Divergence The gradient of a scalar field f is defined as f ( x, y, - - PowerPoint PPT Presentation

Divergence

The gradient of a scalar field f is defined as ∇f(x, y, z) = ∂f ∂x(x, y, z)i + ∂f ∂y (x, y, z)j + ∂f ∂z (x, y, z)k. Definition Suppose that F : D → R3 is a C1-vector field with D ⊂ R3 given by F(x, y, z) := F1(x, y, z)i + F2(x, y, z)j + F3(x, y, z)k. Then the divergence of F is the scalar field div F on the same domain D defined by (div F)(x, y, z) := ∂F1 ∂x (x, y, z) + ∂F2 ∂y (x, y, z) + ∂F3 ∂z (x, y, z), for all (x, y, z) ∈ D.

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Divergence (cont.)

We can define the symbolic vector ∇ as ∇ := ∂ ∂xi + ∂ ∂y j + ∂ ∂z k. Using this vector, one can define div F = ∇ · F. Thus, ∇ · F is an alternative notation for div F (often preferred in more modern books). Remark: A vector field F : D → R3, with D ⊂ R3, is called sole- noidal (incompressible), if ∇ · F = 0. We will see later that the magnetic field B is always solenoidal.

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Curl

Defintion Suppose that F : D → R3 is a C1-vector field with D ⊂ R3 given by F(x, y, z) := F1(x, y, z)i + F2(x, y, z)j + F3(x, y, z)k. Then the curl of F is the vector field curl F on the same domain D defined by (curl F)(x, y, z) := ∂F3 ∂y (x, y, z) − ∂F2 ∂z (x, y, z)

• i+

∂F1 ∂z (x, y, z) − ∂F3 ∂x (x, y, z)

• j+

∂F2 ∂x (x, y, z) − ∂F1 ∂y (x, y, z)

• k.

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Curl (cont.)

An alternative way to define curl F would be as curl F = ∇ × F =

• i

j k

∂ ∂x ∂ ∂y ∂ ∂z

F1 F2 F3

• .

(Physical interpretation: http://mathinsight.org/curl idea.) Theorem Suppose F : D → R3 is a C1-vector field with D = R3. Then the fol- lowing are equivalent

1 F is a conservative vector field; 2

Γ F(r) · dr = 0 for every closed curve Γ in R3;

3 (∇ × F)(x, y, z) = 0 for all (x, y, z) ∈ R3. Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 4/24

Curl (cont.)

Remark A vector field F : R3 → R3 is called irrotational if (∇ × F)(x, y, z) = 0 for all (x, y, z) ∈ R3. Thus, F is conservative iff it is irrotational. Theorem Suppose that f : D → R is a C2-scalar field with D ⊂ R3. Then the vector field ∇f is irrotational, that is (∇ × (∇f))(x, y, z) = 0, ∀(x, y, z) ∈ R3. Theorem Suppose that G : D → R3 is a C2-vector field with D ⊂ R3. Then the vector field ∇ × G is solenoidal, that is (∇ · (∇ × G))(x, y, z) = 0, ∀(x, y, z) ∈ R3.

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Laplacian

Definition Suppose that f : D → R is a C2-scalar field with D ⊂ R3. Then the Laplacian of the scalar field f is another scalar field denoted by ∆f and defined on the same domain D by ∆f(x, y, z) = ∂2f ∂x2 (x, y, z)+ ∂2f ∂y2 (x, y, z)+ ∂2f ∂z2 (x, y, z), ∀(x, y, z) ∈ R3. Definition Suppose that F : D → R3 is a C2-vector field with domain D ⊂ R3 given by F(x, y, z) := F1(x, y, z)i + F2(x, y, z)j + F3(x, y, z)k, ∀(x, y, z) ∈ D. The Laplacian of the vector field F is another vector field ∆F on the same domain D defined by ∆F(x, y, z) := ∆F1(x, y, z)i + ∆F2(x, y, z)j + ∆F3(x, y, z)k.

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Laplacian (cont.)

Theorem Suppose that f : D → R is a C2-scalar field with D ⊂ R3. Then the Laplacian of f is the divergence of the gradient of f, that is ∆f(x, y, z) := ∇ · (∇f)(x, y, z), ∀(x, y, z) ∈ D. Remark Suppose that f : D → R is a C2-scalar field such that its gradient is

• solenoidal. Then,

∆f(x, y, z) := 0, ∀(x, y, z) ∈ D. The above relation is a particular case of Laplace’s equation. Any f that satisfies the relation is called a solution of Laplace’s equation. Laplace’s equation is ubiquitous throughout mathematical phy- sics and engineering.

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Time derivative

Let F(x, y, z, t) be a time-dependant vector field defined on some domain D ⊂ R3. Then, it can be shown that ∂ ∂t [(∇ · F)(x, y, z, t)] = ∇ · ∂F(x, y, z, t) ∂t . Moreover, ∂ ∂t [(∇ × F)(x, y, z, t)] = ∇ × ∂F(x, y, z, t) ∂t .

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Useful identities

∇(cf)(x, y, z, t) = c(∇f)(x, y, z, t) ∇(f + g)(x, y, z, t) = (∇f)(x, y, z, t) + (∇g)(x, y, z, t) ∇(fg)(x, y, z, t) = g(x, y, z, t)(∇f)(x, y, z, t)+f(x, y, z, t)(∇g)(x, y, z, t) ∇(f/g)(x, y, z, t) = g(x, y, z, t)(∇f)(x, y, z, t) − f(x, y, z, t)(∇g)(x, y, z, t) g2(x, y, z, t) , for all (x, y, z, t) such that g(x, y, z, t) = 0. ∆(fg)(x, y, z, t) = g(x, y, z, t)(∆f)(x, y, z, t) + f(x, y, z, t)(∆g)(x, y, z, t)+ + 2(∇f · ∇g)(x, y, z, t)

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Useful identities (cont.)

∇ · (cF)(x, y, z, t) = c(∇ · F)(x, y, z, t) ∇ · (F + G)(x, y, z, t) = (∇ · F)(x, y, z, t) + (∇ · G)(x, y, z, t) ∇ × (cF)(x, y, z, t) = c(∇ × F)(x, y, z, t) ∇ × (F + G)(x, y, z, t) = (∇ × F)(x, y, z, t) + (∇ × G)(x, y, z, t) ∇ × (∇ × F)(x, y, z, t) = ∇(∇ · F)(x, y, z, t) − ∆F(x, y, z, t) ∇ · (F × G)(x, y, z, t) = G(x, y, z, t) · (∇ × F)(x, y, z, t)− −F(x, y, z, t) · (∇ × G)(x, y, z, t)

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Closed surfaces

Definition The surface S is called closed when it is the boundary of some region in R3, or equivalently completely contains some region in R3. Some examples:

1 Close surface: sphere in R3. 2 Open surface: x − y plane. 3 Finite open surface: a disk of radius r < ∞ in R3.

We are going to assume that our finite open surface S always has a boundary Γ which is closed curve. We always give Γ that particular direction which is such that S enclosed by Γ is on your left when you traverse Γ in this direction. Such boundary curve Γ is called positively oriented.

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Theorem of Stokes

Stokes’ theorem Suppose that S is a finite open surface in R3 with closed positively

• riented boundary curve Γ, and F : R3 → R3 is a C1-vector field.

Then

• Γ

F · dr =

• S

(∇ × F) · dA.

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Remark

Suppose that S1 and S2 are two finite open surfaces in R3 having a common closed positively oriented boundary curve Γ. Then,

• Γ

F · dr =

• S1

(∇ × F) · dA. and

• Γ

F · dr =

• S2

(∇ × F) · dA. . Consequently, we obtain

• S1

(∇ × F) · dA =

• S2

(∇ × F) · dA. . This result can be used to replace evaluation of a difficult surface integral with the evaluation of an easier surface integral.

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Remark (cont.)

Suppose we have F : R3 → R3. Let (x, y, z) ∈ R3 be some point and let n be a unit vector pas- sing through (x, y, z). Let Sρ be a flat disc of radius ρ with centre at (x, y, z) and normal direction n. Finally, let Γρ be the closed positively oriented boundary of S .

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Remark (cont.)

From Stokes’ theorem, we have

• Γρ

F · dr =

• Sρ

(∇ × F) · dA. When ρ is very small, we have

• Sρ

(∇ × F) · dA ≈ (∇ × F)(x, y, z) · n area{Sρ}. Consequently, we can write (∇ × F)(x, y, z) · n ≈ 1 area{Sρ}

• Γρ

F · dr, and thus (∇ × F)(x, y, z) · n = lim

ρ→0

1 area{Sρ}

• Γρ

F · dr.

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Remark (cont.)

Suppose that F is an electric field. We have seen that the “turning” of F around Γρ represented by the circulation is actually the electromotive force. In this case, (∇ × F)(x, y, z) · n is the limit of electromotive force per unit area enclosed by Γρ as ρ → 0. This interpretation of ∇ × F, when F is an electric field, is very useful in electromagnetism.

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Outward vs. inward orientation

We say that a finite closed surface has outward orientation when the unit vector normal to the surface points out of the enclosed region at every point on the surface. Analogously, a finite closed surface is said to have inward orien- tation when the unit vector normal to the surface points into the enclosed region at every point on the surface

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Divergence theorem of Gauss-Ostrogradskii

Divergence theorem of Gauss-Ostrogradskii Suppose that S is a closed surface with outward orientation which encloses a finite region Ω ⊂ R3, and F : R3 → R3 is a C1-vector field. Then

• Ω

(∇ · F)dV =

• S

F · dA.

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Remark

Suppose we have a vector field F : R3 → R3 Let Ωρ ⊂ R3 be a sphere with radius ρ centred at some point (x, y, z) ∈ R3. Let Sρ be the outward oriented surface of Ωρ. Then,

• Ωρ

(∇ · F)dV =

• Sρ

F · dA. When ρ is small, we have

• Ωρ

(∇ · F)dV ≈ (∇ · F)(x, y, z)vol{Ωρ}. Therefore, (∇ · F)(x, y, z) ≈ 1 vol{Ωρ}

• Sρ

F · dA.

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Remark (cont.)

In the limit, we have (∇ · F)(x, y, z) = lim

ρ→0

1 vol{Ωρ}

• Sρ

F · dA, ∀(x, y, z) ∈ R3. Thus, ∇ · F(x, y, z) is approximately the flux of F through the small spherical surface Sρ centred at (x, y, z) per volume of the region Ωρ enclosed by Sρ. This approximation becomes exact when the radius ρ becomes “infinitesimally small”.

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Remark (cont.)

When F represents the current density J, we have (∇ · J)(x, y, z) = lim

ρ→0

1 vol{Ωρ}

• Sρ

J · dA, ∀(x, y, z) ∈ R3. where

• Sρ J · dA is the total current flowing out of Ωρ.

If (∇ · J)(x, y, z) > 0, then there must be a source of current located at the point (x, y, z). If (∇ · J)(x, y, z) < 0, then there must be a sink of current located at the point (x, y, z).

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The continuity equation

The charge density scalar field ρ can be used to obtain the total charge Q enclosed in a region Ω ⊂ R3, namely Q =

• Ω

ρ dV =

• Ω

ρ(x, y, z)dxdydz. Suppose that charge is in motion through space. This means that the charge density is now a function of time t for each (x, y, z). Fix some arbitrary region Ω ⊂ R3 having the closed surface S as its boundary. Then the total charge contained within Ω at each instant t is gi- ven by Q =

• Ω

ρ(x, y, z, t)dxdydz =

• Ω

ρ dV.

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The continuity equation (cont.)

The rate of increase of the total charge contained within the re- gion Ω is given by dQ(t) dt = d dt

• Ω

ρ dV =

• Ω

∂ρ ∂t dV. On the other hand, the rate at which charge enters the region Ω through the boundary surface S is given by −

• S

J · dA. The law of conservation of charge says that

• Ω

∂ρ ∂t dV = −

• S

J · dA, i.e., within any region Ω there are never any sources of charge or sinks of charge.

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The continuity equation (cont.)

Using the divergence theorem, we also have

• Ω

∂ρ ∂t dV = −

• Ω

(∇ · J)dV, to equivalently

• Ω
• (∇ · J) + ∂ρ

∂t

• dV = 0.

It is important to realize that the above relation holds for each and every region Ω ⊂ R3, and so by du Bois Reymond theorem (∇ · J) + ∂ρ ∂t = 0, ∀(x, y, z) ∈ Ω, at each t. This is the continuity equation is indispensable in our study of Maxwell’s equations.

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