# Partial Di erential Equations

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1 Partial Di eretial Equatios Partial Di eretial Equatios Much of moder sciece, egieerig, ad mathematics is based o the study of partial di eretial equatios, where a partial di eretial equatio is a equatio ivolvig partial derivatives which implicitly de es a fuctio of or more variables. For example, if u (x; t) is the temperature of a metal bar at a distace x from the iitial ed of the bar, the uder suitable coditios u (x; t) is a solutio to the heat = u where k is a costat. As aother example, cosider that if u (x; t) is the displacemet of a strig a time t; the the vibratio of the strig is likely to satisfy the oe dimesioal wave equatio for a costat, which = u (1) Whe a partial di eretial equatio occurs i a applicatio, our goal is usually that of solvig the equatio, where a give fuctio is a solutio of a partial di eretial equatio if it is implicitly de ed by that equatio. That is, a solutio is a fuctio that satis es the equatio. EXAMPLE 1 Show that if a is a costat, the u (x; y) = si (at) cos (x) is a solutio = u () Solutio: Sice a is costat, the partials with respect to t = a cos (at) cos (x) = a si (at) si (x) (3) Moreover, u x = si (at) si (x) ad u xx = si (at) cos (x) ; so that u = a si (at) cos (x) (4) 1

3 The product of X (x) ad T (t) is the separated solutio of the partial di eretial equatio. EXAMPLE 3 For k costat, d the separated solutio to the Heat = u Solutio: To do so, we substitute u (x; t) = X (x) T (t) ito the equatio @ (X (x) T (t)) = k (X (x) T Sice X (x) does ot deped o t; ad sice T (t) does ot deped o x; we obtai T (t) = kt X (x) which after evaluatig the derivatives simpli es to X (x) T 0 (t) = kt (t) X 00 (x) To separate the variables, we divide throughout by kx (x) T (t): This i tur simpli es to X (x) T 0 (t) kx (x) T (t) = kt (t) X00 (x) kx (x) T (t) T 0 (t) kt (t) = X00 (x) X (x) Thus, there is a costat such that T 0 kt = ad X 00 X = These i tur reduce to the di eretial equatios T 0 = kt ad X 00 = X The solutio to the rst is a expoetial fuctio of the form T (t) = P e kt If > 0; however, the temperature would grow to 1; which is ot physically possible. Thus, we assume that is egative, which is to say that =! for some umber!: As a result, we have X 00 =! X or X 00 +! X = 0 3

4 The equatio X 00 +! X = 0 is a harmoic oscillator, which has a solutio X (x) = A cos (!x) + B si (!x) Cosequetly, the separated solutio for the heat equatio is u (x; t) = X (x) T (t) = P e! kt (A cos (!x) + B si (!x)) It is importat to ote that i geeral a separated solutio to a partial di eretial equatio is ot the oly solutio or form of a solutio. Ideed, i the exercises, we will show that u (x; t) = 1 p kt e x =(4kt) is also a solutio to the heat equatio i example 3. As a simpler example, cosider that F (x; y) = y partial di eretial equatio F x + xf y = 0 This is because substitutig F x = x is a solutio to the x ad F y = 1 ito the equatio yields F x + xf y = x + x 1 = 0 Now let s obtai a di eret solutio by assumig a separated solutio of the form F (x; y) = X (x) Y (y) : EXAMPLE 4 Fid the separated solutio to F x + xf y = 0: Solutio: The separated form F (x; y) = X (x) Y (y) results i which i tur (X (x) Y (y)) + (X (x) Y (y)) = X 0 (x) Y (y) = xx (x) Y 0 (y) Dividig both sides by X (x) Y (y) leads to X 0 (x) xx (x) = Y 0 (y) Y (y) However, a fuctio of x ca be equal to a fuctio of y for all x ad y oly if both fuctios are costat. Thus, there is a costat such that X 0 (x) xx (x) = ad Y 0 (y) Y (y) = 4

5 It follows that Y 0 (y) = Y (y) ; which implies that Y (y) = C 1 e y : However, X 0 (x) = xx (x), so that separatio of variables yields dx dx = xx =) dx X = xdx Thus, R dx=x = R xdx; which yields l jxj = x + C jxj = e x +C X (x) = e C e x Thus, if we let C 3 = e C ; the Y (y) = C 3 exp x = ad the separated solutio is F (x; y) = Ce x e y = Ce (y x ) where C = C 1 C 3 is a arbitrary costat. Notice that there are similarities betwee the separated solutio F (x; y) = Ce (y x ) ad the other solutio we stated earlier, F (x; y) = y solutios are clearly ot the same. x : However, the two Check your Readig: Why is this method called separatio of variables? Boudary Coditios Partial di eretial equatios ofte occur with boudary coditios, which are costraits o the solutio at di eret poits i space. To illustrate how boudary coditios arise i applicatios, let us suppose that u (x; t) is the displacemet at x i [0; l] of a strig of legth l at time t: 5

6 Tesio o a short sectio of the strig over the iterval [x; x + x] is alog the tagets to the edpoits, y = u(x,t) x x+ x dy (x) = u x (x,t) dx x x+ x dy (x+ x) = u x (x+ x,t) dx Thus, the et tesio resposible for pullig the strig toward the x-axis is proportioal to the di erece i the slopes, Net T esio = k ( u x (x + x; t) u x (x; t) ) where k is the tesio costat (see for details). Cosequetly, if is the mass-desity of the strig (mass per uit legth), the mass times acceleratio equal to the force of tesio yields = k ( u x (x + x; t) u x (x; t) ) for arbitrarily small x: Solvig for u tt ad lettig x approach 0 = k lim x!0 u x (x + x; t) u x (x; t) x = k so that if we let a = k=; the the partial di eretial equatio describig the motio of the strig = u (5) which is the oe-dimesioal wave equatio. Moreover, sice the strig is xed at x = 0 ad x = l; we also have the boudary coditios u (0; t) = 0 ad u (l; t) = 0 (6) for all times t: If we avoid the trivial solutio (that of o vibratio, u = 0); the these boudary coditios ca be used to determie some of the arbitrary costats i the separated solutio. EXAMPLE 5 Fid the solutio of the oe dimesioal wave equatio (5) subject to the boudary coditios (6). 6

7 Solutio: To do so, we substitute u (x; t) = X (x) T (t) ito the equatio (X (x) T (t)) = (X (x) T (t)) =) X (x) T 00 (t) = a T (t) X 00 (x) To separate the variables, we the divide throughout by a X (x) T (t): This i tur simpli es to X (x) T 00 (t) a X (x) T (t) = a T (t) X 00 (x) a X (x) T (t) T 00 (t) a T (t) = X00 (x) X (x) As a result, there must be a costat such that T 00 a T = ad X 00 X = These i tur reduce to the di eretial equatios T 00 = a T ad X 00 = X If > 0; however, the oscillatios would become arbitrarily large i amplitude, which is ot physically possible. Thus, we assume that is egative, which is to say that =! for some umber!: As a result, we have T 00 = a! T ad X 00 =! X Both equatios are harmoic oscillators, so that the geeral solutios are T (t) = A 1 cos (a!t)+b 1 si (a!t) ad X (x) = A cos (!x)+b si (!x) where A 1 ; B 1 ; A ; ad B are arbitrary costats. Let s ow cocetrate o X (x) : The boudary coditios (6) imply that u (0; t) = X (0) T (t) = 0 ad u (l; t) = X (l) T (t) = 0 If we let T (t) = 0; the we will obtai the solutio u (x; t) = 0 for all t: This is called the trivial solutio sice it is the solutio correspodig to the strig ot movig at all. To avoid the trivial solutio, we thus assume that X (0) = 0 ad X (l) = 0 7

8 However, X (x) = A cos (!x) + B si (!x) ; so that X (0) = 0 implies that 0 = A cos (0) + B si (0) = A Thus, A = 0 ad X (x) = B si (!x) : The boudary coditio X (l) = 0 the implies that B si (!l) = 0 If we let B = 0; the we agai obtai the trivial solutio. To avoid the trivial solutio, we let si (!l) = 0; which i tur implies that!l = for ay iteger : Thus, there is a solutio for! = =l for each value of ; which meas that X (x) = B si l x is a solutio to the vibratig strig equatio for each : Cosequetly, for each iteger there is a separated solutio of the form h a a i u (x; t) = A 1 cos t + B 1 si t B si l l l x (7) Check your Readig: Where did the a=l come from i the al form of the separated solutio? Liearity ad Fourier Series We say that a partial di eretial equatio is liear if the liear combiatio of ay two solutios is also a solutio. For example, suppose that p (x; t) ad q (x; t) are both solutios to the heat equatio i.e., = = q (8) A liear combiatio of p ad q is of the form u (x; t) = Ap (x; t) + Bq (x; t) where A; B are both costats. Moreover, so that (8) implies (Ap (x; t) + Bq (x; @t = p + q = (Ap (x; t) + Bq (x; 8

9 That is, the liear combiatio u (x; t) = Ap (x; t) + Bq (x; t) is also a solutio to the heat equatio, ad cosequetly, we say that the heat equatio is a liear partial di eretial equatio. Suppose ow that a liear partial di eretial equatio has both boudary coditios ad iitial coditios, where iitial coditios are costraits o the solutio ad its derivatives at a xed poit i time. The a complete solutio to the partial di eretial equatio ca ofte be obtaied from the Fourier series decompositios of the iitial coditios. For example, let us suppose that the vibratig strig i example 5 is plucked at time t = 0, which is to say that it is released from rest at time t = 0 with a iitial shape give by the graph of the fuctio y = f (x): The the iitial coditios for the vibratig strig are u (x; 0) = f (x) ad (x; 0) = Let s apply the iitial coditios to the separated solutio (7). The iitial coditio u t (x; 0) = 0 implies that X (x) T 0 (0) = 0; so that to avoid the trivial solutio we suppose that T 0 (0) = 0. Thus, 0 = T 0 (0) = a!a 1 si (0) + B 1 a! cos (0) = B 1 As a result, we must have T (t) = A 1 cos (at=l) ; ad if we de e b = A 1 B ; the (7) reduces to a u (x; t) = b cos t si l l x (9) As will be show i the exercises, the 1 dimesioal wave equatio is liear. Thus, if u j (x; t) ad u k (x; t) are solutios (9) for itegers j ad k, the u j (x; t)+ u k (x; t) is also a solutio. I fact, the sum all possible solutios, which is the sum of all solutios for ay positive iteger value of ; is a solutio called the 9

10 geeral solutio. That is, the geeral solutio to the 1 dimesioal wave equatio with the give boudary ad iitial coditios is u (x; t) = 1X a b cos t si l l x =1 (10) Hece, the oly task left is that of determiig the values of the costats b : However, (10) implies that u (x; 0) = 1X =1 ad sice u (x; 0) = f (x) ; this reduces to f (x) = b cos (0) si l x 1X =1 b si l x As a result, if f (x) is cotiuous ad if f (0) = f (l) ; the the costats b are the Fourier Sie coe ciets of f (x) o [0; l] ; which are give by b = l Z l 0 f (x) si l x dx (11) For more o Fourier series ad their relatioship to partial di eretial equatios, see the Maple worksheet associated with this sectio. EXAMPLE 6 What is the solutio to the vibratig strig problem for a foot log strig which is iitially at rest ad which has a iitial shape that is the same as the graph of the fuctio u (x; 0) = 1 1 jx 1j 1 Solutio: We begi by dig the Fourier coe ciets b ; which accordig to (11) are for a l = foot log strig give by b = Z jx 1j si 1 x dx :Evaluatig usig the computer algebra system Maple the yields si si () b = 3 10

11 However, sice is a iteger, si () = 0 for all. Thus, b reduces to si b = 3 But si = 0 whe is eve, so that b0 = b = : : : = b = 0: Thus, we oly have odd coe ciets of the form b 1 = si ; b si 3 = 3 3 ; b si 5 = 3 5 ; : : : which simplify to b 1 = (1) 3 1 ; b 3 = ( 1) 3 3 ; b 5 = (1) 3 5 ; : : : Odd umbers are of the form + 1 for = 0; 1; : : : : Thus, we have ( 1) b +1 = 3 ( + 1) ad the solutio (10) is of the form 1X ( 1) a ( + 1) ( + 1) u (x; t) = 3 ( + 1) cos t si x l l =0 The Fourier series (10) is kow as the Harmoic Series i music theory. Ideed, if we write the Fourier series i expaded form a a 3a 3 u (x; t) = b 1 cos l t si l x +b cos t si l l x +b 3 cos t si l l x +: : : the the rst term is kow as the fudametal, which correspods to the strig shape of y = si (x=l) ; which is xed at x = 0 ad x = l, oscillatig at a amplitude of b 1 : The oscillatios themselves have a frequecy of f 1 = a l rad sec 1cycle rad = a l cycles sec A "cycle per secod" is kow as a Hertz ad recall that a = k=; so that f 1 = k l Hz Thus, icreases i tesio k cause the fudametal pitch to rise, while legtheig the strig lowers the pitch. A heavier strig (larger ) has a lower pitch tha a lighter strig. The secod term i the Harmoic Series of the strig oscillates at a amplitude b with twice the frequecy of the fudametal, f = a l 11 = f 1 :

12 It is kow as the rst harmoic or rst overtoe of the strig, ad it correspods to the oscillatio of a strig shape y = si (x=l) that is xed at x = 0, x = l=, ad x = l i.e., half as log as the fudametal. Similarly, the third term is the secod harmoic, which oscillates at a frequecy of f 3 = 3f 1 ad which correspods to oscillatios at amplitude b 3 of siusoidal shapes a third as log as the fudametal. For example, if the strig is at a legth, tesio, ad mass so as to oscillate with a frequecy of 440 hz ("A" above middle "C"), the we also hear a pitch of f = 880 hz (a octave above the fudametal), a pitch of f 3 = 3 (440) hz (a octave ad a fth above the fudametal) ad so o. Exercises Show that the give fuctio is a solutio to the give partial di eretial equatio. Assume that k;!; a; ad c are costats. 1. u (x; y) = x 3 is a solutio to u. u (x; y) = 3x y y 3 is a solutio = 0 = 0 3. u (x; t) = t + x is a solutio u 4. u (x; t) = x + t is a solutio 5. u (x; y) = e x si (y) is a solutio to u = 0 = 0 6. u (x; y) = ta 1 y x is a solutio to 7. u (x; t) = e!kt cos (!x) is a solutio = u 8. u (x; t) = si (!x) si (a!t) is a solutio 9. u (x; t) = f (x + ct) is a solutio 10. u (x; t) = f (x ct) is a solutio = u = u = u Fid the separated solutio to each of the followig partial di eretial equatios. 1

16 41. Laplace s Equatio I: Fid the geeral solutio to the Laplace equatio subject to the boudary coditios = 0 u (0; y) = 0 u (; y) = 0 4. Laplace s Equatio II: If the iitial coditios are u (x; 0) = si (x=) ad u y (x; 0) = 0; the what are the Fourier coe ciets i the geeral solutio foud i exercise 41? 43. Write to Lear: I a short essay, explai i your ow words why a equatio of the form f (x; y) = g (t) implies that both f (x; y) ad g (t) are costat. (x; y; ad t are both idepedet variables). 44. *What is a separated solutio of the -dimesioal wave = u Fid a separated solutio of the followig oliear wave = 16

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