Many oscillators, and the spectral theorem earning its keep

Idea. There are exactly three routes to \(e^{At}\) for a \(2\times2\) matrix, and which one is available is decided by the eigenvalues: distinct real ones allow diagonalisation, a pair \(\pm i\omega\) makes \(A^2\) a multiple of \(I\) so the series folds into sine and cosine, and a repeated defective one makes \(A - \lambda I\) nilpotent.

(a) \(A_a = \begin{pmatrix}3&1\\1&3\end{pmatrix}\). Symmetric, so the spectral theorem applies: \(\det(A_a-\mu I) = (3-\mu)^2-1\) gives \(\mu = 4, 2\) with orthonormal eigenvectors \(\tfrac1{\sqrt2}(1,1)\) and \(\tfrac1{\sqrt2}(1,-1)\). With \(Q = \tfrac1{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix} = Q^{\!\top} = Q^{-1}\), \[ e^{A_at} = Q\begin{pmatrix}e^{4t}&0\\0&e^{2t}\end{pmatrix}Q^{\!\top} = \frac12\begin{pmatrix}e^{4t}+e^{2t} & e^{4t}-e^{2t}\\ e^{4t}-e^{2t} & e^{4t}+e^{2t}\end{pmatrix}. \]

(b) \(A_b = \begin{pmatrix}0&-\omega\\ \omega&0\end{pmatrix}\). Here \(A_b^2 = -\omega^2I\), so the series splits into even and odd parts: \[ e^{A_bt} = \sum_{j}\frac{(-1)^j(\omega t)^{2j}}{(2j)!}I \;+\; \sum_{j}\frac{(-1)^j(\omega t)^{2j+1}}{(2j+1)!}\frac{A_b}{\omega} = \cos(\omega t)\,I + \frac{\sin(\omega t)}{\omega}A_b = \begin{pmatrix}\cos\omega t & -\sin\omega t\\ \sin\omega t & \cos\omega t\end{pmatrix}. \] No diagonalisation over \(\mathbb R\) is possible — the eigenvalues are \(\pm i\omega\) — and none was needed.

(c) \(A_c = \begin{pmatrix}2&1\\0&2\end{pmatrix}\). Write \(A_c = 2I + N\) with \(N = \begin{pmatrix}0&1\\0&0\end{pmatrix}\), \(N^2 = 0\). The two commute, so \[ e^{A_ct} = e^{2t}e^{Nt} = e^{2t}(I+Nt) = e^{2t}\begin{pmatrix}1&t\\0&1\end{pmatrix}. \] Diagonalisation is unavailable: \(\mu = 2\) twice with a one-dimensional eigenspace.

2. The derivative check. (a) Write \(p = e^{4t}+e^{2t}\) and \(q = e^{4t}-e^{2t}\), so \(e^{A_at} = \tfrac12\begin{pmatrix}p&q\\q&p\end{pmatrix}\). Differentiating gives \(\tfrac12\begin{pmatrix}4e^{4t}+2e^{2t}&4e^{4t}-2e^{2t}\\ 4e^{4t}-2e^{2t}&4e^{4t}+2e^{2t}\end{pmatrix}\), while \(A_ae^{A_at} = \tfrac12\begin{pmatrix}3p+q&3q+p\\p+3q&q+3p\end{pmatrix}\), and \(3p+q = 4e^{4t}+2e^{2t}\), \(3q+p = 4e^{4t}-2e^{2t}\): the same matrix. (b) differentiating gives \(\omega\begin{pmatrix}-\sin&-\cos\\ \cos&-\sin\end{pmatrix} = A_be^{A_bt}\). (c) \(\tfrac{\mathrm d}{\mathrm dt}\big[e^{2t}(I+Nt)\big] = e^{2t}\big(2I+2Nt+N\big) = (2I+N)e^{2t}(I+Nt)\), using \(N^2 = 0\). All three give \(I\) at \(t = 0\).

3. \(e^{A_bt}\) is rotation of the plane through angle \(\omega t\), so \(\det = \cos^2 + \sin^2 = 1\) for every \(t\).

Check. The identity \(\det e^{At} = e^{t\,\mathrm{tr}A}\) settles all three at once. (Proof: if \(A = PJP^{-1}\) with \(J\) in Jordan form — cited on the deck — then \(e^{At} = Pe^{Jt}P^{-1}\), and \(e^{Jt}\) is upper triangular with diagonal entries \(e^{\lambda_it}\), so its determinant is \(e^{t\sum\lambda_i} = e^{t\,\mathrm{tr}A}\).) \(\mathrm{tr}A_a = 6\) and indeed \(\tfrac14\big[(e^{4t}+e^{2t})^2 - (e^{4t}-e^{2t})^2\big] = e^{6t}\); \(\mathrm{tr}A_b = 0\) and \(\det = 1\); \(\mathrm{tr}A_c = 4\) and \(\det = e^{4t}\).

Idea. The failure is already visible at order \(t^2\), where \((A+B)^2\) and \(A^2 + 2AB + B^2\) differ by \(AB - BA\). Everything in this problem is that one observation, read forwards and backwards.

1. \(A^2 = B^2 = 0\), so \(e^A = I + A = \begin{pmatrix}1&1\\0&1\end{pmatrix}\) and \(e^B = I + B = \begin{pmatrix}1&0\\1&1\end{pmatrix}\). Hence \[ e^Ae^B = \begin{pmatrix}2&1\\1&1\end{pmatrix}, \qquad e^Be^A = \begin{pmatrix}1&1\\1&2\end{pmatrix}. \] For \(A+B = \begin{pmatrix}0&1\\1&0\end{pmatrix}\) we have \((A+B)^2 = I\), so the even terms sum to \(\cosh1\,I\) and the odd ones to \(\sinh1\,(A+B)\): \[ e^{A+B} = \begin{pmatrix}\cosh1&\sinh1\\ \sinh1&\cosh1\end{pmatrix} \approx \begin{pmatrix}1.543081&1.175201\\ 1.175201&1.543081\end{pmatrix}. \] All three differ; the traces alone, \(3\), \(3\) and \(2\cosh1 = 3.086161\), separate the last from the first two, and the off-diagonal entries separate the first two from each other.

2. Proof. Expand both sides in powers of \(t\), which is legitimate because each series converges absolutely and uniformly on compact intervals: \[ e^{(A+B)t} = I + (A+B)t + \tfrac12(A+B)^2t^2 + O(t^3), \] \[ e^{At}e^{Bt} = \big(I + At + \tfrac12A^2t^2\big)\big(I + Bt + \tfrac12B^2t^2\big) + O(t^3) = I + (A+B)t + \big(\tfrac12A^2 + AB + \tfrac12B^2\big)t^2 + O(t^3). \] Two power series in \(t\) with matrix coefficients agree on an interval only if the coefficients agree, so \(\tfrac12(A^2+AB+BA+B^2) = \tfrac12A^2 + AB + \tfrac12B^2\), i.e. \(\tfrac12(AB+BA) = AB\), i.e. \(BA = AB\). \(\square\)

3. \(At\) and \(As\) commute for any \(A\), since both are multiples of \(A\). So the "if" direction of the commuting case applies and \(e^{A(t+s)} = e^{At}e^{As}\) with no hypothesis. In particular \(e^{At}\) is invertible with inverse \(e^{-At}\).

Check. Part 2's conclusion is consistent with part 1: there \(AB = \begin{pmatrix}1&0\\0&0\end{pmatrix} \ne \begin{pmatrix}0&0\\0&1\end{pmatrix} = BA\), and the \(t^2\) coefficients of the two sides differ by \(\tfrac12(AB-BA) = \tfrac12\begin{pmatrix}1&0\\0&-1\end{pmatrix} \ne 0\) — which is exactly why the traces came out \(3\) and \(3.086\).

Idea. Unit 06 found \(te^{-\omega_0t}\) by trying it and checking. Here it is derived: the matrix of the system is defective at \(\gamma = 2\omega_0\), and a defective eigenvalue is precisely what puts a polynomial in front of the exponential.

1. With \(y = (x,\dot x)\), \(A = \begin{pmatrix}0&1\\-\omega_0^2&-\gamma\end{pmatrix}\) and \(\det(A-\mu I) = \mu^2 + \gamma\mu + \omega_0^2\). At \(\gamma = 2\omega_0\) this is \((\mu+\omega_0)^2\), a double root at \(\mu = -\omega_0\). The eigenvector equation \((A+\omega_0I)w = 0\) reads \(\omega_0w_1 + w_2 = 0\) in both rows — one independent equation — so the eigenspace is spanned by \((1,-\omega_0)\) alone. Geometric multiplicity \(1\) against algebraic multiplicity \(2\): defective.

2. \(N = A + \omega_0I = \begin{pmatrix}\omega_0&1\\-\omega_0^2&-\omega_0\end{pmatrix}\) and \[ N^2 = \begin{pmatrix}\omega_0^2-\omega_0^2 & \omega_0-\omega_0\\ -\omega_0^3+\omega_0^3 & -\omega_0^2+\omega_0^2\end{pmatrix} = 0 . \] Since \(-\omega_0I\) commutes with everything, \[ e^{At} = e^{-\omega_0t}(I+Nt) = e^{-\omega_0t}\begin{pmatrix}1+\omega_0t & t\\ -\omega_0^2t & 1-\omega_0t\end{pmatrix}. \] The \((1,2)\) entry, \(te^{-\omega_0t}\), is the response to a unit initial velocity: unit 06's critically damped Green's function.

3. With \(\omega_d = \sqrt{\omega_0^2-\gamma^2/4} \to 0\) as \(\gamma \to 2\omega_0\), \[ e^{-\gamma t/2}\frac{\sin\omega_dt}{\omega_d} = e^{-\gamma t/2}\,t\,\frac{\sin\omega_dt}{\omega_dt} \longrightarrow e^{-\omega_0t}\,t, \] since \(\sin x/x \to 1\) and \(\gamma/2 \to \omega_0\). The limit is uniform on compact \(t\)-intervals, so the family of solutions is continuous in \(\gamma\) through the degenerate value — nothing jumps.

4. With \(x(0) = 0\), \(\dot x(0) = 1\ \mathrm{m\,s^{-1}}\) and \(\omega_0 = 2\ \mathrm{s^{-1}}\), the second column of \(e^{At}\) gives \(x(t) = te^{-2t}\). Then \(\dot x = (1-2t)e^{-2t}\), zero at \[ t = \frac1{\omega_0} = 0.5\ \mathrm s, \qquad x_{\max} = \frac{1}{\omega_0 e} = 0.5\,e^{-1} = 0.183940\ \mathrm m . \]

Check. Dimensions: \(e^{At}\) has entries \(\begin{pmatrix}1&\mathrm s\\ \mathrm{s^{-2}}&1\end{pmatrix}\), so the \((1,2)\) entry times a velocity gives a length, as used. Limit: as \(\omega_0 \to 0\) the matrix tends to \(\begin{pmatrix}1&t\\0&1\end{pmatrix}\), free motion. Number: \(x_{\max} = 1/(\omega_0e)\) — halve \(\omega_0\) and the peak doubles and arrives twice as late, which is what a softer critically damped system should do.

Idea. With equal masses the system is symmetric under swapping the two coordinates, so the modes must be the symmetric and antisymmetric vectors \((1,1)\) and \((1,-1)\) — and one can read the frequencies off without solving anything. The determinant is done anyway, as a check.

1. \(V = \tfrac12ku_1^2 + \tfrac12k_c(u_2-u_1)^2 + \tfrac12ku_2^2\), so \[ K = \begin{pmatrix}k+k_c & -k_c\\ -k_c & k+k_c\end{pmatrix}, \qquad M = mI_2 . \]

2. \(\det(K-\lambda mI) = (k+k_c-\lambda m)^2 - k_c^2 = 0\) gives \(k+k_c-\lambda m = \pm k_c\), so \[ \lambda_1 = \frac km,\ \ v^{(1)} = (1,1); \qquad \lambda_2 = \frac{k+2k_c}{m},\ \ v^{(2)} = (1,-1). \] The lower mode moves both masses together, leaving the coupling spring unstretched — so it does not feel \(k_c\) at all. The upper mode moves them oppositely, stretching the coupling spring by \(2u\), which is why \(k_c\) enters doubled.

3. At \(k_c = k\): \(\omega_1 = \sqrt{k/m}\) and \(\omega_2 = \sqrt{3k/m}\), which is \(2\sqrt{k/m}\sin(n\pi/6)\) for \(n = 1,2\) — the running example's formula at \(N = 2\). ✓ As \(k_c \to 0\): both frequencies tend to \(\sqrt{k/m}\), a degenerate pair, and indeed the system becomes two independent wall–mass oscillators. ✓ As \(k \to 0\): \(\omega_1 \to 0\), a zero mode (nothing is attached to a wall, so sliding both masses costs nothing), and \(\omega_2 \to \sqrt{2k_c/m}\). Directly: in the opposed motion the midpoint of the coupling spring stays still, so each mass is held by half a spring, and half a spring is twice as stiff, \(2k_c\); hence \(\omega = \sqrt{2k_c/m}\). ✓

4. \(\langle v^{(1)},v^{(2)}\rangle_M = m(1\cdot1 + 1\cdot(-1)) = 0\). It coincides with the Euclidean test because \(M = mI\) is a multiple of the identity, so \(\langle u,v\rangle_M = m\,(u\cdot v)\). The moment the masses differ, the two tests part company — P6.

Check. Trace: \(\lambda_1+\lambda_2 = (2k+2k_c)/m = \mathrm{tr}(M^{-1}K)\). ✓ Determinant: \(\lambda_1\lambda_2 = k(k+2k_c)/m^2 = \det K/\det M\), since \(\det K = (k+k_c)^2-k_c^2 = k^2+2kk_c\). ✓

Idea. Two modes at nearby frequencies added together are one oscillation at the mean frequency multiplied by a slow envelope at half the difference. Which mass is moving is decided by the envelope.

Numbers first. \(\omega_1 = \sqrt{k/m} = \sqrt{40} = 6.324555\ \mathrm{s^{-1}}\), \(\omega_2 = \sqrt{(k+2k_c)/m} = \sqrt{44} = 6.633250\ \mathrm{s^{-1}}\), \(\Delta\omega = \omega_2-\omega_1 = 0.308694\ \mathrm{s^{-1}}\), \(\bar\omega = \tfrac12(\omega_1+\omega_2) = 6.478902\ \mathrm{s^{-1}}\).

1. The initial condition \(u(0) = (A,0)\), \(\dot u(0) = 0\) decomposes as \(u(0) = \tfrac A2(1,1) + \tfrac A2(1,-1)\), so both normal coordinates start at \(A/2\) with zero velocity: \[ u_1 = \tfrac A2\big(\cos\omega_1t + \cos\omega_2t\big) = A\cos\!\Big(\frac{\Delta\omega\,t}{2}\Big)\cos(\bar\omega t), \] \[ u_2 = \tfrac A2\big(\cos\omega_1t - \cos\omega_2t\big) = A\sin\!\Big(\frac{\Delta\omega\,t}{2}\Big)\sin(\bar\omega t), \] using \(\cos P + \cos Q = 2\cos\tfrac{P+Q}2\cos\tfrac{P-Q}2\) and \(\cos P - \cos Q = -2\sin\tfrac{P+Q}2\sin\tfrac{P-Q}2\).

2. The two envelopes are \(A|\cos(\Delta\omega t/2)|\) and \(A|\sin(\Delta\omega t/2)|\): whenever one is at its maximum the other is zero, and \(A^2\) worth of amplitude is conserved between them. Mass 1's envelope first vanishes when \(\Delta\omega t/2 = \pi/2\).

3. \[ t_* = \frac{\pi}{\Delta\omega} = \frac{\pi}{0.308694} = 10.1770\ \mathrm s . \] The number of fast oscillations in that time is \(t_*/(2\pi/\bar\omega) = \bar\omega/(2\Delta\omega) = 6.478902/0.617389 = 10.494\).

4. No. Write the modes as \((1,\alpha)\) and \((1,\beta)\). The initial displacement \((A,0)\) gives mass 1 the two contributions \(c_1 = -A\beta/(\alpha-\beta)\) and \(c_2 = A\alpha/(\alpha-\beta)\), and its motion can only come fully to rest if they are equal in size, \(|\alpha| = |\beta|\). \(M\)-orthogonality, \(m_1 + m_2\alpha\beta = 0\), makes \(\alpha\beta\) negative, so this means \(\beta = -\alpha\). The first row of \(Kv = \lambda Mv\) for the two modes reads \((k+k_c) - k_c\alpha = \lambda_1m_1\) and \((k+k_c) + k_c\alpha = \lambda_2m_1\); adding, \(\lambda_1 + \lambda_2 = 2(k+k_c)/m_1\). But the trace gives \(\lambda_1+\lambda_2 = \mathrm{tr}(M^{-1}K) = (k+k_c)(1/m_1 + 1/m_2)\), and the two agree only if \(m_1 = m_2\). So with unequal masses mass 1's envelope never reaches zero, and some energy always stays with it. (Mass 2's contributions, \(c_1\alpha\) and \(c_2\beta\), are always equal and opposite — it starts at rest at the origin and returns there periodically.) The reason, in words: complete transfer is a tuning condition, and the swap symmetry of equal masses is what tunes it.

Check. Dimensions: \(\Delta\omega\) is a frequency so \(t_*\) is a time. ✓ Limit: as \(k_c \to 0\), \(\Delta\omega \to 0\) and \(t_* \to \infty\) — with no coupling the energy never leaves mass 1. ✓ Number: a direct evaluation of \(u_1, u_2\) at \(t = t_*\) with \(A = 1\) gives \(u_1 = 1.7\times10^{-15}\) and an envelope of \(1\) for \(u_2\). The transfer is complete only up to the small residual velocity \(|\dot u_1(t_*)| \le A\Delta\omega/2\), which carries a fraction at most \((\Delta\omega/2\bar\omega)^2 = 5.7\times10^{-4}\) of the energy: complete to \(0.06\%\), not exactly.

Idea. Everything hinges on one identity: the two symmetric modes have \(v^{(1)}_2v^{(3)}_2 = -\tfrac12\), and the heavy mass multiplies exactly that entry by \(4\). The Euclidean sum misses the \(4\); the \(M\)-sum does not.

1. \(M = \mathrm{diag}(1,4,1)\) and \(K = \begin{pmatrix}2&-1&0\\-1&2&-1\\ 0&-1&2\end{pmatrix}\) in SI units. Expanding along the first row, \[ \det(K-\lambda M) = (2-\lambda)\big[(2-4\lambda)(2-\lambda)-1\big] + \big[-(2-\lambda)\big] = -2(\lambda-2)(2\lambda^2-5\lambda+1). \] So \(\lambda_2 = 2\) and \(\lambda = \tfrac{5\pm\sqrt{17}}4\): \[ \lambda_1 = \tfrac{5-\sqrt{17}}4 = 0.219224, \quad \lambda_2 = 2, \quad \lambda_3 = \tfrac{5+\sqrt{17}}4 = 2.280776 \quad (\mathrm{s^{-2}}), \] \[ \omega_1 = 0.468213,\quad \omega_2 = 1.414214,\quad \omega_3 = 1.510224 \quad (\mathrm{s^{-1}}). \]

2. Rows 1 and 3 of \((K-\lambda M)v = 0\) read \((2-\lambda)v_1 = v_2\) and \((2-\lambda)v_3 = v_2\), so \(v_1 = v_3\) whenever \(\lambda \ne 2\): \[ v^{(1)} = \big(1,\ \tfrac{3+\sqrt{17}}4,\ 1\big) = (1,\ 1.780776,\ 1), \qquad v^{(3)} = \big(1,\ \tfrac{3-\sqrt{17}}4,\ 1\big) = (1,\ -0.280776,\ 1). \] At \(\lambda_2 = 2\) the first row forces \(v_2 = 0\) and the middle row gives \(v_1 + v_3 = 0\): \(v^{(2)} = (1,0,-1)\), the antisymmetric mode in which the heavy mass does not move at all.

3. \[ v^{(1)}\!\cdot v^{(2)} = 1+0-1 = 0, \qquad v^{(2)}\!\cdot v^{(3)} = 1+0-1 = 0, \qquad v^{(1)}\!\cdot v^{(3)} = 1 + \frac{(3+\sqrt{17})(3-\sqrt{17})}{16} + 1 = 2 - \tfrac12 = \tfrac32 . \] The \(M\)-products are \(0\), \(0\) and \(1 + 4\big(-\tfrac12\big) + 1 = 0\). The one nonzero Euclidean product is \(v^{(1)}\!\cdot v^{(3)} = 3/2\).

4. The \(M\)-inner product was right all along: \(M\)-orthogonality is a theorem about any pair of modes with distinct \(\lambda\), while Euclidean orthogonality is an accident of equal masses. What makes the difference visible here is that modes 1 and 3 are both symmetric, so nothing but the mass weighting separates them; modes 1 and 2 and modes 2 and 3 are orthogonal for free, because one is symmetric and the other antisymmetric under the reflection that also preserves \(M\).

Check. Trace: \(\lambda_1+\lambda_2+\lambda_3 = \tfrac52 + 2 = 4.5 = \mathrm{tr}(M^{-1}K) = 2+\tfrac12+2\). ✓ Determinant: \(\lambda_1\lambda_2\lambda_3 = 2\cdot\tfrac{25-17}{16} = 1 = \det K/\det M = 4/4\). ✓ Limit: set \(m_2 = m_1 = m_3\) and the spectrum becomes \(\{2-\sqrt2,\,2,\,2+\sqrt2\}\) — the \(N=3\) case of the chain formula, \(4\sin^2(n\pi/8)\). ✓

Idea. Parts 1 and 2 are the two halves of the unit's main theorem, and they need different hypotheses: part 1 needs only symmetry, part 2 needs \(M\) positive definite. Part 3 shows that the hypothesis of part 2 cannot be dropped, by exhibiting a pencil with no real eigenvalue at all.

1. Proof. Using \(K = K^{\!\top}\) and \(M = M^{\!\top}\), \[ \lambda_i\,v_j^{\!\top}Mv_i = v_j^{\!\top}(Kv_i) = (Kv_j)^{\!\top}v_i = \lambda_j\,(Mv_j)^{\!\top}v_i = \lambda_j\,v_j^{\!\top}Mv_i . \] Hence \((\lambda_i-\lambda_j)v_j^{\!\top}Mv_i = 0\), and since \(\lambda_i \ne \lambda_j\), \(v_j^{\!\top}Mv_i = 0\). Then \(v_j^{\!\top}Kv_i = \lambda_iv_j^{\!\top}Mv_i = 0\) too. Positive definiteness was not used anywhere — only symmetry of both matrices, and the assumption that the two \(\lambda\) are real and distinct.

2. Proof. Since \(M\) is positive definite, \(M^{1/2}\) exists, is positive definite, and is invertible (P8). Put \(S = M^{-1/2}KM^{-1/2}\), symmetric because \(S^{\!\top} = M^{-1/2}K^{\!\top}M^{-1/2} = S\). Then \[ Kv = \lambda Mv \iff M^{-1/2}KM^{-1/2}\big(M^{1/2}v\big) = \lambda\,M^{1/2}v \iff Sw = \lambda w,\ \ w = M^{1/2}v, \] so the pencil's eigenvalues are exactly \(S\)'s, and they are real by the spectral theorem. That theorem also gives an orthonormal basis \(w_1,\dots,w_n\) of \(\mathbb R^n\) of eigenvectors of \(S\); since \(M^{-1/2}\) is invertible, \(v_i = M^{-1/2}w_i\) is a basis of \(\mathbb R^n\), and each solves \(Kv_i = \lambda_iMv_i\). \(\square\)

3. \(K-\lambda M = \begin{pmatrix}-\lambda & 1\\ 1 & \lambda\end{pmatrix}\), so \(\det(K-\lambda M) = -\lambda^2 - 1\), with roots \(\lambda = \pm i\): no real eigenvalue. Suppose some invertible real \(C\) gave \(C^{\!\top}MC = D_1\) and \(C^{\!\top}KC = D_2\), both diagonal. Then \[ \det(K-\lambda M) = \det(C^{-\top})\,\det(D_2-\lambda D_1)\,\det(C^{-1}) = \frac{1}{(\det C)^2}\prod_{i}\big((D_2)_{ii} - \lambda (D_1)_{ii}\big). \] Now \(\det D_1 = (\det C)^2\det M = -( \det C)^2 \ne 0\), so every \((D_1)_{ii} \ne 0\) and each factor has the real root \((D_2)_{ii}/(D_1)_{ii}\). So \(\det(K-\lambda M)\) would have two real roots. It has none. Contradiction. \(\square\)

4. The step that fails is the very first one of part 2: \(M\) is not positive definite, so \(M^{1/2}\) does not exist and the pencil cannot be turned into a symmetric ordinary eigenvalue problem. Everything downstream — reality, completeness of the modes — goes with it.

Check. Consistency: part 1 applies to part 3's pencil too, but vacuously, since there are no two real distinct eigenvalues to be orthogonal. And the physical reading is right: an indefinite "kinetic energy" permits motions that are neither oscillations nor real exponentials, and \(\lambda = \pm i\) says exactly that.

Idea. Three small proofs, each of which is the corresponding definition read carefully. Only the uniqueness of the square root takes more than a line, and it takes the spectral theorem.

1. Bilinearity of \((u,v) \mapsto u^{\!\top}Mv\) is the distributivity of matrix multiplication. Symmetry: \(u^{\!\top}Mv\) is a \(1\times1\) matrix, so it equals its own transpose, \(v^{\!\top}M^{\!\top}u = v^{\!\top}Mu\). Positive definiteness: \(\langle u,u\rangle_M = u^{\!\top}Mu > 0\) for \(u \ne 0\) is the hypothesis on \(M\). If instead \(M\) has an eigenvalue \(\mu < 0\) with eigenvector \(e\), then \(\langle e,e\rangle_M = \mu\lVert e\rVert^2 < 0\): the positive-definiteness axiom fails, and with it linear independence of orthogonal sets and the Cauchy–Schwarz inequality in its usual form.

2. Write \(M = Q\Lambda Q^{\!\top}\) with \(Q\) orthogonal and \(\Lambda = \mathrm{diag}(\mu_i)\), \(\mu_i > 0\) (spectral theorem, and the \(\mu_i\) are positive because \(e_i^{\!\top}\Lambda e_i = \mu_i\) and \(M\) is positive definite). Put \(M^{1/2} = Q\Lambda^{1/2}Q^{\!\top}\). It is symmetric, since \((Q\Lambda^{1/2}Q^{\!\top})^{\!\top} = Q\Lambda^{1/2}Q^{\!\top}\); positive definite, since \(u^{\!\top}M^{1/2}u = (Q^{\!\top}u)^{\!\top}\Lambda^{1/2} (Q^{\!\top}u) = \sum_i\sqrt{\mu_i}\,(Q^{\!\top}u)_i^2 > 0\); and \((M^{1/2})^2 = Q\Lambda Q^{\!\top} = M\). Uniqueness. Let \(R\) be positive definite with \(R^2 = M\). If \(Rx = \rho x\) then \(Mx = \rho^2x\), so every eigenvector of \(R\) is an eigenvector of \(M\) with eigenvalue \(\rho^2\), and \(\rho > 0\). By the spectral theorem \(R\) has an eigenbasis, so \(\mathbb R^n\) is the direct sum of \(R\)'s eigenspaces, each contained in an eigenspace \(E_\mu\) of \(M\) with \(\rho = \sqrt\mu\) determined by \(\mu\). Hence those eigenspaces fill \(E_\mu\) and \(R\) acts on \(E_\mu\) as \(\sqrt\mu\,\mathrm{id}\) — which is how \(M^{1/2}\) acts. So \(R = M^{1/2}\). \(\square\)

3. Symmetry of \(S\) and the equivalence are the computation in P7 part 2. The map \(v \mapsto M^{1/2}v\) is invertible, sends a solution of \(Kv = \lambda Mv\) to an eigenvector of \(S\) with the same \(\lambda\), and its inverse \(w \mapsto M^{-1/2}w\) does the reverse; so it is a bijection between the two eigenspaces for each \(\lambda\).

4. \(\langle v_i,v_j\rangle_M = v_i^{\!\top}Mv_j = v_i^{\!\top}M^{1/2}M^{1/2}v_j = (M^{1/2}v_i)\cdot(M^{1/2}v_j) = w_i\cdot w_j\), using symmetry of \(M^{1/2}\). So the two conditions \(\langle v_i,v_j\rangle_M = \delta_{ij}\) and \(w_i\cdot w_j = \delta_{ij}\) are literally the same equation. \(\square\)

Check. Dimensions: \([M^{1/2}] = \mathrm{kg}^{1/2}\), so \(w = M^{1/2}v\) has units \(\mathrm{kg^{1/2}\,m}\) — the "mass-weighted displacement" in which kinetic energy is \(\tfrac12\lVert\dot w\rVert^2\) with no matrix. Special case: \(M = mI\) gives \(M^{1/2} = \sqrt m\,I\), \(w = \sqrt m\,v\), and the \(M\)-inner product is \(m\) times the Euclidean one — which is why equal masses hide the whole distinction.

Idea. Nothing here needs a determinant: the formula is known, and the point of the problem is to check that it really does satisfy the equations, entry by entry.

1. \(M = mI_4\), \(K = kA_4\), and \(\omega_n = 2\sqrt{k/m}\,\sin(n\pi/10)\): \[ \omega_1 = 0.618034,\quad \omega_2 = 1.175570,\quad \omega_3 = 1.618034,\quad \omega_4 = 1.902113 \qquad \big(\times\sqrt{k/m}\big). \] The first and third are \((\sqrt5-1)/2\) and \((\sqrt5+1)/2\), because \(\sin 18^\circ = (\sqrt5-1)/4\) and \(\sin 54^\circ = (\sqrt5+1)/4\): the angles \(n\pi/10\) are those of the regular pentagon, where the golden ratio lives.

2. \(\lambda_1 = \omega_1^2 = 2-2\cos(\pi/5) = 0.381966\) and \(v = \big(\sin\tfrac\pi5, \sin\tfrac{2\pi}5, \sin\tfrac{3\pi}5, \sin\tfrac{4\pi}5\big) = (0.587785,\ 0.951057,\ 0.951057,\ 0.587785)\). Then, with \(v_0 = v_5 = 0\), \[ \begin{aligned} j=1:&\ 2(0.587785) - 0.951057 = 0.224514, &\ \lambda_1v_1 &= 0.381966\times0.587785 = 0.224514,\\ j=2:&\ -0.587785 + 2(0.951057) - 0.951057 = 0.363271, &\ \lambda_1v_2 &= 0.381966\times0.951057 = 0.363271, \end{aligned} \] and \(j = 3, 4\) repeat these by the symmetry \(v_j = v_{5-j}\). ✓

3. Since \(\tfrac{(5-n)\pi}{10} = \tfrac\pi2 - \tfrac{n\pi}{10}\), \(\sin\big((5-n)\pi/10\big) = \cos(n\pi/10)\), so \[ \omega_n^2 + \omega_{5-n}^2 = \frac{4k}m\Big(\sin^2\frac{n\pi}{10} + \cos^2\frac{n\pi}{10}\Big) = \frac{4k}m . \] Numerically \(0.381966+3.618034 = 4\) and \(1.381966+2.618034 = 4\). ✓ And \(\prod_n\lambda_n = 0.381966\times1.381966\times2.618034\times3.618034 = 5 = \det A_4\). ✓ (In general \(\det A_N = N+1\): expanding along the last row gives \(D_N = 2D_{N-1} - D_{N-2}\) with \(D_1 = 2\), \(D_2 = 3\), whose solution is \(D_N = N+1\) by induction.)

4. \(\sum_{j=1}^4\sin(j\pi/5)\sin(3j\pi/5)\) has terms \(0.559017,\ -0.559017,\ -0.559017,\ 0.559017\), summing to \(0\). Euclidean is the right test because \(M = mI_4\), so \(\langle u,v\rangle_M = m\,(u\cdot v)\) and the two notions of orthogonality differ only by the positive factor \(m\).

Check. Dimensions: all four \(\omega_n\) are quoted in units of \(\sqrt{k/m} = \mathrm{s^{-1}}\). ✓ Ceiling: the largest, \(1.902113\), is below \(2\). ✓ Special case: clamping the two end masses (two linear constraints, \(u_1 = u_4 = 0\)) leaves exactly the \(N = 2\) chain, with \(\lambda' = 1, 3\). Interlacing, from the deck's Rayleigh notes, applied once per constraint gives \(\lambda_p \le \lambda'_p \le \lambda_{p+2}\): indeed \(0.381966 \le 1 \le 2.618034\) and \(1.381966 \le 3 \le 3.618034\). ✓

Idea. Two predictions of the same number — one from a differential equation that does not know the chain is discrete, one exact — and a Taylor series that predicts the gap between them before either is evaluated.

1. \(\sqrt{k/m} = \sqrt{250} = 15.811388\ \mathrm{s^{-1}}\), so \[ c = a\sqrt{k/m} = 0.05\times15.811388 = 0.7905694\ \mathrm{m\,s^{-1}}, \qquad L = 21\times0.05 = 1.05\ \mathrm m, \] \[ \omega_{\text{cont}} = \frac{\pi c}{L} = \frac{\pi\times0.7905694}{1.05} = 2.365378\ \mathrm{s^{-1}} . \]

2. \(\omega_1 = 2\sqrt{k/m}\,\sin\big(\pi/42\big) = 2\times15.811388\times 0.0747301 = 2.363173\ \mathrm{s^{-1}}\), so \[ \frac{\omega_1-\omega_{\text{cont}}}{\omega_{\text{cont}}} = \frac{2.363173-2.365378}{2.365378} = -9.322\times10^{-4} = -0.0932\% . \]

3. \(\kappa_1a = \pi/21 = 0.1495997\), so the leading correction is \(-(\kappa_1a)^2/24 = -9.325\times10^{-4} = -0.0933\%\). Actual \(-0.09322\%\) against predicted \(-0.09325\%\): they agree to three significant figures, a relative discrepancy of \(2.8\times10^{-4}\), which is the size of the next term in the series, \(+(\kappa a)^4/1920\).

4. The deficit \(1 - \mathrm{sinc}(\kappa_na/2)\) rises through \(5\%\) between \(n = 7\) (\(4.507\%\)) and \(n = 8\) (\(5.862\%\)), so the smallest \(n\) is \(\mathbf{8}\). Its wavelength is \(2\pi/\kappa_8 = 2(N+1)a/n = 42a/8 = 5.25a\): about five masses to a wavelength. That is the practical rule — a continuum description is good to a few per cent while a wavelength spans six or more lattice points, and not otherwise.

Check. Dimensions: \(c = a\sqrt{k/m}\) has units \(\mathrm m\times\mathrm{s^{-1}} = \mathrm{m\,s^{-1}}\). ✓ Sign: \(\sin x < x\), so the chain is always softer than the string, and the deficit is negative for every \(n\). ✓ Limit: holding \(L\) fixed and letting \(N \to \infty\) sends \(\kappa_1a = \pi/(N+1) \to 0\) and the deficit to zero like \(1/N^2\) — the continuum theory becomes exact, at second order in the spacing.

Idea. Free ends are not "no boundary condition"; they are the condition that the missing spring has zero extension, which is cleanest to impose with ghost values \(v_0 = v_1\), \(v_{N+1} = v_N\). Then the same sine algebra as in the fixed case runs through, with cosines of half-integer arguments.

1. \(V = \tfrac k2\sum_{j=1}^{N-1}(u_{j+1}-u_j)^2\), so \(\partial_i V = k(2u_i - u_{i-1} - u_{i+1})\) for \(1 < i < N\), and \(\partial_1V = k(u_1-u_2)\), \(\partial_NV = k(u_N-u_{N-1})\). Hence \(K = k\tilde A_N\), which is \(kA_N\) with the two corner entries reduced from \(2\) to \(1\). Applying it to \((1,\dots,1)\): the first row gives \(1-1 = 0\), each interior row gives \(-1+2-1 = 0\), the last gives \(-1+1 = 0\). So \(K(1,\dots,1)^{\!\top} = 0\).

2. \(u^{\!\top}Ku = k\sum_{j=1}^{N-1}(u_{j+1}-u_j)^2\), which vanishes only if every consecutive difference vanishes, i.e. \(u_1 = u_2 = \dots = u_N\). So the null space is exactly the span of \((1,\dots,1)\): one-dimensional, and the zero eigenvalue is simple.

3. With the ghosts, \((\tilde A_Nv)_j = -v_{j-1}+2v_j-v_{j+1}\) for every \(j = 1,\dots,N\). Substituting \(v_j = \cos\big((j-\tfrac12)\theta\big)\) and using \(\cos((j-\tfrac32)\theta) + \cos((j+\tfrac12)\theta) = 2\cos\theta\cos((j-\tfrac12)\theta)\) gives \((\tilde A_Nv)_j = (2-2\cos\theta)v_j\). The left ghost condition is automatic, since \(v_0 = \cos(-\theta/2) = \cos(\theta/2) = v_1\). The right one, \(\cos\big((N+\tfrac12)\theta\big) = \cos\big((N-\tfrac12)\theta\big)\), is \(-2\sin(N\theta)\sin(\theta/2) = 0\), hence \(\sin(N\theta) = 0\) and \(\theta_n = n\pi/N\). Therefore \[ \lambda_n = \frac km\big(2-2\cos\theta_n\big) = \frac{4k}m\sin^2\frac{n\pi}{2N}, \qquad \omega_n = 2\sqrt{\frac km}\,\sin\frac{n\pi}{2N}, \] for \(n = 0,1,\dots,N-1\) (larger \(n\) repeats these vectors up to sign). At \(n = 0\), \(v_j = 1\) for all \(j\): the zero mode of part 1.

4. \(N = 3\): \(K/k = \begin{pmatrix}1&-1&0\\-1&2&-1\\0&-1&1\end{pmatrix}\), with characteristic polynomial \(-\lambda(\lambda-1)(\lambda-3)\) in units of \(k/m\), so \(\omega = 0,\ \sqrt{k/m},\ \sqrt{3k/m}\). The formula gives \(2\sin(0), 2\sin(\pi/6), 2\sin(\pi/3) = 0,\ 1,\ \sqrt3\). ✓ In the \(n = 1\) mode, \(v = \big(\cos\tfrac\pi6, \cos\tfrac\pi2, \cos\tfrac{5\pi}6\big) \propto (1,0,-1)\): the middle mass does not move because the motion is antisymmetric about the centre, and the centre is where that mass sits. Each outer mass is then attached by one spring to an effectively fixed point, giving \(\omega = \sqrt{k/m}\).

Check. Count: \(N\) values of \(n\), hence \(N\) modes, as the theorem requires. ✓ Ceiling: \(\omega_{N-1} = 2\sqrt{k/m}\sin\big((N-1)\pi/2N\big) < 2\sqrt{k/m}\). ✓ Consistency: the nonzero frequencies of the free \(N\)-chain, \(2\sin(n\pi/2N)\) for \(n = 1,\dots,N-1\), are exactly the frequencies of the fixed chain of \(N-1\) masses, \(2\sin(n\pi/2N)\). Removing the walls costs one oscillating mode and buys a drift.

Idea. The zero mode is written down before any algebra, from the symmetry; the algebra then has to produce it, which is the check.

1. \(M = m\,\mathrm{diag}(1,2,1)\) and \(K = k\begin{pmatrix}1&-1&0\\-1&2&-1\\0&-1&1\end{pmatrix}\). Writing \(\lambda\) in units of \(k/m\), \[ \det(K-\lambda M) = -2\lambda(\lambda-1)(\lambda-2), \] so \(\lambda = 0,\ 1,\ 2\) and \(\omega = 0,\ \sqrt{k/m},\ \sqrt{2k/m}\). The modes are \[ v^{(0)} = (1,1,1), \qquad v^{(1)} = (1,0,-1), \qquad v^{(2)} = (1,-1,1). \] Verifying the last: \(Kv^{(2)} = k(2,-4,2)\) and \(\lambda Mv^{(2)} = 2\tfrac km\,m(1,-2,1) = k(2,-4,2)\). ✓

2. The zero mode is \(v^{(0)} = (1,1,1)\). The family is \(\varphi(s) = u^* + s(1,1,1)\), a rigid translation of the whole molecule along the line. \(V\) depends only on the two differences \(u_2-u_1\) and \(u_3-u_2\), which \(\varphi\) leaves unchanged, so \(V(\varphi(s))\) is constant and the theorem gives \(Kv^{(0)} = 0\).

3. \(\langle v^{(0)},v^{(1)}\rangle_M = m(1 + 0 - 1) = 0\); \(\langle v^{(0)},v^{(2)}\rangle_M = m(1 - 2 + 1) = 0\); \(\langle v^{(1)},v^{(2)}\rangle_M = m(1 + 0 - 1) = 0\). The pair that is not Euclidean-orthogonal is \(v^{(0)}\) and \(v^{(2)}\), whose dot product is \(1 - 1 + 1 = 1\). The factor \(2\) on the middle mass is exactly what turns that \(1\) into \(0\).

4. Translations give \(3\) zero modes always. Rotations give \(3\) more for a non-linear molecule, but only \(2\) for a linear one, since rotating about its own axis moves no atom. So:

• non-linear: \(6\) zero modes, \(3n-6\) vibrational;
• linear: \(5\) zero modes, \(3n-5\) vibrational.

Water (\(n = 3\), bent): \(3\times3-6 = \mathbf3\) vibrational modes — symmetric stretch, antisymmetric stretch, bend. Carbon dioxide (\(n = 3\), linear): \(3\times3-5 = \mathbf4\) — the same three plus a second, degenerate bend at right angles to the first.

Check. Dimensions: \(\lambda\) was quoted in units of \(k/m = \mathrm{s^{-2}}\). ✓ Count: three coordinates, three modes, one of them zero — and \(\mathrm{tr}(M^{-1}K) = k/m\,(1 + 1 + 1) = 3k/m = 0 + 1 + 2\) in the same units. ✓ Limit: replace the middle mass by \(\mu\). The mode \((1,0,-1)\) is unaffected, still at \(\lambda = k/m\), and the trace then forces the third root to be \(\lambda_3 = k/m + 2k/\mu\). Setting \(\mu = 2m\) recovers \(2k/m\); letting \(\mu \to \infty\) sends \(\lambda_3 \to k/m\), both outer masses vibrating against an immovable centre — which is right.

Idea. \(R\) is a weighted average of the eigenvalues, with weights fixed by how much of each mode the trial vector contains. An average is never below the smallest entry — that is the whole theorem, and every part of this problem is a corollary.

1. With \(M = mI_5\) and \(u_0 = u_6 = 0\), \(R(u) = \tfrac km\sum_{j=0}^{5}(u_{j+1}-u_j)^2\big/\sum_{j=1}^5u_j^2\).

The exact value is \(\omega_1 = 2\sin(\pi/12)\sqrt{k/m} = \sqrt{2-\sqrt3}\, \sqrt{k/m} = 0.517638\sqrt{k/m}\). All three bounds lie above it, as they must.

2. Proof. Expand \(u = \sum_pc_pv_p\) in the \(M\)-orthonormal modes. Then \(u^{\!\top}Mu = \sum_pc_p^2\) and \(u^{\!\top}Ku = \sum_p\lambda_pc_p^2\), so \[ R(u) = \frac{\sum_p\lambda_pc_p^2}{\sum_pc_p^2} \ \ge\ \lambda_1\frac{\sum_pc_p^2}{\sum_pc_p^2} = \lambda_1 , \] with equality iff \(c_p = 0\) for every \(p\) with \(\lambda_p > \lambda_1\). So every trial vector overestimates, never under. For the quadratic statement, take \(u = v_1 + \varepsilon z\) with \(\langle z,v_1\rangle_M = 0\) and \(\lVert z\rVert_M = 1\); writing \(z = \sum_{p\ge2}c_pv_p\), \[ R(u) = \frac{\lambda_1 + \varepsilon^2\sum_{p\ge2}\lambda_pc_p^2}{1+\varepsilon^2} = \lambda_1 + \varepsilon^2\sum_{p\ge2}(\lambda_p-\lambda_1)c_p^2 + O(\varepsilon^4), \] so the error is \(O(\varepsilon^2)\) in \(\lambda_1\), hence also in \(\omega_1\). \(\square\)

3. Proof. If \(\langle u,v_1\rangle_M = 0\) then \(c_1 = 0\), so the weighted average runs only over \(p \ge 2\): \(R(u) = \sum_{p\ge2}\lambda_pc_p^2\big/\sum_{p\ge2}c_p^2 \ge \lambda_2\). Read as a statement about \(\lambda_2\) this says \(\lambda_2 \le R(u)\) — an upper bound on the second frequency, obtained by the same argument as for the first but on the subspace \(v_1^{\perp_M}\). Take \(u = (1,2,0,-2,-1)\). It is \(M\)-orthogonal to \(v_1 \propto (0.5,0.866,1,0.866,0.5)\) because \(v_1\) is symmetric and \(u\) antisymmetric about the middle mass, so the five products cancel in pairs and the middle term is \(0\). Its differences are \(1,1,-2,-2,1,1\), numerator \(12\), denominator \(10\), so \(R = 1.2\,k/m\) and \[ \omega_2 \le \sqrt{1.2}\,\sqrt{k/m} = 1.095445\sqrt{k/m}. \] The exact \(\omega_2 = 2\sin(2\pi/12)\sqrt{k/m} = \sqrt{k/m}\), so the bound is \(9.54\%\) high.

4. Every argument above compares \(R(u)\) with a minimum, and the minimum is \(\lambda_1\), so evaluating \(R\) anywhere can only produce a number that is at least \(\lambda_1\): an upper bound. To get below \(\lambda_1\) one needs information the trial vector cannot carry. The classic device is Temple's inequality (G. Temple, 1928): for \(\lVert u\rVert_M = 1\) and any \(\beta\) with \(R(u) < \beta \le \lambda_2\), \[ \lambda_1 \ \ge\ R(u) - \frac{\delta^2}{\beta - R(u)}, \qquad \delta^2 = \sum_p(\lambda_p - R)^2c_p^2 = (Ku - RMu)^{\!\top}M^{-1}(Ku - RMu). \] Proof. With \(R = R(u)\) and \(\sum_pc_p^2 = 1\), every term of \(\sum_p(\lambda_p-\lambda_1)(\lambda_p-\beta)c_p^2\) is \(\ge 0\): the \(p = 1\) term vanishes, and for \(p \ge 2\) both factors are \(\ge 0\) since \(\lambda_p \ge \lambda_2 \ge \beta\). Writing \(\lambda_p = (\lambda_p - R) + R\) and using \(\sum_p(\lambda_p - R)c_p^2 = 0\), the sum equals \(\delta^2 + (R-\lambda_1)(R-\beta)\). So \((\lambda_1 - R)(\beta - R) \ge -\delta^2\), i.e. \(\lambda_1 \ge R - \delta^2/(\beta - R)\). The second form of \(\delta^2\) is the first written without the modes, since \(Ku - RMu = \sum_p(\lambda_p - R)c_pMv_p\) and \(\langle Mv_p, M^{-1}Mv_q\rangle = \delta_{pq}\). \(\square\) So a lower bound needs two extra things: a lower estimate \(\beta\) of \(\lambda_2\), and the residual \(\delta\) of the trial vector. For the rounded tent, normalised, with \(\beta = 1\) (here the exact \(\lambda_2\), which the formula supplies; in general it has to be estimated separately), Temple gives \(\lambda_1 \ge 0.255319\,k/m\), so \(0.505291 \le \omega_1/\sqrt{k/m} \le 0.526235\) — a bracket around the true \(0.517638\). Bounds from below cost strictly more than bounds from above.

Check. Dimensions: \(R\) has units \(k/m = \mathrm{s^{-2}}\), so \(\sqrt R\) is a frequency. ✓ Ordering: \(0.517638 < 0.526235 < 0.561951 < 0.632456\), and the better the shape matches \((1,1.732,2,1.732,1)\) the tighter the bound. ✓ Quadratic insensitivity, measured: scaling \((2,4,5,4,2)\) and the true mode \((0.5,0.866,1,0.866,0.5)\) each to unit \(M\)-norm, the shape error is \(\varepsilon = \lVert u - v_1\rVert_M = 0.0718\), so \(\varepsilon^2 = 0.0052\); the excess in \(\lambda_1\) is \(18/65 - (2-\sqrt3) = 0.008974\) (in units of \(k/m\)), which is \(1.74\,\varepsilon^2\) — the constant being \(\sum_{p\ge2}(\lambda_p-\lambda_1)c_p^2/\varepsilon^2\), of order the spectral gaps, which here run from \(0.73\) to \(3.46\). Squared, as part 2 predicts. The corresponding errors for the tent and flat trials are \(\varepsilon = 0.151\) and \(0.270\), in the same order as their bounds.

Another way (P13 part 1)

The three trial vectors can be compared without any arithmetic by noticing that \(R\) penalises differences and rewards size. The flat vector has all its "cost" concentrated in two unit jumps at the walls, which is wasteful; the tent spreads the jumps evenly but is too pointed at the centre, where the true mode is flat; the rounded tent flattens the peak and wins. The exact minimiser is the sampled sine, and every trial vector is being scored by how much of it it contains.