Module 2: First-Order Partial Differential Equations

Similar documents
z x = f x (x, y, a, b), z y = f y (x, y, a, b). F(x, y, z, z x, z y ) = 0. This is a PDE for the unknown function of two independent variables.

Compatible Systems and Charpit s Method

Department of mathematics MA201 Mathematics III

The first order quasi-linear PDEs

MA 201: Partial Differential Equations Lecture - 2

1 The Existence and Uniqueness Theorem for First-Order Differential Equations

First order Partial Differential equations

Propagation of discontinuities in solutions of First Order Partial Differential Equations

( ) and then eliminate t to get y(x) or x(y). Not that the

MATH 19520/51 Class 5

3 Applications of partial differentiation

2.12: Derivatives of Exp/Log (cont d) and 2.15: Antiderivatives and Initial Value Problems

b) The system of ODE s d x = v(x) in U. (2) dt

α u x α1 1 xα2 2 x αn n

MA6351-TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATIONS. Question Bank. Department of Mathematics FATIMA MICHAEL COLLEGE OF ENGINEERING & TECHNOLOGY

1 First Order Ordinary Differential Equation

Lecture Notes on Partial Dierential Equations (PDE)/ MaSc 221+MaSc 225

First Order Partial Differential Equations: a simple approach for beginners

Chapter 3 Second Order Linear Equations

Lecture Notes in Mathematics. A First Course in Quasi-Linear Partial Differential Equations for Physical Sciences and Engineering Solution Manual

Chapter 2. First-Order Partial Differential Equations. Prof. D. C. Sanyal

Math512 PDE Homework 2

LECTURE 15 + C+F. = A 11 x 1x1 +2A 12 x 1x2 + A 22 x 2x2 + B 1 x 1 + B 2 x 2. xi y 2 = ~y 2 (x 1 ;x 2 ) x 2 = ~x 2 (y 1 ;y 2 1

MATH 220: PROBLEM SET 1, SOLUTIONS DUE FRIDAY, OCTOBER 2, 2015

LOWELL JOURNAL. MUST APOLOGIZE. such communication with the shore as Is m i Boimhle, noewwary and proper for the comfort

MA22S3 Summary Sheet: Ordinary Differential Equations

Partial Differential Equations

1 Basic Second-Order PDEs

12. Stresses and Strains

MATH 31BH Homework 5 Solutions

First order wave equations. Transport equation is conservation law with J = cu, u t + cu x = 0, < x <.

(to be used later). We have A =2,B = 0, and C =4,soB 2 <AC and A>0 so the critical point is a local minimum

and verify that it satisfies the differential equation:

MATH 32A: MIDTERM 2 REVIEW. sin 2 u du z(t) = sin 2 t + cos 2 2

THE METHOD OF CHARACTERISTICS FOR SYSTEMS AND APPLICATION TO FULLY NONLINEAR EQUATIONS

g(t) = f(x 1 (t),..., x n (t)).

LESSON 23: EXTREMA OF FUNCTIONS OF 2 VARIABLES OCTOBER 25, 2017

The Maximum and Minimum Principle

Lecture two. January 17, 2019

Math 147, Homework 1 Solutions Due: April 10, 2012

DIFFERENTIAL EQUATIONS

1 Arithmetic calculations (calculator is not allowed)

NPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India

Introduction of Partial Differential Equations and Boundary Value Problems

7a3 2. (c) πa 3 (d) πa 3 (e) πa3

Mathematics 426 Robert Gross Homework 9 Answers

Lecture notes: Introduction to Partial Differential Equations

Solutions to Homework # 1 Math 381, Rice University, Fall (x y) y 2 = 0. Part (b). We make a convenient change of variables:

THEODORE VORONOV DIFFERENTIABLE MANIFOLDS. Fall Last updated: November 26, (Under construction.)

Definition of differential equations and their classification. Methods of solution of first-order differential equations

APPLIED MATHEMATICS. Part 1: Ordinary Differential Equations. Wu-ting Tsai

1MA6 Partial Differentiation and Multiple Integrals: I

Module Two: Differential Calculus(continued) synopsis of results and problems (student copy)

MATH 220: MIDTERM OCTOBER 29, 2015

Solving First Order PDEs

A. H. Hall, 33, 35 &37, Lendoi

How to solve quasi linear first order PDE. v(u, x) Du(x) = f(u, x) on U IR n, (1)

TAM3B DIFFERENTIAL EQUATIONS Unit : I to V

x + ye z2 + ze y2, y + xe z2 + ze x2, z and where T is the

Sébastien Chaumont a a Institut Élie Cartan, Université Henri Poincaré Nancy I, B. P. 239, Vandoeuvre-lès-Nancy Cedex, France. 1.

Second Order Elliptic PDE

Mathematical Analysis II, 2018/19 First semester

INTRODUCTION TO PDEs

Class Meeting # 1: Introduction to PDEs

Final: Solutions Math 118A, Fall 2013

Method of Lagrange Multipliers

Integration - Past Edexcel Exam Questions

ENGI 4430 PDEs - d Alembert Solutions Page 11.01

Lagrange Multipliers

Lecture 13 - Wednesday April 29th

Jim Lambers MAT 280 Summer Semester Practice Final Exam Solution. dy + xz dz = x(t)y(t) dt. t 3 (4t 3 ) + e t2 (2t) + t 7 (3t 2 ) dt

Exam 2 Review Solutions

Existence Theory: Green s Functions

Math 46, Applied Math (Spring 2008): Final

Green s Functions and Distributions

Linearization of Two Dimensional Complex-Linearizable Systems of Second Order Ordinary Differential Equations

Lecture Notes for MATH6106. March 25, 2010

Lecture No 1 Introduction to Diffusion equations The heat equat

Essential Ordinary Differential Equations

A First Course of Partial Differential Equations in Physical Sciences and Engineering

Lecture 9 Approximations of Laplace s Equation, Finite Element Method. Mathématiques appliquées (MATH0504-1) B. Dewals, C.

COMPLETE METRIC SPACES AND THE CONTRACTION MAPPING THEOREM

1 + f 2 x + f 2 y dy dx, where f(x, y) = 2 + 3x + 4y, is

Totally quasi-umbilic timelike surfaces in R 1,2

Comparative Statics. Autumn 2018

4 Introduction to First-Order Partial Differential

MATH H53 : Final exam

AM 205: lecture 14. Last time: Boundary value problems Today: Numerical solution of PDEs

MAT 419 Lecture Notes Transcribed by Eowyn Cenek 6/1/2012

Math 203A - Solution Set 1

Math 147, Homework 5 Solutions Due: May 15, 2012

Q SON,' (ESTABLISHED 1879L

Resolution of Singularities in Algebraic Varieties

The Derivative. Appendix B. B.1 The Derivative of f. Mappings from IR to IR

Section 5. Graphing Systems

UNIVERSITY of LIMERICK OLLSCOIL LUIMNIGH

DO NOT BEGIN THIS TEST UNTIL INSTRUCTED TO START

ORDINARY DIFFERENTIAL EQUATIONS

Definition 5.1. A vector field v on a manifold M is map M T M such that for all x M, v(x) T x M.

ALGEBRAIC GEOMETRY HOMEWORK 3

Transcription:

Module 2: First-Order Partial Differential Equations The mathematical formulations of many problems in science and engineering reduce to study of first-order PDEs. For instance, the study of first-order PDEs arise in gas flow problems, traffic flow problems, phenomenon of shock waves, the motion of wave fronts, Hamilton-Jacobi theory, nonlinear continum mechanics and quantum mechanics etc.. It is therefore essential to study the theory of first-order PDEs and the nature their solutions to analyze the related physical problems. In Module 2, we shall study first-order linear, quasi-linear and nonlinear PDEs and methods of solving these equations. An important method of characteristics is explained for these equations in which solving PDE reduces to solving an ODE system along a characteristics curve. Further, the Charpit s method and the Jacobi s method for nonlinear first-order PDEs are discussed. This module consists of seven lectures. Lecture 1 introduces some basic concepts of first-order PDEs such as formulation of PDEs, classification of first-order PDEs and Cauchy s problem for first-order PDEs. In Lecture 2, we study first-order linear PDEs and the parametric form of solution of first-order PDEs. In Lecture 3, we study a firstorder quasi-linear PDE and discuss the method of characteristics for a first-order quasilinear PDE. Lecture 4 is devoted to nonlinear first-order PDEs and Cauchy s method of characteristics for finding solutions of these equations. Lecture 5 is focused on the compatible system of equations and Charpit s method for solving nonlinear equations. In Lecture 6, we consider some special type of PDEs and method of obtaining their general integrals. Finally, the Jacobi s method for solving nonlinear PDEs is discussed in Lecture 7. 1

MODULE 2: FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 2 Lecture 1 First-Order Partial Differential Equations A first order PDE in two independent variables x, y and the dependent variable z can be written in the form For convenience, we set Equation (1) then takes the form f(x, y, z, x, ) = 0. (1) y p = x, q = y. f(x, y, z, p, q) = 0. (2) The equations of the type (2) arise in many applications in geometry and physics. For instance, consider the following geometrical problem. EXAMPLE 1. Find all functions z(x, y) such that the tangent plane to the graph z = z(x, y) at any arbitrary point (x 0, y 0, z(x 0, y 0 )) passes through the origin characterized by the PDE xz x + yz y z = 0. The equation of the tangent plane to the graph at (x 0, y 0, z(x 0, y 0 )) is z x (x 0, y 0 )(x x 0 ) + z y (x 0, y 0 )(y y 0 ) (z z(x 0, y 0 )) = 0. This plane passes through the origin (0, 0, 0) and hence, we must have z x (x 0, y 0 )x 0 z y (x 0, y 0 )y 0 + z(x 0, y 0 ) = 0. (3) For the equation (3) to hold for all (x 0, y 0 ) in the domain of z, z must satisfy which is a first-order PDE. xz x + yz y z = 0, EXAMPLE 2. The set of all spheres with centers on the z-axis is characterized by the first-order PDE yp xq = 0. The equation x 2 + y 2 + (z c) 2 = r 2, (4) where r and c are arbitrary constants, represents the set of all spheres whose centers lie on the z-axis. Differentiating (4) with respect to x, we obtain ( 2 x + (z c) ) = 2 (x + (z c)p) = 0. (5) x

MODULE 2: FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 3 Differentiate (4) with respect to y to have y + (z c)q = 0. (6) Eliminating the arbitrary constant c from (5) and (6), we obtain the first-order PDE yp xq = 0. (7) Equation (4) in some sense characterized the first-order PDE (7). EXAMPLE 3. Consider all surfaces described by an equation of the form z = f(x 2 + y 2 ), (8) where f is an arbitrary function, described by the first-order PDE. Writing u = x 2 + y 2 and differentiating (8) with respect to x and y, it follows that p = 2xf (u); q = 2yf (u), where f (u) = df du. Eliminating f (u) from the above two equations, we obtain the same first-order PDE as in (7). REMARK 4. The function z described by each of the equations (4) and (8), in some sense, a solution to the PDE (7). Observe that, in Example 2, PDE (7) is formulated by eliminating arbitrary constants from (4) whereas in Example 3, PDE (7) is formed by eliminating an arbitrary function. 1 Formation of first-order PDEs The applications of conservation principles often yield a first-order PDEs. We have seen in the previous two examples that a first-order PDE can be formed either by eliminating arbitrary constants or an arbitrary function involved. Below, we now generalize the arguments of Example 2 and Example 3 to show that how a first-order PDE can be formed. Method I (Eliminating arbitrary constants): Consider two parameters family of surfaces described by the equation F (x, y, z, a, b) = 0, (9) where a and b are arbitrary constants. Equation (9) may be thought of as a generalization of the relation (4).

MODULE 2: FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 4 Differentiating (9) with respect to x and y, we obtain F x + p F = 0 (10) F y + q F = 0. (11) Eliminate the constants a, b from equations (9), (10) and (11) to obtain a first-order PDE of the form f(x, y, z, p, q) = 0. (12) This shows that a family of surfaces described by the relation (9) gives rise to a first-order PDE (12). Method II (Eliminating arbitrary function): Now consider the generalization of Example 3. Let u(x, y, z) = c 1 and v(x, y, z) = c 2 be two known functions of x, y and z satisfying a relation of the form F (u, v) = 0, (13) where F is an arbitrary function of u and v. Differentiating (13) with respect to x and y lead to the equations F u (u x + u z p) + F v (v x + v z p) = 0 F u (u y + u z q) + F v (v y + v z q) = 0. Eliminating F u and F v from the above two equations, we obtain (u, v) (u, v) (u, v) p + q = (y, z) (z, x) (x, y), (14) which is a first-order PDE of the form f(x, y, z, p, q) = 0. Here, (u,v) (x,y) = u xv y u y v x. 2 Classification of first-order PDEs We classify the equation (1) depending on the special forms of the function f. If (1) is of the form a(x, y) + b(x, y) + c(x, y)z = d(x, y) x y then it is called linear first-order PDE. Note that the function f is linear in x, y and z with all coefficients depending on the independent variables x and y only. If (1) has the form a(x, y) + b(x, y) x y = c(x, y, z)

MODULE 2: FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 5 then it is called semilinear because it is linear in the leading (highest-order) terms x and y. However, it need not be linear in z. Note that the coefficients of x functions of the independent variables only. If (1) has the form a(x, y, z) + b(x, y, z) x y = c(x, y, z) and y are then it is called quasi-linear PDE. Here the function f is linear in the derivatives x with the coefficients a, b and c depending on the independent variables x and y as and y well as on the unknown z. Note that linear and semilinear equations are special cases of quasi-linear equations. Any equation that does not fit into one of these forms is called nonlinear. EXAMPLE 5. 1. xz x + yz y = z (linear) 2. xz x + yz y = z 2 (semilinear) 3. z x + (x + y)z y = xy (linear) 4. zz x + z y = 0 (quasilinear) 5. xz 2 x + yz 2 y = 2 (nonlinear) 3 Cauchy s problem or IVP for first-order PDEs Recall the initial value problem for a first-order ODE which ask for a solution of the equation that takes a given value at a given point of R. The IVP for first-order PDE ask for a solution of (2) which has given values on a curve in R 2. The conditions to be satisfied in the case of IVP for first-order PDE are formulated in the classic problem of Cauchy which may be stated as follows: Let C be a given curve in R 2 described parametrically by the equations x = x 0 (s), y = y 0 (s); s I, (15) where x 0 (s), y 0 (s) are in C 1 (I). Let z 0 (s) be a given function in C 1 (I). The IVP or Cauchy s problem for first-order PDE f(x, y, z, p, q) = 0 (16) is to find a function u = u(x, y) with the following properties:

MODULE 2: FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 6 u(x, y) and its partial derivatives with respect to x and y are continuous in a region Ω of R 2 containing the curve C. u = u(x, y) is a solution of (16) in Ω, i.e., f(x, y, u(x, y), u x (x, y), u y (x, y)) = 0 in Ω. On the curve C u(x 0 (s), y 0 (s)) = z 0 (s), s I. (17) The curve C is called the initial curve of the problem and the function z 0 (s) is called the initial data. Equation (17) is called the initial condition of the problem. NOTE: Geometrically, Cauchy s problem may be interpreted as follows: To find a solution surface u = u(x, y) of (16) which passes through the curve C whose parametric equations are x = x 0 (s), y = y 0 (s) z = z 0 (s). (18) Further, at every point of which the direction (p, q, 1) of the normal is such that f(x, y, z, p, q) = 0. The proof of existence of a solution of (16) passing through a curve with equations (18) requires some more assumptions on the function f and the nature of the curve C. We now state the classic theorem due to Kowalewski in the following theorem (cf. [10]). THEOREM 6. (Kowalewski) If g(y) and all its derivatives are continuous for y y 0 < δ, if x 0 is a given number and z 0 = g(y 0 ), q 0 = g (y 0 ), and if f(x, y, z, q) and all its partial derivatives are continuous in a region S defined by x x 0 < δ, y y 0 < δ, q q 0 < δ, then there exists a unique function ϕ(x, y) such that: (a) ϕ(x, y) and all its partial derivatives are continuous in a region Ω : x x 0 < δ 1, y y 0 < δ 2 ; (b) For all (x, y) in Ω, z = ϕ(x, y) is a solution of the equation = f(x, y, z, x y ) (c) For all values of y in the interval y y 0 < δ 1, ϕ(x 0, y) = g(y). We conclude this lecture by introducing different kinds of solutions of first-order PDE.

MODULE 2: FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 7 DEFINITION 7. (A complete solution or a complete integral) Any relation of the form F (x, y, z, a, b) = 0 (19) which contains two arbitrary constants a and b and is a solution of a first-order PDE is called a complete solution or a complete integral of that first-order PDE. DEFINITION 8. (A general solution or a general integral) Any relation of the form F (u, v) = 0 involving an arbitrary function F connecting two known functions u(x, y, z) and v(x, y, z) and providing a solution of a first-order PDE is called a general solution or a general integral of that first-order PDE. It is possible to derive a general integral of the PDE once a complete integral is known. With b = ϕ(a), if we take any one-parameter subsystem f(x, y, z, a, ϕ(a)) = 0 of the system (19) and form its envelope, we obtain a solution of equation (16). When ϕ(a) is arbitrary, the solution obtained is called the general integral of (16) corresponding to the complete integral (19). When a definite ϕ(a) is used, we obtain a particular solution. DEFINITION 9. (A singular integral) The envelope of the two-parameter system (19) is also a solution of the equation (16). It is called the singular integral or singular solution of the equation. NOTE: The general solution of an equation of type (1) can be obtained by solving systems of ODEs. This is not true for higher-order equations or for systems of first-order equations. Practice Problems 1. Classify whether the following PDE is linear, quasi-linear or nonlinear: (a) zz x 2xyz y = 0; (b) z 2 x + zz y = 2; (c) z x + 2z y = 5z; (d) xz x + yz y = z 2. 2. Eliminate the arbitrary constants a and b from the following equations to form the PDE: (a) ax 2 + by 2 + z 2 = 1; (b) z = (x 2 + a)(y 2 + b).

MODULE 2: FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 8 3. Show that z = f(xy), where f is an arbitrary differentiable function satisfies xz x yz y = 0, and hence, verify that the functions sin(xy), cos(xy), log(xy) and e xy are solutions. 4. Eliminate the arbitrary function f from the following and form the PDE: (a) z = x + y + f(xy); (b) z = f( xy z ).