Interpolating the stochastic heat and wave equations with time-independent noise: solvability and exact asymptotics

Abstract

In this article, we study a class of stochastic partial differential equations with fractional differential operators subject to some time-independent multiplicative Gaussian noise. We derive sharp conditions, under which a unique global $L^p(\Omega )$-solution exists for all $p\ge 2$. In this case, we derive exact moment asymptotics following the same strategy as that in a recent work by Balan et al. (Inst Henri Poincaré Probab Stat. To appear, 2021). In the case when there exists only a local solution, we determine the precise deterministic time, $T_2$, before which a unique $L^2(\Omega )$-solution exists, but after which the series corresponding to the $L^2(\Omega )$ moment of the solution blows up. By properly choosing the parameters, results in this paper interpolate the known results for both stochastic heat and wave equations.

Appendix

Proof of Lemma 3.4

In this proof, $\mu (\mathrm {d} x) = \varphi (x)\mathrm {d} x = C_{\alpha ,d}|x|^{-(d-\alpha )}\mathrm {d} x$. By the change of variables $x' = \left( \nu /2\right) ^{1/a} x$ and $y' = \left( \nu /2\right) ^{1/a} y$, we see that

$$\begin{aligned} \rho _{\nu ,a}&\left( |\cdot |^{-\alpha }\right) = \sup _{\left\| f \right\| _{L^2(\mathbb {R}^d)} =1} \int _{\mathbb {R}^d} \left[ \int _{\mathbb {R}^d} \frac{f(x+y)f(y)}{\sqrt{1+\frac{\nu }{2}|x+y|^a } \sqrt{1+\frac{\nu }{2}|y|^a} } \mathrm {d} y \right] ^2 \mu (\mathrm {d} x) \\&= \left( \frac{\nu }{2} \right) ^{-\alpha /a} \sup _{\left\| f \right\| _{L^2(\mathbb {R}^d)} =1} \int _{\mathbb {R}^d} \left[ \int _{\mathbb {R}^d} \frac{f\left( \left( \frac{\nu }{2}\right) ^{-1/a}(x+y)\right) f\left( \left( \frac{\nu }{2}\right) ^{-1/a}y\right) }{\sqrt{1+|x+y|^a } \sqrt{1+|y|^a} } \left( \frac{\nu }{2}\right) ^{-d/a} \mathrm {d} y \right] ^2\\&\quad \varphi (x)\mathrm {d} x. \end{aligned}$$

By setting $f^*(x) =\left( \nu /2\right) ^{-d/(2a)} f\left( \left( \nu /2\right) ^{-1/a} x \right)$, we see that

$$\begin{aligned} \rho _{\nu ,a} \left( |\cdot |^{-\alpha }\right)&= \left( \frac{\nu }{2} \right) ^{-\alpha /a} \sup _{\left\| f \right\| _{L^2(\mathbb {R}^d)} =1} \int _{\mathbb {R}^d} \left[ \int _{\mathbb {R}^d} \frac{f^*\left( x+y\right) f^*\left( y\right) }{\sqrt{1+|x+y|^a } \sqrt{1+|y|^a} } \mathrm {d} y \right] ^2 \varphi (x)\mathrm {d} x \\&= \left( \frac{\nu }{2} \right) ^{-\alpha /a} \sup _{\left\| f^* \right\| _2 =1} \int _{\mathbb {R}^d} \left[ \int _{\mathbb {R}^d} \frac{f^*\left( x+y\right) f^*\left( y\right) }{\sqrt{1+|x+y|^a } \sqrt{1+|y|^a} } \mathrm {d} y \right] ^2 \varphi (x)\mathrm {d} x \\&= \left( \frac{\nu }{2} \right) ^{-\alpha /a} \rho _{2,a}\left( |\cdot |^{-\alpha }\right) , \end{aligned}$$

where the second equality is due to the fact that $\int _{\mathbb {R}^d} f(x)^2 \mathrm {d} x = \int _{\mathbb {R}^d} f^*(x)^2 \mathrm {d} x$. Then an application of (3.16) proves (3.18).

Similarly, for (3.21), by change of variables $\xi _{\sigma (j)}' = \left( \nu /2\right) ^{1/a} \xi _{\sigma (j)}$, we see that

$$\begin{aligned}&\lim _{n \rightarrow \infty } \frac{1}{n} \log \left[ \frac{1}{(n!)^2} \int _{(\mathbb {R}^d)^n} \left( \sum _{\sigma \in \Sigma _n} \prod _{k=1}^n \frac{1}{1+ \frac{\nu }{2}|\sum _{j=k}^n \xi _{\sigma (j)}|^a}\right) ^2 \mu (\mathrm {d} \vec {\xi }\,)\right] \\&\quad = \lim _{n \rightarrow \infty } \frac{1}{n} \log \left[ \frac{1}{(n!)^2} \left( \frac{\nu }{2} \right) ^{-n\alpha /a} \int _{(\mathbb {R}^d)^n} \left[ \sum _{\sigma \in \Sigma _n} \prod _{k=1}^n \frac{1}{1+\left| \sum _{j=k}^n \xi _{\sigma (j)} \right| ^a}\right] ^2 \mu (\mathrm {d} \vec {\xi } )\right] \\&\quad = \log \left[ \left( \frac{\nu }{2} \right) ^{-\alpha /a} \right] + \lim _{n \rightarrow \infty } \frac{1}{n} \log \left[ \frac{1}{(n!)^2} \int _{(\mathbb {R}^d)^n} \left[ \sum _{\sigma \in \Sigma _n} \prod _{k=1}^n \frac{1}{1+\left| \sum _{j=k}^n \xi _{\sigma (j)} \right| ^a}\right] ^2 \mu (\mathrm {d} \vec {\xi } )\right] \\&\quad = \log \left( \left( \frac{\nu }{2} \right) ^{-\alpha /a} \right) + \log \left( \rho _{2,a} \left( |\cdot |^{-\alpha }\right) \right) = \log \left( \rho _{\nu , a} \left( |\cdot |^{-\alpha }\right) \right) , \end{aligned}$$

where we have applied (3.15) and (3.18). This completes the proof of Lemma 3.4. $\square$

Proof of (4.9)

Starting from (3.5), by the change of variables $t_i' = t_i/c$ and the scaling property in (3.7), we have that

$$\begin{aligned} \mathcal {F}f_n(\cdot ,0,ct)(\xi _1,\cdots , \xi _n)&= \int _{[0,ct]^n_<} \prod _{k=1}^n \overline{ \mathcal {F}G(t_{k+1} - t_k, \cdot )\left( \sum _{j=1}^k \xi _j \right) } \mathrm {d} \vec {t} \\&= \int _{[0,t]^n_<} \prod _{k=1}^n \overline{ \mathcal {F}G\left( c(t_{k+1} - t_k), \cdot \right) \left( \sum _{j=1}^k \xi _j \right) } c^n \mathrm {d} \vec {t} \\&= \int _{[0,t]^n_<} \prod _{k=1}^n \overline{ c^{b+r-1} \mathcal {F}G\left( t_{k+1} - t_k, \cdot \right) \left( c^{b/a} \sum _{j=1}^k \xi _j \right) } c^n \mathrm {d} \vec {t} \end{aligned}$$

where in the last line we applied (3.6). Now,

$$\begin{aligned} \mathcal {F}f_n(\cdot ,0,ct)(\xi _1,\cdots , \xi _n)&= \int _{[0,t]^n_<} \prod _{k=1}^n \overline{ c^{b+r-1} \mathcal {F}G\left( t_{k+1} - t_k, \cdot \right) \left( c^{b/a} \sum _{j=1}^k \xi _j \right) } c^n \mathrm {d} \vec {t} \\&= c^{n(b+r)} \mathcal {F}f_n\left( \cdot ,0,t\right) \left( c^{b/a}\xi _1, \cdots , c^{b/a}\xi _n\right) , \end{aligned}$$

from which we see that

$$\begin{aligned}&\int _0^\infty e^{-t} \left\| {\widetilde{f}}_n(\cdot ,0,t) \right\| _{\mathcal {H}^{\otimes n}}^2 \mathrm {d} t = \int _0^\infty e^{-t} \int _{\mathbb {R}^{nd}} \left| \mathcal {F}\widetilde{f_n}(\cdot ,0;t)(\xi _1,\cdots ,\xi _n) \right| ^2 \mu (\mathrm {d} \vec {\xi }\,) \mathrm {d} t \\&\quad = \int _0^\infty e^{-2t} \int _{\mathbb {R}^{nd}} \left| \mathcal {F}\widetilde{f_n}(\cdot ,0;2t)(\xi _1,\cdots ,\xi _n) \right| ^2 \mu (\mathrm {d} \vec {\xi }\,) \, 2\, \mathrm {d} t \\&\quad = 2^{2n(b+r)}\int _0^\infty 2e^{-2t} \int _{\mathbb {R}^{nd}} \left| \mathcal {F}\widetilde{f_n}(\cdot ,0,t)(2^{b/a}\xi _1,\cdots ,2^{b/a}\xi _n) \right| ^2 \mu (\mathrm {d} \vec {\xi }\,) \mathrm {d} t \end{aligned}$$

where in the last line we have applied (3.7). Now,

$$\begin{aligned}&\int _0^\infty e^{-t} \left\| {\widetilde{f}}_n(\cdot ,0,t) \right\| _{\mathcal {H}^{\otimes n}}^2 \mathrm {d} t = 2^{2n(b+r)}\int _0^\infty 2e^{-2t} \\&\quad \quad \int _{\mathbb {R}^{nd}} \left| \mathcal {F}\widetilde{f_n}(\cdot ,0,t)(2^{b/a}\xi _1,\cdots ,2^{b/a}\xi _n) \right| ^2 \mu (\mathrm {d} \vec {\xi }\,) \mathrm {d} t \\&\quad = 2^{2n(b+r)}\int _0^\infty 2e^{-2t} \int _{\mathbb {R}^{nd}} \left| \mathcal {F}\widetilde{f_n}(\cdot ,0;t)(\xi _1,\cdots ,\xi _n) \right| ^2 2^{\frac{-nbd}{a}} 2^{\frac{nb(d-\alpha )}{a}} \mu (\mathrm {d} \vec {\xi }\,) \mathrm {d} t \end{aligned}$$

where the last line follows from a change of variables and recalling that $\mu (\mathrm {d} \vec {\xi }\,)$ in (3.17). Lastly,

$$\begin{aligned}&\int _0^\infty e^{-t} \left\| {\widetilde{f}}_n(\cdot ,0,t) \right\| _{\mathcal {H}^{\otimes n}}^2 \mathrm {d} t = 2^{2n(b+r)}\int _0^\infty 2e^{-2t} \\&\quad \quad \int _{\mathbb {R}^{nd}} \left| \mathcal {F}\widetilde{f_n}(\cdot ,0;t)(\xi _1,\cdots ,\xi _n) \right| ^2 2^{\frac{-nbd}{a}} 2^{\frac{nb(d-\alpha )}{a}} \mu (\mathrm {d} \vec {\xi }\,) \mathrm {d} t \\&\quad = 2^{n(2(b+r) - b\alpha /a)} \int _0^\infty 2e^{-2t} \int _{\mathbb {R}^{nd}} \left| \mathcal {F}\widetilde{f_n}(\cdot ,0;t)(\xi _1,\cdots ,\xi _n) \right| ^2 \mu (\mathrm {d} \vec {\xi }\,) \mathrm {d} t \\&\quad = \frac{2^{n(2(b+r) - b\alpha /a)}}{(n!)^2} \int _0^\infty 2e^{-2t} \int _{\mathbb {R}^{nd}} H_n(t,\vec {x})^2 \mathrm {d} \vec {x} \mathrm {d} t, \end{aligned}$$

which proves (4.9). $\square$