7.1. Simple Eigenfunction Problems. 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 ODE November 20, Simple Eigenfunction Problems Section Objective(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 di erential equation. main ideas to solve that equation. (1) Boundary value problems and eigenfunction problems. (2) Fourier series expansions. In this section we study the first idea: boundary value problems and eigenfunctions.

2 2 G. NAGY ODE 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 satisfying the boundary conditions (BC) y 00 + a 1 (x) y 0 + a 0 (x) y = b(x) b 1 y(x 1 )+b 2 y 0 (x 1 )=y 1, b1 y(x 2 )+ b 2 y 0 (x 2 )=y 2, 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= x 2. (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, b 2 = 0, Boundary Condition: which is the case y(1) = y 2, b1 = 1, b2 = 0. (1b) (1c) (1d) Boundary Condition: Boundary Condition: Boundary Condition: 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, which is the case which is the case which is the case b1 = 1, b 2 = 0, 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 ODE 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. Remarks: The variable t represents time. The variable y represents position. 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. Remarks: The variable x represents position. The variable y may represent temperature. 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 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(l) =y 1, with L 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. (ii) There exists infinitely many solutions. (iii) There exists no solution.

4 4 G. NAGY ODE november 20, 2018 Proof of Theorem 2: The general solution is y(x) =c + e t+x + c - e r-x. The BC are 9 y 0 = y(0) = c + + c - >= y 1 = y(l) =c + e c+l + c - e c-l >; ) e r+l e r-l c = 4 y c y 1 3 This system for c +, c - has a unique solution i 0 6= 1 1 e r+l e r-l ) e r-l e r+l 6=0. Part (A): If r + 6= r -, reals, then e r-l 6= e r+l, hence there is a unique solution c +, c -,which fixes a unique solution y of the BVP. Part (B): If r ± = ± i,then e r+ - L = e ( ±i )L = e L (cos( L) ± i sin( L)), therefore e r-l e r+l = e L cos( L) i sin( L) cos( L) i sin( L) = 2ie L sin( L) =0, L = n. So for L 6= n the BVP has a unique solution, case (Bi). For L = n the BVP has either no solution or infinitely many solutions, cases (Bii) and (Biii).

5 G. NAGY ODE November 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, r ± = ±i. So the general solution of the di erential equation is y(x) =c 1 cos(x)+c 2 sin(x). BC (a): 1=y(0) = c 1 ) c 1 =1. 0=y( ) = c 1 ) c 1 =0. Therefore, there is no solution. BC (b): 1=y(0) = c 1 ) c 1 =1. 1=y( /2) = c 2 ) c 2 =1. So there is a unique solution y(x) = cos(x)+sin(x). BC (c): 1=y(0) = c 1 ) c 1 =1. 1=y( ) = c 1 ) c 2 =1. Therefore, c 2 is arbitrary, so we have infinitely many solutions y(x) = cos(x)+c 2 sin(x), c 2 2 R. C

6 6 G. NAGY ODE november 20, Eigenfunction Problems. Remark: Let us recall the eigenvector problem of a square matrix: Given a square matrix A, findanumber and a nonzero vector v solution of Av = v. Definition 4. An eigenfunction problem is the following: Given a linear operator L(y) =a 2 y 00 + a 1 y 0 + a 0 y,findanumber and a nonzero function y solution of L(y) = y, and homogeneous boundary conditions at x 1 6= x 2, b 1 y(x 1 )+b 2 y 0 (x 1 )=0, b1 y(x 2 )+ b 2 y 0 (x 2 )=0, Remarks: Notice that y =0 is always a solution of the BVP above. Eigenfunctions are the nonzero solutions The eigenfunction problem is a BVP with infinitely many of the BVP above. solutions. So, we look for such that the operator L(y) y has characteristic polynomial with complex roots. So, is such that L(y) y has oscillatory solutions. We focus on the linear operator L(y) = y 00. Example 3: Find all numbers and nonzero functions y solutions of the BVP y 00 = y, with y(0) = 0, y(l) =0, L > 0. Solution: The equation is y 0 + y = 0. We have three cases: (a) <0, (b) = 0, and (c) >0. Case (a): = µ 2 < 0, so the equation is y 00 µ 2 y = 0. The characteristic equation is r 2 µ 2 =0 ) r +- = ±µ.

7 G. NAGY ODE November 20, The general solution is y = c + e µx + c - e µx.thebcimply 0=y(0) = c + + c -, 0=y(L) =c + e µl + c - e µl. So from the first equation we get c + = c -,so 0= c - e µl + c - e µl ) c - (e µl e µl )=0 ) c - =0, c + =0. So we get only the solution y = 0. Case (b): = 0, so the di erential equation is y 00 =0 ) y = c 0 + c 1 x. The BC imply So we get the only solution is y = 0. 0=y(0) = c 0, 0=y(L) =c 1 L ) c 1 =0. Case (c): = µ 2 > 0, so the equation is y 00 + µ 2 y = 0. The characteristic equation is r 2 + µ 2 =0 ) r +- = ±µi. The general solution is y = c + cos(µx)+c - sin(µx). The BC imply 0=y(0) = c +, 0=y(L) =c + cos(µl)+c - sin(µl) =c - sin(µl), therefore, we get c - sin(µl) =0, c - 6=0 ) sin(µl) =0 ) µ n L = n. So we get µ n = n /L, hence the eigenvalue eigenfunction pairs are n = n L 2, n x yn (x) =c n sin. L C

8 8 G. NAGY ODE november 20, 2018 Example 4: Find the numbers and the nonzero functions y solutions of the BVP y 00 = y, y(0) = 0, y 0 (L) =0, L > 0. Solution: The equation is y 00 + y = 0. We have three cases: (a) <0, (b) = 0, and (c) >0. Case (a): Let = µ 2,withµ>0, so the equation is y 00 µ 2 y = 0. The characteristic equation is r 2 µ 2 =0 ) r +- = ±µ, The general solution is y(x) =c 1 e µx + c 2 e µx.thebcimply 9 0=y(0) = c 1 + c 2, >= 0=y 0 (L) = µc 1 e µl + µc 2 e µl >; ) The matrix above is invertible, because c = µe µl µe µl 0 c = µ e µl + e µl 6=0. µe µl µe µl Therefore, the linear system above for c 1, c 2 has a unique solution given by c 1 = c 2 = 0. Hence, we get the only solution y = 0. This means there are no eigenfunctions with negative eigenvalues. Case (b): Let = 0, so the di erential equation is y 00 =0 ) y(x) =c 1 + c 2 x, c 1,c 2 2 R. The boundary conditions imply the following conditions on c 1 and c 2, 0=y(0) = c 1, 0=y 0 (L) =c 2. So the only solution is y = 0. This means there are no eigenfunctions with eigenvalue = 0. Case (c): Let = µ 2,withµ>0, so the equation is y 00 + µ 2 y = 0. The characteristic equation is r 2 + µ 2 =0 ) r +- = ±µi.

9 G. NAGY ODE November 20, The general solution is y(x) =c 1 cos(µx)+c 2 sin(µx). The BC imply 9 0=y(0) = c 1, >= 0=y 0 (L) = µc 1 sin(µl)+µc 2 cos(µl) >; ) c 2 cos(µl) =0. Since we are interested in non-zero solutions y, we look for solutions with c 2 6= 0. This implies that µ cannot be arbitrary but must satisfy the equation cos(µl) =0, µ n L =(2n 1) 2, n > 1. We therefore conclude that the eigenvalues and eigenfunctions are given by n = (2n 1)2 2 (2n 1) x 4L 2, y n (x) =c n sin, n > 1. 2L Since we only need one eigenfunction for each eigenvalue, we choose c n = 1, and we get n = (2n 1)2 2 (2n 1) x 4L 2, y n (x) =sin, n > 1. 2L C