Correction to: Journal of Scientific Computing https://doi.org/10.1007/s10915-022-01885-w

In this article, the Main results in ``Section 3'' contains error. It should read as given below.

For \(n=0, \ldots , N_T\), let \(\mathcal {M}_h^n\) be an approximate value of mass at \(t=t^n\) defined by

$$\begin{aligned} \mathcal {M}_h^n&:=\left\{ \begin{aligned}&\int _\Omega { \frac{1}{2} \bigl ( \phi _h^0+\phi _h^1 \bigr )}~dx, & n = { 0}, \\&\int _\Omega \Bigl ( \frac{3}{2}\phi _h^n-\frac{1}{2}\phi _h^{n-1} \Bigr ) dx, & n\ge { 1}. \end{aligned} \right. \end{aligned}$$
FormalPara Remark 1

The value \(\mathcal {M}_h^n\) is an approximation of \(\int _\Omega \phi ^n dx\) due to the relations \(\frac{3}{2}\phi ^n-\frac{1}{2}\phi ^{n-1} (= \phi ^{n+1/2} + O(\Delta t^2)) = \phi ^n + O(\Delta t)\) and \(\frac{1}{2}(\phi ^0+\phi ^1) (= \phi ^{1/2} + O(\Delta t^2)) = \phi ^0 + O(\Delta t)\) for any smooth function \(\phi \).

FormalPara Theorem 1

(conservation of mass) Suppose that Hypotheses 1 and 2 hold true. Let \(\phi _h = \{\phi _h^n\}_{n=1}^{{ N_T}}\) be a solution to scheme (9) for a given \(\phi _h^0\). Then, we have the following.

  1. (i)

     It holds that, for \(n={ 1},\ldots , N_T\),

    $$\begin{aligned} \mathcal {M}_h^n = \mathcal {M}_h^0 + \Delta t \sum _{i=1}^n \Bigl ( \int _\Omega f^i dx + \int _\Gamma g^i ds \Bigr ). \end{aligned}$$
    (10)
  2. (ii)

     Assume \(f=0\) and \(g=0\) additionally. Then, for the solution to scheme (9), it holds that, for \(n={ 1},\ldots , N_T\),

    $$\begin{aligned} \int _\Omega \phi _h^n dx = \int _\Omega \phi _h^0 dx. \end{aligned}$$
    (11)
FormalPara Remark 2

The identity (10) is equivalent to

$$\begin{aligned} \int _\Omega \Bigl (\frac{3}{2}\phi _h^n - \frac{1}{2}\phi _h^{n-1} \Bigr ) dx&= \int _\Omega { \frac{1}{2}(\phi _h^0 + \phi _h^1)} dx + \Delta t \sum _{i=1}^n \Bigl ( \int _\Omega f^i dx + \int _\Gamma g^i ds \Bigr ), \quad { n\ge 1.} \end{aligned}$$

(In Section 4, Proofs:)

1 Proof of Theorem 1

We first note that due to Proposition 1-(i)

$$\begin{aligned} \int _{\Omega } \rho \circ X_1^n(x) \gamma ^n(x) dx = \int _\Omega \rho (x)\, dx, \quad \int _{\Omega } \rho \circ \tilde{X}_1^n(x) \tilde{\gamma }^n(x) dx = \int _\Omega \rho (x)\, dx \end{aligned}$$
(19)

hold for any \(\rho \in \Psi \) and \(n=1,\dots , N_{ T}\). We substitute \(1 \in \Psi _h\) into \(\psi _h\) in scheme (9) in the following.

We prove (i). Taking into account the case \(n = 1\):

$$\begin{aligned} \mathcal {M}_h^1&= \int _\Omega \Bigl ( \frac{3}{2} \phi _h^1 - \frac{1}{2} \phi _h^0 \Bigr ) dx = \int _\Omega \Bigl [ \frac{1}{2} \bigl ( \phi _h^1 + \phi _h^0 \bigr ) + \bigl ( \phi _h^1 - \phi _h^0 \circ X_1^1 \gamma ^1 \bigr ) \Bigr ] dx \quad \hbox {(by (19))} \nonumber \\&= \mathcal {M}_h^0 + \Delta t \Bigl ( \int _\Omega f^1 dx + \int _\Gamma g^1 ds \Bigr ) \quad \hbox {(by (9a))}, \end{aligned}$$
(E1)

we have, for any \(n \ge 2\),

$$\begin{aligned} \mathcal {M}_h^n&= \int _\Omega \Bigl ( \frac{3}{2} \phi _h^n - \frac{1}{2} \phi _h^{n-1} \Bigr ) dx = \int _\Omega \Bigl ( \frac{3}{2} \phi _h^n - \frac{1}{2} \phi _h^{n-1} \circ X_1^n \gamma ^n \Bigr ) dx \quad \hbox {(by (19))} \\&= \int _\Omega \Bigl ( \frac{3}{2} \phi _h^{n-1} \circ X_1^n \gamma ^n - \frac{1}{2} \phi _h^{n-2} \circ \tilde{X}_1^n \tilde{\gamma }^n \Bigr ) dx + \Delta t \Bigl ( \int _\Omega f^n dx + \int _\Gamma g^n ds \Bigr ) \quad \hbox {(by (9b))} \\&= \int _\Omega \Bigl ( \frac{3}{2} \phi _h^{n-1} - \frac{1}{2} \phi _h^{n-2} \Bigr ) dx + \Delta t \Bigl ( \int _\Omega f^n dx + \int _\Gamma g^n ds \Bigr ) \\&= \mathcal {M}_h^{n-1} + \Delta t \Bigl ( \int _\Omega f^n dx + \int _\Gamma g^n ds \Bigr ) = \mathcal {M}_h^{n-2} + \Delta t \sum _{i=n-1}^n\Bigl ( \int _\Omega f^i dx + \int _\Gamma g^i ds \Bigr ) \\&= \cdots \\&= \mathcal {M}_h^1 + \Delta t \sum _{i=2}^n\Bigl ( \int _\Omega f^i dx + \int _\Gamma g^i ds \Bigr ) = \mathcal {M}_h^0 + \Delta t \sum _{i=1}^n\Bigl ( \int _\Omega f^i dx + \int _\Gamma g^i ds \Bigr ) \quad \hbox {(by (E1)),} \end{aligned}$$

which completes the proof of (i).

We prove (ii). Consider identity (10) with \(f=g=0\). Then, we have \(\int _\Omega \phi _h^1 dx = \int _\Omega \phi _h^0 dx \ ( = \mathcal {M}_h^0 )\) from \(n=1\), and \(\int _\Omega \phi _h^2 dx = \int _\Omega \phi _h^0 dx\) from \(n=2\). Using this argument repeatedly up to \(n=N_T\), we obtain (11). \(\square \)