# TL:DR

How does the following vector calculus identity hold and how is it used to derive the weak form of the heat equation?

$$\nabla \cdot (vp \nabla u) = v\nabla \cdot (p \nabla u) + p(\nabla v) \cdot (\nabla u) \tag{3}$$

# Context and Attempt

I am working through "Finite Element Methods: A Practical Guide" (Whiteley, 2017) and attempting to derive the weak form of the following PDE:

where $$\Omega = [0, 1] \times [0, 1]$$. I understand that I have to multiply the equation by a test function $$v(x, y) \in H_0^1(\Omega)$$ where $$H_0^1$$ is a subset of the Sobolev space of order 1 such that $$v(x, y) = 0$$ at all points where Dirichlet boundary conditions are specified. Since

$$-(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}) = -\nabla \cdot \nabla u = 1$$

then multiplying by $$v(x, y)$$ gives

$$-v(\nabla \cdot \nabla u) = v \tag{1}$$

The book then states that the above equation is equivalent to

$$-\nabla \cdot (v \nabla u) + (\nabla v) \cdot (\nabla u) = v \tag{2}$$

citing the aforementioned vector identity in equation (3). Below I attempt to show how the left handside of equation (1) is the same as the left hand side of equation (2), however, I fail to do so, so please help:

\begin{aligned} v(-\nabla \cdot \nabla u) &\stackrel{?}{=} -\nabla \cdot (v \nabla u) + (\nabla v) \cdot (\nabla u) \\ -(v \frac{\partial^2 u}{\partial x^2} + v \frac{\partial^2 u}{\partial y^2}) &\stackrel{?}{=} \left( - \begin{bmatrix} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \end{bmatrix}^\text{T} \cdot \begin{bmatrix} \frac{v \partial u}{\partial x} \\ \frac{v \partial u}{\partial y} \end{bmatrix} \right) + \left(\begin{bmatrix} \frac{\partial v}{\partial x} \\ \frac{\partial v}{\partial y} \end{bmatrix}^\text{T} \cdot \begin{bmatrix} \frac{\partial u}{\partial x} \\ \frac{\partial u}{\partial y} \end{bmatrix} \right) \\ -(v \frac{\partial^2 u}{\partial x^2} + v \frac{\partial^2 u}{\partial y^2}) &\stackrel{?}= (-\frac{\partial v}{\partial x}\frac{\partial u}{\partial x} - \frac{\partial v}{\partial y}\frac{\partial u}{\partial y}) + (\frac{\partial v}{\partial x}\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}\frac{\partial u}{\partial y}) \\ -(v \frac{\partial^2 u}{\partial x^2} + v \frac{\partial^2 u}{\partial y^2}) &\stackrel{?}{=} 0 \end{aligned}

Edit: The derivation of the weak form via green's theorem shown in weak-formulation-poisson-equation MathSE post is sufficient for my understanding, but it would still be good to know how my question in the current post can be resolved. See prove-a-particular-case-of-product-rule-for-divergence MathSE post for the derivation of the listed identity.

• There is a mistake in the second line of the expansion. You have to derive the term $v\frac{\partial u}{\partial x}$ as a product of functions $v$ and $\frac{\partial u}{\partial x}$. The term with derivatives of $\frac{\partial u}{\partial x}$ is missing and the identity will be ok once you include it. Is your question effectively why the "vector identity in Appendix B.2" holds?
– Korf
Commented Oct 24, 2023 at 8:41

The vector calculus identity is simply the product rule, which written generally is

$$\nabla \cdot (v\vec{F}) = v(\nabla \cdot \vec{F}) + \nabla v \cdot \vec{F} \tag{1}$$

where $$v$$ is a scalar field and $$\vec{F}$$ is a vector field.

To use this in the derivation of the weak form of heat (Poisson's) equation, one simply says that $$\vec{F} = \nabla u$$, and substituting into equation (1),

$$\nabla \cdot (v \nabla u) = \boxed{v(\nabla \cdot \nabla u)} + \nabla v \cdot \nabla u \tag{2}$$

noting that the boxed part of the equation matches the form of the equation described in the question after multiplying by $$v$$,

$$-v(\nabla \cdot \nabla u) = v$$

Solving equation (2) for $$v(\nabla \cdot \nabla u)$$, accounting for the negative sign in front of $$v$$ in the question, and then integrating over the domain $$\Omega$$ gives

$$-\int_{\Omega} v(\nabla \cdot \nabla u)\ d\Omega = -\int_{\Omega} \nabla \cdot (v \nabla u)\ d\Omega + \int_{\Omega} \nabla v \cdot \nabla u\ d\Omega \tag{3}$$

and note that the Divergence theorem

$$\int_{\Omega} \nabla \cdot \vec{A}\ d\Omega = \oint_{\partial \Omega = \Gamma} \vec{A} \cdot \vec{n}\ d\Gamma$$

can be applied to

$$\int_{\Omega} \nabla \cdot (v \nabla u)\ d\Omega = \oint_{\partial \Omega = \Gamma} v \nabla u \cdot \vec{n}\ d\Gamma \tag{4}$$

by assigning $$\vec{A} = v \nabla u$$.

Substituting equation (4) into equation (3) gives the below equation.

$$-\int_{\Omega} v(\nabla \cdot \nabla u)\ d\Omega = - \oint_{\partial \Omega = \Gamma} v \nabla u \cdot \vec{n}\ d\Gamma + \int_{\Omega} \nabla v \cdot \nabla u\ d\Omega$$

Finally, since $$v(x, y) \in H_0^1(\Omega)$$ and thus $$v(x, y) = 0$$ at all points where Dirichlet boundary conditions are specified, and the boundary domain is defined by $$\partial \Omega$$, then

$$\oint_{\partial \Omega = \Gamma} v \nabla u \cdot \vec{n}\ d\Gamma = 0$$

and the weak form of the heat equation specified in the question is

$$\int_{\Omega} \nabla v \cdot \nabla u\ d\Omega = \int_{\Omega} v\ d\Omega$$