APM346H1 Differential Equations. = u x, u = u. y, and u x, y =?. = 2 u t and u xx= 2 u. x,t, where u t. x, y, z,t u zz. x, y, z,t u yy.

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1 INTRODUCTION Types of Prtil Differentil Equtions Trnsport eqution: u x x, yu y x, y=, where u x = u x, u = u y, nd u x, y=?. y Shockwve eqution: u x x, yu x, yu y x, y=. The virting string eqution: u tt x,t=c u xx x,t, where u t = u t nd u xx= u x. The wve eqution: u tt x, y, z,t=c u xx x, y, z,tu yy x, y, z,tu zz x, y, z,t. In generl: u tt x,, x n,t=c u x,, x n,t, where =the plcin= u= u x u x n. Diffusion eqution: u t x,t=c u xx x,t. In generl: u t x,, x n,t=c u x,, x n,t. Stedy stte: u t =. plcin eqution: u= u x u x =. n x x n nd Initil Conditions nd Boundry Vlues for Ordinry Differentil Equtions Consider d y dy =F t, y, dt dt, nd think of yt s the position of the prticle, d y s ccelertion, nd F t, y, dy dt dt s force. The stte/configurtion spce is x t, x t, where x t= yt, x t= dy. Then the system of first order dt dx = dy dt dt =x t equtions is dx dt = d. y dy =F t, y t, dt dt =F t, x t, x t Theorem: Existence nd Uniqueness of Solution There exists one nd only one solution xt= x t,, x n t tht stisfies xt =x t where x t is the given intil condition. Qusi-iner Prtil Differentil Equtions Definition: Qusi-iner Prtil Differentil Eqution x, y, uu x x, y x, y,uu y x, y=cx, y, u (*) where,, c re given functions. Clim et nd e constnt functions, nd c=, so u x u y = (). Then every solution u x, y of () is of the form u x, y= f x y for some function of one vrile (ex: f = u x, y=x y ). of 5

2 Uniqueness nd Initil Conditions For initil condition, we prescrie u long given curve x, so u x, x=u x is given. Note tht when u x, y= f x y, u x, y is constnt long the line x y=c. So if u x= f x x, there is unique f provided tht x x=c is not constnt. Suppose tht x=ax x. Then u x= f x Ax f x=u. In conclusion, A ) The solution u x, y is unique for ny u x over the line y=ax provided tht A. ) When A= then there re infinitely mny solutions provided tht u x is constnt. If u x is not constnt, then there re no solutions. Method of Chrcteristic Define vector field V x, y, z= x, y, z, x, y, z, c x, y, z. Norml direction t x, y, z=u x, y is n=u x x, y,u y x, y,, ut V n=u x u y c = ecuse u x u y =c. So V lies in the tngent plne. dx =xt, yt, z t dt dy dt dz If xt, yt, zt is solution of () = xt, yt, z t such tht x, y, z lies in z=ux, y, ie =c xt, yt, z t dt u x, y=z, then xt, yt, z t lies in z t=u xt, yt. x, x =x x t, s=xt, x s Suppose now tht xt, x, yt, y, z t, z is ny solution of () such tht y, y = y where yt, s= y t, y s. In z, z =z z t, s=zt, z s most situtions, we cn solve for t nd s in terms of x nd y. Then u x, y=zt x, y, s x, y. x x t s Note: When the Jcoin J =det[, then we cn solve for t nd s in terms of x nd y loclly. y t y s ] Note: If J =, then if u x, y =z tht contins u x s, y s=z s stisfies dz s =c x ds s, y s, z s, there re infinitely mny solutions; if not, then there is no solution. Second Order Equtions x, y u xx x, y u xy c x, y u yy d x, y u x e x, y u y f x, y u= (), where,, c, d, e, f re given functions. Cnonicl Types. Hyperolic type: c.. Prolic type: c=. 3. Elliptic type: c. Fct If we mke (one-to-one) chnge in vriles = x, y = x, y nd require tht det [ x y x y] x et y xi y x, then there is trnsformtion such tht () is trnsformed into:. u lower order terms= in the hyperolic type; of 5

3 . u lower order terms= in the prolic type; 3. u u lower order terms= in the elliptic type; Specil Cse:,, c constnts iner chnge of coordintes x, y, given y = x y = x y such tht det[ x xi y x y ] =det [ x xi y x y ] =. Then () ecomes Au Bu Cu lower order terms, where A= c B= c C= c.. In the hyperolic cse, choose = c,= c, ==, = c, ==. Then A=C =, B.. In the prolic cse, choose ==,==. Then B=C=, A or A=B=,C. c 3. In the elliptic cse, choose = c,= c c, =,=, then A=C, B=. THE WAVE EQUATION u tt x,t =c u xx x,t, x with initil conditions u x,= x,u t x,= x. The solution is u x,t = xct x ct x ct c xct z dz. DIFFUSION EQUATION u t x, y, z,t =k u=k u xx u yy u zz. In one dimension, u t x,t =k u xx x,t is prolic type. In One Dimension u t x,t =k u xx x,t, x with given initil conditions u x,= x where x is given function. The solution is u x,t = 4 k t x y 4 kt y e dy. If S x,t = x 4 kt e 4 kt,t, then u x,t = y S x y,t dy. Properties of the Kernel The het kernel/gussin/diffusion kernel S x,t hs the following properties:. Symmetric: S x,t =S x,t.. lim S x,t ={, x=. t, x lim t S x,t dx=, t. x S x,t dx=,. 3 of 5

4 Evlution Techniques Useful formul: ' y S x y,t dy= kt [ If =,'=, then y S x y,t dy=x. So if x =x, then u x,t = y y S x y,t dy x y S x y,t dy]. y S x y,t dy= y S x y,t dy=x nd u x,=x= x. If = y, then y S x y,t dy=x kt. So if x =x, then u x,t =x kt nd u x,=x = x. If = y, then y 3 S x y,t dy=x 3 6 ktx. So if x =x 3, then u x,t =x 3 6 ktx nd u x,=x 3 = x. Theorem Suppose tht x is such tht lim x e x = x y S x y,t dy is solution with u x,= x., then lim t y S x y,t dy= x, x. In tht sense The Mximum Principle et u x,t e solution of u t =k u xx on rectngle x l, t T. The mximum of u x,t occurs only on the prt of the oundry { x,: x l } {,t : t T } {l,t : t T }. Theorem: Uniqueness of Solution Suppose tht we seek solution u x,t tht stisfies u x,= x, x l. Suppose further tht u x,t stisfies u,t = t nd u l,t = t, where t nd t re prescried functions. Then the solution is unique, i.e. there is t most one solution. T l DIFFUSION EQUATION ON HAF INE Eqution: u t x,t =k u xx x,t,x. Initil dt: u x,= x, x. Boundry conditions: Dirichlet Condition: prescrie u,t = t (usully t = ). Neumnn Condition: prescrie u x,t = t (usully t = ). Roin Condition: prescrie u,t u x,t=. 4 of 5

5 Method of Solution: Dirichlet Boundry Condition Tke the cse with u t x,t =k u xx x,t, x, u x,= x, x, u,t =, t. We wnt to extend to the entire line x such tht the solution u x,t induced y this extension stisfies u,t =. Note tht x = x, x. Now, u,t = for ll t iff is n odd function ( x = x ). Then u x,t = y S x y,t dy= y S x y,t S x y,t dy. Method of Solution: Neumnn Boundry Condition Solve u t =k u xx, x, with initil dt u x,= x, x nd Neumnn condition u x,t=. If u x,t is even (i.e. u x,t = u x,t ), then u x x,t is odd (i.e. u x x,t = u x x,t ). The solution is u x,t = y S x y,t dy= y S x y,t S x y,t dy. WAVE EQUATION ON HAF INE Solve u tt =c u xx, x. Initil dt u x,= x, nd u t x,= for simplicity. Dirichlet Boundry Condition Dirichlet condition u,t =. Extend to odd function. Then the solution is u x,t = xct x ct. Note: u x,t = u x,t u,t =. WAVE EQUATION ON FINITE INTERVA Solve: u tt =c u xx,x. Initil dt: x =u x,,x nd x =u t x,,x. Dirichlet Boundry Condition Dirichlet condition: u,t =u,t =. Extend to nd to so tht u x,t is odd out x= (i.e. u x,t = u x,t u x,t = u x,t ). Then the solution is u x,t = xct x ct x ct c xct ) nd odd out x= (i.e. z dz. Seprtion of Vriles nd Boundry Vlue Prolems Method of Seprtion of Vriles The method of seprtion of vriles ssumes tht ny solution u x,t cn e written s u x,t =X x T t. Solutions With the diffusion or wve eqution, we need to solve X ' ' X =, where is n unknown constnt: 5 of 5

6 For, X x =Acos x Bsin x. For, X x =Acosh xbsinh x. For =, X x =AxB. In the Dirichlet cse ( u,t =u,t = ), nd n = n. So X n x =sin n x. In the Neumnn cse ( u x,t =u x,t = ), = so X x =constnt ; or nd n = n, so X n x =cos n x. Dirichlet Boundry Condition For the wve eqution u tt x,t =c u xx x,t, we hve T n t = n cos cn u n x,t =X n x T n t =[ ncos cn nsin t cn ] t sin n x. For the diffusion eqution u t x,t =k u xx x,t, we hve T n t =c n e k n t. So u n x,t =X n x T n t =c n e k n t sin n x. nsin t cn t. So Neumnn Boundry Condition Wve eqution: u n x,t =[ ncos cn Diffusion eqution: u n x,t =c n e k n t nsin cn t ] cos n x. t cos n x. Mixed Boundry Condition Mixed oundry condition u,t =u x,t =, then X =X ' =. n We hve n =. Roin Condition Tke u x,t h u,t = nd u x,t =. We hve X ' ' X =. Assume. Then X x =Acos xb sin x, nd we get tn = h. Setting y= tn y= h y = c y trnscendentl eqution. On y, we get infinitely mny solutions y y, with the difference pproching. Assume. Then X x =Acosh xb sinh x, nd setting y= we get tnh y= c y trnscendentl eqution. We get no solution. So there re infinitely mny eigenvlues with corresponding eigenfunctions X x, X x,., we get VECTOR SPACES: INTRODUCTION TO FOURIER SERIES 6 of 5

7 et V n e the spce of ll liner comintions of f = sin x sin x n sin n x. Define :V n V n f = d f d x = k= n k k sin k. Choose sis: v = { sin =sin x, v n=sin x,, v n x } [ is digonl mtrix since v k = k v k.. Then the mtrix of reltive to this sis ] / n et n nd consider the spce of functions f on x which cn e written s f x = Fourier coefficients k = f x sin k x dx k=,,3, of f reltive to X n. k= k sin k x for FU FOURIER SERIES Definition et x. The full Fourier series of f x is n cos n nsin n.. Coefficients The coefficients re uniquely determined from orthogonlity of functions cos n nd sin n : sin n sin m These reltions imply tht: dx={ n m n=m. sin n cos m dx=. cos n cos m dx={ n m n=m. n = f x cos n dx n=,,,. n = f x sin n dx n=,,,. 7 of 5

8 Reltion To Differentil Equtions Tke x. Dirichlet Condition: Tke f x n odd extension of x. Then n = nd n = f x sin n dx,,. Neumnn Condition: Tke f x n even extension of x. Then n = f x cos n dx n=,,, nd n =. GENERA EIGENVAUES AND EIGENFUNCTIONS X ' ' X = on x.. If X = X =, then n = n nd X n x =sin n x.. If X ' = X ' =, then n = n nd X n x =cos n x. 3. If X ' hx = nd X ' =, then (eigenvlues) nd X, X, (eigenfunctions). 4. If X = X = nd X ' = X ' =, then = nd X x =constnt or n = n nd X n x =A n sin n x B ncos n x where A n nd B n re ritrry constnts. Generl Boundry Conditions Solve X ' ' X =, x suject to the oundry conditions X X 3 X ' 4 X ' = X X 3 X ' 4 X ' = constnts,, n,,, n. for some Definition: Symmetric Boundry Conditions et f nd g e ny functions tht stisfies the ove oundry condition. Then conditions re clled symmetric if f ' x g x f x g ' x x= x= = f ' g f g ' f ' g f g ' =. Fct Conditions to 4 re symmetric. Theorem Suppose tht X n nd X m re eigenfunctions on [, ] tht corresponds to distinct eigenvlues n nd m ( n m ), nd suppose tht the oundry conditions re symmetric. Then X n nd X m re orthogonl in the sense tht X n x X m x dx=. HIBERT SPACE Bsic Spce [, ]={ f : [,]R f x dx}. 8 of 5

9 Fct [, ] is vector spce. Inner Product Tke f nd g in. Then define the inner product to e f, g = f x g x dx. Norm Define f = f x dx = f, f to e the norm of f. Cuchy-Schwrtz Inequlity f, g f g or Note: This implies f f, g f x g x dx f x dx g, so define cos= f g x dx. f, g g. Definition: Convergence { f n } is sid to converge to f if lim f n f = n. Definition: Cuchy Sequence { f n } is clled Cuchy Sequence if f n f m s n, m. Bsic Properties of Inner Product nd Norm. Symmetric: f, g = g, f.. Biliner: f, g h= f, g f,h for, R nd f, g, h. 3. f, f f ; if f, f = f x dx= then f = lmost everywhere. 4. is complete in the sense tht ny Cuchy sequence in converges to n element in. Definition: Hilert Spce Any vector spce H with n inner product, tht stisfies properties to 4 is clled Hilert spce. Theorem If X, X,, X n, re the eigenfunctions corresponding to symmetric oundry prolem, then the Fourier series of ny function f converges to f in norm. EAST SQUARE APPROXIMATION et V n denote the liner spn of X, X,, X n (i.e. f V n f = X X n X n. Prolem 9 of 5

10 Question: et f. For which vlues of,,, n is the distnce f i X i minimum? Answer: i = f, X i X i i=,, n. CONVERGENCE OF FOURIER SERIES Theorem The Fourier series reltive to X, X, of ny element f converges to f. Tht is, if S N = i= lim N S N f =. N f i, X i X i, then Definition: Piecewise Continuous A function is piecewise continuous if it is continuous t ll ut finite numer of points. At point of discontinuity f hs oth right nd left limit (ie f hs jump discontinuity). So if c is point of discontinuity of f, then oth f c + =lim x c xc f x nd f c - =lim x c xc f x exist. Theorem: Point-wise Convergence of Fourier Series Assume f is such tht: f is periodic of period. f nd its derivtive f ' re piecewise continuous. Then lim S N x = n f x+ f x -. Note: If f is continuous t x, then f x + = f x - = f x, so lim S N x = f x n. Auxiliry Results. Bessel's Inequlity: g g, X k where X k= X k, X, re eigenfunctions on [, ] with symmetric oundry vlues nd g [,]. This implies tht lim k. et K N = cos k 3. K N = N k= sin N sin. g, X k X k =. pi. Then K N d = pi K N d =. Definition: Uniform Convergence f n converges to f uniformly if lim mx f n x f x = n x. of 5

11 Hrmonic Functions nd plce's Eqution APACE'S EQUATION In n dimensions, u=div grd u = u x u x =. n Notes: is the plcin, nd it is n opertor tht cts on functions of n vriles. The grdient of sclr function u x,, x n is grd u= u= u x,, u x n. The divergence of vector field V x,, x n =V x,, x n,,v n x,, x n is div V = V x V n x n. When V = u= u x,, u x n, then u=div grd u. Definition: Hrmonic Any solution of plce's eqution is clled hrmonic. Properties : d u d x =, so u x =AxB. n= : Connection to complex functions f z =u x, y iv x, y. Then f nlytic (cn e expressed s Tylor series, i.e. differentile) implies u nd v re hrmonic. MAXIMUM/MINIMUM PRINCIPE Definitions. D is n open suset of R n if for ll x D, there exists r such tht for ll y D, x y r.. D is the oundry of D. A point is oundry point if for ll, B={ x } hs non empty intersection with oth D nd the complement of D in R n. 3. D is connected if there exists polynomil curve joining ny two points in D nd is lying in D. 4. D is ounded if it is contined in some ll B={x x R } R. Mximum/Minimum Principle Assume D to e n open, connected suset of R n such tht D D is ounded. et u e ny solution of (the plce u x u x = eqution) in D such tht u is defined nd continuous in D D. Then: n Mximum Principle: u x mx u x D x D. Minimum Principle: u x min u x D x D. Tht is, u ttins its mximum/minimum on D. BOUNDARY VAUE PROBEMS of 5

12 Dirichlet Prolem u x u x = suject to u D = x (given). n By the Mximum/Minimum Principle, the solution of the Dirichlet prolem is unique. Neumnn Prolem u x u x n = suject to u n D= x (given). Here, n is the externl norml. Roin/Mixed Prolem u x u x = suject to u x u x =k x (given). n n BASIC PROPERTY Solutions to the plce eqution re invrint under rigid motions x =T x R x T x =x is trnsltion, R x =Ax ( A T =A det A=± ) is rottion., where RECTANGUAR HARMONICS x,= f x u x u y = on D={ x, y : x, y } with D={u } u, y =g x where u x,=h x u, y =i x functions. f, g, h, i re given Seprtion of Vriles We ssume u x, y = X x Y y. When D={ u x,= f x u x,=h x When D={ u, y =u, y =}, then u n x, y = ncosh n y nsinh n y sin n x. u, y =g x u, y =i x x,= f x Note: D={u u, y =g x u x,=h x u, y =i x }={ u x,= f x u x,=h x u x,=u x,=}, then u n x, y = n cosh n x n sinh n x sin n y. Note: u x, y = u n x, y. u, y =u, y =} { u, y =g x u, y =i x u x,=u x,=}. Cse Suppose tht we wnt the solution with D= { u x,= f x u x, =u, y =u, y =}. of 5

13 The sine Fourier coefficients of f re n = Since u x,= f z sin n z dz. ncosh n nsinh n sin x= n n cosh n nsinh n =, so n cosh n n = sinh n. Then u x, y = = = ncosh n y nsinh n y sin n x ncosh n n sinh n y sinh y cosh n sinh n sinh n n y sin n x ysin n x. APACE'S EQUATION ON CIRCUAR REGIONS. Annulus: D={r,,, r }, D={,, } {,, }.. Disk: D={r,,, r }, D={,, }. 3. Wedge: D={r,,, r }. Polr Coordintes nd Seprtion of Vriles x=r cos Using polr coordintes y=r sin, u u = xx yy ecomes u rr r u r r u =. Assuming u r,=r r, we get the two equtions r R' 'r R' R= nd ' '=. Annulus Eigenfunctions: When n, n =A n cos nb n sin n nd R n r =C n r n D n r n. When n=, =A n nd R r =c c ln r. So the solution is u r,= n= R n r n =c c ln r C n r n D n r n A n cos nb n sin n. The coefficients re the Fourier coefficients of the oundry conditions u,= f nd u,=g. Disk The usul ssumption is tht u, ounded. This forces c =D n =. So the solution is u r,=c C n r n A n cos nb n sin n= coefficients determined y the oundry condition u,= f. r n n cos n n sin n, where n nd n re the Fourier 3 of 5

14 Wedge Consider the specil cse tht u r,=u r,= nd u,= f. n The solution is u r,= n r sin n, where n = f sin n n/ d is the Fourier coefficient determined y u,= f. POISSON FORMUA AND POISSON KERNE Poisson Formul On the disk D={r,,, r } with u,= f, u r,= [ f = = f P d r f r cos r d r n cos n d ] where P = r n cos n = r r cos r. Poisson Kernel P r,= r r cosr. Bsic Properties. P r, d = u r,= r, lso.. lim P r,={ r =. r d =. In this cse u,= f =, ut r cos r 3. lim f P r, d = f whenever f is continuous function of. r 4. Averging Property of Hrmonic Functions: x= y= r=, so crtesin polr u,=u,= f cos d = f d is the verge vlue of f. Consequences of Poisson Representtion. A hrmonic function u defined on some domin D cnnot ttin mximum (nor minimum) in the interior of D. Here the interior of D re the points p in D such tht there exists disk centered t p tht is entirely contined in D.. u r, hs prtil derivtives of ll orders, even when f is only continuous. 4 of 5

15 u r,= f P r, d u r = f r P r, d. 5 of 5