homework 14 solutions

# homework 14 solutions - Section 9.7 4 Because X ’(0 = X(a...

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Unformatted text preview: Section 9.7 4. Because X ’(0) = X '(a) = 0, we work with the separation of variables _ X"+/1X= 0 = Y"—2,Y. The eigenvalue problem X”+AX= 0, X’(0)=X'(a)=0 has eigenvalues and eigenfunctions 20 = 0, X0(x) = 1 and A" = "7: , Xn(x) = cos—mm a a for n = 1,2, 3, .When n = O, Yo"= 0 yields Yo(y) = Ay+B. Then Yo(0) = 0 gives B = 0, so we take Yo(y) = y. For n > O we have Yn(y)= A cosh——-— mry +B” sinh——- mry, a a and Yn(0) = 0 gives A" = 0, so u(x, y)= Boy + 2B cosmsinh— ”ﬂy. n= 1 a a Finally u(x, b) = Bob + 2 B" cosmsinhﬂ, "=1 a a so we satisfy the condition u(x, b) = ﬁx) by taking Bo = ao/2b and CD B =a {Sinhﬂ} where f(x)= ~29—+ a cosy—72‘: a ":1 a 8. The eigenvalue problem X”+2X = O, X’(0) = X’(a) = yields 20 = 0, X0(x) = 1 and A" = "72’ , Xn(x) = cosﬂ a a for n > 0. Then Y”+/1Y=0 II II yields Yo(y) = Aqy+Bo and Y"(y) = Anemry/u +Bne—mry/a. In order that Y(y) be bounded as y—>oo, we take A0 = 0 and A" = O for n > 0, so u(x,y)= B0 + EB" e'm’y’” cos nrrx "=1 a Finally we satisfy the condition u(x, O) = ﬁx) by choosing Bo = a0/ 2 and B" - where the {an} are the Fourier cosine coefﬁcients of ﬁx) on [0, a]. 13. We start with the periodic polar-coordinate solution u(r, 0) = £33 + Zr" (an cos n0 + b" sin n0) "=1 and choose a" _=. 0 in order to satisfy the conditions u(r, 0) = u(r,7r) = 0. Then u(r, 0) = Zr"c,, sin n0 "=1 satisﬁes the nonhomogeneous boundary condition u(a, 0) = f (0) provided that a"c,, is is the nth Fourier sine coefﬁcient of f (0) on the interval 0 < 0< 71', that is, c =i f(0)sinn0d0. u ”an 16. The only difference between the exterior problem here and the interior problem in the text is that in we must choose Cn = 0 in order that Rn(r) be bounded as r -) oo. 22- With a slight abuse of notation, we say that a function 6) has period T if T is any positive number such that @(0 + T) = (9(0) for all 0. (a) We wish to ﬁnd the a such that (9(0) = Acos(a0) + B sin(a0) has period 271’. As shown on page 185, there exists some C and 6 such that 9(0) : Ccos(a0 -— 6). We thus need to have @(0 + 271”) = @(0). That is7 Ccos(a(0 + 27r) —— 6) = Ccos(a0 - 6), so Ccos(a0 — 6 + 20m) = Ccos(oz0 —— 6). Since the periods of cos are precisely 27771 with n a positive integer, we need a to be a positive integer. (b) We wish to ﬁnd the A and B such that (9(0) = A0 + B is periodic. Let T > 0 be a period of (9. That is, @(0 + T) = 6(0), so A(0 + T) + B = A0 + B. Therefore, AT : 0, which forces A z 0. 23' We wish to ﬁnd the A and B such that 9(0) = Aeo‘e + 38“” is periodic. Let T > 0 be a period of 6). That is, (9(0 + T) = 8(0), so Ae"(9+T) + Be""(9+T) = A609 + Be"°‘”. Taking 0 = 1, we get Ae(e°‘T — 1) + B6_1(€_0T — 1): 0. Taking 0 = —1, we get Ae—1(e“"T — 1) + Be(e“’T — 1) z 0. Since 6 eaT — e- c—aT — 2 _ an _a~ _ elk” 3i) 62;“ _ 1)” = (6 ~ 8 2)(1 — e. 1 — e. 1 + 1) = 2(e2 — e 2)(1 — cosh(aT)), we conclude that the only way to avoid having A = B = 0 is to have 1 — cosh(aT) = 0, which means that (1 = 0. ‘ ...
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