0. The basic idea of Charpit s Method is the introduction of another partial differential equation of the form g(x, y, z, p, q) = 0, and then solve above two equations for p and q. values of p and q in the following equation, In next step we substitute the dz = p(x, y, z, a)dx + q(x, y, z, a)dy. Now the solution of above equation, if it exists, is the complete integral of the given equation. The main idea in Charpit s method is to determine the second order partial differential equation We will use the idea of compatible equations. Assume that the following equation is compatible with the given equation g(x, y, z, p, q) = 0 f(x, y, z, p, q) = 0

33 5.2 Non-linear Partial Differential Equations 27 for which the neccessary and sufficient condition is (f, g) g) (f, g) (f, g) + p (f, + + q (x, p) (z, p) (y, q) (z, q) = 0. The above equation reduces in the form of a quasi-linear equation in g, i.e., g f p x + f g q y + (pf p + qf q ) g z (f x + pf z ) g p (f y + qf z ) g q = 0. from which we can determine g. The auxiliary equations of above quasi-linear equation are dx = dy dz dp = = f p f q pf p + qf q (f x + pf z ) = dq (f y + qf z ). These equations are called Charpit s equations. Note: Any Integral of the above Charpit s equations which involves p and q can be taken as the second equation g(x, y, z,, p, q) = 0. It may also be noted that all Charpit s equations need not to be used, but it is enough to choose the simplest Examples Example Find the complete integral of x 2 p 2 + y 2 q 2 4 = 0. using Charpit s Method. Solution: The complete integral of given equation is z = a ln x + 4 a 2 ln y + b.

34 28 Chapter 5 Lecture 05

35 6 Lecture 06 Review Compatible System of First Order Differential Equations Examples Non-linear Partial Differential Equations Charpit s Method Examples 6.1 Non-linear Partial Differential Equations Charpit s Method Consider a non-linear first order partial differential equation of the form f(x, y, z, p, q) = 0. The basic idea of Charpit s Method is the introduction of another partial differential equation of the form g(x, y, z, p, q) = 0, and then solve equation above two equations for p and q. In next step we substitute the values of p and q in the following equation, dz = p(x, y, z, a)dx + q(x, y, z, a)dy. Now the solution of above equation, if it exists, is the complete integral of the given equation. 29

36 30 Chapter 6 Lecture Examples Example Find the complete integral of the partial differential equation z 2 = pqxy. Solution The solution of the above equation is z = bx 1/a y a. Example Find the solution of the following partial differential equation (p 2 + q 2 )y = qz. Solution: The solution of the above equation is (x + b) 2 = y 2 z2 a. 6.2 Special Types of First Order Non-Linear Partial Differential Equations Type 1 Equations involving p and q only Type 2 Separable equations Type 3 Clairaut s form Equations involving p and q only Consider equation of the type f(p, q) = 0. Let z = ax + by + c = 0 is the solution of the given partial differential equation described by f(p, q) = 0 then p = z z = a, q = x y = b. Substituting these values in the given partial differential equation, we obtain f(a, b) = 0. Solving for b, we get value of b in a, say φ(a). Then z = ax + φ(a)y + c

37 6.2 Special Types of First Order Non-Linear Partial Differential Equations 31 is the complete integral of the given partial differential equation. Example Find a complete integral of the function p + q = 1. Solution: The given partial differential equation of the form f(p, q) = 0. Therefore we assume that its solution in the form z = ax + by + c, where a + b = 1 or b = (1 a) 2. Therefore the solution of the given partial differential equation is Separable Equations z = ax + (1 a) 2 y + c. Definition 6.1. An equation in which z is absent and the terms containing x and p can be separated from those containing y and q is called a separable equation. Then, Or Example Find the complete integral of the following partial differential equation p 2 y(1 + x 2 ) = qx 2. Solution: The given equation can be written as p 2 (1 + x 2 ) x 2 = q y. Assume that this equation is equal to a, an arbitrary constant, such that p 2 (1 + x 2 ) x 2 = q y = a. p 2 (1 + x 2 ) x 2 = a, p = Substituting these values of p and q in ax 1 + x 2, dz = pdx + qdy, q y = a. q = ay. we obtain dz = ax dx + aydy 1 + x 2 On Integration we obtain the solution of the given equation z = a 1 + x 2 + a 2 y2 + b.

38 32 Chapter 6 Lecture Clairaut s Form Definition 6.2. A first order partial differential equation is said to be of Clairaut s form if it can be written as z = px + qy + f(p, q). The corresponding Charpit s equations are dx x + f p = dy y + f p = dz = dp px + qy + pf p + qf q 0 = dq 0 Therefore, dp = 0, dq = 0 which implies p = a, q = b. Substituting the values of p and q in the given partial differential equation we get the solution in the form z = ax + by + f(a, b). Example: The complete solution of the partial differential equation z = px + qy p 2 + q 2 is z = ax + by a 2 + b 2.

39 7 Lecture 07 Review Non-linear partial differential equations Charpit s Method Examples Special types of non-linear partial differential equations are Type 1: Equations involving p and q only Type 2: Separable equations Type 3: Clairaut s form. 7.1 Second Order Partial Differential Equation Definition Definition 7.1. The general form of a second order partial differential equation is Au xx + 2Bu xy + Cu yy + Du x + Eu y + F u + G = 0, where A, B, C, D, E, F, G are the functions of x and y only or in a simple case these coefficients are constant. 7.2 Classification of Second Order Partial Differential Equations The classification of partial differential equation is motivated by the classification of the quadratic equation of the form Ax 2 + 2Bxy + Cy 2 + Dx + Ey + F = 0. 33

40 34 Chapter 7 Lecture 07 The discriminant of the above equation is B 2 AC, then the above equation is Elliptic if B 2 AC < 0 Parabolic if B 2 AC = 0 Hyperbolic if B 2 AC > 0. Generalizing this concept for the second order partial differential equations Au xx + 2Bu xy + Cu yy + Du x + Eu y + F u + G = 0. Then we classify the above above partial differential equation as Elliptic if B 2 AC < 0 Parabolic if B 2 AC = 0 Hyperbolic if B 2 AC > Examples Example Consider the Laplace equation u xx + u yy = 0. Here A = 1, B = 0 and C = 1, and the discriminant of the above equation is B 2 AC = 1 < 0. Therefore the Laplace equation is Elliptic in nature. Example The Diffusion equation is u xx = αu t, Or u xx αu t = 0 Here A = 1, B = 0 and C = 0, and the discriminant of the above equation is B 2 AC = 0. Therefore the Diffusion equation is parabolic. Example Consider the following wave equation u xx = 1 c 2 u tt, Or u xx 1 c 2 u tt = 0. Here A = 1, B = 0 and C = 1 c 2, and now we calculate the discriminant such that B 2 AC = 0 (1)( 1 c 2 ) = 1 c 2 > 0. Therefore it is a hyperbolic equation.

41 7.3 Reduction of Second Order Partial Differential Equation into Canonical Form Reduction of Second Order Partial Differential Equation into Canonical Form Consider the second order partial differential equation Au xx + 2Bu xy + Cu yy + F (x, y, u, u x, u y ) = 0, where A, B and C are functions of x and y only or they are constants and u has continuous first and second order partial derivatives. Also A, B, C do not vanish at the same time. Let us transform the independent variables x, y to a new variables ξ, η, where ξ = ξ(x, y) η = η(x, y). such that the functions ξ, η are continuously differential and the Jacobian (ξ, η) ξ η J = (x, y) = x x 0 In the new coordinates (ξ, η) the second order partial differential equation can be written as where η y η y ( A 2 u x 2 + 2B 2 u x y + C 2 u y 2 + F x, y, u x, u ) = 0 y A 2 u ξ 2 + 2B 2 u ξ η + C 2 u η 2 + F ( ) ξ 2 A = A + 2B ξ ξ x x y + C B = A ξ ( η ξ x x + η x y + ξ y ( ) η 2 C = A + 2B η η x x y + C ( x, y, u ξ, u ) = 0 η ( ) ξ 2, y ) η + C ξ x y ( ) η 2. y η y,

42 36 Chapter 7 Lecture 07

43 8 Lecture 08 Review Second Order Partial Differential Equations Definition Classification of Second Order Partial Differential Equation Examples Reduction of Second Order Partial Differential Equations in Canonical Form 8.1 Reduction of Second Order Partial Differential Equations in Canonical Form The second order partial differential equation A 2 u x 2 + 2B 2 u x y + C 2 u y 2 + F can be reduced in the following canonical form where A 2 u ξ 2 + 2B 2 u ξ η + C 2 u η 2 + F ( x, y, u x, u ) = 0 y ( x, y, u ξ, u ) = 0 η ( ) ξ 2 A = A + 2B ξ ( ) ξ ξ 2 x x y + C, y B = A ξ ( η ξ x x + B η x y + ξ η y x ( ) η 2 C = A + 2B η ( ) η η 2 x x y + C. y 37 ) + C ξ y η y,

44 38 Chapter 8 Lecture Result We know that ( B 2 ξ A C = (B 2 η AC) x y ξ ) η 2, y x where A, B, C are not zero at the same time. Thus the transformation of independent variables does not change the type of equation. Since, ( ξ η x y ξ ) η 2 ξ η 2 = x x > 0. y x Therefore the sign of the discriminant does not change, either we are in (x, y) coordinates or in (ξ, η) coordinates. 8.2 Canonical Form for Hyperbolic Equation η y η y In Hyperbolic case our discriminant B 2 A A = 0 and C = 0. Thus we have C must be positive. For this we set which, on rewriting, become A = A (ξ x ) 2 + 2Bξ x ξ y + C (ξ y ) 2 = 0, C = A (η x ) 2 + 2Bη x η y + C (η y ) 2 = 0, A A ( ξx ξ y ( ηx η y ) 2 + 2B ) 2 + 2B ( ξx ξ y ( ηx η y ) + C = 0 ) + C = 0 These equations are quadratic in (ξ x /ξ y ) and (η x /η y ), so the roots of these equations are ξ x = B ± B 2 AC η x, = B ± B 2 AC. ξ y A η y A Note The condition B 2 AC > 0 ensures that the slopes of the curves ξ(x, y) = c 1 and η(x, y) = c 2 are real. This means that at any point (x, y) there exist two real directions given by the two roots along which the given partial differential equation reduces to the canonical form.

45 8.2 Canonical Form for Hyperbolic Equation 39 or There are two solutions for each quadratic however we will consider only one solution for each otherwise we will end up with the same two coordinates. Thus the chosen solution are of the form ξ x = B + ξ y η x = B η y These are called characteristic equations. Along the curve ξ(x, y) = c 1, we have This implies B 2 AC, A B 2 AC. A dξ = ξ x dx + ξ y dy = 0 ξ y dy = ξ x dx dy dx = ( ξx Similarly, along the curve η(x, y) = c 2, we have or Hence ξ y ) dη = η x dx + η y dy = 0 η y dy = η x dx dy dx = ( ηx Substituting the values of (ξ x /ξ y ) and η x /η y in the following equations such that ( ) ( dy dx = ξx B + ) B = 2 AC, A dy dx = ξ y ( ηx η y ) = ( η y ) B B 2 AC A Integrating the above equations we obtain the equations of family of characteristics ξ(x, y) = c 1 and η(x, y) = c 2, which are called characteristics of the given partial differential equations. Now to obtain the canonical form of the given partial differential equation we substitute the expression of ξ and η in which reduces to A 2 u ξ 2 + 2B 2 u ξ η + C 2 u η 2 + F 2 u ξ η ( x, y, u ξ, u η = Φ(ξ, η, u, u ξ, u η ). ). ) = 0

46 40 Chapter 8 Lecture Examples Example Reduce the following partial differential equation 3u xx + 10u xy + 3u yy = 0 to a canonical form and then find its general solution. Example Reduce the following partial differential equation to a canonical form. y 2 u xx x 2 u yy = 0

47 9 Lecture 09 Review Reduction of Second Order Partial Differential Equations in Canonical Form Result Canonical Form for Hyperbolic Equation Examples 9.1 Reduction of Second Order Partial Differential Equations in Canonical Form The second order partial differential equation A 2 u x 2 + 2B 2 u x y + C 2 u y 2 + F can be reduced in the following canonical form where A 2 u ξ 2 + 2B 2 u ξ η + C 2 u η 2 + F ( x, y, u x, u ) = 0 y ( x, y, u ξ, u ) = 0 η ( ) ξ 2 A = A + 2B ξ ( ) ξ ξ 2 x x y + C, y B = A ξ ( η ξ x x + B η x y + ξ η y x ( ) η 2 C = A + 2B η ( ) η η 2 x x y + C. y ) + C ξ y η y, 41

48 42 Chapter 9 Lecture Result We know that ( B 2 ξ A C = (B 2 η AC) x y ξ ) η 2, y x where A, B, C are not zero at the same time. Thus the transformation of independent variables does not change the type of equation. Since, ( ξ η x y ξ ) η 2 ξ η 2 = x x > 0. y x Therefore the sign of the discriminant does not change, either we are in (x, y) coordinates or in (ξ, η) coordinates. 9.2 Canonical Form for Parabolic Equation In parabolic case our discriminant B 2 A C must be zero, which can be true if B = 0 and A = 0 or C = 0. Suppose we first set A = 0. Then we obtain which, on rewriting, becomes η y η y A = A (ξ x ) 2 + 2Bξ x ξ y + C (ξ y ) 2 = 0, A ( ξx ξ y ) 2 + 2B ( ξx ξ y ) + C = 0 This equation is quadratic in (ξ x /ξ y ), so the roots of the above equation are ξ x = B ± B 2 AC. ξ y A Since B 2 AC = 0, therefore ξ x ξ y = B A. This is the characteristic equation. Also this is a slope along the curve The total differential is ξ(x, y) = c 1. dξ = ξ x dx + ξ y dy = 0 ξ y dy = ξ x dx dy dx = ξ x. ξ y

49 9.2 Canonical Form for Parabolic Equation 43 Or ( ) dy B dx = = B A A. Integration of above equation gives us the implicit solution ξ(x, y) = c 1. Now coming to the second case, when B = 0, in fact with a small calculation, we can prove that if A = 0 then B = 0. We, therefore choose ξ in such a way that both A and B are zero. Then we choose η in any way we like as long as it is not parallel to the ξ- coordinates. In other words, we choose η such that the Jacobian of the transformation is not zero. Thus we can write the canonical equation for parabolic case by simply substituting the values of ξ and η into ( A 2 u ξ 2 + 2B 2 u ξ η + C 2 u η 2 + F x, y, u ξ, u ) = 0 η which reduces either of the forms Examples 2 u ξ 2 = Φ(ξ, η, u, u ξ, u η ), 2 u η 2 = Φ(ξ, η, u, u ξ, u η ). Example Reduce the following partial differential equation x 2 u xx 2xyu xy + y 2 u yy = e x to a canonical form. Example Reduce the following partial differential equation to canonical form. u xx 2xu xy + x 2 u yy 2u y = 0

50 44 Chapter 9 Lecture 09

51 10 Lecture 10 Review Reduction of Second Order Partial Differential Equations in Canonical Form Result Canonical Form for Parabolic Equation Examples 10.1 Reduction of Second Order Partial Differential Equations in Canonical Form The second order partial differential equation A 2 u x 2 + 2B 2 u x y + C 2 u y 2 + F can be reduced in the following canonical form where A 2 u ξ 2 + 2B 2 u ξ η + C 2 u η 2 + F ( x, y, u x, u ) = 0 y ( x, y, u ξ, u ) = 0 η ( ) ξ 2 A = A + 2B ξ ( ) ξ ξ 2 x x y + C, y B = A ξ ( η ξ x x + B η x y + ξ η y x ( ) η 2 C = A + 2B η ( ) η η 2 x x y + C. y ) + C ξ y η y, 45

52 46 Chapter 10 Lecture Result Result We know that ( B 2 ξ A C = (B 2 η AC) x y ξ ) η 2, y x where A, B, C are not zero at the same time. Thus the transformation of independent variables does not change the type of equation. Since, ( ξ η x y ξ ) η 2 ξ η 2 = x x > 0. y x Therefore the sign of the discriminant does not change, either we are in (x, y) coordinates or in (ξ, η) coordinates Canonical Form for Elliptic Equation In elliptic equation the discriminant B 2 AC must be negative. If the coefficients A, B and C are variables and depending on x and y or they are constants then we can find out the reduced canonical form with the following characteristic equation dy dx = B η y η y B 2 AC. A Suppose Φ(x, y) = c is the solution of the characteristic curve which is not real, because B 2 AC < 0 B 2 AC is imaginary. Therefore Φ(x, y) can be written in this form Φ 1 (x, y) + iφ 2 (x, y) = c. Assume that Thus Φ 1 (x, y) = ξ(x, y), Φ 2 (x, y) = η(x, y). Φ(x, y) = ξ(x, y) + iη(x, y). Since Φ(x, y) is the solution of equation dy dx = B B 2 AC, A which basically comes by putting A = 0. In other words Φ(x, y) is the solution of ( ) Φ 2 A + 2B Φ ( ) Φ Φ 2 x x y + C = 0. y

53 10.2 Canonical Form for Elliptic Equation 47 Substituting the values of Φ = ξ+iη in above equation and after a small calculations, we obtain A = C and B = 0. Substituting the values of A, B, and C into A 2 u ξ 2 + 2B 2 u ξ η + C 2 u η 2 + F ( x, y, u ξ, u η ) = 0 which reduces to 2 u ξ u u = Ψ(ξ, η, u, η2 ξ, u η ) Example Example Example Reduce the following partial differential equations to a canonical form. 2 u x 2 + x2 2 u y 2 = 0 Example Reduce the following partial differential equations to a canonical form. (1 + x 2 ) 2 u x 2 + (1 + y2 ) 2 u y 2 + x u x + y u y = 0

54 48 Chapter 10 Lecture 10

55 11 Lecture 11 Review Reduction of Second Order Partial Differential Equations in Canonical Form Result Canonical Form for Elliptic Equation Examples 11.1 Reduction of Second Order Partial Differential Equations in Canonical Form The second order partial differential equation A 2 u x 2 + 2B 2 u x y + C 2 u y 2 + F can be reduced in the following canonical form where A 2 u ξ 2 + 2B 2 u ξ η + C 2 u η 2 + F ( x, y, u x, u ) = 0 y ( x, y, u ξ, u ) = 0 η ( ) ξ 2 A = A + 2B ξ ( ) ξ ξ 2 x x y + C, y B = A ξ ( η ξ x x + B η x y + ξ η y x ( ) η 2 C = A + 2B η ( ) η η 2 x x y + C. y ) + C ξ y η y, 49

56 50 Chapter 11 Lecture Result Result We know that ( B 2 ξ A C = (B 2 η AC) x y ξ ) η 2, y x where A, B, C are not zero at the same time. Thus the transformation of independent variables does not change the type of equation. Since, ( ξ η x y ξ ) η 2 = y x ξ x η y η x η y 2 > 0. Therefore the sign of the discriminant does not change, either we are in (x, y) coordinates or in (ξ, η) coordinates Mixed Type of Partial Differential Equation Mixed Type Partial Differential Equations Definition If a partial differential equation has coefficients that are not constant, it is possible that it will not belong Elliptic, Hyperbolic or Parabolic type but rather be of mixed type Example Example A simple but important example is the EulerTricomi equation 2 u x 2 + u x 2 = 0, x 0, y2 which is called elliptic-hyperbolic because it is elliptic in the region x < 0 and hyperbolic in the region x > 0. Problem Sheet Exercises

57 11.2 Mixed Type of Partial Differential Equation 51 Classify the following equation and reduce them to canonical form 1) y 2 z x 2 + (x + y) 2 z x y + z x 2 y 2 = 0 2) y 2 2 z x 2 2xy 2 z x y + x2 2 z y 2 y2 z x x x2 z y y = 0, 3) 2 z x 2 + x2 2 z y 2 1 z x x = 0 4) 2 z x 2 x2 2 z y 2 = 0 5) 2 z x z x y + 2 z y 2 = 0, 6) x 2 2 z x 2 + 2xy 2 z x y + y2 2 z y 2 = 4x2 7) 2 u x 2 2 sin x 2 u x y cos2 x 2 u cos x u y2 y = 0 Exercises Find the region in the xy-plane in which the following equations are hyperbolic and elliptic in nature 1) (1 x 2 ) 2 u x 2 2 u y 2 = 0 2) 2 u x 2 + xy 2 u y 2 = 0.