Solution to Problems for the 1-D Wave Equation

Transcription

1 Solution to Problems for the -D Wave Equation 8. Linear Partial Differential Equations Matthew J. Hancock Fall 5 Problem (i) Suppose that an infinite string has an initial displacement +, u (, ) = f () =, /, < and > / and zero initial velocity u t (, ) =. Write down the solution of the wave equation u tt = u with ICs u (, ) = f () and u t (, ) = using D Alembert s formula. Illustrate the nature of the solution by sketching the u-profiles y = u (, t) of the string displacement for t =, /,, /. Solution: D Alembert s formula is ( +t ) u (, t) = f ( t) + f ( + t) + g (s) ds In this case g (s) = so that t u (, t) = (f ( t) + f ( + t)) () The problem reduces to adding shifted copies of f () and then plotting the associated u (, t). To determine where the functions overlap or where u (, t) is zero, we plot the characteristics ± t = and ± t = / in the space time plane (t) in Figure. For t =, () becomes u (, ) = (f () + f ()) = f ()

2 .5 +t=/ t= R 4 t.5 +t= t=/ R R.5 R R 5 R Figure : Sketch of characteristics for (a). For t = /, () becomes ( ( ) ( )) u (,t) = f + f + Note that ( ) ( ) ( ) +, ( ) ( ) f =, / ( ), ( ) < and > / +, =,, < and > and similarly, Thus, over the region ( ) f + = +,, < we have to be careful about adding the two

3 functions; in the other regions either one or both functions are zero. We have ( ) ( ( ) ( )) u, = f + f + +, 4 +, For t = /, the forward and backward waves are even further apart, and ( ), f = 4,, < and > 5 5 ( ) +, f + =,, < 5 and > 4 4 = +,,, < and > For t =, your plot of the characteristics shows that f ( ) and f ( + ) do not overlap, so you just have to worry about the different regions. Note that ( + ) +, + f ( + ) = ( + ), + /, + < and + > / +, =, /, < and > /, f ( ) =, /, < and > / so that u (, ) = (f ( ) + f ( + )) +,, / =,, /, <, / < <, and > /

4 u(,).5 u(,/) u(,).5.5 u(,/).5 Figure : Plot of u(,t ) for t =, /,, / for (a). and hence ( ) ( ( ) ( )) u, = f + f + + 5, 5, 4,, =,, 4,, 5, <, < <, and > The solution u (,t ) is plotted at times t =, /,, / in Figure. A D version of u (,t) is plotted in Figure. (ii) Repeat the procedure in (i) for a string that has zero initial displacement but is given an initial velocity, < u t (, ) = g () =,, < and > Solution: D Alembert s formula is ( +t ) u (,t) = f ( t) + f ( + t) + g (s) ds t 4

5 u(,t) t Figure : D version of u(,t) for (a). In this case f (s) = so that u (,t) = +t t g (s) ds The problem reduces to noting where ± t lie in relation to ± and evaluating the integral. These characteristics are plotted in Figure in the notes. You can proceed in two ways. First, you can draw two more characterstics ±t = so you can decide where the integration variable s is with respect to zero, and hence if g (s) = or. The second way is to note that for a < b and a, b <, b a g (s) ds = b a for positive and negative a,b. I ll use the second method; the answers you get from the first are the same. In Region R, ± t 5

6 and hence there are cases: t <, u (,t) = = ( + t t ) In Region R, + t > and < t <, so that ( +t ) u (,t) = + g (s) ds = g (s)ds t t = ( t ) In Region R, t < and < + t <, so that ( +t ) +t u (,t) = + g (s) ds = g (s)ds = ( + t ) t = ( + t ) In Region R 4, + t > and t <, so that ( +t ) u (,t) = + + g (s) ds = g (s) ds = ( + ) = u (, ) = g (s) ds = At t = /, the regions R n are given in the notes and ( ) [ + ( ) (, R ) = [,, R ( ) = [, u, =, R =, +, R 5,R 6 = { > /} eample, for [ /, /, we have / = ( /). Thus, [, R = ( ) [,, R 4 =, [ u, =,, R 4 =, R 5,R = { > /} +t t g (s) ds t In Region R 5, + t < and hence u (,t) =. In region R 6, t >, so that u (,t) =. At t =, The absolute values are easy to resolve (i.e. write without them) in this case. For 6 6

7 At t =, the regions R n are given in the notes and u (, ) = ( ), R = [,, ( + ), R = [,,, R 5,R 6 = { > /}. You could leave your answer like this, or write it without absolute values (have to divide [, and [, into cases): /, [,, ( ), [,, u (, ) = ( + ) = [, /, = [,,, R 5,R 6 = { > /}. At t = /, the regions R n are not given eplicitly, but can be found from Figure in the notes by nothing where the line t = / crosses each region: ( ) [ ( ) (, R ) = [,5 +, R = 5, u, =, R 4,R 5,R 6 = { > 5/ or < /} Again, you could leave your answer like this, or write it without absolute values (have to divide [/, 5/ and [ 5/, / into cases): ( ) [, R ( ) = [, 5 ( ), R ( ) = [,5 5 +, R ( ) = 5 [, u, = +, R =,, R 4,R 5,R 6 = { > 5/ or < /} The solution u (,t ) is plotted at times t =, /,, / in Figure 4. Problem (i) For an infinite string (i.e. we don t worry about boundary conditions), what initial conditions would give rise to a purely forward wave? Epress your answer in terms of the initial displacement u (, ) = f () and initial velocity u t (, ) = g () and their derivatives f (), g (). Interpret the result intuitively. Solution: Recall in class that we write D Alembert s solution as u (,t) = P ( t) + Q ( + t) () 7

8 u(,) u(,) u(,/) u(,/) Figure 4: Plot of u(, t ) for t =, /,, / for (b). where ( ) Q () = f () + g (s) ds + Q () P () () ( ) P () = f () g (s) ds Q () + P () (4) To only have a forward wave, we must have Q () = const = q Substituting () gives ( ) Q () = q = f () + g (s)ds + Q () P () Differentiating in gives Thus ( ) df = + g () d df g () = (5) d 8

9 4 Question [ points total Suppose you shake the end of a rope of dimensionless length at a certain frequency ω. The opposite end of the rope is fied to a wall. We aim to find the special frequencies ω at which certain points along the rope remain fied in mid air. We negelect gravity and friction and model the waves on the rope using the D wave equation: u tt = u, < <, t > () where u is the displacement of the rope away from it s rest state. The end at the wall (at = ) is fied: u (,t) =, t >. () You shake the other end (at = ) sinusoidally, with frequency ω, and give it the displacement u (,t) = sinωt, t >. () We assume the rope has zero initial position and velocity u (, ) =, < <, (4) u t (, ) =, < <. (5) (a) [ points Find a solution of the form u S (,t) = X () sin ωt (6) that satisfies the PDE () and the BCs () and (). (Don t worry about the ICs yet.) Where is the rope stationary (i.e. u s (,t) = )? For what values of ω is your solution invalid? Solution: From (6), we have u S u Stt = X () sin ωt = ω X () sin ωt Thus the PDE () implies Solving gives X + ω X = X () = a cos (ω) + b sin (ω) (7)

10 Substituting () into the BC () gives = X () sin ωt and hence X () =. Thus (7) becomes = X () = a and hence BC () is satisfied if X () = b sin ω = X () = b sin ω Thus b = / sin ω and Rope stationary at ω = nπ, u S (,t) = sin ω sin ω sin ωt = nπ, n =,,... ω such that nπ/ω <. Solution invalid when ω = nπ for some n =,,,... (b) [ points Use u S (,t) from part (a) to find the full solution u (,t) to the PDE (), the BCs () and (), and the ICs (4) and (5). Hint: use u S (,t) and u (,t) to find a wave problem for some quantity v (,t) that has Type I BCs (i.e. zero displacement) at =,. Then write down D Alembert s solution (without derivation) to satisfy the PDE and initial conditions for this problem (don t evalutate the integral in D Alembert s solution). Then adjust D Alembert s solution to handle the BCs at =,. You don t need to evaluate the integral. Solution: Let v = u u S. Then v tt = v, < <, t > v (,t) = = v (,t), t > v (, ) = u (, ) u S (, ) = v t (, ) = u t (, ) u St (, ) = D Alembert s solution for the infinite string is v (,t) = ω sin ω +t t sin (ωs)ds ω sin ω sin ω

11 To satisfy the BCs, we must etend sin (ωs) to an odd periodic function in s. Since sin (ωs) is already odd, we merely need it s periodic etension, ˆf (s) = sin (ω (smod)) where we assume ( s) mod = (smod), so that Thus v (,t) = ω sin ω +t t ˆf (s)ds 5 Question [ points total u (,t) = v (,t) + u S (,t) = ω +t sin (ω (smod))ds sin ω t sin ω + sin ωt sin ω Consider the following quasi-linear PDE, u t where the initial condition is f () = { + ( + u) u, >, = = u; u (, ) = f () (8), < +,, <, > (a) [8 points Find the parametric solution, using r as your parameter along a characteristic and s to label the characteristic (i.e. the initial value of ). First write down the relevant ODEs for t/ r, / r, u/ r. Take the initial conditions t = and = s at r =. Using the initial condition in (8), write down the IC for u at r =, in terms of s. Solve for t, u and (in that order!) as functions of r, s. When integrating for, be careful: u depends on r! Solution: The parametric ODEs are t r =, r = + u, 4 u r = u.

12 The ICs are t (;s) =, (;s) = s, u (;s) = f ( (;s)) = f (s) (9) Solving for t and u gives t = r + c, ln u = r + c where c, are constants of integration. Thus u = c e r Imposing the ICs (9) gives t = r, u = f (s)e r = f (s)e t Substituting u into the equation for gives and integrating yields r = + f (s)e r = r f (s)e r + c 4 Imposing the IC (9) gives = r f (s) ( e r ) + s = t f (s) ( e t ) + s (b) [8 points At what time t sh and position sh does your parametric solution break down? Hint: you might need to consider negative values of f (s). Solution: The solution breaks down when the Jacobian is zero: ( ) r s J = det = t t r t s r s t s r = s = ( f (s) ( e r ) + ) = Since r = t, we solve for t to obtain t sh = ln ( + ) f (s) 5

13 Note that f (s) is either, yielding infinite t sh,, yielding negative t sh, and, yielding ( ) t sh = ln = ln Thus the shock time is at ln. An s value where f (s) = is s =, so that the shock location is sh = t sh f () ( e t sh ) + ( ) = ln + = ln + (c) [4 points Write down in terms of t, s and f (s). For each of s =,,, write down as a function of t. Solution: s = : = t ( e t ) s = : = t 4 ( e t ) s = : = t ( e t ) + (d) [4 points Fill in the table below, using your result from either question (a) or (c) to obtain u and at the s values listed at time t = ln (note that e ln =, and you may use ln =.7): s = Solution: t = ln t = ln u = = s = u = / / = ln ln + ln + (e) [6 points Plot the initial f () vs.. Then, on the SAME plot, plot u (,t) at t = ln by plotting the three points (,u) from the table in part (d). 6