LECTURE 1: DIFFERENTIAL OPERATIONS AND THEOREMS OF THE VECTOR ANALYSIS. Definition of rotation. Let x, y, z be a positive oriented Cartesian coordinate system and v ( x, y, z ) = v 1 ( x, y, z ) i + v 2 ( x, y, z ) j + v 3 ( x, y, z ) k a differentiable vector function. Then the vector function � � � i j k � � � � � ∂ ∂ ∂ curl v = ∇ × v = = � � � ∂x ∂y ∂z � � � v 1 v 2 v 3 � � � � � ∂v 3 � ∂v 1 � ∂v 2 ∂y − ∂v 2 � ∂z − ∂v 3 � ∂x − ∂v 1 � i + j + k ∂z ∂x ∂y is called rotation (or curl) of vector field v . 17 / 143
LECTURE 1: DIFFERENTIAL OPERATIONS AND THEOREMS OF THE VECTOR ANALYSIS. Example 4 Let x, y, z be a positive oriented Cartesian coordinate system. Find curl of the vector field v ( x, y, z ) = yz i + 3 zx j + z k . Solution. The curl of v is calculated according to � � � i j k � � � � � ∂ ∂ ∂ curl v = = � � ∂x ∂y ∂z � � � � yz 3 xz z � � � � � ∂z � ∂ ( yz ) � ∂ (3 xz ) ∂y − ∂ (3 xz ) � − ∂z � − ∂ ( yz ) � i + j + k = − 3 x i + y j + 2 z k . ∂z ∂z ∂x ∂x ∂y 18 / 143
LECTURE 1: DIFFERENTIAL OPERATIONS AND THEOREMS OF THE VECTOR ANALYSIS. Theorem 4 For any twice continuously differentiable scalar function f , curl ( grad f ) = 0 . (1) The potential (or conservative) field is called rotation-free. Proof. � � � � � i j k � � i j k � � � � � � � � � ∂ ∂ ∂ ∂ ∂ ∂ curl ( grad f ) = = = � � � � ∂x ∂y ∂z ∂x ∂y ∂z � � � � � � � � ∂f ∂f ∂f f x f y f z � � � � � ∂x ∂y ∂z � � � � ∂f z � ∂f x � ∂f y ∂y − ∂f y � ∂z − ∂f z � ∂x − ∂f x � i + j + k = ( f zy − f yz ) i + ( f xz − f zx ) j + ( f yx − f xy ) k = 0 . ∂z ∂x ∂y 19 / 143
LECTURE 1: DIFFERENTIAL OPERATIONS AND THEOREMS OF THE VECTOR ANALYSIS. Theorem 5 For any twice continuously differentiable vector function v , div ( curl v ) = 0 . (2) The field of rotation is called divergence-free. Proof. � ∂v 3 � ∂v 1 � ∂v 2 div ( curl v ) = ∂ ∂y − ∂v 2 � + ∂ ∂z − ∂v 3 � + ∂ ∂x − ∂v 1 � = ∂x ∂z ∂y ∂x ∂z ∂y ( v 3 yx − v 2 zx ) + ( v 1 zy − v 3 xy ) + ( v 2 xz − v 1 yz ) = 0 . 20 / 143
LECTURE 1: DIFFERENTIAL OPERATIONS AND THEOREMS OF THE VECTOR ANALYSIS. More vector differential identities: ∇ ( φψ ) = ψ ∇ φ + φ ∇ ψ. ∇ · ( φ F ) = div ( φ F ) = ∇ φ · F + φ ∇ · F . ∇ · ( F × G ) = ∇ × F · G − F · ∇ × G . ∇ × ( ∇ × F ) = ∇ ( ∇ · F ) − ∇ 2 F . (3) 21 / 143
LECTURE 1: DIFFERENTIAL OPERATIONS AND THEOREMS OF THE VECTOR ANALYSIS. Formulate the divergence theorem of Gauss. Theorem 6 Let T be a closed bounded region in space whose boundary is a piecewise smooth orientable surface S . Let F ( x, y, z ) be a vector function that is continuous and has continuous first partial derivatives in some domain containing T . Then � � � � � div F dV = F · n dA. T S In components � ∂F 1 � � � ∂x + ∂F 2 ∂y + ∂F 3 � � � dxdydz = ( F 1 cos α + F 2 cos β + F 3 cos γ ) dA. ∂z T S or � ∂F 1 � � � ∂x + ∂F 2 ∂y + ∂F 3 � � � dxdydz = ( F 1 dydz + F 2 dzdx + F 3 dxdy ) . ∂z T S 22 / 143
LECTURE 1: DIFFERENTIAL OPERATIONS AND THEOREMS OF THE VECTOR ANALYSIS. Example 5 Evaluate � � ( x 3 dydz + x 2 ydzdx + x 2 zdxdy ) , I = (4) S where S is the closed surface consisting of the cylinder x 2 + y 2 = a 2 ( 0 ≤ z ≤ b ) and the circular disks z = 0 and z = b ( x 2 + y 2 ≤ a 2 ). Solution. F 1 = x 3 , F 2 = x 2 y, F 3 = x 2 z. Hence the divergence of F = [ F 1 , F 2 , F 3 ] is div F = ∂F 1 ∂x + ∂F 2 ∂y + ∂F 3 = 3 x 2 + x 2 + x 2 = 5 x 2 . ∂z The form of the surface suggests that we introduce polar coordinates x = r cos θ, y = r sin θ ( cylindriska koordinater r, θ, z ) and dxdydz = rdrdθdz, 23 / 143
LECTURE 1: DIFFERENTIAL OPERATIONS AND THEOREMS OF THE VECTOR ANALYSIS. According to Gauss’s theorem, a surface integral is reduced to a triple integral if the area T is bounded by a cylindrical surface S , � � � � � � � � ( x 3 dydz + x 2 ydzdx + x 2 zdxdy ) = 5 x 2 dxdydz = div F dV = S T T � b � a � 2 π r 2 cos 2 θrdrdθdz = 5 z =0 r =0 θ =0 � a � 2 π � 2 π r 3 cos 2 θdrdθ = 5 b a 4 cos 2 θdθ = 5 b 4 0 0 0 � 2 π 5 b a 4 (1 + 2 cos θ ) dθ = 5 4 πba 4 . 8 0 24 / 143
LECTURE 1: DIFFERENTIAL OPERATIONS AND THEOREMS OF THE VECTOR ANALYSIS. Example 6 Evaluate � � I = F · n dA, F = 7 x i − z k S over the sphere S : x 2 + y 2 + z 2 = 4 . Calculate the integral directly and using Gauss’s theorem. Solution. F ( x, y, z ) = [ F 1 , F 2 , F 3 ] is a differentiable vector function and its components are F = [ F 1 , 0 , F 3 ] , F 1 = 7 x, F 3 = − z. The divergence of F is div F = ∂F 1 ∂x + ∂F 2 ∂y + ∂F 3 = 7 + 0 − 1 = 6 . ∂z Accordingly, dxdydz = 6 · 4 � � � � � � 3 π 2 3 = 64 π. I = div F dV = 6 (5) T, klot T, klot 25 / 143
LECTURE 1: DIFFERENTIAL OPERATIONS AND THEOREMS OF THE VECTOR ANALYSIS. The surface integral of S can be calculated directly. Parametric representation of the sphere of radius 2 S : r ( u, v ) = 2 cos v cos u i + 2 cos v sin u j + 2 sin v k , u, v i rectangle R : 0 ≤ u ≤ 2 π, − π/ 2 ≤ v ≤ π/ 2 . Determine the partial derivatives r u = [ − 2 sin u cos v, 2 cos v cos u, 0] , r v = [ − 2 sin v cos u, − 2 sin v sin u, 2 cos v ] , and the normal vector � � � i j k � � � � � = [4 cos 2 v cos u, 4 cos 2 v sin u, 4 cos v sin v ] . N = r u × r v = − 2 sin u cos v 2 cos v cos u 0 � � � � � � � − 2 sin v cos u − 2 sin v sin u 2 cos v � � � On surface S , x = 2 cos v cos u, z = 2 sin v, and F ( r ( u, v )) = F ( S ) = [7 x, 0 , − z ] = [14 cos v cos u, 0 , − 2 sin v ] . 26 / 143
LECTURE 1: DIFFERENTIAL OPERATIONS AND THEOREMS OF THE VECTOR ANALYSIS. Then F ( r ( u, v )) · N ( u, v ) = (14 cos v cos u )4 cos 2 v cos u + ( − 2 sin v )(4 cos v sin v ) = 56 cos 3 v cos 2 u − 8 cos v sin 2 u. The parameters u, v vary in the rectangle R : 0 ≤ u ≤ 2 π, − π/ 2 ≤ v ≤ π/ 2 . Now, we can write and calculate the surface integral: � 2 π � − π/ 2 � � � � (7 cos 3 v cos 2 u − cos v sin 2 v ) dudv = F · n dA = F ( r ( u, v )) · N ( u, v ) dudv = 8 S R 0 − π/ 2 � 2 π � π/ 2 � π/ 2 � � 7 cos 3 vdv − 2 π cos v sin 2 vdv 8 (1 + cos 2 u ) du = 2 0 − π/ 2 − π/ 2 � π/ 2 � π/ 2 cos 3 vdv − 16 π cos vdv sin 2 vdv = 56 π − π/ 2 − π/ 2 � π/ 2 � π/ 2 � � (1 − sin 2 v ) d sin v − 2 dv sin 2 vd sin v 8 π 7 = − π/ 2 − π/ 2 � 1 � 1 � � (1 − t 2 ) dt − 2 t 2 dt 8 π 7 = 8 π [7 · (2 − 2 / 3) − 4 / 3] = 8 π · 4 / 3 · 6 = 64 π. − 1 − 1 coinciding with the value (5). 27 / 143
LECTURE 1 HARMONIC FUNCTIONS 28 / 143
LECTURE 1: HARMONIC FUNCTIONS A twice continuously differentiable real-valued function u defined on a domain D is called harmonic if it satisfies Laplace’s equation ∆ u = 0 in D, (6) where ∆ u = ∂ 2 u ∂x 2 + ∂ 2 u (7) ∂y 2 is called Laplace operator (Laplacian), the function u = u ( x ) , and x = ( x, y ) ∈ R 2 . We will also use the notation y = ( x 0 , y 0 ) . The function Φ( x , y ) = Φ( x − y ) = 1 1 2 π ln (8) | x − y | is called the fundamental solution of the Laplace equation . For a fixed y ∈ R 2 , y � = x , the function Φ( x , y ) is harmonic, i.e., satisfies Laplace’s equation ∂ 2 Φ ∂x 2 + ∂ 2 Φ ∂y 2 = 0 in D. (9) The proof follows by straightforward differentiation. 29 / 143
LECTURE 1: HARMONIC FUNCTIONS Let D ∈ R 2 be a (two-dimensional) domain bounded by the closed smooth contour Γ and ∂ ∂n y denote the directional derivative in the direction of unit normal vector n y to the boundary Γ directed into the exterior of Γ and corresponding to a point y ∈ Γ . Then for every function u which is once continuously differentiable in the closed domain ¯ D = D + Γ , u ∈ C 1 ( ¯ D ) , and every function v which is twice continuously differentiable in ¯ D , v ∈ C 2 ( ¯ D ) , Green’s first theorem (Green’s first formula) is valid � � � u ∂v ( u ∆ v + grad u · grad v ) d x = dl y , (10) ∂n y D Γ where · denotes the inner product of two vector-functions. For u ∈ C 2 ( ¯ D ) and v ∈ C 2 ( ¯ D ) , Green’s second theorem (Green’s second formula) is valid � � � � u ∂v − v ∂u � ( u ∆ v − v ∆ u ) d x = dl y , (11) ∂n y ∂n y D Γ Let a twice continuously differentiable function u ∈ C 2 ( ¯ D ) be harmonic in the domain D . Then Green’s third theorem (Green’s third formula) is valid � � Φ( x , y ) ∂u − u ( y ) ∂ Φ( x , y ) � u ( x ) = dl y , x ∈ D. (12) ∂n y ∂n y Γ 30 / 143
LECTURE 1: HARMONIC FUNCTIONS Formulate the interior Dirichlet problem : find a function u that is harmonic in a domain D bounded by the closed smooth contour Γ , continuous in ¯ D = D ∪ Γ and satisfies the Dirichlet boundary condition: ∆ u = 0 in D, (13) u | Γ = − f, (14) where f is a given continuous function. Formulate the interior Neumann problem : find a function u that is harmonic in a domain D bounded by the closed smooth contour Γ , continuous in ¯ D = D ∪ Γ and satisfies the Neumann boundary condition ∂u � � = − g, (15) � ∂n � Γ where g is a given continuous function. Theorem 7 The interior Dirichlet problem has at most one solution. Theorem 8 Two solutions of the interior Neumann problem can differ only by a constant. The exterior Neumann problem has at most one solution. 31 / 143
LECTURE 1: HARMONIC FUNCTIONS In the theory of BVPs, the integrals � � ∂ u ( x ) = E ( x , y ) ξ ( y ) dl y , v ( x ) = E ( x , y ) η ( y ) dl y (16) ∂ n y C C are called the potentials . Here, x = ( x, y ) , y = ( x 0 , y 0 ) ∈ R 2 ; E ( x , y ) is the fundamental solution of a second-order elliptic differential operator; ∂ ∂ = ∂ n y ∂n y is the normal derivative at the point y of the closed piecewise smooth boundary C of a domain in R 2 ; and ξ ( y ) and η ( y ) are sufficiently smooth functions defined on C . In the case of Laplace operator ∆ u , 1 1 E ( x , y ) = Φ( x − y ) = 2 π ln | x − y | . (17) 32 / 143
LECTURE 1: HARMONIC FUNCTIONS In the case of the Helmholtz operator L ( k ) = ∆ + k 2 , one can take E ( x , y ) in the form E ( x , y ) = E ( x − y ) = i ( k | x − y | ) = 1 1 4 H (1) 2 π ln | x − y | + h ( k | x − y | ) , (18) 0 where H (1) ( z ) = − 4 i Φ( z ) + ˜ h ( z ) is the Hankel function of the first kind and zero order (one of the so-called 0 1 1 cylindrical functions) and Φ( x − y ) = 2 π ln | x − y | is the kernel of the two-dimensional single layer potential; ˜ h ( z ) and h ( z ) are continuously differentiable and their second derivatives have a logarithmic singularity. 33 / 143
LECTURE 1: HARMONIC FUNCTIONS Theorem 9 Let D ∈ R 2 be a domain bounded by the closed smooth contour Γ . Then the kernel of the double-layer potential V ( x , y ) = ∂ Φ( x , y ) Φ( x , y ) = 1 1 , 2 π ln | x − y | , (19) ∂n y is a continuous function on Γ for x , y ∈ Γ . Gauss formula Let D ∈ R 2 be a domain bounded by the closed smooth contour Γ . For the double-layer potential with a constant density � ∂ Φ( x , y ) Φ( x , y ) = 1 1 v 0 ( x ) = dl y , 2 π ln | x − y | , (20) ∂n y Γ where the (exterior) unit normal vector n of Γ is directed into the exterior domain R 2 \ ¯ D , we have v 0 ( x ) = − 1 , x ∈ D, − 1 v 0 ( x ) = 2 , x ∈ Γ , (21) x ∈ R 2 \ ¯ v 0 ( x ) = 0 , D. 34 / 143
LECTURE 1: HARMONIC FUNCTIONS Corollary. Let D ∈ R 2 be a domain bounded by the closed smooth contour Γ . Introduce the single-layer potential with a constant density � Φ( x , y ) = 1 1 u 0 ( x ) = Φ( x , y ) dl y , 2 π ln | x − y | . (22) Γ For the normal derivative of this single-layer potential ∂u 0 ( x ) � ∂ Φ( x , y ) = dl y , (23) ∂n x ∂n x Γ where the (exterior) unit normal vector n x of Γ is directed into the exterior domain R 2 \ ¯ D , we have ∂u 0 ( x ) = 1 , x ∈ D, ∂n x ∂u 0 ( x ) 1 = 2 , x ∈ Γ , (24) ∂n x ∂u 0 ( x ) x ∈ R 2 \ ¯ = 0 , D. ∂n x 35 / 143
LECTURE 1: HARMONIC FUNCTIONS Theorem 10 Let D ∈ R 2 be a domain bounded by the closed smooth contour Γ . The double-layer potential � ∂ Φ( x , y ) Φ( x , y ) = 1 1 v ( x ) = ϕ ( y ) dl y , 2 π ln | x − y | , (25) ∂n y Γ D and from R 2 \ ¯ D to R 2 \ D with the with a continuous density ϕ can be continuously extended from D to ¯ limiting values on Γ ∂ Φ( x ′ , y ) � ϕ ( y ) dl y ± 1 x ′ ∈ Γ , v ± ( x ′ ) = 2 ϕ ( x ′ ) , (26) ∂n y Γ or v ± ( x ′ ) = v ( x ′ ) ± 1 x ′ ∈ Γ , 2 ϕ ( x ′ ) , (27) where v ± ( x ′ ) = h →± 0 v ( x + hn x ′ ) . lim (28) 36 / 143
LECTURE 1: HARMONIC FUNCTIONS Let D ∈ R 2 be a domain bounded by the closed smooth contour Γ . Introduce the single-layer Corollary. potential � Φ( x , y ) = 1 1 u ( x ) = Φ( x , y ) ϕ ( y ) dl y , 2 π ln | x − y | . (29) Γ with a continuous density ϕ . The normal derivative of this single-layer potential ∂u ( x ) ∂ Φ( x , y ) � = ϕ ( y ) dl y (30) ∂n x ∂n x Γ D and from R 2 \ ¯ D to R 2 \ D with the limiting values on Γ can be continuously extended from D to ¯ ∂u ( x ′ ) ∂ Φ( x ′ , y ) � ϕ ( y ) dl y ∓ 1 x ′ ∈ Γ , 2 ϕ ( x ′ ) , = (31) ∂n x ∂n x ′ ± Γ or ∂u ( x ′ ) = ∂u ( x ′ ) ∓ 1 x ′ ∈ Γ , 2 ϕ ( x ′ ) , (32) ∂n x ∂n x ± where ∂u ( x ′ ) h →± 0 n x ′ · grad v ( x ′ + hn x ′ ) . = lim (33) ∂n x ′ 37 / 143
LECTURE 1: HARMONIC FUNCTIONS Let S Π (Γ) ∈ R 2 be a domain bounded by the closed piecewise smooth contour Γ . We assume that a rectilinear interval Γ 0 is a subset of Γ , so that Γ 0 = { x : y = 0 , x ∈ [ a, b ] } . Let us say that functions U l ( x ) are the generalized single layer (SLP) ( l = 1) or double layer (DLP) ( l = 2) potentials if � U l ( x ) = K l ( x , t ) l ( t ) dt, x = ( x, y ) ∈ S Π (Γ) , Γ where K l ( x , t ) = g l ( x , t ) + F l ( x , t ) ( l = 1 , 2) , g 1 ( x , t ) = g ( x, y 0 ) = 1 1 ∂ [ y 0 = ( t, 0)] , g ( x , y 0 ) π ln | x − y 0 | , g 2 ( x , t ) = ∂y 0 F 1 , 2 are smooth functions, and we shall assume that for every closed domain S 0Π (Γ) ⊂ S Π (Γ) , the following conditions hold i) F 1 ( x , t ) is once continuously differentiable with respect to the variables of x and continuous in t ; ii) F 2 ( x , t ) and t 2 ( x , t ) = ∂ � F 1 q ∈ R 1 , F 2 ( x, s ) ds, ∂y q are continuous. 38 / 143
LECTURE 1: HARMONIC FUNCTIONS Introduce integral operators K 0 and K 1 acting in the space C (Γ) of continuous functions defined on contour Γ � ∂ Φ( x , y ) K 0 ( x ) = 2 ϕ ( y ) dl y , x ∈ Γ (34) ∂n y Γ and ∂ Φ( x , y ) � K 1 ( x ) = 2 ψ ( y ) dl y , x ∈ Γ . (35) ∂n x Γ Theorem 11 The operators I − K 0 and I − K 1 have trivial nullspaces N ( I − K 0 ) = { 0 } , N ( I − K 1 ) = { 0 } , The nullspaces of operators I + K 0 and I + K 1 have dimension one and N ( I + K 0 ) = span { 1 } , N ( I + K 1 ) = span { ψ 0 } with � ψ 0 dl y � = 0 . Γ 39 / 143
LECTURE 1: HARMONIC FUNCTIONS Theorem 12 Let D ∈ R 2 be a domain bounded by the closed smooth contour Γ . The double-layer potential � ∂ Φ( x , y ) Φ( x , y ) = 1 1 v ( x ) = ϕ ( y ) dl y , 2 π ln | x − y | , x ∈ D, (36) ∂n y Γ with a continuous density ϕ is a solution of the interior Dirichlet problem provided that ϕ is a solution of the integral equation � ∂ Φ( x , y ) ϕ ( x ) − 2 ϕ ( y ) dl y = − 2 f ( x ) , x ∈ Γ , (37) ∂n y Γ where f ( x ) is given by (14). Theorem 13 The interior Dirichlet problem has a unique solution. 40 / 143
LECTURE 1: HARMONIC FUNCTIONS Theorem 14 Let D ∈ R 2 be a domain bounded by the closed smooth contour Γ . The double-layer potential ∂ Φ( x , y ) � x ∈ R 2 \ ¯ u ( x ) = ϕ ( y ) dl y , D, (38) ∂n y Γ with a continuous density ϕ is a solution of the exterior Dirichlet problem provided that ϕ is a solution of the integral equation � ∂ Φ( x , y ) ϕ ( x ) + 2 ϕ ( y ) dl y = 2 f ( x ) , x ∈ Γ . (39) ∂n y Γ Here we assume that the origin is contained in D . 41 / 143
LECTURE 1: HARMONIC FUNCTIONS Theorem 15 The exterior Dirichlet problem has a unique solution. Theorem 16 Let D ∈ R 2 be a domain bounded by the closed smooth contour Γ . The single-layer potential � u ( x ) = Φ( x , y ) ψ ( y ) dl y , x ∈ D, (40) Γ with a continuous density ψ is a solution of the interior Neumann problem provided that ψ is a solution of the integral equation ∂ Φ( x , y ) � ψ ( x ) + 2 ψ ( y ) dl y = 2 g ( x ) , x ∈ Γ . (41) ∂n x Γ Theorem 17 The interior Neumann problem is solvable if and only if � ψdl y = 0 (42) Γ is satisfied. 42 / 143
LECTURE 1 LINE INTEGRALS. GREEN’S FORMULA. SURFACE INTEGRALS 43 / 143
LECTURE 1: GREEN’S FORMULA. LINE INTEGRALS. SURFACE INTEGRALS Curves in a parametric form and line integrals. Let xyz be a Cartesian coordinate system in space. We write a spatial curve C using a parametric representation r ( t ) = [ x ( t ) , y ( t ) , z ( t )] = x ( t ) i + y ( t ) j + z ( t ) k ( t ∈ I ) , (43) where variable t is a parameter. As far as a line integral over a spatial curve C is concerned, C is called the path of integration. The path of integration with spatial endpoints A to B goes from A to B (has a certain direction) so that A := r ( a ) is its initial point and B := r ( b ) is its terminal point. C is now oriented. The direction from A to B , in which t increases is called the positive direction on C . Points A and B may coincide, then C is called a closed path. 44 / 143
LECTURE 1: GREEN’S FORMULA. LINE INTEGRALS. SURFACE INTEGRALS Definition of line integral. If C is an oriented curve in a parametric form P = P ( t ) ( x = x ( t ) , y = y ( t ) , z = z ( t )) t ∈ I = ( t 0 , t 1 ) , t : t 0 → t 1 , (44) and f ( P ) and g ( P ) are real (or complex) function defined on C , the line integral of a scalar function is defined as � t = t 1 � f ( P ) dg ( P ) = f ( P ( t )) dg ( P ( t )) , (45) C t = t 0 (if the right-hand side in the equality specifying the integral exists). A line integral of a vector function F ( r ) over a curve C is defined by � b � F ( r ( t )) · d r F ( r ) · d r = dt dt, C a or componentwise � b � � ( F 1 x ′ + F 2 y ′ + F 3 z ′ ) dt ( ′ = d/dt ) . F ( r ) · d r = ( F 1 dx + F 2 dy + F 3 dz ) = C C a 45 / 143
LECTURE 1: GREEN’S FORMULA. LINE INTEGRALS. SURFACE INTEGRALS Example 7 Find the value of the line integral when F ( r ) = [ − y, − xy ] and C is a circular arc from (1 , 0) to (0 , 1) . Solution. We may represent C by r ( t ) = [cos t, sin t ] = cos t i + sin t j , (46) and r ( t ) = [cos t, sin t ] , t : 0 → π/ 2 . The parameter interval is I = ( t 0 , t 1 ) with the initial point t 0 = 0 and endpoint t 1 = π/ 2 . In such an orientation, P (0) = (cos 0 , b sin 0) = (1 , 0) is the initial point and P ( π/ 2) = ( a cos π/ 2 , b sin π/ 2) = (0 , 1) is the endpoint. We have x = cos t, y = sin t and can write vector function F ( r ) on the unit circle F ( r ( t )) = − y ( t ) i − x ( t ) y ( t ) j = [ − sin t, − cos t sin t ] = − sin t i − cos t sin t j . 46 / 143
LECTURE 1: GREEN’S FORMULA. LINE INTEGRALS. SURFACE INTEGRALS Determine r ′ ( t ) = − sin t i + cos t j and calculate the line integral: � π/ 2 � F ( r ) · d r = ( − sin t i − cos t sin t j ) · ( − sin t i + cos t j ) dt = C 0 � π/ 2 � π/ 2 (sin 2 t − cos 2 t sin t ) dt = [(1 / 2)(1 − cos 2 t ) − cos 2 t sin t ] dt = 0 0 � π/ 2 � π/ 2 cos 2 td cos t = π 4 − 1 (1 / 2) [(1 − cos 2 t ) dt + 3 . 0 0 47 / 143
LECTURE 1: GREEN’S FORMULA. LINE INTEGRALS. SURFACE INTEGRALS Example 8 Find the line integral for F ( r ) = [5 z, xy, x 2 z ] when curves C 1 and C 2 have the same initial point A : (0 , 0 , 0) and endpoint B : (1 , 1 , 1) , C 1 is an interval of the straight line r 1 ( t ) = [ t, t, t ] = t i + t j + t k , 0 ≤ t ≤ 1 , and C 2 is a parabola r 2 ( t ) = [ t, t, t 2 ] = t i + t j + t 2 k , 0 ≤ t ≤ 1 . Solution. We have F ( r 1 ( t )) = 5 t i + t 2 j + t 3 k , F ( r 2 ( t )) = 5 t 2 i + t 2 j + t 4 k , r ′ r ′ 1 ( t ) = i + j + j , 2 ( t ) = i + j + 2 t j . Then we can calculate the line integral over C 1 � 1 � 1 � (5 t + t 2 + t 3 ) dt = 5 2 + 1 3 + 1 4 = 37 F ( r 1 ( t )) · r ′ F ( r ) · d r = 1 ( t ) dt = 12 . C 1 0 0 The line integral over C 2 is � 1 � 1 (5 t 2 + t 2 + 2 t 5 ) dt = 5 3 + 1 3 + 2 6 = 28 � F ( r 2 ( t )) · r ′ F ( r ) · d r = 2 ( t ) dt = 12 . C 2 0 0 Thus we have got two different values. 48 / 143
LECTURE 1: LINE INTEGRALS. GREEN’S FORMULA. SURFACE INTEGRALS Theorem 18 The line integral � � F ( r ) · d r = ( F 1 dx + F 2 dy + F 3 dz ) , C C where F 1 , F 2 , F 3 are continuous functions on a domain D in space, is path independent in D , if and only if F = [ F 1 , F 2 , F 3 ] is the gradient of a function f = f ( x, y, z ) in D : F = grad f ; with the components F 1 = ∂f F 2 = ∂f F 3 = ∂f ∂x , ∂y , ∂z . If F is the gradient field and f is a scalar potential of F then the line integral � F ( r ) · d r = f ( B ) − f ( A ) , C where A is the initial point and B the endpoint of C . 49 / 143
LECTURE 1: LINE INTEGRALS. GREEN’S FORMULA. SURFACE INTEGRALS Example 9 Show that the integral � � F ( r ) · d r = (2 xdx + 2 ydy + 4 zdz ) C C is path independent in any domain in space and find its value if integration is performed from A : (0 , 0 , 0) to B : (2 , 2 , 2) . Solution. We have F = [2 x, 2 y, 4 z ] = 2 x i + 2 y j + 4 z k = grad f, and it is easy to check that f ( x, y, z ) = x 2 + y 2 + 2 z 2 . According to Theorem 18, the line integral is path independent in any domain in space. To find its value, we choose the convenient straight path r ( t ) = [ t, t, t ] = t ( i + j + k ) , 0 ≤ t ≤ 2 . Let A : (0 , 0 , 0) , t = 0 , be the initial point and B : (2 , 2 , 2) , t = 2 the endpoint. Then we get r ′ ( t ) = i + j + j . F ( r ) · r ′ = 2 t + 2 t + 4 t = 8 t and � 2 � 2 � F ( r ( t )) · r ′ ( t ) dt = (2 xdx + 2 ydy + 4 zdz ) = 8 tdt = 16 . C 0 0 According to Theorem 18, � F ( r ) d r = f (2 , 2 , 2) − f (0 , 0 , 0) = 4 + 4 + 2 · 4 − 0 = 16 . C 50 / 143
LECTURE 1: LINE INTEGRALS. GREEN’S FORMULA. SURFACE INTEGRALS Theorem 19 The line integral � � F ( r ) · d r = ( F 1 dx + F 2 dy + F 3 dz ) C C where F 1 , F 2 , F 3 are continuous functions on a domain D in space is path independent in D if and only if � F ( r ) · d r = 0 C along every closed path C in D . The differential form F 1 dx + F 2 dy + F 3 dz is called exact in a domain D in space if it is the differential f = ∂f ∂x dx + ∂f ∂y dy + ∂f d ∂z dz of a differentiable function f ( x, y, z ) everywhere in D : F 1 dx + F 2 dy + F 3 dz = d f, where F 1 = ∂f F 2 = ∂f F 3 = ∂f ∂x , ∂y , ∂z . 51 / 143
LECTURE 1: LINE INTEGRALS. GREEN’S FORMULA. SURFACE INTEGRALS Green’s formula. Let C be a closed curve in xy -plane that does not intersect itself and makes just one turn in the positive direction (counterclockwise). Let F 1 ( x, y ) and F 2 ( x, y ) be functions that are continuous and have continuous partial derivatives ∂F 1 and ∂F 2 everywhere in some domain R enclosed by C . Then ∂y ∂x � ∂F 2 � � ∂x − ∂F 1 � � dxdy = ( F 1 dx + F 2 dy ) . ∂y R C Here we integrate along the entire boundary C of R so that R is on the left as we advance in the direction of integration. One can write Green’s formula with the help of curl � � � ( curl F ) · k dxdy = F · d r . R C 52 / 143
LECTURE 1: LINE INTEGRALS. GREEN’S FORMULA. SURFACE INTEGRALS Example 10 Verify Green’s formula for F 1 = y 2 − 7 y , F 2 = 2 xy + 2 x and C being a circle R : x 2 + y 2 = 1 . Solution. Calculate a double integral � ∂F 2 ∂x − ∂F 1 � � � � � � � dxdy = [(2 y + 2) − (2 y − 7)] dxdy = 9 dxdy = 9 π. ∂y R R R Calculate the corresponding line integral. Circle C in the parametric form is given by r ( t ) = [cos t, sin t ] = cos t i + sin t j . r ′ ( t ) = − sin t i + cos t j . On C F 1 = y 2 − 7 y = sin 2 t − 7 sin t, F 2 = 2 xy + 2 x = 2 cos t sin t + 2 cos t, and we get that the line integral in Green’s formula is equal to the double integral: � 2 π � [(sin 2 t − 7 sin t )( − sin t ) + (2 cos t sin t + 2 cos t ) cos t ] dt = 0 + 7 π + 0 + 2 π = 9 π. F ( r ) · d r = C 0 53 / 143
LECTURE 1: LINE INTEGRALS. GREEN’S FORMULA. SURFACE INTEGRALS Surface integral. To define a surface integral, we take a surface S given by a parametric representation r ( u, v ) = [ x ( u, v ) , y ( u, v ) , z ( u, v )] = x ( u, v ) i + y ( u, v ) j + z ( u, v ) k , u, v ∈ R, the normal vector N = r u × r v � = 0 , and unit normal vector 1 n = | N | N . A surface integral of a vector function F ( r ) over a surface S is defined as � � � � F · n dA = F ( r ( u, v )) · N ( u, v ) dudv. (47) S R 54 / 143
LECTURE 1: LINE INTEGRALS. GREEN’S FORMULA. SURFACE INTEGRALS Note that n dA = n | N | dudv = | N | dudv, and we assume that the parameters u, v belongs to a region R in the u, v -plane. Write the equivalent expression componentwise using directional cosine: F = [ F 1 , F 2 , F 3 ] = F 1 i + F 2 j + F 3 k , n = [cos α, cos β, cos γ ] = cos α i + cos β j + cos γ k , N = [ N 1 , N 2 , N 3 ] = N 1 i + N 2 j + N 3 k , and � � � � F · n dA = ( F 1 cos α + F 2 cos β + F 3 cos γ ) dA = S S � � ( F 1 N 1 + F 2 N 2 + F 3 N 3 ) dudv. S 55 / 143
LECTURE 1: LINE INTEGRALS. GREEN’S FORMULA. SURFACE INTEGRALS Example 11 Evaluate a surface integral of the vector function F = [ x 2 , 0 , 3 y 2 ] over a portion of the plane S : x + y + z = 1 , 0 ≤ x, y, z ≤ 1 . Solution. Writing x = u and y = v , we have z = 1 − u − v and can represent S in the form r ( u, v ) = [ u, v, 1 − u − v ] , 0 ≤ v ≤ 1 , 0 ≤ u ≤ 1 − v. We have r u = [1 , 0 , − 1] , r v = [0 , 1 , − 1]; a normal vector is � � � � i j k � � � � N = r u × r v = = i + j + k = [1 , 1 , 1] . 1 0 − 1 � � � � � � � 0 1 − 1 � � � 56 / 143
LECTURE 1: LINE INTEGRALS. GREEN’S FORMULA. SURFACE INTEGRALS The corresponding unit normal vector 1 1 n = | N | N = √ ( i + j + k ) . 3 On surface S , F ( r ( u, v )) = F ( S ) = [ u 2 , 0 , 3 v 2 ] = u 2 i + 3 v 2 k . Hence F ( r ( u, v )) · N ( u, v ) = [ u 2 , 0 , 3 v 2 ] · [1 , 1 , 1] = u 2 + 3 v 2 . Parameters u, v belong to triangle R : 0 ≤ v ≤ 1 , 0 ≤ u ≤ 1 − v . Now we can write and calculate the flux integral: � � � � � � ( u 2 + 3 v 2 ) dudv = F · n dA = F ( r ( u, v )) · N ( u, v ) dudv = S R R � 1 � 1 − v � 1 � 1 − v � 1 � 1 − v ( u 2 + 3 v 2 ) dudv = u 2 du + 3 v 2 dv dv du = 0 0 0 0 0 0 � 1 � 1 � 1 � 1 ( v 2 − v 3 ) dv = (1 − v ) 3 dv + 3 v 2 (1 − v ) dv = (1 / 3) t 3 dt + 3 = (1 / 3) 0 0 0 0 (1 / 3) · (1 / 4) + 3(1 / 3 − 1 / 4) = 1 / 3 . 57 / 143
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LECTURE 1: BASIC ELECTROMAGNETIC THEORY. MAXWELLS AND HELMHOLTZ EQUATIONS. The classical macroscopic electromagnetic field is described by four three-component vector-functions E ( r , t ) , D ( r , t ) , H ( r , t ) , and B ( r , t ) of the position vector r = ( x, y, z ) and time t . The fundamental field vectors E ( r , t ) and H ( r , t ) are called electric and magnetic field intensities . D ( r , t ) and B ( r , t ) which will be eliminated from the description via constitutive relations are called the electric displacement and magnetic induction . The fields and sources are related by the Maxwell equation system ∂ D ∂t − rot H = − J , (48) ∂ B ∂t + rot E = 0 , (49) div B = 0 , (50) div D = ρ, (51) written in the standard SI units. 59 / 143
LECTURE 1: BASIC ELECTROMAGNETIC THEORY. MAXWELLS AND HELMHOLTZ EQUATIONS. The constitutive relations are D = ǫ E , (52) B = µ H , (53) J = σ E . (54) Here ǫ , µ , and σ , which are generally bounded functions of position (the first two are assumed positive), are permittivity, permeability, and conductivity of the medium for J being the conductivity current density. In vacuum, that is, in a homogeneous medium with constant characteristics ǫ = ǫ 0 , µ = µ 0 , and σ = 0 , the Maxwell equation system takes a simpler form ∂ E rot H = ǫ 0 ∂t , (55) ∂ H rot E = − µ 0 ∂t , (56) div H = 0 , (57) div E = ρ. (58) 60 / 143
LECTURE 1: BASIC ELECTROMAGNETIC THEORY. MAXWELLS AND HELMHOLTZ EQUATIONS. In the case of a homogeneous medium, it is reasonable to obtain equations for each vector E ( r , t ) and H ( r , t ) . To this end, assume that ρ = 0 . Applying the operation rot to equation (48) and taking into account the constitutive relations, we have rotrot H = ǫ ∂ ∂t rot E + σ rot E . (59) Using the vector differential identity rotrot A = graddiv A − ∆ A and taking into notice equation (49), we obtain the equation for magnetic field H graddiv H − ∆ H = − ǫµ ∂ 2 H ∂t 2 − σµ∂ H ∂t or ∂ 2 H ∆ H = 1 ∂t 2 + σµ∂ H � a 2 = 1 � (60) a 2 ∂t ǫµ because div H = 0 . 61 / 143
LECTURE 1: BASIC ELECTROMAGNETIC THEORY. MAXWELLS AND HELMHOLTZ EQUATIONS. The same equation holds for electric field E ∂ 2 E ∆ E = 1 ∂t 2 + σµ ∂ E � a 2 = 1 � . (61) a 2 ∂t ǫµ Equations (60) or (61) hold for all field components, ∂ 2 u ∆ u = 1 ∂t 2 + σµ∂u ∂t , (62) a 2 where u is one of the components H x , H y , H z or E x , E y , E z . If the medium is nonconducting, σ = 0 , then (60), (61), or (62) yield a standard wave equation ∂ 2 u ∆ u = 1 ∂t 2 . (63) a 2 This implies that electromagnetic processes are actually waves that propagate in the medium with the speed 1 a = √ ǫµ (the latter holds for vacuum). 62 / 143
LECTURE 1: BASIC ELECTROMAGNETIC THEORY. MAXWELLS AND HELMHOLTZ EQUATIONS. Time-periodic (time-harmonic) fields H ( r , t ) = H ( r ) e − iωt , E ( r , t ) = E ( r ) e − iωt (64) constitute a very important particular case. Functions E and H are the field complex amplitudes; the quantities Re E and Re H have direct physical meaning. Assuming that complex electromagnetic field (64) satisfies Maxwell equations and that the currents are also time-harmonic, J ( r , t ) = J ( r ) e − iωt , substitute (64) into (48)–(51) to obtain rot H = − iω D + J , (65) rot E = iω B , (66) div B = 0 , (67) div D = ρ. (68) Since J = σ E , equation (65) can be transformed by introducing the complex permittivity ǫ ′ = ǫ + i σ ω . As a result, system (65)–(68) takes the form − iωǫ ′ E , rot H = (69) rot E = iωµ H , (70) div ( µ H ) = 0 , (71) div ( ǫ E ) = ρ. (72) In a homogeneous medium and when external currents are absent, equations (71) and (72) follow from the first two Maxwell equations (69) and (70). 63 / 143
LECTURE 1: BASIC ELECTROMAGNETIC THEORY. MAXWELLS AND HELMHOLTZ EQUATIONS. Consider the simplest time-harmonic solutions to Maxwell equations in a homogeneous medium (with constant characteristics), plane electromagnetic waves. In the absence of free charges when div E = 0 , the electric field vector satisfies the equation rotrot E = ω 2 ǫ ′ µ E , (73) or ∆ E + κ 2 E = 0 , (74) where ǫ ′ = ǫ + i σ k = ω √ ǫµ. κ 2 = ω 2 ǫ ′ µ = k 2 + iωµσ, ω , (75) In the cartesian coordinate system, equation (74) holds for every field component, ∆ u + κ 2 u = 0 , (76) where u is one of the components E x , E y , E z . 64 / 143
LECTURE 1: BASIC ELECTROMAGNETIC THEORY. MAXWELLS AND HELMHOLTZ EQUATIONS. The Helmholtz equation (76) has a solution in the form of a plane wave; componentwise, E α = E 0 α e i ( κ x x + κ y y + κ z z ) , κ 2 x + κ 2 y + κ 2 z = κ 2 ( α = x, y, z ) . (77) Here κ is called the wave propagation constant. Therefore, the vector Helmholtz equation (74) has a solution E = E 0 e i ( κ x x + κ y y + κ z z ) = E 0 e i k · r , (78) where the vectors k = ( κ x , κ y , κ z ) , r = ( x, y, z ) , E 0 = const. (79) Since div E = 0 , we have div E = div ( E 0 e i k · r ) = ie i k · r k · E 0 = 0 . Thus, k · E 0 = 0 so that the direction of vector E is orthogonal to the direction of the plane wave propagation governed by vector k . 65 / 143
LECTURE 1: BASIC ELECTROMAGNETIC THEORY. MAXWELLS AND HELMHOLTZ EQUATIONS. Vectors E and H are coupled by the relation rot E = iωµ H . (80) Since rot ( E 0 e i k · r ) = [ grad e i k · r , E 0 ] , we have √ ǫ ′ [ k 0 , E 0 ] = √ µ H 0 , (81) where k 0 = k / | k | is the unit vector in the direction of the wave propagation. Thus, vectors E and H are not only orthogonal to the direction of the wave propagation but also mutually orthogonal: E · H = 0 , E · k = 0 , H · k = 0 . (82) We see that the Maxwell equations have a solution in the form of a plane electromagnetic wave E ( r ) = E 0 e i k · r , H ( r ) = H 0 e i k · r , (83) where √ √ ǫ ′ [ k 0 , E ] = √ µ H , √ µ [ k 0 , H ] = − ǫ ′ E , (84) 66 / 143
LECTURE 1: BASIC ELECTROMAGNETIC THEORY. MAXWELLS AND HELMHOLTZ EQUATIONS. Introduce the dimensionless variables and parameters � k 2 0 = ε 0 µ 0 ω 2 , k 0 x → x, µ 0 /ε 0 H → H , E → E , where ε 0 and µ 0 are permittivity and permeability of vacuum. Propagation of electromagnetic waves along a tube (a waveguide) with cross section Ω (a 2-D domain bounded by smooth curve Γ ) parallel to the x 3 -axis in the cartesian coordinate system x 1 , x 2 , x 3 , x = ( x 1 , x 2 , x 3 ) , is described by the homogeneous system of Maxwell equations (written in the normalized form) with the electric and magnetic field dependence e iγx 3 on longitudinal coordinate x 3 (the time factor e iωt is omitted): rot E = − i H , x ∈ Σ , rot H = iε E , E ( x ) = ( E 1 ( x ′ ) e 1 + E 2 ( x ′ ) e 2 + E 3 ( x ′ ) e 3 ) e iγx 3 , (85) H ( x ) = ( H 1 ( x ′ ) e 1 + H 2 ( x ′ ) e 2 + H 3 ( x ′ ) e 3 ) e iγx 3 , x ′ = ( x 1 , x 2 ) , with the boundary conditions for the tangential electric field components on the perfectly conducting surfaces E τ | M = 0 , (86) 67 / 143
LECTURE 1: BASIC ELECTROMAGNETIC THEORY. MAXWELLS AND HELMHOLTZ EQUATIONS. Write system of Maxwell equations (85) componentwise ∂H 3 ∂E 3 iγH 1 − ∂H 3 − iγH 2 = iεE 1 , − iγE 2 = − iH 1 , = iεE 2 , ∂x 2 ∂x 2 ∂x 1 iγE 1 − ∂E 3 ∂H 2 − ∂H 1 ∂E 2 − ∂E 1 = − iH 2 , = iεE 3 , = − iH 3 , ∂x 1 ∂x 1 ∂x 2 ∂x 1 ∂x 2 and express functions E 1 , H 1 , E 2 , and H 2 via E 3 and H 3 from the first, second, fourth, and fifth equalities, k 2 = ε − γ 2 , denoting ˜ i � γ ∂E 3 − ∂H 3 � i � γ ∂E 3 + ∂H 3 � E 1 = , E 2 = , (87) ˜ ˜ k 2 ∂x 1 ∂x 2 k 2 ∂x 2 ∂x 1 i � ε ∂E 3 + γ ∂H 3 � i � − ε ∂E 3 + γ ∂H 3 � H 1 = , H 2 = . ˜ ˜ k 2 ∂x 2 ∂x 1 k 2 ∂x 1 ∂x 2 Note that this representation is possible if γ 2 � = ε 1 and γ 2 � = ε 2 . It follows from (87) that the field of a normal wave can be expressed via two scalar functions Π ( x 1 , x 2 ) = E 3 ( x 1 , x 2 ) , Ψ ( x 1 , x 2 ) = H 3 ( x 1 , x 2 ) . 68 / 143
LECTURE 1: BASIC ELECTROMAGNETIC THEORY. MAXWELLS AND HELMHOLTZ EQUATIONS. If to look for particular solutions with E 3 ≡ 0 then we have a separate problem for the set of component functions [ E 1 , E 2 , H 3 ] , [ H 1 , H 2 , 0] which are called TE-waves (transverse electric) or the case of H -polarization. For particular solutions with H 3 ≡ 0 we have a problem for the set of component functions [ H 1 , H 2 , E 3 ] , [ E 1 , E 2 , 0] called TM-waves (transverse magnetic) or the case of E -polarization. These two cases constitute two fundamental polarizations of the electromagnetic field associated with a given direction of propagation. For γ = 0 when we consider fields independent of one of the coordinates ( x 3 ) we have two separate problems for the sets of component functions [ E 1 , E 2 , H 3 ] , TE-(H)polarization, and [ H 1 , H 2 , E 3 ] , TM-(E)polarization. Thus the problem on normal waves is reduced to boundary eigenvalue problems for functions Π and Ψ . Namely, from (85) and (86) we have the following eigenvalue problem on normal waves in a waveguide with homogeneous filling: to find γ ∈ C , called eigenvalues of normal waves such that there exist nontrivial solutions of the Helmholtz equations x ′ = ( x 1 , x 2 ) ∈ Ω ∆Π + ˜ k 2 Π = 0 , (88) k 2 = ε − γ 2 , ∆Ψ + ˜ k 2 Ψ = 0 , ˜ (89) satisfying the boundary conditions on Γ 0 � ∂ Ψ � Π | Γ 0 = 0 , = 0 , (90) � ∂n � Γ 0 In fact, it is necessary to determine only one function, H 3 for the TE-polarization or E 3 for the TM-polarization; the remaining components are obtained using differentiation. 69 / 143
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LECTURE 1: STATEMENTS AND ANALYSIS OF THE BVPS FOR MAXWELLS AND HELMHOLTZ EQUATIONS. In the two-dimensional case, the Helmholtz equation L ( k 2 ) u = 0 written in the polar coordinates r = ( r, φ ) has the form ∂ 2 u 1 ∂ � r ∂u � + 1 ∂φ 2 + k 2 u = 0 . (91) r 2 r ∂r ∂r Assume that the function u = u ( r ) satisfies the Helmholtz equation outside a circle of radius r 0 . On any circle of radius r > r 0 function u can be decomposed in a trigonometric Fourier series ∞ � u n ( r ) e inφ u ( r ) = (0 < φ < 2 π ) , (92) n = −∞ where the coefficients � 2 π u n ( r ) = 1 u ( r ) e − inφ dφ (93) 2 π 0 2 π e − inφ and integrate over a circle of 1 are functions of r . In order to find u n ( r ) multiply equations (91) by radius r . 71 / 143
LECTURE 1: STATEMENTS AND ANALYSIS OF THE BVPS FOR MAXWELLS AND HELMHOLTZ EQUATIONS. As a result of integration, we obtain − n 2 1 d � r du n � r 2 u n + k 2 u n = 0 , n = 0 , ± 1 , . . . . (94) r dr dr (94) is a second-order ordinary differential equation with constant coefficients for u n ( r ) which holds for r > r 0 . Equation (94) is actually the Bessel equation of order n . Its general solution can be written as u n ( r ) = A n H (1) n ( kr ) + B n H (2) n ( kr ) , (95) where H (1 , 2) ( z ) are its linearly independent solutions; they are the n th-order Hankel functions of the first and n second kind, respectively. Thus any solution u = u ( r ) to the homogeneous Helmholtz equation (satisfied outside a circle of radius r 0 ) can be represented for r > r 0 in the form of a series ∞ [ A n H (1) n ( kr ) + B n H (2) � n ( kr )] e inφ u ( r ) = (0 < φ < 2 π, r > r 0 ) . (96) n = −∞ 72 / 143
LECTURE 1: STATEMENTS AND ANALYSIS OF THE BVPS FOR MAXWELLS AND HELMHOLTZ EQUATIONS. At infinity, the following asymptotical formulas are valid � 2 � 1 � H (1 , 2) πz e ± i ( z − πn 2 − π 4 ) + O ( z ) = , (97) n z 3 / 2 which yields an asymptotic estimate of the solution to the homogeneous Helmholtz equation at infinity � 1 � u ( r ) = O √ r . (98) For the zero-order Hankel functions of the first and second kind, respectively, the following asymptotical formulas are valid � 2 H (1) πz e i ( z − π 4 ) + . . . , ( z ) = (99) 0 � 2 πz e − i ( z − π H (2) 4 ) + . . . , ( z ) = 0 73 / 143
LECTURE 1: STATEMENTS AND ANALYSIS OF THE BVPS FOR MAXWELLS AND HELMHOLTZ EQUATIONS. Let us recall first that the plane waves propagation along the x -axis have the form t − x t + x � � � � ˆ u = f ˆ , u = f ˆ , (100) a a u and ˆ where ˆ ˆ u are, respectively, the forward wave (propagating in the positive direction of the x -axis) and backward wave (propagating in the negative direction of the x -axis). They satisfy the following first-order partial differential equations ∂x + 1 ∂ ˆ u ∂ ˆ u = 0 , (101) a ∂t ∂ ˆ ∂ ˆ ∂x − 1 u ˆ u ˆ = 0 . (102) a ∂t In the stationary mode u = v ( x ) e iωt (103) For the amplitude function v these relations take the form ∂ ˆ v ∂x + ik ˆ v = 0 , (104) ∂ ˆ ˆ v ∂x − ik ˆ ˆ v = 0 , (105) for the forward and backward waves, respectively, where k = ω a . 74 / 143
LECTURE 1: STATEMENTS AND ANALYSIS OF THE BVPS FOR MAXWELLS AND HELMHOLTZ EQUATIONS. Spherical waves. If a spherical wave is excited by the sources situated in a bounded part of the space (not at infinity), then at large distances from the source, a spherical wave is similar to a plane wave whose amplitude decays as 1 r . This natural physical assumption leads to a conclusion that the outgoing, respectively, incoming, spherical waves must satisfy the relationships � 1 ∂u ∂r + 1 ∂u � = o , (106) a ∂t r � 1 ∂u ∂r − 1 ∂u � = o . (107) a ∂t r For the amplitude functions in the stationary mode we have � 1 ∂v � ∂r + ikv = o for outgoing spherical waves , (108) r � 1 ∂v � ∂r − ikv = o for incoming spherical waves . (109) r 75 / 143
LECTURE 1: STATEMENTS AND ANALYSIS OF THE BVPS FOR MAXWELLS AND HELMHOLTZ EQUATIONS. Let us prove now that at large distances from the source, any outgoing spherical wave decays as 1 r . 1. In the case of a point source at the origin, this statement is trivial because the wave itself has the form u ( r, t ) = e i ( ωt − kr ) = v 0 ( r ) e iωt , (110) r so that � 1 ∂v 0 � ∂r + ikv 0 = o . (111) r Check this relationship. 76 / 143
LECTURE 1: STATEMENTS AND ANALYSIS OF THE BVPS FOR MAXWELLS AND HELMHOLTZ EQUATIONS. 2. Let a spherical wave be excited by a point source situated at a point r 0 . The amplitude of the spherical wave is v 0 ( r ) = e ikR � r 2 + r 2 , R = | r − r 0 | = 0 − 2 rr 0 cos θ. (112) R Calculating the derivative we obtain � 1 ∂R ∂r = r − r 0 cos θ � ∼ 1 + O (113) R r and � 1 ∂v 0 � ∂R + ikv 0 = o . R in view of (111). Next, � 1 � 1 � �� � ∂v 0 ∂r = ∂v 0 ∂R ∂r = ∂v 0 = ∂v 0 1 + O ∂R + o ∂R ∂R r r because � 1 � 1 ∂v 0 � � ∂R · O = o . r r Finally, � 1 � 1 ∂v 0 � � ∂r + ikv 0 + o = o (114) r r what is to be proved. 77 / 143
LECTURE 1: STATEMENTS AND ANALYSIS OF THE BVPS FOR MAXWELLS AND HELMHOLTZ EQUATIONS. 3. Show that the volume potential f ( r 0 ) e − ikR � v ( r ) = dτ r 0 , R = | r − r 0 | , (115) R T satisfies condition (108). Introducing the notation P v = ∂v ∂r + ikv, (116) we obtain � e − ikR � 1 � 1 � � � � � P v = f ( r 0 ) P dτ r 0 = f ( r 0 ) o dτ r 0 = o . (117) R r r T T Volume potential (115) is the amplitude of an outgoing wave excited by the sources distributed arbitrarily in a bounded domain T . Also, function v defined by (115) satisfies the inhomogeneous Helmholtz equation L ( k 2 ) u = − f and decays as 1 r for r → ∞ . In addition, it satisfies the condition � 1 ∂v � ∂r + ikv = o . (118) r 78 / 143
LECTURE 1: STATEMENTS AND ANALYSIS OF THE BVPS FOR MAXWELLS AND HELMHOLTZ EQUATIONS. Theorem 20 There is one and only one solution to the inhomogeneous Helmholtz equation L ( k 2 ) v = (∆ + k 2 ) v = − f ( r ) , (119) where f ( r ) is a function with local support, which satisfies the conditions at infinity � 1 � v = O , (120) r � 1 ∂v � ∂r + ikv = o . r Proof. Assuming that there are two different solutions v 1 and v 2 and setting w = v 1 − v 2 , we see that w satisfies the homogeneous Helmholtz equation L ( k 2 ) w = 0 and the conditions at infinity (120). Let Σ R be a sphere of radius R (later, we will take the limit R → ∞ ). Applying the third Green formula to w ( r ) and the fundamental solution φ 0 ( r 0 ) = e − ikR 4 πR , R = | r 0 − r | , we arrive at the integral representation of w at a point r ∈ Σ R � φ 0 ( r 0 ) ∂w ∂r − w ∂ � � w ( r ) = ∂r ( φ 0 ( r 0 )) dσ r 0 . (121) Σ R 79 / 143
LECTURE 1: STATEMENTS AND ANALYSIS OF THE BVPS FOR MAXWELLS AND HELMHOLTZ EQUATIONS. The conditions at infinity (120) for w ( r ) and φ 0 ( r ) yield � 1 ∂w ∂r − w ∂ � �� φ 0 ∂ν ( φ 0 ) = φ 0 − ikw + o − (122) r � 1 � 1 � 1 � 1 � �� � � � − w − ikφ 0 + o = φ 0 o − wo = o . r 2 r r r Therefore, � 1 � � w ( r ) = o dσ r 0 → 0 , R → ∞ . (123) r 2 Σ R This implies w ( r ) = 0 at any r ∈ Σ R and thus at any spatial r . Conditions (120) are called Sommerfeld radiation conditions . In the two-dimensional case the Sommerfeld radiation conditions at infinity take the form � 1 � v = O √ r , (124) � ∂v √ r � lim ∂r + ikv = 0 . r →∞ 80 / 143
LECTURE 1: STATEMENTS AND ANALYSIS OF THE BVPS FOR MAXWELLS AND HELMHOLTZ EQUATIONS. Theorem 21 Let u 0 ( r ) be a solution to the Helmholtz equation satisfied outside a circle of radius r 0 . If � | u | 2 dl = 0 , lim (125) r →∞ C r where C r is a circle of radius r , then u ≡ 0 for r > r 0 . Proof. Any solution u = u ( r ) to the (homogeneous) Helmholtz equation (satisfied outside a circle of radius r 0 ) can be represented for r > r 0 in the form of series (96) ∞ � u n ( r ) = A n H (1) n ( kr ) + B n H (2) u n ( r ) e inφ , u ( r ) = n ( kr ) (0 < φ < 2 π, r > r 0 ) . (126) n = −∞ Therefore, ∞ � | u | 2 dl = 2 π � r | u n ( r ) | 2 . lim (127) r →∞ C r n = −∞ 81 / 143
LECTURE 1: STATEMENTS AND ANALYSIS OF THE BVPS FOR MAXWELLS AND HELMHOLTZ EQUATIONS. If � | u | 2 dl = 0 , lim r →∞ C r then (127) yields r →∞ r | u n ( r ) | 2 = 0 , lim n = 0 , ± 1 , ± 2 , . . . . (128) Next, according to asymptotical formulas (97) for Hankel functions r | u n ( r ) | 2 are bounded quantities at r → ∞ , namely, � 1 � r | u n ( r ) | 2 = rO = O (1) , n = 0 , ± 1 , ± 2 , . . . , (129) r which, together with (128), implies A n = B n = 0 , n = 0 , ± 1 , ± 2 , . . . , (130) and, consequently, u ≡ 0 for r > r 0 in line with representation (126). 82 / 143
LECTURE 1: STATEMENTS AND ANALYSIS OF THE BVPS FOR MAXWELLS AND HELMHOLTZ EQUATIONS. Theorem 22 Let u 0 ( r ) be a solution to the Helmholtz equation satisfied outside a sphere S r 0 of radius r 0 . If � | u | 2 ds = 0 , lim (131) r →∞ S r then u ≡ 0 for r > r 0 . For the vector solutions of Maxwell equations (69) and (70), electromagnetic field E ( r ) , H ( r ) , the similar statements are valid Theorem 23 Let E ( r ) , H ( r ) be a solution to the Maxwell equation system satisfied outside a sphere of radius r 0 . If � | [ H , e r ] | 2 ds = 0 , lim (132) r →∞ S r or � | [ E , e r ] | 2 ds = 0 , lim (133) r →∞ S r where S r is a sphere of radius r and e r = r /r is the unit position vector of the points on S r , then E ( r ) ≡ 0 , H ( r ) ≡ 0 for r > r 0 . 83 / 143
LECTURE 1: STATEMENTS AND ANALYSIS OF THE BVPS FOR MAXWELLS AND HELMHOLTZ EQUATIONS. Formulate a scalar (acoustical) problem of the wave diffraction by a transparent body Ω 1 . Let Ω 1 be a domain bounded by a piecewise smooth surface Σ . The problem under consideration is reduced to a BVPs for the inhomogeneous Helmholtz equation with a piecewise constant coefficient r ∈ Ω 0 = R 3 \ Ω 1 , ∆ u 0 ( r ) + k 2 0 u 0 ( r ) = − f 0 , (134) ∆ u 1 ( r ) + k 2 1 u 1 ( r ) = − f 1 , r ∈ Ω 1 ; solution u satisfies the conjugation conditions on Σ ∂u 1 ∂n − ∂u 0 u 1 − u 0 = 0 , ∂n = 0 , (135) and the conditions at infinity � 1 � u 0 = O , (136) r � 1 ∂u 0 � ∂r − ik 0 u 0 = o . r 84 / 143
LECTURE 1: STATEMENTS AND ANALYSIS OF THE BVPS FOR MAXWELLS AND HELMHOLTZ EQUATIONS. Theorem 24 The solution to problem (134)–(136) is unique. Proof. Since problem (134)–(136) is linear, it is sufficient to prove that the corresponding homogeneous problem (with f 0 = f 1 = 0 ) has only a trivial solution. Together with u 0 and u 1 consider the corresponding complex conjugate functions u ∗ 0 and u ∗ 1 . They satisfy the same boundary and transmission conditions; however, the condition at infinity takes the form � 1 ∂u ∗ � ∂r + ik 0 u ∗ 0 0 = o . (137) r Applying the second Green formula to u ∗ 1 and u ∗ 1 in domain Ω 1 , we obtain ∂u ∗ � � ∂u 1 � ∂ν − u ∗ 1 u 1 dσ r 0 = 0 , (138) 1 ∂ν Σ where ν denotes the unit normal vector to the boundary Σ directed into the exterior of Ω 1 . Let S R be a sphere of sufficiently large radius R containing domain Ω 1 . Applying the second Green formula to u 0 and u ∗ 0 in the domain Ω S situated between Ω 1 and S R , we obtain ∂u ∗ ∂u ∗ � � ∂u 0 � � � ∂u 0 � 0 − u ∗ ∂r − u ∗ 0 u 0 dσ r 0 + u 0 dσ r 0 = 0 , (139) 0 0 ∂ν 0 ∂ν 0 ∂r Σ S R where ∂∂ν 0 denotes the directional derivative in the direction of the unit normal vector ν to Σ directed into the interior of Ω 1 (external with respect to Ω 0 ). 85 / 143
LECTURE 1: STATEMENTS AND ANALYSIS OF THE BVPS FOR MAXWELLS AND HELMHOLTZ EQUATIONS. Adding up (138) and (139) and taking into account the conjugation conditions on Σ , we have ∂u ∗ � ∂u 0 � � ∂r − u ∗ 0 u 0 dσ r 0 = 0 . (140) 0 ∂r S R Applying the condition at infinity and transferring to the limit R → ∞ in (140) we obtain � | u 0 | 2 ds = 0 , lim (141) R →∞ S R Thus u 0 ≡ 0 outside sphere S R according to Theorem 22. Applying the third Green formula (121) in Ω S we obtain that u 0 ≡ 0 in Ω S . Then applying the third Green formula in Ω 1 we obtain that u 1 ≡ 0 in Ω 1 . Therefore, homogeneous problem (134)–(136) has only a trivial solution. The theorem is proved. 86 / 143
LECTURE 1: STATEMENTS AND ANALYSIS OF THE BVPS FOR MAXWELLS AND HELMHOLTZ EQUATIONS. Formulate a vector (electromagnetic) problem of the wave diffraction by a transparent body Ω 1 . Let Ω 1 be a domain bounded by a piecewise smooth surface Σ and Ω 0 = R 3 \ Ω 1 . The problem under consideration is reduced to a BVP for the inhomogeneous system of Maxwell equations (69) and (70) with a piecewise constant coefficient rot H j = − iωǫ j E j + J j , rot E j = iωµ j H j , j = 0 , 1 , (142) with the transmission conditions stating the continuity of the tangential field components across interface Σ [ H 1 , ν ] = [ H 0 , ν ] , [ E 1 , ν ] = [ E 0 , ν ] , (143) and the Silver–M¨ uller radiation conditions at infinity k 0 = ω √ ǫ 0 µ 0 , r →∞ r ([ H 0 , e r ] − ik 0 E 0 ) = 0 , lim (144) where ν is the unit normal vector to Σ , e r = r /r is the unit position vector of the points on S r and the limit holds uniformly with respect to all directions (specified by e r ). Note that in this case (142) can be written equivalently (in every domain where the parameters are constant) as a one vector equation with respect to i.e. E ( r ) by eliminating H ( r ) : rot rot E j − ω 2 ǫ j µ j E j = ˜ J j , j = 0 , 1 . (145) 87 / 143
LECTURE 1: STATEMENTS AND ANALYSIS OF THE BVPS FOR MAXWELLS AND HELMHOLTZ EQUATIONS. Theorem 25 The solution to problem (142)–(144) is unique. Proof. Since problem (142)–(144) is linear, it is sufficient to prove that the corresponding homogeneous problem (with J j = 0 ) has only a trivial solution. Next, one has to apply Theorem 23 and perform the same steps as in the proof of Theorem 24 using Lorentz lemma instead of the Green formulas. 88 / 143
ELECTROMAGNETIC FIELDS AND WAVES: MATHEMATICAL MODELS AND NUMERICAL METHODS LECTURE 2 89 / 143
LECTURE 2 THE MATHEMATICAL NATURE OF WAVES 90 / 143
LECTURE 2: THE MATHEMATICAL NATURE OF WAVES. Going back to the problems on normal waves we see that the form of solution in (85) E ( x ) = ( E 1 ( x ′ ) e 1 + E 2 ( x ′ ) e 2 + E 3 ( x ′ ) e 3 ) e iγx 3 , H ( x ) = ( H 1 ( x ′ ) e 1 + H 2 ( x ′ ) e 2 + H 3 ( x ′ ) e 3 ) e iγx 3 , (146) x ′ = ( x 1 , x 2 ) , with the dependence e iγx 3 on longitudinal coordinate x 3 specify a wave propagating in the positive direction of x 3 -axis. Problems on normal waves (88)–(90) have nontrivial solutions if k 2 = ε − γ 2 = λ D k 2 = λ N ˜ ˜ or n , n = 1 , 2 , . . . , (147) n so that the eigenvalues of normal waves � � γ = γ D γ = γ N ε − λ D ε − λ N n = or n = n . (148) n We have 0 ≤ λ D,N ≤ λ D,N ≤ . . . ; therefore, that are at most finitely many values of γ D n and γ N n that are real, 1 2 while infinitely many of them are purely imaginary. Consequently, according to (146), there are at most finitely many normal waves that propagate without attenuation (in the positive direction of x 3 -axis) and infinitely many decay exponentially. 91 / 143
LECTURE 2: THE MATHEMATICAL NATURE OF WAVES. Propagation of electromagnetic waves along the waveguide is described by the homogeneous system of Maxwell equations which can be written in the form rot H = − ik E , (149) rot E = ik H , with the boundary conditions for the tangential electric field components on the perfectly conducting walls Σ of the waveguide E τ | Σ = 0 , (150) Look for particular solutions of (149) in the form grad div P + k 2 P , E = (151) = − ik rot P , H using the polarization potential P = [0 , 0 , Π] that has only one nonzero component P 3 = Π . It is easy to see that H 3 = 0 , E = [0 , 0 , E 3 ] , H = [ H 1 , H 2 , 0] , (152) and this case is called TM-polarization or E-polarization Substituting (151) into (149) yields the equations ∆Π + ∂ 2 Π ∆ 3 Π + k 2 Π + k 2 Π = 0 , = 0 or (153) ∂x 2 3 ∂ 2 + ∂ 2 + ∂ 2 ∆ 3 = . ∂x 2 ∂x 2 ∂x 2 1 2 3 92 / 143
LECTURE 2: THE MATHEMATICAL NATURE OF WAVES. Condition (150) is satisfied if we assume that Π | Σ = 0 , (154) because the third components of both P and E are actually tangential components that must vanish on the waveguide wall and they are coupled by the first relation (151). (153) and (154) constitute the Dirichlet BVP for the Helmholtz equation in the tube. We look for the solution to this problem in the form x ′ = ( x 1 , x 2 ) , Π( x ) = Π( x ′ , x 3 ) = ψ ( x ′ ) f ( x 3 ) , ψ ( x ′ ) , f ( x 3 ) � = 0 , (155) using the separation of variables. Namely, substituting (155) into (153) and dividing by nonvanishing product fψ we have + f ′′ ∆ ψ f ∆ ψ + f ′′ ψ + k 2 fψ = 0 = − k 2 , or (156) ψ f which yields f ′′ ∆ ψ = λ − k 2 = − λ, (157) ψ f with a certain constant λ . 93 / 143
LECTURE 2: THE MATHEMATICAL NATURE OF WAVES. Thus ψ must solve the Dirichlet eigenvalue problem for the Laplace equation in cross-sectional domain Ω x ′ ∈ Ω , ∆ ψ + λψ = 0 , (158) ψ | Γ = 0 . Denote by Λ = { λ n } and Ψ = { ψ n } the system of eigenvalues and eigenfunctions of this problem. A particular solution of (153) is Π = Π n ( x ) = ψ n ( x ′ ) f n ( x 3 ) , (159) where f n satisfies the equation n + ( k 2 − λ n ) f n = 0 . f ′′ (160) The general solution of (160) is f n ( x 3 ) = A n e iγ n x 3 + B n e − iγ n x 3 , � k 2 − λ n . γ n = (161) The first and the second terms in (161) correspond, respectively, to the wave propagating in the positive or negative direction of the waveguide axis. 94 / 143
LECTURE 2: THE MATHEMATICAL NATURE OF WAVES. Considering the wave propagating in the positive direction set f n ( x 3 ) = A n e iγ n x 3 . (162) As a result we obtain the solution Π n ( x ′ , x 3 ) = A n ψ n ( x ′ ) e iγ n x 3 . (163) √ k 2 − λ n with k 2 > λ n We have 0 ≤ λ 1 ≤ λ 2 ≤ . . . ; therefore, that are at most finitely many values of γ n = √ λ n − k 2 ( i 2 = − 1 ) with k 2 < λ n , are purely imaginary. that are real, while infinitely many of them, for γ n = i Consequently, there are at most finitely many waves in the waveguide that propagate without attenuation (in the positive direction of x 3 -axis) and infinitely many decay exponentially. 95 / 143
LECTURE 2: THE MATHEMATICAL NATURE OF WAVES. Looking for particular solutions of (149) in the form grad div P + k 2 P , H = (164) = ik rot P , E where the polarization potential P = [0 , 0 , Π] has only one nonzero component P 3 = Π , it is easy to see that E 3 = 0 , H = [0 , 0 , H 3 ] , E = [ E 1 , E 2 , 0] , (165) and this case is called TE-polarization or H-polarization. Substituting (164) into (149) yields the equations ∆Π + ∂ 2 Π ∆ 3 Π + k 2 Π + k 2 Π = 0 , = 0 or (166) ∂x 2 3 ∂ 2 + ∂ 2 + ∂ 2 ∆ 3 = . ∂x 2 ∂x 2 ∂x 2 1 2 3 Condition (150) is satisfied if we assume that � ∂ Π � = 0 , (167) � ∂n � Σ because the third components of P and the first two of E are tangential components that must vanish on the waveguide wall and they are coupled by the first relation (164). 96 / 143
LECTURE 2: THE MATHEMATICAL NATURE OF WAVES. Repeating the above analysis we see that Π = Π n ( x ) = A n ψ n ( x ′ ) e iγ n x 3 , (168) where ψ n solves the Neumann eigenvalue problem for the Laplace equation in cross-sectional domain Ω x ′ ∈ Ω , ∆ ψ + λψ = 0 , (169) ∂ψ � � = 0 . � ∂n � Σ (168) specifies the wave propagating in the positive direction of the waveguide axis. Denote by Λ H = { λ H n } and Ψ H = { ψ H n } the system of eigenvalues and eigenfunctions of this problem. We have 0 ≤ λ H 1 ≤ λ H 2 ≤ . . . ; k 2 − λ H n with k 2 > λ H therefore, that are at most finitely many values of γ H � n = n that are real, while infinitely n − k 2 with k 2 < λ H many of them, for γ H � n = i λ H n are purely imaginary. Consequently, there are at most finitely many waves in the waveguide that propagate without attenuation (in the positive direction of x 3 -axis) and infinitely many decay exponentially. The waves obtained from (151), (152) or (164), (165) are called, respectively, TM-waves or TE-waves. 97 / 143
LECTURE 2: THE MATHEMATICAL NATURE OF WAVES. Diffraction from a dielectric obstacle in a 2D-guide. Introduce the complex magnitude of the stationary electric and magnetic field, E ( r , t ) and H ( r , t ) , respectively, where r = ( x, y, z ) , and consider the problem of diffraction of a TM wave (or mode) E ( r , t ) = E ( r ) exp ( − iωt ) , H ( r , t ) = H ( r ) exp ( − iωt ) , (170) � 1 ∂E x 1 ∂E x � E ( r ) = ( E x , 0 , 0) , H ( r ) = 0 , ∂z , − , (171) iωµ 0 iωµ 0 ∂y by a dielectric inclusion D in a parallel-plane waveguide W = { r : 0 < y < π, −∞ < x, z < ∞} . Figure .1: TE-mode diffraction by a dielectric inclusion in a parallel-plane waveguide 98 / 143
LECTURE 2: THE MATHEMATICAL NATURE OF WAVES. The total field u ( y, z ) = E x ( y, z ) = E inc ( y, z ) + E scat ( y, z ) = u i ( y, z ) + u s ( y, z ) of the diffraction by the D x x of the unit-magnitude TE wave with the only nonzero component is the solution to the BVP [∆ + κ 2 ε ( y, z )] u ( y, z ) = 0 in S = { ( y, z ) : 0 < y < π, −∞ < z < ∞} , u ( ± π, z ) = 0 , (172) ∞ u ( y, z ) = u i ( y, z ) + u s ( y, z ) , u s ( y, z ) = � a ± n exp( i Γ n z ) sin( ny ) , (173) n =1 where ∆ = ∂ 2 ∂y 2 + ∂ 2 ∂z 2 is the Laplace operator, superscripts + and − correspond, respectively, to the domains z > 2 πδ and z < − 2 πδ , ω = κc is the dimensionless circular frequency, κ = ω / c = 2 π / λ is the dimensionless frequency parameter ( λ is the free-space wavelength), c = ( ε 0 µ 0 ) − 1/2 is the speed of light in vacuum, and Γ n = ( κ 2 − n 2 ) 1/2 is the transverse wavenumber satisfying the conditions = ( n 2 − κ 2 ) 1/2 , Im Γ n ≥ 0 , = i | Γ n | , | Γ n | = n > κ. (174) Γ n Im Γ n It is also assumed that the series in (173) converges absolutely and uniformly and allows for double differentiation with respect to y and z . Note that u i ( y, z ) satisfies (172) in S , the boundary condition, and radiation condition (173) only in the positive direction, so that the electromagnetic field with the x -component u i ( y, z ) may be interpreted as a normal wave (a waveguide mode) coming from the domain z < − 2 πδ . 99 / 143
LECTURE 2: THE MATHEMATICAL NATURE OF WAVES. Diffraction from a dielectric obstacle in a 3D-guide. Diffraction of electromagnetic waves by a dielectric body Q in a 3D tube (a waveguide) with cross section Ω (a 2D domain bounded by smooth curve Γ ) parallel to the x 3 -axis in the cartesian coordinate system is described by the solution to the inhomogeneous system of Maxwell equations ǫ E + j 0 rot H = − iω ˆ E (175) rot E = iωµ 0 H , E τ | ∂P = 0 , H ν | ∂P = 0 , (176) admitting for | x 3 | > C and sufficiently large C > 0 the representations (+ corresponds to + ∞ and − to −∞ ) | x 3 | � λ (1) p Π p e 3 − iγ (1) ∇ 2 Π p � E e − iγ (1) � R ( ± ) p � � = p + p H − iωε 0 ( ∇ 2 Π p ) × e 3 p iωµ 0 ( ∇ 2 Ψ p ) × e 3 e − iγ (2) Q ( ± ) | x 3 | � � � + p . (177) p λ (2) p Ψ p e 3 − iγ (2) ∇ 2 Ψ p p p 100 / 143
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