I think that $H_0$ grows like $e^{A}$, which is optimal in light of the example $u(x,t) = e^{-At}\cos(x)$ which solves $u_t - A\Delta u = 0$.

Say we fix $C_i$ and the corresponding Harnack constant is $H$, and that $u$ is positive in $B_2 \times [-4,0]$. The proof is by applying the Harnack inequality to the rescaling you suggested, $\tilde{u}(x,t) = u(x, t/A)$, which is defined on $B_2 \times [-4A,0]$. Iterating the Harnack inequality we have that $\tilde{u}|_{B_1 \times \{-A\}} < H\tilde{u}(0,-A + 1) < H^2 \tilde{u}(0,-A + 2) ... < H^A\tilde{u}(0,0)$. Scaling back gives the Harnack inequality $u|_{B_1 \times \{-1\}} \leq H^Au(0,0)$.

Remark: I think the dependence on the ellipticity ratio $\Lambda = C_1/C_0$ (say we fix $C_0 = 1$ and let $C_1$ get large) can be understood similarly, with dependence going exponentially in $\sqrt{\Lambda}$ as suggested by the stationary example $u = e^{\sqrt{\Lambda}x}\cos(y)$, which solves the elliptic equation $tr(A \cdot D^2u) = 0$ where $A = \text{diag}(1,\Lambda)$. This is because $u$ is harmonic in an ``ellipsoidal geometry'' where the balls are ellipsoids with vertical axis $1$ and horizontal axis $1/\sqrt{\Lambda}$, and the proof by rescaling for constant coefficients works. This is intuitively the worst case scenario, since if the coefficients jump around then the ellipsoids "rotate" and one iterates the Harnack inequality fewer times, but to do things rigorously I think one needs to go through the original proof. (Maybe there is a good probabilistic interpretation of this too.)

I think that $H_0$ grows like $e^{A}$, which is optimal in light of the example $u(x,t) = e^{-At}\cos(x)$ which solves $u_t - A\Delta u = 0$.

Say we fix $C_i$ and the corresponding Harnack constant is $H$, and that $u$ is positive in $B_2 \times [-4,0]$. The proof is by applying the Harnack inequality to the rescaling you suggested, $\tilde{u}(x,t) = u(x, t/A)$, which is defined on $B_2 \times [-4A,0]$. Iterating the Harnack inequality we have that $\tilde{u}|_{B_1 \times \{-A\}} < H\tilde{u}(0,-A + 1) < H^2 \tilde{u}(0,-A + 2) ... < H^A\tilde{u}(0,0)$. Scaling back gives the Harnack inequality $u|_{B_1 \times \{-1\}} \leq H^Au(0,0)$.

Remark: I think the dependence on the ellipticity ratio $\Lambda = C_1/C_0$ (say we fix $C_0 = 1$ and let $C_1$ get large) can be understood similarly, with dependence going exponentially in $\sqrt{\Lambda}$ as suggested by the stationary example $u = e^{\sqrt{\Lambda}x}\cos(y)$, which solves the elliptic equation $tr(A \cdot D^2u) = 0$ where $A = \text{diag}(1,\Lambda)$. This is because $u$ is harmonic in an ``ellipsoidal geometry'' where the balls are ellipsoids with vertical axis $1$ and horizontal axis $1/\sqrt{\Lambda}$, and the proof by rescaling/iterating in overlapping ellipsoids for constant coefficients works. This is intuitively the worst case scenario, since if the coefficients jump around then the ellipsoids "rotate" and one iterates the Harnack inequality fewer times, but to do things rigorously I think one needs to go through the original proof. (Maybe there is a good probabilistic interpretation of this too.)

I think that $H_0$ grows like $e^{A}$, which is optimal in light of the example $u(x,t) = e^{-At}\cos(x)$ which solves $u_t - A\Delta u = 0$.

Say we fix $C_i$ and the corresponding Harnack constant is $H$, and that $u$ is positive in $B_2 \times [-4,0]$. The proof is by applying the Harnack inequality to the rescaling you suggested, $\tilde{u}(x,t) = u(x, t/A)$, which is defined on $B_2 \times [-4A,0]$. Iterating the Harnack inequality we have that $\tilde{u}|_{B_1 \times \{-A\}} < H\tilde{u}(0,-A + 1) < H^2 \tilde{u}(0,-A + 2) ... < H^A\tilde{u}(0,0)$. Scaling back gives the Harnack inequality $u|_{B_1 \times \{-1\}} \leq H^Au(0,0)$.

Remark: I think the dependence on the ellipticity ratio $\Lambda = C_1/C_0$ (say we fix $C_0 = 1$ and let $C_1$ get large) can be understood similarly, with dependence going exponentially in $\sqrt{\Lambda}$ as suggested by the stationary example $u = e^{\sqrt{\Lambda}x}\cos(y)$, which solves the elliptic equation $tr(A \cdot D^2u) = 0$ where $A = \text{diag}(1,\Lambda)$. This is because $u$ is harmonic in an ``ellipsoidal geometry'' where the balls are ellipsoids with short axis $1/\sqrt{\Lambda}$, and the proof by rescaling for constant coefficients works. This is intuitively the worst case scenario, since if the coefficients jump around then the ellipsoids "rotate" and one iterates the Harnack inequality fewer times, but to do things rigorously I think one needs to look at the proof. (Maybe there is a good probabilistic interpretation of this too.)

I think that $H_0$ grows like $e^{A}$, which is optimal in light of the example $u(x,t) = e^{-At}\cos(x)$ which solves $u_t - A\Delta u = 0$.

Say we fix $C_i$ and the corresponding Harnack constant is $H$, and that $u$ is positive in $B_2 \times [-4,0]$. The proof is by applying the Harnack inequality to the rescaling you suggested, $\tilde{u}(x,t) = u(x, t/A)$, which is defined on $B_2 \times [-4A,0]$. Iterating the Harnack inequality we have that $\tilde{u}|_{B_1 \times \{-A\}} < H\tilde{u}(0,-A + 1) < H^2 \tilde{u}(0,-A + 2) ... < H^A\tilde{u}(0,0)$. Scaling back gives the Harnack inequality $u|_{B_1 \times \{-1\}} \leq H^Au(0,0)$.

Remark: I think the dependence on the ellipticity ratio $\Lambda = C_1/C_0$ (say we fix $C_0 = 1$ and let $C_1$ get large) can be understood similarly, with dependence going exponentially in $\sqrt{\Lambda}$ as suggested by the stationary example $u = e^{\sqrt{\Lambda}x}\cos(y)$, which solves the elliptic equation $tr(A \cdot D^2u) = 0$ where $A = \text{diag}(1,\Lambda)$. This is because $u$ is harmonic in an ``ellipsoidal geometry'' where the balls are ellipsoids with vertical axis $1$ and horizontal axis $1/\sqrt{\Lambda}$, and the proof by rescaling for constant coefficients works. This is intuitively the worst case scenario, since if the coefficients jump around then the ellipsoids "rotate" and one iterates the Harnack inequality fewer times, but to do things rigorously I think one needs to go through the original proof. (Maybe there is a good probabilistic interpretation of this too.)