v ( v2)3 2 Solution. Since no s-dens atis Cs occur. sse can sols e this PDE like i u (I. In Sec. it c siould hass obtained it

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1 v -(c.i + a v2). It CHAP. 2 Partial Differential Equations (PDEs) The simple proof of this important theorem is quite similar to that of Theorem in Sec. 2. and is left to the student. Verification of solutions in Probs. 23 proceeds as for ODEs. Problems 623 concern PDEs soh able like ODEs. To help the student with them, we consider to typical examples. Solving u,, u 0 Like an ODE Find solutions it of the PDE ii, u ) depending on x and v. Solution. Since no s-dens atis Cs occur. sse can sols e this PDE like i u (I. In Sec. it c siould hass obtained it Bc n tth constant and B. Here d and B mas be functions of \. so that the aitsil er is lilt. rt VII B vie ssith anbitrans functions. and B. We thus has e a great saneb of solutions. Check the result h differentiation. Solving u Like an ODE Find oluiion it u(s. Vt of this PDti. Solution. Setting o, /. ne have p, integration n ith respect to i. p. p p l. In p v. / rie and h Is. v. (vie vt is here / ( r ci il/v. here. ft i ) and (.r) are arbitrary. v ( 2. Fundamental theorem. Prose tt for second-order 3. (x -- PDEs in two and three independent saniables. Hi,tr.. TEAM PROJECT. Verification of Solutions Prove it by substitutton. (a) ave equation. Verify that it (x. t) V Ft + Ct) - ct) with any twice differentiable functions v an Verifiy (by substitution) that the gisen funclion is a solution se satisfies (I). of the PDE. Sketch ttr graph the solution as a surface in space. (b Poisson equation. Verify that each ii satisfies ( is ith.t(-s. v as indicated. 3 ave Equation () with suitable c 3 2.o ti VX f2v.i 2 i2 it uv t (.v v2)xv 3. u cost2x. u led cos kx 5. it at bx (c) it e2 f dz,2 I c. Laplace equation. Verify that (2 +v2)e 2_ v2)3 2. Heat Equation (2) with suitable t a I + :2 satisfies (6) and 6. it e.v o In a2 v2i sattsfies(3). Is it \ a2 -ot. solution of (3)? Of what Poisson_equation? 7. ii Os wa (d) Verify that it is ith an (sufticientl\ often diifer 8. ii ws 9. ii cos 25x Itt u and ii satisfies the given PDE. I y) Laplace Equation (3) 0. a cos y (V S a a (x 2) + w Ft 2) O u tt rut. a arctan (V r) r un&i value pr Iem.VerifS In Cr satisfies Laplaces equatto 2. a ers h.i. v cosh s it In. v 2 v2) 7 that thejti

2 2.2 Modeling: Vibrating String, Wave Equation 53 St and determine a and b o that u satisfies the 8. 25u,,, u 0 9. u 0 oundarv conditions n 0 on the circle 20. 2it, : 9zi u 3 cos.i 29 s vv2andi0onthecirclev -v 00. 3y - 2. it, - - 6u, - fan 25 PDEs SOLVABLE AS ODEs 22. u,. 0 ur 23. S2ur,- 2xu 2u 0 happens if a PDE involves derisatives sith respect to. Surface o resolution. Shoss that the solutions s,triable univ wr can he transformed to such a form. :s. v of v.,.v represent surface. of i-es olution. GO. c -at the other artahiet s) can be treated as parameteo s). example.. H/itt. Use polar coordinate r. a and shos that e for u it Lx. v: the equation becomes :, 0. ( 7. uc -t- ioir 2u System of PDEs. Sake u. 0, u - 0 Modeling: Vibrating String, Wave Equation In this section ve model a vibrating string, which will lead to our first important PDE. that is, equation (3) which will then be solved in Sec The student should pay close attention to this delicate modelinç process and detailed derivation starting from scratch, as the skills learned can he applied to modeling other phenomena in general and in particular to modeling a vibrating membrane (Sec. 2.7>. We want to derive the PDE modeling small transverse vibrations of an elastic string, such as a violin string. We place the string along the x-axi. stretch it to length L. and fasten it at the ends x 0 and x L. We then distort the string, and at some instant, call it t 0. we release it and alios it to vibrate. The problem is to determine the vibrations of the string. that is. to find its deflection ii Is. fl at any point x and at any time t > 0: see Fig ii (s. t) will be the solution of a PDE that is the model of our physical system to he derived. This PDE should not he too complicated. so that we can solse it. Reasonable simplifying assumptions (just as for ODEs modeling vibrations in Chap. 2> are as follows. Physical Assumptions. The mass of the string per unit length is constant ( homogeneous string ). The string is perfectly elastic and does not offer any resistance to bending. 2. The tension caused by stretching the string before fastening it at the ends is so large that the action of the gravitational force on the string (trying to pull the string down a little) can be neglected. 3. The string performs small transverse motions in a vertical plane; that is, every particle of the string moves strictly vertically and so that the deflection and the slope at every point of the string always remain small in absolute value. Under these assumptions we ina expect solutions u (.i. fl that describe the ph sical reality sufficiently well. /3 T Fig Deflected string at fixed time t. Explanation on p 5

3 the string change if Assumption 3 were violated? ideas more clean,. Assumption 2? The second part of Assumption? The simplification of formulas in this case that may show 5. k 3irx 2xj section for length L ir, to see the very substantial 3. String of length v. Write dossn the derisation in this we double the tension? Why is a contrabass larger than first part? Do we realls need all these assumptions? a violin? sthng? On the mass per unit ]ength? What happens if mode of the sibrating string depend on the length of the. Frequency. Ho does the frequenc of the fundamental f x L x I representation t7i. This leads to the graph shon n in Fig. 29 For graphing the solution we mas use u(s, 0 f(s and the abose interpretation of the two functions in the 2. Physical Assumptions. Hos would the motion of fnx+ ) N // N V I L/5c f(x L) 3L/:5c L/2c zeee_ieezn I 2L/Oc L/5c 0 L )x) u(x, 0) right (dashed) and a wave traveling to the left (left part of the figure) of the figure) obtained as the superposition of a wave traveling to the Fg 29. SoLution u(x, t) in Example for various values of t (right part f)x L f (x+) 6. I itv I say. 0.0 is as follows. Sketch or graph u cx. t) as in the initial velocity is zero and the initial deflection ss tth small Fig. 29 in the text. Find u (x. t) for the string of length L and c when 5 fl DEFLECTION OF THE STRING kinds of graphs. the rt-plane. Explain the connection between these two 2, 3, (. Also graph these solutions as surfaces oser program for graphing tb s ith L r and (2 of your choice similarly as in Fig Apply the program to. CAS PROJECT. Graphing Normal Modes. Write a fix+li f(x3l) Z. / Lui

4 Fig. 293A) cos /3L and I cosh L. El/pA 0. F(x) A cos f3x - /3x B 8. kx2(l 0.5 F G c2 G F 3 const, 5. Substituting u Fx)G(t) into (2). show that area). (Bending of a beam under a load is discussed in velocity and satisfying the boundary conditions (see u(x. 0) f(x) x(l modes of the vibrating string and the vibrating beam? solution of the equation graphically from the intersections of the curves o Determine approximate solutions of (22). for instance 3 x) v-axis in the figure. p density, A cross-sectional 552 CHAP. 2 Partial Differential Equations (PDEs) Sec. 3.3.) -- C cosh /3x Prob. 6 as well as the initial condition (22) cosh/3lcosj3l I. beam (Fig. 292) are modeled by the fourth-order PDE (2) _c (Ref. [Cli]) (ends simply supported for all times t), (zero moments, hence zero curvature, at the ends). basic difference between the frequencies of the normal that F in Prob. 5 satisfies these conditions if /3L is a shown that small free vertical vibrations of a uniform elastic conditions for the clamped beam in Fig. 293W? Show x0 xl 7. Find the solution of (2) that satisfies the conditions in 8. Compare the results of Probs. 7 and 7. What is the 9. Clamped beam in Fig. 293B. What are the boundary u(0, t) 0,u(L, t) 0 F0(x)G() of (2 ) corresponding to zero initial u(0, r) 0. u(l. t) 0 6. Simply supported beam in Fig. 293A. Find solutions Fg 293, Supports of a beam (B) Clamped at both (A) Smply supported G(t) acos c/3 2 t -- h cf3 2 t. h /3x. D I moment of intertia of the cross section with respect to the (F Young s modulus of elasticity, where c2 By the principles used in modeling the string it can be DE. VIBRATING BEAM 3 2( SEPARATION OF A FOURTH-ORDER Fg 292 Elastic beam J to realize experimentally.) ir. (Initial conditions with u (x. 0) 0 are hard if 0xT. ut(x,0)o.0l(7tx) if IT placement and triangular initial velocity Ut(.V, 0) 0.0 lx the string of length L IT and c2 for zero initial dis fi. Nonzero Titial velocity. Find the deflection u(f 3. 2x r 2 if0< x <. 0 if <x< 5 5 x). rightend end, free at the C Clamped at the left

5 9. u + u I cos equation u P. In the elliptic case, i v i, and the characteristics are complex and are of minor interest. For derivation, see Ref. [GenRef3] in App. I. Here. c l(x. v),p l (x, y), F F(v, w. ii. ut,, etc., and we denote ii as PDE is as in d Alembert s solution. In the parabolic case we get just one family of solutions function of v, n again by u, for simplicity. We see that the normal form of a hyperbolic 2 + I ) V x F F Type New VarIables Normal Form 556 CHAP. 2 Partial Differential Equations (PDE5) Sec. 2.3) of the deflection u(x. r) of a vibrating string Ug (3) sketch or graph a figure (similar to Fig. 29 in u5 vg 0 2 L 2u c2u n cu5 Parabolic Elliptic XP) + t((-c F3 P Hyperbolic TRICOMI ( ). Italian mathematician, who orked in integral equations and functional analysis. 2Sir GEORGE BIDELL AIRY , English mathematician. knrm a tot his work in elasticity. -RAsCESC 0 0. it bu 0 your work in detail. Find the type, transform to normal form, and solve. Show I NORMAL FORMS 7. f(x) k 2n x 8. f(x) Lt( 5. f(x) k ra 6. f(x) ( (length L, ends fixed, c ) starting with initial velocity 0 and initial deflection (k small, say, k 0.0). x) 5 GRA II G SOLUTIONS Prob. 3?. What are the frequencies of the eigenfunctiorts in waves? 67. Ib), what is the corresponding speed of transverse Ib) and is stretched by a tensile force of 300 nt (about 3. If a steel wire 2 m in length weighs 0.9 nt (about 0.20 and of period 2L the function fin (3) of this section must be odd 2. Show that, because of the boundary conditions (2), Sec. by ().. Show that c is the speed of each of the two waves given solution u /i(a + Ct) + ci). I Disision b c2 gises u const.thisgivesthenewvariabesvpvxct--xancgc Vsxct-aandd Alembert s (v ) 0. The two families of solutions (characteristics) are (Na. s) v ±x const and Px, v) s 0 in the form 0) h settings Ct. By the chain rule, U u5yt The theory of characteristics gives d Alembert s solution in a systematic fashion. To see this..se rite the vase EXAMPLE I D Alembert s Solution Obtained Systematically Ref. [GenRefli listed in App..) and initial velocity iero is corresponding to initial displacement ncr. 0 u(0, t) 0 and u(l. t) 0. Show that the motion end, x ((, and free at the other, x L, we have (see Tolstoy [C9]. p. 275). If the rod is fastened at one modeled by the wave equation Utt c2u,, c2 E/p These vibrations in the direction of the x-axis are 7. u,j ii A, px cos pact. equation by separation. (For solutions. see p. 6 a Airy equation G 0 from the Tricom equation vucr + u 0 is of mixed type. Obtain tb 20. Tricomi and Airy equations. 2 Show that the Triconi fix) p, 5x dx. m 2L (2ii ± l)7t -- 5u ti 6uy +9Uyy 0 5. XUac VUyy 0 6. UI.r + 2u + lou Longitudinal Vibrations of an Elastic Bar or Rod. (t, + 2u.y c TMyg 0 3. ± + Uyy 0. yii VU 0 x 0, as stated before. Hence the characteristic equation is v 2 ts and u0 c2u.