Use the <weighted Neumann heat kernel with constant drift>. The spatial operator has the <self-adjoint> form
$$
\mathcal Lu=u_{xx}+\beta u_x=e^{-\beta x}(e^{\beta x}u_x)_x,
$$
so its homogeneous <Neumann boundary conditions> give a regular <Sturm-Liouville problem> in the <weighted inner product>
$$
\langle f_1,f_2\rangle_\beta=\int_0^L e^{\beta x}\overline{f_1(x)}f_2(x)\,dx.
$$
The constant <eigenfunction> is $\phi_0=1$, with <eigenvalue> $\lambda_0=0$ and squared <norm> $N_0=(e^{\beta L}-1)/\beta$. For $k\geq1$, put $q_k=k\pi/L$. Substitution of $\phi=e^{-\beta x/2}v$ in $\mathcal L\phi=-\lambda\phi$ gives $v''+(\lambda-\beta^2/4)v=0$ and the <Robin boundary conditions> $v'=\beta v/2$ at both endpoints. Hence
$$
\phi_k(x)=e^{-\beta x/2}\left[\cos(q_kx)+\frac{\beta}{2q_k}\sin(q_kx)\right],
\qquad \lambda_k=q_k^2+\frac{\beta^2}{4}.
$$
Indeed,
$$
\phi_k'(x)=-e^{-\beta x/2}\left(q_k+\frac{\beta^2}{4q_k}\right)\sin(q_kx),
\qquad
N_k=\int_0^Le^{\beta x}\phi_k(x)^2dx=\frac L2\left(1+\frac{\beta^2}{4q_k^2}\right).
$$
The <Sturm-Liouville eigenfunction expansion> is complete and <orthogonal>. Its <heat kernel>, relative to the measure $e^{\beta y}dy$, is
$$
K_\beta(x,y,t)=\frac1{N_0}+\sum_{k=1}^\infty\frac{e^{-\lambda_kt}}{N_k}\phi_k(x)\phi_k(y),\qquad t>0.
$$
For a coefficient $a_k(t)=\langle\phi_k,u(\cdot,t)\rangle_\beta/N_k$, two applications of <integration by parts> yield
$$
a_k'+\lambda_ka_k=\frac{e^{\beta L}\phi_k(L)h(t)-\phi_k(0)g(t)}{N_k},
\qquad
a_k(0)=\frac1{N_k}\int_0^Le^{\beta y}\phi_k(y)u_0(y)\,dy.
$$
Solving these scalar <linear differential equations> and summing gives the <Neumann boundary-forcing formula>:
$$
\boxed{u(x,t)=\int_0^LK_\beta(x,y,t)e^{\beta y}u_0(y)\,dy+\int_0^t\left[e^{\beta L}K_\beta(x,L,t-s)h(s)-K_\beta(x,0,t-s)g(s)\right]ds.}
$$
This depends only on the given data. In particular, the zero <eigenvalue> retains the changing weighted mean:
$$
\frac d{dt}\int_0^Le^{\beta x}u(x,t)\,dx=e^{\beta L}h(t)-g(t).
$$
The endpoint derivatives of the boundary integral are understood as interior limits. Differentiating each homogeneous <Neumann eigenfunction> at the endpoint before summing would incorrectly discard the prescribed boundary forcing: the short-time <heat kernel> makes that interchange invalid. The stated compatibility conditions give the classical boundary traces.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 328 / 2 / Solution

{title2=$K_\beta(x,y,t)$}

On $(0,L)$, the <differential operator> $\mathcal L=\partial_x^2+\beta\partial_x=e^{-\beta x}\partial_x(e^{\beta x}\partial_x)$ with homogeneous <Neumann boundary conditions> is <self-adjoint> in the <weighted inner product> with weight $e^{\beta x}$. Its <eigenfunctions> and nonnegative decay rates are
$$
\phi_0=1,\quad\lambda_0=0,\qquad
\phi_k=e^{-\beta x/2}\left[\cos(q_kx)+\frac\beta{2q_k}\sin(q_kx)\right],\quad
\lambda_k=q_k^2+\frac{\beta^2}4,\quad q_k=\frac{k\pi}L.
$$
Their squared <norms> are $N_0=(e^{\beta L}-1)/\beta$ and $N_k=L[1+\beta^2/(4q_k^2)]/2$. The <Sturm-Liouville eigenfunction expansion> gives
$$
K_\beta(x,y,t)=\sum_{k=0}^\infty\frac{e^{-\lambda_kt}\phi_k(x)\phi_k(y)}{N_k}.
$$
This kernel acts against the measure $e^{\beta y}dy$, not Lebesgue measure alone. It is symmetric in $x,y$; the transition density against Lebesgue measure is $e^{\beta y}K_\beta(x,y,t)$.

Weighted Neumann heat kernel with constant drift

{title2=$\langle f,g\rangle_w$}

For a positive weight $w$, the weighted inner product of <functions> is $\langle f,g\rangle_w=\int\overline f g w$. It is the usual <inner product> in the <L2 space> for the measure $w(x)dx$. A <Sturm-Liouville eigenfunction expansion> uses this weight to make its <eigenfunctions> <orthogonal>.

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 Incoming links: Weighted inner product