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Chapter 12: Boundary-Value Problems in Rectangular Coordinates

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Chapter 12: Boundary-Value Problems in Rectangular Coordinates Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Chapter 12: Boundary-Value Problems in Rectangular Coordinates 王奕翔 Department of Electrical Engineering National Taiwan University [email protected] December 21, 2013 1 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary In this lecture, we focus on solving some classical partial differential equations in boundary-value problems. Instead of solving the general solutions, we are only interested in finding useful particular solutions. We focus on linear second order PDE: (A; ··· ; G: functions of x; y) Auxx + Buxy + Cuyy + Dux + Euy + Fu = G: @2u notation: for example, uxy := @x @y . Method: Separation of variables – convert a PDE into two ODE’s Types of Equations: Heat Equation Wave Equation Laplace Equation 2 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Classification of Linear Second Order PDE Auxx + Buxy + Cuyy + Dux + Euy + Fu = G: @2u notation: for example, uxy := @x @y . 1 Homogeneous vs. Nonhomogeneous Homogeneous () G = 0 Nonhomogeneous () G =6 0: 2 Hyperbolic, Parabolic, and Elliptic: A; B; C; ··· ; G: constants, Hyperbolic () B2 − 4AC > 0 Parabolic () B2 − 4AC = 0 Elliptic () B2 − 4AC < 0 3 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Superposition Principle Theorem If u1(x; y); u2(x; y);:::; uk(x; y) are solutions of a homogeneous linear PDE, then a linear combination Xk u(x; y) := cnun(x; y) n=1 is also a solution. Note: We shall assume without rigorous argument that the linear combination can be an infinite series X1 u(x; y) := cnun(x; y) n=1 4 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary 1 Separation of Variables and Classical PDE’s 2 Wave Equation 3 Laplace’s Equation 4 Summary 5 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Separation of Variables To find a particular solution of an PDE, one method is separation of variables, that is, assume that the solution u(x; y) takes the form of a product of a x-function and a y-function: u(x; y) = X(x)Y(y): Then, with the following, sometimes the PDE can be converted into an ODE of X and an ODE of Y: dX dY u = Y = X 0Y; u = X = XY 0 x dx y dy d2X d2Y u = Y = X 00Y; u = X = XY 00; u = X 0Y 0 xx dx2 yy dy2 xy Note: Derivatives are with respect to different independent variables. dX For example, X 0 := . dx 6 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Convert a PDE into Two ODE’s Example Use separation of variables to convert the PDE below into two ODE’s. 2 x uxx + (x + 1)uy + (x + xy)u = 0 With u(x; y) = X(x)Y(y), the original PDE becomes x2X 00Y + (x + 1)XY 0 + (x + 1)yXY = 0 =) x2X 00Y = −(x + 1)X(Y 0 + yY) x2X 00 Y 0 =) = − − y = λ separation constant (x + 1)X Y Left-hand side is a function of x, independent of y; Right-hand side is a function of y, independent of x. Hence, the above is equal to something independent of x and y 7 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Convert a PDE into Two ODE’s Example Use separation of variables to convert the PDE below into two ODE’s. 2 x uxx + (x + 1)uy + (x + xy)u = 0 With u(x; y) = X(x)Y(y), the original PDE becomes x2X 00Y + (x + 1)XY 0 + (x + 1)yXY = 0 =) x2X 00Y = −(x + 1)X(Y 0 + yY) x2X 00 Y 0 =) = − − y = λ separation constant ((x + 1)X Y x2X 00(x) − λ(x + 1)X(x) = 0 =) Y 0(y) + (y + λ)Y(y) = 0 8 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Some Remarks 1 The method of separation of variables can only solve for some linear second order PDE’s, not all of them. 2 For the PDE’s considered in this lecture, the method works. 3 The method may work for both homogeneous (G = 0) and nonhomogeneous (G =6 0) PDE’s Auxx + Buxy + Cuyy + Dux + Euy + Fu = G: 9 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Three Classical PDE’s In this lecture we focus on solving boundary-value problems of the following three classical PDE’s that arise frequently in physics, mechanics, and engineering: 1 (One-Dimensional) Heat Equation/Diffusion Equation kuxx = ut; k > 0 2 (One-Dimensional) Wave Equation/Telegraph Equation 2 a uxx = utt 3 (Two-Dimensional) Laplace Equation uxx + uyy = 0 10 / 50 王奕翔 DE Lecture 14 92467_12_ch12_p455-492.qxd 2/16/12 11:41 AM Page 461 12.2 CLASSICAL PDEs AND BOUNDARY-VALUE PROBLEMS G 461 variables. If you compare (1)–(3) with the linear form in Theorem 12.1.1 (with t play- ing the part of the symbol y), observe that the heat equation (1) is parabolic, the wave equation (2) is hyperbolic, and Laplace’s equation is elliptic. This observation will be important in Chapter 15. Heat Equation Equation (1) occurs in the theory of heat flow—that is, heat transferred by conduction in a rod or in a thin wire. The function u(x, t) represents Separation of Variables and Classical PDE’s temperature at a point x along the rod at some time t. Problems in mechanical vibra- Wave Equation Laplace’s Equation Summary tions often lead to the wave equation (2). For purposes of discussion, a solution u(x, t) of (2) will represent the displacement of an idealized string. Finally, a solution Heat Transfer within a Thin Rod:u(x, y) of Heat Laplace’s Equation equation (3) can be interpreted as the steady-state (that is, time- independent) temperature distribution throughout a thin, two-dimensional plate. AssumptionsEven: though we have to make many simplifying assumptions, it is worthwhile toHeat see onlyhow flows equations in x-direction. such as (1) and (2) arise. Cross section of area A Suppose a thin circular rod of length L has a cross-sectional area A and coin- No heat escapes from the surface. cides with the x-axis on the interval [0, L]. See Figure 12.2.1. Let us suppose the following:No heat is generated in the rod. 0 x x + ∆x L x Rod is homogeneous with density ρ. FIGURE 12.2.1 One-dimensional • The flow of heat within the rod takes place only in the x-direction. flow of Letheatu(x; t) denote the temperature of the rod• atThe location lateral,x ator timecurved,t. surface of the rod is insulated; that is, no heat Consider the quantity of heat with [x; x + dx]:(escapesγ : 比熱 from) this surface. • No heat is being generated within the rod. ) ) dQ = γ (ρAdx) u = Qx = γρ•AuThe= rodQ xtis =homogeneous;γρAut that is, its mass per unit volume r is a constant. Heat transfer rate through the cross section•= The−KAu specificx, and henceheat g theand net thermal heat conductivity K of the material of the rod are rate inside [x; x + dx] is dQt = −KAux(x; t) − (constants.−KAux(x + dx; t)) dQt = KAu (ux(x + dx; t) −Toux( xderive; t)) = theKAu partialxxdx differential equation satisfied by the temperature u(x, t), we =) Qtx = KAuxxneed two empirical laws of heat conduction: (i) The quantity of heat Q in an element of mass m is 11 / 50 王奕翔 DE Lecture 14 Q ϭ ␥mu,(4) where u is the temperature of the element. (ii) The rate of heat flow Qt through the cross-section indicated in Figure 12.2.1 is proportional to the area A of the cross section and the partial derivative with respect to x of the temperature: Qt ϭϪKAux.(5) Since heat flows in the direction of decreasing temperature, the minus sign in (5) is used to ensure that Qt is positive for ux Ͻ 0 (heat flow to the right) and negative for ux Ͼ 0 (heat flow to the left). If the circular slice of the rod shown in Figure 12.2.1 between x and x ϩ⌬x is very thin, then u(x, t) can be taken as the approximate tem- perature at each point in the interval. Now the mass of the slice is m ϭ r(A ⌬x), and so it follows from (4) that the quantity of heat in it is Q ϭ ␥␳A ⌬x u.(6) Furthermore, when heat flows in the positive x-direction, we see from (5) that heat builds up in the slice at the net rate ϪKAux(x, t) Ϫ [ϪKAux(x ϩ⌬x, t)] ϭ KA [ux(x ϩ⌬x, t) Ϫ ux(x, t)].(7) By differentiating (6) with respect to t, we see that this net rate is also given by Qt ϭ ␥␳A ⌬x ut.(8) Equating (7) and (8) gives K u (x ϩ⌬x, t) Ϫ u (x, t) x x ϭ u .(9) ␥␳ ⌬x t 92467_12_ch12_p455-492.qxd 2/16/12 11:41 AM Page 461 12.2 CLASSICAL PDEs AND BOUNDARY-VALUE PROBLEMS G 461 variables.
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