7.1. Simple Eigenfunction Problems. the. The main idea of this chapter is to solve the heat equation. main ideas to solve that equation.

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1 G. NAGY DE November 20, Simple Eigenfunction Problems Section bjective(s): Two-Point Boundary Value Problems. Comparing IVP vs BVP. Eigenfunction Problems. Remark: The main idea of this chapter is to solve the heat equation. This is a partial We need two Partial two di erential equation. main ideas to solve that equation. (1) Boundary value problems hrs and eigenfunction problems. Boundary Value (2) Fourier series expansions. the Proble In this section we study the first idea: boundary value problems and eigenfunctions. heat equation CBVP Eigenfunction Problems EFP Fourier Series Expansions EE P B P and

2 2 G. NAGY DE november 20, Two-Point Boundary Value Problems. Definition 1. A two-point boundary value problem (BVP) is the following: Find solutions to the di erential equation y 00 + a 1 (x) y 0 + a 0 (x) y = b(x) Y t a Cx Y t clock Y bad d satisfying the boundary conditions (BC) b b 1 y(x 1 )+b 2 y 0 (x 1 )=y 1, Y Xi t be YkX Y b1 To y(x 2 )+ b 2 y 0 (x 2 )=y 2, y Xa t T Y Xa Ya where b 1, b 2, b 1, b 2, y 1, y 2, x 1, x 2 are given and x 1 6= x 2. Remarks: (a) The two boundary conditions are held at di erent points, x 1 6= Xx 2 X2. (b) Both y and y 0 may appear in the boundary condition. Example 1: We now show four examples of boundary value problems that di er only on the boundary conditions: Solve the di erent equation y 00 + a 1 y 0 + a 0 y = b(x) with the boundary conditions at x 1 = 0 and x 2 = 1 given below. (1a) y(0) = y1, b1 = 1 I, b 2 = 0, y(1) = y 2, b1 = 1, b2 = 0. (1b) (1c) (1d) Y y y(0) = y1, y 0 (1) = y 2, y 0 (0) = y 1, y(1) = y 2, y 0 (0) = y 1, y 0 (1) = y 2, I b1 = 1, b 2 = 0, o b1 = 0, b2 = 1. b1 = 0, b 2 = 1, b1 = 1, b2 = 0. b1 = 0, b 2 = 1, b1 = 0, b2 = 1. C

3 G. NAGY DE November 20, Comparing IVP vs BVP. Definition 2. (IVP) Find a solution of y 00 + a 1 y 0 + a 0 y = 0 satisfying the initial condition (IC) y(t 0 )=y 0, y 0 (t 0 )=y 1. Ylto Yo Y to Y Remarks: time position Position The variable t represents time. The variable y represents position. velocity The IC are position and velocity at the initial time. Definition 3. (BVP) Find a solution y of y 00 + a 1 y 0 + a 0 y = 0 satisfying the boundary condition (BC) y(x 0 )=y 0, y(x 1 )=y 1. TIX Yo Y Xi Y Xotxi Remarks: Position for example temperature The variable x represents position. The variable y may represent temperature. temperature positions The BC are temperature at two di erent positions. Theorem 1. The equation y 00 + a 1 y 0 + a 0 y =0withICy(t 0 )=y 0 and y 0 (t 0 )=y 1 has a unique solution y unique solution Y for each choice of the IC. Theorem 2. (BVP) The equation y 00 +a 1 y 0 +a 0 y =0withBCy(0) = y 0 and y() =y 1, with 6= 0 and with r ± roots of p(r) =r 2 + a 1 r + a 0 satisfy the following: (A) If r + 6= r -, reals, then the BVP above has a unique solution. (B) If r ± are complex, then the solution of the BVP above belongs to only one of the following three possibilities: (i) There exists a unique solution. a unique SoC Unique solution Y infinitely many SoCs (ii) There exists infinitely many solutions. (iii) There exists no solution. no solution

4 4 G. NAGY DE november 20, 2018 Proof of Theorem 2: The general solution is Gen y(x) =c + e t+x + c - e r-x. Soc ycx c ee ee The BC are 9 y 0 = y(0) = c + + c - >= y 1 = y() =c + e c+ + c - e c- >; ) B C Yo Y o G te e r+ e r c = 4 y c y 1 3 This system for c +, c - has a unique solution i Y Y q ett a er 0 6= 1 1 e r+ e r- ) e r- e r+ 6=0. 1 ere II f Y It t.sc rte Part (A): If r + 6= r -, reals, then e r- 6= e r+, hence there is a unique solution c +, c -,which fixes a unique solution y of the BVP. 1 Part (B): If r ± = ± i,then Then at therefore e r+ - = e ( ±i ) = e (cos( ) ± i sin( )), III A e r- e r+ = e cos( ) i sin( ) cos( ) i sin( ) dose So for For = 2ie sin( ) =0, = n. ere er to 6= n the BVP has a unique solution, case (Bi). = n the BVP has either no solution or infinitely many solutions, cases (Bii) A and (Biii). rt f E real cord EH A invert B r xt

5 detca e F prt ert ealc.si e4 cosfp tisincf fer le edlcosfs4 isin.g er h oof f2i edsuspd Bi sin Ito unique Sol Bil sin o A Not invertible J Y I Ip No sols Infinitely many Sol E ME

6 G. NAGY DE INovember 20, Example 2: Find all solutions to the BVPs y 00 + y =0withtheBCs: ( ( ( y(0) = 1, y(0) = 1, y(0) = 1, (a) (b) (c) y( ) =0. y( /2) = 1. y( ) = 1. Solution: We first find the roots of the characteristic polynomial r = 0, that is, Y ty o a y e rat to r ± = ±i. So the general solution of the di erential equation is p a BC (a): y(x) =c 1 cos(x)+c 2 sin(x). Ycx c aes x si 1=y(0) = c 1 ) c 1 =1. I Y o c t ca o c c 0=y( ) = c 1 ) c 1 =0. T µ ws Therefore, there is no solution. BC (b): usin T 1=y(0) = c 1 ) c 1 =1. Y IT I 1=y( /2) t = cg 2 ) c 2 = No So there is a unique solution y(x) = cos(x)+sin(x). BC (c): 1=y(0) = c 1 ) c 1 =1. True b I CE 1=y( ) = c 1 ) c 2 =1. Therefore, c 2 is arbitrary, so we have infinitely many solutions o t CzCH E I y(x) = cos(x)+c 2 sin(x), c 2 2 R. cosfxltsin.ci 1 ycxj UnifUes c I JCT I t ca c i C Y x cos t ca Sina f nis i