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Note: This is only a partial answer, and actually I am not sure that I understood the problem correctly. If $x_i(t) = \alpha_i e^{\lambda_i t}$ is the characteristic of the equation, then the left-ha …

(For an actual answer, see the edit below.) Let $\phi$ be smooth near zero and non-negative. Suppose that the Taylor expansion of $\phi$ at zero is non-trivial, and let $P(x)$ be the leading term. The …

Regarding the other question in your post: not quite. If $\phi$ is a bounded function such that $\Delta \phi + \phi = 0$, then $\phi$ is a tempered distribution such that the Fourier transform of $\p …

(This is an extended comment, I suppose). What do we assume about $L$? In complete generality this cannot be true, as shown by the example that follows. Even if we assume, as in the paper that you li …

The problem may have no solution at all! If $\Delta u = f$ in $B$ and $G_B(x,y)$ is the Green's function of a ball, then $$h = u + G_B f$$ is harmonic in $B$ (here $G_B u(x) = \int_B G_B(x,y) u(y) dy …

This is an extended comment, not an answer. Suppose that $a_0$ and $b_0$ satisfy the conditions given in the statement of the problem and $u := b_0-a_0$ is not identically zero. The function $u$ is o …

What about $u(t,x) = x e^{t x^2}$, which is a solution of $$\partial_t u = \Delta u + b(t,x) \partial_x u$$ with $$b(t,x) = \frac{x^3 - 6 t x - 4 t^2 x^3}{1 + 2 t x^2}$$ with initial value $u_0(x) = x …

Yes, one can get a constant better than $1/(2N)$. Simply take a look at the proof of the original bound $\|u\|_\infty \le \|f\|_\infty / (2N)$: one writes $$ u(x) = \int_B G_B(x, y) f(y) dy , $$ estim …