Dirichlet gauge transform for constant drift ($w=e^{ax}u$)

For the advection-diffusion equation $u_t=u_{xx}+2a{u}_x$, the multiplication $w=e^{ax}u$ gives $w_t=w_{xx}-a^{2}w$. A Dirichlet boundary condition remains a prescribed value, multiplied by $e^{ax}$ at the endpoint. This real exponential conjugation removes the first derivative without changing a bounded spatial interval.

For $u_t=u_{xx}+u_{yy}+\beta u_x$ on $(0,L_x)\times (0,L_y)$, impose zero Dirichlet boundary conditions on the two vertical sides and the top, and a zero Neumann boundary condition at the bottom. The Dirichlet gauge transform for constant drift $u=e^{-\beta x/2}v$ gives $v_t=\Delta v-{\beta }^{2}v/4$. Separation of variables produces the displayed modes, with $m,n\ge 1$ and ${\lambda }_{mn}=(m\pi /L_x{)}^2+((n-1/2)\pi /L_y{)}^2+{\beta }^2/4$. Their spatial parts are orthogonal with weight $e^{\beta x}$; equivalently, expand $e^{\beta x/2}{u}_0$ in the ordinary sine-cosine basis.

For $q_t=q_{xx}+\alpha q_x$, the exponential change of unknown gives $u_t=u_{xx}-{\alpha }^{2}u/4$. Its Fourier sine transform satisfies a closed scalar evolution with a boundary forcing term. The initial transform uses $e^{\alpha x/2}{q}_0(x)$; ordinary integrable-transform arguments require sufficient decay of this weighted datum. For $q_0=e^{-ax}$ this requires $a>\alpha /2$. In contrast, the unweighted sine transform of $q$ couples to its cosine transform through the drift term, so it does not directly diagonalize this half-line problem.

After the Dirichlet gauge transform for constant drift, put $q=\sqrt{p+a^2}$. The Dirichlet Green function of $p-{\partial }_x^2+a^2$ isThe original unweighted resolvent kernel is $e^{-ax}{R}_q(x,y)e^{ay}$. Its poles are $p=-a^2-(n\pi /L{)}^2$. The kernel is even in $q$, so the apparent square-root branch point is removable.