Time Discrete Extrapolation in a Riemannian Space of Images

Abstract

The Riemannian metamorphosis model introduced and analyzed in [7, 12] is taken into account to develop an image extrapolation tool in the space of images. To this end, the variational time discretization for the geodesic interpolation proposed in [2] is picked up to define a discrete exponential map. For a given weakly differentiable initial image and a sufficiently small initial image variation it is shown how to compute a discrete geodesic extrapolation path in the space of images. The resulting discrete paths are indeed local minimizers of the corresponding discrete path energy. A spatial Galerkin discretization with cubic splines on coarse meshes for image deformations and piecewise bilinear finite elements on fine meshes for image intensity functions is used to derive a fully practical algorithm. The method is applied to real images and image variations recorded with a digital camera.

A Appendix

Here, we derive Eq. (9) from the system of Euler-Lagrange equations (6), (7) and (8). By using the transformation formula the energy $\mathcal {W}^D$ can be rewritten as follows

$$\begin{aligned} \mathcal {W}^D[{u}_1,{u}_2,\phi _2]=\int _\varOmega |D\phi _2-{\mathbbm {1}}|^2+\gamma |\varDelta ^m\phi _2|^2+\frac{1}{\delta }\frac{({u}_2-{u}_1\circ \phi _2^{-1})^2}{\det (D\phi _2)\circ \phi _2^{-1}}{\,\mathrm {d}}x. \end{aligned}$$

Now we rewrite the Euler-Lagrange equation (8). To this end, we use that

$$\begin{aligned} \partial _{\phi _2}\phi _2^{-1}(\psi )=-((D\phi _2)^{-1}\psi )\circ \phi _2^{-1}, \end{aligned}$$

which follows by differentiating $(\phi _2+\epsilon \psi )\circ (\phi _2+\epsilon \psi )^{-1}={\mathbbm {1}}$ with respect to $\epsilon$, and that $\partial _A\det (A)(B)={{\mathrm{cof}}}(A):B$ for $A\in GL(n)$ and $B\in \mathbb {R}^{n,n}$ with ${{\mathrm{cof}}}A = (\det A)A^{-T}$. Thus, we obtain

$$\begin{aligned}&\int _\varOmega 2D\phi _2:D\psi +2\gamma \varDelta ^m\phi _2\cdot \varDelta ^m\psi +\frac{2}{\delta }({u}_2-{u}_1\circ \phi _2^{-1})\frac{(\nabla {u}_1\cdot (D\phi _2)^{-1}\psi )\circ \phi ^{-1}_2}{\det (D\phi _2)\circ \phi ^{-1}_2} \\&\qquad +\frac{({u}_2-{u}_1\circ \phi _2^{-1})^2}{\delta (\det D\phi _2)^2\circ \phi _2^{-1}}\left( {{\mathrm{cof}}}D\phi _2 \!:\! (D^2\phi _2(D\phi _2)^{-1}\psi )-{{\mathrm{cof}}}D\phi _2\!:\!D\psi \right) \circ \phi _2^{-1}{\,\mathrm {d}}x =0. \end{aligned}$$

A further application of the transformation formula with respect to $\phi _2$ yields

$$\begin{aligned}&\int _\varOmega 2D\phi _2:D\psi +2\gamma \varDelta ^m\phi _2\cdot \varDelta ^m \psi + \frac{2}{\delta }({u}_2\circ \phi _2-{u}_1)\nabla {u}_1\cdot (D\phi _2)^{-1}\psi \nonumber \\&\qquad +\frac{1}{\delta }\frac{({u}_2\circ \phi _2-{u}_1)^2}{\det D\phi _2} \left( {{\mathrm{cof}}}D\phi _2 :(D^2\phi _2(D\phi _2)^{-1}\psi )-{{\mathrm{cof}}}D\phi _2 : D\psi \right) {\,\mathrm {d}}x=0. \end{aligned}$$

(13)

To remove the dependency of the function ${u}_2$ above, we employ the pointwise condition

$$\begin{aligned} {u}_2\circ \phi _2-{u}_1=\frac{{u}_1-{u}_0\circ \phi _1^{-1}}{\det (D\phi _1)\circ \phi _1^{-1}} \end{aligned}$$

for a.e. $x\in \varOmega$, which directly follows from (6). Inserting this in (13) and using the integral transformation formula we achieve

$$\begin{aligned}&\int _\varOmega 2D\phi _2:D\psi +2\gamma \varDelta ^m\phi _2\cdot \varDelta ^m\psi +\frac{2}{\delta }({u}_1\circ \phi _1-{u}_0)(\nabla {u}_1\cdot (D\phi _2)^{-1}\psi )\circ \phi _1\\&\quad +\frac{1}{\delta }\frac{({u}_1\circ \phi _1-{u}_0)^2}{\det D\phi _1}\left( \frac{{{\mathrm{cof}}}D\phi _2:(D^2\phi _2(D\phi _2)^{-1}\psi )-{{\mathrm{cof}}}D\phi _2 : D\psi }{\det D\phi _2}\right) \circ \phi _1{\,\mathrm {d}}x=0. \end{aligned}$$

Here, we take into account the identity ${{\mathrm{cof}}}(A)=\det (A)A^{-T}$. Next, we consider the test function $\zeta :=((D\phi _2)^{-1}\psi )\circ \phi _1$ in (7). To justify this, we need a regularity result for polyharmonic PDEs to show $\zeta \in H^{2m}_0(\varOmega )$. Inserting $\zeta$ into (7) we get

$$\begin{aligned}&-\int _\varOmega \frac{2}{\delta }({u}_1\circ \phi _1-{u}_0)(\nabla {u}_1\cdot (D\phi _2)^{-1}\psi )\circ \phi _1{\,\mathrm {d}}x \\&\quad =\int _\varOmega 2\gamma \varDelta ^m\phi _1\cdot \varDelta ^m(((D\phi _2)^{-1}\psi )\circ \phi _1)+2D\phi _1:D(((D\phi _2)^{-1}\psi )\circ \phi _1){\,\mathrm {d}}x. \end{aligned}$$

By adding this identity to the above equation we finally obtain (9).