Introduction of Partial Differential Equations and Boundary Value Problems 2009
Outline Definition Classification Where PDEs come from? Well-posed problem, solutions Initial Conditions and Boundary Conditions Strategies to solve PDEs Simple PDEs and their general solutions Introduction to 2nd order linear PDEs Classification of 2nd order linear PDEs Change of variables and canonical forms
Definition Definition (Partial Differential Equation) A partial differential equation (PDE) for a function of n variables u(x 1,..., x n ) is an equation of the form F ( k u x k 1,, k u,, x k n u, u, u, x 1,, x n ) = 0 (1) x 1 x n where F is a given function of the independent variables x 1,..., x n and of the unknown function u = u(x 1,, x n ) and of a finite number of its partial derivatives.
Examples In simple words, PDE is an equation that contains an unknown functions of several variables and its partial derivatives. One example is the two dimensional Laplace equation, 2 u(x, y) = 2 u x 2 + 2 u y 2 = 0 or n-dimensional Laplace equation, 2 u(x) = 2 u + 2 u + + 2 u = 0 x 2 1 x 2 2 x 2 n or 3-dimensional Laplace equation in Spherical coordinates 2 u = 1 r 2 r ( r 2 u r ) + 1 r 2 sin θ θ ( sin θ u θ ) + 1 2 u r 2 sin 2 θ φ = 0 2
Linear PDEs Some PDEs are linear u Heat/diffusion eq.: t k 2 u = 0, k > 0. Wave equation: 2 u k 2 u = 0, t 2 k > 0. 2D Laplace equation: 2 u(x, y) = 2 u + 2 u = 0. x 2 y 2 Poisson equation: 2 u = f(x), f(x) 0. Helmholtz equation: 2 u = λu, λ > 0. Telegraph equation: u tt + du t u xx = 0. Schrödinger equation: i ψ t (x, t) = 2 2m 2 ψ(x, t) + V (x)ψ(x, t). Fokker-Planck eq.: u t n i,j=1 (aij u) xi x j n i=1 (bi u) xi = 0.
Non-linear PDEs Some PDEs are non-linear Nonlinear Poisson eq.: 2 u = f(u), f nonlinear. p-laplacian equation: ( u p 2 u) = 0. ( ) Minimal surface eq.: u 1 + u 2 = 0. Hamilton-Jacobi eq.: u t + H(Du, x) = 0. Scalar conservation law: u t F(u) = 0. Scalar reaction-diffusion eq.: u t 2 u = f(u). Porous medium eq.: u t 2 (u γ ) = 0. Nonlinear wave eq.: u tt a(du) = f(u). Korteweg-de Vries (KdV) eq.: u t + uu x + u xxx = 0.
Homogeneous and Non-homogeneous PDEs Some PDEs are homogeneous u Heat/diffusion eq.: t k 2 u = 0, k > 0. Wave equation: 2 u k 2 u = 0, t 2 k > 0. 2D Laplace equation: 2 u(x, y) = 2 u + 2 u = 0. x 2 y 2 and some are non-homogeneous Poisson equation: 2 u = f(x), f(x) 0. Helmholtz equation: 2 u = λu, λ > 0.
Classification of PDEs For the ease of study, we often clasify PDE according to Linearity: linear, semi-linear, quasi-linear, non-linear; Homogeneity: homogeneous and non-homogeneous PDEs; Order: the order of the highest-order derivative present in the PDE. However, what is even more important is we need to understand why we want to study PDEs!
Importance Aspects of PDEs PDEs arise from the modeling of numerous phenomena in physical sciences and engineering: quantum mechanics, thermodynamics, electromagnetism, stock markets, automobile, aerodynamics,.... Different from purely abstract mathematical problems, usually these PDEs represent the variation of physical quantities in space and time. As the quantities are physical, we demand that the mathematical problem must be a well-posed problem (i.e. with a unique solution), and very often these physical quantities are subjected to certain constraints.
Importance Aspects of PDEs Another important aspects of PDEs is the techniques of solving the PDEs. In the next few slides, we will go through these important aspects of PDEs: Modeling; Problems and constraints; Strategies of solving PDEs.
Modeling traveling wave on a vibrating string When x 0 Horizontal component: τ cos α τ cos β Vertical component: τ sin α + τ sin β = ma Divide by τ cos α: tan α + tan β = ρ x 2 u τ cos α t 2 1D wave equation: 2 u t 2 = 2 u c2 x, 2 c2 = τ cos α ρ x
Modeling heat conduction across a bar thermal energy in slice, dq = s(x)u(x, t)ρ(x)a x Energy Conservation: [s(x)u(x, t)ρ(x)a x] = φ(x, t) φ(x + x, t) t u Fourier s law of heat conduction: φ(x, t) = K 0 x Heat equation: u t = K 0 2 u sρ x 2
Well-posed Problem Definition (Well-posed Problem) We say that a given problem for a PDE is well-posed if the problem in fact has a solution; the solution is unique; the solution depends continuously on the data given in the problem.
Solution of PDE Definition (Solution) A solution of a partial differential equation is any function that satisfies the equation. For example if we could verify that ψ(x, y) = cxy is a solution to the Laplace equation, 2 ψ + 2 ψ = 0, by differentiating and x 2 y 2 substituting ψ(x, y) = cxy back into the Laplace equation.
Uniqueness of a Solution However, more often uniqueness of a solution is not to be expected. Unless certain conditions are given, we could have more than one family of solution to a PDE, for example, the following all are solutions to the Laplace equations (in fact, there are many more!) ψ(x, y) = cxy ψ(x, y) = c(x 2 y 2 ) ψ(x, y) = cx x 2 + y 2 ψ(x, y) = ce x cos y ψ(x, y) = c ln(x 2 + y 2 ) ψ(x, y) = c tan 1 (y/x).
In order to identify a unique solution, we will need to have more information, they are: Initial Conditions (the shape of the string, or its velocity at t = 0.) Boundary Condtions (the end points of the vibrated string is fixed.) The PDE, together with the initial conditions and boundary conditions are called Boundary Value Problems.
Boundary conditions We are familiar with the initial (Cauchy) conditions in the Ordinary Differential Equations, now we look at the boundary conditions. The three common types of boundary conditions are: 1. Type 1 BC (Dirichlet condition): u = g; 2. Type 2 BC (Newmann or flux condition): u n = g; 3. Type 3 BC (Mixed or Robin or radiation condition): αu + β u n = g.
Same strategies for solving PDEs Well-Posed Problem = PDE + BC [+ IC] Order of PDE? =1 Transform PDE to ODE Transform BC to Homogeneous BC No =2 Yes Homogeneous BC? SL Problem Eigen-Functions Homogeneous PDE? Yes Rectangular BC? Yes X +kx=0, etc. + ax(a)+bx (A)=0 cx(b)+dx (B)=0 sin, cos, exp, etc. No No Cylindrical BC? Yes Bessel Eq, etc. + A<=r<=B C<=p<=D E<=z<=F sin, cos, J_v, Y_v, etc. No Spherical BC? Yes Legendre Eq, Spherical Bessel Eq, etc. + A<=r<=B C<=p<=D E<=q<=F sin, cos, Y_nm, etc. No Other BC? Yes X +p(x)x +q(x)=0, etc. + ax(a)+bx (A)=0 cx(b)+dx (B)=0 Special Functions Consider the homogeneous part of the PDE and calculate the general solution. Apply the variation of parameter on the general solution. Variation of Parameters X +p(x)x +q(x)=f_n(x) Principle of Superposition General Solution is a Combination of Fourier Series or Fourier Specific Solution Initial Conditions
Solving Simple PDEs Let consider simple PDEs for an unknown functions u(x, y). u x = 0 = u(x, y) = f(y). u = f(x) = u(x, y) = f(x) dx + g(y) x u x = f(y) = u(x, y) = xf(y) + g(y) u = f(x, y) = u(x, y) = f(x, y) dx + g(y) x
Solving Simple PDEs Increase the difficulties (if any at all!) one notch: 2 u x = 0 = u(x, y) = xf(y) + g(y). 2 2 u x y = 0 = u(x, y) = f(x) + g(y)
Cauchy Problems In order to pick up a particular solution we need to supply with initial conditions, the resulting is a Cauchy (initial value) problem. u = y, when x = 1, u(1, y) = 2y = u(x, y) = xy + y. x 2 u x = 0, when x = 0, u = 2 y2, u = 2y sin 2y = x u(x, y) = 2xy x sin 2y + y 2.
Linear 2 nd Order PDEs A general linear 2 nd order can be written as a 2 u x + b 2 u 2 x y + u c 2 y + d u 2 x + e u + fu + g = h(x, y). y if h(x, y) = 0 then it is a homogeneous PDE. the quatity = b 2 4ac is called the discriminant. according to, linear 2 nd order PDE can be classified as if > 0, then it is a hyperbolic PDE if = 0, then it is a parabolic PDE if < 0, then it is a elliptic PDE
Classification of 3 common PDEs Three PDEs that are the main focus of this course are wave equation, heat equation and Laplace equation. These PDEs are linear 2 nd PDEs, and their are classify as Wave equation : 2 u t 2 u 2 c2 x = 0 (hyperbolic); 2 Heat equation : u t c2 2 u x 2 = 0 (parabolic); Laplace equation : 2 u x 2 + 2 u y 2 = 0 (elliptic).
The advantage of the classification is, depending on, we could making a change of variables to convert the PDEs into canonical forms. Sometimes no always these canonical forms can be solved easily, expect solving for elliptic equation is a bit subtle and not straight forward.
Canonical Forms By making a change in variables, s = ax + by, t = cx + dy, linear PDE could convert to its canonical form with proper choice of a, b, c and d. 2 u Hyperbolic equation: s t = F (s, t, u, u s, u t ); Parabolic equation: 2 u s 2 = F (s, t, u, u s, u t )
Example Consider the PDE : 6 2 u x + 5 2 u 2 x y + 2 u y = 0 when 2 y = 0, u = cos x + sin x, and u = cos x + 2 sin x. y > 0 = hyperbolic; By making a change of variables : s = x 2y, t = x 3y; 2 u The PDE will convert to its canonical form: s t = 0; Which has a general solution, u = f(s) + g(t) = f(x 2y) + g(x 3y), and Particular solution u = cos(x 2y) + 4 sin(x 2y) 3 sin(x 3y).