Handbook of mathematics for engineers and scienteists part 97 docx

TABLE 14.12 The Green’s functions of boundary value problems for equations of various types in bounded domains.. The formulas for the Green’s functions presented in Table 14.12 will also

TABLE 14.12 The Green’s functions of boundary value problems for equations of various types in bounded

domains In all problems, the operators Lxand Γx are the same; x ={x1, , x n}

Elliptic equation

–Lx[w] =Φ(x) Γx[w] = g(x) for x

S

(no initial condition required) G(x, y) =

∞



k=1

u k (x)u k(y)

u k 2λ k , λ k≠ 0

Parabolic equation

∂ t w – Lx[w] = Φ(x, t)

w = f (x) at t =0

Γx[w] = g(x, t) for xS G(x, y, t) =

∞



k=1

u k (x)u k(y)

u k 2 exp –λ k t

Hyperbolic equation

∂ tt w – Lx[w] = Φ(x, t)

w = f0(x) at t =0

w = f1(x) at t =0

Γx[w] = g(x, t) for xS

G(x, y, t) =

∞



k=1

u k (x)u k(y)

u k 2√

λ k sin t √

λ k

14.11.1-2 Some remarks and generalizations

Remark 1. Formula (14.11.1.3) can also be used if the domain V is infinite In this case, one should

make sure that the integral on the right-hand side is convergent.

Remark 2. Suppose the equations given in the first column of Table 14.12 contain –Lx[w] – βw instead

of –Lx[w], with β being a free parameter Then the λ kin the expressions of the Green’s function in the third

column of Table 14.12 must be replaced by λ k – β; just as previously, the λ k and u k(x) were determined by

solving the eigenvalue problem (14.11.1.1)–(14.11.1.2).

Remark 3 The formulas for the Green’s functions presented in Table 14.12 will also hold for boundary value problems described by equations of the fourth or higher order in the space variables; provided that the eigenvalue problem for equation (14.11.1.1) subject to appropriate boundary conditions is self-adjoint.

14.11.2 Green’s Functions Admitting Incomplete Separation

of Variables

14.11.2-1 Boundary value problems for rectangular domains

1◦ Consider the parabolic equation

∂w

∂t = L1,t [w] + · · · + L n,t [w] + Φ(x, t), (14.11.2.1)

where each term L m,t [w] depends on only one space variable, x m , and time t:

L m,t [w]≡a m (x m , t) ∂

2w

∂x2

m

+ b m (x m , t) ∂w

∂x m + c m (x m , t)w, m=1, , n.

For equation (14.11.2.1) we set the initial condition of general form

Consider the domain V ={α m≤x m≤β m , m =1, , n}, which is an n-dimensional

parallelepiped We set the following boundary conditions at the faces of the parallelepiped:

s(1)

m ∂x ∂w

m + k

(1)

m (t)w = g(1)m (x, t) at x m = α m,

s(2)

m ∂x ∂w

m + k

(2)

m (t)w = g(2)m (x, t) at x m = β m.

(14.11.2.3)

By appropriately choosing the coefficients s(1)m , s(2)m and functions k m(1)= k(1)m (t), k(2)m = k m(2)(t),

we can obtain the boundary conditions of the first, second, or third kind For infinite

domains, the boundary conditions corresponding to α m= –∞ or β m=∞ are omitted.

2◦ The Green’s function of the nonstationary n-dimensional boundary value problem

(14.11.2.1)–(14.11.2.3) can be represented in the product form

G (x, y, t, τ ) =

n



m=1

G m (x m , y m , t, τ ), (14.11.2.4)

where the Green’s functions G m = G m (x m , y m , t, τ ) satisfy the one-dimensional equations

∂G m

∂t – L m,t [G m] =0 (m =1, , n)

with the initial conditions

G m = δ(x m – y m) at t = τ

and the homogeneous boundary conditions

s(1)

m ∂G ∂x m

m + k

(1)

m (t)G m =0 at x m = α m,

s(2)

m ∂G ∂x m

m + k

(2)

m (t)G m =0 at x m = β m

Here, y m and τ are free parameters (α m ≤y m ≤β m and t≥τ ≥ 0), and δ(x) is the Dirac

delta function

It can be seen that the Green’s function (14.11.2.4) admits incomplete separation of

variables; it separates in the space variables x1, , x n but not in time t.

Example Consider the boundary value problem for the two-dimensional nonhomogeneous heat equation

∂w

∂t =∂

2w

∂x2 +∂

2w

∂x2 +Φ(x1, x2, t)

with initial condition (14.11.2.2) and the nonhomogeneous mixed boundary conditions

w = g1(x2, t) at x1= 0 , w = h1(x2, t) at x1= l1;

∂w

∂x2 = g2(x1, t) at x2 = 0 , ∂w

∂x2 = h2(x1, t) at x2= l2 The Green’s functions of the corresponding homogeneous one-dimensional heat equations with homogeneous boundary conditions are expressed as

Equations and boundary conditions Green’s functions

∂w

∂t = ∂

2w

∂x2, w= 0 at x1 = 0, l1 =⇒ G1 = 2

l

∞



m=1

sin(λ m x1) sin(λ m y1)e–λ2m(t–τ) , λ m= mπ

l ;

∂w

∂t = ∂

2w

∂x2, ∂w

∂x2 = 0 at x2= 0, l2 =⇒ G2 = 1

l + 2

l

∞



n=1

sin(σ n x2) sin(σ n 2)e–σ2n(t–τ) , σ n= nπ

l .

Multiplying G1and G2 together gives the Green’s function for the original two-dimensional problem:

G(x1, x2, y1, y2, t, τ ) = 4

l l

∞

m=1

sin(λ m x1) sin(λ m y1)e–λ2m(t–τ)

1

2 +

∞



n=1

sin(σ n x2) sin(σ n 2)e–σ2n(t–τ)

.

14.11.2-2 Boundary value problems for a arbitrary cylindrical domain.

1◦ Consider the parabolic equation

∂w

∂t = Lx,t [w] + M z,t [w] + Φ(x, z, t), (14.11.2.5)

where Lx,t is an arbitrary second-order linear differential operator in x1, , x nwith

co-efficients dependent on x and t, and M z,t is an arbitrary second-order linear differential

operator in z with coefficients dependent on z and t.

For equation (14.11.2.5) we set the general initial condition (14.11.2.2), where f (x) must be replaced by f (x, z).

We assume that the space variables belong to a cylindrical domain V = {x  D,

z1 ≤z≤z2}with arbitrary cross-section D We set the boundary conditions*

Γ1[w] = g1(x, t) at z = z1 (xD),

Γ2[w] = g2(x, t) at z = z2 (xD),

Γ3[w] = g3(x, z, t) for x∂D (z1 ≤z≤z2),

(14.11.2.6)

where the linear boundary operatorsΓk (k =1, 2,3) can define boundary conditions of the first, second, or third kind; in the last case, the coefficients of the differential operatorsΓk

can be dependent on t.

2◦ The Green’s function of problem (14.11.2.5)–(14.11.2.6), (14.11.2.2) can be represented

in the product form

G (x, y, z, ζ, t, τ ) = G L (x, y, t, τ )G M (z, ζ, t, τ ), (14.11.2.7)

where G L = G L (x, y, t, τ ) and G M = G M (z, ζ, t, τ ) are auxiliary Green’s functions; these can

be determined from the following two simpler problems with fewer independent variables:

⎧

⎪

⎨

⎪

∂G L

∂t = Lx,t [G L] for xD,

G L = δ(x – y) at t = τ ,

Γ3[G L] =0 for x∂D,

⎧

⎪

⎨

⎪

∂G M

∂t = M z,t [G M ] for z1 < z < z2,

G M = δ(z – ζ) at t = τ ,

Here, y, ζ, and τ are free parameters (yD , z1≤ζ ≤z2, t≥τ ≥ 0)

It can be seen that the Green’s function (14.11.2.7) admits incomplete separation of

variables; it separates in the space variables x and z but not in time t.

14.11.3 Construction of Green’s Functions via Fundamental

Solutions

14.11.3-1 Elliptic equations Fundamental solution

Consider the elliptic equation

Lx[w] + ∂

2w

* If z1= –∞ or z2 =∞, the corresponding boundary condition is to be omitted.

where x = {x1, , x n}  Rn , z

 R1, and Lx[w] is a linear differential operator that depends on x1, , x n but is independent of z For subsequent analysis it is significant that

the homogeneous equation (withΦ≡ 0) does not change under the replacement of z by –z and z by z + const.

Let = (x, y, z – ζ) be a fundamental solution of equation (14.11.3.1), which means

that

Lx[ ] + ∂

2

∂z2 = δ(x – y)δ(z – ζ).

Here, y ={y1, , y n}Rn and ζ

R1are free parameters.

The fundamental solution of equation (14.11.3.1) is an even function in the last argument, i.e.,

(x, y, z) = (x, y, –z).

Below, Paragraphs 14.11.3-2 and 14.11.3-3 present relations that permit one to express the Green’s functions of some boundary value problems for equation (14.11.3.1) via its fundamental solution

14.11.3-2 Domain: xRn, 0 ≤z<∞ Problems for elliptic equations.

1◦ First boundary value problem The boundary condition:

w = f (x) at z =0 Green’s function:

G (x, y, z, ζ) = (x, y, z – ζ) – (x, y, z + ζ). (14.11.3.2)

Domain of the free parameters: yRnand0 ≤ζ <∞.

Example 1 Consider the first boundary value problem in the half-space –∞ < x1, x2 <∞,0 ≤x3 <∞

for the three-dimensional Laplace equation

∂2w

∂x2 +∂

2w

∂x2 +∂

2w

∂x2 = 0

under boundary condition

w = f (x1, x2) at x3= 0 The fundamental solution for the Laplace equation has the form

4π

(x1– y1 )2+ (x2– y2 )2+ (x3– y3 )2.

In terms of the notation adopted for equation (14.11.3.1) and its fundamental solution, we have x3= z, y3= ζ,

and = (x1, y1, x2, y2, z – ζ) Using formula (14.11.3.2), we obtain the Green’s function for the first boundary

value problem in the half-space:

G(x1, y1, x2, y2, z, ζ) = (x1, y1, x2, y2, z – ζ) – (x1, y1, x2, y2, z + ζ)

4π

(x1– y1)2+ (x2– y2)2+ (x3– y3)2 –

1

4π

(x1– y1)2+ (x2– y2)2+ (x3+ y3)2.

2◦ Second boundary value problem The boundary condition:

∂ z w = f (x) at z =0

Green’s function:

G (x, y, z, ζ) = (x, y, z – ζ) + (x, y, z + ζ).

Example 2 The Green’s function of the second boundary value problem for the three-dimensional Laplace

equation in the half-space –∞ < x 1, x2<∞,0 ≤x3<∞ is expressed as

G(x1, y1, x2, y2, z, ζ) = 1

4π

(x1– y1 )2+ (x2– y2 )2+ (x3– y3 )2 +

1

4π

(x1– y1 )2+ (x2– y2 )2+ (x3+ y3 )2.

It is obtained using the same reasoning as in Example 1.

3◦ Third boundary value problem The boundary condition:

∂ z w – kw = f (x) at z=0 Green’s function:

G (x, y, z, ζ) = (x, y, z – ζ) + (x, y, z + ζ) –2k

–ks (x, y, z + ζ + s) ds

= (x, y, z – ζ) + (x, y, z + ζ) –2k

z+ζ e

–k(σ–z–ζ) (x, y, σ) dσ.

14.11.3-3 Domain: xRn, 0 ≤z≤l Problems for elliptic equations

1◦ First boundary value problem Boundary conditions:

w = f1(x) at z =0, w = f2(x) at z = l.

Green’s function:

G (x, y, z, ζ) =

∞



n=–∞



(x, y, z – ζ +2nl ) – (x, y, z + ζ +2nl)

(14.11.3.3)

Domain of the free parameters: yRnand0 ≤ζ ≤l

2◦ Second boundary value problem Boundary conditions:

∂ z w = f1(x) at z =0, ∂ z w = f2(x) at z = l.

Green’s function:

G (x, y, z, ζ) =

∞



n=–∞



(x, y, z – ζ +2nl ) + (x, y, z + ζ +2nl)

(14.11.3.4)

3◦ Mixed boundary value problem The unknown function and its derivative are prescribed

at the left and right end, respectively:

w = f1(x) at z =0, ∂ z w = f2(x) at z = l.

Green’s function:

G (x, y, z, ζ) =

∞



n=–∞

(–1)n

(x, y, z – ζ +2nl ) – (x, y, z + ζ +2nl)

(14.11.3.5)

4◦ Mixed boundary value problem The derivative and the unknown function itself are

prescribed at the left and right end, respectively:

∂ z w = f1(x) at z =0, w = f2(x) at z = l.

Green’s function:

G (x, y, z, ζ) =

∞



n=–∞

(–1)n

(x, y, z – ζ +2nl ) + (x, y, z + ζ +2nl)

(14.11.3.6) Remark One should make sure that series (14.11.3.3)–(14.11.3.6) are convergent; in particular, for the three-dimensional Laplace equation, series (14.11.3.3), (14.11.3.5), and (14.11.3.6) are convergent and series (14.11.3.4) is divergent.

14.11.3-4 Boundary value problems for parabolic equations.

Let xRn , z

R1, and t≥ 0 Consider the parabolic equation

∂w

∂t = Lx,t [w] + ∂2w

where Lx,t [w] is a linear differential operator that depends on x1, , x n and t but is independent of z.

Let = (x, y, z – ζ, t, τ ) be a fundamental solution of the Cauchy problem for

equa-tion (14.11.3.7), i.e.,

∂

∂t = Lx,t[ ] + ∂

2

∂z2 for t > τ ,

= δ(x – y) δ(z – ζ) at t = τ

Here, yRn , ζ

R1, and τ ≥ 0are free parameters

The fundamental solution of the Cauchy problem possesses the property

(x, y, z, t, τ ) = (x, y, –z, t, τ ).

Table 14.13 presents formulas that permit one to express the Green’s functions of some nonstationary boundary value problems for equation (14.11.3.7) via the fundamental solution of the Cauchy problem

TABLE 14.13 Representation of the Green’s functions of some nonstationary boundary

value problems in terms of the fundamental solution of the Cauchy problem

Boundary value

First problem

x Rn , z R 1 G= 0 at z =0 G(x, y, z, ζ, t, τ ) = (x, y, z – ζ, t, τ ) – (x, y, z + ζ, t, τ )

Second problem

x Rn , z R 1 ∂ z G= 0 at z =0 G(x, y, z, ζ, t, τ ) = (x, y, z – ζ, t, τ ) + (x, y, z + ζ, t, τ )

Third problem

x Rn , z R 1 ∂ z G – kG =0 at z =0

G(x, y, z, ζ, t, τ ) = (x, y, z – ζ, t, τ ) + (x, y, z + ζ, t, τ )

– 2k

 ∞

0

e–ks (x, y, z + ζ + s, t, τ ) ds

First problem

x Rn, 0 ≤z≤l

G= 0 at z =0 ,

G= 0 at z = l

G(x, y, z, ζ, t, τ ) = ∞

n=–∞



(x, y, z – ζ +2nl, t, τ )

– (x, y, z + ζ +2nl, t, τ )

Second problem

x Rn, 0 ≤z≤l

∂ z G= 0 at z =0 ,

∂ z G= 0 at z = l

G(x, y, z, ζ, t, τ ) = ∞

n=–∞



(x, y, z – ζ +2nl, t, τ )

+ (x, y, z + ζ +2nl, t, τ )

Mixed problem

x Rn, 0 ≤z≤l

G= 0 at z =0 ,

∂ z G= 0 at z = l

G(x, y, z, ζ, t, τ ) = ∞

n=–∞(– 1 )n

(x, y, z – ζ +2nl, t, τ )

– (x, y, z + ζ +2nl, t, τ )

Mixed problem

x Rn, 0 ≤z≤l

∂ z G= 0 at z =0 ,

G= 0 at z = l

G(x, y, z, ζ, t, τ ) = ∞

n=–∞(– 1 )n

(x, y, z – ζ +2nl, t, τ )

+ (x, y, z + ζ +2nl, t, τ )

14.12 Duhamel’s Principles in Nonstationary Problems

14.12.1 Problems for Homogeneous Linear Equations

14.12.1-1 Parabolic equations with two independent variables

Consider the problem for the homogeneous linear equation of parabolic type

∂w

∂t = a(x) ∂2w

∂x2 + b(x)

∂w

with the homogeneous initial condition

w=0 at t=0 (14.12.1.2) and the boundary conditions

s1∂ x w + k1w = g(t) at x = x1, (14.12.1.3)

s2∂ x w + k2w=0 at x = x2 (14.12.1.4)

By appropriately choosing the values of the coefficients s1, s2, k1, and k2 in (14.12.1.3) and (14.12.1.4), one can obtain the first, second, third, and mixed boundary value problems for equation (14.12.1.1)

The solution of problem (14.12.1.1)–(14.12.1.4) with the nonstationary boundary

con-dition (14.12.1.3) at x = x1can be expressed by the formula (Duhamel’s first principle)

w (x, t) = ∂

∂t

0 u (x, t – τ ) g(τ ) dτ =

0

∂u

∂t (x, t – τ ) g(τ ) dτ (14.12.1.5)

in terms of the solution u(x, t) of the auxiliary problem for equation (14.12.1.1) with the initial and boundary conditions (14.12.1.2) and (14.12.1.4), for u instead of w, and the following simpler stationary boundary condition at x = x1:

s1∂ x u + k1u=1 at x = x1 (14.12.1.6) Remark. A similar formula also holds for the homogeneous boundary condition at x = x1 and a

nonho-mogeneous nonstationary boundary condition at x = x2

Example Consider the first boundary value problem for the heat equation

∂w

∂t = ∂

2w

with the homogeneous initial condition (14.12.1.2) and the boundary condition

(The second boundary condition is not required in this case; 0 ≤x<∞.)

First consider the following auxiliary problem for the heat equation with the homogeneous initial condition and a simpler boundary condition:

∂u

∂t = ∂

2u

∂x2, u= 0 at t= 0 , u= 1 at x= 0 This problem has a self-similar solution of the form

w = w(z), z = xt–1/2,

where the function w(z) is determined by the following ordinary differential equation and boundary conditions:

u  zz+12zu  z = 0 , u= 1 at z= 0 , u= 0 at z=∞.

Its solution is expressed as

u(z) = erfc



z

2



=⇒ u(x, t) = erfc



x

2√ t



,

where erfc z = √2

π

 ∞

z exp(–ξ2) dξ is the complementary error function Substituting the obtained expression

of u(x, t) into (14.12.1.5), we obtain the solution to the first boundary value problem for the heat equation

(14.12.1.7) with the initial condition (14.12.1.2) and an arbitrary boundary condition (14.12.1.8) in the form

w(x, t) = 2√ x π

 t

0 exp



2

4(t – τ )



g(τ ) dτ (t – τ )3/2.

in terms of the solution u(x, t) of the auxiliary problem for equation (14.12.1.1) with the initial and boundary conditions (14.12.1.2) and (14.12.1.4), for u instead of w, and the following... x1, , x n and t but is independent of z.

Let = (x, y, z – ζ, t, τ ) be a fundamental solution of the Cauchy problem for

equa-tion (14.11.3.7),...

Table 14.13 presents formulas that permit one to express the Green’s functions of some nonstationary boundary value problems for equation (14.11.3.7) via the fundamental solution of the Cauchy problem