What youβll learn
Formulations β the difference between a strong and a weak formulation, and why weak formulations suit numerical simulations.
Theorems and identities β the vector calculus tools used to derive weak formulations.
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.
β β
( β Γ A ) = 0 \begin{align*}
\nabla \cdot (\nabla \times \boldsymbol{A})
= 0 \tag{1}
\end{align*} β β
( β Γ A ) = 0 β ( 1 ) β
β β
( a b ) = β a β
b + a ( β β
b ) \begin{align*}
\nabla \cdot (a \boldsymbol{b})
= \nabla a \cdot \boldsymbol{b}
+ a(\nabla \cdot \boldsymbol{b}) \tag{2}
\end{align*} β β
( a b ) = β a β
b + a ( β β
b ) β ( 2 ) β
β β
( a Γ b ) = ( β Γ a ) β
b β a β
( β 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*} β β
( a Γ b ) = ( β Γ a ) β
b β a β
( β Γ b ) β ( 3 ) β
a β
( b Γ c ) = c β
( a Γ b ) = b β
( c Γ 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*} a β
( b Γ c ) = c β
( a Γ b ) = b β
( c Γ a ) β ( 4 ) β
β β t ( β Γ A ) = β Γ β A β t \begin{align*}
\frac{\partial}{\partial t} (\nabla \times \boldsymbol{A}) = \nabla \times \frac{\partial \boldsymbol{A}}{\partial t} \tag{5}
\end{align*} β t β β ( β Γ A ) = β Γ β t β A β β ( 5 ) β
Let A \boldsymbol{A} A be a vector field and f f f a scalar function defined on a domain Ξ© \Omega Ξ© with boundary Ξ \Gamma Ξ .
Then let n \boldsymbol{n} n be a unit normal vector pointing outwards on Ξ \Gamma Ξ .
Now
β« Ξ© β
β ( A β
β f ) β
β d Ξ© = β« Ξ β
β f β
β ( A β
n ) β
β d Ξ β β« Ξ© β
β f β
β ( β β
A ) β
β d Ξ© . \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*} β« Ξ© β ( A β
β f ) d Ξ© = β« Ξ β f ( A β
n ) d Ξ β β« Ξ© β f ( β β
A ) d Ξ©. β ( 6 ) β
Let Ξ© \Omega Ξ© be a set with boundary Ξ \Gamma Ξ , let n \boldsymbol{n} n be a unit normal vector pointing outwards on Ξ \Gamma Ξ and let T \boldsymbol{T} T be a vector field or a second-order tensor.
Now
β« Ξ© β β
T β
β d Ξ© = β« Ξ β
β T β
n β
β d Ξ . \begin{align*}
\int_{\Omega} \nabla \cdot \boldsymbol{T}\; d \Omega
= \int_{\Gamma}\; \boldsymbol{T} \cdot \boldsymbol{n}\; d \Gamma. \tag{7}
\end{align*} β« Ξ© β β β
T d Ξ© = β« Ξ β T β
n d Ξ. β ( 7 ) β
A common example of a partial differential equation (PDE) is Poissonβs equation,
Ξ u = f β
β onΒ Ξ© . \begin{align*}
\Delta u
&= f\; \text{on}\ \Omega. \tag{8}
\end{align*} Ξ u β = f on Β Ξ©. β ( 8 ) β
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 Poissonβs equation, multiply both sides of the strong formulation by a test function u β² u^{\prime} u β² and integrate over Ξ© \Omega Ξ© :
β« Ξ© β
β ( Ξ u u β² β f u β² ) β
β d Ξ© = 0 \begin{align*}
\int_{\Omega}\; (\Delta uu^{\prime} - fu^{\prime})\; d \Omega
= 0 \tag{9}
\end{align*} β« Ξ© β ( Ξ u u β² β f u β² ) d Ξ© = 0 β ( 9 ) β
Apply the Leibniz rule for a nabla operator , then use the Divergence theorem on the divergence term:
β« Ξ© β
β Ξ u β
β u β² β
β d Ξ© = β« Ξ© β
β β β
( u β² β u ) β
β d Ξ© β β« Ξ© β
β β u β
β u β² β
β d Ξ© = β« Ξ β
β ( β u β
n ) β
β u β² β
β d Ξ β β« Ξ© β
β β u β
β u β² β
β d Ξ© \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*} β« Ξ© β Ξ u u β² d Ξ© β = β« Ξ© β β β
( u β² β u ) d Ξ© β β« Ξ© β β u β
β u β² d Ξ© = β« Ξ β ( β u β
n ) u β² d Ξ β β« Ξ© β β u β
β u β² d Ξ© β ( 10 ) ( 11 ) β
The resulting weak formulation is
β« Ξ β
β ( β u Β β
n ) β
β u β² β
β d Ξ β β« Ξ© β
β ( β u β
β u β² + f β
β u β² ) β
β d Ξ© = 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*} β« Ξ β ( β u Β β
n ) u β² d Ξ β β« Ξ© β ( β u β
β u β² + f u β² ) d Ξ© = 0. β ( 12 ) β
The regularity requirements for u u u and f f f are now less strict. A solution to (8) \text{(8)} (8) is also a solution to (12) \text{(12)} (12) , and every solution to (12) \text{(12)} (12) that is smooth enough satisfies (8) \text{(8)} (8) as well.