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Formulations and theorems

Allsolve solves simulations using mathematical formulations that model the underlying physics. This page introduces the general approach to formulating physics β€” from strong form to weak form β€” together with the vector calculus identities and integral theorems referenced by every physics derivation on this site.

βˆ‡β‹…(βˆ‡Γ—A)=0\begin{align*} \nabla \cdot (\nabla \times \boldsymbol{A}) = 0 \tag{1} \end{align*} βˆ‡β‹…(ab)=βˆ‡aβ‹…b+a(βˆ‡β‹…b)\begin{align*} \nabla \cdot (a \boldsymbol{b}) = \nabla a \cdot \boldsymbol{b} + a(\nabla \cdot \boldsymbol{b}) \tag{2} \end{align*} βˆ‡β‹…(aΓ—b)=(βˆ‡Γ—a)β‹…bβˆ’aβ‹…(βˆ‡Γ—b)\begin{align*} \nabla \cdot (\boldsymbol{a} \times \boldsymbol{b}) = (\nabla \times \boldsymbol{a}) \cdot \boldsymbol{b} - \boldsymbol{a} \cdot (\nabla \times \boldsymbol{b}) \tag{3} \end{align*} aβ‹…(bΓ—c)=cβ‹…(aΓ—b)=bβ‹…(cΓ—a)\begin{align*} \boldsymbol{a} \cdot (\boldsymbol{b} \times \boldsymbol{c}) = \boldsymbol{c} \cdot (\boldsymbol{a} \times \boldsymbol{b}) = \boldsymbol{b} \cdot (\boldsymbol{c} \times \boldsymbol{a}) \tag{4} \end{align*} βˆ‚βˆ‚t(βˆ‡Γ—A)=βˆ‡Γ—βˆ‚Aβˆ‚t\begin{align*} \frac{\partial}{\partial t} (\nabla \times \boldsymbol{A}) = \nabla \times \frac{\partial \boldsymbol{A}}{\partial t} \tag{5} \end{align*}

Let A\boldsymbol{A} be a vector field and ff a scalar function defined on a domain Ξ©\Omega with boundary Ξ“\Gamma. Then let n\boldsymbol{n} be a unit normal vector pointing outwards on Ξ“\Gamma. Now

βˆ«Ξ©β€…β€Š(Aβ‹…βˆ‡f)β€…β€ŠdΞ©=βˆ«Ξ“β€…β€Šfβ€…β€Š(Aβ‹…n)β€…β€ŠdΞ“βˆ’βˆ«Ξ©β€…β€Šfβ€…β€Š(βˆ‡β‹…A)β€…β€ŠdΞ©.\begin{align*} \int_{\Omega}\; (\boldsymbol{A} \cdot \nabla f)\; d \Omega = \int_{\Gamma}\; f\; (\boldsymbol{A} \cdot \boldsymbol{n})\; d \Gamma - \int_{\Omega}\; f\; (\nabla \cdot \boldsymbol{A})\; d \Omega. \tag{6} \end{align*}

Let Ξ©\Omega be a set with boundary Ξ“\Gamma, let n\boldsymbol{n} be a unit normal vector pointing outwards on Ξ“\Gamma and let T\boldsymbol{T} be a vector field or a second-order tensor. Now

βˆ«Ξ©βˆ‡β‹…Tβ€…β€ŠdΞ©=βˆ«Ξ“β€…β€ŠTβ‹…nβ€…β€ŠdΞ“.\begin{align*} \int_{\Omega} \nabla \cdot \boldsymbol{T}\; d \Omega = \int_{\Gamma}\; \boldsymbol{T} \cdot \boldsymbol{n}\; d \Gamma. \tag{7} \end{align*}

A common example of a partial differential equation (PDE) is Poisson’s equation,

Ξ”u=fβ€…β€ŠonΒ Ξ©.\begin{align*} \Delta u &= f\; \text{on}\ \Omega. \tag{8} \end{align*}

This is the strong formulation β€” a precise mathematical statement with strict regularity requirements for a solution.

The weak formulation relaxes those requirements. It is computationally preferable because it provides the foundation for numerical methods such as the finite element method (FEM).

To derive the weak formulation of Poisson’s equation, multiply both sides of the strong formulation by a test function uβ€²u^{\prime} and integrate over Ξ©\Omega:

βˆ«Ξ©β€…β€Š(Ξ”uuβ€²βˆ’fuβ€²)β€…β€ŠdΞ©=0\begin{align*} \int_{\Omega}\; (\Delta uu^{\prime} - fu^{\prime})\; d \Omega = 0 \tag{9} \end{align*}

Apply the Leibniz rule for a nabla operator, then use the Divergence theorem on the divergence term:

βˆ«Ξ©β€…β€ŠΞ”uβ€…β€Šuβ€²β€…β€ŠdΞ©=βˆ«Ξ©β€…β€Šβˆ‡β‹…(uβ€²βˆ‡u)β€…β€ŠdΞ©βˆ’βˆ«Ξ©β€…β€Šβˆ‡uβ‹…βˆ‡uβ€²β€…β€ŠdΞ©=βˆ«Ξ“β€…β€Š(βˆ‡uβ‹…n)β€…β€Šuβ€²β€…β€ŠdΞ“βˆ’βˆ«Ξ©β€…β€Šβˆ‡uβ‹…βˆ‡uβ€²β€…β€ŠdΞ©\begin{align*} \int_{\Omega}\; \Delta u\; u^{\prime}\; d \Omega &= \int_{\Omega}\; \nabla \cdot (u^{\prime} \nabla u)\; d \Omega - \int_{\Omega}\; \nabla u \cdot \nabla u^{\prime}\; d \Omega \tag{10} \\[5pt] &= \int_{\Gamma}\; (\nabla u \cdot \boldsymbol{n})\; u^{\prime}\; d \Gamma - \int_{\Omega}\; \nabla u \cdot \nabla u^{\prime}\; d \Omega \tag{11} \end{align*}

The resulting weak formulation is

βˆ«Ξ“β€…β€Š(βˆ‡uΒ β‹…n)β€…β€Šuβ€²β€…β€ŠdΞ“βˆ’βˆ«Ξ©β€…β€Š(βˆ‡uβ‹…βˆ‡uβ€²+fβ€…β€Šuβ€²)β€…β€ŠdΞ©=0.\begin{align*} \int_{\Gamma}\; (\nabla u\ \cdot \boldsymbol{n})\; u^{\prime}\; d \Gamma - \int_{\Omega}\; (\nabla u \cdot \nabla u^{\prime} + f\; u^{\prime})\; d \Omega = 0. \tag{12} \end{align*}

The regularity requirements for uu and ff are now less strict. A solution to (8)\text{(8)} is also a solution to (12)\text{(12)}, and every solution to (12)\text{(12)} that is smooth enough satisfies (8)\text{(8)} as well.