Chapter 12: Boundary-Value Problems in Rectangular Coordinates

Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary

王奕翔

Department of Electrical Engineering National Taiwan University

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 ≠ 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

∑k 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 ∑∞ 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 ′Y, u = X = XY ′ x dx y dy d2X d2Y u = Y = X ′′Y, u = X = XY ′′, u = X ′Y ′ xx dx2 yy dy2 xy

Note: Derivatives are with respect to diﬀerent independent variables. dX For example, X ′ := . 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 ′′Y + (x + 1)XY ′ + (x + 1)yXY = 0 =⇒ x2X ′′Y = −(x + 1)X(Y ′ + yY) x2X ′′ Y ′ =⇒ = − − 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 ′′Y + (x + 1)XY ′ + (x + 1)yXY = 0 =⇒ x2X ′′Y = −(x + 1)X(Y ′ + yY) x2X ′′ Y ′ =⇒ = − − y = λ separation constant {(x + 1)X Y x2X ′′(x) − λ(x + 1)X(x) = 0 =⇒ Y ′(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 ≠ 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 temperature 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

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 Hence, escapes from this surface. { • No heat (is being) generated within the rod. Qxt = γρAut • The rod is Khomogeneous; that is, its mass per unit volume r is a constant. =⇒ γρ Aut = K Auxx =⇒ uxx = ut Qtx = KAuxx • The specificγρ heat g and thermal conductivity K of the material of the rod are constants. =⇒ kuxx = ut, k > 0 To derive the partial differential equation satisfied by the temperature u(x, t), we need two empirical laws of heat conduction: (i) The quantity of heat Q in an element of mass m is 12 / 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:

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

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 Equation: Initial and Boundaryu(x, y) of Laplace’s Conditions equation (3) can be interpreted as the steady-state (that is, time- independent) temperature distribution throughout a thin, two-dimensional plate. Initial ConditionEven though: we have to make many simplifying assumptions, it is worthwhile Providesto see the how spatial equations distribution such of as the (1) and (2) arise. Cross section of area A temperatureSuppose at time a tthin= 0 .circular rod of length L has a cross-sectional area A and coin- cides with the x-axis on the interval [0, L]. See Figure 12.2.1. Let us suppose the u(x, 0) = f(x), 0 < x < L following: 0 x x + ∆x L x FIGUREBoundary 12.2.1 ConditionsOne-dimensional: • The flow of heat within the rod takes place only in the x-direction. flow of Atheat the end points x = 0 and x = L, give the• constraintsThe lateral, on or curved, surface of the rod is insulated; that is, no heat escapes from this surface. u: (Dirchlet condition), for example, (u0: constant) • No heat is being generated within the rod. u(L, t) = u Temperature at the right end is held at constant. 0 • The rod is homogeneous; that is, its mass per unit volume r is a constant. ux: (Neumann condition), for example,• The specific heat g and thermal conductivity K of the material of the rod are

ux(L, t) = 0 The right endconstants. is insulated.

ux + hu: (Robin condition), for example,To (deriveh > 0, uthem: partial constants) differential equation satisﬁed by the temperature u(x, t), we

ux(L, t) = −h {u(L, t) − um}needHeat two is empirical lost from thelaws right of heat end. conduction: (i) The quantity of heat Q in an element of mass m is 13 / 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:

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

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 Equation: Boundary-Valueu(x, y Problems) of Laplace’s equation (3) can be interpreted as the steady-state (that is, time- independent) temperature distribution throughout a thin, two-dimensional plate. Even though we have to make many simplifying assumptions, it is worthwhile toA see problem how involvingequations both such initial as (1) and and (2) arise. Cross section of area A boundarySuppose conditions a thin iscircular called a rod of length L has a cross-sectional area A and coin- cidesboundary-value with the xproblem-axis on the interval [0, L]. See Figure 12.2.1. Let us suppose the At the two boundaries x − 0 and x = L, x following: 0 x x + ∆x L one can use different kinds of conditions. FIGURE 12.2.1 One-dimensional • The flow of heat within the rod takes place only in the x-direction. Examples: flow of heat • The lateral, or curved, surface of the rod is insulated; that is, no heat escapes from this surface. ku = u , 0 < x < L, t > 0 Heat xx t • No heat is being generatedequation within the rod. u(0, t) = u , u (L, t) = −h {u(L, t) − u } , t > 0 Boundary 0 x • Them rod is homogeneous;condition that is, its mass per unit volume r is a constant. u(x, 0) = f(x), 0 < x < L Initial • The specific heat conditiong and thermal conductivity K of the material of the rod are constants. Heat kuxx = ut, 0 < x < L, t > 0 equation To derive the partialBoundary differential equation satisfied by the temperature u(x, t), we ux(0, t) = 0, u(L, t) = 0, t > 0 condition need two empirical Initiallaws of heat conduction: u(x, 0) = f(x), 0 < x < L condition (i) The quantity of heat Q in an element of mass m is 14 / 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:

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 462

Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Dynamics462 G CHAPTER of 12 BOUNDARY-VALUE a String PROBLEMS Fixed IN RECTANGULAR at Two COORDINATES Ends: Wave Equation Finally, by taking the limit of (9) as ⌬x : 0, we obtain (1) in the form* (K͞gr)uxx ϭ ut. It is customary to let k ϭ K͞gr and call this positive constant the thermalAssumptions diffusivity. :

u Wave No Equation externalConsider force. a string of length L, such as a guitar string, stretched taut between two points on the x-axis—say, x ϭ 0 and x ϭ L. When the string startsTension to vibrate, forceassume that is largethe motion compared takes place toin the gravity xu-plane and in such is ∆s u(x, t) a manner thatthe each same point on at the all string points. moves in a direction perpendicular to the x-axis (transverse vibrations). As is shown in Figure 12.2.2(a), let u(x, t) denote the vertical displacementSlope of any ofpoint the on the curve string ismeasured very from small the atx-axis all for points. t Ͼ 0. We fur- ther assume the following: 0 x x + ∆xLx ≪ • TheVertical string is perfectly displacement ﬂexible. string length. (a) Segment of string • The string is homogeneous; that is, its mass per unit length r is a constant.String has mass per unit length ρ. u T2 • The displacements u are small in comparison to the length of the string. • The slope of the curve is small at all points. θ 2 Let u(x, t) denote the vertical position (displacement) ∆s • The tension T acts tangent to the string, and its magnitude T is the same at of theall points. string at location x at time t.

θ1 • The tension is large compared with the force of gravity. Consider• No other external the string forces act in on[x the, x string.+ dx]. Net vertical force is T1 x x x + ∆x Now in Figure 12.2.2(b) the tensions T1 and T2 are tangent to the ends of the curve on Tthe( intervalsin θ [x−, x ϩ⌬sinxθ]. For) ≈ smallT u(tanand uθ the− nettan verticalθ ) force acting on (b) Enlargement of segment 2 1 1 22 1 the corresponding element ⌬s of the string is then FIGURE 12.2.2 Flexible string = T {ux(x + dx, t) − ux(x, t)} anchored at x ϭ 0 and x ϭ L T sin ␪2 Ϫ T sin ␪1 Ϸ T tan ␪2 Ϫ T tan ␪1 = Tuxxdx † ϭ T [ux(x ϩ⌬x, t) Ϫ ux(x, t)], 15 / 50 王奕翔 DE Lecture 14 where T ϭ ͉T1 ͉ ϭ ͉T2 ͉. Now r ⌬s Ϸ r ⌬x is the mass of the string on [x, x ϩ⌬x], so Newton’s second law gives

T[ux(x ϩ⌬x, t) Ϫ ux(x, t)] ϭ ␳ ⌬x utt u (x ϩ⌬x, t) Ϫ u (x, t) ␳ or x x ϭ u . ⌬x T tt

If the limit is taken as ⌬x : 0, the last equation becomes uxx ϭ (r͞T)utt. This of Temperature as a course is (2) with a2 ϭ T͞r. function of position Thermometer on the hot plate Laplace’s Equation Although we shall not present its derivation, Laplace’s 220

200 180 equation in two and three dimensions occurs in time-independent problems involv- y 160 140

120 ing potentials such as electrostatic, gravitational, and velocity in ﬂuid mechanics.

100

60 Moreover, a solution of Laplace’s equation can also be interpreted as a steady-state

40 20 temperature distribution. As illustrated in Figure 12.2.3, a solution u(x, y) of (3) 0

–20 ?F could represent the temperature that varies from point to point—but not with time— (x, y) of a rectangular plate. Laplace’s equation in two dimensions and in three dimensions H 2 x W is abbreviated as ٌ u ϭ 0, where O Ѩ2u Ѩ2u Ѩ2u Ѩ2u Ѩ2u 2u ϭ ϩ and ٌ2u ϭ ϩ ϩٌ Ѩx2 Ѩy2 Ѩx2 Ѩy2 Ѩz2

FIGURE 12.2.3 Steady-state are called the two-dimensional Laplacian and the three-dimensional Laplacian, temperatures in a rectangular plate respectively, of a function u.

ux(x ϩ⌬x, t) Ϫ ux(x, t) *The deﬁnition of the second partial derivative is uxx ϭ lim . ⌬x : 0 ⌬x † tan u2 ϭ ux(x ϩ⌬x, t) and tan u1 ϭ ux(x, t) are equivalent expressions for slope. 92467_12_ch12_p455-492.qxd 2/16/12 11:41 AM Page 462

θ1 • The tension is large compared with the force of gravity. Since• No other the external slope forces is small, act on the the string. mass ≈ ρdx. Hence T1 x x x + ∆x Now in Figure 12.2.2(b) the tensions T1 and T2 are tangent to the ends of the curve on the interval [x, x ϩ⌬x]. For small u and u theT net vertical force acting on (b) Enlargement of segment Tu dx = (ρdx) u 1 =⇒2 u = u the corresponding elementxx ⌬s of the stringtt is then ρ xx tt FIGURE 12.2.2 Flexible string anchored at x ϭ 0 and x ϭ L T sin ␪ Ϫ T sin ␪ Ϸ T tan ␪ Ϫ T tan ␪ 2 1 2=⇒ a12u = u xx † tt ϭ T [ux(x ϩ⌬x, t) Ϫ ux(x, t)], 16 / 50 王奕翔 DE Lecture 14 where T ϭ ͉T1 ͉ ϭ ͉T2 ͉. Now r ⌬s Ϸ r ⌬x is the mass of the string on [x, x ϩ⌬x], so Newton’s second law gives

T[ux(x ϩ⌬x, t) Ϫ ux(x, t)] ϭ ␳ ⌬x utt u (x ϩ⌬x, t) Ϫ u (x, t) ␳ or x x ϭ u . ⌬x T tt

200 180 equation in two and three dimensions occurs in time-independent problems involv- y 160 140

120 ing potentials such as electrostatic, gravitational, and velocity in ﬂuid mechanics.

100

60 Moreover, a solution of Laplace’s equation can also be interpreted as a steady-state

40 20 temperature distribution. As illustrated in Figure 12.2.3, a solution u(x, y) of (3) 0

FIGURE 12.2.3 Steady-state are called the two-dimensional Laplacian and the three-dimensional Laplacian, temperatures in a rectangular plate respectively, of a function u.

Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Wave462 G Equation:CHAPTER 12 BOUNDARY-VALUE Initial PROBLEMS and IN RECTANGULAR Boundary COORDINATES Conditions Finally, by taking the limit of (9) as ⌬x : 0, we obtain (1) in the form* (K͞gr)uxx ϭ ut. It is customary to let k ϭ K͞gr and call this positive constant the thermalInitial diffusivity. Conditions: Provide the initial displacement u and velocity ut at u Wave Equation Consider a string of length L, such as a guitar string, stretchedtime tautt = between 0. two points on the x-axis—say, x ϭ 0 and x ϭ L. When the string starts to vibrate, assume that the motion takes place in the xu-plane in such ∆s ( ) a manner that each point on the string moves in a direction perpendicular to the x-axis u x, t u(x, 0) = f(x), ut(x, 0) = g(x), 0 < x < L (transverse vibrations). As is shown in Figure 12.2.2(a), let u(x, t) denote the vertical displacement of any point on the string measured from the x-axis for t Ͼ 0. We fur- ther assume the following: 0 x x + ∆xLx Boundary• The string is Conditions perfectly ﬂexible.: (a) Segment of string At• The the string end is pointshomogeneous;x = that 0 and is, itsx mass= Lper, unit give length the r constraintsis a constant. u on u, ux, or ux + hu. Usually in the scenario of strings, T2 • The displacements u are small in comparison to the length of the string. the• The boundary slope of the conditionscurve is small at areall points. θ 2 ∆s • The tension T acts tangent to the string, and its magnitude T is the same at u(0all, points.t) = 0, u(0, L) = 0, t > 0 Both ends are ﬁxed. θ1 • The tension is large compared with the force of gravity. • No other external forces act on the string. T1 x x x + ∆x Now in Figure 12.2.2(b) the tensions T1 and T2 are tangent to the ends of the curve on the interval [x, x ϩ⌬x]. For small u and u the net vertical force acting on (b) Enlargement of segment ux(0, t) = 0, ux(0, L) = 0, 1t > 02 Free-ends condition the corresponding element ⌬s of the string is then FIGURE 12.2.2 Flexible string anchored at x ϭ 0 and x ϭ L T sin ␪2 Ϫ T sin ␪1 Ϸ T tan ␪2 Ϫ T tan ␪1 † ϭ T [ux(x ϩ⌬x, t) Ϫ ux(x, t)], 17 / 50 王奕翔 DE Lecture 14 where T ϭ ͉T1 ͉ ϭ ͉T2 ͉. Now r ⌬s Ϸ r ⌬x is the mass of the string on [x, x ϩ⌬x], so Newton’s second law gives

T[ux(x ϩ⌬x, t) Ϫ ux(x, t)] ϭ ␳ ⌬x utt u (x ϩ⌬x, t) Ϫ u (x, t) ␳ or x x ϭ u . ⌬x T tt

200 180 equation in two and three dimensions occurs in time-independent problems involv- y 160 140

120 ing potentials such as electrostatic, gravitational, and velocity in ﬂuid mechanics.

100

60 Moreover, a solution of Laplace’s equation can also be interpreted as a steady-state

40 20 temperature distribution. As illustrated in Figure 12.2.3, a solution u(x, y) of (3) 0

FIGURE 12.2.3 Steady-state are called the two-dimensional Laplacian and the three-dimensional Laplacian, temperatures in a rectangular plate respectively, of a function u.

Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Wave462 G Equation:CHAPTER 12 BOUNDARY-VALUE Boundary-Value PROBLEMS IN RECTANGULAR Problems COORDINATES Finally, by taking the limit of (9) as ⌬x : 0, we obtain (1) in the form* (K͞gr)uxx ϭ ut. It is customary to let k ϭ K͞gr and call this positive constant the thermalExamples diffusivity.:

u BothWave ends Equation areConsider ﬁxed: a string of length L, such as a guitar string, stretched taut between two points on the x-axis—say, x ϭ 0 and x ϭ L. When the string starts2 to vibrate, assume that the motion takes place in the xu-plane Wavein such ∆s a uxx = utt, 0 < x < L, t > 0 u(x, t) a manner that each point on the string moves in a direction perpendicular to theequation x-axis (transverseu(0, vibrations).t) = 0, Asu is( shownL, t) in = Figure 0, 12.2.2(a),t > 0 let u(x, t) denote the Boundaryvertical displacement of any point on the string measured from the x-axis for t Ͼ 0. conditionWe fur- Initial ther uassume(x, 0) the = following:f(x), ut(x, 0) = g(x), 0 < x < L 0 x x + ∆xLx condition • The string is perfectly ﬂexible. (a) Segment of string • The string is homogeneous; that is, its mass per unit length r is a constant. u T2 Free• The Ends: displacements u are small in comparison to the length of the string. • The slope of the curve is small at all points. θ 2 ∆s • aThe2u tension= uT acts, tangent0 < tox the< string,L, andt > its 0magnitude T is the sameWave at all points.xx tt equation • The tension is large compared with the force of gravity. Boundary θ1 ux(0, t) = 0, ux(L, t) = 0, t > 0 • No other external forces act on the string. condition T1 Initial x u(x, 0) = f(x), ut(x, 0) = g(x), 0 < x < L x x + ∆x Now in Figure 12.2.2(b) the tensions T1 and T2 are tangent to the endscondition of the curve on the interval [x, x ϩ⌬x]. For small u and u the net vertical force acting on (b) Enlargement of segment 1 2 the corresponding element ⌬s of the string is then FIGURE 12.2.2 Flexible string anchored at x ϭ 0 and x ϭ L T sin ␪2 Ϫ T sin ␪1 Ϸ T tan ␪2 Ϫ T tan ␪1 † ϭ T [ux(x ϩ⌬x, t) Ϫ ux(x, t)], 18 / 50 王奕翔 DE Lecture 14 where T ϭ ͉T1 ͉ ϭ ͉T2 ͉. Now r ⌬s Ϸ r ⌬x is the mass of the string on [x, x ϩ⌬x], so Newton’s second law gives

T[ux(x ϩ⌬x, t) Ϫ ux(x, t)] ϭ ␳ ⌬x utt u (x ϩ⌬x, t) Ϫ u (x, t) ␳ or x x ϭ u . ⌬x T tt

200 180 equation in two and three dimensions occurs in time-independent problems involv- y 160 140

120 ing potentials such as electrostatic, gravitational, and velocity in ﬂuid mechanics.

100

60 Moreover, a solution of Laplace’s equation can also be interpreted as a steady-state

40 20 temperature distribution. As illustrated in Figure 12.2.3, a solution u(x, y) of (3) 0

FIGURE 12.2.3 Steady-state are called the two-dimensional Laplacian and the three-dimensional Laplacian, temperatures in a rectangular plate respectively, of a function u.

462 G CHAPTER 12 BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES

Finally, by taking the limit of (9) as ⌬x : 0, we obtain (1) in the form* (K͞gr)uxx ϭ ut. It is customary to let k ϭ K͞gr and call this positive constant the thermal diffusivity.

u Wave Equation Consider a string of length L, such as a guitar string, stretched taut between two points on the x-axis—say, x ϭ 0 and x ϭ L. When the string starts to vibrate, assume that the motion takes place in the xu-plane in such ∆s u(x, t) a manner that each point on the string moves in a direction perpendicular to the x-axis (transverse vibrations). As is shown in Figure 12.2.2(a), let u(x, t) denote the vertical displacement of any point on the string measured from the x-axis for t Ͼ 0. We fur- ther assume the following: 0 x x + ∆xLx • The string is perfectly ﬂexible. (a) Segment of string • The string is homogeneous; that is, its mass per unit length r is a constant. u T2 • The displacements u are small in comparison to the length of the string. • The slope of the curve is small at all points. θ 2 ∆s • The tension T acts tangent to the string, and its magnitude T is the same at all points.

θ1 • The tension is large compared with the force of gravity. • No other external forces act on the string. T1 x x x + ∆x Now in Figure 12.2.2(b) the tensions T1 and T2 are tangent to the ends of the curve on the interval [x, x ϩ⌬x]. For small u and u the net vertical force acting on (b) Enlargement of segment 1 2 the corresponding element ⌬s of the string is then FIGURE 12.2.2 Flexible string anchored at x ϭ 0 and x ϭ L T sin ␪2 Ϫ T sin ␪1 Ϸ T tan ␪2 Ϫ T tan ␪1 † ϭ T [ux(x ϩ⌬x, t) Ϫ ux(x, t)], Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equationwhere T ϭ ͉T1 ͉ ϭ ͉T2 ͉. Now r ⌬s Ϸ r ⌬x is the mass of the string on [x, x ϩ⌬x], Summaryso Newton’s second law gives

Laplace’s Equation T[ux(x ϩ⌬x, t) Ϫ ux(x, t)] ϭ ␳ ⌬x utt u (x ϩ⌬x, t) Ϫ u (x, t) ␳ or x x ϭ u . ⌬x T tt Laplace’s equation usually occurs in If the time-independentlimit is taken as ⌬xproblems: 0, the last involving equation becomes uxx ϭ (r͞T)utt. This of Temperature as a course is (2) with a2 ϭ T͞r. function of position Thermometer potentials. on the hot plate Laplace’s Equation Although we shall not present its derivation, Laplace’s 220 Its solution can also be interpreted as a 200 180 equation in two and three dimensions occurs in time-independent problems involv- y 160 steady-state temperature distribution. 140

120 ing potentials such as electrostatic, gravitational, and velocity in ﬂuid mechanics.

100

60 Moreover, a solution of Laplace’s equation can also be interpreted as a steady-state

40 20 temperatureTwo-dimensional distribution. LaplaceAs illustrated Equation in Figure 12.2.3, a solution u(x, y) of (3) 0

–20 ?F could represent the temperature that varies from point to point—but not with time— (x, y) 2 of a rectangular plate.∇ u Laplace’s:= uxx + equationuyy = 0in two dimensions and in three dimensions H 2 x W is abbreviated as ٌ u ϭ 0, where O Three-dimensional Laplace Equation Ѩ2u Ѩ2u Ѩ2u Ѩ2u Ѩ2u 2u ϭ ϩ and ٌ2u ϭ ϩ ϩٌ 2 2 2 2 2 2 ∇ u :=Ѩxuxx +Ѩyuyy + uzz = 0 Ѩx Ѩy Ѩz FIGURE 12.2.3 Steady-state are called the two-dimensional Laplacian and the three-dimensional Laplacian, temperatures in a rectangular plate respectively, of a function u.

19 / 50 王奕翔 DE Lecture 14

462 G CHAPTER 12 BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES

Finally, by taking the limit of (9) as ⌬x : 0, we obtain (1) in the form* (K͞gr)uxx ϭ ut. It is customary to let k ϭ K͞gr and call this positive constant the thermal diffusivity.

Laplace’s Equation: Boundary Conditions T[ux(x ϩ⌬x, t) Ϫ ux(x, t)] ϭ ␳ ⌬x utt u (x ϩ⌬x, t) Ϫ u (x, t) ␳ or x x ϭ u . Boundary Conditions: ⌬x T tt In the x-direction, at the end points x = 0 and If the limit is taken as ⌬x : 0, the last equation becomes uxx ϭ (r͞T)utt. This of x = a, give the constraints on u, ux, or ux + hu. Temperature as a course is (2) with a2 ϭ T͞r. function of position Thermometer on the hot plate In theLaplace’sy-direction, Equation at theAlthough end points we yshall= 0 notand present its derivation, Laplace’s 220 200 y = b, give the constraints on u, uy, or uy + hu. 180 equation in two and three dimensions occurs in time-independent problems involv- y 160 140

120 ing potentials such as electrostatic, gravitational, and velocity in ﬂuid mechanics.

100

80 60 Moreover,Examples a: solution of Laplace’s equation can also be interpreted as a steady-state 40 20 temperature distribution. As illustrated in Figure 12.2.3, a solution u(x, y) of (3) 0 –20 Both ends in x are insulated ?F could represent the temperature that varies from point to point—but not with time— (x, y) of a rectangular plate. Laplace’s equation in two dimensions and in three dimensions H ux(02, y) = 0, ux(a, y) = 0 x W is abbreviated as ٌ u ϭ 0, where O TemperaturesѨ of2u twoѨ2 endsu in y are held atѨ2u Ѩ2u Ѩ2u 2u ϭ ϩ and ٌ2u ϭ ϩ ϩٌ diﬀerent distributionsѨx2 Ѩy2 Ѩx2 Ѩy2 Ѩz2

FIGURE 12.2.3 Steady-state are called theu( x two-dimensional, 0) = f(x), u (Laplacianx, b) = g(xand) the three-dimensional Laplacian, temperatures in a rectangular plate respectively, of a function u.

20 / 50 王奕翔 DE Lecture 14

ux(x ϩ⌬x, t) Ϫ ux(x, t) *The deﬁnition of the second partial derivative is uxx ϭ lim . ⌬x : 0 ⌬x † tan u2 ϭ ux(x ϩ⌬x, t) and tan u1 ϭ ux(x, t) are equivalent expressions for slope. Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Laplace’s Equation: Boundary-Value Problems

Example:

Laplace’s uxx + uyy = 0, 0 < x < a, 0 < y < b equation Boundary ux(0, y) = 0, ux(a, y) = 0, 0 < y < b condition Boundary u(x, 0) = f(x), u(x, b) = g(x), 0 < x < a condition

21 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Modiﬁcations of Heat and Wave Equations

In the derivation of the heat equation and the wave equation, we assume that there is no internal or external inﬂuences. For example, no heat escapes from the surface, no heat is generated in the rod, no external force act on the string, etc. Taking external and internal inﬂuences into account, more general forms of the heat equation and the wave equation are the following:

kuxx + G(x, t, u, ux) = ut Heat Equation 2 a uxx + F(x, t, u, ut) = utt Wave Equation

Example:

− − heat transfers from the surface to an environment kuxx h(u um) = ut with constant temperature um 2 a uxx + f(x, t) = utt External force f acts on the string

22 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Homogeneous vs. Nonhomogeneous Boundary Conditions

Homogeneous Boundary Condition:

ux(0, y) = 0, ux(a, y) = 0, u(x, 0) = 0, u(0, L) = 0

Nonhomogeneous Boundary Condition:

ux(0, y) = f(y), ux(a, y) = g(y), u(x, L) = um

Typically, when using separation of variables, start with the independent variable associated with homogeneous boundary conditions, to determine the value of the separation constant.

23 / 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

24 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Wave Equation: a Boundary-Value Problem

Solve u(x, t): auxx = utt, 0 < x < L, t > 0 subject to : u(0, t) = 0, u(L, t) = 0, t > 0

u(x, 0) = f(x), ut(x, 0) = g(x), 0 < x < L

We focus on solving the above BVP (both ends are ﬁxed). Step 1: Separation of variables: Assume that the solution u(x, t) = X(x)T(t), X, T ≠ 0. Then,

′′ ′′ 2 2 ′′ ′′ X T a uxx = utt =⇒ a X T = XT =⇒ = = −λ { X a2T X ′′ + λX = 0 =⇒ T ′′ + a2λT = 0

The 2 homogeneous boundary conditions become X(0) = X(L) = 0.

25 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Solve in the x-Dimension and Find λ

Solve : X ′′ + λX = 0, T ′′ + a2λT = 0 subject to : X(0) = 0, X(L) = 0 X(x)T(0) = f(x), X(x)T ′(0) = g(x), 0 < x < L

Step 2: λ remains to be determined. What values should λ take?

1 λ = 0: X(x) = c1 + c2x. X(0) = X(L) = 0 =⇒ c1 = c2 = 0.

2 −αx αx 2 λ = −α < 0: X(x) = c1e + c2e .

Plug in X(0) = X(L) = 0, we get c1 = c2 = 0.

2 3 λ = α > 0: X(x) = c1 cos(αx) + c2 sin(αx).

Plug in X(0) = X(L) = 0, we get c1 = 0, and c2 sin(αL) = 0. Hence, c2 ≠ 0 only if αL = nπ. n2π2 Since X ≠ 0, pick λ = , n = 1, 2,... =⇒ X(x) = c sin nπ x. L2 2 L

26 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Solve in t-Dimension and Superposition

Solve : X ′′ + λX = 0, T ′′ + a2λT = 0 subject to : X(0) = 0, X(L) = 0 X(x)T(0) = f(x), X(x)T ′(0) = g(x), 0 < x < L

n2π2 Step 3: Once we ﬁx λ = L2 , n = 1, 2,..., we obtain ( ) ( ) ( ) nπ nπa nπa X(x) = c2 sin x , T(t) = c3 cos t + c4 sin t { L( ) ( )}L ( ) L nπa nπa nπ =⇒ u (x, t) = A cos t + B sin t sin x , n n L n L L

(An := c2c3, Bn := c2c4) ∑∞ =⇒ u(x, t) := un(x, t) is a solution, by the superposition principle. n=1

27 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Plug in Initial Condition, Revoke Fourier Series, and Done

Solve u(x, t): auxx = utt, 0 < x < L, t > 0 subject to : u(0, t) = 0, u(L, t) = 0, t > 0

u(x, 0) = f(x), ut(x, 0) = g(x), 0 < x < L

Step 4: Plug in the initial conditions and ﬁnd {An, Bn | n = 1, 2,...}. ∞ { ( ) ( )} ( ) ∑ nπa nπa nπ u(x, 0) = f(x), u(x, t) = A cos t + B sin t sin x n L n L L n=1 ∞ ( ) ∑ nπ =⇒ f(x) = A sin x , 0 < x < L n L n=1 From the Fourier sine series expansion on (0, L), we get ∫ ( ) 2 L nπ An = f(x) sin x dx. L 0 L

28 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Plug in Initial Condition, Revoke Fourier Series, and Done

Solve u(x, t): auxx = utt, 0 < x < L, t > 0 subject to : u(0, t) = 0, u(L, t) = 0, t > 0

u(x, 0) = f(x), ut(x, 0) = g(x), 0 < x < L

Step 4: Plug in the initial conditions and ﬁnd {An, Bn | n = 1, 2,...}. ∞ { ( ) ( )} ( ) ∑ nπa nπa nπ u (x, 0) = g(x), u(x, t) = A cos t + B sin t sin x t n L n L L n=1 ∞ ( ) ∑ nπa nπ =⇒ g(x) = B sin x , 0 < x < L n L L n=1 From the Fourier sine series expansion on (0, L), we get ∫ ( ) ∫ ( ) nπa 2 L nπ 2 L nπ Bn = g(x) sin x dx. =⇒ Bn = g(x) sin x dx. L L 0 L nπa 0 L

29 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Final Solution

Solve u(x, t): auxx = utt, 0 < x < L, t > 0 subject to : u(0, t) = 0, u(L, t) = 0, t > 0

u(x, 0) = f(x), ut(x, 0) = g(x), 0 < x < L

Step 5: The ﬁnal solution is ∞ { ( ) ( )} ( ) ∑ nπa nπa nπ u(x, t) = A cos t + B sin t sin x n L n L L n=1 ∞ ( ) ( ) ∑ nπa nπ = C sin t + ϕ sin x n L n L n=1 ∫ ( ) ∫ ( ) 2 L nπ 2 L nπ An = f(x) sin x dx, Bn = g(x) sin x dx L 0 L nπa 0 L √ 2 2 An Bn Cn = An + Bn, sin ϕn = , cos ϕn = Cn Cn

30 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Standing Waves

The ﬁnal solution ∞ ( ) ( ) ∑ nπa nπ u(x, t) = C sin t + ϕ sin x n L n L n=1 is a linear combination of standing waves or normal modes ( ) ( ) nπa nπ u (x, t) = C sin t + ϕ sin x , n = 1, 2,... n n L n L

For a normal mode n, at a ﬁxed location( )x, the string moves with nπ time-varying amplitude Cn sin L x nπa/L na frequency fn := 2π = 2L πa/L a Fundamental Frequency: f1 := 2π = 2L

31 / 50 王奕翔 DE Lecture 14 aaaaaa.qxd 3/2/12 7:57 AM Page 471

Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary 12.4 WAVE EQUATION G 471

is called the first standing wave, the first normal mode, or the fundamental mode of vibration. The first three standing waves, or normal modes, are shown in 0 L x Figure 12.4.2. The dashed graphs represent the standing waves at various values of time. The points in the interval (0, L), for which sin(np͞L)x ϭ 0, correspond to (a) First standing wave points on a standing wave where there is no motion. These points are called nodes. For example, in Figures 12.4.2(b) and 12.4.2(c) we see that the second standing Node wave has one node at L͞2 and the third standing wave has two nodes at L͞3 and 2L͞3. In general, the nth normal mode of vibration has n Ϫ 1 nodes. The frequency 0 L L x 2 a 1 T f ϭ ϭ (b) Second standing wave 1 2L 2L ␳ B Nodes of the first normal mode is called the fundamental frequency or first harmonic and

32 / 50 王奕翔 DE Lecture 14 is directly related to the pitch produced by a stringed instrument. It is apparent that x the greater the tension on the string, the higher the pitch of the sound. The 0 L 2L L 3 3 frequencies fn of the other normal modes, which are integer multiples of the fundamental frequency, are called overtones. The second harmonic is the ﬁrst overtone, (c) Third standing wave and so on. FIGURE 12.4.2 First three standing waves

EXERCISES 12.4 Answers to selected odd-numbered problems begin on page ANS-22. In Problems 1–8 solve the wave equation (1) subject to the 4. u(0, t) ϭ 0, u(L, t) ϭ 0 given conditions. Ѩu u(x, 0) ϭ 0, ϭ x(L Ϫ x) ͉tϭ0 1. u(0, t) ϭ 0, u(L, t) ϭ 0 Ѩt 1 Ѩu u(x, 0) ϭ x(L Ϫ x), ϭ 0 5. u(0, t) ϭ 0, u(␲, t) ϭ 0 4 Ѩt ͉tϭ0 Ѩu u(x, 0) ϭ 0, ϭ sin x Ѩu Ѩu ͉tϭ0 2. ϭ 0, ϭ 0 Ѩt Ѩx ͉xϭ0 Ѩx ͉xϭL Ѩu 6. u(0, t) ϭ 0, u(␲, t) ϭ 0 u(x, 0) ϭ x, ϭ 0 Ѩu Ѩt ͉tϭ0 1 2 2 u(x, 0) ϭ 6 x(␲ Ϫ x ), ϭ 0 Ѩt ͉tϭ0 This problem could describe the longitudinal displacement u(x, t) of a vibrating elastic bar. The boundary 7. u(0, t) ϭ 0, u(L, t) ϭ 0 conditions at x ϭ 0 and x ϭ L are called free-end conditions. See Figure 12.4.3. 2hx L , 0 Ͻ x Ͻ u(x, t) L 2 Ѩu u(x, 0) ϭ , ϭ 0 x L Ѩt ͉tϭ0 2h 1 Ϫ , Յ x Ͻ L Ά ΂ L΃ 2 x 0 L 8. u(0, t) ϭ 0, u(1, t) ϭ 0 Ѩu FIGURE 12.4.3 Vibrating elastic bar in Problem 2 u(x, 0) ϭ 0.01 sin 3px, ϭ 0 Ѩt ͉tϭ0 3. u(0, t) ϭ 0, u(L, t) ϭ 0 Ѩu u(x, 0) given in Figure 12.4.4, ϭ 0 9. Astring is stretched and secured on the x-axis at x ϭ 0 Ѩt ͉tϭ0 and x ϭ p for t Ͼ 0. If the transverse vibrations take f(x) place in a medium that imparts a resistance proportional to the instantaneous velocity, then the wave equation 1 takes on the form

x Ѩ2u Ѩ2u Ѩu /3 2L/3 LL ϭ ϩ 2␤ , 0 Ͻ ␤ Ͻ 1, t Ͼ 0. Ѩx2 Ѩt2 Ѩt FIGURE 12.4.4 Initial displacement in Problem 3 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

33 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Laplace’s Equation: a Boundary-Value Problem

Solve u(x, y): uxx + uyy = 0, 0 < x < a, 0 < y < b

subject to : ux(0, y) = 0, ux(a, y) = 0, 0 < y < b u(x, 0) = 0, u(x, b) = f(x), 0 < x < a

We focus on solving the above BVP (both ends x = 0 and x = a are insulated).

Step 1: Separation of variables: Assume that the solution u(x, y) = X(x)Y(y), X, Y ≠ 0. Then,

′′ ′′ ′′ ′′ X Y uxx + uyy = 0 =⇒ X Y + XY = 0 =⇒ = − = −λ { X Y X ′′ + λX = 0 =⇒ Y ′′ − λY = 0

The 3 homogeneous boundary conditions become X ′(0) = X ′(a) = Y(0) = 0.

34 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Solve in the x-Dimension and Find λ

Solve : X ′′ + λX = 0, Y ′′ − λY = 0 subject to : X ′(0) = 0, X ′(a) = 0 Y(0) = 0, X(x)Y(b) = f(x), 0 < x < a

Step 2: λ remains to be determined. What values should λ take? ′ ′ 1 λ = 0: X(x) = c1 + c2x. X (0) = X (a) = 0 =⇒ c2 = 0.

2 −αx αx 2 λ = −α < 0: X(x) = c1e + c2e . ′ ′ Plug in X (0) = X (a) = 0, we get c1 = c2 = 0.

2 3 λ = α > 0: X(x) = c1 cos(αx) + c2 sin(αx). ′ ′ Plug in X (0) = X (a) = 0, we get c2 = 0, and c1α sin(αa) = 0. Hence, c1 ≠ 0 only if αa = nπ. n2π2 ( ) Since X ≠ 0, pick λ = , n = 0, 1, 2,... =⇒ X(x) = c cos nπ x . a2 1 a

35 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Solve in y-Dimension and Superposition

Solve : X ′′ + λX = 0, Y ′′ − λY = 0 subject to : X ′(0) = 0, X ′(a) = 0 Y(0) = 0, X(x)Y(b) = f(x), 0 < x < a ( ) n2π2 nπ Step 3: Once we ﬁx λ = a2 , n = 0, 1, 2,..., we obtain X(x) = c1 cos a x { c3 + c4y(, ) ( ) n = 0 ⇒ Y(y) = nπ nπ (Y(0) = 0 = c3 = 0) c3 cosh y + c4 sinh y , n ≥ 1 { a a A y, n = 0 =⇒ u (x, y) = 0 ( ) ( ) , (A := c c ) n nπ nπ ≥ n 1 4 An cos a x sinh a y , n 1 ∑∞ =⇒ u(x, y) := un(x, y) is a solution, by the superposition principle. n=0

36 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Plug in Initial Condition, Revoke Fourier Series, and Done

Solve u(x, y): uxx + uyy = 0, 0 < x < a, 0 < y < b

subject to : ux(0, y) = 0, ux(a, y) = 0, 0 < y < b u(x, 0) = 0, u(x, b) = f(x), 0 < x < a

Step 4: Plug in the initial conditions and ﬁnd {An | n = 1, 2,...}. ∞ ( ) ( ) ∑ nπ nπ u(x, b) = f(x), u(x, y) = A y + A cos x sinh y 0 n a a n=1 ∞ ( ) ( ) ∑ nπ nπ =⇒ f(x) = A b + A cos x sinh b , 0 < x < a 0 n a a n=1 From the Fourier cosine series expansion on (0, a), we get ∫ ( ) ∫ ( ) 2 a nπ 2 a nπ 2A0b = f(x) dx, An sinh b = f(x) cos x dx a 0 a a 0 a

37 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Final Solution

Solve u(x, y): uxx + uyy = 0, 0 < x < a, 0 < y < b

subject to : ux(0, y) = 0, ux(a, y) = 0, 0 < y < b u(x, 0) = 0, u(x, b) = f(x), 0 < x < a

Step 5: The ﬁnal solution is ∞ ( ) ( ) ∑ nπ nπ u(x, y) = A y + A cos x sinh y 0 n a a ∫ n=1 1 a A = f(x) dx 0 ab 0 ∫ ( ) 2 a nπ A = ( ) f(x) cos x dx, n ≥ 1 n nπ a a sinh a b 0

38 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Superposition Principle

(So far) Key steps in solving a boundary-value problem of a PDE using separation of variables: Identify for which “dimension” (independent variable) (in our previous example, x), the given conditions are all homogeneous. Translate these homogeneous conditions into conditions on the single-argument function X(x)). Solve the associated ODE (X ′′ + λX = 0) under these conditions, and ﬁnd the value of the separation constant λ that leads to non-trivial solutions.

Question: What if all dimensions contain some nonhomogeneous condition?

39 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary

Solve u(x, y): uxx + uyy = 0, 0 < x < a, 0 < y < b subject to : u(0, y) = F(y), u(a, y) = G(y), 0 < y < b u(x, 0) = f(x), u(x, b) = g(x), 0 < x < a

u(x, )=g(x) ·

u( ,y)=F (y) 2u =0 u( ,y)=G(y) · r ·

u(x, )=f(x) ·

40 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary

Solve u(x, y): uxx + uyy = 0, 0 < x < a, 0 < y < b subject to : u(0, y) = F(y), u(a, y) = G(y), 0 < y < b u(x, 0) = f(x), u(x, b) = g(x), 0 < x < a

The solution u(x, y) = u1(x, y) + u2(x, y), where u1, u2 are the solutions of the following 2 BVP’s respectively.

Solve u1(x, y): uxx + uyy = 0, 0 < x < a, 0 < y < b subject to : u(0, y) = 0, u(a, y) = 0, 0 < y < b u(x, 0) = f(x), u(x, b) = g(x), 0 < x < a

Solve u2(x, y): uxx + uyy = 0, 0 < x < a, 0 < y < b subject to : u(0, y) = F(y), u(a, y) = G(y), 0 < y < b u(x, 0) = 0, u(x, b) = 0, 0 < x < a

41 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Superposition Principle

u(x, )=g(x) ·

u( ,y)=0 2u =0 u( ,y)=0 · r 1 · u(x, )=g(x) · u(x, )=f(x) · u( ,y)=F (y) 2u =0 u( ,y)=G(y) = + · r · u(x, )=0 · u(x, )=f(x) · u( ,y)=F (y) 2u =0 u( ,y)=G(y) · r 2 ·

u(x, )=0 ·

42 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Superposition Principle

u (x, )=g(x) x ·

2 uy( ,y)=0 u =0 u( ,y)=0 · r 1 · u (x, )=g(x) x · u(x, )=f(x) · 2 uy( ,y)=F (y) u =0 u( ,y)=G(y) = + · r · u (x, )=0 x · u(x, )=f(x) · 2 uy( ,y)=F (y) u =0 u( ,y)=G(y) · r 2 ·

u(x, )=0 ·

43 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Semi-Finte Plate

Solve u(x, y): uxx + uyy = 0, 0 < x < a, y > 0 subject to : u(0, y) = 0, u(a, y) = 0, y > 0 u(x, 0) = f(x), |u(x, ∞)| < ∞, 0 < x < a

2u =0 u =0 r u =0

u = f(x) x

44 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Semi-Finte Plate

Solve u(x, y): uxx + uyy = 0, 0 < x < a, y > 0 subject to : u(0, y) = 0, u(a, y) = 0, y > 0 u(x, 0) = f(x), |u(x, ∞)| < ∞, 0 < x < a

Following the same steps as before (setting u(x, y) = X(x)Y(y)), we can convert the original problem into

Solve u(x, y): X ′′ + λX = 0, Y ′′ − λY = 0 subject to : X(0) = 0, X(a) = 0, y > 0 X(x)Y(0) = f(x), |Y(∞)| < ∞, 0 < x < a ( ) nπ Step 1: First we solve X(x) = c2 sin a x and ﬁnd the possible n2π2 λ = a2 , n = 1, 2,....

45 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Semi-Finte Plate

Solve : X ′′ + λX = 0, Y ′′ − λY = 0 subject to : X(0) = 0, X(a) = 0, y > 0 X(x)Y(0) = f(x), |Y(∞)| < ∞, 0 < x < a ( ) nπ n2π2 Step 1: First we solve X(x) = c2 sin a x (λ = a2 ), n = 1, 2,.... nπ y − nπ y Step 2: Next we solve Y(y) = c3e a + c4c3e a .

By the condition |Y(∞)| < ∞, we have c3 = 0. Hence, ∑∞ ( ) nπ − nπ y u(x, y) = A sin x e a . n a n=1

46 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Semi-Finte Plate

Solve u(x, y): uxx + uyy = 0, 0 < x < a, y > 0 subject to : u(0, y) = 0, u(a, y) = 0, y > 0 u(x, 0) = f(x), |u(x, ∞)| < ∞, 0 < x < a

Final Solution: ∑∞ ( ) nπ − nπ y u(x, y) = A sin x e a . n a n=1

By the condition u(x, 0) = f(x), 0 < x < a, we have

∞ ( ) ∫ ( ) ∑ nπ 2 a nπ f(x) = A sin x =⇒ A = f(x) sin x dx. n a n a a n=1 0

47 / 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

48 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Short Recap

Method of Separation of Variables: Convert PDE into two ODE’s

Solve the ODE with homogeneous boundary conditions ﬁrst, to determine the separation constant

Fourier Series to determine the undetermined coeﬃcients

Heat Equation, Wave Equation, Laplace’s Equation

Superposition Principle

49 / 50 王奕翔 DE Lecture 14 Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Self-Practice Exercises

12-1: 9, 15, 17, 22

12-2: 1, 3, 7, 11

12-4: 3, 7, 9, 11, 14

12-5: 5, 7, 12, 15, 19

50 / 50 王奕翔 DE Lecture 14