4.7 Practice Problems
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Solution. With \(c = 2\), \[ u = \tfrac 12\bigl (\sin (x + 2t) + \sin (x - 2t)\bigr ) + \tfrac 14\bigl (\sin (x + 2t) - \sin (x - 2t)\bigr ) = \sin x\cos 2t + \tfrac 12\cos x\sin 2t . \] Check: \(u_t(x, 0) = \cos x\cos 0 = \cos x\).
Problem 4.7.2. A string with \(c = 1\), at rest and undisplaced, is struck so that \(u_t(x, 0) = 1\) for \(|x| < 1\) and \(0\) otherwise. Find \(u(0, t)\) for all \(t > 0\), and \(u(x, t)\) for large \(t\).
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Solution. \(u(0, t) = \frac 12\int _{-t}^tg = \frac 12\cdot 2\min (t, 1) = \min (t, 1)\): the centre rises steadily until \(t = 1\) and then stays at height \(1\). In general \(u(x, t) = \frac 12\,\bigl |[x - t, x + t] \cap [-1, 1]\bigr |\), which for \(t > 1 + |x|\) is \(1\): the string is left permanently displaced by \(1\) over an expanding interval \(|x| < t - 1\), with ramps at its edges. The disturbance does not pass.
Problem 4.7.3. Use the energy to prove that the problem \(u_{tt} = c^2u_{xx}\) on \(0 < x < L\), \(u(0, t) = u(L, t) = 0\), with given \(u(x, 0)\) and \(u_t(x, 0)\), has at most one \(C^2\) solution.
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Solution. The difference \(w\) of two solutions has zero data. Its energy \(E(t) = \frac 12\int _0^L(w_t^2 + c^2w_x^2)\,dx\) has \(E'(t) = c^2[w_tw_x]_0^L = 0\), since \(w = 0\) at the ends for all \(t\) implies \(w_t = 0\) there. So \(E(t) = E(0) = 0\), hence \(w_t = w_x = 0\), \(w\) is constant, and \(w = 0\) since it vanishes at \(x = 0\).
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Solution. By Duhamel, \[ u = \frac 12\int _0^t\bigl (\cos (x - (t - s)) - \cos (x + (t - s))\bigr )ds = \int _0^t\sin x\sin (t - s)\,ds = \sin x\,(1 - \cos t) . \] Check: \(u_{tt} = \sin x\cos t\) and \(u_{xx} + \sin x = -\sin x(1 - \cos t) + \sin x = \sin x\cos t\).
Problem 4.7.5. A guitar string \(0.65\) m long has fundamental frequency \(196\) Hz. Find the wave speed \(c\), and the tension if the linear density is \(3.1 \times 10^{-3}\) kg/m.
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Solution. From \(c/2L = 196\), \(c = 2 \times 0.65 \times 196 = 254.8\) m/s, and \(T = \rho c^2 = 3.1 \times 10^{-3} \times 254.8^2 = 201\) N.
Problem 4.7.6. For the half-line problem with a fixed end, \(f = 0\) and \(g(x) = 1\) for \(1 < x < 2\), \(0\) otherwise, \(c = 1\), find \(u(\frac 12, t)\) for \(t > 0\).
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Solution. For \(t \leq \frac 12\), \(u = \frac 12\int _{1/2 - t}^{1/2 + t}g = 0\), since the interval lies below \(1\). For \(t > \frac 12\), \(u = \frac 12\int _{t - 1/2}^{t + 1/2}g\), the half-length of \([t - \frac 12, t + \frac 12] \cap [1, 2]\): it is \(\frac 12(t - \frac 12)\) for \(\frac 12 \leq t \leq \frac 32\), \(\frac 12(\frac 52 - t)\) for \(\frac 32 \leq t \leq \frac 52\), and \(0\) after. The point is first reached by the left-moving disturbance at \(t = \frac 12\), and once the wave reflected from the fixed end has passed, it returns to rest: the reflected wave, from the odd extension, cancels the permanent displacement of Problem 4.7.2.
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