2001/02: Lecture 9. Partial differential equations The wave equation; Laplace equation; Diffusion equation; Non-linear reaction diffusion equation; Boundary conditions. The wave equation 2 2u ( x, t ) 2 u ( x, t ) - one dimensional wave equation. c t 2 x 2 The variable t has the significance of time, the variable x is the spatial variable. Unknown function u(x,t) depends both of x and t. For example, in case of vibrating string the function u(x,t) means the string deviation from equilibrium in the point x at the moment t. 2 2u ( x, y, t ) 2u ( x, y, t ) 2 u ( x, y, t ) - two dimensional wave equation. c t 2 x 2 y 2 2u( x, y, t ) 2u( x, y, t ) denotes the Laplacian of u (in one x 2 y 2 2u ( x, t ) dimensional space u ) and the general form of the wave equation is x 2 2u c 2 u . t 2 The symbol u This differential equation occurs in such diverse fields of study as electromagnetic theory, hydrodynamic theory, acoustics and so on. Consider the wave equation for the vibrating string in more details. Applying the 2d Newton’s law to the portion of the string between points x and x x we get the equation 2u(( x x) / 2, t ) u( x x, t ) u( x, t ) and dividing by the x and m kx 2 t x x taking the limit x 0 we obtain the wave equation 2 2u ( x, t ) 2 u ( x, t ) c (1) t 2 x 2 If both ends of the string are fixed then the boundary conditions are u (0, t ) 0, u (1, t ) 0 and the initial conditions are u ( x,0) u ( x,0) f ( x), g ( x), 0 x 1 . t 1 In case of infinite string x , t 0 the wave equation with initial conditions u ( x,0) u ( x,0) f ( x), g ( x), x t has solution (D’Alembert’s solution): x ct 1 1 u( x, t ) [ f ( x ct ) f ( x ct )] g ( s)ds . 2 2c x ct The diffusion (or heat) equation In the Lecture 7 we considered the stationary temperature distribution of the homogeneous rod (i.e. the distribution which does not changes with time). The equation for this distribution is ku ( x ) r ( x ) 0 . Now we consider the temperature distribution u ( x, t ) , which depends on the point x of the rod and time t. u( x, t ) 2 u ( x, t ) k r( x ) , t x 2 In case of diffusion equation r(x)=0. Initial temperature distribution u ( x,0) f ( x) . If the function f(x) is so-called delta function (at initial moment we have the heat source at one point with coordinate ) then the homogeneous (r(x)=0) heat equation u( x, t ) 2 u( x, t ) k t x 2 has the solution u ( x, t ) e ( x ) 2 4 kt 4kt This is so-called fundamental solution. . Non-linear reaction diffusion equation u ( x, t ) 2 u ( x, t ) D1 f (u, v) t x 2 v( x, t ) 2 v ( x, t ) D2 g (u, v) t x 2 here u ( x, t ), v( x, t ) are concentrations of some chemicals (populations) at point x at the moment t. D1 , D2 are the diffusion coefficients, f (u, v), g (u, v) are functions (nonlinear) describing the reaction or interaction between chemicals (populations). 2 Example. The Brusselator (chemical reaction system with two components) on the unit interval u ( x, t ) 2 u ( x, t ) D1 C ( B 1)u u 2 v 2 t x 2 v( x, t ) v ( x, t ) D2 Bu u 2 v t x 2 x [0,1]; D1 , D2 , B, C 0 u (0, t ) u (1, t ) C , v(0, t ) v(1, t ) B / C Laplace’s equation u 0 u ( x, y ) 2u 2u 0. x 2 y 2 u ( x, y , z ) 2u 2u 2u 0. x 2 y 2 z 2 Poisson’s equation u f 2u 2u u ( x, y ) f ( x, y ). x 2 y 2 2u 2u 2u f ( x, y, z ). x 2 y 2 z 2 In the electrostatics, if u is electric potential at some point of the plane or the space then the Poissoin’s equation describes the distribution of the potential u due to a collection of stationary charges, the function f is the charge density. In a region that is free of charges, f=0 and Poisson’s equation reduces to Laplace’s equation. Laplace’s equation plays an important role in potential theory (magnetic potential, gravitational potential, velocity potential in hydrodynamics etc.). For this reason it is often called the potential equation, and functions satisfying it are called potential functions as well as harmonic functions. u ( x, y , z ) Boundary value problem for the Laplace’s equation: u 0, 0 x , 0 y u (0, y ) g ( y ), u ( , y ) h( y ), 0 y , u ( x,0) p ( x), u ( x, ) q ( x), 0 x . Continuity conditions at four vertexes. 3 Example. u 0, 0 x , 0 y u(0, y ) u( , y ) 0, 0 y , u( x,0) f ( x ), u ( x, ) 0, 0 x . Method of separation of variables: u ( x, y ) X ( x)Y ( y ) u xx X Y , u yy XY , X Y XY 0 X Y k. X Y Two-point boundary-value problem: X kX 0, X (0) X ( ) 0 k n 2 , X n ( x) sin nx, n 1,2,3,... Yn n 2Yn 0, Yn ( ) 0, n 1,2,3,... Yn ( y ) an sinh( n ) bn cosh( n ), an sinh( n ) bn cosh( n ) 0, bn cosh( n ) sinh( n ) bn Yn ( y ) sinh( n( y )), sinh( n ) bn u n ( x, y ) sin( nx) sinh( n( y )), n 1,2,3,... sinh( n ) an un ( x,0) bn sin nx f ( x). A square-integrable function f(x) (odd function) can be represented by the series f ( x ) a n sin nx n 1 where an u ( x, y) 2 f ( s) sin( ns)ds , n 1,2,3,... 0 sin( nx) sinh(( n( y)) 0 f (s) sin( ns)ds . 1 sinh( n ) 2 4 bn an Finite difference methods The heat (diffusion) equation u( x, t ) 2 u ( x, t ) R T D , 0 x R, 0 t T , h , , 2 t x K N n uk u( k h, n ), k 0,1,2,..., K , n 0,1,2,..., N . Boundary conditions u (0, t ) a(t ), u (0, R) b(t ), u0n u (0, n ) a(n ), u Kn u ( R, n ) b(n ), n 0,1,2,..., N . Initial conditions u ( x,0) f ( x), uk0 u (k h, 0) f (k h), k 0,1,2,..., K . Explicit form of finite-difference approximation The simplest difference equations are: ukn1 ukn u n 2ukn ukn1 D k 1 , k 1,2,..., K 1, n 0,1,2,..., N 1. h2 t n+1 n k-1 u kn1 k k+1 x D n 2 D D u k 1 (1 2 )u kn 2 u kn1 . 2 h h h (1) This explicit form of the finite-difference approximation is stable iff D 1 / 2. h2 D 1 / 2 then the method (1) is unstable. h2 Let u ( x,0) eisx and we consider the heat equation for infinite spatial case If x . Using Fourie method of separation of variables u( x, t ) eisxet eisxt 5 u kn u (kh, n ) e iskh nt . Substitution into (1) gives e t ukn qu kn e ish (1 2q) cukn e ish , here q D , h2 e t qe ish (1 2q) ce ish , let e 1 2q 2q cos( sh), ukn n 1 4q 1 . If cos( sh ) 1 then for q 1 / 2 1 and 1. It means that u kn grows infinitely when 0 . So, we found that for some particular case the explicit form is unstable if q>1/2. It is possible to prove that for any initial conditions the explicit form for numerical solution of the diffusion equation is stable if q<1/2. Implicit form of finite-difference approximation ukn1 ukn D ukn11 2ukn1 ukn11 , k 1,2,..., K 1, n 0,1,2,..., N 1. h2 t n+1 n k-1 k k+1 x D n1 2 D D uk 1 (1 2 )ukn1 2 ukn11 ukn , k 1,2,..., K 1. 2 h h h Solving this system of linear equations sequentially for n=1,2,…N-1 we will find solution. An advantage of the implicit form is that there are no constraints on the values of h and . 6 The Wave equation 2 2 u ( x, t ) R T 2 u ( x, t ) c , 0 x R, 0 t T , h , , 2 2 t x K N n u k u (k h, n ), k 0,1,..., K , n 0,1,2,..., N . Boundary conditions u (0, t ) a(t ), u (0, R) b(t ), u0n u (0, n ) a(n ), u Kn u ( R, n ) b(n ), n 0,1,2,..., N . Initial conditions u ( x,0) u ( x,0) f ( x), g ( x), t u k0 u k1 0 u k u (k h, 0) f (k h), g (k h) u k1 u k0 g (k h), k 0,1,2,..., K . Explicit form of finite-difference approximation The simplest difference equations are: n n n ukn 1 2ukn ukn 1 2 uk 1 2uk uk 1 c , k 1,2,..., L 1, n 0,1,2,..., N 1. 2 h2 t n+1 n n-1 k-1 u n 1 k k k+1 x c 2 n 2 u k 1 2u kn u kn1 2u kn u kn1 h ukn1 qukn1 (2 2q)ukn qukn1 ukn1 , q c 2 2 , k 1,2,..., K 1, n 0,1,..., N 1, h2 q 1 / 2 - stability!!! 7 Potential Equation 2u 2u 0. x 2 y 2 ukn1 2ukn ukn1 ukn1 2ukn ukn1 0. h2 h2 1 u kn (u kn1 u kn1 u kn1 u kn1 ) . 4 8