1 Introduction
Source condition | \(\Vert \xi (t)-x^\dagger \Vert ^2\) | \(\left\| L\xi (t)-y \right\| ^2\) |
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\(\Vert L^\dag \Vert <\infty \) | \({\mathcal {O}}(\mathrm e^{-\Vert L^\dag \Vert ^{-2}t})\) [27, Theorem 1] | \({\mathcal {O}}(\mathrm e^{-\Vert L^\dag \Vert ^{-2}t})\) |
\(x^\dagger \in {\mathcal {R}}((L^*L)^{\frac{\mu }{2}})\) | \({\mathcal {O}}(t^{-\mu -1})\) Corollary 6 | |
\(x^\dagger \in {\mathcal {N}}(L)^\perp \) |
Source condition | \(\Vert \xi (t)-x^\dagger \Vert ^2\) | \(\left\| L\xi (t)-y \right\| ^2\) |
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\(\Vert L^\dag \Vert <\infty \) | \({\mathcal {O}}(\mathrm e^{\varepsilon t-\beta (\Vert L^\dag \Vert )\frac{bt}{2}})\) [22, Theorem 9.(5)] | \({\mathcal {O}}(\mathrm e^{\varepsilon t-\beta (\Vert L^\dag \Vert )\frac{bt}{2}})\) |
\(x^\dagger \in {\mathcal {R}}((L^*L)^{\frac{\mu }{2}})\) | \({\mathcal {O}}(t^{-\mu -1})\) Corollary 7 | |
\(x^\dagger \in {\mathcal {N}}(L)^\perp \) |
Source condition | Parameters | \(\Vert \xi (t)-x^\dagger \Vert ^2\) | \(\left\| L\xi (t)-y \right\| ^2\) |
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\(\Vert L^\dag \Vert <\infty \) | \(b>3\) | \(o(t^{-2})\) | \(o(t^{-2})\) [4, Theorem 4.16] |
\({\mathcal {O}}(t^{-\frac{2b}{3}})\) [5, Theorem 3.4] | \({\mathcal {O}}(t^{-\frac{2b}{3}})\) [5, Theorem 3.4] | ||
\(\Vert L^\dag \Vert <\infty \) | \(b>2\) | \({\mathcal {O}}(t^{-2})\) | \({\mathcal {O}}(t^{-2})\) [29, Theorem 7] |
\({\mathcal {O}}(t^{-b})\) | \({\mathcal {O}}(t^{-b})\) [7, Theorem 4.2] | ||
\(\Vert L^\dag \Vert <\infty \) | \(0<b<3\) | \({\mathcal {O}}(t^{-\frac{2b}{3}})\) | \({\mathcal {O}}(t^{-\frac{2b}{3}})\) [4, Theorem 4.19] |
\(x^\dagger \in {\mathcal {R}}((L^*L)^{\frac{\mu }{2}})\) | \(0<\mu <\frac{b}{2}\) | \({\mathcal {O}}(t^{-2\mu })\) Corollary 9 | |
\(x^\dagger \in {\mathcal {R}}((L^*L)^{\frac{\mu }{2}})\) | \(0<\mu <\frac{b}{2}-1\) | \({\mathcal {O}}(t^{-2\mu -2})\) Corollary 9 | |
\(x^\dagger \in {\mathcal {N}}(L)^\perp \) | \(b\ge 3\) | \({\mathcal {O}}(t^{-2})\) [5, Theorem 2.7] | |
\(x^\dagger \in {\mathcal {N}}(L)^\perp \) | \(b>0\) |
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In the following sections, we apply the general results of Sect. 2 to regularising flow equations. In Sect. 4 we derive well-known convergence rates results of Showalter’s method and prove optimality of this method. In Sect. 5, we prove regularising properties, optimality, and convergence rates of the heavy ball dynamical flow. In the context of inverse problems, this method has already been analysed by [33], however not in terms of optimality, as it is done here.
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In Sect. 6, we consider the vanishing viscosity flow. We apply the general theory of Sect. 2 and prove optimality of this method. In particular, we prove under source conditions (see for instance [9, 13]) optimal convergence rates (in the sense of regularisation theory) for \(\Vert \xi (t)-x^\dagger \Vert ^2\). These rates (and the resulting ones for the squared norm of the residual) are seen to interpolate nicely between the known rates in the well-posed (finite-dimensional) and those in the ill-posed setting when varying the regularity of the solution \(x^\dagger \) (via changing the parameter \(\mu \) in Table 3).
2 Generalisations of Convergence Rates Results
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the spectral tail of the minimum norm solution \(x^\dagger \) with respect to the operator \(L^*L\), that is, the asymptotic behaviour of \(e(\lambda )\) as \(\lambda \) tends to zero, see [21];
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the error between the minimum norm solution \(x^\dagger \) and the regularised solution \(x_\alpha (y)\) or \(X_\alpha (y)\) for the exact data y called the noise-free regularisation error, that is,respectively, as \(\alpha \) tends to zero;$$\begin{aligned} d(\alpha ) :=\left\| x_\alpha (y)-x^\dagger \right\| ^2 \text { and } D(\alpha ):=\left\| X_\alpha (y)-x^\dagger \right\| ^2, \end{aligned}$$(13)
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the best worst-case error between the minimum norm solution \(x^\dagger \) and the regularised solution \(x_\alpha (\tilde{y})\) or \(X_\alpha (\tilde{y})\) for some data \(\tilde{y}\) with distance less than or equal to \(\delta >0\) to the exact data y under optimal choice of the regularisation parameter \(\alpha \), that is,respectively, as \(\delta \) tends to zero;$$\begin{aligned} \begin{aligned}&{\tilde{d}}(\delta ) :=\sup _{\tilde{y}\in {\bar{B}}_\delta (y)}\inf _{\alpha>0} \left\| x_\alpha ({\tilde{y}})-x^\dagger \right\| ^2 \text { and} \\&{\tilde{D}}(\delta ) :=\sup _{\tilde{y}\in {\bar{B}}_\delta (y)}\inf _{\alpha >0} \left\| X_\alpha ({\tilde{y}})-x^\dagger \right\| ^2, \end{aligned} \end{aligned}$$(14)
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the noise-free residual error, which is the error between the image of the regularised solution \(x_\alpha (y)\) or \(X_\alpha (y)\) and the exact data y, that is,respectively, as \(\alpha \) tends to zero.$$\begin{aligned} q(\alpha ) :=\Vert Lx_\alpha (y)-y\Vert ^2\text { and } Q(\alpha ) :=\Vert LX_\alpha (y)-y\Vert ^2, \end{aligned}$$(15)
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In Lemma 2 and Corollary 3, we show the relations between the convergence rates of the noise-free quantities e, d, and D. For this, we require the function \(\varphi \), which describes the rate of convergence and is the same for all three quantities, to be compatible with the regularisation method, see Definition 4.
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In Lemma 10 and Lemma 11, we derive the relations of the best worst-case errors \({\tilde{d}}\) and \({\tilde{D}}\) to the quantities e and D. The corresponding rate of convergence is hereby of the form \(\varPhi [\varphi ]\), where the mapping \(\varPhi \) is introduced in Definition 5 and some of its elementary properties are shown in Lemma 6, Lemma 7, Lemma 8, and Lemma 9.
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The statements for the residual errors q and Q are then concluded from Theorem 1 by using the identification of q and Q for the minimum norm solution \(x^\dag \) with the noise-free errors d and D for the minimum norm solution \({\bar{x}}^\dag =(L^*L)^{\frac{1}{2}}x^\dag \) of the problem \(L x={\bar{y}}\) with \({\bar{y}}=L{\bar{x}}^\dag \), and they are summarised in Corollary 3, Corollary 4, and Corollary 5.
Abbreviation | Description | References |
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\(r_\alpha \) | Generator | Definition 1 |
\(R_\alpha \) | Envelope generator | Equation 9 |
\(\tilde{r}_\alpha \) | Error function | Equation 7 |
\(\tilde{R}_\alpha \) | Envelope error function | Definition 1 item 3 |
\(x_\alpha ({\tilde{y}})=r_\alpha (L^*L)L^*{\tilde{y}}\) | Regularised solution according to \(r_\alpha \) | Equation 8 |
\(X_\alpha ({\tilde{y}})=R_\alpha (L^*L)L^*{\tilde{y}}\) | Regularised solution according to \(R_\alpha \) | Equation 10 |
\(d(\alpha )=\left\| x_\alpha (y)-x^\dagger \right\| ^2\) | Noise-free regularisation error for \(r_\alpha \) | Equation 13 |
\(D(\alpha )=\left\| X_\alpha (y)-x^\dagger \right\| ^2\) | Noise-free regularisation error for \(R_\alpha \) | Equation 13 |
\(\tilde{d}(\delta )\) | Best worst-case error for \(r_\alpha \) | Equation 14 |
\(\tilde{D}(\delta )\) | Best worst-case error for \(R_\alpha \) | Equation 14 |
\(q(\alpha )=\left\| Lx_\alpha (y)-y \right\| ^2\) | Noise-free residual error for \(r_\alpha \) | Equation 15 |
\(Q(\alpha )=\left\| LX_\alpha (y)-y \right\| ^2\) | Noise-free residual error for \(R_\alpha \) | Equation 15 |
\({\mathbf {E}}_A, {\mathbf {F}}_A\) | Spectral measures of \(L^*L, LL^*\) | Definition 3 |
\(e(\lambda )=\Vert {\mathbf {E}}_{[0,\lambda ]}x^\dagger \Vert ^2\) | Spectral tail of \(x^\dagger \) | Equation 12 |
\(\hat{\varphi }\) | \(\hat{\varphi }(\alpha ) = \sqrt{\alpha \varphi (\alpha )}\) | Definition 5 |
\({\hat{\varphi }}^{-1}\) | Generalised inverse of a function \({\hat{\varphi }}\) | Definition 5 |
\(\varPhi \) | Noise-free to noisy transform | Definition 5 |
2.1 Spectral Representations of the Regularisation Errors
2.2 Bounds for the Noise-Free Regularisation Errors
2.3 Relation Between Convergence Rates for Noise Free and for Noisy Data
2.4 Bounds for the Best Worst-Case Errors
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Assume that \(\delta \in (0,{\bar{\delta }})\) is such that \(\alpha _\delta :={{\hat{e}}}^{-1}(\delta )\in \varvec{\sigma }(L^*L)\setminus \left\{ 0\right\} \). From the continuity of \({\tilde{R}}_{\alpha _\delta }\) and Eq. 29, we find that there exists a parameter \(a_\delta \in (0,\alpha _\delta )\) such thatThen, the assumption \(\alpha _\delta \in \varvec{\sigma }(L^*L)\setminus \left\{ 0\right\} \) implies that the spectral projection \({\mathbf {F}}\) of the operator \(LL^*\) fulfils \({\mathbf {F}}_{[a_\delta ,2\alpha _\delta ]}\ne 0\). To estimate Eq. 28 further, we will choose for given values of \(\alpha >0\) and \(\delta \in (0,{\bar{\delta }})\) a particular point \({\tilde{y}}\). For this choice, we differ again between two cases.$$\begin{aligned} {\tilde{R}}_{\alpha _\delta }(a_\delta )<{\tilde{\sigma }}. \end{aligned}$$(30)Therefore, we end up with
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Ifwe pick$$\begin{aligned} z_{\alpha ,\delta } :={\mathbf {F}}_{[a_\delta ,2\alpha _\delta ]}(r_\alpha (LL^*)LL^*y-y) \ne 0, \end{aligned}$$in Eq. 28 and obtain$$\begin{aligned} \tilde{y}=y+\delta \frac{z_{\alpha ,\delta }}{\left\| z_{\alpha ,\delta } \right\| } \end{aligned}$$Here, we may drop the last term as it is non-negative, which gives us the lower bound$$\begin{aligned}&\left\| x_\alpha \left( y+\delta \tfrac{z_{\alpha ,\delta }}{\left\| z_{\alpha ,\delta } \right\| } \right) - x^\dagger \right\| ^2 = \left\| x_\alpha (y)-x^\dagger \right\| ^2 \\&\quad +\frac{\delta ^2}{\left\| z_{\alpha ,\delta } \right\| ^2}\left\langle z_{\alpha ,\delta },r_\alpha ^2(LL^*)LL^*z_{\alpha ,\delta }\right\rangle +\frac{2\delta }{\left\| z_{\alpha ,\delta } \right\| }\left\langle r_\alpha (LL^*)z_{\alpha ,\delta },z_{\alpha ,\delta }\right\rangle . \end{aligned}$$$$\begin{aligned} \left\| x_\alpha \left( y+\delta \tfrac{z_{\alpha ,\delta }}{\left\| z_{\alpha ,\delta } \right\| } \right) - x^\dagger \right\| ^2 \ge \left\| x_\alpha (y)-x^\dagger \right\| ^2+\delta ^2\min _{\lambda \in [a_\delta ,2\alpha _\delta ]}\lambda r_\alpha ^2(\lambda ). \end{aligned}$$
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Otherwise, ifwe choose \(z_{\alpha ,\delta }\in {\mathcal {R}}({\mathbf {F}}_{[a_\delta ,2\alpha _\delta ]})\setminus \left\{ 0\right\} \) arbitrarily. Then, with \(\tilde{y}=y+\delta \frac{z_{\alpha ,\delta }}{\left\| z_{\alpha ,\delta } \right\| }\), the last term in Eq. 28 vanishes and we find again$$\begin{aligned} {\mathbf {F}}_{[a_\delta ,2\alpha _\delta ]}(r_\alpha (LL^*)LL^*y-y) = 0, \end{aligned}$$$$\begin{aligned} \left\| x_\alpha \left( y+\delta \tfrac{z_{\alpha ,\delta }}{\left\| z_{\alpha ,\delta } \right\| } \right) - x^\dagger \right\| ^2 \ge \left\| x_\alpha (y)-x^\dagger \right\| ^2+\delta ^2\min _{\lambda \in [a_\delta ,2\alpha _\delta ]}\lambda r_\alpha ^2(\lambda ). \end{aligned}$$
Using Eq. 11 and that \(\tilde{R}_\alpha \) is by Definition 1 item 3 monotonically decreasing, we get the inequality$$\begin{aligned} {\tilde{d}}(\delta ) = \sup _{\tilde{y}\in {\bar{B}}_\delta (y)}\inf _{\alpha>0} \left\| x_\alpha ({\tilde{y}})-x^\dagger \right\| ^2 \ge \inf _{\alpha >0}\left( \left\| x_\alpha (y)-x^\dagger \right\| ^2+\delta ^2\min _{\lambda \in [a_\delta ,2\alpha _\delta ]}\lambda r_\alpha ^2(\lambda )\right) . \end{aligned}$$and since we already proved in Lemma 2 that \(d\ge (1-\sigma )^2e\), we can estimate further$$\begin{aligned} \lambda r_\alpha ^2(\lambda ) \ge \frac{1}{\lambda }\left( 1-\tilde{R}_\alpha (\lambda )\right) ^2 \ge \frac{1}{2\alpha _\delta }\left( 1-\tilde{R}_\alpha (a_\delta )\right) ^2\text { for all } \lambda \in [a_\delta ,2\alpha _\delta ], \end{aligned}$$Now, the first term is monotonically increasing in \(\alpha \) and, since \(\alpha \mapsto \tilde{R}_\alpha (\lambda )\) is for every \(\lambda >0\) monotonically increasing, see Definition 1 item 3, the second term is monotonically decreasing in \(\alpha \). Thus, we can estimate the expression for \(\alpha <\alpha _\delta \) from below by the second term at \(\alpha =\alpha _\delta \), and for \(\alpha \ge \alpha _\delta \) by the first term at \(\alpha =\alpha _\delta \):$$\begin{aligned} {\tilde{d}}(\delta ) \ge \inf _{\alpha >0}\left( (1-\sigma )^2e(\alpha )+\frac{\delta ^2}{2\alpha _\delta }\left( 1-\tilde{R}_\alpha (a_\delta )\right) ^2\right) . \end{aligned}$$Recalling that \(\alpha _\delta ={{\hat{e}}}^{-1}(\delta )\) and that the function e is right-continuous, we get from Lemma 6 that \(e(\alpha _\delta )\ge \varPhi [e](\delta )\) and have by Definition 5 that \(\frac{\delta ^2}{\alpha _\delta }=\varPhi [e](\delta )\). Thus, we obtain with Eq. 30 that$$\begin{aligned} {\tilde{d}}(\delta ) \ge \min \left\{ (1-\sigma )^2e(\alpha _\delta ), \frac{\delta ^2}{2\alpha _\delta } \left( 1-{\tilde{R}}_{\alpha _\delta }(a_\delta )\right) ^2\right\} . \end{aligned}$$$$\begin{aligned} {\tilde{d}}(\delta ) \ge c_0\varPhi [e](\delta )\text { with }c_0 :=\min \left\{ (1-\sigma )^2,\tfrac{1}{2}(1-{\tilde{\sigma }})^2\right\} . \end{aligned}$$(31) -
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It remains the case where \(\alpha _\delta :={{\hat{e}}}^{-1}(\delta )\notin \varvec{\sigma }(L^*L)\setminus \left\{ 0\right\} \). We defineSince e is right-continuous and monotonically increasing, the infimum is achieved and we have that \(e(\alpha _0)=e(\alpha _\delta )\). Moreover, \(\alpha _0\in \varvec{\sigma }(L^*L)\), since e is constant on every interval in \((0,\infty )\setminus \varvec{\sigma }(L^*L)\) and so \(\alpha _0\notin \varvec{\sigma }(L^*L)\) would imply that \(e(\lambda )=e(\alpha _\delta )\) for all \(\lambda \in (\alpha _0-\varepsilon ,\alpha _0+\varepsilon )\) for some \(\varepsilon >0\) which would contradict the minimality of \(\alpha _0\).$$\begin{aligned} \alpha _0:=\inf \{\alpha >0\mid e(\alpha )\ge e(\alpha _\delta )\} \in (0,\alpha _\delta ]. \end{aligned}$$Setting \(\delta _0:={{\hat{e}}}(\alpha _0)\) (so \({{\hat{e}}}^{-1}(\delta _0)=\alpha _0\) and, according to Lemma 6, \(e(\alpha _0)=\varPhi [e](\delta _0)\)), we have that \(\delta _0={{\hat{e}}}(\alpha _0)\le {{\hat{e}}}(\alpha _\delta )=\delta \) and we therefore find with the monotonicity of \({\tilde{d}}\), see Corollary 1, Eq. 31, and Lemma 6 that$$\begin{aligned} {\tilde{d}}(\delta ) \ge {\tilde{d}}(\delta _0)\ge c_0\varPhi [e](\delta _0) = c_0e(\alpha _0) = c_0e(\alpha _\delta ) \ge c_0\varPhi [e](\delta ). \end{aligned}$$
2.5 Optimal Convergence Rates
2.6 Optimal Convergence Rates for the Residual Error
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If \({\bar{e}}(\lambda )=0\) for all \(\lambda \in [0,\lambda _0]\) for some \(\lambda _0>0\), then we estimate, using the integral representation for Q from Eq. 36,Since \({\bar{\varphi }}\) is compatible to \((r_\alpha )_{\alpha >0}\), we known from Eq. 22 that$$\begin{aligned} Q(\alpha ) = \int _{\lambda _0}^{\left\| L \right\| ^2}\tilde{R}_\alpha ^2(\lambda )\,\mathrm d{\bar{e}}(\lambda ) \le \tilde{R}_\alpha ^2(\lambda _0)\Vert (L^*L)^{\frac{1}{2}}x^\dagger \Vert ^2. \end{aligned}$$$$\begin{aligned} \lim _{\alpha \rightarrow 0}\frac{Q(\alpha )}{\alpha }= \Vert Lx^\dagger \Vert ^2\lim _{\alpha \rightarrow 0}\frac{\tilde{R}_\alpha ^2(\lambda _0)}{\alpha }= 0. \end{aligned}$$
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If \({\bar{e}}(\lambda )>0\) for all \(\lambda >0\), then we first construct using the compatibility of \({\bar{\varphi }}\), as in the proof of Lemma 3, a monotonically decreasing and integrable function \({\tilde{F}}:[0,\infty )\rightarrow \mathbb {R}\) withNext, we pick a monotonically increasing function \(f:(0,\infty )\rightarrow (0,\left\| L \right\| ^2)\) with$$\begin{aligned} \tilde{R}_\alpha ^2(\lambda ) \le {\tilde{F}}(\tfrac{\lambda }{\alpha })\text { for all }\alpha >0\text { and }0<\lambda \le \left\| L \right\| ^2. \end{aligned}$$and split the integral in Eq. 36 for Q at the point \(f(\alpha )\) into two giving us$$\begin{aligned} \lim _{\alpha \rightarrow 0}f(\alpha ) = 0\text { and }\lim _{\alpha \rightarrow 0}\frac{{\bar{e}}(f(\alpha ))}{\alpha }= \infty \end{aligned}$$(43)We check that both terms decay faster than \(\alpha \).$$\begin{aligned} Q(\alpha ) = \int _0^{\left\| L \right\| ^2}\tilde{R}_\alpha ^2(\lambda )\,\mathrm d{\bar{e}}(\lambda ) \le \int _0^{f(\alpha )}{\tilde{F}}(\tfrac{\lambda }{\alpha })\,\mathrm d{\bar{e}}(\lambda )+\int _{f(\alpha )}^{\left\| L \right\| ^2}{\tilde{F}}(\tfrac{\lambda }{\alpha })\,\mathrm d{\bar{e}}(\lambda ). \end{aligned}$$(44)
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Since \({\bar{e}}\) fulfils by its definition in Eq. 35 thatwe find for every \(\varepsilon >0\) a value \(\alpha _0>0\) such that$$\begin{aligned} \lim _{\lambda \rightarrow 0}\frac{{\bar{e}}(\lambda )}{\lambda }= 0, \end{aligned}$$Therefore, we get for the first term in Eq. 44 with the substitution \(z=\frac{{\bar{e}}(\lambda )}{\varepsilon \alpha }\) that$$\begin{aligned} {\bar{e}}(\lambda ) \le \varepsilon \lambda \text { for all }0<\lambda <f(\alpha _0). \end{aligned}$$(45)And since this holds for arbitrary \(\varepsilon >0\), we see that$$\begin{aligned} \int _0^{f(\alpha )}{\tilde{F}}(\tfrac{\lambda }{\alpha })\,\mathrm d{\bar{e}}(\lambda ) \le \int _0^{f(\alpha )}{\tilde{F}}(\tfrac{{\bar{e}}(\lambda )}{\varepsilon \alpha })\,\mathrm d{\bar{e}}(\lambda )\le \varepsilon \alpha \Vert {\tilde{F}}\Vert _{L^1}\text { for all }\alpha <\alpha _0. \end{aligned}$$$$\begin{aligned} \lim _{\alpha \rightarrow 0}\frac{1}{\alpha }\int _0^{f(\alpha )}{\tilde{F}}(\tfrac{\lambda }{\alpha })\,\mathrm d{\bar{e}}(\lambda ) = 0. \end{aligned}$$
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For the second term in Eq. 44, we remark that Eq. 45 also implies that there exists a constant \(C>0\) withThus, we find with the substitution \(z=\frac{{\bar{e}}(\lambda )}{C\alpha }\) that$$\begin{aligned} {\bar{e}}(\lambda ) \le C\lambda \text { for all }\lambda >0. \end{aligned}$$for all \(\alpha >0\). According to our choice of f, see Eq. 43, the integral converges to zero for \(\alpha \rightarrow 0\) and we therefore obtain$$\begin{aligned} \int _{f(\alpha )}^{\left\| L \right\| ^2}{\tilde{F}}(\tfrac{\lambda }{\alpha })\,\mathrm d{\bar{e}}(\lambda ) \le \int _{f(\alpha )}^{\left\| L \right\| ^2}{\tilde{F}}(\tfrac{{\bar{e}}(\lambda )}{C\alpha })\,\mathrm d{\bar{e}}(\lambda )\le C\alpha \int _{\frac{{\bar{e}}(f(\alpha ))}{C\alpha }}^\infty {\tilde{F}}(z)\,\mathrm dz \end{aligned}$$$$\begin{aligned} \lim _{\alpha \rightarrow 0}\frac{1}{\alpha }\int _{f(\alpha )}^{\left\| L \right\| ^2}{\tilde{F}}(\tfrac{\lambda }{\alpha })\,\mathrm d{\bar{e}}(\lambda ) = 0. \end{aligned}$$
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3 Spectral Decomposition Analysis of Regularising Flows
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We start by showing that the function \(\rho \) defined by Eq. 48 can be extended to a function \(\rho :[0,\infty )\times [0,\infty )\rightarrow \mathbb {R}\) which is N times continuously differentiable with respect to t by settingFor this, we only have to check the continuity of all the derivatives at the points (t, 0), \(t\in [0,\infty )\). We observe that the solution of Eq. 47 for \(\lambda =0\) is given by$$\begin{aligned} \rho (t;0) :=-\partial _\lambda {\tilde{\rho }}(t;0). \end{aligned}$$(50)For the derivatives \(\partial _t^k\rho \), \(k\in \{0,\ldots ,N\}\), we therefore find with the mean value theorem (recall that \(\partial _\lambda \partial _t^k{\tilde{\rho }}=\partial _t^k\partial _\lambda {\tilde{\rho }}\) according to Schwarz’s theorem, see, e.g., [23, Theorem 9.1], since \(\partial _t^\ell {\tilde{\rho }}\in C^1([0,\infty )\times [0,\infty );\mathbb {R})\) for every \(\ell \in \{0,\ldots ,k\}\)) and Eq. 50 that$$\begin{aligned} {\tilde{\rho }}(t;0)=1\text { for every }t\in [0,\infty ). \end{aligned}$$which proves that \(\partial _t^k\rho \) is for every \(k\in \{0,\ldots ,N\}\) continuous in \([0,\infty )\times [0,\infty )\).$$\begin{aligned} \lim _{({\tilde{t}},{\tilde{\lambda }})\rightarrow (t,0)}&\left( \partial _t^k\rho ({\tilde{t}},{\tilde{\lambda }})-\partial _t^k\rho (t,0)\right) \\&= \lim _{({\tilde{t}},{\tilde{\lambda }})\rightarrow (t,0)} \left( \frac{\partial _t^k{\tilde{\rho }}({\tilde{t}},0)-\partial _t^k{\tilde{\rho }}({\tilde{t}},{\tilde{\lambda }})}{{\tilde{\lambda }}} +\partial _t^k\partial _\lambda {\tilde{\rho }}(t;0)\right) \\&= \lim _{({\tilde{t}},{\hat{\lambda }})\rightarrow (t,0)}\left( \partial _t^k\partial _\lambda {\tilde{\rho }}(t;0) -\partial _\lambda \partial _t^k{\tilde{\rho }}({\tilde{t}},{\hat{\lambda }})\right) = 0, \end{aligned}$$
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Next, we are going to show that the function \(\xi \) is N times continuously differentiable with respect to t and that its partial derivatives are for every \(k\in \{0,\ldots ,N\}\) given byTo see this, we assume by induction that Eq. 51 holds for \(k=\ell \) for some \(\ell \in \{0,\ldots ,N-1\}\). Then, we get with the Borel measure \(\mu _{L^*{\tilde{y}}}\) on \([0,\infty )\) defined by \(\mu _{L^*{\tilde{y}}}(A)=\Vert {\mathbf {E}}_AL^*{\tilde{y}}\Vert ^2\) that$$\begin{aligned} \partial _t^k\xi (t;{\tilde{y}}) = \int _{(0,\left\| L \right\| ^2]}\partial _t^k\rho (t;\lambda )\,\mathrm d{\mathbf {E}}_\lambda L^*{\tilde{y}}. \end{aligned}$$(51)Now, since \(\partial _t^{\ell +1}\rho \) is continuous, it is in particular bounded on every compact set \([0,T]\times [0,\left\| L \right\| ^2]\), \(T>0\). And since the measure \(\mu _{L^*{\tilde{y}}}\) is finite, Lebesgue’s dominated convergence theorem implies that$$\begin{aligned} \lim _{h\rightarrow 0}&\left\| \frac{\partial _t^\ell \xi (t+h;{\tilde{y}})-\partial _t^\ell \xi (t;{\tilde{y}})}{h}-\int _{(0,\left\| L \right\| ^2]}\partial _t^{\ell +1}\rho (t;\lambda )\,\mathrm d{\mathbf {E}}_\lambda L^*{\tilde{y}}\right\| ^2 \\&= \lim _{h\rightarrow 0}\left\| \int _{(0,\left\| L \right\| ^2]}\left( \frac{\partial _t^\ell \rho (t+h;\lambda )-\partial _t^\ell \rho (t;\lambda )}{h}-\partial _t^{\ell +1}\rho (t;\lambda )\right) \,\mathrm d{\mathbf {E}}_\lambda L^*{\tilde{y}}\right\| ^2 \\&= \lim _{h\rightarrow 0}\int _{(0,\left\| L \right\| ^2]}\left( \frac{\partial _t^\ell \rho (t+h;\lambda )-\partial _t^\ell \rho (t;\lambda )}{h}-\partial _t^{\ell +1}\rho (t;\lambda )\right) ^2\,\mathrm d\mu _{L^*{\tilde{y}}}(\lambda ). \end{aligned}$$which proves Eq. 51 for \(k=\ell +1\). Since Eq. 51 holds by definition of \(\xi \) for \(k=0\), this implies by induction that Eq. 51 holds for all \(k\in \{0,\ldots ,N\}\).$$\begin{aligned} \lim _{h\rightarrow 0}&\left\| \frac{\partial _t^\ell \xi (t+h;{\tilde{y}})-\partial _t^\ell \xi (t;{\tilde{y}})}{h}\!-\!\int _{(0,\left\| L \right\| ^2]}\partial _t^{\ell +1}\rho (t;\lambda )\,\mathrm d{\mathbf {E}}_\lambda L^*{\tilde{y}}\right\| ^2 \\&\!\!=\! \int _{(0,\left\| L \right\| ^2]}\lim _{h\rightarrow 0}\left( \frac{\partial _t^\ell \rho (t\!+\!h;\lambda )\!-\!\partial _t^\ell \rho (t;\lambda )}{h}\!-\!\partial _t^{\ell +1}\rho (t;\lambda )\right) ^2\,\mathrm d\mu _{L^*{\tilde{y}}}(\lambda )\!=\! 0, \end{aligned}$$Finally, the continuity of the Nth derivative \(\partial _t^N\xi \) follows in the same way directly from Lebesgue’s dominated convergence theorem:$$\begin{aligned} \lim _{{\tilde{t}}\rightarrow t}\left\| \partial _t^N\xi ({\tilde{t}};{\tilde{y}})\!-\!\partial _t^N\xi (t;{\tilde{y}})\right\| ^2 \!=\! \lim _{{\tilde{t}}\rightarrow t}\int _{(0,\left\| L \right\| ^2]}\left( \partial _t^N\rho ({\tilde{t}};\lambda )\!-\!\partial _t^N\rho (t;\lambda )\right) ^2\,\mathrm d\mu _{L^*{\tilde{y}}} \!=\! 0. \end{aligned}$$
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To prove that \(\xi \) solves Eq. 46, we plug the definition of \(\rho \) from Eq. 48 into Eq. 51 and findMaking use of Eq. 47, we get that \(\xi \) fulfils Eq. 46a:$$\begin{aligned} \partial _t^N\xi (t;{\tilde{y}})&+\sum _{k=1}^{N-1}a_k(t)\partial _t^k\xi (t;{\tilde{y}}) \\&= -\int _{(0,\left\| L \right\| ^2]}\frac{1}{\lambda }\left( \partial _t^N{\tilde{\rho }}(t;\lambda )+\sum _{k=1}^{N-1}a_k(t)\partial _t^k{\tilde{\rho }}(t;\lambda )\right) \,\mathrm d{\mathbf {E}}_\lambda L^*{\tilde{y}}. \end{aligned}$$(We remark that we have the relation \({\mathcal {R}}(L^*)\subset {\mathcal {N}}(L)^\perp ={\mathcal {N}}(L^*L)^\perp \) which implies the identity \({\mathbf {E}}_{(0,\Vert L\Vert ^2]}L^*{\tilde{y}}=L^*{\tilde{y}}\).)$$\begin{aligned} \partial _t^N\xi (t;{\tilde{y}})+\sum _{k=1}^{N-1}a_k(t)\partial _t^k\xi (t;{\tilde{y}})&= \int _{(0,\left\| L \right\| ^2]}{\tilde{\rho }}(t;\lambda )\,\mathrm d{\mathbf {E}}_\lambda L^*{\tilde{y}} \\&= \int _{(0,\left\| L \right\| ^2]}(-\lambda \rho (t;\lambda )+1)\,\mathrm d{\mathbf {E}}_\lambda L^*{\tilde{y}} \\&= -L^*L\xi (t;{\tilde{y}})+L^*{\tilde{y}}. \end{aligned}$$And for the initial conditions, we get, in agreement with Eq. 46b, from Eq. 51 that$$\begin{aligned} \partial _t^k\xi (0;{\tilde{y}})&= -\int _{(0,\left\| L \right\| ^2]}\partial _t^k{\tilde{\rho }}(0;\lambda )\,\mathrm d{\mathbf {E}}_\lambda L^*{\tilde{y}} = 0,\;k\in \left\{ 1,\ldots ,N-1\right\} ,\text { and} \\ \xi (0;{\tilde{y}})&= \int _{(0,\left\| L \right\| ^2]}\frac{1-{\tilde{\rho }}(0;\lambda )}{\lambda }\,\mathrm d{\mathbf {E}}_\lambda L^*{\tilde{y}} = 0. \end{aligned}$$
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So assume that we have two different solutions of Eq. 46 and call \(\xi _0\) the difference between the two solutions. We choose an arbitrary \(t_0>0\) and write \(\partial _t^k\xi _0(t_0;{\tilde{y}})=\xi ^{(k)}\) for every \(k\in \left\{ 0,\ldots ,N-1\right\} \). Then, \(\xi _0\) is a solution of the initial value problem$$\begin{aligned}&\partial _t^N\xi _0(t;{\tilde{y}}) + \sum _{k=1}^{N-1}a_k(t)\partial _t^k\xi _0(t;{\tilde{y}}) = - L^*L \xi _0(t;{\tilde{y}}) \text { for all } t \in \left( 0,\infty \right) \end{aligned}$$(52a)We know, for example, from [18, Section II.2.1], that Eq. 52 has a unique solution on every interval \([t_1,t_2]\), \(0<t_1<t_0<t_2\). Thus, we can write \(\xi _0\) in the form$$\begin{aligned}&\partial _t^k\xi _0(t_0;{\tilde{y}}) = \xi ^{(k)} \qquad \qquad \text { for all } k\in \left\{ 0,\ldots ,N-1\right\} . \end{aligned}$$(52b)with the functions \(\rho _\ell \) solving for every \(\lambda \in [0,\infty )\) the initial value problems$$\begin{aligned} \xi _0(t;{\tilde{y}}) = \sum _{\ell =0}^{N-1}\int _{[0,\infty )}\rho _\ell (t;\lambda )\,\mathrm d{\mathbf {E}}_\lambda \xi ^{(\ell )} \end{aligned}$$(Since \(a_k\) is continuous on \((0,\infty )\), Lebesgue’s dominated convergence theorem is applicable to every compact set \([t_1,t_2]\times [0,\left\| L \right\| ^2]\), \(0<t_1<t_0<t_2\).)$$\begin{aligned}&\partial _t^N\rho _\ell (t;\lambda )+\sum _{k=1}^{N-1}&a_k(t)\partial _t^k\rho _\ell (t;\lambda )=-\lambda \rho _\ell (t;\lambda )\text { for all }t \in \left( 0,\infty \right) ,&\\&\partial _t^k\rho _\ell (t_0;\lambda )=\delta _{k\ell }&\text { for all }k,\ell \in \left\{ 0,\ldots ,N-1\right\} .&\end{aligned}$$Now, we have for every measurable subset \(A\subset [0,\infty )\) and every \(k\in \{0,\ldots ,N-1\}\) thatwhere the signed measures \(\mu _{\eta _1,\eta _2}\), \(\eta _1,\eta _2\in {\mathcal {X}}\), are defined by \(\mu _{\eta _1,\eta _2}(A)=\left\langle \eta _1,{\mathbf {E}}_A\eta _2\right\rangle \).$$\begin{aligned} \Vert {\mathbf {E}}_A\partial _t^k\xi _0(t;{\tilde{y}})\Vert ^2 = \sum _{\ell ,m=0}^{N-1}\int _A\partial _t^k\rho _\ell (t;\lambda )\partial _t^k\rho _m(t;\lambda )\,\mathrm d\mu _{\xi ^{(\ell )},\xi ^{(m)}}(\lambda ), \end{aligned}$$The measures \(\mu _{\xi ^{(\ell )},\xi ^{(m)}}\) with \(\ell \ne m\) are absolutely continuous with respect to \(\mu _{\xi ^{(\ell )},\xi ^{(\ell )}}\) and with respect to \(\mu _{\xi ^{(m)},\xi ^{(m)}}\). Moreover, we can use Lebesgue’s decomposition theorem, see, e.g., [24, Theorem 6.10], to split the measures \(\mu _{\xi ^{(\ell )},\xi ^{(\ell )}}\), \(\ell \in \{0,\ldots ,N-1\}\), into measures \(\mu _j\), \(j\in \{0,\ldots ,J\}\), \(J\le N-1\), which are mutually singular to each other, so, explicitly, we writefor some measurable functions \(f_{j\ell m}\) with \(f_{j\ell m}=f_{j m\ell }\). Since then$$\begin{aligned} \mu _{\xi ^{(\ell )},\xi ^{(m)}} = \sum _{j=0}^J f_{j\ell m}\mu _j \end{aligned}$$has to hold for all functions \(g_\ell \in C([0,\infty );\mathbb {R})\), \(\ell \in \{0,\ldots ,N-1\}\), and all measurable sets \(A\subset [0,\infty )\), the matrices \(F_j(\lambda )=(f_{j\ell m}(\lambda ))_{\ell ,m=0}^{N-1}\) are (after possibly redefining \(f_{j\ell m}\) on sets \(A_{j\ell m}\) with \(\mu _j(A_{j\ell m})=0\)) positive semi-definite. Thus, we have for every measurable set \(A\subset [0,\infty )\) that$$\begin{aligned} 0\le \left\| \sum _{\ell =0}^{N-1}\int _Ag_\ell (\lambda )\,\mathrm d{\mathbf {E}}_\lambda \xi ^{(\ell )}\right\| ^2 = \sum _{j=0}^J\int _A\sum _{\ell ,m=0}^{N-1}f_{j\ell m}(\lambda )g_\ell (\lambda )g_m(\lambda )\,\mathrm d\mu _j(\lambda ) \end{aligned}$$where the integrand is a positive semi-definite quadratic form of \(\partial _t^k\rho \), namely \((\partial _t^k\rho )^{\mathrm T}F_j(\partial _t^k\rho )\), where \(\rho =(\rho _\ell )_{\ell =0}^{N-1}\). We can therefore find for every \(j\in \{0,\ldots ,J\}\) and every \(\lambda \) a change of coordinates \(O_j(\lambda )\in \mathrm {SO}_N(\mathbb {R})\) such that the matrix \(O_j^{\mathrm T}(\lambda )F_j(\lambda )O_j(\lambda )=\mathrm {diag}(d_{j\ell }(\lambda ))_{\ell =0}^{N-1}\) is diagonal with non-negative diagonal entries \(d_{j\ell }(\lambda )\). Setting \({\bar{\rho }}_{j\ell }(t;\lambda )=(O_j(\lambda )\rho (t;\lambda ))_\ell \) and \({\bar{\mu }}_{j\ell }=d_{j\ell }\mu _j\), we get$$\begin{aligned} \Vert {\mathbf {E}}_A\partial _t^k\xi _0(t;{\tilde{y}})\Vert ^2 = \sum _{j=0}^J\int _A\sum _{\ell ,m=0}^{N-1}f_{j\ell m}(\lambda )\partial _t^k\rho _\ell (t;\lambda )\partial _t^k\rho _m(t;\lambda )\,\mathrm d\mu _j(\lambda ), \end{aligned}$$Since \(\xi _0:[0,\infty )\rightarrow {\mathcal {X}}\) is N times continuously differentiable, it follows from Eq. 53 that$$\begin{aligned} \Vert {\mathbf {E}}_A\partial _t^k\xi _0(t;{\tilde{y}})\Vert ^2 = \sum _{j=0}^J\sum _{\ell =0}^{N-1}\int _A\left( \partial _t^k{\bar{\rho }}_{j\ell }(t;\lambda )\right) ^2\,\mathrm d{\bar{\mu }}_{j\ell }(\lambda ). \end{aligned}$$(53)and therefore, there exists a set \(\varLambda _{j\ell }\subset [0,\infty )\) with \({\bar{\mu }}_{j\ell }([0,\infty )\setminus \varLambda _{j\ell })=0\) such that$$\begin{aligned} \int _0^{t_0}\int _{[0,\infty )}\left( \partial _t^k{\bar{\rho }}_{j\ell }(t;\cdot )\right) ^2\,\mathrm d{\bar{\mu }}_{j\ell }(\lambda )\,\mathrm dt < \infty \text { for every }k\in \{0,\ldots ,N\}, \end{aligned}$$So, \({\bar{\rho }}_{j\ell }(\cdot ;\lambda )\) is for every \(\lambda \in \varLambda _{j\ell }\) in the Sobolev space \(H^N([0,t_0],{\bar{\mu }}_{j\ell })\). By the Sobolev embedding theorem, see, e.g., [2, Theorem 5.4], we thus have that \(\partial _t^k{\bar{\rho }}_{j\ell }(\cdot ;\lambda )\) extends for every \(\lambda \in \varLambda _{j\ell }\) and every \(k\in \{0,\ldots ,N-1\}\) continuously to a function on \([0,t_0]\).$$\begin{aligned} \int _0^{t_0}\left( \partial _t^k{\bar{\rho }}_{j\ell }(t;\lambda )\right) ^2\,\mathrm dt < \infty \text { for every }\lambda \in \varLambda _{j\ell }\text { and every }k\in \{0,\ldots ,N\}. \end{aligned}$$Since \(\xi _0\) is the difference of two solutions of Eq. 46, we have in particular thatThus, Eq. 53 implies that \(\partial _t^k{\bar{\rho }}_{j\ell }(t;\cdot )\rightarrow 0\) in \(L^2([0,\infty ),{\bar{\mu }}_{j\ell })\) with respect to the norm topology as \(t\rightarrow 0\). Because of the continuity of \(\partial _t^k{\bar{\rho }}_{j\ell }(\cdot ;\lambda )\), this means that there exists a set \({\tilde{\varLambda }}_{j\ell }\) with \({\bar{\mu }}_{j\ell }([0,\infty )\setminus {\tilde{\varLambda }}_{j\ell })=0\) such that we have for every \(k\in \{0,\ldots ,N-1\}\):$$\begin{aligned} \lim _{t\rightarrow 0}\Vert \partial _t^k\xi _0(t;{\tilde{y}})\Vert ^2 = 0\text { for every } k\in \{0,\ldots ,N-1\}. \end{aligned}$$But since Eq. 47 has a unique solution, this implies that \({\bar{\rho }}_{j\ell }(t;\lambda )=0\) for all \(t\in [0,\infty )\), \(\lambda \in {\tilde{\varLambda }}_{j\ell }\), and therefore, because of Eq. 53, that \(\xi _0(t;{\tilde{y}})=0\) for every \(t\in [0,\infty )\), which proves the uniqueness of the solution of Eq. 46.$$\begin{aligned} \lim _{t\rightarrow 0}\partial _t^k{\bar{\rho }}_{j\ell }(t;\lambda ) = 0\text { for every }\lambda \in {\tilde{\varLambda }}_{j\ell }. \end{aligned}$$
4 Showalter’s Method
5 Heavy Ball Dynamics
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We remark that the functionis non-negative and fulfils for arbitrary \(\tau >0\) that$$\begin{aligned} g_\tau :(0,\infty )\rightarrow \mathbb {R},\; g_\tau (\beta )=\cosh (\beta \tau )+\frac{\sinh (\beta \tau )}{\beta }, \end{aligned}$$since \(\tanh (z)\le z\) for all \(z\ge 0\). Thus, writing the function \({\tilde{\rho }}\) for \(\lambda \in (0,\frac{b^2}{4})\) with the function \(\beta _-\) given by Eq. 63 in the form$$\begin{aligned} g_\tau '(\beta )&= \tau \sinh (\beta \tau )+\frac{\tau \cosh (\beta \tau )}{\beta }-\frac{\sinh (\beta \tau )}{\beta ^2} \\&= \tau \sinh (\beta \tau )+\frac{\cosh (\beta \tau )}{\beta ^2}(\beta \tau -\tanh (\beta \tau )) \ge 0, \end{aligned}$$we find that$$\begin{aligned} \tilde{\rho }(t;\lambda )=\mathrm e^{-\frac{b t}{2}}g_{\frac{b t}{2}}(\beta _-(\lambda )), \end{aligned}$$since \(\beta _-'(\lambda )=-\frac{2}{b^2\beta _-(\lambda )}\le 0\). Therefore, the function \(\lambda \mapsto {\tilde{\rho }}(t;\lambda )\) is non-negative and monotonically decreasing on \((0,\frac{b^2}{4})\).$$\begin{aligned} \partial _\lambda \tilde{\rho }(t;\lambda ) = \mathrm e^{-\frac{b t}{2}}g_{\frac{b t}{2}}'(\beta _-(\lambda ))\beta _-'(\lambda ) \le 0, \end{aligned}$$
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Similarly, we consider for \(\lambda \in (\frac{b^2}{4},\infty )\) the functionfor arbitrary \(\tau >0\). Since \(\lim _{\beta \rightarrow 0}G_\tau (\beta )=1+\tau >0\) and since the smallest zero \(\beta _\tau \) of \(G_\tau \) is the smallest non-negative solution of the equation \(\tan (\beta \tau )=-\beta \), implying that \(\beta _\tau \tau \in (\frac{\pi }{2},\pi )\), we have that \(G_\tau (\beta )\ge 0\) for all \(\beta \in (0,\frac{\pi }{2\tau })\subset (0,\beta _\tau )\).$$\begin{aligned} G_\tau :(0,\infty )\rightarrow \mathbb {R},\; G_\tau (\beta )=\cos (\beta \tau )+\frac{\sin (\beta \tau )}{\beta }, \end{aligned}$$Moreover, the derivative of \(G_\tau \) satisfies for every \(\beta \in (0,\tfrac{\pi }{2\tau })\) thatsince \(\tan (z)\ge z\) for every \(z\ge 0\). Therefore, we find for the function \({\tilde{\rho }}\) on the domain \((0,\infty )\times (\frac{b^2}{4},\infty )\), where it has the form$$\begin{aligned} G_\tau '(\beta )&= -\tau \sin (\beta \tau )+\frac{\tau \cos (\beta \tau )}{\beta }-\frac{\sin (\beta \tau )}{\beta ^2} \\&= -\frac{\cos (\beta \tau )}{\beta ^2}\left( (\beta ^2\tau +1)\tan (\beta \tau )-\beta \tau \right) \le 0, \end{aligned}$$with \(\beta _+\) given by Eq. 63, that$$\begin{aligned} \tilde{\rho }(t;\lambda )=\mathrm e^{-\frac{b t}{2}}G_{\frac{b t}{2}}(\beta _+(\lambda )) \end{aligned}$$for \(\beta _+(\lambda )<\frac{\pi }{b t}\), that is, for \(\lambda <\frac{b^2}{4}+\frac{\pi ^2}{4t^2}\); since we have \(\beta _+'(\lambda )=\frac{2}{b^2\beta _+(\lambda )}\ge 0\).$$\begin{aligned} {\tilde{\rho }}(t;\lambda )\ge 0\text { and }\partial _\lambda \tilde{\rho }(t;\lambda ) = \mathrm e^{-\frac{b t}{2}}G_{\frac{b t}{2}}'(\beta _+(\lambda ))\beta _+'(\lambda ) \le 0 \end{aligned}$$
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For \(\lambda \in (0,\frac{b^2}{4})\), we use the two inequalities \(\cosh (z)\le \mathrm e^z\) and \(\frac{\sinh (z)}{z}\le \mathrm e^z\) for all \(z\ge 0\), where the latter follows from the fact that \(f(z)=2z\mathrm e^z(\mathrm e^z-\frac{\sinh (z)}{z})=(2z-1)\mathrm e^{2z}+1\) is because of \(f'(z)=4z\mathrm e^{2z}\ge 0\) monotonically increasing on \([0,\infty )\) and thus fulfils \(f(z)\ge f(0)=0\) for every \(z\ge 0\). With this, we find from Eq. 65 thatSince \(\sqrt{1-z}\le 1-\frac{z}{2}\) for all \(z\in (0,1)\), we then obtain$$\begin{aligned} {\tilde{P}}(t;\lambda ) \le 2\exp \left( \left( \sqrt{1-\frac{4\lambda }{b^2}}-1\right) \frac{b t}{2}\right) . \end{aligned}$$$$\begin{aligned} {\tilde{P}}(t;\lambda ) \le 2\mathrm e^{-\frac{\lambda t}{b}}\text { for every }t>0,\;\lambda \in (0,\tfrac{b^2}{4}). \end{aligned}$$
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For \(\lambda \in [\frac{b^2}{4},\varLambda ]\), we use that \(t\mapsto {\tilde{P}}(t;\lambda )\) is according to Lemma 19 for every \(\lambda \in (0,\infty )\) monotonically decreasing and obtain from Eq. 65 that$$\begin{aligned} {\tilde{P}}(t;\lambda ) \le {\tilde{P}}\left( \frac{\lambda t}{\varLambda };\lambda \right) = \mathrm e^{-\frac{b\lambda t}{2\varLambda }}\left( 1+\frac{b\lambda t}{2\varLambda }\right) \text { for every }t>0. \end{aligned}$$
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For \(\lambda \in (0,\frac{b^2}{4})\), we use that \(\cosh (z)=\mathrm e^z-\sinh (z)\) for every \(z\in \mathbb {R}\) and obtain with the function \(\beta _-\) from Eq. 63 thatSince \(\beta _-(\lambda )\in (0,1)\), we can therefore estimate this with the help of Lemma 14 by$$\begin{aligned} \frac{1}{\lambda }(1-{\tilde{\rho }}(t;\lambda )) = \frac{1}{\lambda }\left( 1-\mathrm e^{-(1-\beta _-(\lambda ))\frac{b t}{2}}-\left( \frac{1}{\beta _-(\lambda )}-1\right) \mathrm e^{-\frac{b t}{2}}\sinh (\beta _-(\lambda )\tfrac{b t}{2})\right) . \end{aligned}$$where \(\sigma _0\in (0,1)\) is the constant found in Lemma 14. Since \(\lambda =\frac{b^2}{4}(1-\beta _-^2(\lambda ))\), this means$$\begin{aligned} \frac{1}{\lambda }(1-{\tilde{\rho }}(t;\lambda )) \le \frac{1}{\lambda }\left( 1-\mathrm e^{-(1-\beta _-(\lambda ))\frac{b t}{2}}\right) \le \frac{\sigma _0}{\lambda }\sqrt{1-\beta _-(\lambda )}\sqrt{\frac{b t}{2}}, \end{aligned}$$$$\begin{aligned} \frac{1}{\lambda }(1-{\tilde{\rho }}(t;\lambda )) \le \frac{\sigma _0}{\sqrt{1+\beta _-(\lambda )}}\sqrt{\frac{2t}{b\lambda }} \le \sigma _0\sqrt{\frac{2t}{b\lambda }}. \end{aligned}$$
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For \(\lambda \in (\frac{b^2}{4},\infty )\), we remark thatwhere the function \(\beta _+\) is given by Eq. 63. Since the function \([0,\infty )\rightarrow \mathbb {R}\), \(z\mapsto (\mathrm e^{-z}\sin (a z))^2\), \(a>0\), attains its maximal value at its smallest non-negative critical point \(z=\frac{1}{a}\arctan (a)\), we have that$$\begin{aligned} \partial _t{\tilde{\rho }}(t;\lambda ) = -\frac{b}{2}\left( \beta _+(\lambda )+\frac{1}{\beta _+(\lambda )}\right) \mathrm e^{-\frac{b t}{2}}\sin \left( \beta _+(\lambda )\frac{b t}{2}\right) , \end{aligned}$$Using that \(\sin (z)=\frac{\tan (z)}{\sqrt{1+\tan ^2(z)}}\) for all \(z\in (-\frac{\pi }{2},\frac{\pi }{2})\), this reads$$\begin{aligned} \left| \partial _t{\tilde{\rho }}(t;\lambda )\right| \le \frac{b}{2}\left( \beta _+(\lambda )+\frac{1}{\beta _+(\lambda )}\right) \mathrm e^{-\frac{\arctan (\beta _+(\lambda ))}{\beta _+(\lambda )}}\left| \sin (\arctan (\beta _+(\lambda )))\right| . \end{aligned}$$We further realise that the function \(f:(0,\infty )\rightarrow \mathbb {R}\), \(f(z)=\frac{1}{\sqrt{1+z^2}}\mathrm e^{-\frac{\arctan (z)}{z}}\), is monotonically decreasing because of$$\begin{aligned} \left| \partial _t{\tilde{\rho }}(t;\lambda )\right| \le \frac{b}{2}\sqrt{1+\beta _+^2(\lambda )}\,\mathrm e^{-\frac{\arctan (\beta _+(\lambda ))}{\beta _+(\lambda )}}. \end{aligned}$$(67)Thus, \(f(z)\le \lim _{z\rightarrow 0}f(z)=\mathrm e^{-1}\) and Eq. 67 therefore implies that$$\begin{aligned} f'(z)&= -\frac{1}{\sqrt{1+z^2}}\mathrm e^{-\frac{\arctan (z)}{z}}\left( \frac{z}{1+z^2}+\frac{1}{z(1+z^2)}-\frac{\arctan (z)}{z^2}\right) \\&= -\frac{1}{z^2\sqrt{1+z^2}}\mathrm e^{-\frac{\arctan (z)}{z}}(z-\arctan (z)) \le 0. \end{aligned}$$With \(\frac{4}{b^2}\lambda =(1+\beta _+^2(\lambda ))\), the mean value theorem therefore gives us$$\begin{aligned} \left| \partial _t{\tilde{\rho }}(t;\lambda )\right| \le \frac{b}{2\mathrm e}(1+\beta _+^2(\lambda )). \end{aligned}$$Since we know from Lemmas 18 and 19 that we can estimate \({\tilde{\rho }}\) with the function \({\tilde{P}}\) from Eq. 65 by$$\begin{aligned} \frac{1}{\lambda }(1-{\tilde{\rho }}(t;\lambda )) = \frac{1}{\lambda }({\tilde{\rho }}(0;\lambda )-{\tilde{\rho }}(t;\lambda )) \le \frac{2t}{\mathrm eb}\text { for all }t>0. \end{aligned}$$we find by using the estimate \(\min \{a,b\}\le \min \{\sqrt{a},\sqrt{b}\}\max \{\sqrt{a},\sqrt{b}\}=\sqrt{ab}\) for all \(a,b>0\) that$$\begin{aligned} |{\tilde{\rho }}(t;\lambda )|\le {\tilde{P}}(t;\lambda )\le {\tilde{P}}(0;\lambda )=1, \end{aligned}$$(68)$$\begin{aligned} \frac{1}{\lambda }(1-{\tilde{\rho }}(t;\lambda ))\le \min \left\{ \frac{2}{\lambda },\frac{2t}{\mathrm eb}\right\} \le \sqrt{\frac{2}{\mathrm e}}\sqrt{\frac{2t}{b\lambda }} \text { for all }t>0. \end{aligned}$$
6 The Vanishing Viscosity Flow
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In particular, the relations in Eq. 76 imply that, for the last terms in Eq. 75, we have with \(\tau =t\sqrt{\lambda }\) asymptotically for \(\tau \rightarrow 0\)Thus, the last terms in Eq. 75 diverge for every \(\kappa \ge 0\) as \(t\rightarrow 0\).
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for \(\kappa =0\):because of the third relation in Eq. 76;$$\begin{aligned} C_{2,0}Y_0(\tau ) = \frac{2}{\pi }C_{2,0}{\mathcal {O}}(\log (\tau )) \end{aligned}$$
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for \(\kappa \in \mathbb {N}\):because of the second relation in Eq. 76; and$$\begin{aligned} C_{2,\kappa }\tau ^{-\kappa } Y_\kappa (\tau )= C_{2,\kappa }\tau ^{-2\kappa } (\tau ^\kappa Y_\kappa (\tau )) = C_{2,\kappa }\left( -\frac{2^\kappa (\kappa -1)!}{\pi }+o(1)\right) \tau ^{-2\kappa } \end{aligned}$$
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for \(\kappa \in (0,\infty )\setminus \mathbb {N}\):because of the first relation in Eq. 76.$$\begin{aligned} C_{2,\kappa }\tau ^{-\kappa }J_{-\kappa }(\tau )=C_{2,\kappa }\tau ^{-2\kappa } (\tau ^\kappa J_{-\kappa }(\tau ))=C_{2,\kappa }\left( \frac{2^\kappa }{\varGamma (1-\kappa )}+{\mathcal {O}}(\tau ^2)\right) \tau ^{-2\kappa } \end{aligned}$$
Since the first terms in Eq. 75 converge according to the first relation in Eq. 76 for \(t\rightarrow 0\), the initial condition \({\tilde{\rho }}(0;\lambda )=1\) can only be fulfilled if the coefficients \(C_{2,\kappa }\), \(\kappa \ge 0\), in front of the singular terms are all zero so that we haveFurthermore, the initial condition \({\tilde{\rho }}(0;\lambda )=1\) implies according to the first relation in Eq. 76 that$$\begin{aligned} {\tilde{\rho }}(t;\lambda ) = C_{1,\kappa }(t\sqrt{\lambda })^{-\kappa }J_\kappa (t\sqrt{\lambda })\text { for all }\kappa \ge 0. \end{aligned}$$which gives the representation of Eq. 72 for the solution \({\tilde{\rho }}\).$$\begin{aligned} C_{1,\kappa } = 2^\kappa \varGamma (\kappa +1)\text { for all }\kappa \ge 0, \end{aligned}$$It remains to check that also the initial condition \(\partial _t{\tilde{\rho }}(0;\lambda )=0\) is for all \(\kappa \ge 0\) fulfilled, which again follows directly from the first relation in Eq. 76:$$\begin{aligned} \partial _t{\tilde{\rho }}(0;\lambda ) = \lim _{t\rightarrow 0}\frac{1}{t}\left( 2^\kappa \varGamma (\kappa +1)(t\sqrt{\lambda })^{-\kappa }J_\kappa (t\sqrt{\lambda })-1\right) = 0\text { for all }\kappa \ge 0. \end{aligned}$$ -
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For \(\kappa \in (-\frac{1}{2},0)\), we have that the first term in \({\tilde{\rho }}(t;\lambda )\) converges for \(t\rightarrow 0\) to 0 because ofwhich follows from the first relation of Eq. 76. Therefore, the initial condition \({\tilde{\rho }}(0;\lambda )=1\) requires that$$\begin{aligned} C_{1,\kappa }(t\sqrt{\lambda })^{|\kappa |}J_{|\kappa |}(t\sqrt{\lambda }) = C_{1,\kappa }(t\sqrt{\lambda })^{2|\kappa |}\left( \frac{1}{2^{|\kappa |}\varGamma (|\kappa |+1)}+{\mathcal {O}}(t^2)\right) , \end{aligned}$$from which we get with the first property in Eq. 76 that$$\begin{aligned} 1 = {\tilde{\rho }}(0;\lambda ) = C_{2,\kappa }\lim _{t\rightarrow 0}(t\sqrt{\lambda })^{|\kappa |}J_{-|\kappa |}(t\sqrt{\lambda })\text { for all }\kappa \in (-\tfrac{1}{2},0), \end{aligned}$$To determine the coefficient \(C_{1,\kappa }\), we remark that the first identity in Eq. 76 then gives us for \(t\rightarrow 0\) the asymptotic behaviour$$\begin{aligned} C_{2,\kappa } = 2^\kappa \varGamma (\kappa +1). \end{aligned}$$Therefore, we have for the first derivative at \(t=0\) the expression$$\begin{aligned} {\tilde{\rho }}(t;\lambda ) = 1+C_{1,\kappa }\frac{(t\sqrt{\lambda })^{2|\kappa |}}{2^{|\kappa |}\varGamma (|\kappa |+1)}+{\mathcal {O}}(t^2). \end{aligned}$$To satisfy the initial condition \(\partial _t{\tilde{\rho }}(0;\lambda )=0\), we thus have to choose \(C_{1,\kappa }=0\) for \(\kappa \in (-\frac{1}{2},0)\), which leaves us again with Eq. 72.$$\begin{aligned} \partial _t{\tilde{\rho }}(0;\lambda ) = \lim _{t\rightarrow 0}C_{1,\kappa }\frac{2|\kappa |\lambda ^{|\kappa |}}{2^{|\kappa |}\varGamma (|\kappa |+1)}\,\frac{1}{t^{1-2|\kappa |}}. \end{aligned}$$
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for every \(\lambda >0\) that the function \(t\mapsto {\tilde{\rho }}(t;\lambda )\) is strictly decreasing on the interval \((0,\frac{1}{\sqrt{\lambda }}j_{\frac{1}{2}(b-1),1})\) and
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for every \(t>0\) that the function \(\lambda \mapsto {\tilde{\rho }}(t;\lambda )\) is strictly decreasing on the interval \((0,\frac{1}{t^2}j_{\frac{1}{2}(b-1),1}^2)\).
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\(|{\tilde{\rho }}(t;\lambda )| \le U(t\sqrt{\lambda })\) for every \(t\ge 0\), \(\lambda >0\),
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\(U(\tau )<1\) for all \(\tau >0\), and
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\(U(\tau ) \le C\tau ^{-\frac{b}{2}}\) for all \(\tau >0\).