= Solution
The two new boundaries surround opposite sides of the same bulk. Their induced orientations are opposite, so their chiral edge modes counterpropagate and $\operatorname{sgn}v_1=-\operatorname{sgn}v_2$.
For $v_1=v$ and $v_2=-v$, define the mass
$$
m(x)=t\cos\frac{\phi(x)}2.
$$
It approaches $+t$ on the left and $-t$ on the right. The zero-energy first-order equations have one normalizable real solution, whose envelope can be chosen as
$$
f(x)=C\exp\left[\frac1{\hbar v}
\int_0^x m(s)\,ds\right]
$$
after choosing the constant Majorana spinor with the appropriate relative sign. The corresponding operator
$$
\gamma=\int dx\,f(x)\,[c_1(x)\mathbin\pm c_2(x)]
$$
is self-adjoint and commutes with the Hamiltonian, so it is a <Majorana zero mode at a mass domain wall>.
Outside the core it decays on
$$
\xi=\frac{\hbar v}{|t|}.
$$
If the phase varies approximately linearly across the core, then $m(x)\simeq-\pi tx/R_c$ near zero and the central envelope is Gaussian with width
$$
\ell_{\rm core}\sim
\sqrt{\frac{\hbar vR_c}{\pi|t|}}.
$$
Thus the spatial extent is of order $\max(\xi,\ell_{\rm core})$, up to constants depending on the detailed vortex profile. When $|v_1|\ne|v_2|$, continuously changing their magnitudes to equality never changes their opposite signs or closes the asymptotic mass gap. The mass domain wall therefore retains its odd, particle-hole-protected zero mode; only its two component amplitudes and localization lengths change.
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