Bounded Domain Problem for the Modified Buckley–Leverett Equation

Abstract

The focus of the present study is the modified Buckley–Leverett (MBL) equation describing two-phase flow in porous media. The MBL equation differs from the classical Buckley–Leverett (BL) equation by including a balanced diffusive–dispersive combination. The dispersive term is a third order mixed derivatives term, which models the dynamic effects in the pressure difference between the two phases. The classical BL equation gives a monotone water saturation profile for any Riemann problem; on the contrast, when the dispersive parameter is large enough, the MBL equation delivers a non-monotone water saturation profile for certain Riemann problems as suggested by the experimental observations. In this paper, we first show that for the MBL equation, the solution of the finite interval $[0,L]$ boundary value problem converges to that of the half line $[0,+\infty )$ boundary value problem exponentially as $L\rightarrow +\infty$. This result provides a justification for the use of the finite interval in numerical studies for the half line problem [Y. Wang and C.-Y. Kao, Central schemes for the modified Buckley–Leverett equation, J. Comput. Sci. 4(1–2), 12 – 23, 2013]. Furthermore, we numerically verify that the convergence rate is consistent with the theoretical derivation. Numerical results confirm the existence of non-monotone water saturation profiles consisting of constant states separated by shocks.

Appendix: Proof of the Lemmas

Proof

(Proof to lemma 2.2) Let $g(u)=\frac{f(u)}{u}=\frac{u}{u^2+M(1-u)^2}$, then

$$\begin{aligned} g^{\prime }(u)=\frac{M-(1+M)u^2}{(u^2+M(1-u)^2)^2}\left\{ \begin{array}{lll} >0{\quad }&{}\mathrm {if}&{}0<u<\sqrt{\frac{M}{M+1}}\\ =0{\quad }&{}\mathrm {if}&{}u=\sqrt{\frac{M}{M+1}}\\ <0{\quad }&{}\mathrm {if}&{}u>\sqrt{\frac{M}{M+1}} \end{array}\right. \end{aligned}$$

and hence $g(u)$ achieves its maximum at $u=\sqrt{\frac{M}{M+1}}$. Therefore, $\frac{f(u)}{u}=g(u)\le D$, where $D=\frac{f(\alpha )}{\alpha }$ and $\alpha =\sqrt{\frac{M}{M+1}}$, and in turn, we have that $f(u)\le Du$ for all $0\le u \le 1$. $\square$

Proof

(Proof to lemma 2.3 (i))

$$\begin{aligned} \int _{0}^{+\infty } \left| e^{-\frac{x+\xi }{\epsilon \sqrt{\tau }}}-e^{-\frac{|x-\xi |}{\epsilon \sqrt{\tau }}}\right| e^{\frac{\lambda x-\lambda \xi }{\epsilon \sqrt{\tau }}}\,d\xi =\epsilon \sqrt{\tau }\frac{-2+2e^{\frac{(\lambda -1)x}{\epsilon \sqrt{\tau }}}}{\lambda ^2-1} \le \frac{2\epsilon \sqrt{\tau }}{1-\lambda ^2} \quad \mathrm {if}\, \lambda \in (0,1). \end{aligned}$$

$\square$

Proof

(Proof to lemma 2.3 (ii))

$$\begin{aligned} \int _{0}^{+\infty } \left| e^{-\frac{x+\xi }{\epsilon \sqrt{\tau }}}-e^{-\frac{|x-\xi |}{\epsilon \sqrt{\tau }}}\right| e^{\frac{\lambda x-\xi }{\epsilon \sqrt{\tau }}}\,d\xi =xe^{\frac{(\lambda -1)x}{\epsilon \sqrt{\tau }}} \le \frac{\epsilon \sqrt{\tau }}{e(1-\lambda )} \quad \mathrm {if}\, \lambda \in (0,1). \end{aligned}$$

$\square$

Proof

(Proof to lemma 2.3 (iii)) Based on the assumption on $u_0$ in (2.1)

$$\begin{aligned} \int _{0}^{+\infty } \left| e^{-\frac{x+\xi }{\epsilon \sqrt{\tau }}}-e^{-\frac{|x-\xi |}{\epsilon \sqrt{\tau }}}\right| e^{\frac{\lambda x}{\epsilon \sqrt{\tau }}} |u_0(\xi )|\,d\xi&\le \int _{0}^{+\infty } e^{-\frac{|x-\xi |}{\epsilon \sqrt{\tau }}}e^{\frac{\lambda x}{\epsilon \sqrt{\tau }}} |u_0(\xi )|\,d\xi \\&\le C_ue^{\frac{\lambda x}{\epsilon \sqrt{\tau }}}\int _{0}^{L_0}e^{-\frac{|x-\xi |}{\epsilon \sqrt{\tau }}} \,d\xi := C_u y_1(x) \end{aligned}$$

Calculating $y_1(x)$ with the assumption that $\lambda \in (0,1)$, we get

$$\begin{aligned} y_1(x)=\left\{ \begin{array}{lll} e^{\frac{\lambda x}{\epsilon \sqrt{\tau }}}\int _0^{L_0}e^{-\frac{|x-\xi |}{\epsilon \sqrt{\tau }}}\,d\xi \le 2\epsilon \sqrt{\tau }e^{\frac{\lambda x}{\epsilon \sqrt{\tau }}}\le 2\epsilon \sqrt{\tau }e^{\frac{\lambda L_0}{\epsilon \sqrt{\tau }}} &{} &{} \mathrm {for}\, x\in [0,L_0]\\ e^{\frac{(\lambda -1)x}{\epsilon \sqrt{\tau }}}\int _0^{L_0}e^{\frac{\xi }{\epsilon \sqrt{\tau }}}\,d\xi \le \epsilon \sqrt{\tau }e^{\frac{(\lambda -1)x+L_0}{\epsilon \sqrt{\tau }}} \le \epsilon \sqrt{\tau }e^{\frac{\lambda L_0}{\epsilon \sqrt{\tau }}} &{}&{}\mathrm {for}\, x\in [L_0,+\infty ) \end{array} \right. \end{aligned}$$

Therefore, we get the desired inequality

$$\begin{aligned} \int _{0}^{+\infty }\left| e^{-\frac{x+\xi }{\epsilon \sqrt{\tau }}}-e^{-\frac{|x-\xi |}{\epsilon \sqrt{\tau }}}\right| e^{\frac{\lambda x}{\epsilon \sqrt{\tau }}} |u_0(\xi )|\,d\xi \le 2C_u\epsilon \sqrt{\tau }e^{\frac{\lambda L_0}{\epsilon \sqrt{\tau }}} . \end{aligned}$$

$\square$

Proof

(Proof to lemma 2.4 (i))

$$\begin{aligned}&\int _{0}^{+\infty } \left| e^{-\frac{x+\xi }{\epsilon \sqrt{\tau }}}+\mathrm {sgn}(x-\xi )e^{-\frac{|x-\xi |}{\epsilon \sqrt{\tau }}}\right| e^{\frac{\lambda x-\lambda \xi }{\epsilon \sqrt{\tau }}}\,d\xi \\&= \frac{\epsilon \sqrt{\tau }}{\lambda ^2-1} \left( -2+2\lambda e^{\frac{(\lambda -1)x}{\epsilon \sqrt{\tau }}}-2(\lambda -1)e^{-\frac{2x}{\epsilon \sqrt{\tau }}}\right) \le \frac{2\epsilon \sqrt{\tau }}{1-\lambda ^2}\qquad \mathrm { if }\, \lambda \in (0,1). \end{aligned}$$

$\square$

Proof

(Proof to lemma 2.4 (ii))

$$\begin{aligned}&\int _{0}^{+\infty } \left| e^{-\frac{x+\xi }{\epsilon \sqrt{\tau }}}+\mathrm {sgn}(x-\xi )e^{-\frac{|x-\xi |}{\epsilon \sqrt{\tau }}}\right| e^{\frac{\lambda x-\xi }{\epsilon \sqrt{\tau }}}\,d\xi \\&= \frac{2e^{{\frac{(\lambda -3)x}{\epsilon \sqrt{\tau }}}}-2e^{{\frac{(\lambda -1)x}{\epsilon \sqrt{\tau }}}}}{\frac{-2}{\epsilon \sqrt{\tau }}} +xe^{\frac{(\lambda -1)x}{\epsilon \sqrt{\tau }}} \le \epsilon \sqrt{\tau }+\frac{\epsilon \sqrt{\tau }}{e(1- \lambda )}\qquad \mathrm { if }\, \lambda \in (0,1). \end{aligned}$$

$\square$

Proof

(Proof to lemma 2.4 (iii)) Based on the assumption on $u_0$ in (2.1)

$$\begin{aligned}&\int _{0}^{+\infty } \left| e^{-\frac{x+\xi }{\epsilon \sqrt{\tau }}}+\mathrm {sgn}(x-\xi )e^{-\frac{|x-\xi |}{\epsilon \sqrt{\tau }}}\right| e^{\frac{\lambda x}{\epsilon \sqrt{\tau }}}|u_0(\xi )|\,d\xi \\&\le C_ue^{\frac{\lambda x}{\epsilon \sqrt{\tau }}}\int _{0}^{L_0}\left| e^{-\frac{x+\xi }{\epsilon \sqrt{\tau }}}+\mathrm {sgn}(x-\xi ) e^{-\frac{|x-\xi |}{\epsilon \sqrt{\tau }}}\right| \,d\xi : = C_u y_3(x) \end{aligned}$$

Calculating $y_3(x)$ with the assumption that $\lambda \in (0,1)$, we get

$$\begin{aligned} y_3(x)\le \left\{ \begin{array}{l} e^{\frac{(\lambda -1)x}{\epsilon \sqrt{\tau }}}\int _0^x(e^{-\frac{\xi }{\epsilon \sqrt{\tau }}}+e^{\frac{\xi }{\epsilon \sqrt{\tau }}})\,d\xi +e^{\frac{(\lambda +1)x}{\epsilon \sqrt{\tau }}}\int _x^{L_0}e^{-\frac{\xi }{\epsilon \sqrt{\tau }}}\,d\xi \le 2\epsilon \sqrt{\tau }e^{\frac{\lambda L_0}{\epsilon \sqrt{\tau }}}, x\in [0,L_0]\\ e^{\frac{(\lambda -1)x}{\epsilon \sqrt{\tau }}}\int _0^{L_0}(e^{-\frac{\xi }{\epsilon \sqrt{\tau }}}+e^{\frac{\xi }{\epsilon \sqrt{\tau }}})\,d\xi \le \epsilon \sqrt{\tau }e^{\frac{(\lambda -1)x+L_0}{\epsilon \sqrt{\tau }}} \le \epsilon \sqrt{\tau }e^{\frac{\lambda L_0}{\epsilon \sqrt{\tau }}}, x\in [L_0,+\infty ) \end{array} \right. \end{aligned}$$

Therefore, we get the desired inequality

$$\begin{aligned} \int _{0}^{+\infty } \left| e^{-\frac{x+\xi }{\epsilon \sqrt{\tau }}}+\mathrm {sgn}(x-\xi )e^{-\frac{|x-\xi |}{\epsilon \sqrt{\tau }}}\right| e^{\frac{\lambda x}{\epsilon \sqrt{\tau }}}|u_0(\xi )|\,d\xi \le 2C_u\epsilon \sqrt{\tau }e^{\frac{\lambda L_0}{\epsilon \sqrt{\tau }}}. \end{aligned}$$

$\square$

Proof

(Proof to lemma 2.5 (i))

$$\begin{aligned} \left| \phi _1(x)-e^{-\frac{x}{\epsilon \sqrt{\tau }}}\right| =e^{-\frac{L}{\epsilon \sqrt{\tau }}}\left| \frac{e^{-\frac{x}{\epsilon \sqrt{\tau }}}-e^{\frac{x}{\epsilon \sqrt{\tau }}}}{e^{\frac{L}{\epsilon \sqrt{\tau }}}-e^{-\frac{L}{\epsilon \sqrt{\tau }}}}\right| =e^{-\frac{L}{\epsilon \sqrt{\tau }}}\left| \phi _2(x)\right| . \end{aligned}$$

$\square$

Proof

(Proof to lemma 2.5 (ii)) Since $\phi _2(x)= \frac{e^{\frac{x}{\epsilon \sqrt{\tau }}}-e^{-\frac{x}{\epsilon \sqrt{\tau }}}}{e^{\frac{L}{\epsilon \sqrt{\tau }}}-e^{-\frac{L}{\epsilon \sqrt{\tau }}}}$, we see that $\phi _2^{\prime }(x)=\frac{1}{\epsilon \sqrt{\tau }}\frac{e^{\frac{x}{\epsilon \sqrt{\tau }}}+e^{-\frac{x}{\epsilon \sqrt{\tau }}}}{e^{\frac{L}{\epsilon \sqrt{\tau }}}-e^{-\frac{L}{\epsilon \sqrt{\tau }}}}>0$ and hence $\phi _2(x)\le \phi _2(L)=1$ for $x\in [0,L]$. $\square$

Proof

(Proof to lemma 2.5 (iii)) $\phi _2^{\prime }(x)=\frac{1}{\epsilon \sqrt{\tau }}\frac{e^{\frac{x}{\epsilon \sqrt{\tau }}}+e^{-\frac{x}{\epsilon \sqrt{\tau }}}}{e^{\frac{L}{\epsilon \sqrt{\tau }}}-e^{-\frac{L}{\epsilon \sqrt{\tau }}}}$ gives that $\phi _2^{\prime \prime }(x)=\frac{1}{\epsilon ^2\tau }\phi _2(x)>0$, and hence $\phi _2^{\prime }(x)\le \phi _2^{\prime }(L)=\frac{1}{\epsilon \sqrt{\tau }}\frac{e^{\frac{L}{\epsilon \sqrt{\tau }}}+e^{-\frac{L}{\epsilon \sqrt{\tau }}}}{e^{\frac{L}{\epsilon \sqrt{\tau }}}-e^{-\frac{L}{\epsilon \sqrt{\tau }}}}=\frac{1}{\epsilon \sqrt{\tau }}\frac{e^{\frac{2L}{\epsilon \sqrt{\tau }}}+1}{e^{\frac{2L}{\epsilon \sqrt{\tau }}}-1}\le \frac{2}{\epsilon \sqrt{\tau }}$    if $\epsilon \ll 1$ for $x\in [0,L]$. $\square$