Math 201 Lecture 25: Partial Differential Equations: Introduction and Overview

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1 Math 201 Lecture 25: Partial Differential Equations: Introduction and Overview Mar. 12, 2012 Many examples here are taken from the textbook. The first number in () refers to the problem number in the UA Custom edition, the second number in () refers to the problem number in the 8th edition. 0. Unfinished Business Example 1. (8.4 26; ) Find at least the first four nonzero terms in a power series expansion about x=0 of a general solution to the given differential equation. Solution. Write y xy +2y= cosx. (1) y(x)= a n x n (2) n=0 and substitute into the equation, recalling we have n ( 1) n=0 cosx= n=0 (2n)! x2n = cosx = y xy +2y = n=2 = n=0 The first few terms of balance is then: From these we have So the solution is ( 1) n (2n)! x2n, (3) a n n(n 1)x n 2 n=1 a n+2 (n+2)(n+1)x n n=1 = (2a 2 +2a 0 )+ n=1 ( 1 y(x) = a 0 +a 1 x+ = ( 1 2 x2 + na n x n + n=0 2a n x n na n x n + n=0 2a n x n [a n+2 (n+2)(n+1) na n +2a n ]x n. (4) x 0 : 1=2a 2 +2a 0 (5) x 1 : 0=6a 3 +a 1 (6) x 2 1 : 2 = 12a 4 (7) a 2 = 1 2 a 0, a 3 = 1 6 a 1. (8) ) 2 a 0 x a 1x 3 + ( )+a 0 (1 x 2 + )+a 1 x 1 6 x3 + Remark 2. From the above we can conlude that the expansions of y p,y 1,y 2 are ). (9) y p (x) = 1 2 x2 + (10) y 1 (x) = 1 x 2 + (11) y 2 (x) = x 1 6 x3 + (12) 1

2 2 Math 201 Lecture 25: Partial Differential Equations: Introduction and Overview (Don t forget that the choices of y p,y 1,y 2 are not unique!) Terminology: 1. From ODE to PDE Ordinary Differential Equation (ODE): A differential equation whose unknown is a single variable function. Partial Differential Equation (PDE): A differential equation whose unknown is a multi-variable function. Therefore the derivatives involved in an ODE are ordinary, while those involved in a PDE are partial derivatives. From ODE to PDE. Let s set the first variable to be time t. Then 1 An ODE describes the physical law governing the motion of one single particle; A system of ODE describes the physical law governing the motion of a finite number of particles; A PDE describes the physical law governing the motion of infinitely many particles which form a continuum. Derivation of heat equation. The essence of heat equation is that it is a mathematical model of the collective behavior of a large number randomly moving particles (from basic thermodynamics we then understand why it can model the change of temperature). Consider the following situation. A large number of particles are distributed between x = 0 and x=l, with initial density distribution f(x), that is the number of particles between x=a and x=b is given by f(x)dx. (13) Now assume that these particles are not static, they are moving. Then the density should change with time t and is then naturally modeled by a function of two variables u(x, t). It turns out that, when these particles are moving randomly, u(x, t) solves the heat equation t (x,t)=β 2 u 2(x,t)+P(x,t). (14) HereP(x,t) is the source term. It models the situation that some particles may spontaneously appear (P(x,t)>0) at location x at time t, or spontaneously disappear (P(x,t)<0). To understand how the heat equation gets involved, let s consider the following simplified situation. Let N be a very large number, and let Now we make the following assumptions: x i = i L. (15) N The particles can only stay on the x i s. That is u(x,t)=0 whenever x x i for some i. There is a small time interval t, such that the particles can only move at t j = j t, and stays put between t j and t j+1. At t j, each particle at x i either jumps left to x i 1, or jumps right to x i+1, with equal probability, 1/2. Note that when the number of particles is large, basically half jumps left and half jumps right. 1. The following categorization is not fully accurate, due to the uncertainty principle of accuracy and clarity/definiteness.

3 Mar. 12, Under these assumptions, let s study how u(x, t) changes. Consider a point x i. According to the above assumptions, the number of particles at x i at time t j+1 is the sum of the number of particles jumped to x i from x i 1 and the number of particles jumped to x i from x i+1. So we have Now add u(x i,t j ) to both sides. We reach u(x i,t j+1 )= 1 2 u(x i 1,t j )+ 1 2 u(x i+1,t j ). (16) If we set u(x i,t j+1 ) u(x i,t j )= 1 2 [u(x i 1,t j )+u(x i+1,t j ) 2u(x i,t j )]. (17) t=t j+1 t j ; x=x i+1 x i =x i x i 1, (18) the above equation can be written as u(x i,t j + t) u(x i,t j ) t = 1 2 x 2 t If we let t and x go to 0, then the above becomes where 2 u(x i x,t j )+u(x i + x,t j ) 2u(x i,t j ) x 2. (19) t =β 2 u 2 (20) 1 x β= lim 2 x, t 0 2 t. (21) Boundary conditions. In the above we didn t describe the boundary condition, which models what happens at the two end points x=0,x=l. In general there are three popular situations. The density is controlled, which means u(0,t) and/or u(l,t) are given functions of t. The flux is controlled, which means (0,t) and/or (L,t) are given functions of t. The whole things is periodic, the two points 0, L are actually one (imagine connecting the two points and make the segment [0,L] a circle). In particular, u(0,t)=u(l,t), and what goes out at x=l comes in at x=0 (think of Pac-Man). As there are two end points, we reach the following four combinations: 1. (Dirichlet boundary condition) 2. (Neumann boundary condition) u(a,t)=a(t); u(b,t)=b(t). (22) (a,t)=a(t); 3. (Mixed bonundary condition) or u(a,t)=a(t); 4. (Periodic boundary condition) (b,t)=b(t). (23) (b,t)=b(t). (24) (a,t)=a(t); u(b,t)=b(t)., (25) u(a,t)=u(b,t); (a,t)= (b,t) (26) 2. We see that to reach heat equation, it is necessary to have t, x 2 related in such a way that the limit exists.

4 4 Math 201 Lecture 25: Partial Differential Equations: Introduction and Overview Remark 3. Of course there can be more esoteric boundary conditions. For example we can require u(a,t)=a(t), u(b,t)+3 (b,t)=b(t). (27) Equation + Initial condition + Boundary condition = Solvable (Well-posed). We have seen that the initial condition tells us what is the starting distribution, the equation tells us how the distribution evolves between 0 and L, while the boundary conditions tell us what happens at the end points 0,L. Therefore we need all three, combined, to be able to solve u(x,t). Equations to solve in this chapter: Equation: Initial condition t (x,t)=β 2 u 2(x,t)+P(x,t), a<x<b,0<t (28) Boundary condition: Any one of the following four: 1. (Dirichlet boundary condition) 2. (Neumann boundary condition) u(x,0)=f(x). (29) u(a,t)=a(t); u(b,t)=b(t). (30) (a,t)=a(t); 3. (Mixed bonundary condition) or u(a,t)=a(t); 4. (Periodic boundary condition) (b,t)=b(t). (31) (b,t)=b(t). (32) (a,t)=a(t); u(b,t)=b(t)., (33) u(a,t)=u(b,t); 2. The Method of Separation of Variables (a,t)= (b,t) (34) Partial differential equations are very difficult to solve. In general we cannot expect an explicit formula for the solutions. The basic equation we will solve using this method is the homogeneous heat equation with homogeneous boundary conditions: Equation: (P(x,t)=0) Initial condition t (x,t)=β 2 u 2(x,t), a<x<b,0<t (35) Boundary condition: Any one of the following four: 1. (Dirichlet boundary condition) u(x,0)=f(x). (36) u(a,t)=0; u(b,t)=0. (37)

5 Mar. 12, (Neumann boundary condition) (a,t)=0; 3. (Mixed bonundary condition) or u(a,t)=0; 4. (Periodic boundary condition) (b,t)=0. (38) (b,t)=0. (39) (a,t)=0; u(b,t)=0., (40) u(a,t)=u(b,t); Basic Idea: The basic idea is as follows: We try to construct u(x,t) as an infinite sum: (a,t)= (b,t) (41) u n (x,t). (42) where each u n (x,t) is simple : a product of a function of t only and a function of x only: u n (x,t)=t n (t)x n (x). (43) We would like this u(x,t) to solve our problem, or more specifically: 1. Satisfy the equation; 2. Satisfy the initial condition; 3. Satisfy the boundary conditions. The hope is that these requirements would give us equations that T n (t),x n (x) must satisfy as they are functions of one variable only, these equations must be ordinary differential equations, and thus presumably easier to solve. How to make u(x,t) satisfy the equation: An obvious thing to do is to require each u n (x,t) to satisfy the equation. As the equation is linear, if such u n (x,t) can be found, we would have t u(x,t)= t u n(x,t)= β 2 2u n(x,t)=β 2 2u(x,t). (44) Substitute u n (x,t)=t n (t)x n (x) into the equation, we have Now we separate the variables : T n (t)x n (x)=βt n (t)x n (x). (45) T n (t) T n (t) =β X n (x) X n (x). (46) The key observation now is that both sides have to be the same constant. To see this, we take of t both sides: ( ) Tn (t) = [ β X ] n (x) T n (t) =0 = constant. (47) T n (t) t X n (x) T n (t) We denote this constant by βk n to be consistent with the textbook. Now we see that requiring T n (t)x n (x) to solve the equation leads to two equations for T n (t) and X n (x) respectively: T n (t) T n (t) =βk n T n (t) βk n T n (t)=0; (48) β X n (x) X n (x) =βk n X n (x)+k n X n (x)=0. (49)

6 6 Math 201 Lecture 25: Partial Differential Equations: Introduction and Overview Note that both equations involve one same parameter K n that is not known a priori. Remark 4. Note that the textbook changed notation from 7th to 8th edition. In 8th edition K n is replaced by λ n. Initial conditions. If u(x,t)= T n (t)x n (x), then necessarily f(x)=u(x,0)= T n (0)X n (x). (50) It turns out that, for appropriate X n (x), f(x) has one and only one expansion in the form of Then f(x)= f n X n (x). (51) T n (0)=f n. (52) Boundary conditions. We take Dirichlet boundary conditions as an example. Other cases are similar. u(a,t)= T n (t)x n (a)=0 X n (a)=0; (53) u(b,t)=0 X n (b)=0. (54) Summarizing, we see that the method of separation of variables should run as follows: 1. Solve the equation + boundary conditions for X n (x) to obtain X n (x) and λ n : X n (x) K n X n (x)=0; X n (a)=x n (b)=0. (55) Note that the boundary conditions for X n changes when the boundary conditions for the original PDE changes. 2. Expand 3. Solve T n (t) using the equation with initial condition T n (0)=f n. 4. The final answer is then f(x)= n f n X n (x). (56) T n (t) βk n T n (t)=0 (57) X n (x)t n (t)= n f n e βknt X n (x). (58) We will understand steps 1 and 3 in 10.2, step 3 in 10.3 and 10.4, and finally put things together in Remark 5. The determination of {X n (x)} (Step 1) is most important. Steps 2,3 can be understood as expanding the equation in X n (x). Remark 6. Note that in the above we do not specify the range for n in the summation n. This is because this range is different for different types of boundary conditions. 3. To General Cases What happens if we have P(x,t) 0? When this happens there is slight change in step 3, the equation for T n (t) is not homogeneous anymore, it has a right hand side p n (t) which comes from the expansion P n (x,t)= n p n (t)x n (x) (59) Recall that {X n } is such that any function can be expanded with respect to them.

7 Mar. 12, What happens if we have non-homogeneous boundary conditions? The difficulty here is that the boundary condition is not homogeneous anymore. If we have each X n (x)t n (t) satisfying, say then clearly the sum of two such functions would have u(0,t)=a(t); u(l,t)=b(t). (60) u(0,t)=2a(t); u(l,t)=2b(t) (61) and does not satisfy the boundary condition anymore. If we form the infinite sum X n (x)t n (t) (62) then u(0,t)= A(t)=, u(l,t)= B(t)= which is nonsensical. The way to fix this is the following. We write X n (x)t n (t)+w(x,t) (63) where n X n(x) T n (t) solves the equation with zero boundary conditions, and w(x, t) is any function that satisfies the non-zero boundary conditions.