Analytic Solutions of Partial Differential Equations

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1 ii Analtic Solutions of Partial Differential Equations 15 credits Taught Semester 1, Year running 3/4 MATH3414 School of Mathematics, Universit of Lee Pre-requisites MATH36 or MATH4 or equivalent. Co-requisites None. Objectives: To provide an understanding of, and metho of solution for, the most important tpes of partial differential equations that arise in Mathematical Phsics. On completion of this module, students should be able to: a use the method of characteristics to solve first-order hperbolic equations; b classif a second order PDE as elliptic, parabolic or hperbolic; c use Green s functions to solve elliptic equations; d have a basic understanding of diffusion; e obtain a priori boun for reaction-diffusion equations. Sllabus: The majorit of phsical phenomena can be described b partial differential equations e.g. the Navier-Stokes equation of fluid dnamics, Mawell s equations of electromagnetism. This module considers the properties of, and analtical metho of solution for some of the most common first and second order PDEs of Mathematical Phsics. In particular, we shall look in detail at elliptic equations Laplace?s equation, describing stead-state phenomena and the diffusion / heat conduction equation describing the slow spread of concentration or heat. The topics covered are: First order PDEs. Semilinear and quasilinear PDEs; method of characteristics. Characteristics crossing. Second order PDEs. Classification and standard forms. Elliptic equations: weak and strong minimum and maimum principles; Green s functions. Parabolic equations: eemplified b solutions of the diffusion equation. Boun on solutions of reaction-diffusion equations. Form of teaching Lectures: 6 hours. 7 eamples classes. Form of assessment One 3 hour eamination at end of semester 1%. Details: Ev Kersalé Office: 9.e Phone: kersale@maths.lee.ac.uk WWW: kersale/ Schedule: three lectures ever week, for eleven weeks from 7/9 to 1/1. Tuesda 13: 14: RSLT 3 Wednesda 1: 11: RSLT 4 Frida 11: 1: RSLT 6 Pre-requisite: elementar differential calculus and several variables calculus e.g. partial differentiation with change of variables, parametric curves, integration, elementar algebra e.g. partial fractions, linear eigenvalue problems, ordinar differential equations e.g. change of variable, integrating factor, and vector calculus e.g. vector identities, Green s theorem. Outline of course: Book list: Introduction: definitions eamples First order PDEs: linear & semilinear characteristics quasilinear nonlinear sstem of equations Second order linear PDEs: classification elliptic parabolic P. Prasad & R. Ravindran, Partial Differential Equations, Wile Eastern, W. E. Williams, Partial Differential Equations, Oford Universit Press, 198. P. R. Garabedian, Partial Differential Equations, Wile, Thanks to Prof. D. W. Hughes, Prof. J. H. Merkin and Dr. R. Sturman for their lecture notes.

2 iv Classification Tpe Canonical form Characteristics Course Summar Definitions of different tpe of PDE linear, quasilinear, semilinear, nonlinear Eistence and uniqueness of solutions Solving PDEs analticall is generall based on finding a change of variable to transform the equation into something soluble or on finding an integral form of the solution. First order PDEs a + b = c. b ac > Hperbolic ξ η +... = b ac = Parabolic η +... = b ac < Elliptic α + u β +... = d d = b ± b ac a d d = b a, η = sa d d = b ± { b ac α = ξ + η a, β = iξ η Elliptic equations: Laplace equation. Maimum Principle. Solutions using Green s functions uses new variables and the Dirac δ-function to pick out the solution. Method of images. Parabolic equations: heat conduction, diffusion equation. Derive a fundamental solution in integral form or make use of the similarit properties of the equation to find the solution in terms of the diffusion variable η = t. First and Second Maimum Principles and Comparison Theorem give boun on the solution, and can then construct invariant sets. Linear equations: change coordinate using η,, defined b the characteristic equation d d = b a, and ξ, independent usuall ξ = to transform the PDE into an ODE. Quasilinear equations: change coordinate using the solutions of d = a, d du = b and = c to get an implicit form of the solution φ,, u = F ψ,, u. Nonlinear waves: region of solution. Sstem of linear equations: linear algebra to decouple equations. Second order PDEs a u + b u + u c + d + e + fu = g. iii

3 ii CONTENTS Contents 1 Introduction Motivation Reminder Definitions Eamples Wave Equations Diffusion or Heat Conduction Equations Laplace s Equation Other Common Second Order Linear PDEs Nonlinear PDEs Sstem of PDEs Eistence and Uniqueness First Order Equations 9.1 Linear and Semilinear Equations Method of Characteristic Equivalent set of ODEs Characteristic Curves Quasilinear Equations Interpretation of Quasilinear Equation General solution: Wave Equation Linear Waves Nonlinear Waves Weak Solution Sstems of Equations Linear and Semilinear Equations Quasilinear Equations Elliptic Equations Definitions Properties of Laplace s and Poisson s Equations Mean alue Propert Maimum-Minimum Principle Solving Poisson Equation Using Green s Functions Definition of Green s Functions Green s function for Laplace Operator Free Space Green s Function Method of Images Etensions of Theor: Parabolic Equations Definitions and Properties Well-Posed Cauch Problem Initial alue Problem Well-Posed Initial-Boundar alue Problem Time Irreversibilit of the Heat Equation Uniqueness of Solution for Cauch Problem: Uniqueness of Solution for Initial-Boundar alue Problem: Fundamental Solution of the Heat Equation Integral Form of the General Solution Properties of the Fundamental Solution Behaviour at large t Similarit Solution Infinite Region Semi-Infinite Region Maimum Principles and Comparison Theorems First Maimum Principle A Integral of e in R 81 3 Second Order Linear and Semilinear Equations in Two ariables Classification and Standard Form Reduction Etensions of the Theor Linear second order equations in n variables The Cauch Problem i

4 1.3 Definitions The usual differentiation identities appl to the partial differentiations sum, product, quotient, chain rules, etc. Chapter 1 Introduction Contents 1.1 Motivation Reminder Definitions Eamples Eistence and Uniqueness Motivation Wh do we stud partial differential equations PDEs and in particular analtic solutions? We are interested in PDEs because most of mathematical phsics is described b such equations. For eample, flui dnamics and more generall continuous media dnamics, electromagnetic theor, quantum mechanics, traffic flow. Tpicall, a given PDE will onl be accessible to numerical solution with one obvious eception eam questions! and analtic solutions in a practical or research scenario are often impossible. However, it is vital to understand the general theor in order to conduct a sensible investigation. For eample, we ma need to understand what tpe of PDE we have to ensure the numerical solution is valid. Indeed, certain tpes of equations need appropriate boundar conditions; without a knowledge of the general theor it is possible that the problem ma be ill-posed of that the method is solution is erroneous. 1. Reminder Partial derivatives: The differential or differential form of a function f of n independent variables, 1,,..., n, is a linear combination of the basis form d 1, d,..., d n n f df = d i = f d 1 + f d f d n, i 1 n i=1 where the partial derivatives are defined b f f 1,,..., i + h,..., n f 1,,..., i,..., n = lim. i h h Notations: I shall use interchangeabl the notations f i i f f i, f i j i j f f i j, for the first order and second order partial derivatives respectivel. We shall also use interchangeabl the notations u u u, for vectors. ector differential operators: in three dimensional Cartesian coordinate sstem i, j, k we consider f,, z : R 3 R and [u,, z, u,, z, u z,, z] : R 3 R 3. Gradient: f = f i + f j + z f k. Divergence: div u u = u + u + z uz. Curl: u = z u u z i + z u u z j + u u k. Laplacian: f f = f + f + z f. Laplacian of a vector: u u = u i + u j + u z k. Note that these operators are different in other sstems of coordinate clindrical or spherical, sa 1.3 Definitions A partial differential equation PDE is an equation for some quantit u dependent variable which depen on the independent variables 1,, 3,..., n, n, and involves derivatives of u with respect to at least some of the independent variables. Note: F 1,..., n, 1 u,..., n u, 1 u, 1 u,..., n 1... n u =. 1. In applications i are often space variables e.g.,, z and a solution ma be required in some region Ω of space. In this case there will be some conditions to be satisfied on the boundar Ω; these are called boundar conditions BCs.. Also in applications, one of the independent variables can be time t sa, then there will be some initial conditions ICs to be satisfied i.e., u is given at t = everwhere in Ω 3. Again in applications, sstems of PDEs can arise involving the dependent variables u 1, u, u 3,..., u m, m 1 with some at least of the equations involving more than one u i. 1

5 Chapter 1 Introduction Eamples The order of the PDE is the order of the highest partial differential coefficient in the equation. As with ordinar differential equations ODEs it is important to be able to distinguish between linear and nonlinear equations. A linear equation is one in which the equation and an boundar or initial conditions do not include an product of the dependent variables or their derivatives; an equation that is not linear is a nonlinear equation. t + c =, + u = Φ,, first order linear PDE simplest wave equation, second order linear PDE Poisson. A nonlinear equation is semilinear if the coefficients of the highest derivative are functions of the independent variables onl = u3, u + + u + u + u = u4. A nonlinear PDE of order m is quasilinear if it is linear in the derivatives of order m with coefficients depending onl on,,... and derivatives of order < m. [ 1 + ] [ ] =. Principle of superposition: A linear equation has the useful propert that if u 1 and u both satisf the equation then so does αu 1 + βu for an α, β R. This is often used in constructing solutions to linear equations for eample, so as to satisf boundar or initial conditions; c.f. Fourier series metho. This is not true for nonlinear equations, which helps to make this sort of equations more interesting, but much more difficult to deal with. 1.4 Eamples Wave Equations Waves on a string, sound waves, waves on stretch membranes, electromagnetic waves, etc. or more generall where c is a constant wave speed. = 1 c t, 1 c t = u 1.4. Diffusion or Heat Conduction Equations or more generall or even t = u κ, t = κ u, t = κ u where κ is a constant diffusion coefficient or thermometric conductivit. Both those equations wave and diffusion are linear equations and involve time t. The require some initial conditions and possibl some boundar conditions for their solution Laplace s Equation Another eample of a second order linear equation is the following. or more generall + u =, u =. This equation usuall describes stead processes and is solved subject to some boundar conditions. One aspect that we shall consider is: wh do the similar looking equations describes essentiall different phsical processes? What is there about the equations that make this the cases? Other Common Second Order Linear PDEs Poisson s equation is just the Lapace s equation homogeneous with a known source term e.g. electric potential in the presence of a densit of charge: u = Φ. The Helmholtz equation ma be regarded as a stationar wave equation: u + k u =. The Schrödinger equation is the fundamental equation of phsics for describing quantum mechanical behavior; Schrödinger wave equation is a PDE that describes how the wavefunction of a phsical sstem evolves over time: u + u = i t.

6 Chapter 1 Introduction Nonlinear PDEs An eample of a nonlinear equation is the equation for the propagation of reaction-diffusion waves: t = u + u1 u nd order, or for nonlinear wave propagation: The equation + u + c t = ; is an eample of quasilinear equation, and is an eample of semilinear equation Sstem of PDEs 1st order. u + + u = u = u3 Mawell equations constitute a sstem of linear PDEs: E = ρ ε, B =, B = µj + 1 c E t, E = B t. In empt space free of charges and currents this sstem can be rearranged to give the equations of propagation of the electromagnetic field, E t = c E, B t = c B. Incompressible magnetohdrodnamic MHD equations combine Navier-Stokes equation including the Lorentz force, the induction equation as well as the solenoidal constraints, U t + U U = Π + B B + ν U + F, B t = U B + η B, U =, B =. Both sstems involve space and time; the require some initial and boundar conditions for their solution Eistence and Uniqueness 1.5 Eistence and Uniqueness Before attempting to solve a problem involving a PDE we would like to know if a solution eists, and, if it eists, if the solution is unique. Also, in problem involving time, whether a solution eists t > global eistence or onl up to a given value of t i.e. onl for < t < t finite time blow-up, shock formation. As well as the equation there could be certain boundar and initial conditions. We would also like to know whether the solution of the problem depen continuousl of the prescribed data i.e. small changes in boundar or initial conditions produce onl small changes in the solution. Illustration from ODEs: Solution: u = e t eists for t < Solution: u = 1/1 t eists for t < 1 du = u, u = 1. dt du dt = u, u = 1. du dt = u, u =, has two solutions: u and u = t /4 non uniqueness. We sa that the PDE with boundar or initial condition is well-formed or well-posed if its solution eists globall, is unique and depen continuousl on the assigned data. If an of these three properties eistence, uniqueness and stabilit is not satisfied, the problem PDE, BCs and ICs is said to be ill-posed. Usuall problems involving linear sstems are well-formed but this ma not be alwas the case for nonlinear sstems bifurcation of solutions, etc. Eample: A simple eample of showing uniqueness is provided b: u = F in Ω Poisson s equation. with u = on Ω, the boundar of Ω, and F is some given function of. Suppose u 1 and u two solutions satisfing the equation and the boundar conditions. Then consider w = u 1 u ; w = in Ω and w = on Ω. Now the divergence theorem gives w w n ds = w w d, Ω Ω = w w + w d where n is a unit normal outwar from Ω. Ω

7 Chapter 1 Introduction Eistence and Uniqueness Ω w d = Ω w w ds =. Now the integrand w is non-negative in Ω and hence for the equalit to hold we must have w ; i.e. w = constant in Ω. Since w = on Ω and the solution is smooth, we must have w in Ω; i.e. u 1 = u. The same proof works if / is given on Ω or for mied conditions.

8 1.1 Linear and Semilinear Equations Chapter First Order Equations Contents.1 Linear and Semilinear Equations Quasilinear Equations Wave Equation Sstems of Equations Linear and Semilinear Equations.1.1 Method of Characteristic We consider linear first order partial differential equation in two independent variables: a, + b, + c, u = f,,.1 where a, b, c and f are continuous in some region of the plane and we assume that a, and b, are not zero for the same,. In fact, we could consider semilinear first order equation where the nonlinearit is present onl in the right-hand side such as a, + b, = κ,, u,. instead of a linear equation as the theor of the former does not require an special treatment as compared to that of the latter. The ke to the solution of the equation.1 is to find a change of variables or a change of coordinates ξ ξ,, η η, which transforms.1 into the simpler equation where wξ, η = uξ, η, ξ, η. w + hξ, ηw = F ξ, η.3 ξ We shall define this transformation so that it is one-to-one, at least for all, in some set D of points in the - plane. Then, on D we can in theor solve for and as functions of ξ, η. To ensure that we can do this, we require that the Jacobian of the transformation does not vanish in D: ξ ξ J = η η = ξ η ξ η {, } for, in D. We begin looking for a suitable transformation b computing derivatives via the chain rule = w ξ ξ + w η and η = w ξ ξ + w η η. We substitute these into equation.1 to obtain w ξ a ξ + w η w ξ + b η ξ + w η η + cw = f. We can rearrange this as a ξ + b ξ w ξ + a η w + b η + cw = f..4 η This is close to the form of equation.1 if we can choose η η, so that a η + b η = for, in D. Provided that η/ we can epress this required propert of η as η η = b a. Suppose we can define a new variable or coordinate η which satisfies this constraint. What is the equation describing the curves of constant η? Putting η η, = k k an arbitrar constant, then dη = η η d + d = implies that d/d = η/ η = b/a. So, the equation η, = k defines solutions of the ODE d b, = d a,..5 Equation.5 is called the characteristic equation of the linear equation.1. Its solution can be written in the form F,, η = where η is the constant of integration and defines a famil of curves in the plane called characteristics or characteristic curves of.1. More on characteristics later. Characteristics represent curves along which the independent variable η of the new coordinate sstem ξ, η is constant. So, we have made the coefficient of w/ η vanish in the transformed equation.4, b choosing η η,, with η, = k an equation defining the solution of the characteristic equation.5. We can now choose ξ arbitraril or at least to suit our convenience, providing we still have J. An obvious choice is ξ ξ, =. 9

9 Chapter First Order Equations Linear and Semilinear Equations Then 1 J = η η = η, and we have alread assumed this on-zero. Now we see from equation.4 that this change of variables, transforms equation.1 to ξ =, η η,, α, w + c, w = f,., ξ where α = a ξ/ + b ξ/. To complete the transformation to the form of equation.3, we first write α,, c, and f, in terms of ξ and η to obtain Aξ, η w + Cξ, ηw = ρξ, η. ξ Finall, restricting the variables to a set in which Aξ, η we have which is in the form of.3 with hξ, η = Cξ, η Aξ, η w ξ + C A w = ρ A, and F ξ, η = ρξ, η Aξ, η. The characteristic method applies to first order semilinear equation. as well as linear equation.1; similar change of variables and basic algebra transform equation. to w ξ = K A, where the nonlinear term Kξ, η, w = κ,, u and restricting again the variables to a set in which Aξ, η = α,. Notation: It is ver convenient to use the function u in places where rigorousl the function w should be used. E.g., the equation here above can identicall be written as / ξ = K/A. Eample: Consider the linear first order equation + + u = 1. This is equation.1 with a, =, b, =, c, = and f, = 1. The characteristic equation is d d = b a =. Solve this b separation of variables 1 1 d = d ln + 1 = k, for >, and. This is an integral of the characteristic equation describing curves of constant η and so we choose η η, = ln + 1. Graphs of ln +1/ are the characteristics of this PDE. Choosing ξ = we have the Jacobian Since ξ =, J = η = 1 as required. η = ln + 1 ξ = eη 1/ξ. Now we appl the transformation ξ =, η = ln + 1/ with wξ, η = u, and we have = w ξ ξ + w η η = w ξ + w η 1 = w ξ 1 w ξ η, = w ξ ξ + w η η = + w η 1 = 1 w e η 1/ξ η. Then the PDE becomes w ξ ξ 1 w ξ + e η 1/ξ 1 w η e η 1/ξ η + ξeη 1/ξ w = 1, which simplifies to ξ w ξ + ξeη 1/ξ w = 1 then to w ξ + 1 ξ eη 1/ξ w = 1 ξ. We have transformed the equation into the form of equation.3, for an region of ξ, η- space with ξ..1. Equivalent set of ODEs The point of this transformation is that we can solve equation.3. Think of w + hξ, ηw = F ξ, η ξ as a linear first order ordinar differential equation in ξ, with η carried along as a parameter. Thus we use an integrating factor method R e hξ,η dξ w R R ξ + hξ, η e hξ,η dξ w = F ξ, η e hξ,η dξ, hξ,η dξ er w = F ξ, η hξ,η dξ er. ξ

10 Chapter First Order Equations Linear and Semilinear Equations Now we integrate with respect to ξ. Since η is being carried as a parameter, the constant of integration ma depend on η hξ,η dξ er w = F ξ, η hξ,η dξ er dξ + gη in which g is an arbitrar differentiable function of one variable. Now the general solution of the transformed equation is wξ, η = e R hξ,η dξ F ξ, η hξ,η dξ er dξ + gη e R hξ,η dξ. We obtain the general form of the original equation b substituting back ξ, and η, to get u, = e α, [β, + gη, ]..6 A certain class of first order PDEs linear and semilinear PDEs can then be reduced to a set of ODEs. This makes use of the general philosoph that ODEs are easier to solve than PDEs. Eample: Consider the constant coefficient equation a + b + cu = where a, b, c R. Assume a, the characteristic equation is with general solution defined b the equation d/d = b/a b a = k, k constant. So the characteristics of the PDE are the straight line graphs of b a = k and we make the transformation with ξ =, η = b a. Using the substitution we find the equation transforms to The integrating factor method gives and integrating with respect to ξ gives w ξ + c a w =. e cξ/a w = ξ e cξ/a w = gη, where g is an differentiable function of one variable. Then and in terms of and we back transform w = gη e cξ/a u, = gb a e c/a. Eercise: erif the solution b substituting back into the PDE. Note: Consider the difference between general solution for linear ODEs and general solution for linear PDEs. For ODEs, the general solution of d + q = p d contains an arbitrar constant of integration. For different constants ou get different curves of solution in --plane. To pick out a unique solution ou use some initial condition sa = to specif the constant. For PDEs, if u is the general solution to equation.1, then z = u, defines a famil of integral surfaces in 3D-space, each surface corresponding to a choice of arbitrar function g in.6. We need some kind of information to pick out a unique solution; i.e., to chose the arbitrar function g..1.3 Characteristic Curves We investigate the significance of characteristics which, defined b the ODE d b, = d a,, represent a one parameter famil of curves whose tangent at each point is in the direction of the vector e = a, b. Note that the left-hand side of equation. is the derivation of u in the direction of the vector e, e u. Their parametric representation is = s, = s where s and s satisf the pair of ODEs d d = a,, = b,..7 The variation of u with respect = ξ along these characteristic curves is given b du d = + d d = + b a, κ,, u = from equation., a, such that, in term of the curvilinear coordinate s, the variation of u along the curves becomes du = du d = κ,, u. d The one parameter famil of characteristic curves is parameterised b η each value of η represents one unique characteristic. The solution of equation. reduces to the solution of the famil of ODEs du = κ,, u or similarl du d = du κ,, u = dξ a, along each characteristics i.e. for each value of η. Characteristic equations.7 have to be solved together with equation.8, called the compatibilit equation, to find a solution to semilinear equation...8

11 Chapter First Order Equations Linear and Semilinear Equations Cauch Problem: Consider a curve Γ in, -plane whose parametric form is = σ, = σ. The Cauch problem is to determine a solution of the equation F,, u, u, u = in a neighbourhood of Γ such that u takes prescribed values u σ called Cauch data on Γ. Γ u o 1 u o ξ=cst η=cst Eample 1: Consider u =. The characteristic equation is d d = 3 and the characteristics are the straight line graphs 3 = c. Hence we take η = 3 and ξ =. ξ=cst η=cst o, o Notes: 1. u can onl be found in the region between the characteristics drawn through the endpoint of Γ.. Characteristics are curves on which the values of u combined with the equation are not sufficient to determine the normal derivative of u. 3. A discontinuit in the initial data propagates onto the solution along the characteristics. These are curves across which the derivatives of u can jump while u itself remains continuous. Eistence & Uniqueness: Wh do some choices of Γ in, -space give a solution and other give no solution or an infinite number of solutions? It is due to the fact that the Cauch data initial conditions ma be prescribed on a curve Γ which is a characteristic of the PDE. To understand the definition of characteristics in the contet of eistence and uniqueness of solution, return to the general solution.6 of the linear PDE: u, = e α, [β, + gη, ]. Consider the Cauch data, u, prescribed along the curve Γ whose parametric form is = σ, = σ and suppose u σ, σ = qσ. If Γ is not a characteristic, the problem is well-posed and there is a unique function g which satisfies the condition qσ = e ασ,σ [β σ, σ + g σ, σ]. If on the other hand = σ, = σ is the parametrisation of a characteristic η, = k, sa, the relation between the initial conditions q and g becomes qσ = e ασ,σ [β σ, σ + G],.9 where G = gk is a constant; the problem is ill-posed. The functions α, and β, are determined b the PDE, so equation.9 places a constraint on the given data function q. If qσ is not of this form for an constant G, then there is no solution taking on these prescribed values on Γ. On the other hand, if qσ is of this form for some G, then there are infinitel man such solutions, because we can choose for g an differentiable function so that gk = G. ξ=cst We can see that an η and ξ cross onl once the are independent, i.e. J ; η and ξ have been properl chosen. This gives the solution u, = e 4 g3 where g is a differentiable function defined over the real line. Simpl specifing the solution at a given point as in ODEs does not uniquel determine g; we need to take a curve of initial conditions. Suppose we specif values of u, along a curve Γ in the plane. For eample, let s choose Γ as the -ais and gives values of u, at points on Γ, sa Then we need and putting t = 3, u, = sin. u, = e 4 g3 = sin i.e. g3 = sine 4, gt = sint/3 e 4t/3. This determines g and the solution satisfing the condition u, = sin on Γ is u, = sin /3 e 8/3. We have determined the unique solution of the PDE with u specified along the -ais. We do not have to choose an ais sa, along =, u, = u, = 4. From the general solution this requires, to give the unique solution satisfing u, = 4. u, = e 4 g = 4, so g = 4 e 4 u, = 3 4 e 8

12 Chapter First Order Equations Linear and Semilinear Equations However, not ever curve in the plane can be used to determine g. Suppose we choose Γ to be the line 3 = 1 and prescribe values of u along this line, sa Now we must choose g so that u, = u, 3 1/ =. e 4 g3 3 1 =. This requires g1 = e 4 for all. This is impossible and hence there is no solution taking the value at points, on this line. Last, we consider again Γ to be the line 3 = 1 but choose values of u along this line to be u, = u, 3 1/ = e 4. Now we must choose g so that e 4 g3 3 1 = e 4. This requires g1 = 1, condition satisfied b an infinite number of functions and hence there is an infinite number of solutions taking the values e 4 on the line 3 = 1. Depending on the initial conditions, the PDE has one unique solution, no solution at all or an infinite number or solutions. The difference is that the -ais and the line = are not the characteristics of the PDE while the line 3 = 1 is a characteristic. Eample : = u with u = on =, 1 Characteristics: d d = d = = c, constant. So, take η = and ξ =. Then the equation becomes w η + w w ξ η = w ξ w ξ w = w ξ ξ =. Finall the general solution is, w = ξ gη or equivalentl u, = g. When = with 1, u = ; so = g g = and the solution is u, =. Γ ξ=const η=const This figure presents the characteristic curves given b = constant. The red characteristics show the domain where the initial conditions permit us to determine the solution. Alternative approach to solving eample : = u with u = on =, 1 This method is not suitable for finding general solutions but it works for Cauch problems. The idea is to integrate directl the characteristic and compatibilit equations in curvilinear coordinates. See also alternative method for solving the characteristic equations for quasilinear equations hereafter. The solution of the characteristic equations d = and d = gives the parametric form of the characteristic curves, while the integration of the compatibilit equation du = u gives the solution us along these characteristics curves. The solution of the characteristic equations is = c 1 e s and = c e s, where the parametric form of the data curve Γ permits us to find the two constants of integration c 1 & c in terms of the curvilinear coordinate along Γ. The curve Γ is described b θ = θ and θ = θ with θ [, 1] and we consider the points on Γ to be the origin of the coordinate s along the characteristics i.e. s = on Γ. So, on Γ s = } } = θ = c 1 s, θ = θ es = θ = c s, θ = θ e s, θ [, 1]. For linear or semilinear problems we can solve the compatibilit equation independentl of the characteristic equations. This propert is not true for quasilinear equations. Along the characteristics u is determined b du = u u = c 3 e s. Now we can make use of the Cauch data to determine the constant of integration c 3, on Γ, at s =, u θ, θ u θ = θ = c 3. Then, we have the parametric forms of the characteristic curves and the solution s, θ = θ e s, s, θ = θ e s and us, θ = θ e s, in terms of two parameters, s the curvilinear coordinate along the characteristic curves and θ the curvilinear coordinate along the data curve Γ. From the two first ones we get s and θ in terms of and. = es s = ln Then, we substitute s and θ in us, θ to find u, = ep ln and = θ θ = θ. = =.

13 Chapter First Order Equations 19. Quasilinear Equations Consider the first order quasilinear PDE a,, u + b,, u = c,, u.1 where the functions a, b and c can involve u but not its derivatives...1 Interpretation of Quasilinear Equation We can represent the solutions u, b the integral surfaces of the PDE, z = u,, in,, z-space. Define the Monge direction b the vector a, b, c and recall that the normal to the integral surface is u, u, 1. Thus quasilinear equation.1 sas that the normal to the integral surface is perpendicular to the Monge direction; i.e. integral surfaces are surfaces that at each point are tangent to the Monge direction, a b c u u = a,, u + b,, u c,, u =. 1 With the field of Monge direction, with direction numbers a, b, c, we can associate the famil of Monge curves which at each point are tangent to that direction field. These are defined b d d dz a b c = c d b dz a dz c d = b d a d d a,, u = d b,, u = dz =, c,, u where dl = d, d, dz is an arbitrar infinitesimal vector parallel to the Monge direction. In the linear case, characteristics were curves in the, -plane see.1.3. For the quasilinear equation, we consider Monge curves in,, u-space defined b d = a,, u, d = b,, u, du = c,, u. Characteristic equations d{, }/ and compatibilit equation du/ are simultaneous first order ODEs in terms of a dumm variable s curvilinear coordinate along the characteristics; we cannot solve the characteristic equations and compatibilit equation independentl as it is for a semilinear equation. Note that, in cases where c, the solution remains constant on the characteristics. The rough idea in solving the PDE is thus to build up the integral surface from the Monge curves, obtained b solution of the ODEs. Note that we make the difference between Monge curve or direction in,, z-space and characteristic curve or direction, their projections in, -space.. Quasilinear Equations.. General solution: Suppose that the characteristic and compatibilit equations that we have defined have two independent first integrals function, f,, u, constant along the Monge curves φ,, u = c 1 and ψ,, u = c. Then the solution of equation.1 satisfies F φ, ψ = for some arbitrar function F equivalentl, φ = Gψ for some arbitrar G, where the form of F or G depen on the initial conditions. Proof: Since φ and ψ are first integrals of the equation, We have the chain rule and then from the characteristic equations And similarl for ψ φ,, u = φs, s, us, = φs = c 1. dφ = φ d + φ d + φ du =, a φ + b φ + c φ =. a ψ + b ψ + c ψ =. Solving for c gives φ ψ a ψ φ φ ψ + b ψ φ = φ or a J[u, ] = b J[, u] where J[ 1, ] = 1 ψ And similarl, solving for a, b J[, ] = c J[u, ]. Thus, we have J[u, ] = J[, ] b/c and J[, u] = J[u, ] a/b = J[, ] a/c. Now consider F φ, ψ = remember F φ,, u,, ψ,, u, and differentiate df = F F d + d = 1 φ ψ Then, the derivative with respect to is zero, F = F φ φ + φ + F ψ ψ + ψ =, as well as the derivative with respect to F = F φ φ + φ + F ψ ψ + ψ. =.

14 Chapter First Order Equations 1. Quasilinear Equations For a non-trivial solution of this we must have φ + φ ψ + ψ ψ + ψ φ + φ =, φ ψ ψ φ φ + ψ ψ φ = φ ψ ψ φ, J[, u] + J[u, ] = J[, ]. Then from the previous epressions for a, b, and c a + b = c, i.e., F φ, ψ = defines a solution of the original equation. Eample 1: + u + = in >, < <, with u = 1 + on = 1. We first look for the general solution of the PDE before appling the initial conditions. Combining the characteristic and compatibilit equations, d = + u,.11 d =,.1 du =.13 we seek two independent first integrals. Equations.11 and.13 give and equation.1 Now, consider d + u d + u = + u, 1 d = 1. = 1 d + u d + u, = + u + u So, + u/ = c 1 is constant. This defines a famil of solutions of the PDE; so, we can choose φ,, u = + u, =. such that φ = c 1 determines one particular famil of solutions. Also, equations.11 and.1 give d = u, and equation.13 Now, consider d = udu. d [ u ] = d [ ] d u, = d du u =. Then, u = c is constant and defines another famil of solutions of the PDE. So, we can take ψ,, u = u. The general solution is + u + u F, u = or u = G, for some arbitrar functions F or G. Now to get a particular solution, appl initial conditions u = 1 + when = = G + 1 G + 1 = 4. Substitute θ = + 1, i.e. = θ 1/, so Gθ = 1 θ. Hence, u = 1 + u = u. We can regard this as a quadratic equation for u: Then, u [ u ] + =, u u =. u ± = 1 ± = 1 ± + 1. Consider again the initial condition u = 1 + on = 1 Hence, u ±, = 1 = 1 ± = 1 ± take the positive root. u, = +.

15 Chapter First Order Equations 3 4. Quasilinear Equations Eample : using the same procedure solve u + + u = + u with u = + 1 on =. Characteristic equations d = u,.14 d = + u,.15 du = + u..16 Again, we seek to independent first integrals. On the one hand, equations.14 and.15 give Now, consider d + d = u + + u = +, = 1 du from equation.16. u 1 d + 1 d = 1 du u Hence, /u = c 1 is constant and φ,, u = u is a first integral of the PDE. On the other hand, d ln =. u d d du = u u = u + =, d + u =. Hence, + u = c is also a constant on the Monge curves and another first integral is given b ψ,, u = + u, so the general solution is = G + u. u Now, we make use of the initial conditions, u = + 1 on =, to determine G: set θ = + 1, i.e. = θ 1, then and finall the solution is 1 + = G + 1; Gθ = θ 1, θ u = + u 1. Rearrange to finish! + u Alternative approach: The characteristic equations are which we can solve to get Solving the characteristic equations. Illustration b an eample, + u = 1, with u = on + = 1. d =, We now parameterise the initial line in terms of θ: and appl the initial data at s =. Hence, d du = u and = 1, = 1 c 1 s,.17 = s + c s + c 3,.18 u = c + s, for constants c 1, c, c θ =, = 1 θ,.17 gives θ = 1 c 1 c 1 = 1 θ,.18 gives 1 θ = c 3 c 3 = 1 θ,.19 gives = c c =. Hence, we found the parametric form of the surface integral, Eliminate s and θ, then = θ 1 sθ, = s + 1 θ and u = s. = θ 1 sθ θ = 1 + s, = u u. Invariants, or first integrals, are from solution.17,.18 and.19, keeping arbitrar c = φ = u / and ψ = /1 + u. Alternative approach to eample 1: Characteristic equations + u + = in >, < <, with u = 1 + on = 1. d = + u,. d =,.1 du =..

16 Chapter First Order Equations Wave Equation Solve with respect to the dumm variable s;.1 gives, Summar: Solving the characteristic equations two approaches.. and. give and. give = c 1 e s, d + u = + u + u = c e s, d = c 1 e s + c e s, so, = c 3 e s + 1 c 1 + c e s and u = c 3 e s + 1 c c 1 e s. Now, at s =, = 1 and = θ, u = 1 + θ parameterising initial line Γ, c 1 = 1, c = 1 + θ and c 3 = 1. Hence, the parametric form of the surface integral is, Then eliminate θ and s: so Finall, = e s θ e s, = e s and u = e s + θ e s. = θ θ = 1 u = u = +, + 1, as before. To find invariants, return to solved characteristics equations and solve for constants in terms of, and u. We onl need two, so put for instance c 1 = 1 and so = e s. Then, Solve for c = c c and u = c c 1. c = + u, so φ = + u, and solve for c 3 c 3 = 1 u, so ψ = u. Observe ψ is different from last time, but this does not as we onl require two independent choices for φ and ψ. In fact we can show that our previous ψ is also constant, u = + u u, = φ ψ, = φ 1ψ which is also constant. 1. Manipulate the equations to get them in a directl integrable form, e.g. 1 d + u = 1 + u and find some combination of the variables which differentiates to zero first integral, e.g. d + u =.. Solve the equations with respect to the dumm variable s, and appl the initial data parameterised b θ at s =. Eliminate θ and s; find invariants b solving for constants..3 Wave Equation We consider the equation where c is some positive constant..3.1 Linear Waves + u + c = with u, = f, t If u is small i.e. u u, then the equation approimate to the linear wave equation t + c = with u, = f. The solution of the equation of characteristics, d/dt = c, gives the first integral of the PDE, η, t = ct, and then general solution u, t = g ct, where the function g is determined b the initial conditions. Appling u, = f we find that the linear wave equation has the solution u, t = f ct, which represents a wave unchanging shape propagating with constant wave speed c. t Γ η= η= ct=cst u,t Note that u is constant where ct =constant, i.e. on the characteristics. f c t= t=t 1 =ct 1 1

17 Chapter First Order Equations Wave Equation.3. Nonlinear Waves For the nonlinear equation, the characteristics are defined b dt = 1, + u + c t =, d du = c + u and =, which we can solve to give two independent first integrals φ = u and ψ = u + ct. So, u = f[ u + ct], according to initial conditions u, = f. This is similar to the previous result, but now the wave speed involves u. However, this form of the solution is not ver helpful; it is more instructive to consider the characteristic curves. The PDE is homogeneous, so the solution u is constant along the Monge curves this is not the case in general which can then be reduced to their projections in the, t-plane. B definition, ψ = c+ut is constant on the characteristics as well as u; differentiate ψ to find that the characteristics are described b These are straight lines, d dt = u + c. = fθ + ct + θ, epressed in terms of a parameter θ. If we make use of the parametric form of the data curve Γ: { = θ, t =, θ R} and solve directl the Cauch problem in terms of the coordinate s = t, we similarl find, u = fθ and = u + ct + θ. The slope of the characteristics, 1/c + u, varies from one line to another, and so, two curves can intersect. t t min Γ {= θ,t=} f θ< θ θ θ u=f & f +ct= and if this is positive it will be in the region of solution. At this point u will not be singlevalued and the solution breaks down. B letting θ θ 1 we can see that the characteristics intersect at t = 1 f θ, and the minimum time for which the solution becomes multi-valued is t min = 1 ma[ f θ], i.e. the solution is single valued i.e. is phsical onl for t < t min. Hence, when f θ < we can epect the solution to eist onl for a finite time. In phsical terms, the equation considered is purel advective; in real waves, such as shock waves in gases, when ver large gradients are formed then diffusive terms e.g. u become vitall important. f u,t t breaking multi valued To illustrate finite time solutions of the nonlinear wave equation, consider fθ = θ1 θ, θ 1, f θ = 1 θ. So, f θ < for 1/ < θ < 1 and we can epect the solution not to remain single-valued for all values of t. ma[ f θ] = 1 so t min = 1. Now, which we can epress as u = f u + ct, so u = [ u + ct] [1 + u + ct], ct 1 + ct, t u t t + ct u + ct + ct + c t =, Consider two crossing characteristics epressed in terms of θ 1 and θ, i.e. = fθ 1 + ct + θ 1, = fθ + ct + θ. These correspond to initial values given at = θ 1 and = θ. intersect at the time t = θ 1 θ fθ 1 fθ, These characteristics and solving for u we take the positive root from initial data Now, at t = 1, u = 1 t t ct 1 + t t 4t ct. u = c c, so the solution becomes singular as t 1 and 1 + c.

18 Chapter First Order Equations Wave Equation.3.3 Weak Solution When wave breaking occurs multi-valued solutions we must re-think the assumptions in our model. Consider again the nonlinear wave equation, + u + c t =, and put w, t = u, t + c; hence the PDE becomes the inviscid Burger s equation or equivalentl in a conservative form w t + w w =, w t + w =, where w / is the flu function. We now consider its integral form, [ w t + ] w d = d w, t d = dt 1 where > 1 are real. Then, d dt 1 w, t d = w 1, t 1 w, t. 1 w d Let us now rela the assumption regarding the differentiabilit of the our solution; suppose that w has a discontinuit in = st with 1 < st <. w,t ws,t + ws,t The velocit of the discontinuit of the shock velocit U = ṡ. If [ ] indicates the jump across the shock then this condition ma be written in the form [ ] w U [w] =. The shock velocit for Burger s equation is U = 1 w s + w s ws + ws = ws+ + ws. The problem then reduces to fitting shock discontinuities into the solution in such a wa that the jump condition is satisfied and multi-valued solution are avoided. A solution that satisfies the original equation in regions and which satisfies the integral form of the equation is called a weak solution or generalised solution. Eample: Consider the inviscid Burger s equation w t + w w =, with initial conditions 1 for θ, w = θ, t = = fθ = 1 θ for θ 1, for θ 1. As seen before, the characteristics are curves on which w = fθ as well as fθ t = θ are constant, where θ is the parameter of the parametric form of the curve of initial data, Γ. For all θ, 1, f θ = 1 is negative f = elsewhere, so we can epect that all the characteristics corresponding to these values of θ intersect at the same point; the solution of the inviscid Burger s equation becomes multi-valued at the time t min = 1/ ma[ f θ] = 1, θ, 1. Then, the position where the singularit develops at t = 1 is = fθ t + θ = 1 θ + θ = 1. 1 st Thus, splitting the interval [ 1, ] in two parts, we have t w 1, t w, t = d dt st 1 = ws, t ṡt + w, t d + d dt st 1 st w, t d, w t d ws+, t ṡt + st w t d, where ws t, t and ws + t, t are the values of w as s from below and above respectivel; ṡ = /dt. w=1 & t=cst w=1 & t= w,t 1 w= & =cst =1 w,t 1 t= weak solution shock wave t>1 U multi valued solution Now, take the limit 1 s t and s + t. Since w/ t is bounded, the two integrals tend to zero. We then have t= t=1 w s, t w s +, t = ṡ ws, t ws +, t.

19 Chapter First Order Equations Sstems of Equations As time increases, the slope of the solution, 1 for t, 1 w, t = for t 1, 1 t for 1, with t < 1, becomes steeper and steeper until it becomes vertical at t = 1; then the solution is multivalued. Nevertheless, we can define a generalised solution, valid for all positive time, b introducting a shock wave. Suppose shock at st = U t + α, with ws, t = 1 and ws +, t =. The jump condition gives the shock velocit, U = ws+ + ws = 1 ; furthermore, the shock starts at = 1, t = 1, so α = 1 1/ = 1/. Hence, the weak solution of the problem is, for t 1, { for < st, w, t = 1 for > st,.4 Sstems of Equations.4.1 Linear and Semilinear Equations These are equations of the form n j=1 a ij u j + b ij u j = c i, i = 1,,..., n where st = 1 t + 1. = u, for the unknowns u 1, u j,..., u n and when the coefficients a ij and b ij are functions onl of and. Though the c i could also involve u k. In matri notation Au + Bu = c, where E.g., a a 1n b b 1n A = a ij =....., B = b ij =....., a n1... a nn b n1... b nn c 1 u 1 c c =. and u = u.. c n u n u 1 u + 3u 1 u = +, u 1 + u 5u1 + u = +, can be epressed as [ or Au + Bu = c where [ 1 A = 1 1 If we multipl b A 1 = ] [ ] u 1 + u ], B = [ 1/3 /3 1/3 1/3 [ [ 3 ] 1 5 ] ] [ ] u 1 u [ ] + = +, [ ] + and c = +. A 1 A u + A 1 B u = A 1 c, we obtain u + Du = d, [ ] [ ] [ ] 1/3 / /3 1 where D = A 1 B = = and d = A 1/3 1/3 5 8/3 1 1 c. We now assume that the matri A is non-singular i.e., the inverse A 1 eists at least there is some region of the, -plane where it is non-singular. Hence, we need onl to consider sstems of the form u + Du = d. We also limit our attention to totall hperbolic sstems, i.e. sstems where the matri D has n distinct real eigenvalues or at least there is some region of the plane where this hol. D has the n distinct eigenvalues λ 1, λ,..., λ n where detλ i I D = i = 1,..., n, with λ i λ j i j and the n corresponding eigenvectors e 1, e,..., e n so that De i = λ i e i. The matri P = [e 1, e,..., e n ] diagonalises D via P 1 DP = Λ, λ λ... Λ = λ n λ n We now put u = P v, which is of the form where The sstem is now of the form then P v + P v + DP v + DP v = d, and P 1 P v + P 1 P v + P 1 DP v + P 1 DP v = P 1 d, v + Λv = q, q = P 1 d P 1 P v P 1 DP v. v i + λ i v i = q i i = 1,..., n, where q i can involve {v 1, v,..., v n } and with n characteristics given b d d = λ i. This is the canonical form of the equations.

20 Chapter First Order Equations Sstems of Equations Eample 1: Consider the linear sstem u u =, u + 9 u 1 =, Here, u + Du = with D = Eigenvalues: Eigenvectors: } with initial conditions u = [, 3] T on =. [ ] 4. 9 detd λi = λ 36 = λ = ±6. [ ] [ ] [ ] [ ] [ ] 6 4 = = for λ = 6, [ ] [ ] [ ] [ ] [ ] 6 4 = = for λ = 6. Then, [ ] P =, P 1 = 1 [ ] [ ] 3 and P 1 6 DP = [ ] [ ] 6 So we put u = v and v v 6 =, which has general solution i.e. Initial conditions give so, f = /6 and g = ; then Eample : v 1 = f6 and v = g6 +, u 1 = v 1 + v and u 1 = 3v 1 3v. = f6 + g6, 3 = 3f6 3g6, u 1 = 1 6, 3 u = 1 6. Reduce the linear sstem [ ] 4 u + u 4 = to canonical form in the region of the, -space where it is totall hperbolic. Eigenvalues: [ ] 4 λ det = λ {3, 3}. 4 λ The sstem is totall hperbolic everwhere epect where =. Eigenvalues: So, [ ] 1 P =, P 1 = [ ] 1 Then, with u = v we obtain 1.4. Quasilinear Equations λ 1 = 3 e 1 = [1, ] T, λ = 3 e = [, 1] T. v + [ ] 1 1 [ ] 3 v 3 =. and P 1 DP = [ ] 3. 3 We consider sstems of n equations, involving n functions u i, i = 1,..., n, of the form u + Du = d, where D as well as d ma now depend on u. We have alread shown how to reduce a more general sstem Au +Bu = c to that simpler form. Again, we limit our attention to totall hperbolic sstems; then Λ = P 1 DP D = P ΛP 1, using the same definition of P, P 1 and the diagonal matri Λ, as for the linear and semilinear cases. So, we can transform the sstem in its normal form as, P 1 u + ΛP 1 u = P 1 d, such that it can be written in component form as n j=1 Pij 1 uj + λ i uj = n j=1 P 1 ij d j, i = 1,..., n where λ i is the ith eigenvalue of the matri D and wherein the ith equation involves differentiation onl in a single direction the direction d/d = λ i. We define the ith characteristic, with curvilinear coordinate s i, as the curve in the, -plane along which d i = 1, d i = λ i or equivalentl Hence, the directional derivative parallel to the characteristic is d u j = i uj + λ i uj, d d = λ i. and the sstem in normal form reduces to n ODEs involving different components of u n j=1 P 1 ij d n u j = i j=1 P 1 ij d j i = 1,..., n.

21 Chapter First Order Equations Sstems of Equations Eample: Unstead, one-dimensional motion of an inviscid compressible adiabatic gas. Consider the equation of motion Euler equation and the continuit equation t + u = 1 P ρ, ρ t + ρ + u ρ =. If the entrop is the same everwhere in the motion then P ρ γ = constant, and the motion equation becomes t + u + c ρ ρ =, where c = dp/dρ = γp/ρ is the sound speed. We have then a sstem of two first order quasilinear PDEs; we can write these as with w = [ ] u ρ w t + D w =, [ u c and D = ] /ρ. ρ u The two characteristics of this hperbolic sstem are given b d/dt = λ where λ are the eigenvalues of D; detd λi = u λ c /ρ ρ u λ = u λ = c and λ ± = u ± c. The eigenvectors are [c, ρ] T for λ and [c, ρ] T for λ +, such that the usual matrices are [ ] c c T =, T 1 = 1 [ ] [ ] ρ c, such that Λ = T 1 u c DT =. ρ ρ ρc ρ c u + c Put α and β the curvilinear coordinates along the characteristics d/dt = u c and d/dt = u + c respectivel; then the sstem transforms to the canonical form dt dα = dt dβ = 1, d dα = u c, d dβ = u + c, ρ du dα c dρ du = and ρ dα dβ + c dρ dβ =.