First Order Equations

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1 10 1 Linear and Semilinear Equations Chapter First Order Equations Contents 1 Linear and Semilinear Equations 9 Quasilinear Equations 19 3 Wave Equation 6 4 Sstems of Equations 31 1 Linear and Semilinear Equations 11 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 w(ξ, η = u((ξ, η, (ξ, η + h(ξ, 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 {0, } for (, in D We begin looking for a suitable transformation b computing derivatives via the chain rule = ξ ξ + η η and = ξ ξ + η η We substitute these into equation (1 to obtain ( ξ a ξ + ( η ξ + b η ξ + η η + cw = f We can rearrange this as ( a ξ + b ξ ( ξ + a η + 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 = 0 for (, in D Provided that η/ 0 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 = 0 implies that d/ = η/ η = b/a So, the equation η(, = k defines solutions of the ODE d b(, = a(, (5 Equation (5 is called the characteristic equation of the linear equation (1 Its solution can be written in the form F (,, η = 0 (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 / 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 0 An obvious choice is ξ ξ(, = 9

2 Chapter First Order Equations Linear and Semilinear Equations Then 1 0 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(, + 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 A(ξ, η + C(ξ, ηw = ρ(ξ, η ξ Finall, restricting the variables to a set in which A(ξ, η 0 we have which is in the form of (3 with h(ξ, η = C(ξ, η A(ξ, η ξ + C A w = ρ A, and F (ξ, η = ρ(ξ, η A(ξ, 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 ξ = K A, where the nonlinear term K(ξ, η, w = κ(,, u and restricting again the variables to a set in which A(ξ, η = α(, 0 Notation: It is ver convenient to use the function u in places where rigorousl the function w should be used Eg, 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 = b a = Solve this b separation of variables 1 1 d = ln + 1 = k, for > 0, and 0 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 0 as required η = ln + 1 ξ = eη 1/ξ Now we appl the transformation ξ =, η = ln + 1/ with w(ξ, η = u(, and we have = ξ ξ + η η = ξ + ( η 1 = ξ 1 ξ η, = ξ ξ + η η = 0 + η 1 = 1 e η 1/ξ η Then the PDE becomes ( ξ ξ 1 ξ + e η 1/ξ 1 η e η 1/ξ η + ξeη 1/ξ w = 1, which simplifies to ξ ξ + ξeη 1/ξ w = 1 then to 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 0 1 Equivalent set of ODEs The point of this transformation is that we can solve equation (3 Think of + h(ξ, 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 h(ξ,η dξ R R ξ + h(ξ, η e h(ξ,η dξ w = F (ξ, η e h(ξ,η dξ, ( h(ξ,η dξ er w = F (ξ, η h(ξ,η dξ er ξ

3 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 h(ξ,η dξ er w = F (ξ, η h(ξ,η dξ er dξ + g(η in which g is an arbitrar differentiable function of one variable Now the general solution of the transformed equation is w(ξ, η = e R h(ξ,η dξ F (ξ, η h(ξ,η dξ er dξ + g(η e R h(ξ,η dξ We obtain the general form of the original equation b substituting back ξ(, and η(, to get u(, = e α(, [β(, + g(η(, ] (6 A certain class of first order PDEs (linear and semilinear PDEs can then be reced 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 = 0 where a, b, c R Assume a 0, the characteristic equation is with general solution defined b the equation 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 ξ + c a w = 0 ( e cξ/a w = 0 ξ e cξ/a w = g(η, where g is an differentiable function of one variable Then and in terms of and we back transform w = g(η e cξ/a u(, = g(b a e c/a Eercise: Verif 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( 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 ( 0 = 0 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; ie, to chose the arbitrar function g 13 Characteristic Curves We investigate the significance of characteristics which, defined b the ODE d b(, = 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 = a(,, = b(, (7 The variation of u with respect = ξ along these characteristic curves is given b = + 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 = = κ(,, u The one parameter famil of characteristic curves is parameterised b η (each value of η represents one unique characteristic The solution of equation ( reces to the solution of the famil of ODEs = κ(,, u ( or similarl = κ(,, u = dξ a(, along each characteristics (ie 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

4 Chapter First Order Equations Linear and Semilinear Equations Cauch Problem: Consider a curve Γ in (, -plane whose parametric form is ( = 0 (σ, = 0 (σ The Cauch problem is to determine a solution of the equation F (,, u, u, u = 0 in a neighbourhood of Γ such that u takes prescribed values u 0 (σ called Cauch data on Γ Γ u o (1 u o ( ξ=cst η=cst Eample 1: Consider u = 0 The characteristic equation is 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 e 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(, + g(η(, ] Consider the Cauch data, u 0, prescribed along the curve Γ whose parametric form is ( = 0 (σ, = 0 (σ and suppose u 0 ( 0 (σ, 0 (σ = q(σ If Γ is not a characteristic, the problem is well-posed and there is a unique function g which satisfies the condition q(σ = e α(0(σ,0(σ [β( 0 (σ, 0 (σ + g( 0 (σ, 0 (σ] If on the other hand ( = 0 (σ, = 0 (σ is the parametrisation of a characteristic (η(, = k, sa, the relation between the initial conditions q and g becomes q(σ = e α(0(σ,0(σ [β( 0 (σ, 0 (σ + G], (9 where G = g(k 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 q(σ 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 q(σ 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 g(k = G ξ=cst (We can see that an η and ξ cross onl once the are independent, ie J 0; η and ξ have been properl chosen This gives the solution u(, = e 4 g(3 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(, 0 = sin( u(, 0 = e 4 g(3 = sin( ie g(3 = sin(e 4, g(t = sin(t/3 e 4t/3 This determines g and the solution satisfing the condition u(, 0 = 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(

5 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 g(3 (3 1 = This requires g(1 = 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 g(3 (3 1 = e 4 This requires g(1 = 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( = 0 = c, constant So, take η = and ξ = Then the equation becomes η + ξ η = w ξ ξ w = 0 w ξ ξ = 0 Finall the general solution is, w = ξ g(η 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 = and d = gives the parametric form of the characteristic curves, while the integration of the compatibilit equation = u gives the solution u(s 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 0 (θ = θ and 0 (θ = θ with θ [, 1] and we consider the points on Γ to be the origin of the coordinate s along the characteristics (ie s = 0 on Γ So, on Γ (s = 0 } } 0 = θ = c 1 (s, θ = θ es 0 = θ = c (s, θ = θ e s, θ [0, 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 = 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 = 0, u 0 ( 0 (θ, 0 (θ u 0 (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 u(s, 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 u(s, θ to find ( u(, = ep ln and = θ θ = (θ 0 = =

6 Chapter First Order Equations 19 Quasilinear Equations Consider the first order quasilinear PDE a(,, u + b(,, u = c(,, u (10 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 (10 sas that the normal to the integral surface is perpendicular to the Monge direction; ie 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 = 0 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 dz a b c = c d b dz a dz c = 0 b a d a(,, u = d b(,, u = dz (=, c(,, u where dl = (, d, dz is an arbitrar infinitesimal vector parallel to the Monge direction In the linear case, characteristics were curves in the (, -plane (see 13 For the quasilinear equation, we consider Monge curves in (,, u-space defined b = a(,, u, d = b(,, u, = c(,, u Characteristic equations (d{, }/ and compatibilit equation (/ are simultaneous first order ODEs in terms of a mm 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 0, 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 0 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 (10 satisfies F (φ, ψ = 0 for some arbitrar function F (equivalentl, φ = G(ψ 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 = φ((s, (s, u(s, = φ(s = c 1 dφ = φ + φ d + φ = 0, a φ + b φ + c φ = 0 a ψ + b ψ + c ψ = 0 Solving for c gives ( φ ψ a ψ ( φ φ ψ + b ψ φ = 0 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 = 0 remember F (φ(,, u(,, ψ(,, u(, and differentiate df = F F + d = 0 1 φ ψ Then, the derivative with respect to is zero, F = F ( φ φ + φ + F ( ψ ψ + ψ = 0, as well as the derivative with respect to F = F ( φ φ + φ + F ( ψ ψ + ψ = 0

7 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 = 0, ( 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, ie, F (φ, ψ = 0 defines a solution of the original equation Eample 1: ( + u + = in > 0, < <, 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, = + u, (11 d =, (1 = (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, = 0 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 ( = u d [ ( u ] = d [ ( ] d ( u, = ( d ( u = 0 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 = 0 or ( u = G, for some arbitrar functions F or G Now to get a particular solution, appl initial conditions (u = 1 + when = 1 ( 1 ( + 1 = G( + 1 G( + 1 = 4 Substitute θ = + 1, ie = (θ 1/, so G(θ = (1 θ Hence, ( ( u = 1 + u = ( u We can regard this as a quadratic equation for u: Then, u [ u ] + ( = 0, u u ( = 0 u ± = 1 ( ± = 1 ( ± + 1 Consider again the initial condition u = 1 + on = 1 Hence, u ± (, = 1 = 1 ± ( = 1 ± take the positive root u(, = +

8 Chapter First Order Equations 3 4 Quasilinear Equations Eample : using the same procere solve ( u + ( + u = ( + u with u = + 1 on = Characteristic equations = ( u, (14 d = ( + u, (15 = ( + u (16 Again, we seek to independent first integrals On the one hand, equations (14 and (15 give Now, consider + d = u + + u = ( +, = 1 from equation (16 u d = 1 u Hence, /u = c 1 is constant and φ(,, u = u is a first integral of the PDE On the other hand, d ( ln = 0 u d = u u = u( + =, d ( + u = 0 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, ie = θ 1, then and finall the solution is 1 + = G( + 1; G(θ = 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 = 0 on + = 1 =, We now parameterise the initial line in terms of θ: and appl the initial data at s = 0 Hence, d = u and = 1, = 1 c 1 s, (17 = s + c s + c 3, (18 u = c + s, for constants c 1, c, c 3 (19 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 0 = c c = 0 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 = 0 φ = u / and ψ = /(1 + u Alternative approach to eample 1: Characteristic equations ( + u + = in > 0, < <, with u = 1 + on = 1 = + u, (0 d =, (1 = (

9 Chapter First Order Equations Wave Equation Solve with respect to the mm variable s; (1 gives, Summar: Solving the characteristic equations two approaches (0 and ( give and (0 give = c 1 e s, d ( + u = + u + u = c e s, = 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 = 0, = 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 + (1 + θ e s, = e s and u = e s + θ e s = 1 + (1 + θ θ = 1 u = ( + 1 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 (1 + 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, eg 1 d ( + u = 1 + u and find some combination of the variables which differentiates to zero (first integral, eg ( d + u = 0 Solve the equations with respect to the mm variable s, and appl the initial data (parameterised b θ at s = 0 Eliminate θ and s; find invariants b solving for constants 3 Wave Equation We consider the equation where c is some positive constant 31 Linear Waves + (u + c = 0 with u(0, = f(, t If u is small (ie u u, then the equation approimate to the linear wave equation t + c = 0 with u(, 0 = f( The solution of the equation of characteristics, /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(, 0 = 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=0 η= ct=cst u(,t Note that u is constant where ct =constant, ie on the characteristics f( c t=0 t=t 1 =ct 1 1

10 Chapter First Order Equations Wave Equation 3 Nonlinear Waves For the nonlinear equation, the characteristics are defined b dt = 1, + (u + c t = 0, = c + u and = 0, 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(, 0 = 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 reced 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, dt = u + c = (f(θ + ct + θ, epressed in terms of a parameter θ (If we make use of the parametric form of the data curve Γ: { = θ, t = 0, θ R} and solve directl the Cauch problem in terms of the coordinate s = t, we similarl find, u = f(θ 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=0} f ( θ<0 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], ie the solution is single valued (ie is phsical onl for 0 t < t min Hence, when f (θ < 0 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 (eg u become vitall important f( u(,t t breaking multi valued To illustrate finite time solutions of the nonlinear wave equation, consider f(θ = θ(1 θ, (0 θ 1, f (θ = 1 θ So, f (θ < 0 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 + (1 + t t + ct u + ( ct + ct + c t = 0, Consider two crossing characteristics epressed in terms of θ 1 and θ, ie = (f(θ 1 + ct + θ 1, = (f(θ + ct + θ (These correspond to initial values given at = θ 1 and = θ intersect at the time t = θ 1 θ f(θ 1 f(θ, 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 + (1 + t 4t( ct u = (c c, so the solution becomes singular as t 1 and 1 + c

11 Chapter First Order Equations Wave Equation 33 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 = 0, and put w(, t = u(, t + c; hence the PDE becomes the inviscid Burger s equation or equivalentl in a conservative form t + w = 0, t + ( w = 0, where w / is the flu function We now consider its integral form, [ t + ( ] w = 0 d w(, t = dt 1 where > 1 are real Then, d dt 1 w(, t = w ( 1, t 1 w (, t 1 ( w Let us now rela the assumption regarding the differentiabilit of the our solution; suppose that w has a discontinuit in = s(t with 1 < s(t < w(,t w(s,t + w(s,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 w(s + w(s = w(s+ + w(s The problem then reces 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 t + w = 0, with initial conditions 1 for θ 0, w( = θ, t = 0 = f(θ = 1 θ for 0 θ 1, 0 for θ 1 As seen before, the characteristics are curves on which w = f(θ as well as f(θ t = θ are constant, where θ is the parameter of the parametric form of the curve of initial data, Γ For all θ (0, 1, f (θ = 1 is negative (f = 0 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 (0, 1 Then, the position where the singularit develops at t = 1 is = f(θ t + θ = 1 θ + θ = 1 1 s(t Thus, splitting the interval [ 1, ] in two parts, we have t w ( 1, t w (, t = d dt s(t 1 = w(s, t ṡ(t + w(, t + d dt s(t 1 s(t w(, t, t w(s+, t ṡ(t + s(t t, where w(s (t, t and w(s + (t, t are the values of w as s from below and above respectivel; ṡ = /dt w=1 & t=cst w=1 & t=0 w(,t 1 w=0 & =cst =1 w(,t 1 t=0 weak solution (shock wave t>1 U multi valued solution Now, take the limit 1 s (t and s + (t Since / t is bounded, the two integrals tend to zero We then have t=0 t=1 w (s, t w (s +, t = ṡ ( w(s, t w(s +, t

12 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 0 for 1, with 0 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 introcting a shock wave Suppose shock at s(t = U t + α, with w(s, t = 1 and w(s +, t = 0 The jump condition gives the shock velocit, U = w(s+ + w(s = 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, { 0 for < s(t, w(, t = 1 for > s(t, 4 Sstems of Equations 41 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 s(t = 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 Eg, a 11 a 1n b 11 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( 5u(1 + 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 (ie, 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, ie 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 det(λ i I D = 0 (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 0 0 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 = λ i This is the canonical form of the equations

13 Chapter First Order Equations Sstems of Equations Eample 1: Consider the linear sstem u (1 + 4 u ( = 0, u ( + 9 u (1 = 0, Here, u + Du = 0 with D = Eigenvalues: Eigenvectors: Then, [ ] P = 3 3 [ So we put u = 3 3 ie Initial conditions give } so, f( = /6 and g( = 0; then Eample : with initial conditions u = [, 3] T on = 0 [ ] det(d λi = 0 λ 36 = 0 λ = ±6 [ ] [ ] [ ] [ ] [ ] = = for λ = 6, [ ] [ ] [ ] [ ] [ ] = = for λ = 6, P 1 = 1 [ ] [ ] 3 and P DP = ] [ ] 6 0 v and v + v 0 6 = 0, which has general solution v (1 = f(6 and v ( = g(6 +, u (1 = v (1 + v ( and u (1 = 3v (1 3v ( = f(6 + g(6, 3 = 3f(6 3g(6, u (1 = 1 (6, 3 u ( = 1 (6 Rece the linear sstem [ ] 4 u + u 4 = 0 to canonical form in the region of the (, -space where it is totall hperbolic Eigenvalues: [ ] 4 λ det = 0 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 0 v 0 3 = 0 and P 1 DP = [ ] 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 rece 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 u(j + λ i u(j = 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/ = λ i We define the ith characteristic, with curvilinear coordinate s i, as the curve in the (, -plane along which i = 1, d i = λ i or equivalentl Hence, the directional derivative parallel to the characteristic is d u (j = i u(j + λ i u(j, d = λ i and the sstem in normal form reces 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

14 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 = 0 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 = 0, 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 t + D = 0, [ u c and D = ] /ρ ρ u The two characteristics of this hperbolic sstem are given b /dt = λ where λ are the eigenvalues of D; det(d λi = u λ c /ρ ρ u λ = 0 (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 0 DT = ρ ρ ρc ρ c 0 u + c Put α and β the curvilinear coordinates along the characteristics /dt = u c and /dt = u + c respectivel; then the sstem transforms to the canonical form dt dα = dt dβ = 1, dα = u c, dβ = u + c, ρ dα c dρ = 0 and ρ dα dβ + c dρ dβ = 0