MATH 41 Practice Second Midterm Exam - Fall 1 1. Let f(x = { 1 x for x 1 for 1 x (a Compute the Fourier sine series of f(x. The Fourier sine series is b n sin where b n = f(x sin dx = 1 = (1 x cos = 4 1 n π sin if n is even = 4 n π if n = 4k + 1 + 4 rm if n = 4k + 3 n π (1 x sin dx cos dx (b Draw a careful graph of the function to which your series converges for 6 x 6. (c Compute the Fourier cosine series of f(x. The Fourier cosine series is a n cos n=
where and a = 1 a n = f(x dx = 1 f(x cos dx = 1 = (1 x sin 1 = 4 n π cos 4 if n is odd n π = 8 if n = 4k + n π rm if n = 4k 1 x dx = 1 + (x x 1 = 1 4 (1 x cos dx sin dx (d Draw a careful graph of the function to which your series converges for 6 x 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 4L.
We have to show that f(x + 4L = f(x for all x. Well, f(x + 4L = f(l + (x + 3L getting ready to use the second condition = f(l (x + 3L by the second condition = f( x L duh = f(x + L by the first condition = f(l + (x + L getting ready to use the second condition again = f(l (x + L by the second condition = f( x = f(x by the first condition (b Draw the graph of a non-zero oddly odd function for L x L (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? In this figure, L = 1. The function is even around x = L (i.e., f(l + x = f(l x and odd around x = L (i.e., f(l + x = f(l x. (c Show that the Fourier series of an oddly odd function is of the form f(x = Give a formula for the coefficients b n. b n sin (n 1πx. L Since we know the function is odd, we expect only sines. And since it has period 4L, we know that we only have to use its values on the interval x L, and the function should have a series expansion of the form f(x = b n sin L where b n = 1 L L f(x sin dx L
But since f(l x = f(x (from our symmetry observation in part (b and the middle of our chain of equalities in part (a, we can change variables (let u = L x so du = dx etc and get L L f(x sin dx = L = L L = ( (L u f(l u sin du L ( f(u sin u du L ( u f(u( 1 n sin du L L Therefore b n = 1 L L f(x sin (xl dx = 1 ( L L f(x sin dx + f(x sin L L L L = 1 ( L L f(x sin dx ( 1 n f(x sin L L = 1 L ( ( 1n x f(x sin dx L L So we have b n = if n is even, and if n is odd, say n = k 1, b k 1 = L ( (k 1πx f(x sin dx L L dx dx L and so with these coefficients. f(x = ( (k 1πx b k 1 sin L k=1 3. Solve the initial-boundary value problem for the wave equation: u tt = 4u xx, < x < 1, t > where u(x, = sin πx, u t (x, =, u(, t =, u(1, t = 1. (Hint: What is the equilibrium solution? We have to use the equilibrium solution u eq (x in order to satisfy the non-homogeneous boundary condition at x = 1. Since the equilibrium solution has u tt =, it will be a linear function in x satisfying u eq ( = and u eq (1 = 1, so u eq (x = x.
Now let v(x, t = u(x, t u eq = u(x, t x. Then v satisfies the same differential equation as u, so v tt = 4v xx, but the boundary conditions for v are v(, t = v(1, t =, and the initial conditions for v will be v(x, = sin πx x and v t (x, =. By the usual separation of variables, let v(x, t = X(xT (t, then we ll have T 4T = X X = λ Since X( = X(1 =, we will have X = sin x with n = 1,, 3,... and λ = n π. Then T + 4n π T = so T = a n cos t + b n sin t. Since v t (x, = we know that there are no sin t terms in the answer. We thus have v(x, t = a n sin x cos t where, since v(x, = sin πx x, a 1 = 1 x sin πx dx and a n = Since x sin x dx = 1 x cos x we have and so v(x, t = sin πx cos πt + 1 + ( 1 n u(x, t = u eq (x, t + v(x, t = x + sin πx cos πt + x sin x dx for n. cos x dx = ( 1n+1, sin x cos t ( 1 n sin x cos t 4. (a Find the Fourier-Bessel series of the form a m J (z m r for the function f(r = 1 for r 1. From the text or the notes, we know that the coefficients will be of the form a m = rj (z m r dr. r(j (z m r dr
From the bottom of page of the notes on the wave equation on the disk we recall that x(j m (z nm x dx = 1 J m+1(z nm, so the denominator is 1 J 1(z m. As for the numerator, formula (1 from those notes, with n = 1, says d dx (xj 1(x = xj (x. Therefore xj (x dx = xj 1 (x + C so (making the substitution s = z m r rj (z m r dr = zm s J (s ds = 1 zm sj z m z m zm (s ds = J 1(z m. z m So we conclude that on the interval < r < 1. 1 = z m J 1 (z m J (z m r (b Use this series to solve the initial-boundary value problem for the heat equation: u t = 3 u on the unit disk (x + y 1, or equivalently r 1 with initial condition u(r, θ, = 1 for r 1, θ π and zero boundary values, u(1, θ, t = for θ π, t >. Since neither the initial nor the boundary condition contain θ, we know that the solution will be independent of θ. So we can assume u(r, θ, t = u(r, t and do u = R(rT (t in the separation of variables. We obtain T 3T = (rr rr = λ, so T + 3λT = (which will give us T = e 3λt and rr + R + λrr =. Multiplying the last equation by r gives us Bessel s equation of order zero: r R + rr + λr R =
. Therefore R = J (z m r and λ = z m. We conclude that u(r, t = where by the initial condition we need u(, t = a m e 3z m t J (z m r a m J (z m r = 1. But we found these coefficients in part (a, so the solution to the problem is u(r, t = z m J 1 (z m e 3z m t J (z m r.. Find the steady-state temperature distribution on the solid cylinder with radius 3 (so r 3 and height (so z if the temperature at the ends is held at zero and the temperature on the lateral side is given by u(3, θ, z = z( z sin 3θ. This is the case of Part III in the notes on the Laplacian on the cylinder. Since the top and bottom are held at zero temperature, we have that u(r, θ, z is what we called there u 3 (r, θ, z. So with H =, so u(r, θ, z = u(r, θ, z = n= n= ( mπr I n sin H ( mπr I n sin n= [e nm cos nθ + f nm sin nθ], H We have to choose the coefficients e nm and f nm so that u(3, θ, z = z( z sin 3θ = sin I n ( 3mπ [e nm cos nθ + f nm sin nθ], [e nm cos nθ + f nm sin nθ], Since the boundary data is a multiple of sin 3θ, we know immediately that e mn = for all m and n and f mn = for all m and all n except n = 3. So we can simplify things to finding the coefficients f m3 so that ( 3mπ z( z = f m3 I 3 sin
for z. This is an ordinary Fourier sine series, so we know that f m3 = 1 I 3 ( 3mπ Work the integral by parts as follows: (z z sin dz = mπ (z z cos z( z sin dz. = m π ( z sin = m 3 π cos 3 = (1 ( 1m m 3 π 3 + mπ m π ( z cos dz ( sin dz which is zero if m is even and /(m 3 π 3 if m is odd. So let m = k + 1 and get f (k+13 = I 3 ( 3(k + 1π (k + 1 3 π. 3 So the solution of the problem is u(r, θ, z = k= (k + 1 3 π 3 I 3 ( 3(k + 1π ( (k + 1πr I 3 sin ( (k + 1πz sin 3θ 6. Find the first three non-zero terms of the power series (centered at zero for the solution of the equation x d y dx (x + y = that is bounded at x =. Does the solution oscillate (like sines and cosines and J n (x or does it increase steadily to infinity (like e x and hyperbolic sines and cosines and I n (x? The point x = is a regular singular point for this equation, so we must assume that y = a n x n+s n=
with a and discover what values of s work. If we substitute this assumption into the differential equation, we get: (n + s(n + s 1a n x n+s a n x n+s+ a n x n+s =. n= n= Put m = n + in the middle term (so n = m and m will go from to. Then we can rewrite the whole thing as a single sum from to with the n = and n = 1 terms from the first and third sums separated out as follows: [s(s 1 ]a x s +[(s+1s ]a 1 x 1+s + [((n + s(n + s 1 a n a n ] x n+s = n= In order for a we must have s(s 1 = s s = (s (s + 1 =, which gives s = or s = 1. The solution with s = 1 will start with 1/x and so cannot be bounded at x =, so we look only at the s = solution, which will start with x. Next, the x 1+s term has coefficient [(s + 1s ]a 1, which is 4a 1 for s =. This tells us that a 1 =. And from the summation, we get the recurrence relation (using s = straightaway [(n + (n + 1 ]a n a n = or a n = a n n(n + 3. Since a 1 =, we have a n = for all odd values of n. For the first few even values of n we have a = a a 4 = a 4 7 = a ( 4( 7 a 6 = a 4 6 9 = a ( 4 6( 7 9 If you think about it long enough, you can see that a n = 6(n + 1!a n!(n + 3! In any case, the first three non-zero terms are: x + 1 1 x4 + 1 8 x6 + This function is always positive and increasing and concave up for x > because all the coefficients in its Maclaurin series are positive (and by the ratio test, the series converges for all x. n=