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Analysis and computation of some tumor growth models with nutrient: From cell density models to free boundary dynamics

  • * Corresponding author: Zhennan Zhou

    * Corresponding author: Zhennan Zhou
J. Liu is partially supported by KI-Net NSF RNMS grant No.11-07444, NSF grant DMS-1812573 and NSF grant DMS-1514826. M. Tang is supported by Science Challenge Project No. TZZT2017-A3-HT003-F and NSFC 91330203. Z. Zhou is partially supported by RNMS11-07444 (KI-Net) and the start up grant from Peking University. L. Wang is partially supported by the start up grant from SUNY Buffalo and NSF grant DMS-1620135.
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  • In this paper, we study a tumor growth equation along with various models for the nutrient component, including a in vitro model and a in vivo model. At the cell density level, the spatial availability of the tumor density $ n$ is governed by the Darcy law via the pressure $ p(n) = n^{γ}$. For finite $ γ$, we prove some a priori estimates of the tumor growth model, such as boundedness of the nutrient density, and non-negativity and growth estimate of the tumor density. As $ γ \to ∞$, the cell density models formally converge to Hele-Shaw flow models, which determine the free boundary dynamics of the tumor tissue in the incompressible limit. We derive several analytical solutions to the Hele-Shaw flow models, which serve as benchmark solutions to the geometric motion of tumor front propagation. Finally, we apply a conservative and positivity preserving numerical scheme to the cell density models, with numerical results verifying the link between cell density models and the free boundary dynamical models.

    Mathematics Subject Classification: 35K55, 35B25, 76D27, 92C50.

    Citation:

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  • Figure 1.  Example 1: expanding disk with constant nutrient and initial data (5.55). Left: plot of solution at time $t = 0.5$ with different $\gamma = 20,\ 40, \ 80$. Here $\Delta r = 0.05$, and $\Delta t = 5\times 10^{-5}$ for $\gamma = 20, \ 40$ and $\Delta t = 2.5\times 10^{-5}$ for $\gamma = 80$. Right: comparison of the numerical solution with $\gamma = 80$ with the analytical solution (5.56) at different times $t = 0.0975$, $t = 0.2975$, $t = 0.4975$. Here the black solid curve is the numerical solution and the red dashed curve is the analytical solution

    Figure 2.  Example 2: a single annulus with constant nutrient and initial data (5.57). Left: plot of solution at time $t = 0.6$ with different $\gamma = 20,\ 40, \ 80$. Here $\Delta r = 0.05$, and $\Delta t = 2.5\times 10^{-5}$. Right: comparison of the numerical solution with $\gamma = 80$ with the analytical solution (5.58) at different times $t = 0.2494$, $t = 0.4994$, $t = 0.8$. Here we use $\Delta r = 0.025$ and $\Delta t = 6.25 \times 10^{-6}$. The black solid curve is the numerical solution and the red dashed curve is the analytical solution

    Figure 3.  Example 3: a double annulus with constant nutrient and initial data (5.59). Here we compare the numerical solution (black solid curve) and analytical solution (red dashed curve) at time $t = 0.2495$ (left) and $t = 0.6$ (right). Here we use $\Delta r = 0.025$ and $\Delta t = 5\times 10^{-6}$

    Figure 4.  Example 4: a 1D in vitro model with linear growing function. Left: plots of $n$ at time $t = 0.5$ with various $\gamma = 20, \ 40, \ 80$. The red curve is the analytical solution (5.61). Here $\Delta x = 0.05$ and $\Delta t = 2.5e-5$

    Figure 5.  Example 5: a 1D in vivo model with linear growing function. Left: plots of $n$ at time $t = 0.5$ with various $\gamma = 20, \ 40, \ 80$. The red curve is the analytical solution (5.61). Here $\Delta x = 0.05$ and $\Delta t = 2.5\times 10^{-5}$

    Figure 6.  A comparison of the front propagation speed for 1D in vitro model and in vivo model. The dots represent the position of the right boundary at each time, and the curves are computed via (3.37) and (3.39). Here $\Delta x = 0.05$, $\Delta t = 2.5 \times 10^{-5}$, $\gamma = 80$

    Figure 7.  A comparison of the front propagation speed in the 2D radial symmetric in vitro model and in vivo model. The dots represent the the position of the right boundary at each time, and the curve are computed via (3.43) and (3.44). Here $\Delta x = 0.05$, $\Delta t = 2.5\times 10^{-5}$, $\gamma = 80$

    Figure 8.  Plot of $n$ at four different times with initial data (5.63). From left to right, up to down, $t = 0$, $t = 0.0177$, $t = 0.0311$, $t = 0.05$

    Figure 9.  Plot of $n$ at four different times with initial data (5.64). From left to right, up to down, $t = 0$, $t = 0.0177$, $t = 0.0311$, $t = 0.05$

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