# Does the symbol $\nabla^2$ has the same meaning in Laplace Equation and Hessian Matrix?

we know that the Laplace Equation can be written in the form: $$\nabla^2 \Phi=0$$ while in this equation,the symbol $\nabla^2 \Phi$ stand for $\sum_{i=1}^n\frac{\partial^2 \Phi}{\partial x_i^2}$.

At the same time, the Hessian Matrix is also denoted as $$H=\nabla^2f$$

My question is what does $\nabla^2$ realy mean?

Both of them are second order differential operators.

Laplacian: first outer product then inner product

Hessian : first outer product then outer product again

Usually use $$\Delta$$ for Laplacian to avoid confusion.

Example: outer product of outer product in 2D:

$${\bf H} = \left(\begin{array}{c} \frac{\partial}{\partial x}\\\frac{\partial}{\partial y}\end{array}\right)\left(\begin{array}{cc} \frac{\partial}{\partial x}&\frac{\partial}{\partial y}\end{array}\right) = \left(\begin{array}{cc} \frac{\partial^2}{\partial x^2} & \frac{\partial^2}{\partial xy} \\\frac{\partial^2}{\partial yx}& \frac{\partial^2}{\partial y^2}\end{array}\right)$$

$${\bf L} = \left(\begin{array}{cc} \frac{\partial}{\partial x}&\frac{\partial}{\partial y}\end{array}\right)\left(\begin{array}{c} \frac{\partial}{\partial x}\\\frac{\partial}{\partial y}\end{array}\right) = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$$

In matrix language you see that Laplacian is the trace of the Hessian $${\bf L} = \rm tr({\bf H})$$, i.e. that $$\bf L$$ is equal to the sum of the diagonal elements of $$\bf H$$. This is a "contraction". Reducing the number of indices of the operator from 2 to 0.

The problem with writing $$\nabla^2$$ is that we do not know if each application is supposed to be an inner or an outer product. For vectors there are many types of products so it is important to make sure we know which kind of product we are talking about. Therefore one should try and not use $$\nabla^2$$ when there can be confusion.

• Would inner or outer products be connected matrix language, like gradient meaning column derivative vector? Feb 12, 2021 at 9:18
• @MathArt a gradient is outer product with $\nabla$. Gradient on a scalar becomes a vector. Gradient on a vector becomes a 2-tensor or matrix. A Jacobian could be an example. So the number of indexes increase by one. Divergence in vector analysis is an inner product with $\nabla$. Divergence of a vector becomes a scalar. Number of indices reduce by one. Feb 12, 2021 at 9:25

As pointed out by the other answer, the notation $$\nabla^2$$ is somewhat ambiguous (though it most commonly refers to the Laplacian). The reason is that $$\nabla^2$$ could be interpreted either as \begin{align} \nabla^2 &= \nabla \cdot \nabla \\ &= \left( \sum_i \mathbf{e}_i \nabla_i \right) \cdot \left( \sum_i \mathbf{e}_i \nabla_i \right) \\ &= \sum_i \nabla_i \nabla_i \\ \end{align} where $$\cdot$$ is the inner product, which is the Laplacian (scalar), or as \begin{align} \nabla^2 &= \nabla \otimes \nabla \\ &= \left( \sum_i \mathbf{e}_i \nabla_i \right) \otimes \left( \sum_i \mathbf{e}_i \nabla_i \right) \\ &= \sum_i \sum_j \mathbf{e}_i \mathbf{e}_j \nabla_i \nabla_j \end{align} where $$\otimes$$ is the outer product, which is the Hessian (matrix).

• Thank you for a great answer. This tradition is seriously messed up, though. Dec 6, 2021 at 17:26