# Cauchy problem for a Schrödinger-type equation related to the Riemann zeta function

Mohamed Bensaid

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Mohamed Bensaid. Cauchy problem for a Schrödinger-type equation related to the Riemann zeta function. 2025. hal-05369105

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# Cauchy problem for a Schrödinger-type equation related to the Riemann zeta function

Mohamed Bensaid*

November 17, 2025

## Abstract

We study the Cauchy problem in the space $H^1$ for a nonlinear damped Schrödinger equation of the form

$$iu_t + \Delta u + i\lambda u \zeta(|u| + 1) = 0, \quad u(0, x) = u_0,$$

where $\zeta$ denotes the Riemann zeta function. This equation involves a damping term inspired by a fundamental object from number theory. We first establish the uniqueness of solutions in the sense of distributions. Then, by considering a regularized problem, we prove the existence of a global solution in $H^1$, using uniform energy estimates and compactness arguments. Finally, we show that the limiting solution indeed satisfies the original equation in the weak sense.

## Introduction

The Schrödinger equation plays a central role in mathematical physics, particularly in the modeling of quantum phenomena. When nonlinear damping terms are added, it also allows the description of dissipative effects, leading to rich and complex dynamical behaviors. In this work, we study a Cauchy problem associated with a damped version of the Schrödinger equation, where the damping depends on a function related to the famous Riemann zeta function.

The Riemann zeta function, initially defined for $\mathrm{Re}(s) > 1$ by

$$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s},$$

plays a fundamental role in number theory, notably through its deep connection with the distribution of prime numbers. Its analytic continuation, as well as the famous Riemann hypothesis concerning the location of its nontrivial zeros, constitute one of the most important problems in modern mathematics.

In our study, we consider a damping term defined in terms of the function $\zeta(x)$. We first establish the uniqueness of solutions to the problem in the sense of distributions, before employing a regularization approach to prove the existence of weak solutions in $H^1$.

Furthermore, we consider a Riemannian manifold $\Sigma$, which we assume to be compact and of finite dimension $d$. This is necessary because, if we work in $\mathbb{R}^d$, the function $x \mapsto x \zeta(x + 1)$ no longer belongs to $L^p(\mathbb{R}^d)$, making the control of the nonlinear term more delicate.

<footnote>* Université de Lille, France, Professional E-mail adresse : mohamed.bensaid@univ-lille.fr</footnote>

<footnote>Personnel E-mail adresse : bensaidmed00@gmail.com</footnote>
As $x$ approaches 0, we have $\zeta(x+1) \sim \frac{1}{x}$. In this context, the Cauchy theory in $H^1(\Sigma)$ for the equation

$$ (E0) \qquad iv_t + \Delta v + i\lambda \frac{v}{|v|} = 0, \quad v(0, x) = v_0, $$

was established in [3] by Rémi Carles and Clément Gallo.

## Construction of the Solution in $H^1(\Sigma)$

Let $(\Sigma, g)$ be a compact Riemannian surface. We consider the Cauchy problem

$$ (E) \qquad \begin{cases} i\partial_t u + \Delta u + i\lambda u \zeta(|u|+1) = 0, & (t,x) \in \mathbb{R}_+ \times \Sigma, \\ u(0,x) = u_0(x) \in H^1(\Sigma), \end{cases} $$

where $\lambda > 0$ and $\zeta: \mathbb{R}_+ \to \mathbb{R}$ denotes the Riemann zeta function.

We construct solutions via a regularization by adding a parameter $\varepsilon > 0$:

$$ (E_\varepsilon) \qquad \begin{cases} i\partial_t u_\varepsilon + \Delta u_\varepsilon + i\lambda u_\varepsilon \zeta(|u_\varepsilon| + 1 + \varepsilon) = 0, \\ u_\varepsilon(0) = u_0. \end{cases} $$

**Proposition 1** (Uniqueness). Let $u_1, u_2 \in L^\infty(\mathbb{R}_+; H^1(\Sigma))$ be two (distributional) solutions of $(E)$ with the same initial data. If $\lambda > 0$, then $u_1 = u_2$.

*Proof.* Set $w = u_1 - u_2$. Since $u_1$ and $u_2$ are solutions, we have

$$ i\partial_t w + \Delta w + i\lambda(u_1\zeta(|u_1|+1) - u_2\zeta(|u_2|+1)) = 0. $$

We take the $L^2$ inner product with $w$ and consider the real part:

$$ \frac{1}{2} \frac{d}{dt} \|w(t)\|_{L^2(\Sigma)}^2 = \operatorname{Re} \langle \partial_t w, w \rangle = \operatorname{Re} \langle -i\Delta w - \lambda(u_1\zeta(|u_1|+1) - u_2\zeta(|u_2|+1)), w \rangle. $$

The term containing $\Delta$ is purely imaginary, so its real part vanishes. Hence,

$$ \frac{d}{dt} \|w(t)\|_{L^2(\Sigma)}^2 = -2\lambda \operatorname{Re}\langle u_1\zeta(|u_1|+1) - u_2\zeta(|u_2|+1), w \rangle. $$

To conclude, we use the following monotonicity lemma (with $f(x) = x \mapsto x\zeta(x+1)$):

$$ \operatorname{Re} \left( (uf(|u|) - vf(|v|))(\bar{u} - \bar{v}) \right) \geq 0 \quad \forall u, v \in \mathbb{C}. $$

This lemma (proved below) implies that the right-hand side is $\le 0$, hence

$$ \frac{d}{dt} \|w(t)\|_{L^2(\Sigma)}^2 \le 0. $$

Since $w(0) = 0$ (same initial data), it follows that $\|w(t)\|_{L^2(\Sigma)}^2 = 0$, and therefore $u_1 = u_2$. $\square$

**Lemma 2.** Let $f: \mathbb{R}_+ \to \mathbb{R}_+$ be a function such that $xf(x)$ is non-decreasing. Then, for all $u, v \in \mathbb{C}$,

$$ \operatorname{Re} \left( (uf(|u|) - vf(|v|))(\bar{u} - \bar{v}) \right) \geq 0. $$
Proof. Expanding the real part gives

$$
\begin{align*}
\mathrm{Re}((uf(|u|) - v f(|v|))(\bar{u} - \bar{v})) &= |u|^2 f(|u|) + |v|^2 f(|v|) - \mathrm{Re}(u\bar{v})(f(|u|) + f(|v|)) \\
&\geq |u|^2 f(|u|) + |v|^2 f(|v|) - |u||v|(f(|u|) + f(|v|)) \\
&= (|u| - |v|)(|u|f(|u|) - |v|f(|v|)).
\end{align*}
$$

The product above is non-negative since $x \mapsto xf(x)$ is non-decreasing, which proves the inequality. $\square$

Now, we construct the solution in $H^1(\Sigma)$. Consider the following regularized problem:

$$
(E_{\epsilon}) : \begin{cases} i \partial_t u_{\epsilon} + \Delta u_{\epsilon} + i \lambda u_{\epsilon} \zeta(|u_{\epsilon}| + 1 + \epsilon) = 0, \\ u_{\epsilon}(0, x) = u_0. \end{cases}
$$

For a fixed $\epsilon > 0$, equation $(E_\epsilon)$ is a Schrödinger equation with a locally Lipschitz nonlinearity on $H^1(\Sigma)$. Therefore, by a fixed-point argument (see Chapter 3 of [2]), there exists a maximal solution

$$
u_{\epsilon} \in C([0, T_{\epsilon}); H^{1}(\Sigma)) \cap C^{1}([0, T_{\epsilon}]; H^{-1}(\Sigma)).
$$

The estimates below show that $T_\epsilon = +\infty$ and provide uniform bounds with respect to $\epsilon$.
Multiplying $(E_\epsilon)$ by $\bar{u}_\epsilon$, integrating over $\Sigma$, and taking the real part yields

$$
\frac{1}{2} \frac{d}{dt} \|u_{\epsilon}(t)\|_{L^2(\Sigma)}^2 = -\lambda \int_{\Sigma} |u_{\epsilon}|^2 \zeta(|u_{\epsilon}| + 1 + \epsilon) dx.
$$

Under the assumption $\lambda > 0$ and if $\zeta(\cdot) \ge 0$ (or if the product is non-negative), we obtain in particular

$$
(1) \qquad \|u_{\epsilon}(t)\|_{L^2(\Sigma)} \le \|u_0\|_{L^2(\Sigma)}, \quad \forall t \ge 0.
$$

Thus, the mass is uniformly bounded in $\epsilon$.

**Lemma 3.**

$$
(2) \qquad \| \nabla u_{\epsilon}(t) \|_{L^2(\Sigma)} \le \| \nabla u_0 \|_{L^2(\Sigma)}, \quad \forall t \ge 0.
$$

Proof.

$$
\frac{1}{2} \frac{d}{dt} \| \nabla u_{\epsilon} \|_{L^2(\Sigma)}^2 = \mathrm{Re} \langle \partial_t u_{\epsilon}, -\Delta u_{\epsilon} \rangle.
$$

Replacing $\partial_t u_\epsilon = -i\Delta u_\epsilon - \lambda u_\epsilon \zeta(|u_\epsilon| + 1 + \epsilon)$, the term involving $\Delta$ is purely imaginary and vanishes, giving

$$
\frac{1}{2} \frac{d}{dt} \| \nabla u_{\epsilon} \|_{L^2(\Sigma)}^2 = -\lambda \operatorname{Re} \langle u_{\epsilon} \zeta(|u_{\epsilon}| + 1 + \epsilon), \Delta u_{\epsilon} \rangle.
$$

By integration by parts,

$$
\mathrm{Re}\langle u_{\epsilon}\zeta(|u_{\epsilon}| + 1 + \epsilon), \Delta u_{\epsilon} \rangle = - \int_{\Sigma} \nabla(u_{\epsilon}\zeta(|u_{\epsilon}| + 1 + \epsilon)) \cdot \overline{\nabla u_{\epsilon}} \, dx.
$$

Expanding the gradient and taking the real part, we obtain

$$
\frac{1}{2} \frac{d}{dt} \| \nabla u_{\epsilon} \|_{L^2(\Sigma)}^2 = -\lambda \int_{\Sigma} \zeta(|u_{\epsilon}| + 1 + \epsilon) |\nabla u_{\epsilon}|^2 - \lambda \int_{\Sigma} |u_{\epsilon}| \zeta'(|u_{\epsilon}| + 1 + \epsilon) \operatorname{Re} \frac{u_{\epsilon} \nabla |u_{\epsilon}| \cdot \overline{\nabla u_{\epsilon}}}{|u_{\epsilon}|} \, dx.
$$
By Lemma 19, the function $f: x \mapsto x\zeta(x+1+\epsilon)$ is non-decreasing, so $-f'(x) \le 0$. In particular,

$$-|u_{\epsilon}|\zeta'(|u_{\epsilon}| + 1 + \epsilon) \le 0, \quad \text{and} \quad \zeta'(|u_{\epsilon}| + 1 + \epsilon) \le 0.$$

Thus,

$$\frac{1}{2} \frac{d}{dt} \|\nabla u_{\epsilon}\|_{L^2(\Sigma)}^2 \le \lambda \int_{\Sigma} |u_{\epsilon}| \zeta'(|u_{\epsilon}| + 1 + \epsilon) \left( |\nabla u_{\epsilon}|^2 - \operatorname{Re} \frac{u_{\epsilon} \nabla |u_{\epsilon}| \cdot \overline{\nabla u_{\epsilon}}}{|u_{\epsilon}|} \right) dx \le 0.$$

**Remark 1.** Here we used that

$$\left| \operatorname{Re} \frac{u_{\epsilon} \nabla |u_{\epsilon}| \cdot \overline{\nabla u_{\epsilon}}}{|u_{\epsilon}|} \right| \le |\nabla u_{\epsilon}|^2, \quad \zeta'(|u_{\epsilon}| + 1 + \epsilon) \le 0.$$

By the fundamental theorem of calculus, the lemma follows:

$$\|\nabla u_{\epsilon}(t)\|_{L^2(\Sigma)} \le \|\nabla u_0\|_{L^2(\Sigma)}, \quad \forall t \ge 0.$$

□

**Remark 2.** The inequalities (1) and (2) show that

$$\sup_{t \ge 0} \|u_{\epsilon}(t)\|_{H^1(\Sigma)} \le C(\|u_0\|_{H^1(\Sigma)}),$$

which allows the solution to be extended globally in time: $T_{\epsilon} = +\infty$.

**Lemma 4.** Up to extraction of a subsequence, the family $(u_{\epsilon})_{\epsilon}$ converges in $C([0, T]; L^2(\Sigma))$, and the limit is denoted by $u$.

*Proof.* From the previous estimates, the family $\{u_{\epsilon}\}_{0<\epsilon<1}$ is uniformly bounded in $L^{\infty}(\mathbb{R}_{+}; H^1(\Sigma))$. Moreover, from equation $(E_{\epsilon})$,

$$\partial_t u_{\epsilon} = -i\Delta u_{\epsilon} - \lambda u_{\epsilon} \zeta(|u_{\epsilon}| + 1 + \epsilon).$$

The first term belongs to $L^{\infty}(\mathbb{R}_{+}; H^{-1}(\Sigma))$ (since $\Delta u_{\epsilon} \in H^{-1}$) and the second term is uniformly bounded in $L^{\infty}(\mathbb{R}_{+}; L^2(\Sigma))$. Since $L^2(\Sigma) \hookrightarrow H^{-1}(\Sigma)$, we have

$$\sup_{\epsilon} \|\partial_t u_{\epsilon}\|_{L^{\infty}(0,T;H^{-1}(\Sigma))} < \infty.$$

We then apply the Aubin-Lions lemma (compactness in time): since

$$H^1(\Sigma) \xrightarrow{\text{compact}} L^2(\Sigma) \xrightarrow{\text{continuous}} H^{-1}(\Sigma),$$

and $(u_{\epsilon})$ is bounded in $L^{\infty}(0, T; H^1(\Sigma))$ while $(\partial_t u_{\epsilon})$ is bounded in $L^{\infty}(0, T; H^{-1}(\Sigma))$, there exists a subsequence (still denoted $u_{\epsilon}$) and a function

$$u \in C([0, T]; L^2(\Sigma))$$

such that

$$u_{\epsilon} \to u \quad \text{in} \quad C([0, T]; L^2(\Sigma)).$$

□
**Lemma 5.** The function $u$, defined in Lemma 4, satisfies:

1. $u \in L^{\infty}(\mathbb{R}_+, H^1(\Sigma))$.

2. $u_{\epsilon}(t) \to u(t)$ weakly in $H^1(\Sigma)$, for all $t \in \mathbb{R}_+$.

*Proof.*

1. For any $\phi \in C_c^{\infty}(\Sigma)$, by integration by parts and the Cauchy-Schwarz inequality:

$$ \left| \int_{\Sigma} u_{\epsilon} \nabla \phi \, dx \right| \leq M \|\phi\|_{2}. $$

Passing to the limit, we deduce:

$$ \left| \int_{\Sigma} u \nabla \phi \, dx \right| \leq M \|\phi\|_{2}, $$

which shows that $u(t) \in H^1(\Sigma)$ and $\|\nabla u(t)\|_2 \leq M$ for all $t \in \mathbb{R}_+$.

2. To show weak convergence in $H^1(\Sigma)$: for any $\phi \in C_c^{\infty}(\Sigma)$,

$$ \int_{\Sigma} \nabla u_{\epsilon}(t) \phi \, dx \to_{\epsilon} \int_{\Sigma} \nabla u(t) \phi \, dx. $$

Since $C_c^{\infty}(\Sigma)$ is dense in $L^2(\Sigma)$, for any $\phi \in L^2(\Sigma)$ and $\delta > 0$, choose $\phi_n \in C_c^{\infty}(\Sigma)$ such that

$$ \|\phi_n - \phi\|_2 < \delta. $$

Then,

$$ \begin{aligned} \left| \int_{\Sigma} \nabla u_{\epsilon} \phi - \int_{\Sigma} \nabla u \phi \right| & \leq \left| \int_{\Sigma} \nabla u_{\epsilon}(\phi - \phi_n) \right| + \left| \int_{\Sigma} \nabla u_{\epsilon} \phi_n - \int_{\Sigma} \nabla u \phi_n \right| + \left| \int_{\Sigma} \nabla u(\phi_n - \phi) \right| \\ & \leq M\delta + \left| \int_{\Sigma} \nabla u_{\epsilon} \phi_n - \int_{\Sigma} \nabla u \phi_n \right| + \|\nabla u\|_2 \delta. \end{aligned} $$

Taking the limit superior as $\epsilon \to 0$ gives

$$ \limsup_{\epsilon \to 0} \left| \int_{\Sigma} \nabla u_{\epsilon} \phi - \int_{\Sigma} \nabla u \phi \right| \leq (M + \|\nabla u\|_2) \delta. $$

Since $\delta > 0$ is arbitrary, this proves weak convergence in $H^1(\Sigma)$.

□

**Lemma 6.** For all $t \in \mathbb{R}_+$,

$$ u_{\epsilon}(t) \zeta(|u_{\epsilon}(t)| + 1 + \epsilon) \to u(t) \zeta(|u(t)| + 1) \quad \text{in } L^2(\Sigma). $$
Proof. For all $(t, x) \in \mathbb{R}_+ \times \Sigma$, we have

$$|u_{\epsilon}(t, x)\zeta(|u_{\epsilon}| + 1 + \epsilon)|^2 \leq 2(1 + |u_{\epsilon}(t, x)|^2).$$

Since $\Sigma$ is bounded and $\|u_{\epsilon}(t)\|_2 \leq \|u_0\|_2$, the sequence $u_{\epsilon}\zeta(|u_{\epsilon}| + 1 + \epsilon)$ is uniformly bounded in $L^{\infty}(\mathbb{R}_+, L^2(\Sigma))$. By Banach-Alaoglu, there exists $V$ such that

$$u_{\epsilon}\zeta(|u_{\epsilon}| + 1 + \epsilon) \xrightarrow{\ast} V \in L^{\infty}(\mathbb{R}_+, L^2(\Sigma)).$$

Moreover, up to a subsequence,

$$u_{\epsilon} \to u \text{ in } C([0, T], L^2(\Sigma)).$$

Hence, for $(t, x) \in [0, T] \times \Sigma$, up to a subsequence,

$$u_{\epsilon}\zeta(|u_{\epsilon}| + 1 + \epsilon)(t, x) \to u\zeta(|u| + 1)(t, x),$$

which shows that

$$V(t, x) = u\zeta(|u| + 1)(t, x).$$

□

**Theorem 7.** Let $u$ be the function defined in Lemma 4. Then

$$u \in L^{\infty}(\mathbb{R}_+; H^1(\Sigma)) \cap W^{1, \infty}(\mathbb{R}_+; H^{-1}(\Sigma)).$$

Moreover,

$$(3) \qquad iu_t + \Delta u + i\lambda u\zeta(|u| + 1) = 0 \quad \text{in } H^{-1}(\Sigma), \quad \forall t \in \mathbb{R}_+.$$

Proof. Let $\phi \in C_c^1(\mathbb{R}_+^*)$ and $\psi \in C_c^{\infty}(\Sigma)$. Then

$$\int_{\mathbb{R}_+} (iu_{\epsilon}, \psi)_2 \phi'(t) dt = \int_{\mathbb{R}_+} [- (\nabla u_{\epsilon}, \nabla \psi)_2 + \lambda(u_{\epsilon}\zeta(|u_{\epsilon}| + 1 + \epsilon), \psi)_2] \phi(t) dt.$$

By Cauchy-Schwarz,

$$|(\nabla u_{\epsilon}, \nabla \psi)_2| \leq M \|\nabla \psi\|_2, \quad |(u_{\epsilon}\zeta(|u_{\epsilon}| + 1 + \epsilon), \psi)_2| \leq (M + |\Sigma|^{1/2}) \|\psi\|_2.$$

Passing to the limit and using the dominated convergence theorem together with Lemmas 6 and 5, we obtain

$$\int_{\mathbb{R}_+} (iu, \psi)_2 \phi'(t) dt = \int_{\mathbb{R}_+} [- (\nabla u, \nabla \psi)_2 + \lambda(u\zeta(|u| + 1), \psi)_2] \phi(t) dt.$$

□

**Lemma 8.** Let $u \in L^{\infty}(\mathbb{R}_+, H^1(\Sigma))$ be the solution obtained in Theorem 7. Then, for all $t \in \mathbb{R}_+$,

$$\|u(t)\|_{L^2(\Sigma)} \leq e^{-\lambda t} \|u_0\|_{L^2(\Sigma)}.$$

In particular,

$$\|u(t)\|_{L^2(\Sigma)} \to 0 \quad \text{as } t \to +\infty.$$
*Proof.* We have

$$ \frac{1}{2} \frac{d}{dt} \|u(t)\|_2^2 = -\lambda \int_{\Sigma} |u(t)|^2 \zeta(|u(t)| + 1) \, dx \leq -\lambda \int_{\Sigma} |u(t)|^2 \, dx, $$

where the last inequality follows from $\zeta(|u| + 1) \geq 1$. By Grönwall's lemma, it follows that

$$ \|u(t)\|_2 \leq e^{-\lambda t} \|u_0\|_2. $$

□

**Lemma 9.** There exists $c > 0$ such that, for all $z, s \in \mathbb{C}^*$,

$$ \mathrm{Re}((f(z) - f(s))(\bar{z} - \bar{s})) \geq c|z - s|^2. $$

*Proof.* We write

$$ f(z) - f(s) = \int_{0}^{1} Df(s + t(z - s)) \cdot (z - s) \, dt, $$

where

$$ Df(w) \cdot h = \zeta(|w| + 1)h + w \zeta'(|w| + 1) \frac{\mathrm{Re}(\overline{wh})}{|w|}. $$

Set $\delta_t = s + t(z - s)$ and $h = z - s$. Then

$$ \mathrm{Re}((f(z) - f(s))(\bar{z} - \bar{s})) = \mathrm{Re} \int_{0}^{1} \left( \zeta(|\delta_t| + 1)|h|^2 + \zeta'(|\delta_t| + 1) \frac{(\mathrm{Re}(\overline{\delta_t h}))^2}{|\delta_t|} \right) dt. $$

Since $\zeta'(\cdot) < 0$, using $(\mathrm{Re}(\overline{\delta_t h}))^2 \leq |\delta_t|^2|h|^2$, we get

$$ \mathrm{Re}((f(z) - f(s))(\bar{z} - \bar{s})) \geq \int_{0}^{1} (\zeta(|\delta_t| + 1) + |\delta_t|\zeta'(|\delta_t| + 1))|h|^2 dt. $$

On the other hand, for $r > 0$,

$$ \zeta(r + 1) + r\zeta'(r + 1) = 1 + \sum_{n \geq 2} \frac{1 - r \ln n}{n^{r+1}} \geq 1 + \int_{2}^{\infty} \frac{1 - r \ln t}{t^{r+1}} dt = 1 - \frac{\ln 2}{2r} \geq 1 - \ln 2. $$

Hence,

$$ \mathrm{Re}((f(z) - f(s))(\bar{z} - \bar{s})) \geq c|z - s|^2, $$

with $c := 1 - \ln 2 > 0$.

□

**Lemma 10.** For all $z \in \mathbb{C}^*$, we have

$$ |f(z) - \frac{z}{|z|}| \leq |z|. $$

*Proof.* Let $z \in \mathbb{C}^*$. Then

$$ |f(z) - \frac{z}{|z|}| = |z| |\zeta(|z| + 1) - \frac{1}{|z|}| \leq |z|. $$

□
**Proposition 11.** Let $u_0, v_0 \in H^1(\Sigma)$, and denote by $u$ and $v$ the solutions of (E) and (E0), respectively. Suppose that $u_0 = v_0$. Then there exists $\alpha > 0$ such that, for all $t \ge 0$,

$$ \|u(t) - v(t)\|_{L^2(\Sigma)} \lesssim t e^{-\alpha t}. $$

*Proof.* We compute

$$
\begin{align*}
\frac{1}{2} \frac{d}{dt} \|u(t) - v(t)\|_2^2 &= -\lambda \operatorname{Re} \int \left(f(u) - \frac{v}{|v|}\right) (\bar{u} - \bar{v}) dx \\
&= -\lambda \operatorname{Re} \int \left(f(u) - f(v)\right) (\bar{u} - \bar{v}) dx - \lambda \operatorname{Re} \int \left(f(v) - \frac{v}{|v|}\right) (\bar{u} - \bar{v}) dx \\
&\leq -c\lambda \|u(t) - v(t)\|_2^2 + \lambda \|v(t)\|_2 \|u(t) - v(t)\|_2 \\
&\leq -c\lambda \|u(t) - v(t)\|_2^2 + \lambda \|v_0\|_2 \|u(t) - v(t)\|_2.
\end{align*}
$$

It follows that

$$ \frac{d}{dt} \|u(t) - v(t)\|_2 \leq -c\lambda \|u(t) - v(t)\|_2 + \lambda \|v_0\|_2. $$

Integrating from 0 to $t$ and using $u_0 = v_0$, we obtain

$$ \|u(t) - v(t)\|_2 \leq -c\lambda \int_0^t \|u(\tau) - v(\tau)\|_2 d\tau + \lambda \|v_0\|_2 t. $$

By Grönwall's lemma, it follows that

$$ \|u(t) - v(t)\|_2 \leq \lambda \|v_0\|_2 t e^{-c\lambda t}. $$

□

**Corollary 12.** Let $v_0 \in H^1(\Sigma)$ and denote by $v$ the solution of (E0). Then, for all $t \ge 0$,

$$ \|v(t)\|_2 \lesssim (t+1) e^{-\alpha t}, $$

for some $\alpha > 0$.

*Proof.* Using Proposition 11 with $u_0 = v_0$, we have

$$ \|v(t)\|_2 \le \|u(t)\|_2 + \lambda \|v_0\|_2 t e^{-c\lambda t} \le \|u_0\|_2 e^{-\lambda t} + \lambda \|v_0\|_2 t e^{-c\lambda t} \le \|v_0\|_2 (e^{-\lambda t} + t e^{-c\lambda t}) \lesssim (t+1) e^{-c\lambda t}. $$

□

## Dissipative Schrödinger Equation with Logarithmic Perturbation

We now consider the dissipative Schrödinger equation perturbed by a logarithmic term:

$$ (E) \qquad \begin{cases} iu_t + \Delta u + i\lambda u \zeta(|u| + 1) + \mu u \log(|u|) = 0, & (t, x) \in \mathbb{R}_+ \times \Sigma, \\ u(0, x) = u_0(x) \in H^1(\Sigma), \end{cases} $$

where $\lambda > 0$ and $\mu \in \mathbb{R}$.

We aim to establish the Cauchy theory for (E). As before, we consider a regularized problem to avoid singularities near $u = 0$:

$$ (E_\varepsilon) \qquad \begin{cases} iu_{\varepsilon t} + \Delta u_{\varepsilon} + i\lambda u_{\varepsilon} \zeta(|u_{\varepsilon}| + 1 + \varepsilon) + \mu u_{\varepsilon} \log(|u_{\varepsilon}| + \varepsilon) = 0, \\ u_{\varepsilon}(0, x) = u_0(x). \end{cases} $$
**Proposition 13** (Uniqueness). Let $u_1, u_2 \in L^\infty(0, T; H^1(\Sigma))$ be two solutions of $(E)$ (in the distributional sense). If $\lambda > 0$, then $u_1 = u_2$.

*Proof.* Set $w = u_1 - u_2$. Then

$$i\partial_t w + \Delta w + i\lambda(u_1\zeta(|u_1| + 1) - u_2\zeta(|u_2| + 1)) + \mu(u_1 \log(|u_1|) - u_2 \log(|u_2|)) = 0.$$

Taking the $L^2$ inner product with $w$ and the real part, we get

$$\frac{1}{2} \frac{d}{dt} \|w\|_{L^2(\Sigma)}^2 = -\lambda \operatorname{Re}\langle u_1 \zeta(|u_1| + 1) - u_2 \zeta(|u_2| + 1), w \rangle - \mu \operatorname{Im}\langle u_1 \log(|u_1|) - u_2 \log(|u_2|), w \rangle.$$

The first term is nonpositive by the monotonicity lemma for $x \mapsto x\zeta(x + 1)$. The second term can be estimated as

$$|\operatorname{Im}\langle u_1 \log(|u_1|) - u_2 \log(|u_2|), w \rangle| \le |w|,$$

so by Grönwall's lemma,

$$\|w(t)\|_{L^2(\Sigma)} \le e^{|\mu|t} \|w(0)\|_{L^2(\Sigma)}.$$

If $u_1(0) = u_2(0)$, it follows that $w \equiv 0$.

□

Multiplying $(E_\varepsilon)$ by $\overline{u_\varepsilon}$ and taking the real part gives

$$\frac{1}{2} \frac{d}{dt} \|u_\varepsilon\|_{L^2(\Sigma)}^2 = -\lambda \int_\Sigma |u_\varepsilon|^2 \zeta(|u_\varepsilon| + 1 + \varepsilon) \, dx,$$

so that, as before,

$$(4) \qquad \|u_\varepsilon(t)\|_{L^2(\Sigma)} \le \|u_0\|_{L^2(\Sigma)}, \quad \forall t \ge 0.$$

For the gradient, we compute

$$\frac{1}{2} \frac{d}{dt} \|\nabla u_\varepsilon\|_{L^2(\Sigma)}^2 = -\lambda \operatorname{Re} \int_\Sigma \nabla(u_\varepsilon \zeta(|u_\varepsilon| + 1 + \varepsilon)) \cdot \overline{\nabla u_\varepsilon} - \mu \operatorname{Im} \int_\Sigma \nabla(u_\varepsilon \log(|u_\varepsilon| + \varepsilon)) \cdot \overline{\nabla u_\varepsilon}.$$

The first term is handled as before and is nonpositive. The second term can be estimated by

$$|\operatorname{Im} \int_{\Sigma} \nabla (u_{\varepsilon} \log (|u_{\varepsilon}| + \varepsilon)) \cdot \overline{\nabla u_{\varepsilon}}| \le \int_{\Sigma} |\nabla u_{\varepsilon}|^2 dx.$$

Applying Grönwall's inequality, we obtain

$$(5) \qquad \|\nabla u_\varepsilon(t)\|_{L^2(\Sigma)} \le e^{|\mu|t} \|\nabla u_0\|_{L^2(\Sigma)}.$$

The estimates (4)-(5) give

$$\sup_{t \in [0, T]} \|u_\varepsilon(t)\|_{H^1(\Sigma)} \le C_T,$$

so that the family $(u_\varepsilon)_{0<\varepsilon<1}$ is uniformly bounded in $L^\infty(0, T; H^1(\Sigma))$. Repeating the compactness argument as in the NLS-Zeta case allows us to prove the following theorem:

**Theorem 14.** Given $u_0 \in H^1(\Sigma)$, there exists a unique solution $u$ satisfying

$$u \in L_{\text{loc}}^{\infty}(\mathbb{R}_+; H^1(\Sigma)) \cap W_{\text{loc}}^{1,\infty}(\mathbb{R}_+; H^{-1}(\Sigma)).$$

Moreover,

$$(6) \qquad iu_t + \Delta u + i\lambda u \zeta(|u| + 1) + \mu u \log(|u|) = 0 \quad \text{in } H^{-1}(\Sigma), \quad \forall t \in \mathbb{R}_+.$$
# Estimates for the Riemann Zeta Function

**Lemma 15.** For all $x > 1$, we have

$$ \frac{1}{x-1} \le \zeta(x) \le \frac{1}{x-1} + 1. $$

*Proof.* Recall that for any continuous decreasing function $f$, we have

$$ f(n) \le \int_{n-1}^{n} f(x) dx \le f(n-1). $$

Applying this to $f(x) = 1/x^s$ for $s > 1$, which is decreasing, gives the desired bounds. $\square$

**Lemma 16.** For all $x > 1$,

$$ |\zeta'(x)| \le \frac{1}{\ln(2)} \int_{1}^{\infty} \frac{\ln(z)}{z^x} dz. $$

In particular,

$$ |\zeta'(x)| \le \frac{1}{\ln(2)(x-1)^2}. $$

*Proof.* The same technique as in the previous lemma applies, except here $f(z) = \ln(z)/z^x$ is not decreasing on $(1, 2]$.

We write

$$ |\zeta'(x)| \le \frac{\ln 2}{2^x} + \int_2^\infty \frac{\ln(z)}{z^x} dz. $$

Moreover, for $z \in (1, 2)$, $\lim_{x \to \infty} \frac{2^x}{z^x} = \infty$, so there exists $B > 0$ such that for all $x > B$,

$$ \int_1^2 \frac{\ln(z)}{z^x} dz \ge \frac{\ln 2}{2^x}. $$

For $x \in (1, B]$, the inequality also holds. $\square$

**Remark 3.** For any $\epsilon > 0$ and $z \ge 1$,

$$ \ln(z) \le \frac{1}{\epsilon} z^{\epsilon}. $$

Hence, in particular,

$$ |\zeta'(x)| \le \frac{1}{\epsilon} \cdot \frac{1}{x - \epsilon - 1}, \quad \forall x > \epsilon + 1. $$

**Remark 4.** We also have

$$ |\zeta'(x)| \le \frac{1}{(x-1)^2} + 1. $$

**Lemma 17.** There exists a constant $C > 0$ such that for all $x, y > 0$,

$$ |\zeta(x+1) - \zeta(y+1)| \le C \frac{|x-y|}{xy}. $$
*Proof.* We have

$$|\zeta(x+1) - \zeta(y+1)| = \left|\int_x^y \zeta'(s+1) ds\right| \le C \left|\int_x^y \frac{1}{s^2} ds\right| \le C \frac{|x-y|}{xy}.$$

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**Lemma 18.** There exists a constant $C > 0$ such that for all $x, y > 0$ and all $\epsilon, \delta > 0$, we have

$$|x\zeta(x+1+\epsilon) - y\zeta(y+1+\delta)| \le C \frac{|x-y| + |\epsilon - \delta|}{\min(x, y)}.$$

*Proof.* Using the fundamental theorem of calculus along a linear path:

$$\begin{aligned} & |x\zeta(x+1+\epsilon) - y\zeta(y+1+\delta)| \\ &= \left|\int_0^1 \frac{d}{dt}((tx+(1-t)y)\zeta(t(x+\epsilon)+(1-t)(y+\delta)+1))dt\right| \\ &\le \int_0^1 |x-y|\zeta(\dots)dt + \int_0^1 (tx+(1-t)y)(|x-y|+(\epsilon-\delta))|\zeta(\dots)|dt \\ &\le \frac{|x-y|}{\min(x,y)} + C \frac{|x-y| + |\epsilon - \delta|}{\min(x,y)}. \end{aligned}$$

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**Lemma 19.** For all $\epsilon \ge 0$, the function

$$x \mapsto x\zeta(x+1+\epsilon)$$

is increasing on $\mathbb{R}_{>0}$.

*Proof.* Set $f(x) = x\zeta(x+1+\epsilon)$, then

$$f'(x) = \sum_{n \ge 1} \frac{1 - x \ln(n)}{n^{1+x+\epsilon}}.$$

For $t \in [n, n+1)$,

$$\frac{1 - x \ln(n)}{n^{1+x+\epsilon}} \ge \frac{1 - x \ln(t)}{t^{1+x+\epsilon}}.$$

Hence

$$f'(x) \ge \int_1^\infty \frac{1 - x \ln(t)}{t^{1+x+\epsilon}} dt = \frac{\epsilon}{(x+\epsilon)^2} \ge 0.$$

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## References

[1] Masayuki Hayashi and Tohru Ozawa, The Cauchy Problem for the Logarithmic Schrödinger Equation Revisited https://arxiv.org/pdf/2309.01695

[2] Thierry Cazenave, *Semilinear Schrödinger Equations*, Providence, RI: American Mathematical Society; New York, N.Y.: Courant Institute of Mathematical Sciences, New York University, 2003.
[3] Rémi Carles, Clément Gallo, Finite time extinction by nonlinear damping for the Schrodinger equation https://hal.science/hal-00496616v2/document

[4] L. Nirenberg, On elliptic partial differential equations, Ann. Scuola Norm. Sup. Pisa (3), 13 (1959), pp. 115-162