Solve the Sturm-Liouville Problem y''-2y'+λy=0 y(0)=y'(1)=0?

Question

Yea... Eigenvalues and explanation of each step would be greatly appreciated.

somebody please help

Answer 1

Start by substituting y = e^(m*x) as usual for an equation like this. You get the indicial equation

m² - 2m + λ = 0

whose solutions are

m = 1 ± √(1 - λ)

As usual, there are three cases to consider:

(i) 1 - λ > 0 . Let 1 - λ = α² with α > 0. So m = 1 ± α

Solution is

y = A e^[(1+α)x] + B e^[(1-α)x]
y' = A(1+α) e^[(1+α)x] + B (1-α) e^[(1-α)x]

y(0) = 0 → A + B = 0
y'(1) = 0 → A [(1+α) e] + B[(1-α) e] = 0

The only solution is A = B = 0, so no eigenvalues here.

(ii) 1 - λ = 0 (i.e., λ = 1) so m = 1 (double root)

Solution is

y = e^x (A + Bx)

y(0) = 0 → A = 0 so y = Bx e^x
y' = (B + Bx) e^x = B(1+x) e^x
y'(1) = 0 → 2Be = 0 → B = 0

So no eigenvalues here.

(iii) 1 - λ < 0 Let 1 - λ = -α² with α > 0. So m = 1 ± iα

Solution:

y = e^x [A cos(α x) + B sin(α x)]

y(0) = 0 → A = 0 so y = B e^x sin(α x)
y' = Be^x [sin(α x) + α cos(α x)]
y'(1) = 0 → B = 0 or sin(α) + α cos(α) = 0

We don't want B = 0 so we must have sin(α) + α cos(α) = 0, or

α = -tan(α)

Graph z = α and z = -tan(α) on the same set of axes. You will see that there are an infinite number of solutions. The solution α=0 gives only y=0 so we discard it. Similarly, the negative solutions can be discarded because they contribute no solutions for y that are not also contributed by positive solutions for α. These values of α are the eigenvalues.

The first few positive solutions for α are

α ≈ 2.02875784
α ≈ 4.91318044
α ≈ 7.97866571
α ≈ 11.0855384

and it can be shown that

α ≈ (2n - 1)π/2 + 2/[(2n - 1)π] for n "large" (say, n > 4)

With this estimate as a first guess, solving for α = -tan(α) (or better, α = -atan(α) + nπ) using Newton's method converges very fast.

So

y_n(x) = A_n sin((α_n) x) where α_n solves α_n = -tan(α_n), n=1,2,3,... are the nontrivial solutions of the given S-L problem.

Answer 2

Solve the BVP for λ > 0

y +λy=0, y(0) − y (0)=0, y(1) + y (1)=0