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I'm following this paper to create my own version of Poisson image editing. However, I don't quite understand what equation 1 and 2 are asking me.

The simplest interpolant $f$ of $f^*$ over $\Omega$ is the membrane interpolant defined as the solution of the minimization problem: $$\underset{f}{\text{min}}\iint_\Omega\left|\nabla f\right|^2\ \text{with}\ f|_{\partial\Omega}=f^*|_{\partial\Omega},\tag1$$ where $\nabla.=[\frac{\partial.}{\partial x}\frac{\partial.}{\partial y}]$ is the gradient operator. The minimizer must satisfy the associated Euler-Lagrange equation $$\Delta f=0\ \text{over}\ \Omega\ \text{with}\ f|_{\partial\Omega}=f^*|_{\partial\Omega},\tag2$$ where $\Delta.=\frac{\partial^2.}{\partial x^2}+\frac{\partial^2.}{\partial y^2}$ is the Laplacian operator. [...]

Poisson Image Editing by Pérez, Gangnet, Blake from Microsoft Research UK. http://www.irisa.fr/vista/Papers/2003_siggraph_perez.pdf

Andrew Chin
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1 Answers1

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For equation $(1)$, the problem asks you to consider functions $f:\Omega\subset \Bbb R^2\longrightarrow \Bbb R$ that agree with $f^*$ on the boundary $\partial \Omega$ of $\Omega$, that is, $f=f^*$ on $\partial\Omega$. There are many such functions. For each of them, we may consider the value

$$I(f) = \int_\Omega{\lVert\nabla f\rVert}^2$$

Because $\Omega\subset\Bbb R^2$, we have that $f$ is a function of two variables, say $(x,y)$. Then, $\nabla f(x,y)$ is the vector of partial derivatives $\left(\frac\partial{\partial x}f(x,y), \frac\partial{\partial y}f(x,y)\right)$. Hence,

$${\lVert\nabla f(x,y)\rVert}^2 = {\left(\frac\partial{\partial x}f(x,y)\right)}^2 + {\left(\frac\partial{\partial y}f(x,y)\right)}^2,$$

which is a real number, so the integral $I(f)$ is a real number too. Equation $(2)$ tells you that the 'simplest interpolant $f$ of $f^*$ over $\Omega$' is the $f$ which minimizes $I(f)$; call it $f_0$.

Moreover, equation$(2)$ tells you that $f_0$ satisfies

$$\Delta f_0 = 0$$

on $\Omega$, meaning that for all $(x,y)$ on $\Omega$

$$\Delta f_0(x,y) = \frac{\partial^2}{\partial x^2}f(x,y) + \frac{\partial^2}{\partial y^2}f(x,y) = 0.$$

Fimpellizzeri
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