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.