# Tutorial¶

This tutorial assumes knowedge of the Python programming language, as well as familiarity with the third party modules Numpy and Matplotlib. It also assumes knowledge of advanced calculus, Einstein notation, weak formulations, and the Finite Element Method.

We will introduce fundamental Nutils concepts based on the 1D homogeneous Laplace problem,

with boundary conditions \(u(0) = 0\) and \(u'(1) = 1\). Even though the solution is trivially found to be \(u(x) = x\), the example serves to introduce many key concepts in the Nutils paradigm, concepts that can then be applied to solve a wide class of physics problems.

## A little bit of theory¶

Before turning to code we must first formally cast the problem into weak form.

Let \(Ω\) be the unit line \([0,1]\) with boundaries \(Γ_\text{left}\) and \(Γ_\text{right}\), and let \(H_0(Ω)\) be a suitable function space such that any \(u ∈ H_0(Ω)\) satisfies \(u = 0\) in \(Γ_\text{left}\). The Laplace problem is solved uniquely by the element \(u ∈ H_0(Ω)\) for which \(R(v, u) = 0\) for all test functions \(v ∈ H_0(Ω)\), with \(R\) the bilinear functional

We next restrict ourselves to a finite dimensional subspace, to which end we adopt a set of Finite Element basis functions \(φ_n ∈ H_0(Ω)\). In this space, the Finite Element solution is established by solving the linear system of equations \(R_n(\hat{u}) = 0\), with residual vector \(R_n(\hat{u}) := R(φ_n, \hat{u})\), and discrete solution

Note that discretization inevitably implies approximation, i.e. \(u ≠ \hat{u}\) in general. In this case, however, we choose \(\{φ_n\}\) to be the space of piecewise linears, which contains the exact solution. We therefore expect our Finite Element solution to be exact.

## Whetting your appetite¶

The computation can be set up in about 20 lines of Nutils code, including visualization. The entire script is presented below, in copy-pasteable form suitable for interactive exploration using for example ipython. In the sections that follow we will go over these lines ones by one and explain the relevant concepts involved.

```
>>> from nutils import function, mesh, solver
>>> import numpy
>>> from matplotlib import pyplot as plt
>>> topo, geom = mesh.rectilinear([numpy.linspace(0, 1, 5)])
>>> ns = function.Namespace()
>>> ns.x = geom
>>> ns.basis = topo.basis('spline', degree=1)
>>> ns.u = 'basis_n ?lhs_n'
>>> sqr = topo.boundary['left'].integral('u^2 J(x)' @ ns, degree=2)
>>> cons = solver.optimize('lhs', sqr, droptol=1e-15)
optimize > constrained 1/5 dofs
optimize > optimum value 0.00e+00
>>> res = topo.integral('d(basis_n, x_i) d(u, x_i) J(x)' @ ns, degree=0)
>>> res -= topo.boundary['right'].integral('basis_n J(x)' @ ns, degree=0)
>>> lhs = solver.solve_linear('lhs', residual=res, constrain=cons)
solve > solving 4 dof system to machine precision using arnoldi solver
solve > solver returned with residual 9e-16
>>> bezier = topo.sample('bezier', 32)
>>> nanjoin = lambda array, tri: numpy.insert(array.take(tri.flat, 0).astype(float), slice(tri.shape[1], tri.size, tri.shape[1]), numpy.nan, axis=0)
>>> sampled_x = nanjoin(bezier.eval(ns.x[0]), bezier.tri)
>>> def plot_line(func, **arguments):
... plt.plot(sampled_x, nanjoin(bezier.eval(func, **arguments), bezier.tri))
... plt.xlabel('x_0')
... plt.xticks(numpy.linspace(0, 1, 5))
>>> plot_line(ns.u, lhs=lhs)
```

You are encouraged to execute this code at least once before reading on, as the
code snippets that follow may assume certain products to be present in the
namespace. In particular the `plot_line`

function is used heavily in the
ensuing sections.

## Topology vs Geometry¶

Rather than having a single concept of what is typically referred to as the
‘mesh’, Nutils maintains a strict separation of *topology* and *geometry*. The
`nutils.topology.Topology`

represents a collection of elements and
inter-element connectivity, along with recipes for creating bases. It has no
(public) notion of position. The geometry takes the
`Topology`

and positions it in space. This separation
makes it possible to define multiple geometries belonging to a single
`Topology`

, a feature that is useful for example in
certain Lagrangian formulations.

While not having mesh objects, Nutils does have a `nutils.mesh`

module,
which hosts functions that return tuples of topology and geometry. Nutils
provides two builtin mesh generators: `nutils.mesh.rectilinear()`

, a
generator for structured topologies (i.e. tensor products of one or more
one-dimensional topologies), and `nutils.mesh.unitsquare()`

, a unit square
mesh generator with square or triangular elements or a mixture of both. The
latter is mostly useful for testing. In addition to generators, Nutils also
provides the `nutils.mesh.gmsh()`

importer for gmsh-generated meshes.

The structured mesh generator takes as its first argument a list of element vertices per dimension. A one-dimensional topology with four elements of equal size between 0 and 1 is generated by

```
>>> mesh.rectilinear([[0, 0.25, 0.5, 0.75, 1.0]])
(StructuredTopology<4>, Array<1>)
```

Alternatively we could have used `numpy.linspace()`

to generate a sequence
of equidistant vertices, and unpack the resulting tuple:

```
>>> topo, geom = mesh.rectilinear([numpy.linspace(0, 1, 5)])
```

We will use this topology and geometry throughout the remainder of this tutorial.

Note that the argument is a list of length one: this outer sequence lists the dimensions, the inner the vertices per dimension. To generate a two-dimensional topology, simply add a second list of vertices to the outer list. For example, an equidistant topology with four by eight elements with a unit square geometry is generated by

```
>>> mesh.rectilinear([numpy.linspace(0, 1, 5), numpy.linspace(0, 1, 9)])
(StructuredTopology<4x8>, Array<2>)
```

Any topology defines a boundary via the `Topology.boundary`

attribute. Optionally, a topology can
offer subtopologies via the getitem operator. The rectilinear mesh generator
automatically defines ‘left’ and ‘right’ boundary groups for the first
dimension, making the left boundary accessible as:

```
>>> topo.boundary['left']
StructuredTopology<>
```

Optionally, a topology can be made periodic in one or more dimensions by
passing a list of dimension indices to be periodic via the keyword argument
`periodic`

. For example, to make the second dimension of the above
two-dimensional mesh periodic, add `periodic=[1]`

:

```
>>> mesh.rectilinear([numpy.linspace(0, 1, 5), numpy.linspace(0, 1, 9)], periodic=[1])
(StructuredTopology<4x8p>, Array<2>)
```

Note that in this case the boundary topology, though still available, is empty.

## Bases¶

In Nutils, a *basis* is a vector-valued function object that evaluates, in any
given point \(ξ\) on the topology, to the full array of basis function
values \(φ_0(ξ), φ_1(ξ), \dots, φ_{n-1}(ξ)\). It must be pointed out that
Nutils will in practice operate only on the basis functions that are locally
non-zero, a key optimization in Finite Element computations. But as a concept,
it helps to think of a basis as evaluating always to the full array.

Several `Topology`

objects support creating bases via
the `Topology.basis()`

method. A
`StructuredTopology`

, as generated by
`nutils.mesh.rectilinear()`

, can create a spline basis with arbitrary
degree and arbitrary continuity. The following generates a degree one spline
basis on our previously created unit line topology `topo`

:

```
>>> basis = topo.basis('spline', degree=1)
```

The five basis functions are

```
>>> plot_line(basis)
```

We will use this basis throughout the following sections.

Change the `degree`

argument to `2`

for a quadratic spline basis:

```
>>> plot_line(topo.basis('spline', degree=2))
```

By default the continuity of the spline functions at element edges is the
degree minus one. To change this, pass the desired continuity via keyword
argument `continuity`

. For example, a quadratic spline basis with
\(C^0\) continuity is generated with

```
>>> plot_line(topo.basis('spline', degree=2, continuity=0))
```

\(C^0\) continuous spline bases can also be generated by the `'std'`

basis:

```
>>> plot_line(topo.basis('std', degree=2))
```

The `'std'`

basis is supported by topologies with square and/or triangular
elements without hanging nodes.

Discontinuous basis functions are generated using the `'discont'`

type, e.g.

```
>>> plot_line(topo.basis('discont', degree=2))
```

## Functions¶

A *function* in Nutils is a mapping from a topology onto an n-dimensional
array, and comes in the form of a functions: `nutils.function.Array`

object. It is not to be confused with Python’s own function objects, which
operate on the space of general Python objects. Two examples of Nutils
functions have already made the scene: the geometry `geom`

, as returned by
`nutils.mesh.rectilinear`

, and the bases generated by `Topology.basis()`

. Though seemingly different, these two
constructs are members of the same class and in fact fully interoperable.

The `Array`

functions behave very much like
`numpy.ndarray`

objects: the functions have a
`shape`

, `ndim`

and a
`dtype`

:

```
>>> geom.shape
(1,)
>>> basis.shape
(5,)
>>> geom.ndim
1
>>> geom.dtype
<class 'float'>
```

The functions support numpy-style indexing. For example, to get the first
element of the geometry `geom`

you can write `geom[0]`

and to select the
first two basis functions you can write

```
>>> plot_line(basis[:2])
```

The usual unary and binary operators are available:

```
>>> plot_line(geom[0]*(1-geom[0])/2)
```

Several trigonometric functions are defined in the `nutils.function`

module. An example with a sine function:

```
>>> plot_line(function.sin(2*geom[0]*numpy.pi))
```

The dot product is available via `nutils.function.dot()`

. To contract
the basis with an arbitrary coefficient vector:

```
>>> plot_line(function.dot(basis, [1,2,0,5,4]))
```

Recalling the definition of our discrete solution (2), the above is precisely the way to evaluate the resulting function. What remains now is to establish the coefficients for which this function solves the Laplace problem.

## Namespace¶

Nutils functions behave entirely like Numpy arrays, and can be manipulated as
such, using a combination of operators, object methods, and methods found in
the `nutils.function`

module. Though powerful, the resulting code is often
lengthy, littered with colons and brackets, and hard to read. *Namespaces*
provide an alternative, cleaner syntax for a prominent subset of array
manipulations.

A `nutils.function.Namespace`

is a collection of
`Array`

functions. An empty
`Namespace`

is created as follows:

```
>>> ns = function.Namespace()
```

New entries are added to a `Namespace`

by assigning an
`Array`

to an attribute. For example, to assign the
geometry `geom`

to `ns.x`

, simply type

```
>>> ns.x = geom
```

You can now use `ns.x`

where you would use `geom`

. Similarly, to assign a
linear basis to `ns.basis`

, type

```
>>> ns.basis = topo.basis('spline', degree=1)
```

You can also assign numbers and `numpy.ndarray`

objects:

```
>>> ns.a = 1
>>> ns.b = 2
>>> ns.c = numpy.array([1,2])
>>> ns.A = numpy.array([[1,2],[3,4]])
```

### Expressions¶

In addition to inserting ready objects, a namespace’s real power lies in its
ability to be assigned string expressions. These expressions may reference any
`Array`

function present in the
`Namespace`

, and must explicitly name all array
dimensions, with the object of both aiding readibility and facilitating high
order tensor manipulations. A short explanation of the syntax follows; see
`nutils.expression.parse()`

for the complete documentation.

A *term* is written by joining variables with spaces, optionally preceeded by a
single number, e.g. `2 a b`

. A *fraction* is written as two terms joined by
`/`

, e.g. `2 a / 3 b`

, which is equivalent to `(2 a) / (3 b)`

. An
*addition* or *subtraction* is written as two terms joined by `+`

or `-`

,
respectively, e.g. `1 + a b - 2 b`

. *Exponentation* is written by two
variables or numbers joined by `^`

, e.g. `a^2`

. Several trigonometric
functions are available, e.g. `0.5 sin(a)`

.

Assigning an expression to the namespace is then done as follows.

```
>>> ns.e = '2 a / 3 b'
>>> ns.e = (2*ns.a) / (3*ns.b) # equivalent w/o expression
```

The resulting `ns.e`

is an ordinary `Array`

. Note
that the variables used in the expression should exist in the namespace, not
just as a local variable:

```
>>> localvar = 1
>>> ns.f = '2 localvar'
Traceback (most recent call last):
...
nutils.expression.ExpressionSyntaxError: Unknown variable: 'localvar'.
2 localvar
^^^^^^^^
```

When using arrays in an expression all axes of the arrays should be labelled
with an index, e.g. `2 c_i`

and `c_i A_jk`

. Repeated indices are summed,
e.g. `A_ii`

is the trace of `d`

and `A_ij c_j`

is the matrix-vector
product of `d`

and `c`

. You can also insert a number, e.g. `c_0`

is the
first element of `c`

. All terms in an expression should have the same set of
indices after summation, e.g. it is an error to write `c_i + 1`

.

When assigning an expression with remaining indices to the namespace, the indices should be listed explicitly at the left hand side:

```
>>> ns.f_i = '2 c_i'
>>> ns.f = 2*ns.c # equivalent w/o expression
```

The order of the indices matter: the resulting `Array`

will have its axes ordered by the listed indices. The following three
statements are equivalent:

```
>>> ns.g_ijk = 'c_i A_jk'
>>> ns.g_kji = 'c_k A_ji'
>>> ns.g = ns.c[:,numpy.newaxis,numpy.newaxis]*ns.A[numpy.newaxis,:,:] # equivalent w/o expression
```

Operator `d`

returns the gradient of a variable with respect to a geometry.
e.g. the gradient of the basis is `d(basis_n, x_i)`

and the laplacian
`d(basis_n, x_i, x_i)`

. This works with expressions as well, e.g. ```
d(2
basis_n + basis_n^2, x_i)
```

is the gradient of `2 basis_n + basis_n^2`

.

### Manual evaluation¶

Sometimes it is useful to evaluate an expression to an
`Array`

without inserting the result in the namespace.
For scalar or vector expressions, this can be done using the ```
<expression> @
<namespace>
```

notation. An example with a scalar expression:

```
>>> '2 a / 3 b' @ ns
Array<>
>>> (2*ns.a) / (3*ns.b) # equivalent w/o `... @ ns`
Array<>
```

An example with a vector expression:

```
>>> '2 c_i' @ ns
Array<2>
>>> 2*ns.c # equivalent w/o `... @ ns`
Array<2>
```

If an expression has more than one remaining index, the order of the indices
must be specified explicitly. For this situation there is the
`<namespace>.eval_<indices>(<expression>)`

notation. An example:

```
>>> ns.eval_ijk('c_i A_jk')
Array<2,2,2>
>>> ns.c[:,numpy.newaxis,numpy.newaxis]*ns.A[numpy.newaxis,:,:] # equivalent w/o `ns.eval_...(...)`
Array<2,2,2>
```

### Arguments¶

A discrete model is often written in terms of an unknown, or a vector of
unknowns. In Nutils this translates to a function argument,
`nutils.function.Argument`

. In an expression an
`Argument`

is denoted by a `?`

folowed by an
identifier. For example, the discrete solution (2) can be
written as

```
>>> ns.u = 'basis_n ?lhs_n'
```

with argument `lhs`

the vector of unknowns \(\hat{u}_n\). The shape of
the argument `lhs`

is resolved from the expression. In the above example,
the argument `lhs`

has the same shape as `ns.basis`

.

## Integrals¶

A central operation in any Finite Element application is to integrate a
function over a physical domain. In Nutils, integration starts with the
topology, in particular the `integral()`

method.

The integral method takes a `Array`

function as first
argument and the degree as keyword argument. The function should contain the
Jacobian of the geometry against which the function should be integrated, using
either `nutils.function.J()`

or the `J`

operator in a namespace
expression. For example, the following integrates `1`

against geometry `x`

:

```
>>> I = topo.integral('1 J(x)' @ ns, degree=0)
>>> I
Array<>
```

The resulting `nutils.function.Array`

object is a representation of the
integral, as yet unevaluated. To compute the actual numbers, call the
`Array.eval()`

method:

```
>>> I.eval()
1.0
```

Be careful with including the Jacobian in your integrands. The following two integrals are different:

```
>>> topo.integral('(1 + 1) J(x)' @ ns, degree=0).eval()
2.0
>>> topo.integral('1 + 1 J(x)' @ ns, degree=0).eval()
5.0
```

Like any other `Array`

, the integrals can be added or
subtracted:

```
>>> J = topo.integral('x_0 J(x)' @ ns, degree=1)
>>> (I+J).eval()
1.5
```

Recall that a topology boundary is also a `Topology`

object, and hence it supports integration. For example, to integrate the
geometry `x`

over the entire boundary, write

```
>>> topo.boundary.integral('x_0 J(x)' @ ns, degree=1).eval()
1.0
```

To limit the integral to the right boundary, write

```
>>> topo.boundary['right'].integral('x_0 J(x)' @ ns, degree=1).eval()
1.0
```

Note that this boundary is simply a point and the integral a point evaluation.

Integrating and evaluating a 1D `Array`

results in a 1D
`numpy.ndarray`

:

```
>>> topo.integral('basis_i J(x)' @ ns, degree=1).eval()
array([0.125, 0.25 , 0.25 , 0.25 , 0.125])
```

Since the integrals of 2D `Array`

functions are usually
sparse, the `Array.eval()`

method does
not return a dense `numpy.ndarray`

, but a Nutils sparse matrix object: a
subclass of `nutils.matrix.Matrix`

. Nutils interfaces several linear
solvers (more on this in Section Solvers below) but if you want to use a
custom solver you can export the matrix to a dense, compressed sparse row or
coordinate representation via the `Matrix.export()`

method. An example:

```
>>> M = topo.integral(ns.eval_nm('d(basis_n, x_i) d(basis_m, x_i) J(x)'), degree=1).eval()
>>> M.export('dense')
array([[ 4., -4., 0., 0., 0.],
[-4., 8., -4., 0., 0.],
[ 0., -4., 8., -4., 0.],
[ 0., 0., -4., 8., -4.],
[ 0., 0., 0., -4., 4.]])
>>> M.export('csr') # (data, column indices, row pointers)
(array([ 4., -4., -4., 8., -4., -4., 8., -4., -4., 8., -4., -4., 4.]),
array([0, 1, 0, 1, 2, 1, 2, 3, 2, 3, 4, 3, 4]),
array([ 0, 2, 5, 8, 11, 13]))
>>> M.export('coo') # (data, (row indices, column indices))
(array([ 4., -4., -4., 8., -4., -4., 8., -4., -4., 8., -4., -4., 4.]),
(array([0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4]),
array([0, 1, 0, 1, 2, 1, 2, 3, 2, 3, 4, 3, 4])))
```

## Solvers¶

Using topologies, bases and integrals, we now have the tools in place to start performing some actual functional-analytical operations. We start with what is perhaps the simplest of its kind, the least squares projection, demonstrating the different implementations now available to us and working our way up from there.

Taking the geometry component \(x_0\) as an example, to project it onto the basis \(\{φ_n\}\) means finding the coefficients \(\hat{u}_n\) such that

for all \(φ_n\), or \(A_{nm} \hat{u}_m = f_n\). This is implemented as follows:

```
>>> A = topo.integral(ns.eval_nm('basis_n basis_m J(x)'), degree=2).eval()
>>> f = topo.integral('basis_n x_0 J(x)' @ ns, degree=2).eval()
>>> A.solve(f)
solve > solving 5 dof system to machine precision using arnoldi solver
solve > solver returned with residual 3e-17
array([0. , 0.25, 0.5 , 0.75, 1. ])
```

Alternatively, we can write this in the slightly more general form

```
>>> res = topo.integral('basis_n (u - x_0) J(x)' @ ns, degree=2)
```

Taking the derivative of \(R_n\) to \(\hat{u}_m\) gives the above
matrix \(A_{nm}\), and substituting for \(\hat{u}\) the zero vector
yields \(-f_n\). Nutils can compute those derivatives for you, using the
method `Array.derivative()`

to
compute the derivative with respect to an `Argument`

,
returning a new `Array`

.

```
>>> A = res.derivative('lhs').eval()
>>> f = -res.eval(lhs=numpy.zeros(5))
>>> A.solve(f)
solve > solving 5 dof system to machine precision using arnoldi solver
solve > solver returned with residual 3e-17
array([0. , 0.25, 0.5 , 0.75, 1. ])
```

The above three lines are so common that they are combined in the function
`nutils.solver.solve_linear()`

:

```
>>> solver.solve_linear('lhs', res)
solve > solving 5 dof system to machine precision using arnoldi solver
solve > solver returned with residual 3e-17
array([0. , 0.25, 0.5 , 0.75, 1. ])
```

We can take this formulation one step further. Minimizing

for \(\hat{u}\) is equivalent to the above two variants. The derivative of \(S\) to \(\hat{u}_n\) gives \(2 R_n\):

```
>>> sqr = topo.integral('(u - x_0)^2 J(x)' @ ns, degree=2)
>>> solver.solve_linear('lhs', sqr.derivative('lhs'))
solve > solving 5 dof system to machine precision using arnoldi solver
solve > solver returned with residual 6e-17
array([0. , 0.25, 0.5 , 0.75, 1. ])
```

The optimization problem can also be solved by the
`nutils.solver.optimize()`

function, which has the added benefit that
\(S\) may be nonlinear in \(\hat{u}\) — a property not used here.

```
>>> solver.optimize('lhs', sqr)
optimize > solve > solving 5 dof system to machine precision using arnoldi solver
optimize > solve > solver returned with residual 0e+00
optimize > optimum value 0.00e+00
array([0. , 0.25, 0.5 , 0.75, 1. ])
```

Nutils also supports solving a partial optimization problem. In the Laplace problem stated above, the Dirichlet boundary condition at \(Γ_\text{left}\) minimizes the following functional:

```
>>> sqr = topo.boundary['left'].integral('(u - 0)^2 J(x)' @ ns, degree=2)
```

By passing the `droptol`

argument, `nutils.solver.optimize()`

returns an
array with `nan`

(‘not a number’) for every entry for which the optimization
problem is invariant, or to be precise, where the variation is below
`droptol`

:

```
>>> cons = solver.optimize('lhs', sqr, droptol=1e-15)
optimize > constrained 1/5 dofs
optimize > optimum value 0.00e+00
>>> cons
array([ 0., nan, nan, nan, nan])
```

Consider again the Laplace problem stated above. The residual (1) is implemented as

```
>>> res = topo.integral('d(basis_n, x_i) d(u, x_i) J(x)' @ ns, degree=0)
>>> res -= topo.boundary['right'].integral('basis_n J(x)' @ ns, degree=0)
```

Since this problem is linear in argument `lhs`

, we can use the
`nutils.solver.solve_linear()`

method to solve this problem. The
constraints `cons`

are passed via the keyword argument `constrain`

:

```
>>> lhs = solver.solve_linear('lhs', res, constrain=cons)
solve > solving 4 dof system to machine precision using arnoldi solver
solve > solver returned with residual 9e-16
>>> lhs
array([0. , 0.25, 0.5 , 0.75, 1. ])
```

For nonlinear residuals you can use `nutils.solver.newton`

.

## Sampling¶

Having obtained the coefficient vector that solves the Laplace problem, we are
now interested in visualizing the function it represents. Nutils does not
provide its own post processing functionality, leaving that up to the
preference of the user. It does, however, facilitate it, by allowing
`Array`

functions to be evaluated in samples. Bundling
function values and a notion of connectivity, these form a bridge between
Nutils’ world of functions and the discrete realms of matplotlib, VTK, etc.

The `Topology.sample(method, ...)`

method generates a collection of points on the
`Topology`

, according to `method`

. The `'bezier'`

method generates equidistant points per element, including the element
vertices. The number of points per element per dimension is controlled by the
second argument of `Topology.sample()`

. An example:

```
>>> bezier = topo.sample('bezier', 2)
```

The resulting `nutils.sample.Sample`

object can be used to evaluate
`Array`

functions via the `Sample.eval(func)`

method. To evaluate the geometry `ns.x`

write

```
>>> x = bezier.eval('x_0' @ ns)
>>> x
array([0. , 0.25, 0.25, 0.5 , 0.5 , 0.75, 0.75, 1. ])
```

The first axis of the returned `numpy.ndarray`

represents the collection
of points. To reorder this into a sequence of lines in 1D, a triangulation in
2D or in general a sequence of simplices, use the `Sample.tri`

attribute:

```
>>> x.take(bezier.tri, 0)
array([[0. , 0.25],
[0.25, 0.5 ],
[0.5 , 0.75],
[0.75, 1. ]])
```

Now, the first axis represents the simplices and the second axis the vertices of the simplices.

If an `Array`

function has arguments, those arguments
must be specified by keyword arguments to `Sample.eval()`

. For example, to evaluate `ns.u`

with argument
`lhs`

replaced by solution vector `lhs`

, obtained using
`nutils.solver.solve_linear()`

above, write

```
>>> u = bezier.eval('u' @ ns, lhs=lhs)
>>> u
array([0. , 0.25, 0.25, 0.5 , 0.5 , 0.75, 0.75, 1. ])
```

We can now plot the sampled geometry `x`

and solution `u`

using
matplotlib, plotting each line in `Sample.tri`

with a different color:

```
>>> plt.plot(x.take(bezier.tri.T, 0), u.take(bezier.tri.T, 0))
[...]
```

Recall that we have imported `matplotlib.pyplot`

as `plt`

above. The
`plt.plot()`

function takes an array of x-values
and and array of y-values, both with the first axis representing vertices and
the second representing separate lines, hence the transpose of `bezier.tri`

.

The `plt.plot()`

function also supports plotting
lines with discontinuities, which are represented by `nan`

values. We can
use this to plot the solution as a single, but possibly discontinuous line.
The function `numpy.insert()`

can be used to prepare a suitable array. An
example:

```
>>> nanjoin = lambda array, tri: numpy.insert(array.take(tri.flat, 0).astype(float), slice(tri.shape[1], tri.size, tri.shape[1]), numpy.nan, axis=0)
>>> nanjoin(x, bezier.tri)
array([0. , 0.25, nan, 0.25, 0.5 , nan, 0.5 , 0.75, nan, 0.75, 1. ])
>>> plt.plot(nanjoin(x, bezier.tri), nanjoin(u, bezier.tri))
[...]
```

Note the difference in colors between the last two plots.

## Two-dimensional Laplace problem¶

All of the above was written for a one-dimensional example. We now extend the Laplace problem to two dimensions and highlight the changes to the corresponding Nutils implementation. Let \(Ω\) be a unit square with boundary \(Γ\), on which the following boundary conditions apply:

The 2D homogeneous Laplace solution is the field \(u\) for which \(R(v, u) = 0\) for all v, where

Adopting a Finite Element basis \(\{φ_n\}\) we obtain the discrete solution \(\hat{u}(x) = φ_n(x) \hat{u}_n\) and the system of equations \(R(φ_n, \hat{u}) = 0\).

Following the same steps as in the 1D case, a unit square mesh with 10x10
elements is formed using `nutils.mesh.rectilinear()`

:

```
>>> nelems = 10
>>> topo, geom = mesh.rectilinear([numpy.linspace(0, 1, nelems+1), numpy.linspace(0, 1, nelems+1)])
```

Recall that `nutils.mesh.rectilinear()`

takes a list of element vertices
per dimension. Alternatively you can create a unit square mesh using
`nutils.mesh.unitsquare()`

, specifying the number of elements per dimension
and the element type:

```
>>> topo, geom = mesh.unitsquare(nelems, 'square')
```

The above two statements generate exactly the same topology and geometry. Try
replacing `'square'`

with `'triangle'`

or `'mixed'`

to generate a unit
square mesh with triangular elements or a mixture of square and triangular
elements, respectively.

We start with a clean namespace, assign the geometry to `ns.x`

, create a
linear basis and define the solution `ns.u`

as the contraction of the basis
with argument `lhs`

.

```
>>> ns = function.Namespace()
>>> ns.x = geom
>>> ns.basis = topo.basis('std', degree=1)
>>> ns.u = 'basis_n ?lhs_n'
```

Note that the above statements are identical to those of the one-dimensional example.

The residual (3) is implemented as

```
>>> res = topo.integral('d(basis_n, x_i) d(u, x_i) J(x)' @ ns, degree=2)
>>> res -= topo.boundary['right'].integral('basis_n cos(1) cosh(x_1) J(x)' @ ns, degree=2)
```

The Dirichlet boundary conditions are rewritten as a least squares problem and
solved for `lhs`

, yielding the constraints vector `cons`

:

```
>>> sqr = topo.boundary['left'].integral('u^2 J(x)' @ ns, degree=2)
>>> sqr += topo.boundary['top'].integral('(u - cosh(1) sin(x_0))^2 J(x)' @ ns, degree=2)
>>> cons = solver.optimize('lhs', sqr, droptol=1e-15)
optimize > solve > solving 21 dof system to machine precision using arnoldi solver
optimize > solve > solver returned with residual 3e-17
optimize > constrained 21/121 dofs
optimize > optimum value 4.32e-10
```

To solve the problem `res=0`

for `lhs`

subject to `lhs=cons`

excluding
the `nan`

values, we can use `nutils.solver.solve_linear()`

:

```
>>> lhs = solver.solve_linear('lhs', res, constrain=cons)
solve > solving 100 dof system to machine precision using arnoldi solver
solve > solver returned with residual 2e-15
```

Finally, we plot the solution. We create a `Sample`

object from `topo`

and evaluate the geometry and the solution:

```
>>> bezier = topo.sample('bezier', 9)
>>> x, u = bezier.eval(['x', 'u'] @ ns, lhs=lhs)
```

We use `plt.tripcolor`

to plot the sampled
`x`

and `u`

:

```
>>> plt.tripcolor(x[:,0], x[:,1], bezier.tri, u, shading='gouraud', rasterized=True)
<...>
>>> plt.colorbar()
<...>
>>> plt.gca().set_aspect('equal')
>>> plt.xlabel('x_0')
Text(...)
>>> plt.ylabel('x_1')
Text(...)
```

This two-dimensional example is also available as script: laplace.py.