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\noindent Math 285J, L. Vese. {\bf Assignment 1:} Due on Wednesday, November 6
\medbreak
\noindent {\bf [1]} Consider in two dimensions the functional minimization problem
$$\inf_{u} F(u)=F_1(u)+\lambda F_2(u_0-Ku),$$ where $u_0:\Omega\rightarrow \R$ is a given degraded version of a true (unknown) image $u:\Omega\rightarrow \R$, $\Omega\subset \R^2$, and $K$ is a linear and continuous operator on $L^2(\Omega)$.
Here, $F_1$ represents the regularization term while $F_2$ represents the data fidelity term. Recall that $|\nabla u|=\sqrt{(u_x)^2+(u_y)^2}$.
\medbreak
Assume $u_0,u\in L^2(\Omega)$,
$F_1(u)=\int_{\Omega}\phi_1(|\nabla u|)dxdy$, $F_2(u_0-Ku)=\int_{\Omega}\phi_2(u_0-Ku)dxdy$,
where $\nabla u=(u_x,u_y)$ is the spatial gradient operator,
$\phi_i:\R\rightarrow \R$ are functions of class $C^1$ ($i=1,2$), and that $\phi_2'(u_0-Ku)\in L^2(\Omega)$, as long as $u_0-Ku\in L^2(\Omega)$.
{\bf (i)} For $u\in W^{1,1}(\Omega)$, obtain the Euler-Lagrange equation associated with the minimization problem in $u$, in the stationary and time-dependent cases, together with the appropriate boundary conditions in $u$ on $\partial\Omega$. For the time-dependent case, show that the energy $E(t)=F(u(x,y,t))$ is decreasing in time.
\footnote{We may need to formally assume, in addition, that $(Ku)_t=K(u_t)$;
this is natural for a linear and continuous operator $K$ that does not depend on $t$, for instance if $Ku=k*u$ and $k=k(x,y)$ does not depend on $t$.}
{\bf (ii)} Show that, if $\phi_i$, $i=1,2$ are both convex, and $\phi_1$ is in addition non-decreasing from $[0,\infty)$ to $[0,\infty)$, then the functional $F(u)$ is convex.
\bigbreak
\noindent{\bf [2]} Consider in two dimensions $f\in L^2(\Omega)$, and $u(\cdot,\lambda)$ the minimizer of
$$F(u)=\lambda\int_{\Omega}|\nabla u|dxdy+\frac{1}{2}\int_{\Omega}(u-f)^2dxdy,$$
with $\lambda>0$. Recall that $|\nabla u|=\sqrt{(u_x)^2+(u_y)^2}$ can be made differentiable substituting it by $\sqrt{\epsilon^2+(u_x)^2+(u_y)^2}$.
{\bf (i)} Using the result from the previous problem, give the associated
Euler-Lagrange equation with the corresponding boundary conditions for a minimizer $u=u(\cdot,\lambda)\in W^{1,1}(\Omega)$.
{\bf (ii)} Show that the $L^2$-norm of $u(\cdot,\lambda)$, given by $\sqrt{\int_{\Omega}(u(x,y,\lambda))^2dxdy}$ is bounded by a constant independent of $\lambda$.
{\bf (iii)} Show (e.g. using the obtained stationary E.-L. equation and associated boundary condition), that
$$\int_{\Omega}u(x,y,\lambda)dxdy=\int_{\Omega}f(x,y)dxdy.$$
{\bf (iv)} Show that $u(\cdot,\lambda)$ converges in the $L^1(\Omega)-strong$ topology
to the average of the initial data. In other words, show that
$$\lim_{\lambda\rightarrow\infty}\int_{\Omega}\Big|u(x,y,\lambda)-\frac{\int_{\Omega}f(x,y)dxdy}{|\Omega|}\Big|dxdy=0.$$
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\noindent{\bf [3]} Discretize and implement the stationary or the non-stationary E.-L. equation from {\bf [2]} by the method of your choice using finite differences, for the denoising case. More details will be discussed in class. Choose an original true image $\hat u$, and define a noisy version $f=\hat u+noise$ (see matlab sample codes on the class web-page, or in matlab you can add noise of zero mean to an image using ``imnoise''). Give the optimal $\lambda$ (may be different in each case) and the RMSE between the original clean image $\hat u$ and the reconstructed image $u$:
$$RMSE=\sqrt{\frac{\sum_{i=1,j=1}^{i=M,j=N}(\hat u(i,j)-u(i,j))^2}{MN}}.$$
Plot the energy versus iterations.
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\noindent{\it Optional:} You can make additional tests by substituting the data fidelity term $\|f-u\|^2_{L^2(\Omega)}$ above by $\|f-u\|_{L^2(\Omega)}$ or by $\|f-u\|_{L^1(\Omega)}$, and compare the results. Each method may require different $\lambda$ for the same image. $\lambda$ can also be automatically selected if we know the noise variance in the form $\|f-u\|^2=\sigma^2$. Using a norm, instead of the norm square for the data fidelity term avoids the intensity loss drawback of the ROF model.
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