= Solution
The <Debye model> approximates a three-dimensional crystal as an elastic continuum supporting quantized <normal modes>, or <phonons>. At long <wavelengths>, the <acoustic phonon branches> have approximately linear <dispersion relations> $\omega_s(k)=v_s k$: one longitudinal branch and two transverse branches. Replace their complicated actual spectra by these linear forms, then introduce a cutoff chosen to preserve the total $3N$ vibrational degrees of freedom for $N$ atoms. This is an approximation to the spectrum, not a claim that a real crystal has no non-acoustic branches.
For <periodic boundary conditions> in <volume> $V$, each allowed <wavevector> occupies $(2\pi)^3/V$ in the space of <wavevectors>. Counting a spherical shell for each branch gives the <density of states> per unit <angular frequency>
$$
g(\omega)=\frac{V\omega^2}{2\pi^2}\left(\frac1{v_l^3}+\frac2{v_t^3}\right).
$$
An effective <speed of sound> $v_D$ can be defined by $3/v_D^3=1/v_l^3+2/v_t^3$. In the single-speed <Debye model>, choose $\omega_D=v_Dk_D$ so that
$$
\int_0^{\omega_D}g(\omega)\,d\omega=3N.
$$
Thus the <Debye wavevector>, <Debye frequency> and <Debye temperature> satisfy
$$
\boxed{k_D=\left(\frac{6\pi^2N}{V}\right)^{1/3},\qquad
\omega_D=v_Dk_D,\qquad \Theta_D=\frac{\hbar\omega_D}{k_B},\qquad
g(\omega)=\frac{9N\omega^2}{\omega_D^3}\quad(0<\omega<\omega_D).}
$$
Here $g$ has units of seconds when <angular frequency> is measured in $\mathrm{s^{-1}}$; its integral is the dimensionless number of <normal modes>. The spherical cutoff replaces the real <Brillouin zone> by an equal-count continuum construction.
Each <phonon> mode is a <quantum harmonic oscillator>. Since <phonon> number is not conserved at <thermal equilibrium>, the <chemical potential> is zero and its <Bose-Einstein distribution> gives mean thermal <energy> $\hbar\omega/[e^{\hbar\omega/(k_BT)}-1]$. The <zero-point energy> is independent of <temperature> in the harmonic, fixed-volume approximation and does not contribute to the <heat capacity at constant volume>. Excluding that constant term,
$$
U(T)=\int_0^{\omega_D}\frac{g(\omega)\hbar\omega}{e^{\hbar\omega/(k_BT)}-1}\,d\omega
=9Nk_BT\left(\frac{T}{\Theta_D}\right)^3
\int_0^{\Theta_D/T}\frac{x^3}{e^x-1}\,dx.
$$
Differentiating the first <integral> at fixed <volume> avoids losing the temperature-dependent upper limit in the second one. The resulting <Debye heat capacity> is
$$
\boxed{C_V=9Nk_B\left(\frac{T}{\Theta_D}\right)^3
\int_0^{\Theta_D/T}\frac{x^4e^x}{(e^x-1)^2}\,dx.}
$$
For $T\ll\Theta_D$, extend the upper limit to infinity. Expanding $1/(e^x-1)=\sum_{r\ge1}e^{-rx}$ and integrating term by term gives $\int_0^\infty x^3/(e^x-1)\,dx=6\zeta(4)=\pi^4/15$. Therefore
$$
U\simeq\frac{3\pi^4}{5}Nk_B\frac{T^4}{\Theta_D^3},\qquad
\boxed{C_V\simeq\frac{12\pi^4}{5}Nk_B\left(\frac{T}{\Theta_D}\right)^3.}
$$
This $T^3$ law follows from the three-dimensional $\omega^2$ <density of states> and the availability only of low-<frequency> modes at low <temperature>. It is a central success of the <Debye model>.
For $T\gg\Theta_D$, the <Bose-Einstein distribution> reduces to the classical <equipartition theorem>: every mode contributes thermal <energy> $k_BT$. Consequently
$$
\boxed{U\simeq3Nk_BT,\qquad C_V\longrightarrow3Nk_B,}
$$
the <Dulong-Petit law>, or $3R$ per mole of atoms. The simple model omits detailed <phonon dispersion relation>, distinct <optical phonon branches>, and anharmonic effects such as <thermal expansion>. In a <metal>, the low-<temperature> total <heat capacity> additionally contains a term approximately proportional to $T$ from <Electrons>, so the lattice $T^3$ term should not be mistaken for the whole measurement.
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