Chapter 12 Partial Differential Equations

Chapter 12 Partial DifferentialEquationsAdvanced Engineering MathematicsWei-Ta ChuNational Chung Cheng Universitywtchu@cs.ccu.edu.tw1

12.1 Basic Concepts of PDEs2

Partial Differential EquationA partial differential equation (PDE) is an equationinvolving one or more partial derivatives of an(unknown) function, call it u, that depends on two ormore variables, often time t and one or severalvariables in space.The order of the highest derivative is called the orderof the PDE. Just as was the case for ODEs, secondorder PDEs will be the most important ones inapplications.3

Partial Differential EquationWe say that a PDE is linear if it is of the first degree inthe unknown function u and its partial derivatives.We call a linear PDE homogeneous if each of its termscontains either u or one of its partial derivatives.4

Example 15(1)One-dimensional wave equation(2)One-dimensional heat equation(3)Two-dimensional Laplace equation(4)Two-dimensional Poisson equation(5)Two-dimensional wave equation(6)Three-dimensional Laplace equationHere c is a positive constant, t is time, x, y, z are Cartesian coordinates, anddimension is the number of these coordinates in the equation.

Partial Differential EquationA solution of a PDE in some region R of the space ofthe independent variables is a function that has all thepartial derivatives appearing in the PDE in somedomain D containing R, and satisfies the PDEeverywhere in R.In general, the totality of solutions of a PDE is verylarge. For example, the functions(7)which are entirely different from each other, aresolutions of (3).6

Partial Differential EquationWe shall see that the unique solution of a PDEcorresponding to a given physical problem will beobtained by the use of additional conditions arisingfrom the problem.For instance, this may be the condition that thesolution u assume given values on the boundary of theregion R (“boundary conditions”). Or, when time t isone of the variables, u (oror both) may beprescribed at t 0 (“initial conditions”).7

Theorem 1Fundamental Theorem on SuperpositionIf u1 and u2 are solutions of a homogeneous linear PDEin some region R, thenwith any constants c1 and c2 is also a solution of thatPDE in the region R.8

Example 2Find solutions u of the PDEdepending onx and y.Solution. Since no y-derivatives occur, we can solvethis PDE like. In Sec. 2.2, we would haveobtainedwith constant A and B. HereA and B may be functions of y, so that the answer iswith arbitrary functions A and B. We thus have a greatvariety of solutions. Check the result by differentiation.9

Example 3Find solutions u u(x,y) of the PDESolution. Setting, we have,,,and by integration withrespect to x,here, f(x) and g(y) are arbitrary.10

12.2 Modeling: Vibrating String,Wave Equation11

ModelingWe want to derive the PDE modeling small transversevibrations of an elastic string, such as a violin string.We place the string along the x-axis, stretch it to lengthL, and fasten it at the ends x 0 and x L. We thendistort the string, and at some instant, call it t 0, werelease it and allow it to vibrate. The problem is todetermine the vibrations of the string, that is, to find itsdeflection u(x,t) at any point x and at any time t 0; seeFig. 286.12

Modelingu(x,t) will be the solution of a PDE that is the model ofour physical system to be derived. This PDE shouldnot be too complicated, so that we can solve it.Reasonable simplifying assumptions are as follows.13

ModelingPhysical AssumptionsThe mass of the string per unit length is constant(“homogeneous string”). The string is perfectly elasticand does not offer any resistance to bending.The tension caused by stretching the string beforefastening it at the ends is so large that the action of thegravitational force on the string can be neglected.The string performs small transverse motions in a verticalplane; that is, every particle of the string moves strictlyvertically and so that the deflection and the slope at everypoint of the string always remain small in absolute value.14

Derivation of the PDE of the ModelWe consider the forces acting on a small portion of thestring (Fig. 286). This method is typical of modeling inmechanics and elsewhere.Since the string offers no resistance to bending, thetension is tangential to the curve of the string at eachpoint. Let T1 and T2 be the tension at the endpoints Pand Q of that portion. Since the points of the stringmove vertically, there is no motion in the horizontaldirection. Hence the horizontal components of thetension must be constant.15

Derivation of the PDE of the ModelUsing the notation shown in Fig. 286, we thus obtain(1)In the vertical direction we have two forces, namely,the vertical componentsandof T1and T2; here the minus sign appears because thecomponent that P is directed downward.16

Derivation of the PDE of the ModelBy Newton’s second law the resultant of these twoforces is equal to the massof the portion timesthe acceleration, evaluated at some pointbetween and; here is the mass of theundeflected string per unit length, andis the lengthof the portion of the undeflected string. ( is generallyused to denote small quantities)Hence17

Derivation of the PDE of the ModelUsing (1), we can divide this by(2)Noware slopes of the string atandHere we have to write partial derivatives because ualso depends on time t. Dividing (2) by, we thushave18

Derivation of the PDE of the ModelIf we letapproach zero, we obtain the linear PDEThis is called the one-dimensional wave equation.We see that it is homogeneous and of the second order.The physical constantis denoted by (instead ofc) to indicate that this constant is positive, a fact thatwill be essential to the form of the solutions. “Onedimensional” means that the equation involves onlyone space variable, x.19

12.3 Solution by SeparatingVariables. Use of Fourier Series20

One-Dimensional Wave EquationThe model of a vibrating elastic string consists of theone-dimensional wave equation(1)for the unknown deflection u(x,t) of the string, a PDEthat we have just obtained, and some additionalconditions, which we shall now derive.21

One-Dimensional Wave EquationSince the string is fastened at the ends x 0 and x L,we have the two boundary conditions(2)Furthermore, the form of the motion of the string willdepend on its initial deflection (deflection at time t 0),call it f(x), and on its initial velocity (velocity at t 0),call it g(x).22

One-Dimensional Wave EquationWe thus have two initial conditions(3)where. We now have to find a solution ofthe PDE (1) satisfying the conditions (2) and (3). Thiswill be the solution of our problem.(1)(2)23

One-Dimensional Wave EquationStep 1. By the “method of separating variables” orproduct method, setting u(x,t) F(x)G(t), we obtainfrom (1) two ODEs, one for F(x) and the other one forG(t).Step 2. We determine solutions of these ODEs thatsatisfy the boundary conditions (2).Step 3. Finally, using Fourier series, we compose thesolutions found in Step 2 to obtain a solution of (1)satisfying both (2) and (3), that is, the solution of ourmodel of the vibrating string.24

(1)Step 1In the method of separating variables, or productmethod, we determine solutions of the wave equation(1) of the form(4)which are a product of two functions, each dependingon only one of the variables x and t. This is a powerfulgeneral method that has various applications inengineering mathematics.Differentiating (4), we obtainwhere dots denote derivatives with respect to t.25

(1)Step 1By inserting this into the wave equation (1) we haveDividing byand simplifying givesThe variables are now separated, the left sidedepending only on t and the right side only on x.Hence both sides must be constant because, if theywere variable, then changing t or x would affect onlyone side, leaving the other unaltered. Thus, say26

Step 1Multiplying by the denominators gives immediatelytwo ordinary DEs(5)(6)Here, the separation constant k is still arbitrary.27

(5)Step 2We now determine solutions F and G of (5) and (6) sothat u FG satisfies the boundary conditions (2), that is(7)We first solve (5). Ifis of no interest. Hence, then, whichand then by (7),(8)We show that k must be negative. For k 0 the generalsolution of (5) is, and from (8) we obtaina b 0, so that F 0 and u FG 0, which is of nointerest.28

(5)Step 2For positivea general solution of (5) isand from (8) we obtain F 0 as before. Hence we areleft with the possibility of choosing k negative, say,. Then (5) becomesand has as ageneral solutionFrom this and (8) we have29

(8)Step 2(9)We must takeThussince otherwise F 0. Hence(n integer)Setting B 1, we thus obtain infinitely many solutions, where(10)(n 1, 2, )These solutions satisfy (8). [For negative integer n weobtain essentially the same solutions, except for aminus sign, because]30

(6)Step 2We now solve (6) withfrom (9), that is,resulting(11*)A general solution isHence solutions of (1) satisfying (2) are, written out(11)(n 1, 2, )(1)31(2)

Step 2(11)(n 1, 2, )These functions are called the eigenfunctions, orcharacteristic functions, and the valuesarecalled the eigenvalues, or characteristic values, of thevibrating string. The setis called thespectrum.32

(11)(n 1, 2, )Step 2Discussion of Eigenfunctions. We see that each unrepresents a harmonic motion having the frequencycycles per unit time. This motion iscalled the nth normal mode of the string. The firstnormal mode is known as the fundamental mode (n 1),and the others are known as overtones; musically theygive the octave, octave plus fifth, etc.Since in (11)atthe nth normal mode has n-1 nodes, that is, points ofthe string that do not move; see Fig. 287.33

Step 2Figure 288 shows the second normal mode for variousvalues of t. At any instant the string has the form of asine wave. When the left part of the string is movingdown, the other half is moving up, and conversely. Forthe other modes the situation is similar.34

Step 2Tuning is done by changing the tension T. Our formulafor the frequencyof un with[see (3), Sec. 12.2] confirms that effect because itshows that the frequency is proportional to the tension.T cannot be increased indefinitely, but can you seewhat to do to get a string with a high fundamentalmode? (Think of both L and .) Why is a violinsmaller than a double-bass?35

(11)(n 1, 2, )Step 3The eigenfunctions (11) satisfy the wave equation (1)and the boundary conditions (2) (string fixed at theends). A single un will generally not satisfy the initialconditions (3). But since the wave equation (1) islinear and homogeneous, it follows from FundamentalTheorem 1 in Sec. 12.1 that the sum of finitely manysolutions un is a solution of (1). To obtain a solutionthat also satisfies the initial conditions (3), we considerthe infinite series (withas before)(12)36

(3)Step 3Satisfying Initial Condition (3a) (Given InitialDisplacement). From (12) and (3a) we obtain(13)Hence we must choose the Bn’s so that u(x,0) becomesthe Fourier sine series of f(x). Then, by (4) in Sec.11.3,(14)37

(3)Step 3Satisfying Initial Conditions (3b) (Given InitialVelocity). Similarly, by differentiating (12) withrespect to t and using (3b), we obtainHence we must choose the ’s so that for t 0 thederivativebecomes the Fourier sine series ofg(x). Thus, again by (4) in Sec. 11.3,38

Step 3Since, we obtain by division(15)Result. Our discussion shows that u(x,t) given by (12)with coefficients (14) and (15) is a solution of (1) thatsatisfies all the conditions in (2) and (3), provided theseries (12) converges and so do the series obtained bydifferentiating (12) twice termwise with respect to xand t and have the sumsand,respectively, which are continuous.39

Step 3Solution (12) Established. According to ourderivation, the solution (12) is at first a purely formalexpression, but we shall now establish it. For the sakeof simplicity we consider only the case when the initialvelocity g(x) is identically zero. Then theare zero,and (12) reduces to(16)It is possible to sum this series, that is, to write theresult in a closed or finite form.40