f( x) = f(x), f(l + x) = f(l x)

Transcription

1 Math 5 / AMCS 55 Dr. DeTurck Practice problems for midterm March 3, 6. Say that a function is oddly odd if it satisfies both the conditions f( x) f(x), f(l + x) f(l x) (a) Show that such a function is periodic with period L. (b) Draw the graph of a non-zero oddly odd function for 5L x 5L (pick one that is interesting but not too interesting, perhaps have the graph consist mostly of line segments). What (if any) kind of symmetry does it have around the line x L?... around the line x L? (c) Show that the Fourier series of an oddly odd function is of the form Give a formula for the coefficients b n. f(x) b n sin n (n )πx. L (a) Repeatedly use the given properties. We have to show that f(x + L) f(x) for all x. Well, f(x + L) f(l + (x + 3L)) getting ready to use the second condition f(l (x + 3L)) by the second condition f( (x + L)) f(x + L) by the first condition f(l + (x + L)) getting ready to use the second condition f(l (x + L) by the second condition f( x) f(x) by the first condition (b) In this figure, L. The function is even around x L (i.e., f(l + x) f(l x)) odd around x L (i.e., f(l + x) f(l x)).

2 (c) We have b n if n is even, if n is odd, say n k, b k L L ( ) (k )πx f(x) sin dx L so with these coefficients. f(x) ( ) (k )πx b k sin L k. Let f(x) { x < x < x < x < Solve the heat equation u t u xx for x, t, ) with initial condition u(x, ) f(x) boundary conditions u(, t) u(, t). Draw a sketch of the graph of u(x, ɛ) for a fixed, very small value of ɛ x. Because of the boundary conditions, we know we have to use sines in our Fourier expansion. So the solution will be u(x, t) b n e n π t/ sin nπx, where the inner product is given by b n f, g n nπx f(x), sin sin nπx, sin nπx f(x) sin nπx dx. Since f is even around, you can see that b n if n is even. For odd n, the integral from to will be twice what it is from to. So we get b k+ ( )k (k + ) π, so u(x, t) k ( ) k (k + ) π e (k+) π t/ sin (k + )πx.

3 3 In this figure: the red curve is the initial data the blue curve is the solution at time t. you can see that the corner gets around off the whole curve goes down a little bit. 3. (a) Find the Fourier (cosine) series of the function f(x) x, π < x < π. (b) Draw the graph of the function to which your series converges. Explain how you know the series converges pointwise to this function. Does it converge uniformly? (c) Use the series to show that n + π ( )n+ n + π (d) Use the results in part (c) to deduce (n ) + π

4 (a) Since x is even, extends to be continuous as a periodic function with period π, we ll have for π < x < π where x a n f, g a n cos nx n x, cos nx cos nx, cos nx π f(x)g(x) dx. (Using an integral from to π will double both the numerator the denominator in a n, so won t affect its value). Thus a n π x cos nx dx ( )n π n for n > So a π π x π 3 + n x dx π 3. ( ) n n cos nx. (b) The series does converge uniformly (by the theorem since the function is continous piecewise differentiable, or by the Weierstrass M-test. (c) Plug in x π x to get the results in this part. Add the two together divide by to get the result in (d).. Solve the initial-boundary value problem for the wave equation: u tt c u xx, < x <, t > where u(x, ) sin πx, u t (x, ), u(, t), u(, t). The wrinkle in this problem is the inhomogeneous boundary conditions u(, t). So we begin by writing u(x, t) v(x, t) + w(x, t), where v takes care of the inhomogeneous boundary condition, u picks up the slack. The simplest function that satisfies the wave equation together with v(, t) v(, t) is v(x, t) x. So we let w(x, t) u(x, t) x. Then w will solve the problem: w tt c w xx together with the initial condition w(x, ) sin πx x the boundary conditions w(, t) w(, t). By the usual separation of variables shtick, we have w(x, t) b n cos(cnπt) sin(nπx), n where b n sin πx x, sin nπx sin nπx, sin nπx

5 5 f, g f(x)g(x) dx. Thus + ( )n b n (sin πx x) sin nπx dx nπ ( ) n nπ Putting it all together, we get that if n if n > u(x, t) x + cos(cπt) sin(πx) + n ( ) n nπ cos(cnπt) sin(nπx). 5. (a) Find the eigenvalues eigenfunctions of the boundary-value problem: for u(x) defined on the interval,3. u + λu, u(), u (3) + u(3) (b) If we number the eigenvalues in increasing order, so that λ < λ < λ 3 <..., find A B so that ( lim λn (An + B) ). n (a) First check to see if λ could be negative: If λ <, because u(), we d have u(x) c sinh λ x. But then ( λ ) u (3) + u(3) c cosh(3 λ) + sinh(3 λ) this can never be zero, since cosh x > sinh x > if x >. Likewise, λ can t be zero, since no linear function other than u satisfies the conditions. Thus λ is positive u(x) c sin λ x (because u() ). To find the precise eigenvalues, we need to solve for λ: u (3) + u(3) λ cos 3 λ + sin λ. Rewrite this as tan 3 λ λ. Graphing both sides, with λ on the horizontal axis the values of the two sides on the vertical, we see that there are infinitely many solutions, places where the line y λ crosses the graph of y tan 3 λ, one for each branch of the tangent function. This gives us the eigenvalues, the corresponding eigenfunctions are sin λ x. (b) As λ gets larger larger, the places where the graph of y λ crosses the graph of y tan 3 λ get closer closer to the vertical asymptotes of the tangent function, which occur for 3 (n + )π λ.

6 6 In other words, the larger n is, the closer λ n is to ( ) (n + )π. 6 So A π 3 B π This problem shouldn t require any integration, but (b) especially (c) will require some thinking. (a) Solve the Laplace equation u rr + u r r + u θθ r on the inside of the disk r < with boundary condition for < θ < π. u(, θ) sin(3θ) (b) Solve the Laplace equation on the outside of the circle r (that is, for r > ) with boundary condition u(, θ) sin(θ). Assume we want the solution to remain bounded as r. How does this change the form of the solution? (c) Solve the Laplace equation in the annulus inside the circle r but outside the circle r, i.e., for < r < with boundary conditions u(, θ) sin(θ) u(, θ) sin(3θ). Since there is neither a condition at r nor at infinity, both parts of R(r) in the separated solutions come into play. (a) On the inside of the disk, the solution is u(r, θ) a + ) n (an cos nθ + b n sin nθ). n To match the boundary condition when r, we all the a n s to be zero as well as all the b n s except for n 3, b 3. So ) 3 u(r, θ) sin 3θ r 3 sin 3θ. (b) On the outside of the disk, the solution is u(r, θ) a + n ( ) n (a n cos nθ + b n sin nθ). r

7 7 This time, to match the boundary conditions we just need b the rest zero. So u(r, θ) sin θ r. (c) For this one, you might think we just need to add the previous two solutions, but that wouldn t be right, since neither solution is zero on the other s boundary. But now, we have both the interior exterior parts of the solution, so u is of the form: u(r, θ) a + ) n (an cos nθ + b n sin nθ) + n ( ) n (c n cos nθ + d n sin nθ). r Given all the data, the only non-zero coefficients will be b, b 3, d d 3. So we have u(r, θ) b When r, the boundary data says: ) sin θ + b3 ) 3 sin 3θ + d sin θ r + d 3 sin 3θ r 3 sin θ b sin θ + b 3 sin 3θ + d sin θ + d 3 sin 3θ which tells us that b + d b 3 + d 3. Likewise, when r, we have sin 3θ b sin θ + b 3 sin 3θ + d sin θ + d 3 sin 3θ which tells us that b + d b 3 + d 3. So there is a -by- system for b d, another for b 3 d 3. We have b d b3 d Therefore u(r, θ) ) 5 ) 3 6 sin θ sin θ + sin 3θ r 6 (sorry about the numbers!) sin 3θ r 3 7. For the exam, be sure you know the integrals: cos ax cos bx dx, sin ax cos bx dx.