PDE Notes. Paul Carnig. January ODE s vs PDE s 1

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1 PDE Notes Pul Crnig Jnury 2014 Contents 1 ODE s vs PDE s 1 2 Section 1.2 Het diffusion Eqution Fourier s w of Het Conduction Energy Conservtion Idelistions Infinitesiml Approch: Integrl Approch: Higher Dimensions of the Het Eqution: Section Wve Phenomenon: Section Potentil Eqution Section Clssifiction of PDE s Generl Solution PDE Clssifictions Section 3.3 Introduction to Fourier Series 7 7 Section 3.4 Determining Fourier Coefficients Period ODE s vs PDE s ODE s ODE s re Ordinry Differentil Equtions. Exmple 1

2 d 2 u dt 2 = t2 + cost here u = ut which only hs one independent vrible, t. PDE s PDE s re Prtil Differentil Equtions. Exmple 2 u t 2 = 2 u x 2 here u = ut, x which hs two independent vribles t nd x. PDE s must hve t lest two independent vribles. 2 Section 1.2 Het diffusion Eqution Flux: Consider the flow of physicl quntity mss, het, energy. The flux q x, t of this flow is vector in the direction of the flow t x, t whose mgnitude is the mount of quntity crossing unit re t x norml i.e. orthogonl to the norml flow in unit time. So t = time nd x = So q x, t = lim s 0 t 0 x y z Quntity pssing through s in time[t, t] s t Here s is smll surfce re t x tht is norml to the flow. So, the quntity pssing through s in time t is Q x, t, s, t = q x, t s t Suppose wter is flowing in river with velocity v x, t et s be smll surfce re norml to v x, t Then the mount of the wter flowing through s in time [t, t + ] is: Mss = Q x, t, s, t = ρ v s t where ρ = density so q x, t = ρ v A chnge Q in the mount of het in body of mss is ccompnied by chnge u of temperture. The reltionship is: Q = cm u where m = mss nd c is the specific het of the mteril. Normlize to Qcmu so Q nd u re proportionl. Het is trnsported in the direction opposite to the temperture grdient t rte proportionl to it. So the het flux t q x, t is q x, t = κ u x, t where κ is the therml conductivity. 2

3 2.1 Fourier s w of Het Conduction q x, t = κ u x u y u z 2.2 Energy Conservtion The rte of chnge of het = rte het in-rte het out 2.3 Idelistions First ssume we re working with thin rod. 1 Rod is homogeneous i.e. c, κ nd ρ re ll constnt 2 ength of the rod is constnt i.e. no expnding or contrcting 3 Rod is perfectly insulted so het flows only horizontlly 4 Het flows from left to right, so the left side is wrmer thn the right side. Now tht we hve the proper frmework we cn derive the het eqution with 2 pproches. 2.4 Infinitesiml Approch: First let use strt with the following: qx, ta qx + xa u = temperture = ux, t volume = A x mss = ρa x Now the mount of het is Qx, t, x = cρa xu dq dt = cρa xu t By our conservtion lw we get dq dt cρa xu t = A[qx + x, t qx, t] cρ xu t = [qx + x, t qx, t] = qx, ta qx + x, ta [qx + x, t qx, t] cρu t = here let x 0 x cρu t = q x 3

4 q = κu x so q x = κu xx cρu t = κu xx Now let the 1 k = cρ κ which is the therml diffusivity This give us 1 k u t = u xx which is the 1D het eqution. 2.5 Integrl Approch: Consider portion of the rod between x = nd x = b. So the totl mount of het in this section of the rod cn be given by: Qt,, b = dq dt = cρaux, tdx cρa du dt dx now by the conservtion lw we hve : dq dt = dq = Aq, t Aqb, t dt cρa du dt dx = A dq dx dx Now recll Fourier s lw tht sttes q = κ du dx dq dx = κu xx cρau t dx = A κu xx dx cρau t dx Aκu xx dx = 0 cρau t Aκu xx dx = 0 A dq dx dx cρau t Aκu xx = 0 which is wht we used the Infinitesiml Approch. 2.6 Higher Dimensions of the Het Eqution: 1D: 1 k u t = u xx 2D: 1 k u t = u xx + u yy 3D: 1 k u t = u xx + u yy + u zz ny D: 1 k u t = 2 u where 2 u is the plcin of u i.e. u x0 x u xn x n 4

5 3 Section Wve Phenomenon: On guitr string See pge 11 for criteri. On curve let T 1 be the tngent force 1 nd α 1 be the ngle of T 1 nd the sme with T 2 nd β let b nd let u 1 so we get tht s x nd the ccelertion of the string given by u tt Newton s second lw sttes tht F = m Now the sum of the verticl forces cting on the segment of the string is Tsinβ-Tsinα = F where the mss is the liner density given by; mss = ρ s So ρ su tt = Tsinβ-Tsinα with smll slopes sinα tnα nd the sme for β. tnα = u t x, t tnβ = u t x + x, t so ρ su tt = Tu x x + x, t u x x, t ρu tt = T u xx + x, t u x x, t note tht s x s ρu tt = T u xx + x, t u x x, t let x 0 x ρu tt =Tu xx Now let c 2 = T ρ 1 c 2 u tt = u xx Which is the eqution for the 1D sitution. 4 Section Potentil Eqution Recll tht the 2D Het Eqution is: 1 u k t = 2 u x u y 2 If we hve stedy stte temperture distribution then the temperture does not chnge with respect to time. So in the cse of the 2D Het Eqution we hve tht u t = 0. From which we get: Which is the 2D Potentil Eqution. 2 u x u y 2 = 0 5 Section Clssifiction of PDE s 1. Order The order of the PDE is the order of its highest derivtive. 5

6 Exmple 1. u t u xx = 0 is second order PDE. Exmple 2. u t 2 u = 0 is first order PDE. 2. inerity A PDE in u is liner iff it nd its derivtives re liner in u. Exmple 1. u t u xx = cosx is liner. Exmple 2. sinu+u t =0 is non liner Exmple 3. uu t = 0 is non liner. 3. Homogeneous A PDE is homogeneous homog iff the terms not contining u or its derivtives sum to 0. Exmple 1. u t u xx = 0 is homogeneous. Exmple 2. u t u xx = tx is non-homogeneous. Exmple 3. u t u xx = 1 is non-homogeneous. Theorem 1. If u 1 nd u 2 re both solution of liner homogeneous PDE then so is: w = c 1 u 1 + c 2 u 2 where c 1 nd c 2 re both constnts. Exmple et u 1 nd u 2 both stisfy u t u xx = 0 We see here tht both u 1 nd u 2 re liner nd homogeneous nd becuse they re both solution to the PDE we get: u 1 t u 1 xx = 0 nd u 2 t u 2 xx = 0 et w = c 1 u 1 + c 2 u 2 Now tking its derivtives we get: w t = c 1 u 1 t + c 2 u 2 t nd w xx = c 1 u 1 xx + c 2 u 2 xx Now consider w t w xx. We will be done when we show tht w t w xx = 0 6

7 w t w xx = c 1 u 1 t + c 2 u 2 t [c 1 u 1 xx + c 2 u 2 xx ] w t w xx = c 1 [u 1 t c 1 u 1 xx ] + c 2 [u 2 t c 2 u 2 xx ] w t w xx = c c 2 0 w t w xx = 0 Theorem 2. If u 1 nd u 2 re both solution of liner non-homogeneous PDE then u 1 -u 2 is solution of the corresponding liner homogeneous PDE. Exmple et the non-homogeneous liner PDE u t u xx = f x, t cll it 1. Now let u 1 nd u 2 be solutions to 1. so we hve: u 1 t u 1 xx = f x, t nd u 2 t u 2 xx = f x, t We will be done when we show tht u 1 u 2 t u 1 u 2 xx = 0 now tke u 1 u 2 t u 1 u 2 xx = [u 1 t u 1 xx ] [u 2 t u 2 xx ] [u 1 t u 1 xx ] [u 2 t u 2 xx ] = f x, t f x, t = Generl Solution Now to find the generl solution to PDE you dd the homogeneous solution to the nonhomogeneous solution i.e. if w = u 1 u 2 then u 1 = w + u PDE Clssifictions Consider the PDE 11 u xx + 12 u xy + 22 u yy +[OT]= f x, y note tht [OT]= lower order terms. et = if < 0 then we sy the PDE is Elliptic. if = 0 then we sy the PDE is Prbolic if > 0 then we sy the PDE is Hyperbolic 7

8 Exmple 1. 1D Het Eqution 1 k u t = u xx 1 k u t u xx = 0 so we hve 11 = 1, 12 = 0 nd 22 = 0 so we hve =0 so the 1D Het Eqution is Prbolic Exmple 2. 1D Wve Eqution 1 c 2 u tt = u xx 1 c 2 u tt u xx = 0 so we hve 11 = 1, 12 = 0 nd 22 = 1 c 2 so we hve =0 = 0 1 c 2 = 1 so the 1D Wve Eqution is Hyperbolic c2 Exmple 3. 2D Potentil Eqution u xx + u yy = 0 so 11 = 1, 12 = 0 nd 22 = 1 so = -1. So the 2D Potentil Eqution is Elliptic. 6 Section 3.3 Introduction to Fourier Series Given function f x we wnt to express in the form: f x = n cos nπx So the gol is to find the n nd b n terms, given. Definition: Orthogonlity + b nsin nπx which is the Fourier Series. et f x nd gx both be not identicl 0 nd f x = gx. Now let f x, gx, nd wx be integrble on the intervl, b nd wx>0. Now if we hve: f xgxwx dx = 0 then we sy tht f x nd gx re orthogonl with respect to w on the intervl [, b]. In Fourier Series we hve wx = 1 nd the intervl [, b] is [, ] Fct 8

9 The set of functions {1, sin nπx nπx, cos } is orthogonl. To prove this we need to show the following Proof of 1 sin nπx dx = 0 2 sin nπx cos nπx mπx cos dx = 0 4 mπx cos dx = 0 cos nπx dx = 0 sin nπx mπx sin dx = 0 sin nπx dx = nπx cos nπ = cosnπ cos nπ note here tht the cos nπ is even so we get: nπ cosnπ cosnπ = nπ nπ 0 = 0 Definition: Even Function f x is n even function provided f x = f x x domx Even functions re symmetric to the y-xis for y = f x Even functions lso hve the property tht if f x is n even function then: f xdx = 2 f xdx 0 Definition: Odd Function f x is n odd function provided f x = f x x domx Odd functions re symmetric bout the origin. Odd functions lso hve the property tht if f x is n odd function then: f xdx = 0 7 Section 3.4 Determining Fourier Coefficients Recll tht: 9

10 f x = n cos nπx + b nsin nπx is the Fourier Series. First find 0. 0 = = 0 f xdx = 2 dx + 0 n = 0 2 dx + Here we should note tht n 2 + n cos nπx + b nsin nπx n cos nπx dx + cos nπx dx + b n b n sin nπx dx sin nπx dx cos nπx dx nd b n sin nπx dx re equl to zero due to them being orthogonl on [, ] with respect to wx where wx = 1. Now we hve 0 = = 0 2 Thus 0 = 0 2 dx + 0 dx = = 0 f x dx Now tht we hve 0 to find the other terms consider: f xcos mπx dx = n cos nπx 0 mπx cos 2 + n cos nπx cos mπx dx + b nsin nπx + b nsin nπx cos mπx dx mπx cos dx Here note tht just like in the cse where we found 0 ll of the terms will go to zero due to orthogonlity except for when n = m. Which gives us: m cos 2 mπx dx = m so m = f xcos mπx nlogously we get b m = dx f xsin mπx dx 7.1 Period The Fourier series is helpful when f is periodic with period 2. 10

11 Exmple: sinx sinx is periodic with its period equl to 2π i.e. sinx = sinx 2π = sinx 4π nd so on. so we cn sy tht sinx = sinx p cll it 1 where p is the smllest positive number tht mkes 1 true. So here we hve 2 = 2π giving us = π in this exmple. The point is tht it is importnt to identify correctly Exmple: Find the Fourier Series for f x given its grph Clerly f is periodic with period 10 which gives us =5 nd our f x is defined s: f x = { 3 if n < x < 10n 3 if 10n < x < n for n Z 11

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