From N masses to a field: the wave equation

Idea. A symmetric difference is an even function of the step, and the whole accuracy story follows from that one observation.

1. Proof. For \(f \in C^6\), Taylor's theorem with the Peano remainder gives

\[ f(x\pm a) = \sum_{k=0}^{6}\frac{(\pm a)^k}{k!}f^{(k)}(x) + o(a^6). \]

Adding the two, the terms with odd \(k\) cancel identically and the even ones double:

\[ f(x+a) + f(x-a) - 2f(x) = 2\Big[\frac{a^2}{2!}f'' + \frac{a^4}{4!}f^{(4)} + \frac{a^6}{6!}f^{(6)}\Big] + o(a^6). \]

Dividing by \(a^2\) and using \(2/4! = 1/12\) and \(2/6! = 1/360\) gives the statement. \(\square\) Only even powers appear because the operator \(f \mapsto f(x+a) - 2f(x) + f(x-a)\) is invariant under \(a \mapsto -a\): it is an even function of \(a\), and an even smooth function has only even Taylor coefficients.

2. For \(f = \sin(qx)\),

\[ \sin(q(x+a)) + \sin(q(x-a)) - 2\sin(qx) = 2\sin(qx)\big[\cos(qa) - 1\big] = -4\sin(qx)\sin^2\!\frac{qa}2 . \]

So the chain equation \(\ddot y_n = (c^2/a^2)\times(\text{second difference})\) applied to \(y_n = \sin(qna)e^{-\mathrm i\omega t}\) gives \(-\omega^2 = -(4c^2/a^2)\sin^2(qa/2)\), that is \(\omega = (2c/a)\sin(qa/2)\) ✓. Expanding \(\sin z = z - z^3/6 + O(z^5)\) at \(z = qa/2\):

\[ \omega = \frac{2c}{a}\Big[\frac{qa}{2} - \frac{1}{6}\Big(\frac{qa}{2}\Big)^3 + \cdots\Big] = cq\Big[1 - \frac{(qa)^2}{24} + O\big((qa)^4\big)\Big] ✓. \]

Consistency with part 1: \(f'' = -q^2f\) and \(f^{(4)} = q^4f\), so the second-difference estimate predicts \(-q^2 + q^4a^2/12\) in place of \(-q^2\), i.e. \(\omega^2 = c^2q^2(1 - q^2a^2/12)\), and the square root of that is \(cq(1 - q^2a^2/24 + \cdots)\) ✓ — the same coefficient by a different route.

3. With \(N = 20\) and \(k = 5\), \(q_ka/2 = k\pi/2(N+1) = 5\pi/42 = 0.3740\), so \(q_ka = 0.7480\).

• Exact: \(\omega_k/(cq_k) = \sin(0.3740)/0.3740 = 0.9768\) (to four places, \(0.9768\); more fully \(0.976850\)).
• Two-term: \(1 - (0.7480)^2/24 = 1 - 0.023313 = 0.9767\) (\(0.976687\)).
• Deficit: \(1 - 0.976850 = 0.023150\), i.e. the fifth mode of a twenty-bead chain is 2.3150 per cent flat compared with the continuum string; the two-term formula predicts \(2.3313\) per cent, in error by less than two parts in ten thousand of the ratio itself.

4. The scheme's phase speed is \(\omega/q = c\sin(z)/z\) with \(z = qa/2\). Setting \(\sin(z)/z = 0.99\): expanding, \(1 - z^2/6 = 0.99\) gives \(z^2 = 0.06\), \(z = 0.2449\); solving the exact equation gives \(z = 0.24532\). Hence \(qa = 0.49064\). Since \(q = 2\pi/\lambda\), the number of grid points per wavelength is

\[ \frac{\lambda}{a} = \frac{2\pi}{qa} = \frac{2\pi}{0.49064} = 12.81 . \]

So about 13 points per wavelength buys 1 per cent phase accuracy for this scheme — the rule of thumb every numerical wave calculation carries, derived here from a sine.

Check. Dimensions: \(qa\) is dimensionless ✓; \(\omega/q\) is a speed ✓. Limits: \(a \to 0\) at fixed \(q\) gives \(\omega \to cq\) and the deficit \(\to 0\) ✓; \(qa \to \pi\) (two points per wavelength, the Nyquist limit) gives \(\omega/cq = 2/\pi = 0.6366\), a 36 per cent error, and the mode with \(qa = \pi\) does not move at all in group velocity \(\mathrm d\omega/\mathrm dq = c\cos(qa/2) = 0\) ✓. Number: \(k = 20\) in part 3's family has \(q_ka/2 = 1.4960\) and ratio \(0.6666\), a 33 per cent deficit, consistent with the Nyquist figure since \(k = N = 20\) is nearly the shortest wave the chain supports ✓.

Idea. d'Alembert's formula does all of it. The only thought required is that the first term reads the data at exactly two points while the second reads it over a whole interval — which is why "shape, no velocity" is transient and "velocity, no shape" is permanent.

1. \(u(0,t) = \tfrac12[f(-ct) + f(ct)] = f(2t)\) since \(f\) is even.

• \(t = 0.25\): \(f(0.5) = 0.5\) m.
• \(t = 0.5\): \(f(1) = 0\) m.
• \(t = 1\): \(f(2) = 0\) m.

At \(t = 1\) s the string is two separate triangular humps of height \(\tfrac12\) m and half-width 1 m, centred at \(x = -2\) m and \(x = +2\) m, with everything between them flat and motionless. The original hump has split into two half-height copies travelling apart at \(2\ \mathrm{m\,s^{-1}}\).

2. \(u(0,t) = \dfrac{1}{2c}\displaystyle\int_{-2t}^{2t} g = \dfrac{3}{4}\times \text{length}\big([-2t,2t]\cap(-1,1)\big)\). For \(2t \le 1\) the overlap is the whole interval, of length \(4t\); for \(2t \ge 1\) it is all of \((-1,1)\), of length 2. So

\[ u(0,t) = \begin{cases} 3t, & 0 \le t \le \tfrac12\ \mathrm{s},\\[2pt] \tfrac32\ \mathrm{m}, & t \ge \tfrac12\ \mathrm{s}.\end{cases} \]

The midpoint rises at \(3\ \mathrm{m\,s^{-1}}\) — its initial velocity — until the news that the kicked region is finite arrives from the two edges at \(t = 1/c = 0.5\) s, and then stops, permanently displaced by \(\tfrac32\) m.

It does not come back because nothing in the equation pulls it back: the string is infinite, has no equilibrium shape it prefers, and the only restoring force is curvature — which is zero on the flat raised plateau. Momentum \(\mu\int g\) was injected and is conserved; the raised region simply spreads sideways at speed \(c\) for ever.

3. Domain of dependence of \((3, 1)\) is \([3 - 2\cdot1,\ 3 + 2\cdot1] = [1, 5]\) m. In part 1, \(f\) vanishes on \([1,5]\) except at the single endpoint \(x = 1\), where \(f(1) = 0\) anyway, and d'Alembert's first term reads \(f\) only at \(x \mp ct = 1\) and \(5\): both zero. In part 2 the integral of \(g\) over \([1,5]\) is zero as well, since \(g\) vanishes there. So \(u(3,1) = 0\) in both cases.

Check. Dimensions: part 2's \(g/2c\) times a length is \((\mathrm{m\,s^{-1}}/\mathrm{m\,s^{-1}})\times\mathrm m = \mathrm m\) ✓. Limit: in part 1, setting \(t = 0\) recovers \(u(0,0) = f(0) = 1\) m ✓; in part 2, \(t \to 0^+\) gives \(u \to 0\) with \(u_t \to 3\ \mathrm{m\,s^{-1}} = g(0)\) ✓. Number: \(\int u_t\,\mathrm dx\) is conserved and equals \(\int g = 6\ \mathrm{m^2\,s^{-1}}\); and the raised region is a single plateau of height \(\tfrac32\) m whose two edges move apart at \(2\ \mathrm{m\,s^{-1}}\), so \(\int u\,\mathrm dx\) grows at \(\tfrac32\times(2+2) = 6\ \mathrm{m^2\,s^{-1}}\) ✓.

Another way

Part 2 without any integration: \(g\) is the derivative of the ramp \(G(x) = 3(x+1)\) clipped to \([0,6]\), and d'Alembert with \(f=0\) is \(\tfrac1{2c}[G(x+ct) - G(x-ct)]\). Reading off \(G\) at \(\pm2t\) gives the same two cases with no overlap bookkeeping.

Idea. A boundary condition is a statement that some combination vanishes at one point for all time. Put an imaginary pulse on the far side of that point, chosen so the combination cancels by symmetry, and the boundary condition holds automatically. Odd image for a fixed end, even image for a free one — the continuous version of the odd and even extensions in the deck.

1. For \(u = F(x-ct)\): \(u(x,0) = F(x)\) and \(u_t(x,0) = -cF'(x)\). Setting \(F = f\) gives exactly the stated data, and \(F(x-ct)\) solves the equation since \(u_{tt} = c^2F'' = c^2u_{xx}\). Conversely d'Alembert's formula on that data gives \(\tfrac12[f(x-ct)+f(x+ct)] + \tfrac1{2c}\int_{x-ct}^{x+ct}(-cf') = \tfrac12[f(x-ct)+f(x+ct)] - \tfrac12[f(x+ct) - f(x-ct)] = f(x-ct)\) ✓. With \(+cf'\) the same computation leaves \(f(x+ct)\), the left-mover.

2. The second term is \(f\big(-(x+ct) + 2L\big)\), a function of \(x + ct\) alone, so it solves the equation; the sum of two solutions solves it (linearity, unit 06). At \(x = L\): \(f(L - ct) - f(2L - L - ct) = f(L-ct) - f(L-ct) = 0\) ✓. Its initial data on \(x \le L\) is \(f(x) - f(2L-x)\) and \(-cf'(x) - cf'(2L-x)\); since \(f\) is supported in \([x_0 - w, x_0 + w] = [0.25, 0.35]\) and \(2L - x \ge 2L - L = L = 0.6477\) there, the image terms vanish on the support ✓.

Flatness. Put \(s = x - L\) and \(ct_* = L - x_0\). Then \(x - ct_* = x_0 + s\) and \(2L - x - ct_* = L - x + x_0 = x_0 - s\). Since \(f\) is symmetric about \(x_0\), \(f(x_0+s) = f(x_0-s)\), so \(u(x,t_*) \equiv 0\) for every \(x\): a photograph at that instant shows a perfectly straight string.

\[ t_* = \frac{L - x_0}{c} = \frac{0.6477 - 0.30}{427.0} = 8.143\times10^{-4}\ \mathrm s = 0.8143\ \mathrm{ms}. \]

The energy has not gone: \(u_t(x,t_*) = -c\big[f'(x_0+s) + f'(x_0-s)\big]\), and \(f'\) is odd about \(x_0\)… which would make this vanish too. It does not, because the second term differentiates with respect to \(x\) through a minus sign: explicitly \(u_t = -cf'(x-ct) + cf'(2L-x-ct)\), which at \(t_*\) is \(-c[f'(x_0+s) - f'(x_0-s)] = -2cf'(x_0+s)\), nonzero. All the energy is kinetic at that instant.

3. \(u = f(x-ct) + f(2L-x-ct)\) gives \(u_x(L,t) = f'(L-ct) - f'(L-ct) = 0\) ✓. At \(t_*\), \(u(x,t_*) = f(x_0+s) + f(x_0-s) = 2f(x_0+s)\): the pulse at doubled height, so \(u(L,t_*) = 2f(x_0) = 2h = 4\) mm. And \(u_t(x,t_*) = -c[f'(x_0+s) + f'(x_0-s)] = 0\) by the oddness of \(f'\) about \(x_0\): the string is instantaneously motionless, with all its energy potential.

4. For \(t\) large the first term has left the region \(x \le L\) entirely and what remains is \(\mp f(2L - x - ct)\), a left-moving pulse of the same shape and height, arriving back at \(x_0\) at \(t = 2(L-x_0)/c = 1.629\) ms. So: a fixed end returns the pulse inverted, a free end returns it upright; both reverse its direction and neither changes its size.

Energy: by equipartition for a travelling wave, \(E = \int(\tfrac12\mu u_t^2 + \tfrac12Tu_x^2) = T\int f'^2\), and \(|f'| = h/w\) over a total length \(2w\), so

\[ E = T\,\frac{h^2}{w^2}\,2w = \frac{2Th^2}{w} = \frac{2(72.53)(2\times10^{-3})^2}{0.05} = 1.160\times10^{-2}\ \mathrm J = 11.60\ \mathrm{mJ}, \]

and the reflected pulse has the same \(|f'|\) and the same support length, hence the same energy — as the conservation theorem requires, since \(S = -Tu_tu_x\) vanishes at a fixed end (\(u_t = 0\)) and at a free end (\(u_x = 0\)) alike.

Check. Dimensions: \(Th^2/w\) is \(\mathrm{N\,m^2/m} = \mathrm{N\,m} = \mathrm J\) ✓. Limit: \(x_0 \to L\) gives \(t_* \to 0\), the pulse already at the wall ✓; \(w \to \infty\) at fixed \(h\) gives \(E \to 0\), a vanishingly gentle ramp ✓. Number: the maximum slope is \(h/w = 0.04\), so the small-slope approximation is wrong here by about \(2(0.04)^2 = 3.2\times10^{-3}\), i.e. 0.32 per cent ✓ — the linear model is safe.

Another way

Parts 2 and 3 also follow from the deck's extension theorem without inventing the image: extend \(f\) and \(-cf'\) oddly (resp. evenly) about \(L\) and feed the pair to d'Alembert's formula. The image form is the same solution with the bookkeeping done in advance, and it is the form that generalises — unit 23 uses image charges for exactly this reason, and for exactly this reason it only works for flat boundaries.

Idea. Nothing here is an approximation or a series. The initial profile is piecewise linear; the extension is piecewise linear; the average of two piecewise-linear functions is piecewise linear; so the string is, at every instant, a small number of straight segments whose joints travel at exactly \(\pm c\). Track the joints and interpolate.

1. On \([0,L]\), \(f = 3hx/L\) for \(0 \le x \le L/3\) and \(f = \tfrac32 h(1 - x/L)\) for \(L/3 \le x \le L\). On \([-L, 0]\), \(\tilde f(x) = -f(-x)\), which is \(3hx/L\) for \(-L/3 \le x \le 0\) and \(-\tfrac32 h(1 + x/L)\) for \(-L \le x \le -L/3\). Then extend \(2L\)-periodically. Its corners are exactly at \(x = \pm L/3 + 2nL\): there is no corner at \(x = 0\) (the two branches there share the slope \(3h/L\)) and none at \(x = \pm L\) (oddness about \(L\) continues the slope \(-\tfrac32 h/L\) straight through).

2. At \(t = L/4c\). Corners at \(x = L/3 - L/4 = L/12\) and \(x = L/3 + L/4 = 7L/12\). Evaluate:

• \(x = L/12\): \(x - ct = -L/6\) and \(x + ct = L/3\). \(\tilde f(-L/6) = -f(L/6) = -h/2\); \(\tilde f(L/3) = h\). So \(u = \tfrac12(h - \tfrac h2) = \tfrac h4\).
• \(x = 7L/12\): \(x - ct = L/3\) and \(x + ct = 5L/6\). \(\tilde f(L/3) = h\); \(\tilde f(5L/6) = \tfrac32h(1 - \tfrac56) = \tfrac h4\). So \(u = \tfrac12(h + \tfrac h4) = \tfrac{5h}8\).

Slopes: on \([0, L/12]\), \((h/4)/(L/12) = 3h/L\) — still the original left slope, because the left-going corner has not yet reached the wall. On \([L/12, 7L/12]\), \((\tfrac58 - \tfrac14)h/(L/2) = \tfrac34 h/L\), which is the mean \(\tfrac12\big(3 - \tfrac32\big)h/L\) ✓. On \([7L/12, L]\), \(-(\tfrac58 h)/(\tfrac{5}{12}L) = -\tfrac32 h/L\) ✓.

3. At \(t = L/2c\). The right-going corner is at \(x = L/3 + L/2 = 5L/6\). The left-going one left \(x = 0\) at \(t = a/c = L/3c\) and has travelled back to \(x = c(t - L/3c) = L/2 - L/3 = L/6\). Evaluate:

• \(x = L/6\): \(x - ct = -L/3\), \(\tilde f = -h\); \(x + ct = 2L/3\), \(\tilde f = \tfrac32h(1 - \tfrac23) = \tfrac h2\). So \(u = \tfrac12(-h + \tfrac h2) = -\tfrac h4\).
• \(x = 5L/6\): \(x - ct = L/3\), \(\tilde f = h\); \(x + ct = 4L/3\), and by oddness about \(L\), \(\tilde f(4L/3) = -\tilde f(2L/3) = -\tfrac h2\). So \(u = \tfrac12(h - \tfrac h2) = +\tfrac h4\).

Slopes: \(-\tfrac32h/L\) on \([0, L/6]\), then \(\tfrac34h/L\) on \([L/6, 5L/6]\), then \(-\tfrac32h/L\) on \([5L/6, L]\). The corner that bounced is the left-going one, at \(t = L/3c\), and the bounce inverted it: the segment adjoining \(x=0\) has flipped from slope \(+3h/L\) to \(-\tfrac32h/L\).

4. Energy. At rest initially, so all of it is elastic:

\[ E = \tfrac12T\!\int_0^L\! f'^2 = \tfrac12T\Big[\frac{h^2}{a} + \frac{h^2}{L-a}\Big] = \frac{Th^2L}{2a(L-a)} \;\xrightarrow{\ a = L/3\ }\; \frac{Th^2 L}{2\cdot\frac L3\cdot\frac{2L}3} = \frac{9Th^2}{4L}. \]

The potential energy at \(t = L/2c\) is \(\tfrac12T\int u_x^2\). With \(a = L/3\) the three segments have lengths \(L/6,\ 2L/3,\ L/6\) and slopes \(-\tfrac32,\ \tfrac34,\ -\tfrac32\) in units of \(h/L\):

\[ \int_0^L u_x^2 = \frac{h^2}{L^2}\Big[\tfrac94\cdot\tfrac L6 + \tfrac9{16}\cdot\tfrac{2L}3 + \tfrac94\cdot\tfrac L6\Big] = \frac{h^2}{L}\Big[\tfrac38 + \tfrac38 + \tfrac38\Big] = \frac{9h^2}{8L}. \]

and \(\int f'^2 = h^2L/\big(a(L-a)\big) = 9h^2/(2L)\). The potential fraction is therefore \(\tfrac{9}{8}\big/\tfrac92 = \tfrac14 = \mathbf{25}\) per cent, the remaining 75 per cent being kinetic. (For \(a = L/5\) the same arithmetic on lengths \(0.3L,\ 0.4L,\ 0.3L\) and slopes \(-1.25,\ 1.875,\ -1.25\) gives \(2.344/6.25 = 37.5\) per cent, the deck's number ✓.)

Check. Dimensions: \(Th^2/L\) is \(\mathrm{N\,m^2/m} = \mathrm J\) ✓. Limits: at \(a = L/2\) the middle slope \(\tfrac12(h/a - h/(L-a))\) is zero — the genuinely flat top ✓; as \(a \to 0\), \(E \to \infty\) ✓. Consistency: every displacement computed above is a rational multiple of \(h\), and at \(t = L/c\) the formula gives \(u(x) = -f(L-x)\), the mirrored inverted triangle, so the motion has period \(2L/c\) ✓. Number: for the E string with \(h = 3\) mm and \(a = L/3\), \(E = 9(72.53)(9\times10^{-6})/(4\times0.6477) = 2.268\) mJ, less than the \(3.149\) mJ of the \(a = L/5\) pluck, as \(1/\big(a(L-a)\big)\) being smaller at \(L/3\) demands ✓.

Idea. Write the matrix, take \(-\det\), read the sign. The verdict on data then comes from the table: Cauchy data on a non-characteristic curve for hyperbolic, one-sided initial data plus boundary conditions for parabolic, whole-boundary data for elliptic.

1. \(\begin{pmatrix}1&0\\0&-c_0^2\end{pmatrix}\), \(D = c_0^2 \gt 0\): hyperbolic everywhere. Takes \(u\) and \(u_t\) on \(t = 0\) (non-characteristic, since the characteristics are \(x \mp c_0t = \)const), in both time directions.
2. \(\begin{pmatrix}0&0\\0&-k\end{pmatrix}\) in the ordering \((t,x)\), \(D = 0\): parabolic everywhere. Takes \(u\) at one instant plus a boundary condition at each spatial end, for \(t\) increasing only.
3. \(\begin{pmatrix}1&0\\0&1\end{pmatrix}\), \(D = -1 \lt 0\): elliptic everywhere. Takes one condition — \(u\), or its normal derivative, or a combination — on the entire boundary of a bounded region, and nothing else.
4. \(\begin{pmatrix}1&-1\\-1&1\end{pmatrix}\), \(D = 1 - 1 = 0\): parabolic everywhere. The single characteristic family solves \(m^2 + 2m + 1 = 0\), \(m = -1\), so the characteristics are the lines \(x + y = \) const, and data goes on a curve crossing them. (P6 reduces a very similar equation to canonical form.)
5. \(\begin{pmatrix}1&0\\0&4\end{pmatrix}\), \(D = -4 \lt 0\): elliptic everywhere; the substitution \(\beta = y/2\) makes it \(u_{\alpha\alpha} + u_{\beta\beta} + u_\alpha = 0\). Boundary data on a closed curve.
6. \(\begin{pmatrix}x^2&0\\0&-y^2\end{pmatrix}\), \(D = x^2y^2\): hyperbolic in each of the four open quadrants, and degenerate on the axes — rank 1 there (parabolic) except at the origin, where the symbol is identically zero and the equation says nothing at all at second order.

For (6) the type is a pointwise property, so a global verdict is meaningless: one may pose a Cauchy problem inside any one quadrant, on a curve not tangent to the characteristics \(m = \pm y/x\) (integrating \(\mathrm dy/\mathrm dx = \pm y/x\) gives \(xy = \)const and \(y/x = \)const), but a curve or region straddling an axis crosses into degeneracy, where the statement "Cauchy data determines the solution" has no content because the leading coefficient vanishes.

Check. Consistency: (1) and (6) restricted to the quadrant \(x,y \gt 0\) are both hyperbolic, and indeed the substitution \(\sigma = \ln x\), \(\tau = \ln y\) turns (6) into \(u_{\sigma\sigma} - u_{\tau\tau} - u_\sigma + u_\tau = 0\), the wave equation with first-order terms ✓. Special case: putting \(k = 0\) in (2) leaves \(u_t = 0\), whose symbol is the zero form — the degenerate limit, and correspondingly the equation no longer propagates or smooths anything ✓. Sign check on (4): with \(b = +1\) instead, \(u_{xx} + 2u_{xy} + u_{yy}\) is also parabolic, with characteristics \(x - y\) — the sign of \(b\) tilts the characteristics, it cannot change the type ✓.

Idea. The characteristic ODE hands you the coordinates in which the pure second derivatives are guaranteed to vanish — because each new coordinate's differential is a characteristic covector, and the transformation theorem says the diagonal entries of the new symbol are exactly \(\sigma\) evaluated on those differentials.

1. \(a = 3\), \(b = 5\), \(c = 3\); \(D = 25 - 9 = 16 \gt 0\): hyperbolic. \(3m^2 - 10m + 3 = 0\) gives \(m = 3,\ \tfrac13\), so the characteristics are \(y - 3x = \)const and \(y - \tfrac13x = \)const. Take

\[ \xi = y - 3x, \qquad \eta = 3y - x \]

(the second scaled by 3 to keep integers). Then \(\partial_x = -3\partial_\xi - \partial_\eta\) and \(\partial_y = \partial_\xi + 3\partial_\eta\), so

\[ u_{xx} = 9u_{\xi\xi} + 6u_{\xi\eta} + u_{\eta\eta}, \quad u_{xy} = -3u_{\xi\xi} - 10u_{\xi\eta} - 3u_{\eta\eta}, \quad u_{yy} = u_{\xi\xi} + 6u_{\xi\eta} + 9u_{\eta\eta}. \]

Combining with weights \(3, 10, 3\): the \(u_{\xi\xi}\) coefficient is \(27 - 30 + 3 = 0\) ✓, the \(u_{\eta\eta}\) coefficient is \(3 - 30 + 27 = 0\) ✓, and the \(u_{\xi\eta}\) coefficient is \(18 - 100 + 18 = -64\). So the canonical form is \(u_{\xi\eta} = 0\) and

\[ \boxed{\,u(x,y) = F(y - 3x) + G(3y - x)\,}. \]

Verify. For \(u = F(y-3x)\): \(u_{xx} = 9F''\), \(u_{xy} = -3F''\), \(u_{yy} = F''\), and \(27 - 30 + 3 = 0\) ✓. For \(u = G(3y-x)\): \(u_{xx} = G''\), \(u_{xy} = -3G''\), \(u_{yy} = 9G''\), and \(3 - 30 + 27 = 0\) ✓.

2. \(a = 1\), \(b = -2\), \(c = 4\); \(D = 4 - 4 = 0\): parabolic. \(m^2 + 4m + 4 = 0\) gives the repeated root \(m = -2\), so the one characteristic family is \(y + 2x = \)const. Take \(\xi = y + 2x\) and \(\eta = y\). Then \(\partial_x = 2\partial_\xi\), \(\partial_y = \partial_\xi + \partial_\eta\), and

\[ u_{xx} = 4u_{\xi\xi},\quad u_{xy} = 2u_{\xi\xi} + 2u_{\xi\eta},\quad u_{yy} = u_{\xi\xi} + 2u_{\xi\eta} + u_{\eta\eta}, \]

so \(u_{xx} - 4u_{xy} + 4u_{yy} = (4 - 8 + 4)u_{\xi\xi} + (-8 + 8)u_{\xi\eta} + 4u_{\eta\eta} = 4u_{\eta\eta}\). Canonical form \(u_{\eta\eta} = 0\), hence \(u = A(\xi) + \eta B(\xi)\):

\[ \boxed{\,u(x,y) = A(y+2x) + y\,B(y+2x)\,}. \]

Still two arbitrary functions, but they enter differently: one profile is transported along the characteristics and the other grows linearly in \(\eta\) while being transported. The count of arbitrary functions is two for any second-order equation in two variables; what the repeated root changes is that there is only one direction of transport, so the second function has to ride on the first family rather than on a second one.

3. On \(y = 0\): \(\xi = -3x\) and \(\eta = -x\), so \(u = F(-3x) + G(-x) = 0\) for all \(x\). Differentiating in \(y\): \(u_y = F'(\xi) + 3G'(\eta)\), so \(F'(-3x) + 3G'(-x) = \phi(x)\). From the first relation, \(F(-3x) = -G(-x)\); differentiating in \(x\), \(-3F'(-3x) = G'(-x)\). Substituting, \(F'(-3x) - 9F'(-3x) = \phi(x)\), so \(F'(-3x) = -\phi(x)/8\), that is \(F'(s) = -\phi(-s/3)/8\); and \(G'(-x) = 3\phi(x)/8\), i.e. \(G'(s) = 3\phi(-s)/8\). Integrating,

\[ u(x,y) = -\tfrac18\!\int_0^{\,y-3x}\!\!\phi\!\big(-\tfrac s3\big)\mathrm ds + \tfrac38\!\int_0^{\,3y-x}\!\!\phi(-s)\,\mathrm ds , \]

the two constants of integration cancelling because \(F(0) + G(0) = 0\) at \(x = 0\).

The line \(y = 0\) is non-characteristic: its conormal is \(\mathrm dy = (0,1)\) and \(\sigma(0,1) = c = 3 \ne 0\). Had it been characteristic, the two relations above would have been proportional: one of them would give no new information, and the system would be either inconsistent (no solution) or under-determined (infinitely many). That is P12.

Check. Dimensions play no role here — these are pure mathematics problems with dimensionless coefficients — so check instead by special case. Part 3 with \(\phi \equiv 1\): the formula gives \(u = -\tfrac18(y-3x) + \tfrac38(3y-x) = -\tfrac y8 + \tfrac{3x}8 + \tfrac{9y}8 - \tfrac{3x}8 = y\), and indeed \(u = y\) has \(u = 0\), \(u_y = 1\) on \(y=0\) and is annihilated by every second-order operator ✓. Part 2 special case \(B \equiv 0\) reduces to pure transport along \(y + 2x\) ✓. Part 1 limit: replacing \(10u_{xy}\) by \(6u_{xy}\) gives \(D = 9 - 9 = 0\), the parabolic boundary, and the two roots \(m = 1\) merge — the hyperbolic answer degenerates continuously into the parabolic one ✓.

Idea. An elliptic operator has no real characteristic covector, so a short wave in \(x\) cannot simply propagate in \(y\): it must grow or decay exponentially instead, at a rate equal to its own wavenumber. The Cauchy problem, which fixes the function and its normal derivative, has no way to select the decaying branch, so it selects a mixture — and the growing part wins.

1. Symbol matrix \(\mathrm{diag}(1,1)\), \(D = -1 \lt 0\): elliptic. The table says elliptic equations take one condition on the whole boundary of a bounded region. Cauchy data is two conditions on an unbounded curve that is not a boundary of anything — wrong in both respects.

2. \(\partial_x^2u_n = -n^2\cdot n^{-p-1}\sin(nx)\sinh(ny)\) and \(\partial_y^2u_n = +n^2\cdot n^{-p-1}\sin(nx)\sinh(ny)\); they cancel, so \(u_n\) is harmonic ✓. At \(y = 0\): \(\sinh 0 = 0\) so \(u_n(x,0) = 0\) ✓, and \(\partial_yu_n = n^{-p}\sin(nx)\cosh(ny)\), which at \(y = 0\) is \(n^{-p}\sin(nx)\) ✓. Uniqueness for this problem is guaranteed by Cauchy–Kovalevskaya, the data being analytic.

3. \(\partial_x^k\big[n^{-p}\sin(nx)\big]\) has supremum \(n^{k-p}\), which tends to 0 for every \(k \le p-1\). So the data goes to zero in \(C^{p-1}\). Meanwhile

\[ \sup_x|u_n(x,1)| = \frac{\sinh n}{n^{p+1}} \sim \frac{e^{n}}{2n^{p+1}} \longrightarrow \infty \]

for any fixed \(p\), since the exponential beats every power. Existence holds (we exhibited the solution) and uniqueness holds (Cauchy–Kovalevskaya): what fails is Hadamard's third condition, continuous dependence on the data. And it fails in the strongest possible sense — data tending to zero with as many derivatives as one likes, solutions tending to infinity.

4. With data \(\epsilon\sin(nx)\) the solution is \((\epsilon/n)\sin(nx)\sinh(ny)\), of supremum \((\epsilon/n)\sinh n\) at \(y = 1\). Requiring this below \(10^{-3}\) with \(n = 50\):

\[ \epsilon \le \frac{10^{-3}\times 50}{\sinh 50} = \frac{5\times10^{-2}}{2.5924\times10^{21}} = 1.93\times10^{-23}. \]

Double precision resolves relative differences of about \(2\times10^{-16}\), so a datum of order 1 carries a representation error some seven orders of magnitude larger than the tolerance. One cannot even write down data accurate enough, let alone measure it. This is what "ill posed" costs in practice, and it is why the diagnosis must precede the computation.

5. \(u = \dfrac{1}{n^2c}\sin(nx)\sin(nct)\) has \(u_{tt} = -n^2c^2u = c^2u_{xx}\) ✓, \(u(x,0) = 0\) ✓, and \(u_t(x,0) = \tfrac1n\sin(nx)\cdot\cos 0 = \tfrac1n\sin(nx)\) ✓ — the same data as part 2 with \(p = 1\). Its supremum is \(1/(n^2c)\), for all time. With \(n = 50\) and \(c = 427.0\ \mathrm{m\,s^{-1}}\) that is \(9.4\times10^{-7}\), against the elliptic answer \(\sinh(50)/50^2 = 1.0\times10^{18}\) at unit distance: a ratio of about \(10^{24}\).

Check. Dimensions: parts 1–4 are dimensionless; in part 5 the data \(n^{-1}\sin(nx)\) is a velocity and \(1/(n^2c)\) is \(\mathrm{m^2}\cdot(\mathrm{m\,s^{-1}})^{-1}\cdot\mathrm{m^{-1}}\cdot\mathrm{m\,s^{-1}} = \mathrm m\) once \(n\) is given its proper \(\mathrm m^{-1}\) ✓. Limit: as \(n \to 1\) the elliptic solution is \(\sinh(1) = 1.175\), perfectly tame — the failure is a statement about short waves only, which is why the principal symbol, which sees only short waves, is the right diagnostic ✓. Special case: replace \(\sinh(ny)\) by \(e^{-ny}\), also harmonic, and the problem becomes benign — but that is the Dirichlet problem on a half-plane with decay imposed, not the Cauchy problem, and it is exactly the extra condition the Cauchy formulation refuses to supply ✓.

Another way

One can avoid exhibiting \(\sinh\) at all: the general separated solution is \(\sin(nx)(\alpha e^{ny} + \beta e^{-ny})\), and any Cauchy datum that is not exactly tuned to \(\alpha = 0\) excites the growing branch. The tuning required is \(\alpha = 0\) to within \(e^{-n}\), which is precisely the statement that the map from data to solution has operator norm \(e^{n}\).

Idea. Type is a pointwise property of the symbol, so a coefficient that changes sign changes the question the equation is willing to answer. The Euler–Tricomi equation does this across a line, and the geometry of its characteristics near that line is the whole difficulty of transonic aerodynamics in miniature.

1. \(a = y\), \(b = 0\), \(c = 1\): \(D = b^2 - ac = -y\).

• \(y \gt 0\): \(D \lt 0\), elliptic.
• \(y = 0\): \(D = 0\), parabolic (the symbol is \(\xi_y^2\), rank 1).
• \(y \lt 0\): \(D \gt 0\), hyperbolic.

2. \(y\,m^2 + 1 = 0\) gives \(m = \mathrm dy/\mathrm dx = \pm(-y)^{-1/2}\) for \(y \lt 0\). Separating, \(\sqrt{-y}\,\mathrm dy = \pm\mathrm dx\). Put \(s = -y \gt 0\), so \(\mathrm dy = -\mathrm ds\):

\[ -\int\sqrt s\,\mathrm ds = \pm\int\mathrm dx \;\Longrightarrow\; -\tfrac23 s^{3/2} = \pm x + C \;\Longrightarrow\; x \pm \tfrac23(-y)^{3/2} = \mathrm{const}. \]

3. \(\mathrm dy/\mathrm dx = \pm(-y)^{-1/2} \to \pm\infty\) as \(y \to 0^-\), so the tangent becomes vertical: each characteristic meets \(y = 0\) perpendicularly, in a cusp. Physically, a signal travelling along a characteristic arrives at the parabolic line moving purely in the \(y\)-direction — it strikes the line head-on rather than grazing it — and beyond it there is no characteristic to continue along, because the region \(y \gt 0\) has none. Information stops propagating and starts being determined globally.

4. Through \((0,-1)\), \((-y)^{3/2} = 1\), so the constants are \(0 \pm \tfrac23 = \pm\tfrac23\):

\[ x + \tfrac23(-y)^{3/2} = \tfrac23 \qquad\text{and}\qquad x - \tfrac23(-y)^{3/2} = -\tfrac23 . \]

At \(y = 0\) they give \(x = +\tfrac23\) and \(x = -\tfrac23\). The interval influenced is \([-\tfrac23, \tfrac23]\), of width \(\tfrac43 = 1.3333\).

5. For steady two-dimensional potential flow, linearising about a uniform stream gives \((1 - M^2)\varphi_{xx} + \varphi_{yy} = 0\), where \(M\) is the local Mach number. Subsonic flow (\(M \lt 1\)) makes the coefficient positive — elliptic, no characteristics, every point felt everywhere, which is why a subsonic aerofoil's upstream flow "knows" it is coming. Supersonic flow (\(M \gt 1\)) makes it negative — hyperbolic, with Mach lines as characteristics and a genuine upstream-silent zone. A transonic flow has both regions at once, separated by the sonic line \(M = 1\), and near it the coefficient vanishes linearly in the distance: which is exactly \(u_{yy} + y\,u_{xx} = 0\) with \(y\) measuring signed distance from sonic. So \(y = 0\) is the sonic line, and the mixed type is the reason transonic flow resisted both the elliptic and the hyperbolic methods of the 1940s and had to wait for numerical ones.

Check. Dimensions are absent (the equation is a model form; in the aerodynamic version the coefficient \(1-M^2\) is dimensionless ✓). Limit: at \(y = -1\) the characteristic slopes are \(\pm1\), matching the ordinary wave equation with unit speed, and as \(y \to -\infty\) they flatten to \(\pm(-y)^{-1/2} \to 0\) — signals travel ever faster in \(x\), which is the hyperbolic region becoming ever more "supersonic" ✓. Special case: replacing \(y\) by the constant \(-1\) gives \(u_{yy} - u_{xx} = 0\) with straight characteristics \(x \pm y\), and indeed the semicubical parabolas reduce to straight lines in that limit if one rescales ✓. Number: the influenced width \(\tfrac43\) exceeds the \(2\times1 = 2\) one would get from straight unit-slope characteristics only if the curves bent outward — they bend inward, and \(\tfrac43 \lt 2\) ✓.

Idea. A conserved quantity that is a sum of squares is a uniqueness theorem in disguise: it is zero at \(t=0\) for the difference of two solutions, stays zero, and being a sum of squares can only be zero if each square is.

1. Proof. Suppose \(u_1, u_2\) both solve the problem, and set \(w = u_1 - u_2\). By linearity \(\mu w_{tt} = Tw_{xx}\), and \(w(0,t) = 0\), \(w_x(L,t) = 0\), \(w(x,0) = 0\), \(w_t(x,0) = 0\).

Define \(E(t) = \int_0^L\big(\tfrac12\mu w_t^2 + \tfrac12Tw_x^2\big)\mathrm dx\). Since \(w\) is \(C^2\), the integrand and its \(t\)-derivative are continuous on the compact rectangle \([0,L]\times[0,\tau]\) for every \(\tau\); uniform continuity on that compact set is what licenses Leibniz's rule, so

\[ \dot E = \int_0^L\big(\mu w_tw_{tt} + Tw_xw_{xt}\big)\mathrm dx = -\int_0^L\partial_xS\,\mathrm dx = S(0,t) - S(L,t), \qquad S = -Tw_tw_x, \]

the middle equality being the local conservation law, which is one line of algebra using the equation. At \(x = 0\), \(w(0,t) = 0\) for all \(t\) so \(w_t(0,t) = 0\) and \(S(0,t) = 0\). At \(x = L\), \(w_x(L,t) = 0\) so \(S(L,t) = 0\). Hence \(\dot E \equiv 0\).

At \(t = 0\): \(w_t(\cdot,0) = 0\), and \(w(\cdot,0) = 0\) identically in \(x\) so \(w_x(\cdot,0) = 0\) too (differentiating along the initial line). Thus \(E(0) = 0\), so \(E \equiv 0\).

The integrand \(\tfrac12\mu w_t^2 + \tfrac12Tw_x^2\) is continuous and nonnegative with zero integral, hence identically zero: if it were \(\epsilon \gt 0\) at some \(x_0\), continuity would give a neighbourhood of width \(\delta\) on which it exceeds \(\epsilon/2\), forcing \(E \ge \epsilon\delta/2 \gt 0\). So \(w_t \equiv w_x \equiv 0\) on the connected set \([0,L]\times[0,\infty)\), making \(w\) constant; and \(w(0,0) = 0\) makes the constant zero. Therefore \(u_1 = u_2\). \(\square\)

Where continuity and compactness were used: (i) uniform continuity of the integrand's \(t\)-derivative on a compact rectangle, for Leibniz's rule; (ii) continuity of the integrand, for the "nonnegative, continuous, zero integral" step; (iii) connectedness, to pass from vanishing gradient to constancy.

2. The images construction extends \(f, g\) oddly about \(x = 0\) and evenly about \(x = L\). Composing the two reflections: odd about 0 then even about \(L\) gives \(\tilde f(x + 2L) = -\tilde f(x)\), an antiperiodicity of period \(2L\), hence periodicity of period \(4L\). So \(u(x, t + 4L/c) = u(x,t)\), and \(u(x, t + 2L/c) = -u(x,t)\), which is not periodicity unless \(u \equiv 0\). Any frequency present must therefore satisfy \(e^{\mathrm i\omega\cdot 2L/c} = -1\), i.e. \(\omega\cdot2L/c = (2k-1)\pi\), giving \(\omega_k = (2k-1)\pi c/2L\): the odd harmonics only.

3. Proof. With the damping term, \(e_t + S_x = u_t\big(\mu u_{tt} - Tu_{xx}\big) = u_t\big(-\gamma u_t\big) = -\gamma u_t^2\). Integrating over \([0,L]\) and using Leibniz's rule as before,

\[ \dot E = -\big[S\big]_0^L - \gamma\int_0^Lu_t^2\,\mathrm dx = -\gamma\int_0^Lu_t^2\,\mathrm dx \le 0, \]

both ends fixed killing the boundary term. For uniqueness, take \(w = u_1 - u_2\) again: it solves the damped equation with zero data, so \(E_w(0) = 0\) and \(E_w\) is non-increasing and nonnegative, hence \(E_w \equiv 0\), and the argument of part 1 finishes identically. \(\square\) Note that monotonicity, not conservation, is all the method ever needed.

4. Nothing breaks. The argument in part 1 is applied to the difference \(w = u_1 - u_2\) of two solutions of the same problem, and the difference satisfies the homogeneous condition \(w_x(L,t) = \alpha(t) - \alpha(t) = 0\) whatever \(\alpha\) is. So \(S(L,t) = 0\) for \(w\) and uniqueness holds verbatim. What is not true any more is that the energy of a solution is conserved: \(\dot E = -S(L,t) = T u_t(L,t)\alpha(t)\), which is the power fed in by whatever is driving the end — correctly so, since a driven string gains and loses energy. The lesson is that the energy method proves uniqueness for the homogeneous problem and simultaneously accounts for the energy budget of the inhomogeneous one.

Check. Dimensions: \(\gamma u_t\) must be a force per unit length, \(\mathrm{N\,m^{-1}}\), so \([\gamma] = \mathrm{kg\,m^{-1}s^{-1}}\), and \(\gamma\int u_t^2\) is \(\mathrm{kg\,m^{-1}s^{-1}}\cdot\mathrm{m^2s^{-2}}\cdot\mathrm m = \mathrm{J\,s^{-1}} = \mathrm W\) ✓, matching \([\dot E]\). Limit: \(\gamma \to 0\) recovers part 1's conservation ✓. Special case: for a string fixed at both ends, part 2's composition of two odd reflections gives period \(2L/c\) and all harmonics — the guitar, not the clarinet ✓. Number: with the E string's \(L\) and \(c\), a fixed–free string of the same length sounds \(c/4L = 164.8\) Hz, one octave below the fixed–fixed \(329.6\) Hz ✓.

Idea. A travelling wave is the one case in which the local conservation law becomes trivial — everything moves rigidly at \(c\), so the flux is the density times the speed, exactly as for a river.

1. Write \(\theta = q(x-ct)\). Then \(u_x = qA\cos\theta\) and \(u_t = -cqA\cos\theta\). So

\[ e = \tfrac12\mu c^2q^2A^2\cos^2\theta + \tfrac12Tq^2A^2\cos^2\theta = Tq^2A^2\cos^2\theta \]

using \(\mu c^2 = T\), and \(S = -Tu_tu_x = cTq^2A^2\cos^2\theta = c\,e\) ✓. Directly: \(\partial_te = Tq^2A^2\cdot2\cos\theta(-\sin\theta)(-cq) = 2cq^3TA^2\sin\theta\cos\theta\), while \(\partial_xS = cTq^2A^2\cdot2\cos\theta(-\sin\theta)(q) = -2cq^3TA^2\sin\theta\cos\theta\). They cancel ✓.

2. Kinetic density \(\tfrac12\mu u_t^2 = \tfrac12\mu c^2q^2A^2\cos^2\theta\); potential density \(\tfrac12Tu_x^2 = \tfrac12Tq^2A^2\cos^2\theta\). They are equal because \(\mu c^2 = T\), identically in \(x\) and \(t\) — equipartition, pointwise.

For the standing wave \(u = A\sin(qx)\cos(\omega t)\) with \(\omega = cq\): \(u_t = -\omega A\sin(qx)\sin(\omega t)\) and \(u_x = qA\cos(qx)\cos(\omega t)\), so

\[ \tfrac12\mu u_t^2 = \tfrac12 T q^2A^2\sin^2(qx)\sin^2(\omega t),\qquad \tfrac12 T u_x^2 = \tfrac12Tq^2A^2\cos^2(qx)\cos^2(\omega t). \]

These agree nowhere except on isolated lines: at a node (\(\sin qx = 0\)) the energy is all potential, at an antinode all kinetic, and twice per period the whole string is momentarily at rest with all its energy potential. A standing wave is a superposition of two travelling waves going opposite ways, and their fluxes cancel — \(\bar S = 0\) — which is exactly why energy no longer moves and equipartition no longer holds locally. (It does still hold when averaged over both \(x\) and \(t\): both averages are \(\tfrac18Tq^2A^2\).)

3. \(\overline{\cos^2} = \tfrac12\), so

\[ \bar e = \tfrac12 T q^2A^2, \qquad \bar P = c\,\bar e = \tfrac12 cTq^2A^2 = \tfrac12\,\mu c^3 q^2 A^2 = \boxed{\ \tfrac12\,\mu\,c\,\omega^2A^2\ } \]

using \(T = \mu c^2\) and \(q = \omega/c\). Power goes as amplitude squared and as frequency squared: the reason a high note of the same amplitude is louder, and the reason acoustic intensities are quoted in decibels.

4. \(\omega = 2\pi(329.63) = 2071.1\ \mathrm{rad\,s^{-1}}\) and

\[ q = \omega/c = 2071.1/427.0 = 4.8504\ \mathrm m^{-1} \quad(\lambda = 2\pi/q = 1.295\ \mathrm m = 2L\ ✓), \] \[ \bar e = \tfrac12(72.53)(4.8504)^2(5\times10^{-4})^2 = 2.133\times10^{-4}\ \mathrm{J\,m^{-1}}, \] \[ \bar P = \tfrac12(3.978\times10^{-4})(427.0)(2071.1)^2(5\times10^{-4})^2 = 9.107\times10^{-2}\ \mathrm W. \]

So \(\bar e = 0.213\ \mathrm{mJ\,m^{-1}}\) and \(\bar P = 91.1\) mW.

Check. Dimensions: \(\mu c\omega^2A^2\) is \(\mathrm{kg\,m^{-1}}\cdot\mathrm{m\,s^{-1}}\cdot\mathrm{s^{-2}}\cdot\mathrm{m^2} = \mathrm{kg\,m^2s^{-3}} = \mathrm W\) ✓. Cross-check: \(c\bar e = 427.0\times2.133\times10^{-4} = 9.11\times10^{-2}\) W, agreeing with the other route ✓. Limit: \(A \to 0\) gives \(\bar P \to 0\) ✓; \(q \to 0\) (a uniform displacement) carries no energy, correctly, since a straight string is unstressed ✓. Magnitude: the whole E string at this amplitude holds \(\bar e\,L = 0.138\) mJ, a factor 23 below the 3.149 mJ of the 3 mm pluck. Energy goes as amplitude squared, and \((3/0.5)^2 = 36\), so the pluck holds rather less than a pure fundamental of its peak height would — correctly, since a triangle is not a sinusoid and its peak exceeds its root-mean-square ✓.

Idea. Adding a term in \(u\) itself — with no derivatives — to the Lagrangian density leaves the principal symbol untouched, so it changes neither the type nor the characteristics nor the speed at which signals travel. What it does change is everything about individual sinusoids, which now disperse.

1. Take \(\eta\) vanishing on the boundary of the spacetime rectangle and set \(\Phi(\epsilon) = \mathcal S[u + \epsilon\eta]\). Then

\[ \Phi'(0) = \iint\big[\mu u_t\eta_t - Tu_x\eta_x - \kappa u\,\eta\big]\mathrm dx\,\mathrm dt = \iint\big[-\mu u_{tt} + Tu_{xx} - \kappa u\big]\eta\ \mathrm dx\,\mathrm dt, \]

integrating the first term by parts in \(t\) and the second in \(x\), both boundary terms vanishing because \(\eta\) does. The two-variable fundamental lemma then forces the bracket to vanish:

\[ \mu u_{tt} - Tu_{xx} + \kappa u = 0 \iff u_{tt} - c^2u_{xx} + \omega_0^2u = 0, \quad c^2 = T/\mu,\ \omega_0^2 = \kappa/\mu. \]

Equivalently, straight from the formula: \(\mathcal L_{u_t} = \mu u_t\), \(\mathcal L_{u_x} = -Tu_x\), \(\mathcal L_u = -\kappa u\), so \(\partial_t(\mu u_t) + \partial_x(-Tu_x) + \kappa u = 0\) ✓.

2. The principal symbol is unchanged, \(\mathrm{diag}(1, -c^2)\) in \((t,x)\): \(D = c^2 \gt 0\), still hyperbolic, with the same characteristics \(x \mp ct\) and therefore the same finite propagation speed \(c\). The mass term is a lower-order term and the classification does not see it. This is the same point the heat equation made in reverse: the classification tells you about signal propagation and nothing about what the lower-order terms do to the solution's shape.

3. Substituting \(u = A\cos(qx - \omega t)\): \(u_{tt} = -\omega^2u\), \(u_{xx} = -q^2u\), so \(-\omega^2 + c^2q^2 + \omega_0^2 = 0\):

\[ \boxed{\ \omega^2 = c^2q^2 + \omega_0^2\ } \]

Hence \(\omega \gt cq\) for every \(q\), so \(v_{\text{ph}} = \omega/q \gt c\). Differentiating, \(2\omega\,\mathrm d\omega = 2c^2q\,\mathrm dq\), so \(v_{\text{gr}} = \mathrm d\omega/\mathrm dq = c^2q/\omega = c^2/v_{\text{ph}} \lt c\). And

\[ v_{\text{ph}}\,v_{\text{gr}} = \frac{\omega}{q}\cdot\frac{c^2q}{\omega} = c^2 \ \text{exactly}. \]

No signal travels faster than \(c\): part 2 already guaranteed that, and the phase velocity is not the speed of anything — it is the speed of a pattern in an infinite wave that carries no information. The same relation \(v_{\text{ph}}v_{\text{gr}} = c^2\) holds for a relativistic particle in unit 27, where \(\omega_0\) is \(mc^2/\hbar\); the equation here is its classical ancestor, and \(\omega_0\) is a genuine cut-off: below it, \(q^2 \lt 0\), and the solutions do not propagate at all but decay exponentially in \(x\).

4. Take the Legendre transform \(e = u_t\mathcal L_{u_t} - \mathcal L = \mu u_t^2 - \tfrac12\mu u_t^2 + \tfrac12Tu_x^2 + \tfrac12\kappa u^2\):

\[ e = \tfrac12\mu u_t^2 + \tfrac12Tu_x^2 + \tfrac12\kappa u^2, \qquad S = -T\,u_tu_x \ \text{(unchanged)}. \]

Proof. \(e_t = \mu u_tu_{tt} + Tu_xu_{xt} + \kappa u\,u_t\) and \(S_x = -Tu_{xt}u_x - Tu_tu_{xx}\). Adding, the \(Tu_xu_{xt}\) terms cancel and

\[ e_t + S_x = u_t\big(\mu u_{tt} - Tu_{xx} + \kappa u\big) = 0 \]

by the field equation. \(\square\) The flux is unchanged because the new term has no derivatives: it can store energy locally but cannot transport it.

5. \(\omega_0 = 2\pi(100) = 628.32\ \mathrm{rad\,s^{-1}}\), so

\[ \kappa = \mu\omega_0^2 = (3.978\times10^{-4})(628.32)^2 = 157.0\ \mathrm{N\,m^{-2}}. \]

The fixed–fixed fundamental still has \(q_1 = \pi/L = 4.8504\ \mathrm m^{-1}\) — the images construction and the boundary conditions are untouched — but now

\[ \omega_1 = \sqrt{c^2q_1^2 + \omega_0^2} = \sqrt{(2071.1)^2 + (628.32)^2} = 2164.3\ \mathrm{rad\,s^{-1}}, \quad f_1 = 344.5\ \mathrm{Hz}, \]

up from \(329.63\) Hz — the note sharpens by \(76\) cents, about three quarters of a semitone. At that mode \(v_{\text{ph}} = \omega_1/q_1 = 446.2\ \mathrm{m\,s^{-1}}\) and \(v_{\text{gr}} = c^2/v_{\text{ph}} = 408.6\ \mathrm{m\,s^{-1}}\), whose product is \(1.8233\times10^{5} = c^2\) ✓.

Check. Dimensions: \([\kappa] = \mathrm{N\,m^{-1}}\) per metre \(= \mathrm{N\,m^{-2}}\), and \(\kappa/\mu\) is \(\mathrm{N\,m^{-2}}/(\mathrm{kg\,m^{-1}}) = \mathrm s^{-2}\) ✓, so \(\omega_0\) is a frequency; \(\tfrac12\kappa u^2\) is \(\mathrm{N\,m^{-2}\,m^2} = \mathrm{J\,m^{-1}}\), matching the other energy densities ✓. Limit: \(\kappa \to 0\) recovers every result of the deck — the wave equation, \(\omega = cq\), \(v_{\text{ph}} = v_{\text{gr}} = c\), and the two-term energy density ✓. Also \(q \to \infty\) gives \(\omega \to cq\): short waves do not feel the medium, exactly as the principal-symbol argument predicted ✓. Number: the shift \(344.5/329.63 = 1.0451\), and \(1200\log_2(1.0451) = 76.4\) cents ✓.

Idea. On a characteristic, one of the two travelling waves is frozen — its argument does not vary along the curve — so the data cannot see it at all, and must instead over-determine the other one. That is what "non-characteristic" is protecting against, and the failure is not subtle: it is a first-order ODE with two incompatible right-hand sides.

1. Proof. By the factorisation theorem every \(C^2\) solution on the plane is \(u = F(x-ct) + G(x+ct)\). On \(x = ct\) the arguments are \(x - ct = 0\) and \(x + ct = 2ct\). The data therefore read

\[ \text{(i)}\quad \phi(t) = F(0) + G(2ct), \qquad \text{(ii)}\quad \psi(t) = F'(0) + G'(2ct), \]

since \(u_x = F'(x-ct) + G'(x+ct)\).

Differentiate (i) in \(t\): \(\phi'(t) = 2c\,G'(2ct)\), so \(G'(2ct) = \phi'(t)/(2c)\). Substituting into (ii),

\[ \psi(t) = F'(0) + \frac{\phi'(t)}{2c} \quad\Longleftrightarrow\quad \psi(t) - \frac{\phi'(t)}{2c} = F'(0) = \text{constant}. \]

If \(\psi - \phi'/2c\) is not constant, no \(F, G\) can satisfy both conditions, so no solution exists. If it is constant, call it \(k\); then (i) determines \(G\) on \([0,\infty)\) up to the single number \(F(0)\), namely \(G(s) = \phi(s/2c) - F(0)\), and (ii) is satisfied as soon as \(F'(0) = k\). Every \(F\) with those two values at the origin then works, and \(u = F(x-ct) + G(x+ct)\) is a solution. \(\square\)

2. The family is

\[ u(x,t) = F(x - ct) + \phi\!\left(\frac{x+ct}{2c}\right) - F(0), \qquad F \in C^2,\quad F'(0) = k , \]

with \(F\) otherwise completely arbitrary — one arbitrary function of one variable, constrained only by two numbers at a point. That is an infinite-dimensional family, and it is the same size as the whole set of right-moving waves, which is exactly the half of the solution space the data never touched. (Away from \(t \ge 0\) or for \(x+ct \lt 0\) one continues \(G\) arbitrarily as well, so the statement is local to the region the characteristic bounds.)

3. The Goursat problem. Data on both characteristics through the origin. Now

\[ \phi(t) = F(0) + G(2ct), \qquad \chi(t) = F(-2ct) + G(0). \]

Normalise by absorbing a constant into \(F\) and \(G\) so that \(F(0) = \tfrac12\phi(0)\) and \(G(0) = \tfrac12\chi(0)\); since \(\phi(0) = \chi(0)\) these are equal, call the common value \(\tfrac12\phi(0)\). Then \(G(s) = \phi(s/2c) - \tfrac12\phi(0)\) for \(s \ge 0\) and \(F(s) = \chi(-s/2c) - \tfrac12\phi(0)\) for \(s \le 0\). Both are fully determined, so

\[ \boxed{\ u(x,t) = \chi\!\left(\frac{ct - x}{2c}\right) + \phi\!\left(\frac{x + ct}{2c}\right) - \phi(0)\ } \]

on the wedge \(|x| \le ct\), and it is the only solution there. Note the compatibility condition \(\phi(0) = \chi(0)\) is exactly the requirement that the two data agree where the characteristics meet; nothing further is needed, and no derivative data may be given.

4. A characteristic covector \(\xi\) is one on which the principal symbol vanishes, which means the operator's leading part does not differentiate across the corresponding curve — so the equation cannot convert data on that curve into a second derivative off it, and the normal derivative is not free to be prescribed: it is either already determined by the tangential data (part 1's compatibility condition) or meaningless.

Check. Dimensions: \(\phi\) and \(\chi\) are displacements, \((x+ct)/2c\) is a time, and the Goursat formula returns a displacement ✓. Special case of part 3: \(\phi = \chi \equiv 0\) gives \(u \equiv 0\), the unique solution ✓; \(\phi(t) = \chi(t) = vt\) gives \(u = v(ct-x)/2c + v(x+ct)/2c = vt\), which indeed solves \(u_{tt} = c^2u_{xx}\) and equals \(vt\) on both characteristics ✓. Consistency with part 1: in the Goursat problem \(u_x\) on \(x = ct\) is \(F'(0) + G'(2ct) = F'(0) + \phi'(t)/2c\), with \(F'(0)\) now fixed by \(\chi\) — so the number \(k\) that part 1 left free is precisely what the second characteristic's data supplies ✓. Counting: the wave equation's solution space is two arbitrary functions of one variable; ordinary Cauchy data fixes both, Goursat data fixes both, and characteristic Cauchy data fixes one and a half ✓.