Question

4. (This is problem #4 on p. 318 of the text). Solve via the method of images: Utt = cum ur(0,t) = 0 u(x,0) = 23 ut(3,0) = 0 (3 > 0, t > 0) (t > 0) (x > 0) (x > 0). (Note: if f(x) is defined for x > 0 by f(x) > 0 by f(x) = x3, then the even extension of f to the entire real line is given by feven(x) = |x|3.)

Answer 1

Solution The given - c2 Uan given P. D.E is Utt (no, to) Ux (ort) = 0 (t > U (4,0) = x3 U (2,0) = 0 (>0 Let the solution of - (i) is of the form U (x, t) = x® T+) Now, Uxx = X'T and Utt TIX Putting the value of Uxx and utt in (1) have Се T"X crylat T" Х 1 ст (say) where dis Constant.

Now TI-ACT X = 2x (iii (iv) Let T= emt (where m being Constant) be a solution of the Coresponding homogenious us equation of - (ii) Then its auxiliary equation is m? de 2 (v) (where in being Let x=emn constant) be a solution of the Coresp ponding equation of -(iv), Then lious homogenia its auxiliary equation is m2=0 (vi) Arises three cases as Now, we follows.

Case 1 for do cwe have have from (v) หา 2 O :-) m Thus the General solution of (iii) is T(t) = (A + Bit) e oit T (t) = Ait Bit. where AI, Bi are arbitrary Constants. Also, for x=0, we have from (vi) m2 = m 00 is 0.X Thus the General solution of (iv) X (n) = (A2+ B2 x) e =) X(X) = (A2 + B2²) where A2,B2 are arbitrary constants

We have Now from (ii) U (t) = X T(4) =) u (x, t) = (A2 + B2 %) (A, +3,1 Now differentiating (vii) writix we have Ux (e,t) = (A + B, t) B2 (vii) wort Also differentiating 'Z' we have U ų (x, t) = (A2+ B27) B Now using the given Condition Ut (x,0) = 0 and Ux (0,t) = 0 we have U Ų (x, 0) = B. (A₂ + B2 x) = 0 Un (0,t) = B₂ (A + B, t) zo and

A2 AI = x 3 Now if Az - Bzx=0 As we Then from (vii) u (not) = 0 which is at rivial solution of (i) are interested to non trivial solution of (1) we must have (A2 +B2 ) # 0.B,o Also if A, FBit A, FB, t = 0 Then from (vii) u (r, t) = 0 which is again a trivial solution of -(i).so interested in non trivial solution of -() (A + B, t) to we are we must have B2 = 0 Now U using the given condition U (,0) = x3 we have from -(vii) U (0,0) = (A 2 + B2 M) A, = x3 [: BL B₂ = 0

Now from (vii) Az Ai ulurt) (A2 +6 2 + B2n) (A, FB, 2) = u(x, t) = =) u (nt) [: A, A2 = x :: u(x, t) = x² is a solution of 23 -C) Case 2 For t= M2 (M)0) , we have from (v) m2 - C2M2 =/ = 0 =) m² - C²M² (m - CM) (me CM) = 0 = m = f cu Thus the General solution of (ii) is cut T (t) = Aze -cut +Bze where A3, B3 are arbitrary constants

Also For I we have from dur (v) mr mr_mr=0 - (m_M) (mfm) = 0 =) m = fm Thus the General solution of -(iv) X (n) = Aue be MA MX & Bye where Ay, By arbitrary are constants. (ii) we have Now from u(uit) - X) T(t) MX & Bye -cart ebze > U (",t) = (Aue Now differentiating-(vii) wonit 'x' we have Up (x, t) = (A 3e caut abge-Cut) (awet - Buce any)

Again differentiating (viii) worit t' Се have MR Up (wit) a (emanent cu oge med (Ane Bu e-mx) Using the given Condition ut (4,0) = 0 and Un (ort) = 0 Now We have U (+,0) = (CMA 3 - CMB3) (Auer MU e Bye rewall As we are interested in non trivial solution of (i) so we must have Ane Mr & By e un to CM A3-CM B3=0 =) A3 = B3 [:em to М И - Ми If Aue + Bye Then from (viii) Uw,t) = 0 which is trivial solution of

also Un (ort) = (Azeem Abré e-ent) (A, M-Bu )- AS We are in interested getting a non trivial solution of ☺ we must have Az eemt & Bz e emt to O Aum-Bum = 0 = Au = By | 1 If Azeemt Then [Mto + Bz e-ent=0 from (viii) u(x,t)=0 which is trivial solution of -i] Also using the given Condition U (2,0) = x3 we have from (viii) U (4,0) = (Ane Mn + Buemn) (Az + 3) = 2* Au Cemnte A) 2 Az = x²

A3 (ecut te 13 =) A 4 A 3 = 2 (enx tema) : from City U (x, t) = Au (emn peme) Femt e evan) (ecmutue.cent (emri pe- may (eemt te-ent u(a,t) - 23 (ecantee-ent) is a solution of 2 3 거 2 Case 3 or = - M² (m)o) we have from m² = Mic2 = mrt Mc2 = 0 (m + Mci) (m-mci) = 0 = m = I Mci

Thus the General solution of (111) is T(E)= Ascosuct+ Bg sinuct where A5, B5 are arbitrary Constants. have we A d=-M"(m>o) Also for from (vi) m = M2 =) mrem²=0 (m + wi) (m-mi) = 0 =) үй I mi General solution of in Thus the is X (n) = A6 Cosmat Be Sinux where A6, B6 are arbitrary Constants

A6 Cosun of B. Sin mal Now from (ii), we have U (wit) = Xa) T(t) 7) u (x, t) = (As Cosmet + Bg Sinmet) sin mx) Now differentiating tix). Wirlit we have U (2,t) = (Ascosmet +Bg sinmet) (- MAC Sinun + M Bc Cosun) Again differentiating (ix) worit 't we have

ut (xt) =(- MCA s sinuet & MCBs Cosmet) As cosmat Besin mix) Now using using the given Condition U (4,0) = 0 and Ux (0,t=0 we have t (a,0) = MC B5 (-MAG sin un + MB Cosmn) = 0 As Се are interested on Non trivial solution of - (i) so we must have -MAG Sin ma AMBE.COSMnto .: MCBS = B5 о [ Mcto If - MA Sin Mr & MBC Cosun = 0 Then from from - ix we have U(nt)=0 which is trivial solution of - 0]

Also, Un (o,t) = MB (As Cosmet +Be sinmet) = 0 in te nested in getting non trivial solution of (i) so As we are We must have As Cosmet & Bg sin met to MBG=0 B6 [ .. m to If. As cosmet &Bs sinuet - o Then from -Cix) ut) = 0 trivial solution of which is of (1) Again using the given Condition U (4,0) = x3 we have from – (ix) As (AC Cosma it Be Sindix) = x3 => A5 A6 Cosmr=x3

A5 A6 2 3 T Cosma U (w; t) = (as 5 cosuct & Bg sinact Now from = (A 5 Cosmo in met (A6 Cosmo AG Sinan = (As Cosmed) (A scossex) - = A5 A6 As A Cosmet Cosua x 3 11 Cosuct COSMR Cosun 23 Cosmet u (art) = x3 cosuct 3 Cosuct is a solution of