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
2 Sice (3) ad (4) are the same, u (x; t) = si (at) cos (x) is a solutio to (). EXAMPLE Show that u (x; t) = e y si (x) is a solutio to Laplace s Equatio, = 0 Solutio: To begi with, u x = e y cos (x) ad u xx = e y si (x) : Moreover, u y = e y si (x) ad u yy = e y si (x) ; so that = ey si (x) + e y si (x) = 0 Check your Readig: Why are u; u y, ad u yy the same as u i example? Separatio of Variables Solutios to may (but ot all!) partial di eretial equatios ca be obtaied usig the techique kow as separatio of variables. It is based o the fact that if f (x) ad g (t) are fuctios of idepedet variables x; t respectively ad if f (x) = g (t) the there must be a costat for which f (x) = ad g (t) = : ( The proof is straightforward, f g (t) = 0 =) f 0 (x) = 0 =) f g f (x) = 0 =) g0 (t) = 0 =) g (x) costat) I separatio of variables, we rst assume that the solutio is of the separated form u (x; t) = X (x) T (t) We the substitute the separated form ito the equatio, ad if possible, move the xterms to oe side ad the tterms to the other. If ot possible, the this method will ot work; ad correspodigly, we say that the partial di eretial equatio is ot separable. Oce separated, the two sides of the equatio must be costat, thus requirig the solutios to ordiary di eretial equatios. A table of solutios to commo di eretial equatios is give below: Equatio Geeral Solutio y 00 +! y = 0 y (x) = A cos (!x) + B si (!x) y 0 = ky y (t) = P e kt y 00! y = 0 y (x) = A cosh (!x) + B sih (!x)
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 xaxis 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 massdesity 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 oedimesioal 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
13 Assume that k; a; c; ad are costat. = = 0 = 15. F x + e x F y = F x + 3x F y = u x + u t = = 19. u = 0 0. u t = u xx @t V = 0. u t = u xx = 0 + = 0 5. Show that u (x; t) = 1 p t e x =(4t) is a solutio to the heat equatio u t = u xx : 6. Show that u (x; y; z) = x + y + z 1= is a solutio to the 3 dimesioal Laplace equatio = 0 7. Let i = 1 ad suppose that u (x; y) ad v (x; y) are such that (x + iy) = u (x; y) + i v (x; y) Fid u ad v ad show that both satisfy Laplace s equatio that is, that = 0 v = 0 I additio, show that u ad v satisfy the CauchyRiema Equatios u x = v y ; u y = v x 8. Let i = 1 ad suppose that u (x; y) ad v (x; y) are such that (x + iy) 4 = u (x; y) + i v (x; y) Fid u ad v ad show that both satisfy Laplace s equatio that is, that = 0 v = 0 I additio, show that u ad v satisfy the CauchyRiema Equatios u x = v y ; u y = v x 9. Suppose that a large populatio of microorgaisms (e.g., bacteria or plakto) is distributed alog the xaxis. If u (x; t) is the populatio per uit legth at locatio x ad at time t; the u (x; t) satis es a di usio equatio of the = + ru 13
14 where is the rate of dispersal ad r is the birthrate of the microorgaisms. If ad r are positive costats, the what is a separated solutio of this di usio equatio? (adapted from Mathematical Models i Biology, Leah Edelstei Keshet, Radom House, 1988, p. 441). 30. Suppose that t deotes time ad x deotes the age of a cell ia give populatio of cells, ad let u (x; t) dx = umber of cells whose age at time t is betwee x ad x + dx The u (x; t) is the cell desity per uit age at time t; ad give appropriate assumptios, it satis + v = d 0 where v 0 ad d 0 are positive costats. What is the separated solutio to this equatio? (adapted from Mathematical Models i Biology, Leah Edelstei Keshet, Radom House, 1988, p. 466). 31. Fid the separated solutio of the telegraph equatio with zero self iductace: + RSu Here u (x; t) is the electrostatic potetial at time t at a poit x uits from oe ed of a trasmissio lie, ad R, C; ad S are the resistace, capacitace, ad leakage coductace per uit legth, respectively. 3. If V (x; t) is the membrae voltage at time t i secods ad at a distace x from the distal (i.e., iitial) ed of a uiform, cylidrical, ubrached sectio of a dedrite, the V (x; t) satis es d 4R V = + 1 V (1) R m where d is the diameter of the cylidrical dedritic sectio, R i is the resistivity of the itracellular uid, C m is the membrae capacitace, ad R m is the membrae resistivity. Fid a separated solutio to (1) give that C m ; R m ; ad R i are positive costats. 33. I Quatum mechaics, a particle movig i a straight lie is said to be i a state (x; t) if Z b a j (x; t)j dx represets the probability of the particle beig i the iterval [a; b] o the lie at time t: If a subatomic particle is travelig i a straight lie close to the speed of light, the it s state satis es the oe dimesioal @t = 14
15 where > 0 is costat. Fid the separated solutio of the oe dimesioal KleiGordo equatio. 34. If a subatomic particle is travelig i a straight lie much slower tha the speed of light ad o forces are actig o that particle, the its state (as explaied i problem 33) satis es the oe dimesioal Schrödiger equatio of a sigle = (13) where i = 1: Fid the separated solutio of (13) (Hit: you will eed to use Euler s idetity e it = cos (t) + i si (t) 35. Show that if u (x; t) ad v (x; t) are both solutios to the oe dimesioal wave = u the so also is the fuctio w (x; t) = Au (x; t) + Bv (x; t) where A ad B are costats. What does this say about the wave equatio? 36. Show that if u (x; y) ad v (x; y) are both solutios to Laplace s equatio = 0 the so also is the fuctio w (x; y) = Au (x; y) + Bv (x; y) where A ad B are costats. What does this tell us about Laplace s equatio? 37. Suppose that the iitial coditios for the guitar strig i example 6 are x u (x; 0) = si ad (x; 0) = What are the coe ciets b i the solutio (10) for these iitial coditios? 38. Solve the vibratig strig problem for the boudary (0; t) = 0 ad (l; t) = ad for the iitial coditios u (x; 0) = f (x) ad u t (x; 0) = 0: 39. Heat Equatio I: Fid the geeral solutio to the heat equatio subject to the boudary = u u (0; t) = 0 u (; t) = Heat Equatio II: If the iitial coditio is u (x; 0) = x x ; the what are the Fourier coe ciets i the geeral solutio foud i exercise 39? 15
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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