Put Equations in Divergence Form

Coefficient Matching for Divergence Form

As explained in Equations You Can Solve Using PDE Toolbox, Partial Differential Equation Toolbox™ solvers address equations of the form

$- \nabla \cdot (c \nabla u) + a u = f$

or variants that have derivatives with respect to time, or that have eigenvalues, or are systems of equations. These equations are in divergence form, where the differential operator begins $\nabla \cdot$ . The coefficients a, c, and f are functions of position (x, y, z) and possibly of the solution u.

However, you can have equations in a form with all the derivatives explicitly expanded, such as

$(1 + x^{2}) \frac{\partial^{2} u}{\partial x^{2}} - 3 x y \frac{\partial^{2} u}{\partial x \partial y} + \frac{(1 + y^{2})}{2} \frac{\partial^{2} u}{\partial y^{2}} = 0$

In order to transform this expanded equation into toolbox format, you can try to match the coefficients of the equation in divergence form to the expanded form. In divergence form, if

$c = (\begin{matrix} c_{1} & c_{3} \\ c_{2} & c_{4} \end{matrix})$

then

$\begin{matrix} \nabla \cdot (c \nabla u) = c_{1} u_{x x} + (c_{2} + c_{3}) u_{x y} + c_{4} u_{y y} \\ + (\frac{\partial c_{1}}{\partial x} + \frac{\partial c_{2}}{\partial y}) u_{x} + (\frac{\partial c_{3}}{\partial x} + \frac{\partial c_{4}}{\partial y}) u_{y} \end{matrix}$

Matching coefficients in the u_xx and u_yy terms in $- \nabla \cdot (c \nabla u)$ to the equation, you get

$\begin{array}{l} c_{1} = - (1 + x^{2}) \\ c_{4} = - (1 + y^{2}) / 2 \end{array}$

Then looking at the coefficients of u_x and u_y, which should be zero, you get

$\begin{array}{l} (\frac{\partial c_{1}}{\partial x} + \frac{\partial c_{2}}{\partial y}) = - 2 x + \frac{\partial c_{2}}{\partial y} \\ so \\ c_{2} = 2 x y . \\ (\frac{\partial c_{3}}{\partial x} + \frac{\partial c_{4}}{\partial y}) = \frac{\partial c_{3}}{\partial x} - y \\ so \\ c_{3} = x y \end{array}$

This completes the conversion of the equation to the divergence form

$- \nabla \cdot (c \nabla u) = 0$

Boundary Conditions Can Affect the c Coefficient

The c coefficient appears in the generalized Neumann condition

$\vec{n} \cdot (c \nabla u) + q u = g$

So when you derive a divergence form of the c coefficient, keep in mind that this coefficient appears elsewhere.

For example, consider the 2-D Poisson equation –u_xx – u_yy = f. Obviously, you can take c = 1. But there are other c matrices that lead to the same equation: any that have c(2) + c(3) = 0.

$\begin{matrix} \nabla \cdot (c \nabla u) = \nabla \cdot ((\begin{matrix} c_{1} & c_{3} \\ c_{2} & c_{4} \end{matrix}) (\begin{matrix} u_{x} \\ u_{y} \end{matrix})) \\ = \frac{\partial}{\partial x} (c_{1} u_{x} + c_{3} u_{y}) + \frac{\partial}{\partial y} (c_{2} u_{x} + c_{4} u_{y}) \\ = c_{1} u_{x x} + c_{4} u_{y y} + (c_{2} + c_{3}) u_{x y} \end{matrix}$

So there is freedom in choosing a c matrix. If you have a Neumann boundary condition such as

$\vec{n} \cdot (c \nabla u) = 2$

the boundary condition depends on which version of c you use. In this case, make sure that you take a version of c that is compatible with both the equation and the boundary condition.

Some Equations Cannot Be Converted

Sometimes it is not possible to find a conversion to a divergence form such as

$- \nabla \cdot (c \nabla u) + a u = f$

For example, consider the equation

$\frac{\partial^{2} u}{\partial x^{2}} + \frac{\cos (x + y)}{4} \frac{\partial^{2} u}{\partial x \partial y} + \frac{1}{2} \frac{\partial^{2} u}{\partial y^{2}} = 0$

By simple coefficient matching, you see that the coefficients c₁ and c₄ are –1 and –1/2 respectively. However, there are no c₂ and c₃ that satisfy the remaining equations,

$\begin{matrix} c_{2} + c_{3} = \frac{- \cos (x + y)}{4} \\ \frac{\partial c_{1}}{\partial x} + \frac{\partial c_{2}}{\partial y} = \frac{\partial c_{2}}{\partial y} = 0 \\ \frac{\partial c_{3}}{\partial x} + \frac{\partial c_{4}}{\partial y} = \frac{\partial c_{3}}{\partial x} = 0 \end{matrix}$

Partial Differential Equation Toolbox Documentation

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