Linear gradient structures and discrete gradient methods for conservative/dissipative differential-algebraic equations

Abstract

In this paper, the use of discrete gradients is considered for differential-algebraic equations (DAEs) with a conservation/dissipation law. As one of the most popular numerical methods for conservative/dissipative ordinary differential equations, the framework of the discrete gradient method has been intensively developed over recent decades. Although discrete gradients have been applied to several specific DAEs, no unified framework has yet been constructed. In this paper, the author moves toward the establishment of such a framework, and introduces concepts including an appropriate linear gradient structure for DAEs. Then, it is revealed that the simple use of discrete gradients does not imply the discrete conservation/dissipation laws. Fortunately, however, for the case of index-1 DAEs, an appropriate reformulation and a new discrete gradient enable us to successfully construct a novel scheme, which satisfies both of the discrete conservation/dissipation law and the constraint. This first attempt may provide an indispensable basis for constructing a unified framework of discrete gradient methods for DAEs.

Appendices

Appendix A: Continuity of ${\overline{\nabla }}_{\mathrm {P}}V$

• (i) V is quadratic, i.e., $V(z) = (1/2) z^{\top } B z$ for a symmetric matrix B:

  in this case, since ${\overline{\nabla }}_{\mathrm {P}}V(z,z') = B (z+z')/2$ holds, it is clearly continuous map. Note that, in this case, ${\overline{\nabla }}_{\mathrm {P}}V = {\overline{\nabla }}_{\mathrm {AVF}} V$.

• (ii) V is strictly convex:

  in this case, the inequality

  $$\begin{aligned} 0< V(z) - V(z') - \langle \nabla V(z') , z - z' \rangle < \left\langle \nabla V(z) - \nabla V(z'), z - z' \right\rangle \end{aligned}$$

  holds for any $z \ne z'$. It implies that $\theta (z,z') \in (0,1)$ holds for $z \ne z'$. As ${\overline{\nabla }}_{\mathrm {P}}V$ can also be written in the form

  $$\begin{aligned} {\overline{\nabla }}_{\mathrm {P}}V(z,z') = \nabla V(z') + \theta (z,z') \left( \nabla V(z) - \nabla V(z') \right) , \end{aligned}$$

  (A.1)

  the boundedness of $\theta$ proves the continuity of ${\overline{\nabla }}_{\mathrm {P}}V$.

• (iii) V is convex and $L_V$-smooth:

  in this case, we also use (A.1), but we prove that $\theta (z,z')$ is bounded when $\nabla V(z) \ne \nabla V(z')$. Since V is a convex function, we see

  $$\begin{aligned} 0 \le V(z) - V(z') - \langle \nabla V(z') , z - z' \rangle \le \left\langle \nabla V(z) - \nabla V(z'), z - z' \right\rangle . \end{aligned}$$

  This implies that $\theta (z,z') \in [0,1]$ holds when the denominator of $\theta (z,z')$ does not vanish. Moreover, $L_V$-smoothness provides us with the inequality

  $$\begin{aligned} \left\langle \nabla V(z) - \nabla V(z'), z - z' \right\rangle \ge \frac{1}{L_V} \left\| \nabla V(z) - \nabla V(z') \right\| ^2 \end{aligned}$$

  (note that the conjugate function $V^{*}$ is $1 / L_V$-strong convex). Summing up, we see that $\theta \in [0,1]$ holds when $\nabla V(z) \ne \nabla V(z')$.

Appendix B: Non-convex case: sine-Gordon equation

Here we consider the sine-Gordon equation $u_{tx} = \sin u$ which has the conservation law ${\mathscr {H}} (u) = - \int _{{\mathbb {S}}} \cos u \, {\mathrm {d}}x = \text {const.}$ and the implicit constraint ${\mathscr {F}} (u) = \int _{{\mathbb {S}}} \sin u \, {\mathrm {d}}x = 0$. The spatial discretization $D {\dot{u}} = M \sin u$ satisfies the discrete counterparts $H(u) = - \sum _{k=1}^K \cos u_k$ and $F(u) = \sum _{k=1}^K \sin u_k$ of the conservation law and the implicit constraint.

However, since H is not convex (and not quadratic), we do not know whether ${\overline{\nabla }}_{\mathrm {P}}H$ is continuous. Therefore, we use a trick to construct a discrete gradient which is compatible with properness, i.e., belongs to ${\mathrm {car}}(D)$. Let us consider the expression $H(u) = H_1 (u) + H_2 (u)$, where $H_1 (u) = H(u) + (\alpha / 2) \langle u , P u \rangle$ and $H_2 (u) = - (\alpha / 2) \langle u , P u \rangle$ ($\alpha \in {\mathbb {R}}$ is a constant and P is the orthogonal projector on the set of zero-mean vectors). In the expression, (a) $H_1$ and $H_2$ are proper, and (b) $H_2$ is quadratic so that ${\overline{\nabla }}_{\mathrm {P}}H_2$ is continuous. Thus, roughly speaking, it is sufficient to choose the constant $\alpha$ such that $H_1$ is convex. This itself cannot be done since P is a singular matrix, but a sufficiently large $\alpha$ provides us with the similar result as shown in Lemma B.1. To this end, we assume $H_0 := H(u_0) < 0$ (the case $H_0 > 0$ can be treated similarly).

Lemma B.1

Let $H_0 < 0$ be a constant. Then, if $\alpha > 1 - K / H_0$, the Hessian $\nabla ^2 H_1 (u ) = \mathrm {diag} ( \cos u_k ) + \alpha P$ of ${\widetilde{H}}$ is positive definite on the domain $\{ u \in {\mathbb {R}}^K \mid H(u) = H_0 \}$.

Proof

It is sufficient to prove $\langle {\widetilde{v}} , \nabla ^2 H_1 (u) {\widetilde{v}} \rangle > 0$ holds for any ${\widetilde{v}} \in \{ w \mid \Vert w \Vert = 1 \} = \{ v \pm \sqrt{ (1-x^2)/K } {\mathbf {1}} \mid v \in {\mathbb {R}}^K , \ \Vert v \Vert = x, \, \langle v,\ {\mathbf {1}} \rangle = 0, x \in [0,1] \}$. Since

$$\begin{aligned} 2 x \sqrt{1-x^2}&= 2 \sqrt{ \left( - \frac{1-x^2}{K} H_0 \right) \left( -\frac{K}{H_0} x^2 \right) } \le - \frac{1-x^2}{K} H_0 - \frac{K}{H_0} x^2 \end{aligned}$$

holds due to the inequality of arithmetic and geometric means, we see

$$\begin{aligned} \langle {\widetilde{v}} , \nabla ^2 H_1 (u) {\widetilde{v}} \rangle&= \sum _{k=1}^K (\cos u_k )(v_k)^2 \pm 2 \sqrt{ \frac{1-x^2}{K} } \langle v, \cos u \rangle - \frac{1-x^2}{K} H_0 + \alpha x^2 \\&\ge - \Vert v \Vert ^2 - 2 \sqrt{ \frac{1-x^2}{K} } \Vert v \Vert \Vert \cos u \Vert - \frac{1-x^2}{K} H_0 + \alpha x^2\\&\ge - x^2 - 2 x \sqrt{1-x^2} - \frac{1-x^2}{K} H_0 + \alpha x^2 \\&\ge - x^2 + \frac{1-x^2}{K} H_0 + \frac{K}{H_0} x^2 - \frac{1-x^2}{K} H_0 \\&\quad + \alpha x^2 = \left( - 1 + \frac{K}{H_0} + \alpha \right) x^2. \end{aligned}$$

Thus, for $x \in (0,1]$, $\langle {\widetilde{v}} , \nabla ^2 H_1 (u) {\widetilde{v}} \rangle > 0$ holds under the assumptions of this lemma. On the other hand, for the case $x = 0$, i.e., ${\widetilde{v}} = \sqrt{1/K} {\mathbf {1}}$, $\langle {\widetilde{v}} , \nabla ^2 H_1 (u) {\widetilde{v}} \rangle = - H_0/K > 0$ holds. $\square$

By using $H_1$ and $H_2$, we can construct

$$\begin{aligned} {\widetilde{\nabla }} H(u,u') = {\overline{\nabla }}_{\mathrm {P}}H_1 (u,u') - \alpha P \frac{u + u'}{2} \end{aligned}$$

(B.1)

which can be used like as a discrete gradient (see Proposition B.1).

Proposition B.1

Suppose that $\alpha > 1 - K / H_0$. The function ${\widetilde{\nabla }} H :{\mathbb {R}}^K \times {\mathbb {R}}^K \rightarrow {\mathbb {R}}^K$ defined by (B.1) satisfies

• $H(u) - H(u') = \langle {\widetilde{\nabla }} H (u,u'), u - u' \rangle$,

• ${\widetilde{\nabla }} H(u,u) = \nabla H(u)$,

• ${\widetilde{\nabla }} H(u,u+\epsilon )$ is continuous in the neighborhood of $\epsilon = 0$ as a function of $\epsilon$ for each $u \in \{ u \in {\mathbb {R}}^K \mid H(u) = H_0 \}$.

Proof

The first two conditions are immediate. The third condition also holds since the denominator of $\theta$ with respect to $H_1$ is positive in the neighborhood due to the positive definiteness of the Hessian. $\square$

Since we do not know the continuity of ${\widetilde{\nabla }} H$ on the whole domain ${\mathbb {R}}^K$, it cannot be referred to as a discrete gradient (recall Definition 1.1). Fortunately, however, we can construct a consistent numerical scheme by using ${\widetilde{\nabla }} H$, because the continuity of the discrete gradient is just a sufficient condition for the consistency of the resulting numerical scheme. In fact, the scheme (6.3) using ${\widetilde{\nabla }} H$ defined by (B.1) was found to be solvable by “fsolve” of MATLAB R2016b. Since the behavior is quite similar to the case of sinh-Gordon equation, we omit the numerical results.