SET 5

MATH 543: GREEN’S FUNCTIONS

(References: DK, Sadri Hassan and Hildebrandt) Generalized Green’s Identity Let L_{x} be a differential operator and u , v are some functions in L_{w}^{2}(a, b) then the adjoint operator L^{†} is defined through the Lagrange identity

w[v̄L_{x}u − u \overline{\{L_{x}^{†}v\}}] = \frac{d}{dx}Q(u, v̄),  (1)

where w > 0 is the weight function and Q(u, v̄) has in general the following form

Q(u, v̄) = Auv̄ + Bv̄\frac{du}{dx} + Cu\frac{dv̄}{dx} + D\frac{du}{dx} \frac{dv̄}{dx},  (2)

Integrating the above identity over the interval [a, b] we get the generalized Green’s identity

∫_{a}^{b} w(x) [v̄ L_{x}u − u \overline{\{L_{x}^{†} v\}}]dx = Q(u(b), v̄(b)) − Q(u(a), v̄(a)), (3)

The right hand side of this equation is called the surface term.

Adjoint Boundary Conditions: Let the function u satisfy the boundary conditions

B_{1}(u) = α_{1}u(a) + β_{1}u′(a) + γ_{1}u(b) + δ_{1}u′(b) = 0,  (4)
B_{2}(u) = α_{2}u(a) + β_{2}u′(a) + γ_{2}u(b) + δ_{2}u′(b) = 0,  (5)

Here α_{i}, β_{i}, γ_{i}, and δ_{i} are some given constants. If in the above generalized Green’s identtiy (3) the function u satisfies the above boundary condtions (4) and (5) then the function v satisfies the adjoint boundary conditions when the surface term vanishes. Hence for future purposes we can define the Green’s identity

∫_{a}^{b} w[v̄ L_{x}u − u \overline{\{L_{x}^{†} v\}}]dx = 0,  (6)

for the functions u satsfying the boundary conditions (4) and (5) and v satisfying the adjoint boundary condtions

Self Adjoint Operators: We sall be interested in second order differential operators

L_{x}u(x) = a(x)\frac{d^{2}u}{dx^{2}} + b(x)\frac{du}{dx} + c(x)u,  (7)

where the coefficients a, b and c are some functions of x. Here we assume that Re(a) > 0. It is easy to show that the adhoint operator is given by

L^{†}v = \frac{d^{2}}{dx^{2}} (ā v) − \frac{d}{dx}(b̄v) + c̄v,  (8)

and the function Q in the surface term is found as

Q(u, v̄) = av̄\frac{du}{dx} − u\frac{d}{dx}(av̄) + buv̄,  (9)

If the coefficient functions are real and

b(x) = \frac{da(x)}{dx},  (10)

Then L_{x}^{†} = L, becomes a self-adjoint operator and the Q term becomes.

Q(u, v̄) = a(v̄\frac{du}{dx} − u\frac{dv̄}{dx}),  (11)

With a suitable weight factor all second order differential operators are selfadjoint

Theorem 1. Every linear second order differential operator with real coefficients is self adjoint provided the weight function w(x) is choosen properly

w(x) = \frac{p}{a}, \frac{p′}{p} = \frac{b(x)}{a(x)},  (12)

where the diferential operator takes the form

L_{x}u = \frac{1}{w}(\frac{d}{dx} p(x)\frac{du}{dx}) + cu,  (13)

and the function Q is given by

Q(u, v̄) = p(x) (v̄\frac{du}{dx} − u\frac{dv̄}{dx}),  (14)

Problems:

1. Prove that the following boundary conditions on u(x) , L_{x} defines a Hermitian differential operator L

(i) u(a) = u(b) = 0 (Dirichlet conditons), (ii) \frac{du}{dx}(a) = \frac{du}{dx}(b), (Neumann conditions), (iii) α u(a) − \frac{du}{dx}(a) = βu(b) − \frac{du}{dx}(b), (α, β real) (general unmixed conditions), (iv) u(a) = u(b) and \frac{du}{dx}(a) = \frac{du}{dx}(b), (periodic conditions).

2. Determine the formal adjoint of each of the operators in (a) throuh (d) below (i) os a differential operator, and (ii) as an opeartor, that is, including the boundary conditions (Determine the functions spaces of L and L^{†}). Which opeartors are formally self-adjoint? Which opeartors are self- adjoint?

(a) L_{x} = \frac{d^{2}}{dx^{2}} + 1 in [0, 1] with u(0) = u(1) = 0,

(b) L_{x} = \frac{d^{2}}{dx^{2}} in [0, 1] with u(0) = u′(0) = 0,

(c) L_{x} = \frac{d}{dx} in [0, ∞) with u(0) = 0,

(d) L_{x} = \frac{d^{3}}{dx^{3}} −sin x\frac{d}{dx} +3 in [0, π] with u(0) = u′(0) = 0 and u′′(0)−4u(π) = 0

Boundary value problems Let u satisfy the inhomogenous differential equation with homogenous boundary conditions

L_{x} u(x) = f(x), B_{1}(u) = 0, B_{2}(u) = 0,  (15)

where f(x) is any integrable function. The function v satisfying the adjoint differential and the homogenous adjoind boundary conditions

L_{x}^{†} v(x) = h(x), adB_{1}(v) = 0, adB_{2}(v) = 0,  (16)

Green’s Functions: Green’s functions G(x, y) and g(x, y) of the above boundary value problems are, respectively given by

L_{x} G(x, y) = \frac{δ(x − y)}{w(x)},  (17)
L_{x}^{†} g(x, y) = \frac{δ(x − y)}{w(x)},  (18)

using the Gren’s identity for the function u, v, G, and g pairwise (u, v), (u, g), (v, G) and (G, g) we obtain the following important identities

u(x) = ∫_{a}^{b} w(y)ḡ(y, x) f(y)dy, ,  (19)
v(x) = ∫_{a}^{b} w(y)Ḡ(y, x) h(y)dy,  (20)

and

< f, g >=< G, h >, G(x, y) = ḡ(y, x),  (21)

Since the solutions are given in (19) and (20), then the boundary value problems (15) and (16) reduces to the determinattion of the Green’s functions G and g

Second order operators and Green’s functions: For a second order opeartors with real coefficients we have

\frac{d}{dx}(p(x)\frac{du}{dx}) + cwu = f(x), B_{1}(u) = 0, B_{2}(u) = 0,  (22)
\frac{d}{dx}(p(x)\frac{dv}{dx}) + cwv = h(x), adB_{1}(v) = 0, adB_{2}(v) = 0, (23)

and the Green’s functions satisfy the following equations

\frac{d}{dx}(p(x)\frac{dG(x, y)}{dx}) + cwG(x, y) = δ(x − y),  (24)
\frac{d}{dx}(p(x)\frac{dg(x, y)}{dx}) + cwg(x, y) = δ(x − y),  (25)

Since the Greens’s fuctions satisfy the homogenous equation (L_{x}G(x, y) = 0) for x ≠ y the we have

G(x, y) = \{ \begin{matrix}c_{1}u_{1}(x) + c_{2}u_{2}(x) & a ≤ x < y \\ d_{1}u_{1}(x) + d_{2}u_{2}(x) & x < y ≤ b \end{matrix}  (26)

where u_{1} and u_{2} are the independent solutions of the homogenous equation L_{x}u = 0 and c_{1}, c_{2}, d_{1}, d_{2} are constants (wrt x) to be determined by the following conditions on G. First two are the homogenous boundary conditions and the rest two are the continuity of G and jump condition for the derivative of G

B_{1}(G) = 0, B_{2}(G) = 0,  (27)
G(x, y)|_{x=y_{+}} = G(x, y)|_{x=y_{+}},  (28)
\frac{dG(x, y)}{dx}|_{x=y_{−}} − \frac{dG(x, y)}{dx}|_{x=y_{−}} = \frac{1}{p(y)},  (29)

We have the following important teorem.

Theorem 2. Consider the boundary value problem L_{x}u = f(x) with the homogenous boundary conditons B_{1}(u) = 0 and B_{2}(u) = 0. Here we assume that L_{x}is a second order differential operator. Provided that the homogenous equation has no notrivial solution satisfying the above boundary conditions the Green’s function associated with the boundary value problem exist and unique. The solution is given by

u(x) = ∫_{a}^{b} w(y) G(x, y) f(y)dy  (30)

Problems:

3. Let B_{1}(u) = 0 and B_{2}(u) = 0 are defined at x = a and x = b respectively. Solve the Green’s function defined in (26). Solution: The continuity condtions (28) and (29) are solved easily (see DK page 281)

G(x, y) = \{ \begin{matrix}\frac{U_{<}(x)U_{>}(y)}{p(y)W} & a ≤ x < y \\ \frac{U_{<}(y)U_{>}(x)}{p(y)W} & y < x ≤ b \end{matrix}  (31)

where W = U_{<}U′_{>} − U_{>}U′_{<} is the Wronksian of U_{<}(x) = u_{1}(x) − α u_{2}(x) and U_{>}(x) = u_{1}(x) − βu_{2}(x). Here u_{1} and u_{2} are independent solutions of the homogenous DE (the fundamenatal set). Prove that p(x)W (U_{<}, U_{>}) is constant for all x. The boundary conditions (27) mus also be satisfied, hence

B_{1}(U_{<}) = B_{1}(u_{1}) − αB_{1}(u_{2}) = 0,
B_{2}(U_{>}) = B_{2}(u_{1}) − βB_{2}(u_{2}) = 0

Hence α and β are determined. As we discussed above the product pW is assumed to be a nonvanishing constant. If it is zero, then U_{<}(x) = σU_{>}(x), where σ is an arbitaray constant. Using the definitions of U_{<} and U_{>} and α and β found above this relation gives ,

B_{1}(u_{1} − ρu_{2}) = 0, B_{2}(u_{1} − ρu_{2}) = 0

We obtain this by letting σ = ρ \frac{B_{2}(u_{2})}{B_{2}(u_{1})} , where ρ is another constant. The above relation implies the existance of a nontrivial solution u_{1} − ρu_{2} of the homogenous DE satisfying the homogenous boundary condions, but this is contardiction with our assumtion.

4. Solve the Green’s function when both boundary conditions are given at the same point x = a. As an example take for instantce u(a) = 0 and u′(a) = 0.

Boundary value problems with inhomogenous boundary conditions: Let B_{1}(u) = γ_{1} and B_{2}(u) = γ_{2}, where γ_{1} and γ_{2} are some given constants. In this case the construction of the Gree’s function is exactly similar as above. We determine the Green’s function as if the problem is with homogenous bondary condtions. Here the expression for (30) u changes. To obtain the expression for u we use the Green’s identity, but in this case since the boundary condtions are homogenmous we cannot assum that the surface term is zero. For the operator L_{x} given in (13) the surface term is given in (14). Now using the generalized Green’s identity we get

u(x) = ∫_{a}^{b} w(y)G(x, y)f(y)dy +\{p(y)[u(y)\frac{∂G(x, y)}{∂y} −G(x, y)\frac{du}{dy}]\}|_{y=a}^{y=b}, (32)

here we use B_{1}(u) = γ_{1} and B_{2}(u) = γ_{2}. In the above expression (32) we have u(a), u(b) and u′(a), u′(b) terms in the surface term. Two of them are insterted from the inhomogenous boundary condtions B_{1}(u) = γ_{1} and B_{2}(u) = γ_{2} and the rest dissapears from the expressions due to adjoint homogenous boundary conditions (for self dual operators due to the same homogenous boundary conditions)

Problems:

5. In problem 3 , if B_{1}(u) = γ_{1} and B_{2}(u) = γ_{2} then the complete solution will be

u(x) = ∫_{a}^{b} w(y) G(x, y) f(y)dy + \frac{U_{<}(x)}{U_{<}(b} γ_{2} + \frac{U_{>}(x)}{U_{>}(a)} γ_{1}

6. Solve u′′ + u = f(x) with u(0) = u′(0) = 0.

7. Find the Green’s function for L_{x} = \frac{d^{2}}{dx^{2}} + k^{2} with u(0) = u(a) = 0.

8. Find the Green’s function for L_{x} = \frac{d^{2}}{dx^{2}} − k^{2} with u(∞) = u(−∞) = 0.

9. Find the Geen’s function for L_{x} = \frac{d}{dx} x\frac{d}{dx} given the condition that G(x, y) is finite at x = 0 and vanishes at x = 1.

10. Problem 5 can be generalized further. Let L_{x}u = f with boundary condition B_{1}(u) = γ_{1} and B_{2}(u) = γ_{2}. Here L_{x} is a second order differential operator with real coefficients. Prove that the solution is given by

u(x) = ∫_{a}^{b} w(y) G(x, y) f(y)dy + \frac{1}{Δ}[B_{2}(u_{2}) u_{1}(x) − B_{2}(u_{1})u_{2}(x)] γ_{1} +

\frac{1}{Δ}[B_{1}(u_{1})u_{2}(x) − B_{1}(u_{2})u_{1}(x)] γ_{2} (33)

where Δ = B_{1}(u_{1})B_{2}(u_{2})−B_{1}(u_{2})B_{2}(u_{1}) and the Green’s function is found as if the boundary conditions are homogenous. Discuss the case where Δ = 0.

When there exist a nontrivial solution of the homogenous equation satisfying the homogenous boundary conditions: Let v_{i} be the nontrivial solutions of the homogenous adjoin DE satisfying adjoint boundary conditions . Hence using the pair (v_{i}, G) we obtain v_{i} = 0 , from (20) by letting h = 0 (v satisfies the homogenous equation). To resolve this we must modify the differential equations for G and g. They are

L_{x}G(x, y) = \frac{δ(x − y)}{w(x)} − ∑_{k=1} v̄_{i}(x) v_{i}(y),  (34)
L_{x}^{†}g(x, y) = \frac{δ(x − y)}{w(x)} − ∑_{k=1}^{N} ū_{i}(x) u_{i}(y)  (35)

where u_{i}, (i ≤ 2), v_{i}, (i ≤ 2) are orthonormalized solutions of homogenous L_{x}u = 0 and L_{x}^{†}v = 0 satisfying homogenous boundary and adjoint boundary conditons respectively. To restore the uniquenes of the solutions we have the conditions

< u_{i}, G >= 0, (i ≤ 2) < v_{i}, g >= 0(i ≤ 2)  (36)

For the self adjoint operators and for the case there exists only one homogenous solution satisfying the homogenous boundary condtions we have

L_{x}G(x, y) = −ē(x)e(y), x ≠ y  (37)
∫_{a}^{b} G(x, y) e(x)dx = 0,  (38)

here e is normalized solution of the homogenous equation satisfying both homogenous boundary conditions.

See the example in DK page 284. For the solutions to exist we must have (from (21, since h = 0)

< f, e >= ∫_{a}^{b} f(x)e(x)dx = 0

If this condition is satisfied the solution of boundary value proble is given by

u(x) = αe(x) + ∫_{a}^{b} w(y)G(x, y) f(y)dy

where e(x) is any solution of the homogenous equation satisfying the homogenous boundary conditions. Hence we have the following theorem

Theorem 3. Assume that there exists nontrivial solutions of the homogenous DE Lu = 0 satisfying the both boundary condtions B_{1}(u) = and B_{2}(u) = 0 boundary. Then, either the solutions of the boundary value problem Lu = f with B_{1}(u) = 0, B_{2}(u) = 0 do not exist or (if it exists) they are not unique.

Problems:

11. The solution of the example in DK (page 284) is given as follows: Solution does not exist if ∫_{−a}^{a} f(y)dy ≠ 0. If the equality holds then there are infinitely many solutions of the boundary value problem.

u(x) = α + \frac{x}{2a} ∫_{−a}^{a} yf(y)dy + ∫_{−a}^{x}(x − y)f(y)dy

where α is any constant.

12. Solve u′′ + u = f(x) with u(0) = u(π) = 0

13. Solve u′′ = f(x) with u(0) = 0 and u′(1) = 1

Eigenfunction expansion of the Green,s functions Theorem 4. Let L_{x} be second order differential operator and let L_{x}u = f(x), x ∈ [a, b] with the boundary conditions cos αu(a) − sin αu′(a) = 0 and cos βu(b) + sin βu′(b) = 0. The we have the following: (i) the spectrum consists entirely of eigenvalues (ii) The eigenvalues are countable and can be listed in a sequence λ_{1} < λ_{2} < ⋯ < λ_{n} < ⋯ with lim_{n→∞}λ_{n} = 0 (iii) the set of normalized eigenfunctions {u_{i}} is an orthonormal basis for L_{w}^{2}(a, b) (iv) For the equation L_{x}u − λu = f(x) exactly one of the following hold:

(a) If λ is not an eigenvalue of L then the solution is unique for every f ∈ L_{w}^{2}(a, b). This solution is given by

u(x) = ∑_{i=1}^{∞} \frac{(u_{i}, f >}{λ_{i} − λ} u_{i}(x)

(b) If λ = λ_{j} is an eigenvalue then

u(x) = Cu_{j}(x) + ∑_{i≠j}^{∞} \frac{< u_{i}, f >}{λ_{i} − λ} u_{i}(x)

provided < f, u_{j} >= 0 otherwise there is no solution. Here C is an arbitrary constant

Problems

14. Consider the Sturm-Liouville eigenvalue problem consisting of the differential equations (L_{x}u + λu) = 0 and the bounadry conditions B_{1}(u) = 0 and B_{2}(u) = 0

(i) u′′ + λu = 0, u(0) = 0, cos β u(1) + sin βu′(1) = 0 (ii) u′′ + λu = 0, u′(0) = 0, cos β u(1) + sin β u′(1) = 0 (iii) (xu′)′ + λxu = 0, u(a) = u(b) = 0

Higher dimensional Green’s functions: Let ∇^{2} denote the Laplace operator in n− dimensions. Then ∇^{2} G(x) = δ(x) where x ∈ R^{n} has solution (see Set 4, problem 8)

G(x) = \{ \begin{matrix}c_{n} r^{2−n} & n ≥ 3 \\ c_{2} ln r & n = 2 \end{matrix}

where

c_{n} = −\frac{1}{(n − 2)A_{n}} , n ≥ 3
c_{2} = \frac{1}{2π}

Here A_{n} is the surface area of the unit sphere in n dimensions given by A_{n} = \frac{2π^{n/2}}{Γ(n/2)} More problems about the Green’s function technique:

1. Evaluate the Green’s function and solutions for each of the following differential equations in the inetrval [0, 1].

(a) u′′ − k^{2}u = f, u(0) − u′(0) = a, u(1) = b,

(b) u′′ = f, u(0) = u′(0) = 0,

(c) u′′ + 6u′ + 9u = 0, u(0) = 0, u′(0) = 1,

(d) u′′ + w^{2}u = f, for x > 0 u(0) = a, u′(0) = 1,

(e) u^{4} = f, u(0) = 0, y′(0) = 2u′(1), u(1) = a, u′′(0) = 0

2. (From Hildebrand Advanced Calculus for Applications) Consider u′′ + u = f(x) with u(0) = 0 and u(a) = 0 with sin a ≠ 0

(a) Show that

u(x) = ∫_{0}^{x} f(y) sin(x − y)dy + c sin x

where c is such that

c sin a = − ∫_{0}^{a} f(y) sin(a − y)dy

Hence

u(x) = \frac{1}{sin a} [∫_{0}^{x} f(y) sin(x − y) sin ady + ∫_{0}^{a} f(y) sin(a − y) sin x dy]

(b) Prove that

G(x, y) = \{ \begin{matrix}\frac{sin y sin(x−a)}{sin a} & y ≤ x \\ \frac{sin(y−a) sin x}{sin a} & x ≤ y \end{matrix}

when sin a ≠ 0

(c) If sin a = 0, show that the above equations has no solution unless f(x) satisfies the condition

∫_{0}^{a} f(x) sin(a − x)dx = 0

in which case there are infinitely many solutions , each of the form

u(x) = ∫_{0}^{x} f(y) sin(x − y)dy + C sin x

where C is an arbitrary constant

3. Solve u′′ = f(x) with u(0) = α and u′(a) = β and show that u(x) = α + βx + ∫_{0}^{a} G(x, y)f(y)dy

with

G(x, y) = \{ \begin{matrix}−x & x ≤ y \\ −y & x ≥ y \end{matrix}

4. Solve u′′ − u = f(x) with u(−∞) = 0 and u(∞) = 0 and show that

u(x) = \frac{1}{2} ∫_{−∞}^{∞} e^{−|x−y|} f(y)dy

5. Solve u′′ − \frac{1}{x}u′ = f(x) with u(0) = 0 and u(1) = 0 and show that

u(x) = ∫_{0}^{1} G(x, y)f(y)dy

with

G(x, y) = \{ \begin{matrix}−\frac{(1−y^{2})x^{2}}{2y} & x ≤ y \\ −\frac{y(1−x^{2})}{2} & x ≥ y \end{matrix}