General Introduction to PDE's Notation and Terminology Consider a Function U = U(X 1,X2,...,Xn). the Partial Derivative Of

General Introduction to PDE’s

Notation and terminology

Consider a function u = u(x1, x2, . . . , xn). The partial derivative of u at (x1, x2, . . . , xn) with respect to xi is deﬁned by ∂u u(x , x , . . . , x + h, . . . , x ) − u(x , x , . . . , x ) = lim 1 2 i n 1 2 n (1) ∂xi h→0 h provided that the limit exists. In many practical applications u = u(x, y, z, t). Notational convention

∂u ∂u ∂2u ∂2u = u , = u , = u , = u ,... (2) ∂x x ∂t t ∂x2 xx ∂x∂t xt A partial diﬀerential equation (PDE) is an equation that involves partial derivatives of an unknown function of several variables.

F (x, y, z, t; u, ux, uy, uz, ut, uxx,...) = 0 (3)

The order of the PDE is the order of the highest partial derivative in the equation. The PDE (3) is said to be linear if the function F is linear in each of the variables u, ux, uy,..., and if the coeﬃcients of u and its derivatives are functions of the independent variables only.

Example: second order linear equation in two variables u = u(x, t)

Auxx + Buxt + Cutt + Dux + Eut + F u = G (4)

The coeﬃcients A, B, C, D, E, F, G may be constants or given functions of x, t. Equation (4) is called homogeneous if G(x, t) ≡ 0 for all x, t. If G(x, t) is not identically zero, equation (4) is called nonhomogeneous.

Examples

st ut + 3ux = 0 (1 order, linear, homogeneous, const. coeﬀ.) nd ut = k(x)uxx + 1 (2 order, linear, nonhomogeneous, var. coeﬀ.) nd ut + uux = uxx (2 order, nonlinear)

Basic types of linear PDEs, in terms of the coeﬃcients:

2 (a) Parabolic: B − 4AC = 0. Example: u(x, t) such that ut − uxx = 1.

2 (b) Hyperbolic: B − 4AC > 0. Example: u(x, t) such that utt − uxx = 0.

2 (c) Elliptic: B − 4AC < 0. Example: u(x, y) such that uxx + uyy = 0.

1 Practical applications:

Example 1. Equation for conservation of mass in a one-dimensional gas dynamics problem

Consider the ﬂow in a tube where the properties of the gas such as density and velocity are assumed to be constant across each cross section of the tube. Let ρ(x, t) be the density of the gas at point x and time t.

mass in [x , x ] at time t = R x2 ρ(x, t)dx 1 2 x1

Assume that mass can change only because of gass ﬂowing across the endpoints x1 or x2 and let v(x, t) be the velocity of the gas. The rate of ﬂow, or flux of gas past point x at time t is

mass ﬂux at (x, t) = ρ(x, t)v(x, t)

The rate of change in mass is

d Z x2 ρ(x, t)dx = ρ(x1, t)v(x1, t) − ρ(x2, t)v(x2, t) (5) dt x1 Equation (5) represents the integral form of the mass conservation law. By integrating (5) in the time interval [t1, t2]

2 Z x2 Z x2 Z t ρ(x, t2)dx − ρ(x, t1)dx = [ρ(x1, t)v(x1, t)dt − ρ(x2, t)v(x2, t)]dt (6) x1 x1 t1 In addition we require that ρ(x, t), v(x, t) are diﬀerentiable functions. Using

Z t2 ∂ ρ(x, t2) − ρ(x, t1) = ρ(x, t)dt (7) t1 ∂t

Z x2 ∂ ρ(x2, t)v(x2, t) − ρ(x1, t)v(x1, t) = (ρ(x, t)v(x, t))dx (8) x1 ∂x in (6) it results Z t2 Z x2 ∂ ∂ ρ(x, t) + (ρ(x, t)v(x, t)) dxdt = 0 (9) t1 x1 ∂t ∂x

Since [x1, x2], [t1, t2] were arbitrary selected, continuity of the integrand implies

ρt + (ρv)x = 0 (10) which is the diﬀerential form of the mass conservation law.

2 Example 2. Conduction of heat in a one-dimensional rod

Consider a rod of constant cross-sectional area A, oriented in the x-direction, from x = 0 to x = L. Consider

e(x, t) ≡ thermal energy density which represents the amount of thermal energy per unit of volume. In addition, assume that e(x, t) is constant across a section and no thermal energy can pass through the lateral surface. In any finite segment [a, b] of the rod, the total heat energy is R b total heat energy in [a, b] = a e(x, t)Adx Introduce the heat flux as the amount of thermal energy per unit time flowing to the right per unit surface area.

φ(x, t) = heat ﬂux

Heat energy ﬂowing through the side edges at time t: φ(a, t)A − φ(b, t)A. In addition, we consider heat sources

Q(x, t) = heat energy per unit volume generated per unit time

R b Heat energy generated inside the rod region [a, b] at time t: a Q(x, t)dx A. Conservation of heat energy

d Z b Z b e(x, t)dx = φ(a, t) − φ(b, t) + Q(x, t)dx (11) dt a a Equation (11) represents the integral conservation law of heat energy. If in addition, e, φ are diﬀerentiable, from (11) it follows that

Z b ∂e ∂φ + − Q dx = 0 (12) a ∂t ∂x Continuity of the integrand in (12) implies ∂e ∂φ = − + Q (13) ∂t ∂x which represents the diﬀerential form of the conservation law of heat energy.

Temperature and thermal energy Notations: u(x, t) = temperature of the rod c(x) = speciﬁc heat (the heat energy that must be supplied to a unit mass of a substance to raise its temperature one unit) ρ(x) = mass density (mass per unit volume) The thermal energy is the energy it takes to raise the temperature from a reference temperature 0◦ to its actual temperature u(x, t).

e(x, t) = c(x)ρ(x)u(x, t) (14)

3 Replacing (14) in (13), it follows

∂u ∂φ c(x)ρ(x) = − + Q (15) ∂t ∂x Fourier’s law of heat conduction

∂u φ = −K (16) 0 ∂x where the coeﬃcient K0 is the thermal conductivity of the material.

Heat equation

By replacing (16) in (15), it follows

∂u ∂ ∂u cρ = K + Q (17) ∂t ∂x 0 ∂x

If c, ρ, K0 are all constants and there are no heat sources (Q = 0), then (17) becomes

∂u ∂2u = k (18) ∂t ∂x2 where the constant K k = 0 (19) cρ is called the thermal diﬀusivity. Equation (18) is called the heat equation and is also known as the diﬀusion equation.

Initial Conditions u(x, 0) = f(x) (20)

Boundary Conditions

1. Prescribed temperature u(0, t) = uB(t) (21)

2. Prescribed heat ﬂux ∂u −K (0) (0, t) = φ(t) (22) 0 ∂x ∂u (0, t) = 0 perfectly insulated boundary (23) ∂x 3. Newton’s law of cooling ∂u −K (0) (0, t) = −H[u(0, t) − u (t)] (24) 0 ∂x R ∂u −K (L) (L, t) = H[u(L, t) − u (t)] (25) 0 ∂x L

4 Q: Show that the problem

∂u ∂2u = k , 0 < x < L, t > 0 (26) ∂t ∂x2 u(x, 0) = f(x) (27)

u(0, t) = T1(t) (28)

u(L, t) = T2(t) (29) has at most one solution.

Steady-state solutions; heat equation in 2-D and 3-D

Steady-state solutions

Consider the 1-D heat equation ∂u ∂2u = k , 0 < x < L, t > 0 (30) ∂t ∂x2 with the initial condition

u(x, 0) = f(x) (31) We assume that the temperature at the boundaries remains constant in time

u(0, t) ≡ T1 (32)

u(L, t) ≡ T2 (33) where T1 and T2 are given constants. A steady-state (equilibrium) solution of the problem (30-33) is a temperature distribution that is time independent

u(x, t) = u(x), 0 ≤ x ≤ L, t > 0 (34)

∂u(x) Since u is a function of the spatial variable x only, ∂t ≡ 0 and (30) becomes an ordinary diﬀerential equation d2u = 0 (35) dx2 with the general solution u(x) = C1x + C2 (36)

where C1,C2 are arbitrary constants. To determine C1,C2 we use the boundary conditions

u(0) = T1 ⇒ C2 = T1 (37) T − T u(L) = T ⇒ C = 2 1 (38) 2 1 L Replacing these values in (36) we obtain the temperature distribution at equilibrium T − T u(x) = T + 2 1 x (39) 1 L

5 Notice that the initial condition (31) was not used in the derivation of (39). In general (39) does not satisfy (31), but the steady-state solution may be used to describe the behaviour of the solution u(x, t) of the problem (30),(31),(32),(33) after a very long period of time.

T2 − T1 lim u(x, t) = T1 + x (40) t→∞ L

Insulated Boundaries

Next we want to ﬁnd the steady-state solutions of the 1-D heat equation (30) with insulated boundary conditions

∂u (0, t) = 0, t > 0 (41) ∂x ∂u (L, t) = 0, t > 0 (42) ∂x

Since the equilibrium temperature is time independent (ut ≡ 0), the steady-state solution must satisfy the ODE d2u = 0 (43) dx2 and the boundary values

du (0) = 0 (44) dx du (L) = 0 (45) dx The solution of the problem (43), (44),(45) is

u(x) = C2 (46)

where C2 is an arbitrary constant. To determine C2 we use the initial condition and the fact that the total thermal energy in the rod remains constant in time (why?)

Z L E = cρudx = cρC2L = constant (47) 0 At any future time t > 0, the total thermal energy in the rod must be equal to the total thermal energy at the initial time t = 0 which is

Z L E0 = cρ f(x)dx (48) 0 From (47) and (48) we obtain the steady-state temperature distribution

1 Z L u(x) = C2 = f(x)dx (49) L 0

6 Gradient, directional derivatives, divergence, Laplacian

To derive the heat equation in two or three dimensions, first we need to recall some concepts from vector calculus and integral identities. You may (should) be familiar with the concepts of gradient, divergence and Laplacian. Let’s recall them briefly: If u = u(x, y, z) is a C1 function defined on a region Ω ⊂ R3, the gradient of u is defined by ∂u ∂u ∂u ∇u = î + ˆj + kˆ (50) ∂x ∂y ∂z 3 If α = α1î + α2ˆj + α3kˆ denotes a unit vector in R ,(|α| = 1), the directional derivative of u in direction α is given by ∂u ∂u ∂u ∂u = α · ∇u = α + α + α (51) ∂α 1 ∂x 2 ∂y 3 ∂z If θ denotes the angle between the vectors α and ∇u ∂u = |∇u| cos θ (52) ∂α The largest rate of change of u (the largest directional derivative) is |∇u| > 0, and occurs in the direction of the gradient (θ = 0).

1 If w = w1(x, y, z)î + w2(x, y, z)ˆj + w3(x, y, z)kˆ is a C vector field on Ω, the divergence of w is defined to be the scalar field ∂w ∂w ∂w div w ≡ ∇ · w = 1 + 2 + 3 (53) ∂x ∂y ∂z In particular, for w = ∇u, we have

∂2u ∂2u ∂2u div grad u ≡ ∇ · ∇u ≡ ∇2u = + + = u + u + u (54) ∂x2 ∂y2 ∂z2 xx yy zz

The expression ∇2u is called the Laplacian of u.

The Divergence Theorem

Let Ω be a bounded domain with piecewise smooth boundary surface S. Let n be the unit outward normal vector to S and let w be a vector ﬁeld that is C1 in Ω and C0 on Ω ∪ S. Then Z Z ∇ · wdΩ = w · ndS (55) Ω S

7 Heat equation in 2D or 3D

We consider an arbitrary bounded region Ω ⊂ R3, with a boundary S. At a given time t, the heat energy in Ω is given by the volume integral

E(t) = heat energy = y e(x, y, z, t)dxdydz = y cρudV (56) Ω Ω The rate of change in the heat energy in Ω is given by the heat flowing across the boundary S per unit time. In three dimensions, the heat flux φ is a vector and the magnitude |φ| represents the heat energy flowing per unit time per unit of surface area. At any point (x, y, z) on the boundary S we consider the unit outward normal vectorn ˆ. The amount of energy flowing out of the region Ω per unit surface area per unit time is given by the outward normal component of the heat flux vector φ · nˆ. Notice that if φ · nˆ > 0 then the heat flux is directed outward (energy flows out of Ω). The total energy flowing out over the boundary surface S per unit time is

x φ · ndSˆ (57) S In addition, there may be sources of energy inside the region Ω. Let Q denote the heat energy generated per unit time per unit volume. The total energy generated per unit time is then

y QdV (58) Ω The conservation of heat energy in the region Ω implies

d cρudV = − φ · ndSˆ + QdV (59) dt y x y Ω S Ω Next we use the divergence theorem (55) in the right side of (59) to obtain

d cρudV = − ∇ · φdV + QdV (60) dt y y y Ω Ω Ω which may be written as ∂u cρ + ∇ · φ − Q dV = 0 (61) y ∂t Ω Since Ω was arbitrary, from (61) we obtain

∂u cρ = −∇ · φ + Q (62) ∂t Fourier’s law of heat conduction φ = −K0∇u (63) Replacing (63) in (62) we obtain the PDE for the temperature

∂u cρ = ∇ · (K ∇u) + Q (64) ∂t 0

8 If Q ≡ 0 and the coeﬃcients c, ρ, K0 are constants, then (64) becomes (see deﬁnition (54)) ∂u = k∇2u (65) ∂t

where k = K0/cρ. Equation (65) is the heat (diﬀusion) equation in 3D.

Initial conditions u(x, y, z, 0) = f(x, y, z) (66)

Boundary values

1. Prescribed temperature: u(x, y, z, t) = T (x, y, z, t) on some region of S.

2. Prescribed ﬂux: −K0∇u · nˆ = φ on some region of S.

In particular, insulated boundary means : −K0∇u · nˆ = 0

3. Newton’s law of cooling: −K0∇u · nˆ = H(u − ub) on some region of S.

Steady-state solution of the heat equation

∇ · (K0∇u) + Q = 0 (67)

If in addition, K0 is a constant we obtain Poisson’s equation Q ∇2u = − (68) K0 If in addition, Q ≡ 0, we obtain Laplace’s equation

∇2u = 0 (69)

Changing the coordinates In many practical applications (e.g. engineering, geophysical problems) it is more useful to work in polar coordinates (2D)

x = r cos θ (70) y = r sin θ (71)

spherical coordinates (3D)

x = r sin γ cos θ (72) y = r sin γ sin θ (73) z = r cos γ (74)

9 or cylindrical coordinates (3D)

x = r cos θ (75) y = r sin θ (76) z = z (77)

Q: Derive the expression of the Laplacian ∇2u in polar, spherical, and cylindrical coordinates.