= Solution
Take $0\le\sigma\le\pi$ and split the coordinates into $a=0,\ldots,p$ along the branes and $I=p+1,\ldots,25$ transverse to them. In <conformal gauge> the <wave equation> permits a separated mode $e^{-i\omega\tau}f(\sigma)$ with $f''+\omega^2f=0$. For <Neumann boundary conditions> at both ends, $f'(0)=f'(\pi)=0$ selects $\cos m\sigma$ and <integer> $m$. The zero-frequency solution is independent of $\sigma$, with a constant position and a term linear in time. Therefore the <open-string mode expansion> is
$$
X^a=x^a+2\alpha'k^a\tau+i\sqrt{2\alpha'}\sum_{m\ne0}\frac{\alpha_m^a}{m}e^{-im\tau}\cos m\sigma.
$$
The <canonical momentum> density integrated over this interval is $k^a$.
For <Dirichlet boundary conditions>, subtract the static interpolation between fixed endpoint positions $y_0^I$ and $y_\pi^I$. The remaining field vanishes at both ends, selecting $\sin m\sigma$ with <integer> $m$ and excluding a <momentum> <worldsheet zero mode>. The expansion is
$$
\boxed{X^I=y_0^I+\frac{y_\pi^I-y_0^I}{\pi}\sigma+\sqrt{2\alpha'}\sum_{m\ne0}\frac{\alpha_m^I}{m}e^{-im\tau}\sin m\sigma.}
$$
Here $\alpha_m^{I\dagger}=\alpha_{-m}^I$. There is no factor $i$ in the <sine series> with this reality convention: changing $m$ to $-m$ reverses both the sine and the denominator. These are the <integer-mode expansion with parallel D-brane endpoints>; they are not the half-integer modes of a coordinate with a Neumann condition at one endpoint and a Dirichlet condition at the other.
The endpoints can move on $(p+1)$-dimensional <worldvolumes> but are fixed in the transverse directions. Thus these are <open strings> attached to <D-branes>, or stretched between parallel <D-branes> at $y_0$ and $y_\pi$. The tangential <momenta> describe propagation on the brane; the transverse positions describe its embedding. Let $L=|y_\pi-y_0|$. The linear spatial <worldsheet zero mode> contributes $L^2/(4\pi^2\alpha')$ to $L_0$. Quantization with <string intercept> one gives
$$
\boxed{M^2=\frac{L^2}{4\pi^2\alpha'^2}+\frac{N-1}{\alpha'}.}
$$
The first term is also the square of the classical stretched-string energy $TL$, since the <string tension> is $T=1/(2\pi\alpha')$.
For two coincident <D3-branes>, label the ends of an oriented <open string> by $i,j\in\{1,2\}$. The four endpoint combinations give <Chan-Paton factors> $\lambda_{ij}$, which form the full algebra of $2\times2$ <matrices>. Under a change of brane basis these transform as $\lambda\mapsto U\lambda U^{-1}$, the <adjoint action> of $U(2)$. At $L=0$, the first oscillator level is massless. Its tangential polarization supplies a four-dimensional vector $A_a$, while the 22 <transverse polarizations> supply real adjoint <scalar fields> $\Phi^I$. This is the <bosonic D3-brane low-energy field content>.
The endpoint <matrices> multiply in their boundary order in disk amplitudes. The difference between the two orders in the three-vector interaction gives the <matrix commutator>; the kinematic factor is the one-derivative <Yang-Mills theory> cubic vertex. Factorization and the vector <gauge redundancy> then supply the quartic interaction with the same <commutator>. Equivalently, the leading low-energy action is the <dimensional reduction> of 26-dimensional <Yang-Mills theory>, with
$$
F_{ab}=\partial_aA_b-\partial_bA_a-i[A_a,A_b],\qquad D_a\Phi^I=\partial_a\Phi^I-i[A_a,\Phi^I],
$$
$$
S_{\mathrm{low}}=\frac1{g_{\mathrm{YM}}^2}\int d^4x\,\operatorname{Tr}\left(-\frac14F_{ab}F^{ab}-\frac12D_a\Phi^I D^a\Phi^I+\frac14[\Phi^I,\Phi^J][\Phi^I,\Phi^J]\right)+\cdots.
$$
For Hermitian <scalar fields> the last displayed term corresponds to a nonnegative potential, because their <commutators> are anti-Hermitian. Higher-derivative terms are suppressed at energies well below $1/\sqrt{\alpha'}$. \b[Ignoring the <tachyon>, the gauge sector is four-dimensional $U(2)$ <Yang-Mills theory>, accompanied by 22 adjoint <scalar fields>.] It is not pure <Yang-Mills theory>, nor the supersymmetric six-scalar <D3-brane> theory of a ten-dimensional superstring.
Separate the two branes and use the <D-brane scalar-position normalization>
$$
\langle\Phi^I\rangle=\frac1{2\pi\alpha'}\begin{pmatrix}y_1^I&0\\0&y_2^I\end{pmatrix}.
$$
A <Yang-Mills gauge transformation> preserves this <expectation value> only when it is diagonal if $y_1\ne y_2$. Thus the unbroken <group> is $U(1)\times U(1)$. The <scalar field> <kinetic term> supplies a mass for the off-diagonal vectors, since
$$
[A_a,\langle\Phi^I\rangle]_{12}=\frac{y_2^I-y_1^I}{2\pi\alpha'}(A_a)_{12}.
$$
Summing over $I$ gives $m_{12}^2=L^2/(4\pi^2\alpha'^2)$, exactly the mass of the stretched $N=1$ string. The two diagonal gauge vectors remain massless. This identifies brane separation with an adjoint <Higgs mechanism>, rather than an explicit deletion of the off-diagonal strings.
For the <ground state> of an <open string> joining the two different branes, $N=0$, so the <bosonic stretched-string tachyon threshold> is
$$
\boxed{M_0^2<0\quad\Longleftrightarrow\quad L<2\pi\sqrt{\alpha'}.}
$$
At equality it is massless, and above it the stretched <ground state> has positive mass squared. The same-brane ground strings still have $M_0^2=-1/\alpha'$; this calculation removes only the stretched-string instability, not the original instability of the bosonic branes.
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