# pdeCoefficients

Extract coefficients of partial differential equation

## Description

extracts the coefficients of a partial differential equation (PDE) as a structure of
double-precision numbers and function handles, which can be used as input of the
`coeffs`

= pdeCoefficients(`pdeeq`

,`u`

)`specifyCoefficients`

function in Partial Differential Equation Toolbox™.

`pdeeq`

is a scalar PDE or a PDE system in symbolic form that is a
function of `u`

. The `pdeCoefficients`

function converts
`pdeeq`

into a system of equations of the form

$$m\frac{{\partial}^{2}u}{\partial {t}^{2}}+d\frac{\partial u}{\partial t}-\nabla \xb7\left(c\otimes \nabla u\right)+au=f$$

and returns the structure `coeffs`

that contains the
coefficients `m`

, `d`

, `c`

,
`a`

, and `f`

. For more information, see Equations You Can Solve Using PDE Toolbox (Partial Differential Equation Toolbox).

If `pdeCoefficients`

cannot convert a PDE into the divergence form
above, then it issues a warning message and writes all remaining gradients to the
`f`

coefficient. PDE Toolbox will be unable to solve this type of
PDE.

## Examples

### Extract Coefficients From a Symbolic PDE

Create a symbolic PDE that represents the Laplacian of the variable `u(x,y)`

.

```
syms u(x,y)
pdeeq = laplacian(u,[x y]) == -3
```

pdeeq(x, y) =$$\frac{{\partial}^{2}}{\partial {x}^{2}}\mathrm{}u\left(x,y\right)+\frac{{\partial}^{2}}{\partial {y}^{2}}\mathrm{}u\left(x,y\right)=-3$$

Extract the coefficients of the PDE.

coeffs = pdeCoefficients(pdeeq,u)

`coeffs = `*struct with fields:*
a: 0
c: [4x1 double]
m: 0
d: 0
f: -3

`pdeCoefficients`

converts the symbolic PDE into a scalar PDE equation of the form

$$m\frac{{\partial}^{2}u}{\partial {t}^{2}}+d\frac{\partial u}{\partial t}-\nabla \cdot (c\nabla u)+au=f$$

and extracts the coefficients `m`

, `d`

, `c`

, `a`

, and `f`

into the structure `coeffs`

. For 2-D systems of $$N$$ equations, `c`

is a tensor with 4$${N}^{2}$$ elements. For more information, see c Coefficient for specifyCoefficients (Partial Differential Equation Toolbox). In this case, $$N=1$$, so the coefficient `c`

is a 4-by-1 column vector that represents `c`

$${}_{xx}$$, `c`

$${}_{xy}$$, `c`

$${}_{yx}$$, and `c`

$${}_{yy}$$.

coeffs.c

`ans = `*4×1*
-1
0
0
-1

### Solve Homogenous Laplace's Equation

Solve a 2-D homogenous Laplace's equation in the domain bounded by a unit circle. Use the `pdeCoefficients`

function to extract the coefficients of the Laplace's equation.

Create the PDE model and include the geometry.

model = createpde; geometryFromEdges(model,@circleg);

Plot the geometry and display the edge labels for use in the boundary condition definition.

figure pdegplot(model,'EdgeLabels','on') axis equal

Create a symbolic expression `pdeeq`

that represents the Laplace's equation.

```
syms u(x,y)
pdeeq = laplacian(u,[x y])
```

pdeeq(x, y) =$$\frac{{\partial}^{2}}{\partial {x}^{2}}\mathrm{}u\left(x,y\right)+\frac{{\partial}^{2}}{\partial {y}^{2}}\mathrm{}u\left(x,y\right)$$

Extract the coefficients of the Laplace's equation.

coeffs = pdeCoefficients(pdeeq,u)

`coeffs = `*struct with fields:*
a: 0
c: [4x1 double]
m: 0
d: 0
f: 0

Specify the coefficients of the PDE model.

specifyCoefficients(model,'m',coeffs.m,'d',coeffs.d, ... 'c',coeffs.c,'a',coeffs.a,'f',coeffs.f);

Apply the Dirichlet boundary condition $$u(x,y)={x}^{4}+{y}^{4}$$ at all 4 edges that form the circle.

applyBoundaryCondition(model,'dirichlet','Edge',1:4,'u',@(region,state) region.x.^4 + region.y.^4);

Generate the default mesh for the geometry.

generateMesh(model);

Solve the PDE and plot the solution.

```
results = solvepde(model);
pdeplot(model,'XYData',results.NodalSolution)
```

### Solve Poisson's Equation with Nonconstant `f`

Coefficient

Solve a 2-D Poisson's equation in the domain bounded by a unit circle. The divergence form of the PDE has nonconstant `f`

coefficient. You can solve the PDE by extracting the coefficients using `pdeCoefficients`

and specifying the coefficients in the PDE model using `specifyCoefficients`

.

Create the PDE model and include the geometry.

model = createpde; geometryFromEdges(model,@circleg);

Create a symbolic expression `pdeeq`

that represents Poisson's equation.

```
syms u(x,y)
pdeeq = diff(u,x,x) + diff(u,y,y) - 1/(x^2 + y^2)
```

pdeeq(x, y) =$$\frac{{\partial}^{2}}{\partial {x}^{2}}\mathrm{}u\left(x,y\right)-\frac{1}{{x}^{2}+{y}^{2}}+\frac{{\partial}^{2}}{\partial {y}^{2}}\mathrm{}u\left(x,y\right)$$

Extract the coefficients of the equation.

coeffs = pdeCoefficients(pdeeq,u)

`coeffs = `*struct with fields:*
a: 0
c: [4x1 double]
m: 0
d: 0
f: @makeCoefficient/coefficientFunction

The `f`

coefficient is not a constant. It is displayed as `@makeCoefficient/coefficientFunction`

, which indicates that it has been returned from the workspace of some internal function. To show the formula behind the function handle, use `coeffs.f('show')`

.

`coeffs.f('show')`

ans =$$\frac{1}{{x}^{2}+{y}^{2}}$$

Specify the coefficients of the PDE model.

specifyCoefficients(model,'m',coeffs.m,'d',coeffs.d, ... 'c',coeffs.c,'a',coeffs.a,'f',coeffs.f);

Apply the Dirichlet boundary condition $$u(x,y)=0$$ at all 4 edges that form the circle.

applyBoundaryCondition(model,'dirichlet','Edge',1:4,'u',0);

Generate the default mesh for the geometry.

generateMesh(model);

Solve the PDE. Plot the nodal solution using the option `'XYData'`

in the `pdeplot`

function.

```
results = solvepde(model);
pdeplot(model,'XYData',results.NodalSolution)
```

Plot the gradient of the solution at the nodal locations using the option `'FlowData'`

.

`pdeplot(model,'FlowData',[results.XGradients results.YGradients])`

### PDE With No Divergence Form

Construct a PDE with no divergence form.

```
syms u(x,y)
pdeeq = diff(u,x,x) + cos(x+y)/4*diff(u,x,y) + 1/2*diff(u,y,y)
```

pdeeq(x, y) =$$\frac{{\partial}^{2}}{\partial {x}^{2}}\mathrm{}u\left(x,y\right)+\frac{\mathrm{cos}\left(x+y\right)\hspace{0.17em}\frac{\partial}{\partial y}\mathrm{}\frac{\partial}{\partial x}\mathrm{}u\left(x,y\right)}{4}+\frac{\frac{{\partial}^{2}}{\partial {y}^{2}}\mathrm{}u\left(x,y\right)}{2}$$

Extract its coefficients. When `pdeCoefficients`

cannot convert a PDE into the divergence form

$$m\frac{{\partial}^{2}u}{\partial {t}^{2}}+d\frac{\partial u}{\partial t}-\nabla \cdot (c\otimes \nabla u)+au=f$$,

it issues a warning message, but it still produces output.

coeffs = pdeCoefficients(pdeeq,u)

Warning: After extracting m, d, and c, some gradients remain. Writing all remaining terms to f.

`coeffs = `*struct with fields:*
a: 0
c: @makeCoefficient/coefficientFunction
m: 0
d: 0
f: @makeCoefficient/coefficientFunction

To show the function handles in the extracted coefficients `c`

and `f`

, use the option `'show'`

. All remaining gradients in the PDE are written to the `f`

coefficient.

`coeffs.c('show')`

ans =$$\left(\begin{array}{c}-1\\ \frac{1}{8}-\frac{\mathrm{cos}\left(x+y\right)}{4}\\ \frac{1}{8}-\frac{\mathrm{cos}\left(x+y\right)}{4}\\ -\frac{1}{2}\end{array}\right)$$

`coeffs.f('show')`

ans =$$\frac{\mathrm{sin}\left(x+y\right)\hspace{0.17em}\frac{\partial}{\partial x}\mathrm{}u\left(x,y\right)}{4}+\frac{\mathrm{sin}\left(x+y\right)\hspace{0.17em}\frac{\partial}{\partial y}\mathrm{}u\left(x,y\right)}{4}-\frac{\mathrm{cos}\left(x+y\right)\hspace{0.17em}\frac{\partial}{\partial y}\mathrm{}\frac{\partial}{\partial x}\mathrm{}u\left(x,y\right)}{4}+\frac{\frac{\partial}{\partial y}\mathrm{}\frac{\partial}{\partial x}\mathrm{}u\left(x,y\right)}{4}$$

Since the PDE has no divergence form required by the PDE Toolbox, the toolbox will be unable to solve this PDE.

### Solve Time-Dependent Wave Equation

Solve a time-dependent wave equation in the domain bounded by a unit circle. The wave equation is a PDE with a scalar function $$u(t,x,y)$$ that depends on time $$t$$ and coordinates $$x$$ and $$y$$.

Create the PDE model and include the geometry.

model = createpde(1); geometryFromEdges(model,@circleg);

Create a symbolic PDE that represents the wave equation.

```
syms u(t,x,y)
pdeeq = diff(u,t,t) - laplacian(u,[x y])
```

pdeeq(t, x, y) =$$\frac{{\partial}^{2}}{\partial {t}^{2}}\mathrm{}u\left(t,x,y\right)-\frac{{\partial}^{2}}{\partial {x}^{2}}\mathrm{}u\left(t,x,y\right)-\frac{{\partial}^{2}}{\partial {y}^{2}}\mathrm{}u\left(t,x,y\right)$$

Extract the coefficients of the PDE.

coeffs = pdeCoefficients(pdeeq,u)

`coeffs = `*struct with fields:*
a: 0
c: [4x1 double]
m: 1
d: 0
f: 0

Specify the coefficients of the PDE model.

specifyCoefficients(model,'m',coeffs.m,'d',coeffs.d, ... 'c',coeffs.c,'a',coeffs.a,'f',coeffs.f);

Set the initial conditions of the time-dependent problem on the entire geometry.

setInitialConditions(model,0,1);

Apply the Dirichlet boundary condition $$u(x,y)=0$$ at all 4 edges that form the circle.

applyBoundaryCondition(model,'dirichlet','Edge',1:4,'u',0);

Generate the default mesh for the geometry.

generateMesh(model);

Find the solutions of the time-dependent PDE at a time range from 0s to 50s with a 2s interval.

results = solvepde(model,linspace(0,2,50));

Plot the solution of the wave equation for each 5s interval.

figure; for k = 1:10 subplot(5,2,k); pdeplot(model,'XYData',results.NodalSolution(:,k*5)) axis equal end

### System of Several PDEs

Solve a system of two second-order PDEs. You can solve the PDE system by extracting the PDE coefficients symbolically using `pdeCoefficients`

, converting the coefficients to double-precision numbers using `pdeCoefficientsToDouble`

, and specifying the coefficients in the PDE model using `specifyCoefficients`

.

The system of PDEs represents the deflection of a clamped structural plate under a uniform pressure load. The system of PDEs with the dependent variables $${u}_{1}$$ and $${u}_{2}$$ is given by

$$-{\nabla}^{2}{u}_{1}+{u}_{2}=0$$,

$$-D{\nabla}^{2}{u}_{2}=p$$,

where $$D$$ is the bending stiffness of the plate given by

$$D=\frac{E{h}^{3}}{12(1-{\nu}^{2})}$$,

and $$E$$ is the modulus of elasticity, $$\nu $$ is Poisson's ratio, $$h$$ is the plate thickness, $${u}_{1}$$ is the transverse deflection of the plate, and $$p$$ is the pressure load.

Create a PDE model for the system of two equations.

model = createpde(2);

Create a square geometry. Specify the side length of the square. Then include the geometry in the PDE model.

len = 10.0; gdm = [3 4 0 len len 0 0 0 len len]'; g = decsg(gdm,'S1',('S1')'); geometryFromEdges(model,g);

Specify the values of the physical parameters of the system. Let the external pressure $$p$$ be a symbolic variable `pres`

that can take any value.

```
E = 1.0e6;
h_thick = 0.1;
nu = 0.3;
D = E*h_thick^3/(12*(1 - nu^2));
syms pres
```

Declare the PDE system as a system symbolic equations. Extract the coefficients of the PDE and return them in symbolic form.

syms u1(x,y) u2(x,y) pdeeq = [-laplacian(u1) + u2; -D*laplacian(u2) - pres]; symCoeffs = pdeCoefficients(pdeeq,[u1 u2],'Symbolic',true)

`symCoeffs = `*struct with fields:*
m: 0
a: [2x2 sym]
c: [4x4 sym]
f: [2x1 sym]
d: 0

Display the coefficients `m`

, `a`

, `c`

, `f`

, and `d`

.

structfun(@disp,symCoeffs)

`$$0$$`

$$\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$$

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& \frac{25000}{273}& 0\\ 0& 0& 0& \frac{25000}{273}\end{array}\right)$$

$$\left(\begin{array}{c}0\\ \mathrm{pres}\end{array}\right)$$

`$$0$$`

Substitute a value for `pres`

using the `subs`

function. Since the outputs of `subs`

are symbolic objects, use the `pdeCoefficientsToDouble`

function to convert the coefficients to the `double`

data type, which makes them valid inputs for the PDE Toolbox.

symCoeffs = subs(symCoeffs,pres,2); coeffs = pdeCoefficientsToDouble(symCoeffs)

`coeffs = `*struct with fields:*
a: [4x1 double]
c: [16x1 double]
m: 0
d: 0
f: [2x1 double]

Specify the PDE coefficients for the PDE model.

specifyCoefficients(model,'m',coeffs.m,'d',coeffs.d, ... 'c',coeffs.c,'a',coeffs.a,'f',coeffs.f);

Specify spring stiffness. Specify boundary conditions by defining distributed springs on all four edges.

k = 1e7; bOuter = applyBoundaryCondition(model,'neumann','Edge',(1:4), ... 'g',[0 0],'q',[0 0; k 0]);

Specify the mesh size of the geometry and generate a mesh for the PDE model.

```
hmax = len/20;
generateMesh(model,'Hmax',hmax);
```

Solve the model.

res = solvepde(model);

Access the solution at the nodal locations.

u = res.NodalSolution;

Plot the transverse deflection of the plate.

numNodes = size(model.Mesh.Nodes,2); figure; pdeplot(model,'XYData',u(1:numNodes),'contour','on') title 'Transverse Deflection'

Find the transverse deflection at the plate center.

wMax = min(u(1:numNodes,1))

wMax = -0.2763

Compare the result with the deflection at the plate center computed analytically.

pres = 2; wMax = -.0138*pres*len^4/(E*h_thick^3)

wMax = -0.2760

## Input Arguments

`pdeeq`

— PDE in symbolic form

symbolic equation | symbolic expression | symbolic vector

PDE in symbolic form, specified as a symbolic equation, expression, or a symbolic vector.

`u`

— Variable of PDE

symbolic function

Variable of PDE, specified as a symbolic function. `u`

must be a
stationary or time-dependent variable in two or three dimensions. For example, create
the variable `u`

using one of these statements:

`syms u(x,y)`

`syms u(t,x,y)`

`syms u(x,y,z)`

`syms u(t,x,y,z)`

## Output Arguments

`coeffs`

— Coefficients of PDE

structure of doubles and function handles

Coefficients of PDE, returned as a structure of double-precision numbers and
function handles as required by the `specifyCoefficients`

function. The
fields of the structure are `a`

, `c`

,
`m`

, `d`

, and `f`

. For details on
interpreting the coefficients in the format required by the PDE Toolbox, see:

c Coefficient for specifyCoefficients (Partial Differential Equation Toolbox)

m, d, or a Coefficient for specifyCoefficients (Partial Differential Equation Toolbox)

f Coefficient for specifyCoefficients (Partial Differential Equation Toolbox)

Any coefficient can be a double-precision number or function handle. For example,
the coefficient `coeffs.f`

can be a function handle that represents
some internal function in the workspace. It takes two structures as input arguments,
`location`

and `state`

, and returns a double.

Function handles are displayed as
`@makeCoefficient/coefficientFunction`

in the Command Window. To
display the formula of a function handle `coeffs.f`

in symbolic form,
use `coeffs.f('show')`

.

`symCoeffs`

— Coefficients of PDE in symbolic form

structure of symbolic expressions

Coefficients of PDE in symbolic form, returned as a structure of symbolic expressions.

## See Also

`syms`

| `diff`

| `laplacian`

| `pdeCoefficientsToDouble`

| `specifyCoefficients`

(Partial Differential Equation Toolbox)

**Introduced in R2021a**

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