Handbook of mathematics for engineers and scienteists part 97 docx

640 LINEAR PARTIAL DIFFERENTIAL EQUATIONS TABLE 14.12 The Green’s functions of boundary value problems for equations of various types in bounded domains. In all problems, the operators L x and Γ x are the same; x = {x 1 , ,x n } Equation Initial and boundary conditions Green’s function Elliptic equation –L x [w]=Φ(x) Γ x [w]=g(x)forx S (no initial condition required) G(x, y)= ∞  k=1 u k (x)u k (y) u k  2 λ k , λ k ≠ 0 Parabolic equation ∂ t w – L x [w]=Φ(x, t) w = f(x)att = 0 Γ x [w]=g(x, t)forx S G(x, y, t)= ∞  k=1 u k (x)u k (y) u k  2 exp  –λ k t  Hyperbolic equation ∂ tt w – L x [w]=Φ(x, t) w = f 0 (x)att = 0 w = f 1 (x)att = 0 Γ x [w]=g(x, t)forx S 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 –L x [w]–βw instead of –L x [w], with β being a free parameter. Then the λ k in 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 = L 1,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) ∂ 2 w ∂x 2 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 w = f(x)att = 0.(14.11.2.2) 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 ∂w ∂x m + k (1) m (t)w = g (1) m (x, t)atx m = α m , s (2) m ∂w ∂x m + k (2) m (t)w = g (2) m (x, t)atx m = β m . (14.11.2.3) 14.11. CONSTRUCTION OF THE GREEN’S FUNCTIONS.GENERAL FORMULAS AND RELATIONS 641 By appropriately choosing the coefficients s (1) m , s (2) m and functions k (1) m = k (1) m (t), k (2) m = k (2) m (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 )att = τ and the homogeneous boundary conditions s (1) m ∂G m ∂x m + k (1) m (t)G m = 0 at x m = α m , s (2) m ∂G m ∂x 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)istheDirac 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 x 1 , , x n but not in time t. Example. Consider the boundary value problem for the two-dimensional nonhomogeneous heat equation ∂w ∂t = ∂ 2 w ∂x 2 1 + ∂ 2 w ∂x 2 2 + Φ(x 1 , x 2 , t) with initial condition (14.11.2.2) and the nonhomogeneous mixed boundary conditions w = g 1 (x 2 , t)atx 1 = 0, w = h 1 (x 2 , t)atx 1 = l 1 ; ∂w ∂x 2 = g 2 (x 1 , t)atx 2 = 0, ∂w ∂x 2 = h 2 (x 1 , t)atx 2 = l 2 . 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 = ∂ 2 w ∂x 2 1 , w = 0 at x 1 = 0, l 1 =⇒ G 1 = 2 l 1 ∞  m=1 sin(λ m x 1 )sin(λ m y 1 )e –λ 2 m (t–τ ) , λ m = mπ l 1 ; ∂w ∂t = ∂ 2 w ∂x 2 2 , ∂w ∂x 2 = 0 at x 2 = 0, l 2 =⇒ G 2 = 1 l 2 + 2 l 2 ∞  n=1 sin(σ n x 2 )sin(σ n y 2 )e –σ 2 n (t–τ ) , σ n = nπ l 2 . Multiplying G 1 and G 2 together gives the Green’s function for the original two-dimensional problem: G(x 1 , x 2 , y 1 , y 2 , t, τ )= 4 l 1 l 2  ∞  m=1 sin(λ m x 1 )sin(λ m y 1 )e –λ 2 m (t–τ )  1 2 + ∞  n=1 sin(σ n x 2 )sin(σ n y 2 )e –σ 2 n (t–τ )  . 642 LINEAR PARTIAL DIFFERENTIAL EQUATIONS 14.11.2-2. Boundary value problems for a arbitrary cylindrical domain. 1 ◦ . Consider the parabolic equation ∂w ∂t = L x,t [w]+M z,t [w]+Φ(x, z, t), (14.11.2.5) where L x,t is an arbitrary second-order linear differential operator in x 1 , , x n with co- efficients dependent on x and t,andM 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, z 1 ≤ z ≤ z 2 } with arbitrary cross-section D. We set the boundary conditions* Γ 1 [w]=g 1 (x, t)atz = z 1 (x D), Γ 2 [w]=g 2 (x, t)atz = z 2 (x D), Γ 3 [w]=g 3 (x, z, t)forx ∂D (z 1 ≤ z ≤ z 2 ), (14.11.2.6) where the linear boundary operators Γ k (k = 1, 2, 3) can define boundary conditions of the first, second, or third kind; in the last case, the coefficients of the differential operators Γ k 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)andG 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: Problem on the cross-section D: Problem on the interval z 1 ≤ z ≤ z 2 : ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ∂G L ∂t = L x,t [G L ]forx D, G L = δ(x – y)att = τ, Γ 3 [G L ]=0 for x ∂D, ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ∂G M ∂t = M z,t [G M ]forz 1 < z < z 2 , G M = δ(z – ζ)att = τ, Γ k [G M ]=0 at z = z k (k = 1, 2). Here, y, ζ,andτ are free parameters (y D, z 1 ≤ ζ ≤ z 2 , 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 L x [w]+ ∂ 2 w ∂z 2 = Φ(x, z), (14.11.3.1) *Ifz 1 =–∞ or z 2 = ∞ , the corresponding boundary condition is to be omitted. 14.11. CONSTRUCTION OF THE GREEN’S FUNCTIONS.GENERAL FORMULAS AND RELATIONS 643 where x = {x 1 , , x n } R n , z R 1 ,andL x [w] is a linear differential operator that depends on x 1 , , x n but is independent of z. For subsequent analysis it is significant that the homogeneous equation (with Φ ≡ 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 L x [ ]+ ∂ 2 ∂z 2 = δ(x – y)δ(z – ζ). Here, y = {y 1 , , y n } R n and ζ R 1 are 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: x R n , 0 ≤ z < ∞. Problems for elliptic equations. 1 ◦ . First boundary value problem. The boundary condition: w = f(x)atz = 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: y R n and 0 ≤ ζ < ∞. Example 1. Consider the fi rst boundary value problem in the half-space –∞ < x 1 , x 2 < ∞, 0 ≤ x 3 < ∞ for the three-dimensional Laplace equation ∂ 2 w ∂x 2 1 + ∂ 2 w ∂x 2 2 + ∂ 2 w ∂x 2 3 = 0 under boundary condition w = f(x 1 , x 2 )atx 3 = 0. The fundamental solution for the Laplace equation has the form = 1 4π  (x 1 – y 1 ) 2 +(x 2 – y 2 ) 2 +(x 3 – y 3 ) 2 . In terms of the notation adopted for equation (14.11.3.1) and its fundamental solution, we have x 3 = z, y 3 = ζ, and = (x 1 , y 1 , x 2 , y 2 , 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(x 1 , y 1 , x 2 , y 2 , z, ζ)= (x 1 , y 1 , x 2 , y 2 , z – ζ)– (x 1 , y 1 , x 2 , y 2 , z + ζ) = 1 4π  (x 1 – y 1 ) 2 +(x 2 – y 2 ) 2 +(x 3 – y 3 ) 2 – 1 4π  (x 1 – y 1 ) 2 +(x 2 – y 2 ) 2 +(x 3 + y 3 ) 2 . 2 ◦ . Second boundary value problem. The boundary condition: ∂ z w = f(x)atz = 0. 644 LINEAR PARTIAL DIFFERENTIAL EQUATIONS 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 , x 2 < ∞, 0 ≤ x 3 < ∞ is expressed as G(x 1 , y 1 , x 2 , y 2 , z, ζ)= 1 4π  (x 1 – y 1 ) 2 +(x 2 – y 2 ) 2 +(x 3 – y 3 ) 2 + 1 4π  (x 1 – y 1 ) 2 +(x 2 – y 2 ) 2 +(x 3 + y 3 ) 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)atz = 0. Green’s function: G(x, y, z, ζ)= (x, y, z – ζ)+ (x, y, z + ζ)–2k  ∞ 0 e –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: x R n , 0 ≤ z ≤ l. Problems for elliptic equations. 1 ◦ . First boundary value problem. Boundary conditions: w = f 1 (x)atz = 0, w = f 2 (x)atz = 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: y R n and 0 ≤ ζ ≤ l. 2 ◦ . Second boundary value problem. Boundary conditions: ∂ z w = f 1 (x)atz = 0, ∂ z w = f 2 (x)atz = 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 = f 1 (x)atz = 0, ∂ z w = f 2 (x)atz = 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 = f 1 (x)atz = 0, w = f 2 (x)atz = 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. CONSTRUCTION OF THE GREEN’S FUNCTIONS.GENERAL FORMULAS AND RELATIONS 645 14.11.3-4. Boundary value problems for parabolic equations. Let x R n , z R 1 ,andt ≥ 0. Consider the parabolic equation ∂w ∂t = L x,t [w]+ ∂ 2 w ∂z 2 + Φ(x, z, t), (14.11.3.7) where L x,t [w] is a linear differential operator that depends on x 1 , , 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 = L x,t [ ]+ ∂ 2 ∂z 2 for t > τ , = δ(x – y) δ(z – ζ)att = τ . Here, y R n , ζ R 1 ,andτ ≥ 0 are 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 problems Boundary conditions Green’s functions First problem x R n , 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 R n , 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 R n , 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 R n , 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 R n , 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 R n , 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 R n , 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)  646 LINEAR PARTIAL DIFFERENTIAL EQUATIONS 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) ∂ 2 w ∂x 2 + b(x) ∂w ∂x + c(x)w (14.12.1.1) with the homogeneous initial condition w = 0 at t = 0 (14.12.1.2) and the boundary conditions s 1 ∂ x w + k 1 w = g(t)atx = x 1 ,(14.12.1.3) s 2 ∂ x w + k 2 w = 0 at x = x 2 .(14.12.1.4) By appropriately choosing the values of the coefficients s 1 , s 2 , k 1 ,andk 2 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 = x 1 can be expressed by the formula (Duhamel’s first principle) w(x, t)= ∂ ∂t  t 0 u(x, t – τ) g(τ)dτ =  t 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,andthe following simpler stationary boundary condition at x = x 1 : s 1 ∂ x u + k 1 u = 1 at x = x 1 .(14.12.1.6) Remark. A similar formula also holds for the homogeneous boundary condition at x = x 1 and a nonho- mogeneous nonstationary boundary condition at x = x 2 . Example. Consider the first boundary value problem for the heat equation ∂w ∂t = ∂ 2 w ∂x 2 (14.12.1.7) with the homogeneous initial condition (14.12.1.2) and the boundary condition w = g(t)atx = 0. (14.12.1.8) (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 = ∂ 2 u ∂x 2 , 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 + 1 2 zu  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(–ξ 2 ) 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)= x 2 √ π  t 0 exp  – x 2 4(t – τ)  g(τ )dτ (t – τ ) 3/2 . . values of the coefficients s 1 , s 2 , k 1 ,andk 2 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. OF THE GREEN’S FUNCTIONS.GENERAL FORMULAS AND RELATIONS 643 where x = {x 1 , , x n } R n , z R 1 ,andL x [w] is a linear differential operator that depends on x 1 , , x n but is independent of. fundamental solution for the Laplace equation has the form = 1 4π  (x 1 – y 1 ) 2 +(x 2 – y 2 ) 2 +(x 3 – y 3 ) 2 . In terms of the notation adopted for equation (14.11.3.1) and its fundamental