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Quiz 7 (Notes, books, and calculators are not authorized)
Show all your work in the blank space you are given on the exam sheet. Always justify your answer. Answers with
no justification will not be graded.
Question 1: Consider the equation −∂x ((1 + x)∂x u(x)) + ∂x u(x) = f (x), x ∈ (0, 1) with u(0) = α and −2∂x u(1) +
u(1) = β. Let G(x, x0 ) be the Green’s function. Give the integral representation of u(x0 ) for all x0 ∈ (0, 1) in
terms of G, f , α and β and give the equation and boundary conditions that G must satisfy. Do not compute G.
(Hint: The differential operator is not self-adjoint).
We multiply the PDE by G(x, x0 ) and integrate by parts,
1
Z
1
Z
((1 + x)∂x u(x)∂x G(x, x0 ) − u(x)∂x G(x, x0 )) dx + [(−(1 + x)∂x u(x) + u(x))G(x, x0 )]10
f (x)G(x, x0 )dx =
0
0
1
Z
((1 + x)∂x u(x)∂x G(x, x0 ) − u(x)∂x G(x, x0 )) dx + βG(1, x0 ) − (−∂x u(0) + α)G(0, x0 ).
=
0
Since ∂x u(0) is not known, we must have G(0, x0 ) = 0. Then
Z
1
Z
1
(−u(x)∂((1 + x)∂x G(x, x0 )) − u(x)∂x G(x, x0 )) dx + βG(1, x0 ) + [(1 + x)u(x)∂x G(x, x0 )]10
f (x)G(x, x0 )dx =
0
0
Z
1
−u(x) (∂((1 + x)∂x G(x, x0 )) + ∂x G(x, x0 )) dx + βG(1, x0 ) + 2u(1)∂x G(1, x0 ) − α∂x G(0, x0 ).
=
0
Since u(1) is not known, we must have ∂x G(1, x0 ) = 0. Finally G must satisfy
−∂x ((1 + x)∂x G(x, x0 )) − ∂x G(x, x0 ) = δ(x − x0 ),
G(0, x0 ) = 0,
∂x G(1, x0 ) = 0,
and we have the following representation for u(x0 ),
Z
1
f (x)G(x, x0 )dx − βG(1, x0 ) + α∂x G(0, x0 ).
u(x0 ) =
0
Question 2: (a) Compute all the solutions set of the equation
u00 + u = 0, x ∈ (0, 2π)
with u(0) = u(2π), u0 (0) = u0 (2π).
clearly u(x) = a cos(x) + b sin(x) where a and b are arbitrary numbers. The solution set is a two-dimensional vector
space spanned by cos(x) and sin(x).
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Quiz 7, November 6, 2014
(b) Do the following equation have solution(s)? (Hint: think of the Fredholm alternative)
u00 + u = sin(2x), x ∈ (0, 2π)
with u(0) = u(2π), u0 (0) = u0 (2π).
We are in the second case of the Fredholm alternative, i.e., the null space of the operator is not {0}. We have to verify
that sin(2x) is orthogonal to cos(x) and sin(x), which is clearly true. In conclusion the equation has solutions.
(c) Do the following equation have solution(s)? (Hint: think of the Fredholm alternative)
u00 + u = sin(x), x ∈ (0, 2π)
with u(0) = u(2π), u0 (0) = u0 (2π).
We are in the second case of the Fredholm alternative, i.e., the null space of the operator is not {0}. We have to verify
that sin(x) is orthogonal to cos(x) and sin(x), which is clearly wrong. In conclusion the equation has no solution.
Question 3: Consider the following Cauchy-Euler problem: Let f be a smooth function in [0, π2 ] and find u such
π
π
that 13u − 5x∂x u + x2 ∂xx u = f (x), x ∈ (1, e 2 ), u(1) = 0, u(e 2 ) = 0. (i) Under which condition does this problem
have a solution. (Hint: The solution to the homogeneous equation (i.e., f = 0) is v(x) = ax3 cos(2 log(|x|)) +
bx3 sin(2 log(|x|)).)
Let us compute the null space of the operator
π
L : {v ∈ C 2 (0, π)| v( ) = 0, v(0) = 0} 3 u 7−→ 13u − 5x∂x u + x2 ∂xx u ∈ C 0 (0, π).
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Let N (L) be the null space. Let v ∈ N (L), then
13v − 5x∂x v + x2 ∂xx v = 0
which means that v = ax3 cos(2 log(|x|)) + bx3 sin(2 log(|x|)). The boundary conditions imply that
0 = a cos(2 log(1)) + bx3 sin(2 log(1)) = a,
π
0 = b(π/2)3 sin(2 log(e 2 )) = b(π/2)3 sin(π),
a = 0 and b is an arbitrary real number; as a result N (L) = span(x3 sin(2 log(|x|))), i.e., N (L) is the one-dimensional
vector space spanned by the function x3 sin(2 log(|x|)). Fredholm’s alternative implies that the above problem has a
Rπ
solution only if 12 x3 sin(2 log(|x|))f (x)dx = 0.