Strauss PDEs e: Sectio 4.1 - Exercise 4 Page 1 of 5 Exercise 4 Cosider waves i a resistat medium that satisfy the probem u tt = c u xx ru t for < x < u = at both eds ux, ) = φx) u t x, ) = ψx), where r is a costat, < r < πc/. Write dow the series expasio of the soutio. Soutio The PDE ad its boudary coditios, u, t) = u, t) =, are iear ad homogeeous, so the method of separatio of variabes ca be appied to sove it. Assume a product soutio of the form, ux, t) = Xx)T t), ad pug it ito the PDE ad the boudary coditios. Separate variabes ow. u tt = c u xx ru t XT = c T rxt u, t) = X)T t) = X) = u, t) = X)T t) = X) = XT + rxt = c T T + rt = X X Note that c is a costat ad ca go o either side. The fia aswer wi be the same regardess. We have a fuctio of t o the eft side ad a fuctio of x o the right side. The oy way both fuctios ca be equa is if they are equa to a costat. T + rt = X X = k Vaues of k for which X) = ad X) = are satisfied are caed the eigevaues, ad the otrivia fuctios Xx) associated with them are caed the eigefuctios. Determiatio of Positive Eigevaues: k = µ Assumig k is positive, the differetia equatio for X becomes X = µ = µ X
Strauss PDEs e: Sectio 4.1 - Exercise 4 Page of 5 The geera soutio ca be writte i terms of hyperboic sie ad hyperboic cosie. Xx) = C 1 cosh µx + C sih µx Now use the boudary coditios to determie C 1 ad C. X) = C 1 = X) = C 1 cosh µ + C sih µ = We see that C 1 = ad C =. Hece, oy the trivia soutio Xx) = resuts from cosiderig positive vaues for k, ad there are o positive eigevaues. Determiatio of the Zero Eigevaue: k = Assumig k is zero, the differetia equatio for X becomes X =. = The geera soutio is a iear fuctio. Xx) = C 3 x + C 4 Now use the boudary coditios to determie C 3 ad C 4. X) = C 4 = X) = C 3 + C 4 = We see that C 3 = ad C 4 =. Hece, oy the trivia soutio Xx) = resuts from cosiderig k =, ad zero is ot a eigevaue. Determiatio of Negative Eigevaues: k = λ Assumig k is egative, the differetia equatio for X becomes X = λ. = λ X The geera soutio ca be writte i terms of sie ad cosie. Xx) = C 5 cos λx + C 6 si λx Now use the boudary coditios to determie C 5 ad C 6. X) = C 5 = X) = C 5 cos λ + C 6 si λ =
Strauss PDEs e: Sectio 4.1 - Exercise 4 Page 3 of 5 The secod equatio simpifies to C 6 si λ =. To avoid gettig the trivia soutio, we isist that C 6. Doig so yieds a equatio for the eigevaues. Sove for λ. si λ = λ = π, = 1,,... So the λ = λ = π, = 1,,.... The eigefuctios associated with these eigevaues are Xx) = C 6 si λx X x) = si λ x, = 1,,.... Now sove the differetia equatio for T t). Mutipy both sides by. Brig c λ T to the eft side. T + rt = λ T + rt = c λ T T + rt + c λ T = This is a ODE with costat coefficiets, so its soutio is of the form Substitute this ito the ODE to determie s. Divide both sides by e st. T = e st. s e st + rse st + c λ e st = s + rs + c λ = This is a quadratic equatio for s, so use the quadratic formua to sove for it. s = r ± r 4c λ Now substitute λ = π/. s = r ± r 4c ) π = r ± r ) πc Sice < r < πc/, the quatity uder the square root is egative for every vaue that takes. Factor out 1 ad brig it out of the square root as i. s = r ± i πc ) r = r ± i π c r 4 = r 4 ± i π c r
Strauss PDEs e: Sectio 4.1 - Exercise 4 Page 4 of 5 Thus, the geera soutio to the ODE for T is ) ) T t) = C 7 e r 4 t cos π c r t + C 8 e r 4 t si π c r t. Accordig to the pricipe of iear superpositio, the soutio to the PDE for ux, t) is a iear combiatio of a products T t)x x) over a the eigevaues. [ ) ) ux, t) = A e r 4 t cos π c r t + B e r 4 t si π c r t si πx Therefore, ux, t) = =1 ) ) e r 4 [A t cos π c r 4 t + B si π c r t si πx. =1 The fia task is to use Fourier s method to express the coefficiets, A ad B, i terms of the provided iitia data, φx) ad ψx). ux, ) = =1 A si πx = φx) Mutipy both sides by si λ m x, where m is a positive iteger. ux, ) = =1 A si πx si mπx = φx) si mπx Itegrate both sides with respect to x over the domai the PDE is defied. ˆ =1 A si πx Brig the itegra iside the sum o the eft. ˆ A =1 Sice ad m are itegers, ˆ si πx si πx si mπx si mπx si mπx ˆ ˆ φx) si mπx φx) si mπx { = m m, as ca be verified with trigoometric idetities. This impies that every term i the ifiite series vaishes except for the = m term. Therefore, the first coefficiet is A = ˆ φx) si πx A = ˆ φx) si πx.
Strauss PDEs e: Sectio 4.1 - Exercise 4 Page 5 of 5 I order to use the secod iitia coditio, differetiate the boxed soutio for u with respect to t. u t x, t) = r ) ) ) e r 4 [A t cos π c r 4 t + B si π c r t si πx =1 [ ) + e r 4 t A π c r 4 si π c r t =1 ) 4 + B π c r 4 cos π c r t si πx Now pug i t =. u t x, ) = =1 =1 r ) A si πx =1 + =1 B 4 π c r Brig the costat i frot of the first sum. r ) A si πx 4 + B π c r The first sum is ux, ), ad ux, ) is equa to φx). r ) 4 φx) + B π c r =1 Brig the first term to the right side. 4 B π c r =1 =1 si πx si πx si πx si πx = ψx) = r φx) + ψx) = ψx) = ψx) Now mutipy both sides by si λ m x, where m is a positive iteger. 4 B π c r si πx si mπx = φx) + ψx) si mπx Itegrate both sides with respect to x over the domai the PDE is defied. ˆ 4 B π c r si πx si mπx ˆ φx) + ψx) si mπx =1 =1 Brig the itegra iside the sum o the eft. 4 B π c r ˆ si πx si mπx Every term i this ifiite series vaishes except for the oe whe = m. Therefore, B 4 π c r = ˆ ˆ 4 B = 4 π c r ˆ φx) + ψx) si mπx φx) + ψx) si πx φx) + ψx) si πx.