Poisson Equation Solution and Its Gradient Vector Field to Geometric Features Detection

Abstract

In this paper we solve the Poisson partial differential equation (PDE) with a free right side, which is a function of the image and its gradient. We call such a PDE Poisson Image (PI) equation. Further, we define the function $\phi ={u+\left\| \nabla u\right\| }^2$, where u is the PI’s solution. Then, we generate the Poisson gradient vector fields (PGVFs) $\nabla u$ and $\nabla \phi$ and study the patterns of their trajectories in the vicinity of the singular points (SPs). Next, we use the critical points (CPs) of u and $\phi$, the SPs of $\nabla u$ and $\nabla \phi$, and their relations, to determine the image objects’ concavities and convexities, and use them for automatic objects partitioning. We validated the theoretical concepts with experiments on above 80 synthetic and real-life images, and show some of them in the paper. At the end we compare the new method with contemporary methods in the field and list its contributions, advantages and bottlenecks.

Appendix. Proof Theorem 1

Consider that $u_x=u_y=0$, and $\phi _x, \phi _y, \phi _{xx}, \phi _{yy}, \phi _{xy}$ and

$$\begin{aligned} \begin{aligned} D_{{\phi }}&={\phi }_{xx}{\phi }_{yy}-{\phi }_{xy}^2 =(u_{xx}+2{u^2_{xx}}+2{u^2_{xy}})(u_{yy}+2{u^2_{yy}}+2{u^2_{xy}})-\\&-(u_{xy}+2({u_{xx}}+{u_{yy}}){u_{xy}})^2. \end{aligned} \end{aligned}$$

(5)

Part I.1. Given that u(c, d) is a local maximum at some $(c,d) \in \varOmega$. Hence $u_{xx}<0$, $u_{yy}<0$, and $D_u(c,d)=u_{xx} u_{yy} -u_{xy}^2 >0$ at (c, d).

Case (a): Assume $u_{xx}-u_{yy}=0$. Follows that

$$\begin{aligned} u_{xx} (c,d)=u_{yy} (c,d) = {f(c,d)} /{2}<0; \end{aligned}$$

(6)

$$\begin{aligned} u_{xx} (c,d)u_{yy} (c,d) = {f^2(c,d)}/ {4}>u^2_{xy} (c,d). \end{aligned}$$

(7)

$$\begin{aligned} \begin{aligned} D_{{\phi }}(c,d)= {f^2} (f+1)^2 / {4} - u_{xy}^2(2f^2+2f+1)+4u_{xy}^4. \end{aligned} \end{aligned}$$

(8)

Solving $D_{{\phi }} (u_{xy}^2)=0$ we receive $u_{xy}^2= {f^2} /{4}$ or ${(f+1)^2}/ {4}$. Now, if:

• (i) $f \in [- {1} /{2}, 0)$, and $\phi _{xx}(c,d)<0$, $\phi _{yy}(c,d)<0$, and $D_{{\phi }}(c,d)>0$, follows that ${\phi }(c,d)$ is a local maximum.

• (ii) $f \in (-1, - {1} /{2})$, and if: 1). ${u}_{xy}^2 (c,d) \in ( {(f+1)^2}/ {4}, {f^2}/ {4})$, then $D_{{\phi }}(c,d)<0$. So (c, d) is a saddle point; 2). ${u}_{xy}^2 (c,d) \in (0, {(f+1)^2}/ {4})$, then $\phi _{xx}(c,d)<0$, $\phi _{yy}(c,d)<0$, and $D_{{\phi }}(c,d)>0$. Therefore, $\phi (c,d)$ is a local maximum.

• (iii) $f=-1$, that is, $u_{xx}=u_{yy}=- {1} /{2}$, $u_{xy}^2(c,d) \in (0, {1}/{4})$, follows that $D_{{\phi }}(c,d)=4u_{xy}^4-u_{xy}^2=u_{xy}^2(4u_{xy}^2-1)<0$. So (c, d) is a saddle point.

• (iv) $f \in (-\infty , -1)$, then $\phi _{xx}>0$, $\phi _{yy}>0$. And if: 1). ${u}_{xy}^2 \in ( {(f+1)^2} /{4}, {f^2}/ {4})$, then $D_{{\phi }}(c,d)<0$, so (c, d) is a saddle point. 2). ${u}_{xy}^2 \in (0, {(f+1)^2}/ {4})$, then $\phi _{xx}>0$, $\phi _{yy}>0$, and $D_{{\phi }}(c,d)>0$. Therefore, $\phi (c,d)$ is a local minimum.

Case (b): Assume now that $(a',b') \in \varOmega$ is a local maximum and $u_{xx}-u_{yy} \ne 0$. Rotate the coordinate system by $\theta$ to make the mixed derivative 0:

$$\begin{aligned} u_{x'x'} (a',b')<0, u_{y'y'} (a',b') <0, u_{x'y'} (a',b') =0; \end{aligned}$$

(9)

Then one can compute the following second derivatives and $D_{{\phi }}(a',b')$:

$$\begin{aligned} {\phi }_{x'y'} (a',b')=0; {\phi }_{x'x'} (a',b')=u_{x'x'} (a',b')(1+2u_{x'x'} (a',b')); \end{aligned}$$

(10)

$$\begin{aligned} {\phi }_{y'y'} (a',b')=u_{y'y'} (a',b')(1+2u_{y'y'} (a',b')); \end{aligned}$$

(11)

$$\begin{aligned} \begin{aligned} D_{{\phi }}(a',b')={\phi }_{x'x'} {\phi }_{y'y'} -{\phi }_{x'y'}^2 ={u}_{x'x'} {u}_{y'y'}(1+2f+(4f){u}_{x'x'}-4{u}_{x'x'}^2). \end{aligned} \end{aligned}$$

(12)

We denote $g({u}_{x'x'})=1+2f+(4f){u}_{x'x'}-4{u}_{x'x'}^2$, and solve $g({u}_{x'x'})=0$ to receive ${u}_{x'x'}=-\frac{1}{2}$ or $f+\frac{1}{2}$. Now, if:

• (i) $f \in [-{1}/ {2}, 0)$, then ${u}_{x'x'} \in (- {1} /{2}, 0)$ and ${u}_{y'y'} \in (- {1} /{2}, 0)$. One may find $\phi _{x'x'}(a',b')<0$, $D_{{\phi }}(a',b')>0$, so ${\phi }(a',b')$ is a local maximum.

• (ii) $f \in (-1, - {1}/ {2})$, then ${u}_{x'x'} \in (-1, 0)$ and ${u}_{y'y'} \in (-1, 0)$. One may find if: 1). ${u}_{x'x'} (a',b') \in (- {1}/ {2}, f+ {1}/ {2})$, then $\phi _{x'x'} (a',b')<0$, $D_{{\phi }}(a',b')>0$, so ${\phi }(a',b')$ is a local maximum; 2). ${u}_{x'x'} (a',b') \in (-1, - {1}/ {2}) \bigcup (f+{1}/{2}, 0)$, then $D_{{\phi }}(a',b')<0$, so $(a',b')$ is a saddle point.

• (iii) $f=-1$, one may found that $D_{{\phi }}(a',b')<0$ except ${u}_{x'x'}=- {1} /{2}$, which belongs to the third case in case (a). So $(a',b')$ is a saddle point.

• (iv) $f \in (-\infty , -1)$, one may find if: 1). ${u}_{x'x'} (a',b') \in (f+ {1}/ {2}, -{1}/ {2})$, then $\phi _{x'x'} (a',b')>0$, $D_{{\phi }}(a',b')>0$, so ${\phi }(a',b')$ is a local minimum; 2). ${u}_{x'x'} (a',b') \in (-\infty , f+{1}/{2}) \bigcup (- {1}/ {2}, 0)$, then $D_{{\phi }}(a',b')<0$, so $(a',b')$ is a saddle point.

Part I.2. Since u has a saddle point $(c,d) \in \varOmega$, then $D_u=u_{xx}u_{yy}-u_{xy}^2<0$.

Case (a): Assume $u_{xx}-u_{yy}=0$. Follows that

$$\begin{aligned} u_{xx} (c,d)u_{yy} (c,d) = {f^2(c,d)}/ {4}<u^2_{xy} (c,d). \end{aligned}$$

(13)

• (i) If $f \in (- {1}/ {2}, 0)$, and if: 1). ${u}_{xy}^2 (c,d) \in ( {f^2}/{4}, {(f+1)^2}/ {4})$, then $D_{{\phi }}(c,d)<0$, so (c, d) is a saddle point; 2). ${u}_{xy}^2 (c,d) \in ( {(f+1)^2}/ {4}, \infty )$, then $\phi _{xx}(c,d)>0$, $\phi _{yy}(c,d)>0$, and $D_{{\phi }}(c,d)>0$. Therefore, $\phi (c,d)$ is a local minimum.

• (ii) If $f \in (-1, - {1}/ {2}]$, then $\phi _{xx}(c,d)>0$, $\phi _{yy}(c,d)>0$, and $D_{{\phi }}(c,d)>0$, so ${\phi }(c,d)$ is a local minimum.

• (iii) If $f=-1$, that is, $u_{xx}=u_{yy}=-\frac{1}{2}$, $u_{xy}^2(c,d) > \frac{1}{4}$, then $\phi _{xx}=\phi _{yy}=2{u_{xy}}^2>0$. According to Eq. 8, $D_{{\phi }}(c,d)>0$. So $\phi (c,d)$ is a local minimum.

• (iv) If $f \in (-\infty , -1)$, then $\phi _{xx}(c,d)>0$, $\phi _{yy}(c,d)>0$, and $D_{{\phi }}(c,d)>0$. Therefore, (c, d) is a local minimum point.

Case (b): Assume that for $(a',b') \in \varOmega$ $u_{xx}-u_{yy} \ne 0$. Analogously to part I.1 Case (b), we rotate the coordinate system by $\theta$ and prove

• (i) If $f \in [- {1}/ {2},0)$, and if: 1). ${u}_{x'x'} \in (-\frac{1}{2}, f+\frac{1}{2})$, then $D_{{\phi }} (a',b')<0$. So $(a',b')$ is a saddle point; 2). ${u}_{x'x'} (a',b') \in (-\infty , -\frac{1}{2}) \bigcup (- f+\frac{1}{2}, \infty )$, then $\phi _{x'x'} (a',b')>0$, $\phi _{y'y'} (a',b')>0$, $D_{{\phi }} (a',b')>0$, so ${\phi } (a',b')$ is a local minimum.

• (ii) If $f \in (-\infty , -\frac{1}{2}) \phi _{x'x'}(a',b')>0, \phi _{y'y'}(a',b')>0$, then $D_{{\phi }}(a',b') >0$, so ${\phi } (a',b')$ is a local minimum.

Part II. u has saddle points. Hence $D_u=u_{xx}u_{yy}-u_{xy}^2<0$, that is, $u_{xy}^2>u_{xx}u_{yy}>0$. Analogously to I.1 Case (b) we rotate by $\theta$ and determine

$$\begin{aligned} D_{{\phi }}(a',b')={u}_{x'x'}^2 (a',b')(2{u}_{x'x'} (a',b')-1)(2{u}_{x'x'} (a',b')+1). \end{aligned}$$

(14)

• (i) If ${u}_{x'x'} \in (- {1}/ {2}, 0) \bigcup (0, {1}/{2})$, $D_{{\phi }}(a',b')<0$, so $(a',b')$ is a saddle point.

• (ii) If ${u}_{x'x'} (a',b') \in (-\infty , - {1}/ {2}) \bigcup ({1}/{2}, \infty )$, one may find that $\phi _{xx}(a',b')>0$, $\phi _{yy}(a',b')>0$, $D_{{\phi }}(a',b')>0$, so ${\phi }(a',b')$ is a local minimum.

Consider ${u}_{xx}=- {1}/ {2}$ or 1 / 2. Without rotation of the coordinate system, we calculate $\phi _{xx}>0$, $\phi _{yy}>0$, $D_{{\phi }} >0$. Therefore, ${\phi }$ attains a local minimum.

Hence all cases of the theorem are exhausted and proven.    $\diamond$