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8 Greens function
Question 101: Let Ω be a three-dimensional domain and consider the PDE
∇^2 u = f (x), x ∈ Ω, with u(x) = h(x) on the boundary of Ω, say Γ.
Let G(x, x 0 ) be the Green’s function of this problem (the exact expression of G does not matter; just assume that G is known). Give a representation^1 of u(x) in terms of G, f and h.
Solution: By definition
∇^2 xG(x, x 0 ) = δ(x − x 0 ), x ∈ Ω, with G(x, x 0 ) = 0 x ∈ Γ.
Then using the integration by parts formula, we obtain ∫
Ω
u(x)∇^2 x(G(x, x 0 ))dx =
Ω
∇^2 x(u(x))G(x, x 0 )dx+
Γ
u(x)∂n(G(x, x 0 ))dx−
Γ
∂n(u(x))G(x, x 0 )dx.
which can also be rewritten
u(x 0 ) =
Ω
f (x)G(x, x 0 )dx +
Γ
h(x)∂n(G(x, x 0 ))dx.
Question 102: Let f be a smooth function in [0, 1]. Consider the PDE
u − ∂xxu = f (x), x ∈ (0, 1), ∂xu(1) + u(1) = 2, −∂xu(0) + u(0) = 1.
What PDE and which boundary conditions must satisfy the Green function, G(x, x 0 ), (DO NOT compute the Green function)? Give the integral representation of u assuming G(x, x 0 ) is known. Fully justify your answer.
Solution: Multiply the equation by G(x, x 0 ) and integrate over (0, 1):
∫ (^1)
0
f (x)G(x, x 0 )dx =
0
(u(x) − ∂xxu(x))G(x, x 0 )dx
0
u(x)G(x, x 0 ) + ∂xu(x)∂xG(x, x 0 )dx − ∂xu(1)G(1, x 0 ) + ∂xu(0)G(0, x 0 )
0
u(x)(G(x, x 0 ) − ∂xxG(x, x 0 ))dx + u(1)∂xG(1, x 0 ) − u(0)∂xG(0, x 0 )
− ∂xu(1)G(1, x 0 ) + ∂xu(0)G(0, x 0 )
=
0
u(x)(G(x, x 0 ) − ∂xxG(x, x 0 ))dx + u(1)∂xG(1, x 0 ) − u(0)∂xG(0, x 0 )
(u(1) − 2)G(1, x 0 ) + (u(0) − 1)G(0, x 0 )
=
0
u(x)(G(x, x 0 ) − ∂xxG(x, x 0 ))dx
- u(1)(G(1, x 0 ) + ∂xG(1, x 0 )) + u(0)(G(0, x 0 ) − ∂xG(0, x 0 )) − 2 G(1, x 0 ) − G(0, x 0 )
If we define G(x, x 0 ) so that
G(x, x 0 ) − ∂xxG(x, x 0 ) = δ(x − x 0 ), G(1, x 0 ) + ∂xG(1, x 0 ) = 0, G(0, x 0 ) − ∂xG(0, x 0 ) = 0,
then u(x 0 ), x 0 ∈ (0, 1), has the following representation
u(x 0 ) =
0
f (x)G(x, x 0 )dx + 2G(1, x 0 ) + G(0, x 0 ), ∀x 0 ∈ (0, 1).
(^1) Hint: use ∫ Ω ψ∇ (^2) (φ) = ∫ Ω ∇ (^2) (ψ)φ + ∫ Γ ψ∂n(φ) − ∫ Γ ∂n(ψ)φ
Question 103: Consider the equation u′(x) + u = f (x) for x ∈ (0, 1) with u(0) = a. Let G(x, x 0 ) be the associated Green’s function. (Pay attention to the number of derivatives). (a) Give the equation and boundary condition defining G and give an integral representation of u(x 0 ) in terms of G, f and the boundary data a. (Do not compute G.)
Solution: The Green’s function is defined by
−G′(x, x 0 ) + G(x, x 0 ) = δ(x − x 0 ), G(1, x 0 ) = 0.
We multiply the equation by u and we integrate over (0, 1) (in the distribution sense),
∫ (^1)
0
−G′(x, x 0 )u(x)dx +
0
G(x, x 0 )u(x)dx = u(x 0 ).
We integrates by parts and we obtain,
u(x 0 ) =
0
G(x, x 0 )(u′(x) + u(x))dx − G(1, x 0 )u(1) + G(0, x 0 )u(0)
Then, using the fact that u′^ + u = f and using the boundary conditions for G and u, we obtain
u(x 0 ) =
G(x, x 0 )f (x)dx + 2G(0, x 0 ). ∀x 0 ∈ (0, 1).
(ii) Compute G(x, x 0 ) for all x, x 0 ∈ (1, 2).
Solution: For all x 6 = x 0 we have
−∂x(x∂xG(x, x 0 )) = 0.
The solution is
G(x, x 0 ) =
a log(x) + b if 1 < x < x 0 c log(x) + d if x 0 < x < 2
The boundary conditions give b = 0 and d = −c log(2); as a result,
G(x, x 0 ) =
a log(x) if 1 < x < x 0 c log(x/2) if x 0 < x < 2
G must be continuous at x 0 ,
a log(x 0 ) = c log(x 0 ) − c log(2)
and must satisfy the gap condition
∫ (^) x 0 +
x 0 −
∂x(x∂xG(x, x 0 ))dx = 1, ∀ > 0.
This gives
−x 0
∂xG(x+ 0 , x 0 ) − G(x− 0 , x 0 )
−x 0 (
c x 0
a x 0
This gives a − c = 1.
In conclusion log(x 0 ) = −c log 2 and
c = − log(x 0 )/ log(2), a = 1 − log(x 0 )/ log(2) = log(2/x 0 )/ log(2).
This means
G(x, x 0 ) =
{ (^) log(2/x 0 ) log(2) log(x)^ if^1 < x < x^0 log(x 0 ) log(2) log(2/x)^ if^ x^0 < x <^2
Question 105: Consider the equation u′(x) + u = f (x) for x ∈ (0, 1) with u(0) = a. Let G(x, x 0 ) be the associated Green’s function. (Pay attention to the number of derivatives). (a) Give the equation and boundary condition defining G and give an integral representation of u(x 0 ) in terms of G, f and the boundary data a. (Do not compute G.)
Solution: The Green’s function is defined by
−G′(x, x 0 ) + G(x, x 0 ) = δ(x − x 0 ), G(1, x 0 ) = 0.
We multiply the equation by u and we integrate over (0, 1) (in the distribution sense),
∫ (^1)
0
−G′(x, x 0 )u(x)dx +
0
G(x, x 0 )u(x)dx = u(x 0 ).
We integrates by parts and we obtain,
u(x 0 ) =
G(x, x 0 )(u′(x) + u(x))dx − G(1, x 0 )u(1) + G(0, x 0 )u(0)
Then, using the fact that u′^ + u = f and using the boundary conditions for G and u, we obtain
u(x 0 ) =
0
G(x, x 0 )f (x)dx + aG(0, x 0 ). ∀x 0 ∈ (0, 1).
(b) Compute G(x, x 0 ).
Solution: For x < x 0 and x 0 > x we have
−G′(x, x 0 ) + G(x, x 0 ) = 0.
The solution is
G(x, x 0 ) =
αex^ for x < x 0 βex^ for x > x 0.
The boundary condition G(1, x 0 ) = 0 implies β = 0.
For every > 0 we have
∫ (^) x 0 +
x 0 −
(−G′(x, x 0 ) + G(x, x 0 ))dx
= G(x 0 − , x 0 ) − G(x 0 + , x 0 ) +
∫ (^) x 0 +
x 0 −
G(x, x 0 )dx
The term R =
∫ (^) x 0 + x 0 − G(x, x^0 )dx^ can be bounded as follows: |R| ≤ 2 max x∈[0,1] |G(x, x 0 )| = 2αex^0.
Clearly R goes to 0 with . As a result we obtain the jump condition
1 = G(x− 0 , x 0 ) − G(x+ 0 , x 0 ) = αex^0.
This implies α = e−x^0.
Finally
G(x, x 0 ) =
ex−x^0 for x < x 0 0 for x > x 0.
Question 106: Consider the equation −∂x(x∂xu(x)) = f (x) for all x ∈ (1, 2) with u(1) = a and u(2) = b. Let G(x, x 0 ) be the associated Green’s function. (i) Give the equation and boundary conditions satisfied by G and give the integral representation of u(x 0 ) for all x 0 ∈ (1, 2) in terms of G, f , and the boundary data. (Do not compute G in this question).
Solution: We have a second-order PDE and the operator is clearly self-adjoint. The Green’s function solves the equation
−∂x(x∂xG(x, x 0 )) = δ(x − x 0 ), G(1, x 0 ) = 0, G(2, x 0 ) = 0.
We multiply the equation by u and integrate over the domain (1, 2) (in the distribution sense).
〈δ(x − x 0 ), u〉 = u(x 0 ) = −
1
∂x(x∂xG(x, x 0 ))u(x)dx.
We integrate by parts and we obtain,
u(x 0 ) =
1
x∂xG(x, x 0 )∂xu(x)dx − [x∂xG(x, x 0 )u(x)]^21
G(x, x 0 )∂x(x∂xu(x))dx − 2 ∂xG(2, x 0 )u(2) + ∂xG(1, x 0 )u(1).
This implies that G(x, x 0 ) = cx + d in (x 0 , L). The boundary condition ∂xG(L, x 0 ) = 0 gives c = 0. The continuity of G(x, x 0 ) at x 0 implies that ax 0 = d. The condition
∫ (^)
−
∂xxG(x, x 0 )dx = 1, ∀ > 0 ,
gives the so-called jump condition: ∂xG(x+ 0 , x 0 ) − ∂xG(x− 0 , x 0 ) = 1. This means that 0 − a = 1, i.e., a = − 1 and d = −x 0. In conclusion
G(x, x 0 ) =
−x if ≤ x ≤ x 0 , −x 0 otherwise.
(b) Give the integral representation of u using the Green’s function.
Solution: Let x 0 be a point in (0, L). The definition of the Dirac measure at x 0 is such that
u(x 0 ) = 〈δx 0 , u〉 = 〈∂xxG(·, x 0 ), u〉
= −
∫ L
0
∂xG(x, x 0 )∂xu(x)dx + [∂xG(x, x 0 )u(x)]L 0
∫ L
0
G(x, x 0 )∂xxu(x)dx − [G(x, x 0 )∂xu(x)]L 0 + [∂xG(x, x 0 )u(x)]L 0
∫ L
0
G(x, x 0 )f (x)dx − G(L, x 0 )∂xu(L) + G(0, x 0 )∂xu(0) + ∂xG(L, x 0 )u(L) − ∂xG(0, x 0 )u(0).
This finally gives the following representation of the solution:
u(x 0 ) =
∫ L
G(x, x 0 )f (x)dx − G(L, x 0 )b − ∂xG(0, x 0 )a
