= Solution
For the <inflow transport boundary condition>, the <weak formulation> is
$$
\int_0^\infty\!\int_0^\infty u(\varphi_t+a\varphi_x)\,dx\,dt+\int_0^\infty u_0(x)\varphi(0,x)\,dx+a\int_0^\infty f(t)\varphi(t,0)\,dt=0.
$$
Here $\varphi$ is a smooth compactly supported <test function> on the closed quadrant. Put $z=x-at$ and $\psi(t,z)=\varphi(t,z+at)$. The transformed region is $t>\tau(z):=\max(0,-z/a)$. The formula in the preceding solution is $v(t,z)=b(z)$, with $b(z)=u_0(z)$ for $z\ge0$ and $b(z)=f(-z/a)$ for $z<0$. Consequently the first integral equals $-\int b(z)\psi(\tau(z),z)\,dz$. Its positive-$z$ portion cancels the initial integral; the substitution $z=-at$ in its negative-$z$ portion cancels the inflow integral, including the factor $a$. This proves weak existence for arbitrary bounded data without corner compatibility.
For uniqueness, subtract two <weak solutions>. Interior <test functions> in these <characteristic coordinates> give $\partial_t v=0$ distributionally, so $v(t,z)=b(z)$ on each vertical ray. This follows first on rectangles compactly contained in $t>\tau(z)$ using tensor <test functions>, and then on the whole region by overlapping rectangles. In the full <weak formulation> the remaining lower-boundary term is $-\int b(z)\psi(\tau(z),z)\,dz=0$. Arbitrary smooth traces supported separately on $z>0$ and $z<0$ force $b=0$ there. The single point $z=0$ has zero measure. Thus \b[the two-branch formula is the unique bounded weak solution].
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