y =3x2 y 2 x 5 siny x y =6xy2 5x 4 siny

Transcription

1 Skeibare PDVs/ Separable PDEs SLIDE 1/9 Parsiële afgeleides f ( x, y) Partial derivatives y f y x f x Voorbeeld/Example:f(x,y)=x 2 y 3 +x 5 cosy f x =2xy3 +5x 4 cosy; f y =3x2 y 2 x 5 siny 2 f x 2=2y3 +20x 3 cosy; 2 f x y =6xy2 5x 4 siny

2 Skeibare PDVs/ Separable PDEs SLIDE 2/9 Hfst 13: Parsiële DVs Chapter 13: Partial DEs Gewone DVs het net een onafhanklike veranderlike, en het oplossings soos/ Ordinary DEs have one independent variable only, and have solutions like y=y(x) en/and y=y(t) Parsiële DVs het meer as 1 onafhanklike veranderlike, en het oplossings soos Partial DEs have more than 1 independent variable only, and have solutions like y=y(x,t) en/and u=u(x,y,z) 13.1: Skeibare PDVs/ Separable PDEs(p 689) Meesalgemenelineêre2deordePDV/Mostgenerallinear2ndorderPDE: A 2 u x 2+B 2 u x y +C 2 u y 2+D u x +E u y +Fu=G Die koëffisiënte, A=A(x,y),B=B(x,y),...,G=G(x,y), is almal gegee (dikwelsisa,b,...,gkonstant)/thecoefficients,a=a(x,y),b=b(x,y),...,g=g(x,y),areallgiven(oftena,b,...,gareconstant) Homogeen as/homogeneous if G = 0; Nie-homogeen as/non-homog s if G 0 Onswildusdieoplossingvind/Wethereforewanttofindthesolution u=u(x,y)

3 Skeibare PDVs/ Separable PDEs SLIDE 3/9 Voorbeelde Examples Hitte- of diffusievergelyking(lineêr) Heat or diffusion equation(linear) u t =c2 2 u x 2 (c 2 diffusie-konstante/diffusionconstant) Golfvergelyking(lineêr) 2 u t 2 =c2 2 u x 2 (c konstante(spoed)/constant(speed)) Wave equation(linear) Laplace se vergelyking(lineêr) Burger se vergelyking(nie-lineêr) 2 u u x 2+ 2 y =0 2 u t +u u x =0 Laplace s equation(linear) Burger s equation(non-linear)

4 Skeibare PDVs/ Separable PDEs SLIDE 4/9 Klassifikasie van PDVs/ Classification of PDEs Hiperbolies as B 2 4AC>0 Parabolies as B 2 4AC=0 Ellipties as B 2 4AC<0 Hyperbolic if B 2 4AC>0 Parabolic if B 2 4AC=0 Elliptic if B 2 4AC<0 Toon nou aan dat die hittevergelyking parabolies, die golfvergelyking hiperbolies en Laplace se vergelyking ellipties is / Now show that the heat equation is parabolic, that the wave equation is hyperbolic, and that Laplace s equation is elliptic Produk-oplossings, skeiding van veranderlikes/ Product solutions, separation of variables Soekoplossingsvandievorm/Seeksolutionsoftheform: u(x,y)=x(x)y(y) u x = X (x)y(y) 2 u x 2 = X (x)y(y) u y = X(x)Y (y) 2 u y 2 = X(x)Y (y) 2 u x y = X (x)y (y) StelinPDV skeipdvin2gewonedvs,eeninxeneeniny! /Substitute intopde separatepdeintotwoordinarydes,oneinx andoneiny!

5 Skeibare PDVs/ Separable PDEs SLIDE 5/9 Voorbeeld 1/Example 1: (p 692, Nr 4) Gebruik skeiding van veranderlikes en vind (indien moontlik) produkoplossings vir die PDV / Use separation of variables and find(if possible) product solutions for the PDE: u x =u y +u Soekoplossingsvandievorm/Seeksolutionsoftheform: u(x,y)=x(x)y(y) u x =X Y en/and u y =XY X Y =XY +XY DeelmetXY /DividebyXY: X (x) (y) X(x) =Y Y(y) +1 (1) LetopdatdieLKvan(1)netafhanklikisvanx,endatdieRKvan(1)net afhanklik is van y. Vgl (1) moet egter geldig wees vir alle x en y. Dit is net moontlik as die LK en RKgelyk aan n konstante, sê λ, is! Ons noem λdieskeidingskonstante. /NotethattheLHSof(1)isonlydependantonx,and thattherhsof(1)isonlydependantony. Eqn(1)musthoweverbevalidforallx andy. ThisisonlypossibleiftheLHSandRHSequalsaconstant,sayλ! Wecallλ the separation constant.

6 Skeibare PDVs/ Separable PDEs SLIDE 6/9 X Y =λ en/and X Y +1=λ Ons het dus 1 PDV met 2 gewone DVs vervang! / Wehavethusreplaced1 PDEwith2ordinaryDEs! Oplossingvir/Solutionfor X X =λ: X(x)=ceλx Oplossingvir/Solutionfor Y +1=λ: Y(y)=de(λ 1)y Y Die produkoplossing word dus gegee deur / The product solution is therefore given by u(x,y)=x(x)y(y)=ae λ(x+y) y, vir alle konstantes/for all constants a en/and λ. ) Bevestig!/Verify! u x =λ (ae λ(x+y) y =λu u y =(λ 1)u=λu u RK/RHS=u y +u=(λu u)+u=λu=u x =LK/LHS Produkoplossing bevredig die PDV!/ Product solution satisfies the PDE!

7 Skeibare PDVs/ Separable PDEs SLIDE 7/9 Voorbeeld 2/Example 2: Gebruik skeiding van veranderlikes en vind (indien moontlik) produkoplossings vir die PDV/ Use separation of variables and find(if possible) product solutions for the PDE: u xx +u yy =0 Soekoplossingsvandievorm/Seeksolutionsoftheform: u(x,y)=x(x)y(y) u xx =X Y en/and u yy =XY X Y +XY =0 DeelmetXY /DividebyXY: X X +Y Y =0 X (x) (y) X(x) = Y Y(y) Let op dat die LK van (2) net afhanklik is van x, en dat die RK van (2) net afhanklik is van y. Vgl (2) moet egter geldig wees vir alle x en y. Dit is net moontlik as die LK en RK gelyk aan n konstante, sê ±λ 2, is! (Die kwadraatisgerieflik.) /NotethattheLHSof(2)isonlydependantonx,andthat therhsof(2)isonlydependantony. Eqn(2)musthoweverbevalidforallxandy. ThisisonlypossibleiftheLHSandRHSequalsaconstant,say±λ 2! (The squared is convenient.) (2)

8 Skeibare PDVs/ Separable PDEs SLIDE 8/9 Vir skeidingskonstante/for separation constant λ = 0 X =0 en/and Y =0 X=ax+b en/and Y =cy+d u(x,y)=(ax+b)(cy+d) Bevestig nou dat dit die PDV bevredig/now verify that this satisfies the PDE Virskeidingskonstante/Forseparationconstant +λ 2 X =+λ 2 X en/and Y = λ 2 Y X=acosh(λx)+bsinh(λx) en/and Y =ccos(λy)+dsin(λy) dus.../therefore... u(x,y)=[acosh(λx)+bsinh(λx)][ccos(λy)+dsin(λy)] Bevestig nou dat dit die PDV bevredig/now verify that this satisfies the PDE

9 Skeibare PDVs/ Separable PDEs SLIDE 9/9 Virskeidingskonstante/Forseparationconstant λ 2 X = λ 2 X en/and Y =+λ 2 Y X=acos(λx)+bsin(λx) en/and Y =ccosh(λy)+dsinh(λy) dus.../therefore... u(x,y)=[acos(λx)+bsin(λx)][ccosh(λy)+dsinh(λy)] Bevestig nou dat dit die PDV bevredig/now verify that this satisfies the PDE Die produkoplossing bevredig dus die PDV ongeag van watter reële skeidingskonstante gekies word / The product solution therefore satisfies the PDE irrespective of which real separation constant is chosen Let op: Sommige lineêre PDVs is nie skeibaar nie, bv / Note: Certainlinear PDEs are not separable, e.g. u xx u y =x Die aanname van u(x,y) = X(x)Y(y) lei nie na n oplossing nie! / The assumptionofu(x,y)=x(x)y(y)doesnotleadtoasolution!