First order Partial Differential equations
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1 First order Partial Differential equations 0.1 Introduction Definition A Partial Deferential equation is called linear if the dependent variable and all its derivatives have degree one and not multiple together. Example: Classify which of the following P DE is linear or nonlinear. u x u y = x ln y. u xy + u yy = 0. u yy = uu xxxx + exp t. u y u zzy + u x u yy u 2 z + xy 2 = z. (b). (d). (a). (c). Definition : A Partial Deferential equation is called almost linear if the highest order derivatives have degree one and their coefficients are function of the independent variables only. Example: Classify which of the following P DE: u xx + exp x + yu yy + uu x = 0. uu xx u yy = 0. (b). (a). Definition : A partial Deferential equation is called quasilinear if the highest order derivatives have degree one and the derivatives not multiplied by each other. Examples: Classify which of the following P DE: uu x + u y = 0. u t u t t + sin u u x = 0. 1
2 (b). (a). Definition The general solution G.S. of the nonhomogeneous PD equations is u = u h +u p where u h is the solution of the corresponding homogeneous equation and u p is a particular solution of the nonhomogeneous equation. Example: Find the G. S. of the P DE u x + u = y. 2
3 0.2 G.S. of Linear first order P DE with constant coefficient A P DE of the form Au x + Bu y + Cu = f(x, y) Making the change of variables r = x, s = Bx Ay, then the P DE becomes, Au r + Cu = f(r, s) which is ordinary DE in r keeping s constant. Examples: Find the G.S. of the following equations. (1) u + u x + u y = 0 (2) 2u x + 3u y = x 2 (3) 3u x + 6u y + u = x + 2 exp y 3
4 0.3 G.S. of Linear first order P DE with nonconstant coefficient A(x, y)u x + B(x, y)u y + C(x, y)u = f(x, y) Let r = x, s satisfies the solution of ODE dy = B in which the G.S. is dx A s(x, y) = constant where A 0. Then the transformed equation is Au r + Cu = h(r, s). Which is ODE in r keeping s constant, solving this equation we get the solution of the P DE. Examples: Find the G.S. of the following equations: (1) x 2 u x xyu y + yu = 0. (2) xu x + yu y = 2u + x 2. 4
5 0.4 Quasilinear equations General form P (x, y, u)u x + Q(x, y, u)u y = R(x, y, u). First: Method of Lagrange: Consider the system of first order ODE, dx P (x, y, u) = dy Q(x, y, u) = du R(x, y, u This system is called subsidiary equations and it is equivalent to the system dy dx = Q P and du dx = R P where x is the independent variable. The G.S. of this system has the form y = y(x, c 1, c 2 ), u = u(x, c 1, c 2 ) where c 1, c 2 are arbitrary constants. If these equations are solved for c 1, c 2, then the G.S. of the subsidiary equations can be written in the form v(x, y, u) = c 1, w(x, y, u) = c 2 and the G.S. of the P DE is F (v, w) = 0 or w = f(v) or v = g(w) where F, f, g are arbitrary functions. Examples: Find the G.S. of (1). xuu x + yuu y = (x 2 + y 2 ) (2). xu x + yu y + u = 0 5
6 Second:Method of Multipliers: A useful technique for integrating a system of first order equations is that of multipliers. Recall that if a b = c d then λa + µc λb + µd = a b = c d for arbitrary values of multipliers λ, µ. Examples: Solve the P DE (1) uu x + yu y = x (2) (y x)u x + (y + x)u y = x2 +y 2 u. (3) x 2 u x + y 2 u y = xy 6
7 0.5 Cauchy problem for Quasilinear equations In ODE to determine a solution of first order equations y = f(x, y) which passes through a point in xy plane under a general condition I.C. a unique solution to the problem exists. In the study of first order P DE in two independent variables (x, y) we determine an integral surface such that the surface passes through a curve in xyu space. such a problem is called Cauchy problem. Before proceeding with the general case, consider the following examples. Examples: Solve the Cauchy problems (a) yu x xu y = 0,, u(x, 0) = x 4 (b) u x + u y = u,, u(x, 0) = sin x 7
8 Cauchy problem with parameterized curve : Let Γ be a given smooth curve defined perimetrically by ; x = f(t), y = g(t), u = h(t), a < t < b To construct an integral surface of P DE which contains Γ, we proceed as follows : Let v(x, y, u) = c 1, w(x, y, u) = c 2 be two independent integrals of the subsidiary equations. Wright v(f(t), g(t), h(t)) = c 1 and w(f(t), g(t), h(t)) = c 2 Eliminate t from these equations to derive the relation, F (c 1, c 2 ) = 0 Then the solution of the Cauchy problem is F (v, w) = 0. Examples: (a) Find an integral surface of (y + xu)u x + (x + yu)u y = u 2 1 which passes through the parabola x = t, y = 1, u = t 2. (b) Solve the Cauchy problem uu x + uu y = y + x,, x = 1, y = t, u = t 2. 8
9 Chapter 1 Second order Partial Differential equations 1.1 Classification of linear P DE s of second order The general form of this equation is: A(x, y)u xx + B(x, y)u xy + C(x, y)u yy + D(x, y)u x + E(x, y)u y + F (x, y)u = G(x, y). Definition Let = B 2 4AC 1) If > 0 at a point (x, y), then the P DE above is hyperbolic at (x, y). 2) If = 0 at a point (x, y), then the P DEaboveisparabolicat (x,y). If < 0 at a point (x, y), then the P DE above is elliptic at (x, y). Example: Classify the P DE as hyperbolic, parabolic,or elliptic 3) 1) u xx + 2u xy + u yy u x + xu y = 0. 2) x 2 u xx y 2 u yy = xy. 3) u xx + u xy + 2u yy = u x 3yu. 9
10 1.2 Initial Value problem Definition : initial value problem (IV P ) Problems involving time t as one of the independent variables of P DE, may have a condition given at time t = t 0, such a problem is called initial value problem (IV P ). Example: u t = a 2 u xx, 0 x 1, t > 0, u(x, 0) = f(x), 0 x 1. Definition : Boundary value problem (BV P ) Problems composed of a P DE and conditions on the boundary of the domain is called BV P. Example: u t = a 2 u xx, 0 x 1, t > 0, u(0, t) = u(1, t) = 0, t 0. Definition : Initial Boundary value problem (IBV P ) Problems that has both initial and boundary conditions are called IBV P. Example: u t = a 2 u xx, 0 x 1, t > 0, u(0, t) = u(1, t) = 0, t 0. u(x, 0) = f(x), 0 x 1. Theorem : Cauchy- Kovalevsky theorem: Let u tt = F (t, x, u, u x, u tx, u xx ) be the P DE with IC u(0, x) = f(x), u t (0, x) = g(x) where, f, g are functions defined on an interval of the x-axis containing the origin. Assume that f, g are analytic in a neighborhood of the origin and, F is analytic in a neighborhood of the point (0, 0, f(0), g(0), f (0), g (0), f (0)),Then the problem above has a unique analytic solution u(x, t) in a neighborhood of the origin. Definition : A mathematical problem is well possed if it has a unique solution that depends continuously on initial or boundary data. and the solution is called Stable. If the solution does not depend on initial or boundary conditions the solution is unstable. 10
11 1.3 Second Order Linear P DE with constant coefficients Type I: One of the simplest equation is a second order partial derivative equal to a function of the independent variables. Examples:Solve the P DE 1) u xy = xy 2 2) u yy = e y, u y (x, 0) = x 3, u(x, 0) = e x. Type II: Equations with second partial derivatives only. Au xx + Bu xy + Cu yy = 0, A, B, C are constants. Let u = f(y + mx) be a solution We need to find m such that u satisfied the P DE. u x = f (y + mx).m, u xx = m 2 f, u xy = mf, u y = f, u yy = f. Substitute into P DE we get ; Am 2 f + Bmf + Cf = 0 if f 0 then Am 2 + Bm + C = 0 which is called the characteristic equation. If the roots are real and distinct: Let m 1 m 2 then f(y + m 1 x), g(y + m 2 x) are solutions and u = f(y + m 1 x) + g(y + m 2 x) is the general solution. Example: Solve the wave equation: u tt = c 2 u xx, c > 0. 11
12 Note: the characteristic equation has real and distinct roots if B 2 4AC > 0 this is the same as here > 0 the equation is hyperbolic So hyperbolic equations have real and distinct roots, Parabolic equations have double roots, and Elliptic equations have complex conjugate roots. If the roots are real and equal: If m 1 = m 2 then B 2 4AC = 0 and m = B 2A. A second solution for P DE is u = xg(y B x), and 2A u = f(y + mx) + xg(y + mx) is a general solution. Also u = f(y + mx) + yg(y + mx) is a general solution. Example: Find the general solution of u xx + 4u xy + 4u yy = 0. If the roots are Complex: Let m, m are complex conjugate roots then the G.S. is u = f(y + mx) + g(y + mx). Example: Determine a solution for u xx + 4u yy = 0 Exponential type solution: Comparison with ODE, we may look for solutions in the form u(x, y) = e αx+βy for the homogeneous P DE Au xx + Bu xy + Cu yy + Du x + Eu y + F u = 0 then u x = αe αx+βy, u xx = α 2 e αx+βy, u xy = αβe αx+βy, u y = βe αx+βy, u yy = β 2 e αx+βy Substitute into the equation,we get; e αx+βy [Aα 2 + Bαβ + Cβ 2 + Dα + Eβ + F ] = 0, e αx+βy 0. So Aα 2 + Bαβ + Cβ 2 + Dα + Eβ + F = 0. We can solve for α, β one is a function of the other. 12
13 Assume that β = β(α) solving the characteristic equation, we get β 1 (α), β 2 (α). A particular solution is u = K 1 e αx+β 1y + K 2 e αx+β 2y. where K 1, K 2 are arbitrary constants. Example: Find an exponential type solution of the PDE then find the G.S. 1) u xx u yy 2u x + u = 0 2) u xx 2u xy + u yy 2u y + 2u x + u = 0. 3) u xx + u yy 2u y + u = 0. 4) u xx + u yy + 4u = 0. 13
14 Other Solvable Cases: Some equations can be solved like ODE if only partial derivatives with respect to one variable appear. Examples: Solve the P DE : 1) u yy 4u y + 3u = 0. 2) xu xy + 2u y = y 2. 14
15 1.4 Separation of variables In this method, the solution of the P DE can be expressed in the form of a product of functions of single independent variables. Example: Find a solution for the P DE using method of separation of variables: 1) u t = 4 u xx. 2) u t = a 2 (u xx + u yy ). 3) x 2 u xx 2xu x = u tt. 15
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