Variational Image Restoration and Segmentation with Rician Noise

Abstract

Restoring and segmenting images corrupted by Rician noise are now challenging issues in the field of medical image processing. Our previously proposed restoration model, which is based on the statistical property of Rician noise, was proven efficient only when the standard variation of Rician noise in the image is greater than a certain positive number. The present paper further theoretically proves that this certain positive number can be replaced by zero, i.e., the standard variation of Rician noise can be any positive value. This broadens its application range. In addition, the data-fidelity term in the proposed restoration model can be applied into the famous two-stage segmentation method for segmenting images corrupted by Rician noise. In the first stage, a new variant of modified Mumford–Shah model is established with whose data-fidelity term is designed to manipulate Rician noise in the image. The strict convexity holds for this optimization model and linearized primal-dual algorithm with theoretical convergence analysis can be implemented for achieving the global optimal solution. For the second stage, partition on the optimal smooth cartoon image is done simply by thresholding. Such two-stage segmentation method is apparently more suitable for image with Rician noise compared to other state-of-art algorithms. Numerical experiments are conducted on both synthetic and real images. The results suggest that the proposed method is more favorable for image segmentation task with Rician noise.

Appendix

1.1 Appendix 1: Proof of Lemma 1

Proof

Based on the recurrence relation (equation (10.29.1) in [36]) for the modified Bessel function $I_{\nu }$ as follows:

$$\begin{aligned} I_{\nu -1}(t)-I_{\nu +1}(t) = (2\nu /t)I_{\nu }(t), \end{aligned}$$

(26)

we derive $I_2(t)=I_0(t)-\frac{2}{t}I_1(t)$. Then h(t) (when $t\ne 0$) can be expressed as follows:

$$\begin{aligned} h(t)=2t^{\frac{3}{2}}\left\{ 1-\frac{I_1(t)}{I_0(t)}\left[ \frac{1}{t}+\frac{I_1(t)}{I_0(t)}\right] \right\} . \end{aligned}$$

(27)

Combing equation (10.34.3) (taking $m=0$) and equation (10.40.10) in [36],

$$\begin{aligned} I_{\nu }(t)=\frac{e^t}{(2\pi t)^{1/2}}\left\{ \sum _{s=0}^{n-1}(-)^s \frac{a_s(\nu )}{t^s}+\delta _n(\nu ,t)\right\} -i e^{-\nu \pi i}\frac{e^{-t}}{(2 \pi t)^{1/2}}\left\{ \sum _{s=0}^{n-1} \frac{a_s(\nu )}{t^s}+\gamma _n(\nu ,t)\right\} , \end{aligned}$$

where $|\gamma _n(\nu ,t)|$ is bounded by

$$\begin{aligned}&2\exp \left\{ \left| \left( \nu ^2-\frac{1}{4}\right) t^{-1}\right| \right\} |a_n(\nu )t^{-n}| \quad \left( |\mathrm {ph} t| \le \frac{1}{2}\pi \right) ,\\&2\chi (n)\exp \left\{ \frac{1}{2}\pi \left| \left( \nu ^2-\frac{1}{4}\right) t^{-1}\right| \right\} \left| a_n(\nu )t^{-n}\right| \quad \left( \frac{1}{2}\pi \le \left| \mathrm {ph} t\right| \le \pi \right) ,\\&4\chi (n)\exp \left\{ \pi \left| \left( \nu ^2-\frac{1}{4}\right) \left( \mathrm {Re} t\right) ^{-1}\right| \right\} \left| a_n(\nu )(\mathrm {Re} t)^{-n}\right| \quad \left( \pi \le |\mathrm {ph} t|<\frac{3}{2} \pi \right) ;\\ \end{aligned}$$

$|\delta _n(\nu ,t)|$ is subject to the same bounds, except that the applicable sectors are respectively changed to $-\frac{3}{2}\pi \le \mathrm {ph} t \le -\frac{1}{2}\pi ,$       $-\frac{1}{2}\pi \le \mathrm {ph} t \le 0,$       $0 \le \mathrm {ph} t <\frac{1}{2} \pi$; ph denotes phase;

$$\begin{aligned} \chi (n)=\pi ^{\frac{1}{2}}\frac{\Gamma (\frac{1}{2}l+1)}{\Gamma (\frac{1}{2}l+\frac{1}{2})}; \quad a_n(\nu )=\frac{(4\nu ^2-1)(4\nu ^2-3^2)\dots (4\nu ^2-(2n-1)^2)}{n!8^n}. \end{aligned}$$

Hence, when $\nu =0,1$,

$$\begin{aligned} I_1(t)= & {} \frac{e^t}{(2\pi t)^{1/2}}\left\{ 1-\frac{3}{8t}-\frac{15}{128t^2}+\delta _3(1,t)+i e^{-2t}[1+\gamma _1(1,t)]\right\} := \frac{e^t}{(2\pi t)^{1/2}}P_1(t),\\ I_0(t)= & {} \frac{e^t}{(2\pi t)^{1/2}}\left\{ 1+\frac{1}{8t}+\frac{9}{128t^2}+\delta _3(0,t)-i e^{-2t}[1+\gamma _1(0,t)]\right\} := \frac{e^t}{(2\pi t)^{1/2}}P_0(t), \end{aligned}$$

with $\chi (1)=\pi /2$, $\chi (3)=3\pi /4$,

$$\begin{aligned} |\delta _3(\nu ,t)|\le \frac{3\pi |a_3(\nu )|}{2t^3}\exp \left\{ |\nu ^2-\frac{1}{4}|\frac{\pi }{2t}\right\} \quad \mathtt {and} \quad |\gamma _1(\nu ,t)|\le \frac{2|a_1(\nu )|}{t}\exp \left\{ |\nu ^2-\frac{1}{4}|\frac{1}{t}\right\} . \end{aligned}$$

Therefore,

$$\begin{aligned} \frac{I_1(t)}{I_0(t)}= & {} \frac{1-\frac{3}{8t}-\frac{15}{128t^2}+\delta _3(1,t)+i e^{-2t}[1+\gamma _1(1,t)]}{1+\frac{1}{8t}+\frac{9}{128t^2}+\delta _3(0,t)-i e^{-2t}[1+\gamma _1(0,t)]}\\= & {} 1-\frac{1}{2t}-\frac{1}{8t^2}+\frac{1}{t^2}R, \end{aligned}$$

where

$$\begin{aligned} R:= & {} t^2\left( \frac{I_1(t)}{I_0(t)}-1+\frac{1}{2t}+\frac{1}{8t^2}\right) =t^2\left[ \frac{P_1(t)}{P_0(t)}-1+\frac{1}{2t}+\frac{1}{8t^2}\right] \\= & {} \frac{t^2 P_1(t)-(t^2-\frac{t}{2}-\frac{1}{8})P_0(t)}{P_0(t)}. \end{aligned}$$

When $t>4.3$,

$$\begin{aligned} |\delta _3(1,t)|<\frac{0.1478}{t^2}, \quad |\delta _3(0,t)|<\frac{0.0879}{t^2}, \quad \gamma _1(1,t)|>0.2077 \quad \mathrm {and} \quad |\gamma _1(0,t)|>0.0616. \end{aligned}$$

Hence, we have

$$\begin{aligned} |P_0(t)|= & {} \Bigg |1+\frac{1}{8t}+\frac{9}{128t^2}+\delta _3(0,t)-i e^{-2t}[1+\gamma _1(0,t)]\Bigg |\\> & {} 1+\frac{1}{8t}+\frac{9}{128t^2}-|\delta _3(0,t)|- e^{-2t}(1+|\gamma _1(0,t)|)\\> & {} 1 \quad \mathrm {for} \qquad t>4.3, \end{aligned}$$

and then

$$\begin{aligned} |R|< & {} \left| t^2 P_1(t)-\left( t^2-\frac{t}{2}-\frac{1}{8}\right) P_0(t)\right| \\< & {} t^2|\delta _3(1,t)|+t^2 \left( 1-\frac{1}{2t}-\frac{1}{8t^2}\right) |\delta _3(0,t)|+\frac{52t+9}{1024t^2}\\&+\,t^2 e^{-2t}\left\{ 1+|\gamma (1,t)|+\left( 1-\frac{1}{2t}-\frac{1}{8t^2}\right) (1+|\gamma (0,t)|)\right\} \\< & {} 0.2558 \quad \mathrm {for} \,\,t>4.3. \end{aligned}$$

So we obtain that when $t>4.3$,

$$\begin{aligned} h(t)= & {} 2t^{\frac{3}{2}}\left\{ 1-\frac{I_1(t)}{I_0(t)}\left[ \frac{1}{t}+\frac{I_1(t)}{I_0(t)}\right] \right\} \\= & {} t^{-\frac{1}{2}}\left( 1-4R-\frac{1-16R+64R^2}{32t^2}\right) ,\\< & {} t^{-\frac{1}{2}}\left( 1+4|R|+\frac{1+16|R|+64R^2}{32t^2}<2.0389t^{-\frac{1}{2}}<0.9832\right) . \end{aligned}$$

In addition, according to Lemma 3 in [12], we know

$$\begin{aligned} (I_0(t)+I_2(t))I_0(t)-2I_1^2(t)\ge 0, \end{aligned}$$

for any $t\ge 0$, i.e, $h(t)\ge 0$.

So when $t>4.3$, $0\le h(t)<1$.

Based on Lemma 2 in our previous work [12] indicating $0 \le h(t)<1$ on [0, 3902], we conclude that $0 \le h(t)<1$ on $[0, +\infty )$.$\square$

1.2 Appendix 2: Proof of Lemma 2

Proof

As $g(t)= -\log I_0(t)-2\sqrt{t}$, the gradient of g(t) is given by

$$\begin{aligned} \nabla g(t) = -\frac{I_1(t)}{I_0(t)}-\frac{1}{\sqrt{t}}. \end{aligned}$$

Then

$$\begin{aligned} |\nabla g(t)-\nabla g(t')| = \Bigg |\frac{I_1(t')}{I_0(t')}+\frac{1}{\sqrt{t'}}-\frac{I_1(t)}{I_0(t)}-\frac{1}{\sqrt{t}}\Bigg |. \end{aligned}$$

We have proved that g(t) is strictly convex on $[0,+\infty )$. That is, $\nabla g(t)$ is strictly increasing on $[0,+\infty )$. Then it’s sufficient to prove the following inequality

$$\begin{aligned} \frac{I_1(t')}{I_0(t')}+\frac{1}{\sqrt{t'}}-\frac{I_1(t)}{I_0(t)}-\frac{1}{\sqrt{t}}\le \frac{1}{2}(t-t') \quad \text {if} \quad t>t', t,t' \in [1,+\infty ). \end{aligned}$$

Denote $J(t): = \frac{I_1(t)}{I_0(t)}+\frac{1}{\sqrt{t}}+\frac{1}{2}t$. It is sufficient to prove J(t) is increasing on $[1,+\infty )$. Since

$$\begin{aligned} J'(t)= & {} \frac{1}{2}-\frac{1}{2}t^{-\frac{3}{2}}+\frac{(I_0(t)+I_2(t))I_0(t)-2I_1(t)^2}{2I_0(t)^2} \\= & {} \frac{1}{2}t^{-\frac{3}{2}}\left[ t^{\frac{3}{2}}-1+t^{\frac{3}{2}}\frac{(I_0(t)+I_2(t))I_0(t)-2I_1(t)^2}{I_0(t)^2}\right] \\\ge & {} \frac{1}{2}t^{-\frac{3}{2}}\left( t^{\frac{3}{2}}-1\right) \\\ge & {} 0, \end{aligned}$$

where the last second $\ge$ is based on Lemma 1 in which $t^{\frac{3}{2}}\frac{(I_0(t)+I_2(t))I_0(t)-2I_1(t)^2}{I_0(t)^2}\ge 0$, J(t) is increasing and we finish the proof.$\square$