Partial Solutions to 18.03 Problem Set 5

Part I. 20. (c) The general solution to y′′ +2y′ +2y = 0 is of the form y = e^{−x}(A cos x+ B sin x). If y(0) = 0, then A = 0, so y(x) = e^{−x}B sin x. If furthermore x_{0} is not a multiple of π (so sin x_{0} ≠ 0), then if B = e^{x_{0}}/ sin x_{0}, then y(x_{0}) = 1. But if x_{0} is a multiple of 2π, no such B exists. Hence the condition is “x_{0} is not a multiple of π”.

Part II.

20. (a) (All derivatives in this solution will be with respect to t.) Let’s look for a solution of the form y = ca(αt) for some α. We know that ca(t) is a solution of x′′ + tx = 0, so ca′′(t) = −tca(t). Then y′ = αca′(αt), and y′′ = α^{2}ca′′(αt). But ca′′(z) = −zca(z), so taking z = αt, ca′′(αt) = −αtca(αt). Hence y′′ = −α^{3}tca(αt). If y′′ + tλy = 0, then −α^{3}tca(αt) + tλca(αt) = 0. This is true if α^{3} = λ, i.e. α = ^{3} \sqrt{λ}, so y = ca( ^{3} \sqrt{λ}t). Note that y(0) = 1 and y′(0) = 0.

Similarly, y = sa(αt) = sa( ^{3} \sqrt{λ}t) is a solution; in this case, y(0) = 0, y′(1) = ^{3} \sqrt{λ}.

Hence the normalized solutions are y_{1}(t) = ca( ^{3} \sqrt{λ}t), y_{2}(t) = \frac{1}{\sqrt[3]{λ}}sa( ^{3} \sqrt{λ}t).

(b) As before, y_{1}(t) = ca( ^{3} \sqrt{λ}t). From the Airy handout, ca(t) ∼ at^{−1/4} cos (\frac{2t^{3/2}}{3} − \frac{π}{12}) asymptotically, so

ca( ^{3} \sqrt{λ}t) ∼ a( ^{3} \sqrt{λ}t)^{−1/4} cos (\frac{2( ^{3} \sqrt{λ}t)^{3/2}}{3} − \frac{π}{12}) .

This is 0 when ^{\frac{2( ^{3} \sqrt{λ}t)^{3/2}}{3}} − \frac{π}{12} = (n + 1/2)π where n is some integer, i.e.

t = \frac{1}{\sqrt[3]{λ}} (\frac{3}{2} (n + \frac{7}{12}) π)^{2/3} .

We can do the same thing with y_{2}(t) = \frac{1}{\sqrt[3]{λ}}sa( ^{3} \sqrt{λ}t), which has zeros roughly when

\frac{2( ^{3} \sqrt{λ}t)^{3/2}}{3} + \frac{π}{12} = nπ

(using the Airy handout again), which leads to zeros when t = \frac{1}{\sqrt[3]{λ}} (\frac{3}{2} (n − \frac{1}{12}) π)^{2/3} .

21. (a) R′′ − aR′ − bcR = 0. Since x′′ + px′ + qx = 0 is stable exactly when p, q > 0, stability occurs precisely when a < 0 and bc < 0. The equation is underdamped if p^{2} < 4q, i.e. |a|^{2} < 4|bc|. The roots of r^{2} − ar − bc = 0 are of the form s ± ωi for s = a/2, ω = \sqrt{|bc| − (a/2)^{2}}. As |a| grows towards 2\sqrt{|bc|}, the frequency ω shrinks to 0. As |bc| increases, ω increases.

(b) R = R(0)e^{at/2} cos(ωt), so J = (R′ − aR)/b = −\frac{1}{b}R(0)e^{at/2}(\frac{a}{2} cos(ωt) + ω sin(ωt)). Consider a right triangle with sides a/2 and ω and hypotenuse r, with angle ϕ opposite side ω, so a/2 = r cos ϕ and ω = r sin ϕ. Then r = \sqrt{(a/2)^{2} + ω^{2}} = \sqrt{|bc|}. Thus J(t) = −\frac{1}{b}R(0)r cos(ωt − ϕ).

ρ = \frac{−\frac{1}{b}R(0)r}{R(0)} = −r/b = ±\sqrt{|c/b|}.

(c) If |a|^{2} > 4|bc| the roots are real; by stability the solutions fall to 0 without oscillation.

(d) If R and J are constant, then R′ = J′ = 0, so we are to solve aR + bJ = −r_{1}, cR = −r_{2}. Since

det ( \begin{matrix}a & b \\ c & d \end{matrix} ) = −bc ≠ 0

this has a unique solution.

22. (a)

[ \begin{matrix}R \\ J \end{matrix} ]′ = [ \begin{matrix}a & b \\ c & 0 \end{matrix} ] [ \begin{matrix}R \\ J \end{matrix} ] .

The characteristic polynomial of [ \begin{matrix}a & b \\ c & 0 \end{matrix} ] is λ^{2}−aλ−bc with eigenvalues \frac{a}{2}±\sqrt{(a/2)^{2} + bc}.

(Look familiar?)

(b) A = [ \begin{matrix}−3 & −1 \\ 2 & 0 \end{matrix} ]. f_{A}(λ) = λ^{2} + 3λ + 2 = (λ + 1)(λ + 2).

With eigenvalue λ_{1} = −1, A − λ_{1}I = [ \begin{matrix}−2 & −1 \\ 2 & 1 \end{matrix} ], and α⃗_{1} = [ \begin{matrix}1 \\ −2 \end{matrix} ] (or any nonzero

multiple thereof).

With eigenvalue λ_{2} = −2, A − λ_{2}I = [ \begin{matrix}−1 & −1 \\ 2 & 2 \end{matrix} ], and α⃗_{2} = [ \begin{matrix}1 \\ −1 \end{matrix} ] (or any nonzero

multiple thereof).

The normal modes are e^{−t} [ \begin{matrix}1 \\ −2 \end{matrix} ] and e^{−2t} [ \begin{matrix}1 \\ −1 \end{matrix} ].

The general solution is c_{1}e^{−t} [ \begin{matrix}1 \\ −2 \end{matrix} ] + c_{2}e^{−2t} [ \begin{matrix}1 \\ −1 \end{matrix} ].

(c) [ \begin{matrix}−1 \\ 0 \end{matrix} ] = x⃗(0) = c_{1} [ \begin{matrix}1 \\ −2 \end{matrix} ] + c_{2} [ \begin{matrix}1 \\ −1 \end{matrix} ]. c_{1} = 1, c_{2} = −2, so x⃗ = e^{−t} [ \begin{matrix}1 \\ −2 \end{matrix} ] −

2e^{−2t} [ \begin{matrix}1 \\ −1 \end{matrix} ]. Hence R = e^{−t} − 2e^{−2t}, J = −2e^{−t} + 2e^{−2t}.

Romeo comes to like Juliet when R(t) = 0 i.e. e^{−t} = 2e^{−2t}, i.e. t = ln(2). Thereafter Romeo likes Juliet. Romeo is fondest when 0 = R′(t) = −e^{−t} + 4e^{−2t}, i.e. t = 2 ln(2). For t > 0, e^{−t} > e^{−2t}, so J < 0 always. Graphs: