MATH 3150: PDE FOR ENGINEERS FINAL EXAM (VERSION A)

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1 MAH 35: PDE FOR ENGINEERS FINAL EXAM VERSION A). Draw the graph of 2. y = tan x labelling all asymptotes and zeros. Include at least 3 periods in your graph. What is the period of tan x? See figure. Asymptotes are at π/2 + πn for any integer n, and zeros at πn for any integers n. fx) = { x 2 x < π x 2 π x < and fx) has period 2π. a) Draw the graph of fx).include at least two periods in your graph. Date: December 4, 2. 2 y x 2 Figure. Graph of the tangent function

2 2 MAH 35: PDE FOR ENGINEERS FINAL EXAM VERSION A) b) Calculate the real Fourier amplitudes a m and b m of fx). Hint: x 2 sin x dx = x 2 cos x + 2 cos x + 2x sin x. c) Find the energy of fx). d) In the form of a fraction, find the total percentage of energy contained among the amplitudes a, a, b, a 2, b 2. Do not use any decimal approximations. e) Show that about 6% of the energy is stored among the amplitudes a, a, b, a 2, b 2. Hint: π 2 π 4 π 6. So collect all terms over a common denominator of π 6. All a m = because is it odd. So b m = 2 L = 2 π = 2 π L π mπ = 2 πm 3 fx) sin πmx ) dx L x 2 sin mx) dx mπ u/m) 2 sin u du/m) u 2 sin u du = 2 u 2 πm 3 cos u + 2 cos u + 2u cos u ) u=mπ u= = 2 [ mπ) 2 πm 3 ) m + 2 ) m] [2] ) = 2 )m+ π m 4 + ) m+) πm 3. b = 2π 8 π b 2 = π. he energy in these two is b b 2 ) 2 = π 5π ) π 2 he total energy in the function is π π fx) 2 dx = 2 π x 4 dx = 2π5 5.

3 MAH 35: PDE FOR ENGINEERS FINAL EXAM VERSION A) y.5.5 x.5 Figure 2. A round peak he percentage of energy in these two amplitudes is π ) 5π π 2 2π 5 = 5 5π 4 32π 2 64 /5 2 π = 6. So 6 per cent. 3. Suppose that ux, y, t) satisfies t = u 2 u 2 ) 3 ) ) 2 x 2 + u y 2 exp + x ) ) 4 + y which is so horrible an equation that no one can solve it). Suppose that the function u at time t = looks like a round peak as in figure 2. Does the top of the peak move up or down? Explain. It goes up, because at the top of the peak so that u > x = y = 2 u x 2 < 2 u y 2 < t = )+) ) + )) e + = +). 4. Solve the heat equation t = c2 2 u x 2

4 4 MAH 35: PDE FOR ENGINEERS FINAL EXAM VERSION A) x Figure 3. A circular ring, with x axis wound around it in a circular ring of radius r see figure 3) by finding a kernel Kx, t) so that ux, t) = 2πr us, )Kx s, t) ds. Note that ux, t) has period 2πr in x, so you can use complex Fourier series not Fourier transforms, since it isn t infinitely long). Writing ux, t) = k a k t)e ikx/r we find that the heat equation becomes ) 2 da k cik = a k dt r so that a k t) = exp ) 2 ck t) a k ) r and where ux, t) = k = 2πr exp Kx, t) = 2πr ) 2 ck t) a k )e ikx/r r us, )Kx s, t) ds e ck/r)2t+ikx/r. k 5. Find the Fourier transform of the function { if x fx) = otherwise

5 MAH 35: PDE FOR ENGINEERS FINAL EXAM VERSION A) 5 ˆfω) = 2π You might not recognize this: it is e iωx dx = ) e iω 2π iω eiω iω = e iω e iω ). 2π iω ˆfω) = 2 sinω) π ω. 6. Show that according to the heat equation, heat propagates with infinite speed. o do this, take initial temperature ux, ) which is positive near x =, and not negative anywhere, and zero everywhere except in a little region near x =. Now apply the heat kernel in the equation ux, t) = Gx, t) = e x2 /4c 2 t 2c πt us, )Gx s, t) ds giving final temperature at time t. How does this show you that a small amount of heat spreads out to all points of space at as small a time as you like? Because the heat kernel is positive everywhere, and the function ux, ) is positive somewhere, negative nowhere, the function us, )Gx s, t) is positive somewhere, negative nowhere, so has positive integral over s, for any value of x and t. herefore ux, t) is positive everywhere, no matter how small t is. So the temperature is suddenly positive everywhere. 7. ake a disk of unit radius and heat the top half of the edge of the disk to o and the bottom half of the edge to o. Let the disk sit with these temperatures on its edges for a long time, until it reaches a steady state. Center the disk at origin of coordinates. ry to estimate without making any decimal approximations just use fractions) the temperature at the point x, y) =, π ).

6 6 MAH 35: PDE FOR ENGINEERS FINAL EXAM VERSION A) Hint: recall that if we write the temperature of the edge of the disk as fθ), and write the Fourier amplitudes of fθ) as a = 2π a m = π b m = π 2π 2π 2π fθ) dθ fθ) cosmθ) dθ fθ) sinmθ) dθ then the steady state temperature inside the plate, in polar coordinates, is ur, θ) = a + r m a m cos mθ) + b m sin mθ)) m= We calculate: so that u = 2 + = π a = 2 a m = b m = + )m+ πm m= r m + )m+ πm k= sin mθ) r 2k+ sin 2k + ) θ) 2k + On the positive y axis, we find θ = π/2, so that and therefore sin 2k + ) π) = ) k u = π ) r r3 3 + r In particular, at r = π/, we find u = ) π π π2 3 + so u = = 7 8. Suppose that I have a function y = fx) which is periodic with period. Suppose that it has energy Ef) = f 2. If I rescale the x variable by an amount α, making a new variable X = αx and a new function F X) = fx), a) what is the period of F X)? and

7 MAH 35: PDE FOR ENGINEERS FINAL EXAM VERSION A) 7 b) by what amount do I have to rescale the y variable to keep the same energy as f? a) he new period is α. b) E αf ) = Ef). 9. Suppose that fx) has period and average value. a) Using complex Fourier series, show that the energy in the derivative df/dx satisfies 2 ) df 2 2π dx f 2. b) Which functions fx) satisfy 2 ) df 2 2π dx = f 2? a) If fx) = a k e 2πikx/ k then df dx = ) 2πik a k e 2πikx/ k so that the energy of df/dx is df dx while the energy of f is 2 = k a k 2 2πk ) 2 f 2 = k a k 2. Clearly each term in the first sum is at least 2π/ ) 2 larger than in the second, except the a term. But the a term is the average, which is zero. So we have the required inequality 2 ) df 2 2π dx f 2. b) Equality can only occur in the a and a terms, because all of the others have a factor of ) 2 ) 2 2πk 2π >. So fx) = a e 2πix/ + a e 2πix/.. Write down the general solution of the heat equation for temperature ux, t) in a wire of length L with left end at o and insulated right end. Explain how you got your answer. You don t have to explain how to dig out the amplitudes at time t =.)

8 8 MAH 35: PDE FOR ENGINEERS FINAL EXAM VERSION A) where he final result is ux, t) = n λ n = π L a n e c2 λ 2 n t sin λ n x) n + ) 2