Math 641 — HW 8 Examples

Problem 3.3.2(d)

From the definition of the kernel k(x,y), we see that
Ku(x) = ∫₀^xy u(y)dy + x∫_x^π u(y)dy.
We want to solve Ku = λ u. Here is how to do that. We will differentiate λ u = Ku twice.
λ du/dx = x u(x) + ∫_x^π u(y)dy - x u(x) = ∫_x^π u(y)dy.
λ d²u/dx² = - u.
Note that this equation shows that λ ≠ 0. So, λu(0) = Ku(0) = 0 implies that u(0) = 0. Similarly, we see that du/dx(π) = 0.
The differential-equation version of the eigenvalue problem is then
d²u/dx² + λ⁻¹u = 0, u(0) = 0, du/dx(π) = 0,
which is easily solved.

Problem 3.5.1(a)

The parameter λ is extraneous here. Ignore it. We will however solve the problem
u(x) = f(x) + ∫₀^xu(y)dy.
We first convert this to a differential equation:
u′ = f′ + u.
this is a simple first order linear differential equation; it is solved using the integrating factor exp(−x),
(u e^−x)′ = f′(x)e^−x. Hence,
u(x) = u(0)e^x + ∫₀^xe^x−t f′(t)dt
If we now integrate this by parts we get
u(x) = f(x) + ∫₀^xe^x−t f(t)dt.
The resolvent operator is Rf(x) = ∫₀^xe^x−t f(t)dt.

Updated 11/22/2010.