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For the discretization of gradient, if I set $$ \mathcal \nabla := \begin{bmatrix}1&-1&\\&1&-1\end{bmatrix},$$

then $$ \mathcal \nabla^T\mathcal\nabla := \begin{bmatrix}1&0\\-1&1\\0&-1\end{bmatrix}\begin{bmatrix}1&-1&\\&1&-1\end{bmatrix}=\begin{bmatrix}1&-1&0\\-1&2&-1\\0&-1&1\end{bmatrix}$$

It seems that $\mathcal \nabla^T\mathcal\nabla=-\Delta$, not $\mathcal \nabla^T\mathcal\nabla=\Delta$. Is this true? If yes, could you explain why it holds?

jakeoung
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1 Answers1

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This is true, and is a good observation. Since $\nabla^T\nabla $ is a positive semidefinite matrix by construction, its eigenvalues are nonnegative. On the other hand, the eigenvalues of $\dfrac{d^2}{dx^2}$ (with zero boundary values) are nonpositive, they come from $(\sin \beta x)''=-\beta^2 \sin\beta x$, etc. So, it makes sense that your positive semidefinite matrix represents $-\Delta$ rather than $\Delta$.

On the level of details, a look at the rows of $\nabla$ shows it's a discretization of $-\dfrac{d}{dx}$, not of $\dfrac{d}{dx}$.

On the level of concepts, this happens because the transpose matrix represents the adjoint operator, and the adjoint of $\dfrac{d}{dx}$ (with zero boundary values) is $-\dfrac{d}{dx}$.

  • Then, is the Dirichlet boundary condition assumed? If I want to impose Neumann boundary condition (derivative=0), how to discretize? Also, could you recommend a book or branch where I can study in more detail? For me, it is hard to find book to study. Many thanks. – jakeoung Sep 27 '15 at 10:29
  • The first and third rows of the transpose matrix represent differences where one of two terms is zero. But a similar conclusion would hold for many other boundary conditions... Usually, one discretizes the Laplacian directly, using its symmetric difference approximation. Look up "finite difference" Laplacian "boundary condition". –  Sep 27 '15 at 14:08
  • Why is the adjoint of $d/dx$ is $-d/dx$? Also, if I discretize $-d/dx$, not $d/dx$, it seems to derive the same result. Is it right? – jakeoung Sep 28 '15 at 03:44
  • Because integration by parts has a minus sign in it. If you need more details, search or ask a new question. Of course, changing the sign of $\nabla$ does not change the product, the sign cancels out. –  Sep 28 '15 at 03:53
  • Actually, I don't know why $\nabla$ shows $-d/dx$. Anyway, I posted a new question: http://math.stackexchange.com/questions/1463983/why-is-the-adjoint-of-d-dx-d-dx-with-zero-boundary-condition – jakeoung Oct 04 '15 at 15:38
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    @topy Apply it to a vector of increasing values, like $(1,2,3)^T$, and see what the result is. –  Oct 04 '15 at 16:06