A Convergent Lagrangian Discretization for a Nonlinear Fourth-Order Equation

Abstract

A fully discrete Lagrangian scheme for numerical solution of the nonlinear fourth-order DLSS equation in one space dimension is analyzed. The discretization is based on the equation’s gradient flow structure in the $L^2$-Wasserstein metric. By construction, the discrete solutions are strictly positive and mass conserving. A further key property is that they dissipate both the Fisher information and the logarithmic entropy. Our main result is a proof of convergence of fully discrete to weak solutions in the limit of vanishing mesh size. Convergence is obtained for arbitrary nonnegative, possibly discontinuous initial data with finite entropy, without any CFL-type condition. The key estimates in the proof are derived from the dissipations of the two Lyapunov functionals. Numerical experiments illustrate the practicability of the scheme.

Appendices

Appendix 1: Some technical lemmas

Lemma 28

For each $p>1$ and $\vec {\mathrm {x}}\in \mathfrak {x}_\delta$ with $\vec {\mathrm {z}}=\mathbf {z}[\vec {\mathrm {x}}]$, one has that

$$\begin{aligned} \sum _{\kappa \in {\mathbb {I}_K^{1/2}}}\left( \frac{\delta }{z_\kappa }\right) ^p =\sum _{\kappa \in {\mathbb {I}_K^{1/2}}}\left( x_{\kappa +\frac{1}{2}}-x_{\kappa -\frac{1}{2}}\right) ^p \le (b-a)^p. \end{aligned}$$

(99)

Proof

The first equality is simply the definition (22) of $z_\kappa$. Since trivially $x_{\kappa +\frac{1}{2}}-x_{\kappa -\frac{1}{2}}<b-a$ for each $\kappa \in {\mathbb {I}_K^{1/2}}$, and since $p-1>0$, it follows that

$$\begin{aligned} \sum _{\kappa \in {\mathbb {I}_K^{1/2}}}\left( x_{\kappa +\frac{1}{2}}-x_{\kappa -\frac{1}{2}}\right) ^p\le (b-a)^{p-1}\sum _{\kappa \in {\mathbb {I}_K^{1/2}}}\left( x_{\kappa +\frac{1}{2}}-x_{\kappa -\frac{1}{2}}\right) = (b-a)^p. \end{aligned}$$

$\square$

Lemma 29

For each $\vec {\mathrm {x}}\in \mathfrak {x}_\delta$ with $\vec {\mathrm {z}}=\mathbf {z}[\vec {\mathrm {x}}]$, one has that

$$\begin{aligned} \frac{\delta }{b-a}\le z_\kappa \le M^{1-1/q}\left( \delta \sum _{k\in {\mathbb {I}_K^+}}\left| \frac{z_{k+\frac{1}{2}}-z_{k-\frac{1}{2}}}{\delta }\right| ^q\right) ^{1/q} + \frac{M}{b-a} \quad \text {for all}\, \kappa \in {\mathbb {I}_K^{1/2}}, \end{aligned}$$

(100)

and consequently,

$$\begin{aligned} z_\kappa \le \big (2M\mathbf {F}_\delta [\vec {\mathrm {x}}]\big )^{1/2} + \frac{M}{b-a} \quad \text {for all }\,\kappa \in {\mathbb {I}_K^{1/2}}. \end{aligned}$$

(101)

Proof

The first estimate in (100) is an immediate consequence of the definition of $z_\kappa$ in (22). To prove the second estimate, let $\kappa ^*\in {\mathbb {I}_K^{1/2}}$ be such that $z_{\kappa ^*}=\max z_k$. Observe that there exists a $\kappa _*\in {\mathbb {I}_K^{1/2}}$ such that

$$\begin{aligned} z_{\kappa _*} \le \frac{M}{b-a} \le z_{\kappa ^*}. \end{aligned}$$

(102)

Writing out $z_{\kappa ^*}-z_{\kappa _*}$ as a sum over differences of adjacent values of $z_k$ and applying the triangle and Cauchy Schwarz inequality, one obtains

$$\begin{aligned} z_{\kappa ^*}-z_{\kappa _*} \le \sum _{k\in {\mathbb {I}_K^+}}|z_{k+\frac{1}{2}}-z_{k-\frac{1}{2}}| \le \left( \delta \sum _{k\in {\mathbb {I}_K^+}}1\right) ^{1-1/q}\left( \delta \sum _{k\in {\mathbb {I}_K^+}}\left| \frac{z_{k+\frac{1}{2}}-z_{k-\frac{1}{2}}}{\delta }\right| ^q\right) ^{1/q}. \end{aligned}$$

Now combine this with (102). $\square$

Lemma 30

With $\widehat{u}$ and $\bar{u}$ being, respectively, the piecewise linear and the piecewise constant densities associated with a given vector $\vec {\mathrm {x}}$, then

$$\begin{aligned} \mathbf {H}_\delta (\vec {\mathrm {x}}) = \mathcal {H}(\bar{u})\le \mathcal {H}(\widehat{u}). \end{aligned}$$

(103)

Proof

First observe that

$$\begin{aligned} \int _0^1 \ln \big (p(1-\lambda )+q\lambda \big )\,\mathrm {d}\lambda = \frac{p\ln p-q\ln q}{p-q} -1 \ge \frac{1}{2}(\ln p+\ln q), \end{aligned}$$

(104)

which is an easy consequence of a Taylor expansion for the function $s\mapsto (1+s)\ln s$ around $s=1$, substituting $s=p/q$. On the one hand, we have that

$$\begin{aligned} \int _a^b\bar{u}(x)\ln \bar{u}(x)\,\mathrm {d}x = \delta \sum _{\kappa \in {\mathbb {I}_K^{1/2}}}\ln z_\kappa = \delta \frac{\ln z_{1/2}+\ln z_{K-1/2}}{2} + \delta \sum _{k\in {\mathbb {I}_K^+}}\frac{\ln z_{k+\frac{1}{2}}+\ln z_{k-\frac{1}{2}}}{2}, \end{aligned}$$

and on the other hand,

$$\begin{aligned} \int _a^b\widehat{u}(x)\ln \widehat{u}(x)\,\mathrm {d}x&= \int _0^M \ln \widehat{z}(\xi )\,\mathrm {d}\xi \\&= \frac{\delta }{2}\big (\ln z_0+\ln z_K) + \delta \sum _{k\in {\mathbb {I}_K^+}}\int _0^1 \ln \big (z_{k-\frac{1}{2}}(1-\lambda )+z_{k+\frac{1}{2}}\lambda \big )\,\mathrm {d}\lambda \\&\ge \delta \frac{\ln z_{1/2}+\ln z_{K-1/2}}{2} + \delta \sum _{k\in {\mathbb {I}_K^+}}\frac{\ln z_{k+\frac{1}{2}}+\ln z_{k-\frac{1}{2}}}{2}, \end{aligned}$$

where we have used (104). This clearly implies (103). $\square$

Lemma 31

(Gargliardo–Nirenberg inequality). For each $f\in H^1([a,b])$, one has that

$$\begin{aligned} \Vert f\Vert _{C^{1/6}([a,b])} \le (9/2)^{1/3} \Vert f\Vert _{H^1([a,b])}^{2/3} \Vert f\Vert _{L^2([a,b])}^{1/3}. \end{aligned}$$

(105)

Proof

Assume first that $f\ge 0$. Then, for arbitrary $a<x<y<b$, the fundamental theorem of calculus and Hölder’s inequality imply that

$$\begin{aligned} \big |f(x)^{3/2}-f(y)^{3/2}\big |&\le \frac{3}{2}\int _x^y 1\cdot f(z)^{1/2}|f'(z)|\,\mathrm {d}z \\&\le \frac{3}{2}|x-y|^{1/4}\Vert f\Vert _{L^2([a,b])}^{1/2}\Vert f'\Vert _{L^2([a,b])}. \end{aligned}$$

Since $f\ge 0$, we can further estimate

$$\begin{aligned} |f(x)-f(y)|&\le \big |f(x)^{3/2}-f(y)^{3/2}\big |^{2/3} \\&\le (3/2)^{2/3}|x-y|^{1/6}\Vert f\Vert _{L^2([a,b])}^{1/3}\Vert f\Vert _{H^1([a,b])}^{1/3}. \end{aligned}$$

This shows (105) for nonnegative functions f. A general f can be written in the form $f=f_+-f_-$, where $f_\pm \ge 0$. By the triangle inequality, and since $\Vert f_\pm \Vert _{H^1([a,b])}\le \Vert f\Vert _{H^1([a,b])}$,

$$\begin{aligned} \Vert f\Vert _{C^{1/6}([a,b])} \le \Vert f_+\Vert _{C^{1/6([a,b])}}+\Vert f_-\Vert _{C^{1/6([a,b])}} \le 2(3/2)^{2/3}\Vert f\Vert _{L^2([a,b])}^{1/3}\Vert f\Vert _{H^1([a,b])}^{1/3}. \end{aligned}$$

This proves the claim. $\square$

Appendix 2: Proof of Lemma 23

Proof of estimate (76)

First, observe that by definition of $\widehat{z}$,

$$\begin{aligned} \int _0^M \left[ \partial _\xi \widehat{z}^n_\Delta (\xi )\right] ^2\rho ''\circ \mathrm {X}^n_\Delta (\xi )\,\mathrm {d}\xi = \sum _{k\in {\mathbb {I}_K^+}} \Big (\frac{z^n_{k+\frac{1}{2}}-z^n_{k-\frac{1}{2}}}{\delta }\Big )^2 \int _{\xi _{k-\frac{1}{2}}}^{\xi _{k+\frac{1}{2}}}\rho ''\circ \mathrm {X}^n_\Delta (\xi )\,\mathrm {d}\xi , \end{aligned}$$

and therefore, by Hölder’s inequality,

$$\begin{aligned} R_1 \le R_{1,\alpha }^{1/2}R_{1,\beta }^{1/2}, \end{aligned}$$

(106)

with, recalling (41),

$$\begin{aligned} R_{1,\alpha }&= \tau \sum _{n=1}^N\delta \sum _{k\in {\mathbb {I}_K^+}}\left( \frac{z^n_{k+\frac{1}{2}}-z^n_{k-\frac{1}{2}}}{\delta }\right) ^4 \le \tau \sum _{n=1}^\infty \Vert \widehat{z}_\Delta ^n\Vert _{L^4([a,b])}^4\le 9\overline{\mathcal {H}}, \\ \nonumber R_{1,\beta }&= \tau \sum _{n=1}^N\delta \sum _{k\in {\mathbb {I}_K^+}} \left[ \frac{z^n_{k+\frac{1}{2}}+z^n_{k-\frac{1}{2}}}{2}\frac{\rho '(x^n_{k+1}) -\rho '(x^n_{k-1})}{\delta }-\frac{2}{\delta }\int _{\xi _{k-\frac{1}{2}}}^{\xi _{k+\frac{1}{2}}} \rho ''\circ \mathrm {X}^n_\Delta \,\mathrm {d}\xi \right] ^2. \end{aligned}$$

(107)

To simplify $R_{1,\beta }$, let us fix n, and introduce $\tilde{x}_k^+\in (x_k^n,x_{k+1}^n)$ and $\tilde{x}_k^-\in (x_{k-1}^n,x_k^n)$ such that

$$\begin{aligned} \frac{\rho '(x_{k+1}^n)-\rho '(x_{k-1}^n)}{\delta }&= \frac{\rho '(x_{k+1}^n)-\rho '(x_k^n)}{\delta }+ \frac{\rho '(x_k^n)-\rho '(x_{k-1}^n)}{\delta }\\&= \rho ''(\tilde{x}_k^+)\frac{x_{k+1}^n-x_k^n}{\delta }+ \rho ''(\tilde{x}_k^+)\frac{x_{k+1}^n-x_k^n}{\delta }\\&= \frac{\rho ''(\tilde{x}_k^+)}{z_{k+\frac{1}{2}}^n} + \frac{\rho ''(\tilde{x}_k^-)}{z_{k-\frac{1}{2}}^n}. \end{aligned}$$

For each $k\in {\mathbb {I}_K^+}$, we have that—recalling (70)—

$$\begin{aligned}&\frac{z_{k+\frac{1}{2}}^n+z_{k-\frac{1}{2}}^n}{2} \left( \frac{\rho ''(\tilde{x}_k^+)}{z_{k+\frac{1}{2}}^n} + \frac{\rho ''(\tilde{x}_k^-)}{z_{k-\frac{1}{2}}^n}\right) - \frac{2}{\delta }\int _{\xi _{k-\frac{1}{2}}}^{\xi _{k+\frac{1}{2}}}\rho ''\circ \mathrm {X}_\Delta ^n\,\mathrm {d}\xi \\&= \frac{1}{2}\left[ \left( \frac{z_{k-\frac{1}{2}}^n}{z_{k+\frac{1}{2}}^n}+1\right) \rho ''(\tilde{x}_k^+) + \left( \frac{z_{k+\frac{1}{2}}^n}{z_{k-\frac{1}{2}}^n}+1\right) \rho ''(\tilde{x}_k^-)\right] - \frac{2}{\delta }\int _{\xi _{k-\frac{1}{2}}}^{\xi _{k+\frac{1}{2}}}\rho ''\circ \mathrm {X}_\Delta ^n\,\mathrm {d}\xi \\&= \frac{1}{2}\left[ \left( \frac{z_{k-\frac{1}{2}}^n}{z_{k+\frac{1}{2}}^n}-1\right) \rho ''(\tilde{x}_k^+) + \left( \frac{z_{k+\frac{1}{2}}^n}{z_{k-\frac{1}{2}}^n}-1\right) \rho ''(\tilde{x}_k^-)\right] \\&\qquad - \frac{2}{\delta }\int _{\xi _k}^{\xi _{k+\frac{1}{2}}}\big [\rho ''\circ \mathrm {X}_\Delta ^n-\rho ''(\tilde{x}_k^+)\big ]\,\mathrm {d}\xi - \frac{2}{\delta }\int _{\xi _{k-\frac{1}{2}}}^{\xi _k}\big [\rho ''\circ \mathrm {X}_\Delta ^n-\rho ''(\tilde{x}_k^-)\big ]\,\mathrm {d}\xi . \end{aligned}$$

Since $\mathrm {X}_\Delta ^n(\xi )\in [x_k^n,x_{k+\frac{1}{2}}^n]$ for each $\xi \in [\xi _k,\xi _{k+\frac{1}{2}}]$, and $\tilde{x}_k^+\in [x_k^n,x_{k+1}^n]$, it follows that $|\mathrm {X}_\Delta ^n(\xi )-\tilde{x}_k^+|\le x_{k+1}^n-x_k^n$, and therefore,

$$\begin{aligned} \frac{2}{\delta }\int _{\xi _k}^{\xi _{k+\frac{1}{2}}}\big |\rho ''\circ \mathrm {X}_\Delta ^n(\xi )-\rho ''(\tilde{x}_k^+)\big |\,\mathrm {d}\xi \le B(x_{k+1}^n-x_k^n). \end{aligned}$$

(108)

A similar estimate holds for the other integral. Thus,

$$\begin{aligned} R_{1,\beta } \le B^2\tau \sum _{n=1}^N\delta \sum _{k\in {\mathbb {I}_K^+}}\left[ \left( \frac{z^n_{k-\frac{1}{2}}}{z^n_{k+\frac{1}{2}}}-1\right) ^2 + \left( \frac{z^n_{k+\frac{1}{2}}}{z^n_{k-\frac{1}{2}}}-1\right) ^2 + 2(x^n_{k+1}-x^n_{k-1})^2\right] . \end{aligned}$$

Combining the estimate (43) with inequality (99) from the “Appendix,” we further conclude that

$$\begin{aligned} R_{1,\beta } \le B^2\big (6(b-a)^2(\overline{\mathcal {H}}T\delta )^{1/2}+4T(b-a)^2\delta \big ). \end{aligned}$$

(109)

In combination with (106) and (107), this proves the claim. $\square$

Proof of estimate (77)

The proof is almost identical to (and even easier than) the one for estimate (76) above. Again, we have a decomposition of the form

$$\begin{aligned} R_2 \le R_{2,\alpha }^{1/2}R_{2,\beta }^{1/2}, \end{aligned}$$

where $R_{2,\alpha }$ equals $R_{1,\alpha }$ from (107), and

$$\begin{aligned} R_{2,\beta }&= \tau \sum _{n=1}^N\delta \sum _{k\in {\mathbb {I}_K^+}}\left[ \frac{z^n_{k+\frac{1}{2}}+z^n_{k-\frac{1}{2}}}{2 z^n_{k+\frac{1}{2}}z^n_{k-\frac{1}{2}}}\rho ''(x^n_k)-\frac{1}{\delta }\int _{\xi _{k-\frac{1}{2}}}^{\xi _{k+\frac{1}{2}}}\rho ''\circ \mathrm {X}^n_\Delta \,\mathrm {d}\xi \right] ^2. \end{aligned}$$

By writing

$$\begin{aligned} \frac{(z^n_{k+\frac{1}{2}})^2+(z^n_{k-\frac{1}{2}})^2}{2 z^n_{k+\frac{1}{2}}z^n_{k-\frac{1}{2}}} = \frac{1}{2}\left( \frac{z^n_{k-\frac{1}{2}}}{z^n_{k+\frac{1}{2}}}-1\right) + \frac{1}{2}\left( \frac{z^n_{k+\frac{1}{2}}}{z^n_{k-\frac{1}{2}}}-1\right) + 1, \end{aligned}$$

and observing—in analogy to (108)—that

$$\begin{aligned} \frac{1}{\delta }\int _{\xi _{k-\frac{1}{2}}}^{\xi _{k+\frac{1}{2}}}\big |\rho ''\circ \mathrm {X}_\Delta ^n(\xi )-\rho ''(x_k^n)\big |\,\mathrm {d}\xi \le B\left( x_{k+\frac{1}{2}}^n-x_{k-\frac{1}{2}}^n\right) , \end{aligned}$$

we obtain the same bound on $R_{2,\beta }$ as the one on $R_{1,\beta }$ from (109). $\square$

Proof of estimate (78)

Arguing like in the previous proofs, we first deduce—now by means of Hölder’s inequality instead of the Cauchy–Schwarz inequality—that

$$\begin{aligned} R_3 \le R_{3,\alpha }^{1/4}R_{3,\beta }^{3/4}, \end{aligned}$$

where $R_{3,\alpha }=R_{1,\alpha }$, and

$$\begin{aligned} R_{3,\beta }&= \tau \sum _{n=1}^N \delta \sum _{k\in {\mathbb {I}_K^+}}\Bigg |\left( \frac{\left( z^n_{k+\frac{1}{2}}\right) ^2+\left( z^n_{k-\frac{1}{2}}\right) ^2}{2}\right) \\&\quad \times \left( \frac{\rho '(x^n_{k+1})-\rho '(x^n_k)-(x^n_{k+1}-x^n_k)\rho ''(x^n_k)}{\delta ^2}\right) \\&\quad - \,\frac{1}{2\delta }\int _{\xi _{k-\frac{1}{2}}}^{\xi _{k+\frac{1}{2}}}\rho '''\circ \mathrm {X}^n_\delta \,\mathrm {d}\xi \Bigg |^{4/3}. \end{aligned}$$

Introduce intermediate values $\tilde{x}_k^+$ such that

$$\begin{aligned} \rho '(x^n_{k+1})\!-\!\rho '(x^n_k)\!-\!(x^n_{k+1}\!-\!x^n_k)\rho ''(x^n_k) \!=\!\frac{1}{2}(x^n_{k+1}\!-\!x^n_k)^2\rho '''(\tilde{x}_k^+) \!=\!\frac{\delta ^2}{2\left( z^n_{k+\frac{1}{2}}\right) ^2}\rho '''(\tilde{x}_k^+). \end{aligned}$$

Thus, we have that

$$\begin{aligned}&\left( \frac{(z^n_{k+\frac{1}{2}})^2+(z^n_{k-\frac{1}{2}})^2}{2}\right) \left( \frac{\rho '(x^n_{k+1})-\rho '(x^n_k)-(x^n_{k+1}-x^n_k)\rho ''(x^n_k)}{\delta ^2}\right) \\&\quad - \,\frac{1}{2\delta }\int _{\xi _{k-\frac{1}{2}}}^{\xi _{k+\frac{1}{2}}}\rho '''\circ \mathrm {X}^n_\delta \,\mathrm {d}\xi \\&\quad = \frac{1}{4}\left( \left( \frac{z^n_{k-\frac{1}{2}}}{z^n_{k+\frac{1}{2}}}\right) ^2+1\right) \rho '''(\tilde{x}_k^+) -\frac{1}{2\delta }\int _{\xi _{k-\frac{1}{2}}}^{\xi _{k+\frac{1}{2}}}\rho '''\circ \mathrm {X}^n_\delta \,\mathrm {d}\xi \\&\quad = \frac{1}{4}\left( \frac{z^n_{k-\frac{1}{2}}}{z^n_{k+\frac{1}{2}}}+1\right) \left( \frac{z^n_{k-\frac{1}{2}}}{z^n_{k+\frac{1}{2}}}-1\right) \rho '''(\tilde{x}_k^+) -\frac{1}{2\delta }\int _{\xi _{k-\frac{1}{2}}}^{\xi _{k+\frac{1}{2}}}\big [\rho '''\circ \mathrm {X}^n_\Delta -\rho '''(\tilde{x}_k^+)\big ]\,\mathrm {d}\xi . \end{aligned}$$

By the analogue of (108), it follows further that

$$\begin{aligned} R_{3,\beta }&\le 2B^{4/3}\tau \sum _{n=1}^N \delta \sum _{k\in {\mathbb {I}_K^+}}\left[ \left( \frac{z^n_{k-\frac{1}{2}}}{z^n_{k+\frac{1}{2}}}+1\right) ^{4/3}\left( \frac{z^n_{k-\frac{1}{2}}}{z^n_{k+\frac{1}{2}}}-1\right) ^{4/3}+(x^n_{k+1}-x^n_{k-1})^{4/3}\right] \\&\le 2B^{4/3}\left( \tau \sum _{n=1}^N \delta \sum _{k\in {\mathbb {I}_K^+}}\left( \frac{z^n_{k-\frac{1}{2}}}{z^n_{k+\frac{1}{2}}}+1\right) ^4\right) ^{1/3} \left( \tau \sum _{n=1}^N\delta \sum _{k\in {\mathbb {I}_K^+}}\left( \frac{z^n_{k-\frac{1}{2}}}{z^n_{k+\frac{1}{2}}}-1\right) ^2\right) ^{2/3} \\&\quad + \, 2B^{4/3}T(b-a)^{4/3}\delta , \end{aligned}$$

where we have used (99) from the “Appendix.” At this point, the estimates (42) and (43) are used to control the first and the second sum, respectively. $\square$

Proof of estimate (79)

Here, one proceeds in full analogy to the proof of estimate (78) above. $\square$