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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

`$-\nabla \cdot \left(c\nabla u\right)+au=f$`

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 $\nabla ·$. 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

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

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

`$c=\left(\begin{array}{cc}{c}_{1}& {c}_{3}\\ {c}_{2}& {c}_{4}\end{array}\right)$`

then

`$\begin{array}{c}\nabla ·\left(c\nabla u\right)={c}_{1}{u}_{xx}+\left({c}_{2}+{c}_{3}\right){u}_{xy}+{c}_{4}{u}_{yy}\\ +\left(\frac{\partial {c}_{1}}{\partial x}+\frac{\partial {c}_{2}}{\partial y}\right){u}_{x}+\left(\frac{\partial {c}_{3}}{\partial x}+\frac{\partial {c}_{4}}{\partial y}\right){u}_{y}\end{array}$`

Matching coefficients in the uxx and uyy terms in $-\nabla \cdot \left(c\nabla u\right)$ to the equation, you get

`$\begin{array}{l}{c}_{1}=-\left(1+{x}^{2}\right)\\ {c}_{4}=-\left(1+{y}^{2}\right)/2\end{array}$`

Then looking at the coefficients of ux and uy, which should be zero, you get

`$\begin{array}{l}\left(\frac{\partial {c}_{1}}{\partial x}+\frac{\partial {c}_{2}}{\partial y}\right)=-2x+\frac{\partial {c}_{2}}{\partial y}\\ \text{so}\\ {c}_{2}=2xy.\\ \left(\frac{\partial {c}_{3}}{\partial x}+\frac{\partial {c}_{4}}{\partial y}\right)=\frac{\partial {c}_{3}}{\partial x}-y\\ \text{so}\\ {c}_{3}=xy\end{array}$`

This completes the conversion of the equation to the divergence form

`$-\nabla \cdot \left(c\nabla u\right)=0$`

Boundary Conditions Can Affect the c Coefficient

The `c` coefficient appears in the generalized Neumann condition

`$\stackrel{\to }{n}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\left(c\nabla u\right)+qu=g$`

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 uxx – uyy = 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.

`$\begin{array}{c}\nabla \text{\hspace{0.17em}}·\text{\hspace{0.17em}}\left(c\nabla u\right)=\nabla \text{\hspace{0.17em}}·\text{\hspace{0.17em}}\left(\left(\begin{array}{cc}{c}_{1}& {c}_{3}\\ {c}_{2}& {c}_{4}\end{array}\right)\left(\begin{array}{c}{u}_{x}\\ {u}_{y}\end{array}\right)\right)\\ =\frac{\partial }{\partial x}\left({c}_{1}{u}_{x}+{c}_{3}{u}_{y}\right)+\frac{\partial }{\partial y}\left({c}_{2}{u}_{x}+{c}_{4}{u}_{y}\right)\\ ={c}_{1}{u}_{xx}+{c}_{4}{u}_{yy}+\left({c}_{2}+{c}_{3}\right){u}_{xy}\end{array}$`

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

`$\stackrel{\to }{n}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\left(c\nabla u\right)=2$`

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

`$-\nabla \cdot \left(c\nabla u\right)+au=f$`

For example, consider the equation

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

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,

`$\begin{array}{c}{c}_{2}+{c}_{3}=\frac{-\mathrm{cos}\left(x+y\right)}{4}\\ \frac{\partial {c}_{1}}{\partial x}+\frac{\partial {c}_{2}}{\partial y}=\frac{\partial {c}_{2}}{\partial y}=0\\ \frac{\partial {c}_{3}}{\partial x}+\frac{\partial {c}_{4}}{\partial y}=\frac{\partial {c}_{3}}{\partial x}=0\end{array}$`

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