Math 316/22: Solutions to Assignment 7 1.8.6(a) Using separation of variables, we write u(r, θ) = R(r)Θ(θ), where Θ() = Θ(π) =. The Laplace equation in polar coordinates (equation 19) becomes R Θ + 1 r R Θ + 1 r 2 RΘ =, so r2 R + rr + Θ =. The second term must be constant. The R Θ boundary value problem Θ + λθ =, Θ() = Θ(π) = has solutions Θ(θ) = sin(nθ) for positive integers n, with λ = n 2. Then the Euler equation r 2 R + rr n 2 R = has solutions R = c 1 r n + c 2 r n. We must reject r n because u(r, θ) must be bounded as r. So our basic solutions are u n (r, θ) = r n sin(nθ), and we look for a series solution of the form u(r, θ) = a n r n sin(nθ). The remaining boundary condition u(a, θ) = f(θ) is satisfied if a n a n sin(nθ) is the Fourier sine series of f(θ) on the interval [, π]. Thus a n = 2 π f(θ) sin(nθ) dθ. πa n (b) With f(θ) = θ(π θ) we have a n = 2 πa n ( θ cos (nθ) ( π + θ) n sin (nθ) ( π + 2 θ) n 2 2 cos (nθ) n 3 ) π = 4(1 ( 1)n ) πa n n 3 i.e. 8/(πa n n 3 ) if n is odd and if n is even. So u(r, θ) = 8 r π a sin θ + 8 ( ) r 3 sin 3θ + ( ) 3 3 π a 8 r 5 sin 5θ +.... 5 3 π a (c) Here is u versus r for θ = π/8, π/4, 3π/8 and π/2 in red, green, blue and magenta respectively. Note that u(π θ) = u(θ). This was produced by the following Maple code: u:= add(8/pi/(2*n-1)^3*(r/2)^(2*n-1)*sin((2*n-1)*theta),..4): plot([seq(u,theta=[pi/8,pi/4,3*pi/8,pi/2])], r=..2, colour=[red,green,blue,magenta],labels=[r, u ]); 1
Here is u versus θ for r = 1/2, 1, 3/2 and 2 in red, green, blue and magenta respectively. This was produced by the following Maple code: plot([seq(u,r=[1/2,1,3/2,2])], theta=..pi, colour=[red,green,blue,magenta],labels=[theta, u ]); 2
Here is a plot of the three-dimensional surface z = u(r, θ). It was produced by the following Maple code: plot3d([r*cos(theta),r*sin(theta),u], r=..2, theta=..pi, axes=box, scaling=constrained, orientation=[-45,5], shading=zhue); 3
Here is a contour plot. It was produced by the following Maple code: plots[contourplot]([r*cos(theta),r*sin(theta),u], r=..2, theta=..pi, axes=frame, scaling=constrained, labels=[, ], view=[-2..2,..2]); 4
1.8.1(a) We use separation of variables with u(x, y) = X(x)Y (y), X () = Y () = Y (b) =. We must have X Y + XY = so X = Y is constant. As we ve seen X Y in the heat equation, the equation Y + λy = with Y () = Y (b) = has solutions Y n (y) = cos(nπy/b) for nonnegative integers n, with λ n = (nπ/b) 2. Then the equation X (nπ/b) 2 X = has solutions X(x) = c 1 exp(nπx/b) + c 2 exp( nπx/b). From the boundary condition X () = (nπ/b)(c 1 c 2 ) =, so c 1 = c 2. We can take c 1 = c 2 = 1/2, so that X(x) = cosh(nπx/b). Thus we have the fundamental set of solutions u n (x, y) = cosh(nπx/b) cos(nπy/b) for nonnegative integers n, where u (x, y) = 1. (b) Writing the formal series solution ( ) ( ) nπx nπy u(x, y) = c n cosh cos n= b b we have ( ) ( ) nπ nπa nπy u x (a, y) = c n b sinh cos = f(y) b b which must be the Fourier cosine series of f on the interval [, b], except that there is b no constant term. Thus we must have f(y) dy =, and for n 1 c n = 2 ( ) nπa 1 b ( ) nπy nπ sinh f(y) cos dy b b 5
while c is arbitrary (you can always add a constant to a solution and it remains a solution). 1.8.11 The separation of variables is similar to 1.8.6, except that the boundary conditions on Θ are Θ(2π) = Θ() and Θ (2π) = Θ () (periodic boundary conditions). Solutions of Θ + λθ = with these boundary conditions are Θ = 1 for λ = and Θ = cos(nθ) and Θ = sin(nθ) for integers n 1. Again from the R equation we take R(r) = r n. Thus we have the formal series solution We then need u(r, θ) = a + r n (a n cos(nθ) + b n sin(nθ)) u r (a, θ) = na n 1 (a n cos(nθ) + b n sin(nθ)) = g(θ) This must be the full Fourier series of g(θ) on the interval [, 2π]. The constant term must be, i.e. a n = 2π while a is arbitrary. g(θ) dθ =, and for n 1 1 2π g(θ) cos(nθ) dθ and b nπa n 1 n = 1 2π g(θ) sin(nθ) dθ nπa n 1 11.1.9 For λ =, y = has solutions y = c 1 + c 2 x; y() y () = c 1 c 2 = and y(1) + y (1) = c 1 + 2c 2 = imply c 1 = c 2 =, so is not an eigenvalue. If λ = σ 2 < with σ >, the solutions of the differential equation y σ 2 y = are y = c 1 e σx + c 2 e σx ; y() y () = c 1 + c 2 σ(c1 c 2 ) = (1 σ)c 1 + (1 + σ)c 2 = y(1) + y (1) = c 1 e σ + c 2 e σ + σ(c 1 e σ c 2 e σ ) = (1 + σ)e σ c 1 + (1 σ)e σ c 2 = The determinant of the coefficient matrix of this system of two equations in two unknowns is (1 σ) 2 e σ (1 + σ) 2 e σ. To have an eigenvalue, this must be. But (1 + σ) 2 > (1 σ) 2 and e σ > e σ, so the determinant <, and we have no eigenvalue in this case. If λ = σ 2 > with σ >, the solutions of the differential equation are c 1 cos(σx) + c 2 sin(σx); y() y () = c 1 σc 2 y(1) + y (1) = c 1 cos σ + c 2 sin σ c 1 σ sin σ + c 2 σ cos σ = c 1 (cos σ σ sin σ) + c 2 (sin σ + σ cos σ) 6
The determinant of the coefficient matrix is (sin σ + σ cos σ) + σ(cos σ σ sin σ) = 2σ cos σ + (1 σ 2 ) sin σ This is if tan σ = 2σ/(1 σ 2 ). Graphically, we see that there is one solution in every interval nπ < σ < (n + 1/2)π for nonnegative integers n. The two smallest solutions are approximately σ 1 = 1.3654 and σ 2 = 3.67319, corresponding to λ 1 = σ 2 1 = 1.775 and λ 2 = σ 2 2 = 13.49236 respectively. Since 2σ/(1 σ 2 ) as σ, as n we have σ n nπ and λ n (nπ) 2. 11.1.16 XT + cxt + kxt = a 2 X T, so T + ct + kt T T + ct + kt = λt and a 2 X = λx. = a 2 X. The ODE s are X E.7 The temperature u in the meatball should depend only on ρ (the distance from ) the centre) and t (the time). As discussed in class, the PDE is u t = α 2 (u ρρ + 2 ρ u ρ The boundary conditions are u(ρ, ) =, u(1, t) = 1, and u should be continuous at ρ =. By taking v(ρ, t) = 1 u(ρ, t), we have v(ρ, ) = 1 and v(1, t) =, and v satisfies the same PDE as u. We get the series solution where v(ρ, t) = a n ρ sin(nπρ)e (nπα)2 t ρv(ρ, ) = 1ρ = a n sin(nπρ) The coefficients in this Fourier sine series are a n = 2( 1) n+1 /(nπ). Note that lim sin(nπρ)/ρ = ρ nπ, so v(, t) = 2( 1) n+1 e (nπα)2 t and we want to find t when this is 1 71 = 29, where α 2 =.38. Using 2 terms in the sum, and Maple s fsolve function, we find v(, t) = 29 when t 51.456 seconds. Actually, by this value of t if you take just one term of the series the temperature would be accurate to within less than.1 degree. If you solve 2 exp( (πα) 2 t) = 29 you get t = (πα) 2 ln(2/29) 51.4877 seconds, which is within about.8 seconds of the exact answer.. 7