Solutions to Exercises 8.1

Transcription

1 Section 8. Partial Differential Equations in Physics and Engineering 67 Solutions to Exercises 8.. u xx +u xy u is a second order, linear, and homogeneous partial differential equation. u x (,y) is linear and homogeneous. 5. u t u x + u xt u is second order and nonlinear because of the term u t u x. u(,t)+u x (,t)is linear and homogeneous. 9. (a) Let u(x, y) e ax e by. Then So because e ax e by for all x and y. (b) By (a), in order to solve u x ae ax e by u y be ax e by u xx a e ax e by u yy b e ax e by u xy abe ax e by. Au xx +Bu xy + Cu yy + Du x + Eu y + Fu Aa e ax e by +Babe ax e by + Cb e ax e by +Dae ax e by + Ebe ax e by + Fe ax e by e ax e by( Aa +Bab + Cb + Da + Eb + F ) Aa +Bab + Cb + Da + Eb + F, u xx +u xy + u yy +u x +u y + u, we can try u(x, y) e ax e by, where a and b are solutions of a +ab + b +a +b +. But a +ab + b +a +b +(a + b +). So a + b +. Clearly, this equation admits infinitely many pairs of solutions (a, b). Here are four possible solutions of the partial differential equation: a,b u(x, y) e x e y a,b u(x, y) e y a /, b / u(x, y) e x/ e y/ a 3/, b/ u(x, y) e 3x/ e y/ 3. We follow the outlined solution in Exercise. We have A(u) ln(u), φ(x) e x, A(u(x(t)), t)) A(φ(x())) ln(e x() )x(). So the characteristic lines are x tx() + x() x() L(xt) x t +.

2 Partial Differential Equations in Rectangular Coordinates ( So u(x, t) f(l(x, t)) f x t+ Check: u t e x t+ x (t+),u x e x t+ t+, 7. We have So the characteristic lines are ). The condition u(x, ) e x implies that f(x) e x and so u t + ln(u)u x e x t+ u(x, t) e x t+. x (t +) + x t + e x t+ t +. A(u) u, φ(x) x, A(u(x(t)), t)) A(φ(x())) x(). x tx() + x() x()(t +) x. Solving for x(), we find and so Now So x() u(x, t) f x t +, ( ) x. t + u(x, ) f(x) x. x u(x, t) t +.

3 Section 8. Wave Equation, the Method of Separation of Variables 69 Solutions to Exercises 8.. The solution is u(x, t) sin nπx L ( b n cos c nπt L + b n sin c nπt ), L where b n are the Fourier sine coefficients of f and b n are cnπ times the Fourier coefficients of g. In this exercise, b n, since g,b.5; and b n for all n>, because f is already given by its Fourier sine series (period ). So u(x, t).5 sin πx cos t. 5. (a) The solution is u(x, t) sin(nπx)(b n cos(4nπt)+b n sin(4nπt)), where b n is the nth sine Fourier coefficient of f and b n is L/(cn) times the Fourier coefficient of g, where L andc 4. Since g, we have b n for all n. As for the Fourier coefficients of f, we can get them by using Exercise 7, Section.4, with p,h, and a /. We get b n 8 nπ π sin n. Thus u(x, t) 8 sin nπ π n sin(nπx) cos(4nπt) 8 π k ( ) k sin((k +)πx) cos(4(k +)πt). (k +) (b) Here is the initial shape of the string. Note the new Mathematica command that we used to define piecewise a function. (Previously, we used the If command.) Clear f f x_ : x ; x f x_ : x ; x Plot f x, x,, L Because the period of cos(4(k +)πt) is/, the motion is periodic in t with period /. This is illustrated by the following graphs. We use two different ways to plot the graphs: The first uses simple Mathematica commands; the second one is more involved and is intended to display the graphs in a convenient array.

4 Partial Differential Equations in Rectangular Coordinates Clear partsum partsum x_, t_ : 8 Pi^ Sum Sin ^k k Pi x Cos 4 k Pi t k ^, k,, Plot Evaluate partsum x,, f x, x,, Here is the motion in an array.

5 Section 8. Wave Equation, the Method of Separation of Variables 7 tt Table Plot Evaluate partsum x, t, x,,, PlotRange,,,, Ticks.5,,.5,.5,, DisplayFunction Identity, t,,, ; Show GraphicsArray Partition tt, The solution is u(x, t) sin(nπx)(b n cos(nπt)+b n sin(nπt)), where b π and all other b n. The Fourier coefficients of f are b n x( x) sin(nπx) dx. To evaluate this integral, we will use integration by parts to derive first the formula: for a, and x sin(ax) dx x cos(ax) a x sin(ax) dx cos(ax) a 3 x cos(ax) a + sin(ax) a + C, + x sin(ax) a + C;

6 Partial Differential Equations in Rectangular Coordinates thus x( x) sin(ax) dx cos(ax) a 3 x cos(ax) a Applying the formula with a nπ, weget x( x) sin(nπx) dx + x cos(ax) a + sin(ax) a x sin(ax) a + C. cos(nπ x) x cos(nπ x) (nπ) 3 + x cos(nπ x) sin(nπ x) x sin(nπ x) + nπ nπ (nπ) (nπ) (( )n ) (nπ) 3 { 4 (nπ) 3 ( )n nπ if n is odd, if n is even. + ( )n nπ (( )n ) (nπ) 3 Thus b n { 8 (nπ) 3 if n is odd, if n is even, and so 3. To solve u(x, t) 8 π 3 k sin((k +)πx) cos((k +)πt) (k +) 3 + sin(πx) sin(πt). π u t + u t u x, u(,t)u(π, t), u u(x, ) sin x, (x, ), t we follow the method of the previous exercise. We have c,k.5, L π, f(x) sin x, and g(x). Thus the real number Lk cπ.5 is not an integer and we have n>kl π for all n. So only Case III from the solution of Exercise needs to be considered. Thus u(x, t) e.5t sin nx ( a n cos λ n t + b n sin λ n t, where Setting t, we obtain λ n (.5n). sin x a n sin nx.

7 Section 8. Wave Equation, the Method of Separation of Variables 73 Hence a and a n for all n>. Now since b n ka n λ n + λ n L L g(x) sin nπ L x dx, n,,..., it follows that b n for all n> and and the solution takes the form u(x, t) e.5t sin x ( cos λ t + b sin λ t ), where λ (.5).75 3 and b ka λ 3. So u(x, t) e.5t sin x ( cos( 3 t)+ 3 sin( 3 t)). 7. (a) That G is even follows from its Fourier series representation that we derive in a moment. That G is L-periodic follows from the fact that g is L-periodic and its integral over one period is, because it is odd (see Section 7., Exercise 5). Since G is an antiderivative of g, to obtain its Fourier series, we apply Exercise 33, Section 7.3, and get G(x)A L b n (g) π n cos nπ L x, where b n (g) is the nth Fourier sine coefficient of g, b n (g) L L g(x) sin nπ L xdx and In terms of b n, we have A L π b n (g) n. L b n (g) π n nπ L g(x) sin nπ L xdx cb n, and so G(x) L π cb n b n (g) n L π b n (g) n ( cos( nπ ) L x). cos nπ L x

8 Partial Differential Equations in Rectangular Coordinates Since the Fourier series of G contains only cosine terms, it follows that G is even. (b) From (a), it follows that G(x + ct) G(x ct) cb n [( cos( nπ ) L (x + ct)) ( cos( nπ )] L (x ct)) cb n [cos( nπ ] L (x + ct)) cos(nπ L (x ct)) (c) Continuing from (b) and using the notation in the text, we obtain c x+ct x ct g (s) ds [G(x + ct) G(x ct)] c b [ n cos( nπ ] (x + ct)) cos(nπ L L (x ct)) b n sin(nπ L x) sin(nπ L ct) b n sin( nπ L x) sin(λ nt). (d) To derive d Alembert s solution, proceed as follows: u(x, t) b n sin( nπ L x) cos(λ nt)+ b n sin(nπ L x) sin(λ nt) ( f (x ct)+f (x + ct) ) + [G(x + ct) G(x ct)], c where in the last equality we used Exercise 6 and part (c).

9 Section 8.3 The One Dimensional Heat Equation 75 Solutions to Exercises 8.3. Multiply the solution in Example by 78 u(x, t) 3 π k to obtain e (k+) t k + sin(k +)x. 5. We have where So b n u(x, t) b n e (nπ)t sin(nπx), [ x sin(nπx) dx x cos(nπx) + sin(nπx) ] nπ n π cos nπ nπ u(x, t) π ( )n+. nπ ( ) n+ e (nπ)t sin(nπx). n 9. (a) The steady-state solution is a linear function that passes through the points (, ) and (, ). Thus, u(x) x. (b) The steady-state solution is a linear function that passes through the points (, ) and (, ). Thus, u(x). This is also obvious: If you keep both ends of the insulated bar at degrees, the steady-state temperature will be degrees. 3. We have u (x) 5 π x +. We use (3) and the formula from Exercise, and get (recall the Fourier coefficients of f from Exercise 3) u(x, t) 5 π x + [ 3 sin(n π + ) ( ( ) n )] π n e nt sin nx. nπ 7. Fix t > and consider the solution at time t t : u(x, t ) b n sin nπ L xe λ n t. We will show that this series converges uniformly for all x (not just x L) by appealing to the Weierstrass M-test. For this purpose, it suffices to establish the following two inequalities: (a) b n sin nπ L xe λ nt M n for all x; and

10 Partial Differential Equations in Rectangular Coordinates (b) M n <. To establish (a), note that b n L f(x) sin nπ L L xdx L f(x) sin nπ L L x dx (The absolute value of the integral is the integral of the absolute value.) L L f(x) dx A (because sin u for all u). Note that A is a finite number because f is bounded, so its absolute value is bounded and hence its integral is finite on [, L]. We have b n sin nπ L xe λ n t Ae λ n t Ae c π t L n ( ) A e c π t n L Ar n, where r e c π t L <. Take M n Ar n. Then a holds and M n is convergent because it is a geometric series with ratio r<.

11 Section 8.4 Heat Conduction in Bars: Varying the Boundary Conditions 77 Solutions to Exercises 8.4. Since the bar is insulated and the temperature inside is constant, there is no exchange of heat, and so the temperature remains constant for all t>. Thus u(x, t) for all t>. This is also a consequence of (), since in this case all the a n s are except a. 5. The solution is given by (4) with c L, and where the coefficients c n are the Fourier cosine coefficients of the function f(x) cos πx. This function is clearly its own cosine series with a n for all n and a. Thus u(x, t) e λ t cos πx e πt cos πx. 9. This is a straightforward application of Exercise 7. For Exercise the average is. For Exercise the average is a. 3. The solution is given by (8), where c n is given by (). We have sin µ n xdx ( cos(µ n x) dx ( x ) sin(µ n x) µ ( ) sin(µ n ). n µ n Since µ n is a solution of tan µ µ, we have sin µ n µ n cos µ n,so sin µ n sin µ n cos µ n µ n cos µ n, and hence sin µ n xdx ( + cos ) µ n. Also, Applying (), we find / sin µ n xdx ( µ n ) cos. µ n c n / / sin µ n xdx sin µ n xdx ( µ n ) / ( cos +cos ) µ n µ n ( cos µn µ n ( + cos µ n ). ) Thus the solution is u(x, t) ( ) cos µn nt µ n ( + cos µ n ) e µ sin µ n x.

12 Partial Differential Equations in Rectangular Coordinates 7. Part (a) is straightforward as in Example. We omit the details that lead to the separated equations: X kx, T kt, X () X(), X () X(), where k is a separation constant. (b) If k then So k leads to trivial solutions. (c) If k α >, then X X ax + b, X () X() a b X () X() a (a + b) a b; a b. X µ X X c cosh µx + c sinh µx; X () X() µc c X () X() µc sinh µ + µc cosh µ c cosh µ c sinh µ µc sinh µ c cosh µ c cosh µ c sinh µ µc sinh µ c sinh µ µc sinh µ c sinh µ. µ Since µ, sinh µ. Take c and divide by sinh µ and get µc c µ µ k. So X c cosh x + c sinh x. But c c,so X c cosh x + c sinh x c cosh x c sinh x c e x. Solving the equation for T, we find T (t) e t ; thus we have the product solution c e x e t, where, for convenience, we have used c as an arbitrary constant. (d) If k α <, then X + µ X X c cos µx + c sin µx; X () X() µc c X () X() µc sin µ + µc cos µ c cos µ c sin µ µc sin µ c cos µ c cos µ c sin µ µc sin µ c sin µ µc sin µ c sin µ. µ

13 Section 8.4 Heat Conduction in Bars: Varying the Boundary Conditions 79 Since µ, take c (otherwise you will get a trivial solution) and divide by c and get µ sin µ sin µ sin µ µ nπ, where n is an integer. So X c cos nπx + c sin nπx. But c c µ,so X c ( nπ cos nπx sin nπx ). Call X n nπ cos nπx sin nπx. (e) To establish the orthogonality of the X n s, treat the case k separately. For k µ, we refer to the boundary value problem X + µ nx,x() X (), X() X (), that is satisfied by the X n s, where µ n nπ. We establish orthogonality using a trick from Sturm-Liouville theory (Chapter 6, Section 6.). Since X m µ mx m and X n µ nx n, multiplying the first equation by X n and the second by X m resulting equations, we obtain and then subtracting the X n X m µ mx m X n and X m X n µ nx n X m X n X m X m X n (µ n µ m)x m X n ( Xn X m X m X n ) (µ n µ m)x m X n where the last equation follows by simply checking the validity of the identity X n X m X m X n ( X n X m X m X n).so (µ n µ m) X m (x)x n (x) dx ( Xn (x)x m(x) X m (x)x n(x) ) dx X n (x)x m(x) X m (x)x n(x), because the integral of the derivative of a function is the function itself. Now we use the boundary conditions to conclude that X n (x)x m(x) X m (x)x n(x) X n ()X m() X m ()X n() X n ()X m() + X m ()X n() X n ()X m () + X m ()X n () + X n ()X m () X m ()X n (). Thus the functions are orthogonal. We still have to verify the orthogonality when one of the X n s is equal to e x. This can be done by modifying the argument that we just gave. (f) Superposing the product solutions, we find that u(x, t) c e x e t + c n T n (t)x n (x).

14 Partial Differential Equations in Rectangular Coordinates Using the initial condition, it follows that u(x, ) f(x) c e x + c n X n (x). The coefficients in this series expansion are determined by using the orthogonality of the X n s in the usual way. Let us determine c. Multiplying both sides by e x and integrating term by term, it follows from the orthogonality of the X n that f(x)e x dx c e x dx + c n {}}{ X n (x)e x dx. Hence f(x)e x dx c e x e dx c. Thus c e e f(x)e x dx. In a similar way, we prove that where c n f(x)x n (x) dx κ n κ n X n(x) dx. This integral can be evaluated as we did in Exercise 5 or by straightforward computations, using the explicit formula for the X n s, as follows: X n(x) dx ( nπ cos nπx sin nπx ) dx ( n π cos nπx + sin nπx nπ cos(nπx) sin(nπx) ) dx (n π )/ {}}{ nπ n π cos nπx dx + / {}}{ sin nπx dx { }}{ cos(nπx) sin(nπx) dx n π +.

15 Section 8.5 The Two Dimensional Wave and Heat Equations 8 Solutions to Exercises We proceed as in Exercise 3. We have u(x, y, t) (B mn cos λ mn t + Bmn sin λ mn t) sin mπx sin nπy, m where λ mn m + n, B mn, and B mn 4 m + n sin mπx sin nπy dx dy 4 sin mπx dx sin nπy dy m + n { 6 m if m and n are both odd, +n (mn)π otherwise. Thus u(x, y, t) 6 sin((k +)πx) sin((l +)πy) (k +) +(l +) (k + )(l +)π sin (k +) +(l +) t k l

16 Partial Differential Equations in Rectangular Coordinates Solutions to Exercises 8.6. The solution is given by u(x, y) where B n sin(nπx) sinh(nπy), B n x sin(nπx) dx sinh(nπ) [ x cos(nπx) + sin(nπx) ] sinh(nπ) nπ n π ( ) n sinh(nπ) nπ sinh(nπ) ( ) n+. nπ Thus, u(x, y) π ( ) n+ sin(nπx) sinh(nπy). n sinh(nπ) 5. Start by decomposing the problem into four subproblems as described by Figure 3. Let u j (x, y) denote the solution to problem j (j,, 3, 4). Each u j consists of only one term of the series solutions, because of the orthogonality of the sine functions. For example, to compute u, we have u (x, y) A n sin nπx sinh[nπ( y)], where A n sinh nπ sin 7πx sin nπx dx. Since the integral is unless n 7 and, when n 7, A 7 sinh 7π sin 7πx dx sinh 7π. Thus u (x, y) sin 7πx sinh[7π( y)]. sinh 7π

17 Section 8.6 Laplace s Equation in Rectangular Coordinates 83 In a similar way, appealing to the formulas in the text, we find u (x, y) u 3 (x, y) sin πx sinh(πy) sinh π sinh[3π( x)] sin(3πy) sinh 3π u 4 (x, y) sinh 6πx sin(6πy); sinh 6π u(x, y) u (x, y)+u (x, y)+u 3 (x, y)+u 4 (x, y) sin 7πx sinh[7π( y)] + sin(πx) sinh(πy) sinh 7π sinh π + sinh 3π sinh[3π( x)] sin(3πy)+ sinh(6πx) sin(6πy) sinh 6π 3. Consider the function f(z) ia cos[α(z z )], where A and α and the given real constants and z β + iγ, where β and γ are the given real constants. Clearly, f is entire (analytic for all z). Appealing to (5), Section.6, we find that f(z) ia {cos[ Re (α(z z ))] cosh[ Im (α(z z ))] sin[ Re (α(z z ))] sinh[ Im (α(z z ))]} ia {cos[α(x β)] cosh[α(y γ)] sin[α(x β)] sinh[α(y γ)]} A sin[α(x β)] sinh[α(y γ)] + ia cos[α(x β)] cosh[α(y γ)]. Thus the real part of f is and its imaginary part is φ(x, y) A sin[α(x β)] sinh[α(y γ)] ψ(x, y) A cos[α(x β)] cosh[α(y γ)]. This completes the solution, since the real and imaginary parts are harmonic functions and the imaginary part is a harmonic conjugate of the real part.

18 Partial Differential Equations in Rectangular Coordinates Solutions to Exercises 8.7. We apply (), with a b : where Thus E mn u(x, y) 8 π 4 u(x, y) 4 π (m + n ) m E mn sin mπx sin nπy, x sin mπx sin nπy dx dy ( )n nπ {}}{ 4 π (m + n ) x sin mπx dx 4 ( ) n π 4 (m + n ) n 4 ( ) n ( ) m π 4 (m + n ) n m. k m sin nπy dy ( x cos(mπx) m + sin(mx) m π ) ( ) m (m +(k +) sin mπx sin((k +)πy). )m(k +) 5. We will use an eigenfunction expansion based on the eigenfunctions φ(x, y) sin mπx sin nπy, where π(x, y) π ( m + n ) sin mπx sin nπy. So plug u(x, y) E mn sin mπx sin nπy m into the equation u 3u, proceed formally, and get ( m E mn sin mπx sin nπy) 3 m E mn sin mπx sin nπy m E mn (sin mπx sin nπy) 3 m E mn sin mπx sin nπy m E mnπ ( m + n ) sin mπx sin nπy 3 m E mn sin mπx sin nπy ( m 3+π ( m + n )) E mn sin mπx sin nπy. Thinking of this as the double sine series expansion of the function identically, it follows that ( 3+π ( m +n )) E mn are double Fourier sine coefficients, given by (see (8), Section 3.7) ( 3+π ( m + n )) E mn 4 sin mπx sin nπy dx dy 4 ( )m ( ) n { mπ nπ if either m or n is even, if both m and n are even. 6 π mn.

19 Section 8.7 Poisson s Equation: The Method of Eigenfunction Expansions 85 Thus and so { if either m or n is even, E mn 6 )) (3+π (m +n if both m and n are even, u(x, y) 6 π π mn k l sin((k +)πx) sin((l +)πy) (k + ) (l +) ( 3+π ( (k +) +(l +) )).

20 Partial Differential Equations in Rectangular Coordinates Solutions to Exercises 8.8. We use a combination of solutions from () and (3) and try a solution of the form u(x, y) sin mx [ A m cosh m( y)+b m sinh my ]. (If you have tried a different form of the solution, you can still do the problem, but your answer may look different from the one derived here. The reason for our choice is to simplify the computations that follow.) The boundary conditions on the vertical sides are clearly satisfied. We now determine A m and B m so as to satisfy the conditions on the other sides. Starting with u(, ), we find that A m cosh m sin mx. m Thus A m cosh m is the sine Fourier coefficient of the function f(x). Hence A m cosh m π π sin mx dx A m πm cosh m [ ( )m ]. Using the boundary condition u y (x, ), we find sin mx [ A m ( m) sinh[m( y)] + mb m cosh my ] y. Thus m mb m sin mx cosh m. m By the uniqueness of Fourier series, we conclude that mb m cosh m for all m. Since m cosh m, we conclude that B m and hence u(x, y) π 4 π m k [ ( ) m ] m cosh m sin mx cosh[m( y)] sin[(k +)x] cosh[(k + )( y)]. (k + ) cosh(k +) 5. We combine solutions of different types from Exercise 4 and try a solution of the form u(x, y) A + B y + cos mπ a x[ A m cosh[ mπ a (b y)] + B m sinh[ mπ a y]]. m Using the boundary conditions on the horizontal sides, starting with u y (x, b), we find that mπ B + a B m cos mπ a x cosh [ mπ a b]. m

21 Section 8.8 Neumann and Robin Conditions 87 Thus B and B m for all m and so A + A m cos mπ a x cosh[mπ (b y)]. a m Now, using u(x, ) g(x), we find g(x)a + m Recognizing this as a cosine series, we conclude that and equivalently, for m, A a A m cosh[ mπ a b] a A m a cosh[ mπ a b] A m cosh[ mπ mπ b] cos a a x. a a a 9. We follow the solution in Example 3. We have g(x) dx g(x) cos mπ a xdx; g(x) cos mπ a x dx. u(x, y) u (x, y)+u (x, y), where with and with Hence 4 π B m A m u(x, y) π k u (x, y) B m sin mx sinh my, m π sin mx dx πm cosh(mπ) u (x, y) πm cosh(mπ) ( ( )m ); A m sin mx cosh[m(π y)], m π sin mx dx π cosh(mπ) ( ( ) m [ ) sinh my sin mx m cosh(mπ) m [ sinh[(k +)y] (k +) m sin(k +)x (k + ) cosh[(k + )π] πm cosh(mπ) ( ( )m ). ] + cosh[m(π y)] ] + cosh[(k + )(π y)].