Question

+ dac 1. Consider the following BVP day dy d.x2 + Ag = 0, g(0) = g(2) = 0. (a) Find eigenvalues and eigenfunctions of the problem; (b) Put the equation in self-adjoint form, and give an orthogonality relation; (c) Show that each eigenfunction of the problem can not correspond to two different eigenvalues.

Answer 1

tea o day duz + dy dx + ay = 0, y los = y(2) =0 7 case 1: when aco : le-a = - el Then the given egh becames m2 tm+2=0 ma .-1IJI_AX 2 ma - 1 I ST-49 1 tj It All 2 2 a 2 llow By t=0 Ci (6,=0 los hlo C Ils =0 - The general solution is y(x) = c, cosh ( 11 SIHA) x + G sinh (1+51 +41 using boundary Condition. y(0) = y (2)=0 = Casinh (1+Jital) ( sinh (1+JI+AM) & 0) the general Case Is 1=0 dzy -> m=o, mlm t)=o 0 it C2=0 solb. m?+mo i + dy dat =0 du2 m=1 Geox y (1) = e,+ caed y (n)= By applying Condition oundevey y (o)= 0, y(a)=0 &=cit ce? and C+C2 3 ditex = 0 Ci+c₂e²=0 C₂(1-e2)=0 a) C2=0) =a q=0 1y =o) is the is the general soln

Case 3: when nxo lel-a=ll Its characteristic en is IST-44 m2 + m + el=0 (: el=aso - -41> l-au <o. Wa 2 ME - + ż Ji-All Z Dat ca sin ( 1-44)x yan) - a e-%2C c, cos (Ji-All given condition, 3 By using the ylos=0 = g(2) =) o = 1-C to) a cascos e-/ Cz sin (Si a X. -) c=0 y CH3 = (Jaqu)* y(2) =0 e-1 sin (J1 - All C2 4 sin (S7-44)) = 0 sin (JI-AUJO 0 = C2 Sin =) I-4 l O, ł, zn, 3 Ji-áun = na n1,2,- =) I- Aung n272 1-42A2 Alln = ) I n2n2 n=1,2, 3-6,0,0,0, ul 4 1 - 1212 n=1,2,- An ане eigenvalues Hence the Thus, the eigenfunctions ане sin (J1 - 4(1-12 m2) FO Yn A yn = sin ( Fitn23 It n2a2)x n=1,2,

१ P(x) = 1, An es op de eside لا when M(M)=en @ Now fore self adjoint form, y" + y' + 264) = 0 b $(2) = (by comparing with the standard equation) integrating factor is given by To F - The required self adjoint form is en ly") + exyl + xety =o (eng) + reny = 0 (6My')'t relyzo Xelly = -(eny')' p(x) = en let us suppose that each eigenfunction of the two different eigenvalues preoblem correspond to i + s Bay nis Xe ri dj ; 2x² > Corresponding eigen function 9 = sin (Jitra2)x , If = sin(Jith my 92= 9; ( since eigenfunction is unique) Sin (J1 + 12A2) x = sin (J1+; }72) =Jl+j2 2² j² 1 + j²42 iaj Hente which is a contradiction it is Our assumption is wrong wrong fence [nizli where = I- & 4 q are But n > Siti?a2 -) Itih? =)

że cach eigen function cannot cannot corresponds to tulo different eigenvalues. Hence proved