<br>

<i>
This program was contributed by Natasha Sharma, Guido Kanschat, Timo
Heister, Wolfgang Bangerth, and Zhuoran Wang.

The first author would like to acknowledge the support of NSF Grant
No. DMS-1520862.
Timo Heister and Wolfgang Bangerth acknowledge support through NSF
awards DMS-1821210, EAR-1550901, and OAC-1835673.
</i>

<a name="step-47-Intro"></a>
<h1>Introduction</h1>

This program deals with the <a
href="https://en.wikipedia.org/wiki/Biharmonic_equation">biharmonic
equation</a>,
@f{align*}{
  \Delta^2 u(\mathbf x) &= f(\mathbf x)
  \qquad \qquad &&\forall \mathbf x \in \Omega.
@f}
This equation appears in the modeling of thin structures such as roofs
of stadiums. These objects are of course in reality
three-dimensional with a large aspect ratio of lateral extent to
perpendicular thickness, but one can often very accurately model these
structures as two dimensional by making assumptions about how internal
forces vary in the perpendicular direction. These assumptions lead to the
equation above.

The model typically comes in two different kinds, depending on what
kinds of boundary conditions are imposed. The first case,
@f{align*}{
  u(\mathbf x) &= g(\mathbf x) \qquad \qquad
  &&\forall \mathbf x \in \partial\Omega, \\
  \Delta u(\mathbf x) &= h(\mathbf x) \qquad \qquad
  &&\forall \mathbf x \in \partial\Omega,
@f}
corresponds to the edges of the thin structure attached to the top of
a wall of height $g(\mathbf x)$ in such a way that the bending forces
that act on the structure are $h(\mathbf x)$; in most physical
situations, one will have $h=0$, corresponding to the structure simply
sitting atop the wall.

In the second possible case of boundary values, one would have
@f{align*}{
  u(\mathbf x) &= g(\mathbf x) \qquad \qquad
  &&\forall \mathbf x \in \partial\Omega, \\
  \frac{\partial u(\mathbf x)}{\partial \mathbf n} &= j(\mathbf x) \qquad \qquad
  &&\forall \mathbf x \in \partial\Omega.
@f}
This corresponds to a "clamped" structure for which a nonzero
$j(\mathbf x)$ implies a certain angle against the horizontal.

As with Dirichlet and Neumann boundary conditions for the Laplace
equation, it is of course possible to have one kind of boundary
conditions on one part of the boundary, and the other on the
remainder.


<h3> What's the issue? </h3>

The fundamental issue with the equation is that it takes four
derivatives of the solution. In the case of the Laplace equation
we treated in step-3, step-4, and several other tutorial programs,
one multiplies by a test function, integrates, integrates by parts,
and ends up with only one derivative on both the test function and
trial function -- something one can do with functions that are
continuous globally, but may have kinks at the interfaces between
cells: The derivative may not be defined at the interfaces, but
that is on a lower-dimensional manifold (and so doesn't show up
in the integrated value).

But for the biharmonic equation, if one followed the same procedure
using integrals over the entire domain (i.e., the union of all cells),
one would end up with two derivatives on the test functions and trial
functions each. If one were to use the usual piecewise polynomial
functions with their kinks on cell interfaces, the first derivative
would yield a discontinuous gradient, and the second derivative with
delta functions on the interfaces -- but because both the second
derivatives of the test functions and of the trial functions yield a
delta function, we would try to integrate the product of two delta
functions. For example, in 1d, where $\varphi_i$ are the usual
piecewise linear "hat functions", we would get integrals of the sort
@f{align*}{
  \int_0^L (\Delta \varphi_i) (\Delta \varphi_j)
  =
  \int_0^L
  \frac 1h \left[\delta(x-x_{i-1}) - 2\delta(x-x_i) + \delta(x-x_{i+1})\right]
  \frac 1h \left[\delta(x-x_{j-1}) - 2\delta(x-x_j) + \delta(x-x_{j+1})\right]
@f}
where $x_i$ is the node location at which the shape function
$\varphi_i$ is defined, and $h$ is the mesh size (assumed
uniform). The problem is that delta functions in integrals are defined
using the relationship
@f{align*}{
  \int_0^L \delta(x-\hat x) f(x) \; dx
  =
  f(\hat x).
@f}
But that only works if (i) $f(\cdot)$ is actually well defined at
$\hat x$, and (ii) if it is finite. On the other hand, an integral of
the form
@f{align*}{
\int_0^L \delta(x-x_i) \delta (x-x_i)
@f}
does not make sense. Similar reasoning can be applied for 2d and 3d
situations.

In other words: This approach of trying to integrate over the entire
domain and then integrating by parts can't work.

Historically, numerical analysts have tried to address this by
inventing finite elements that are "C<sup>1</sup> continuous", i.e., that use
shape functions that are not just continuous but also have continuous
first derivatives. This is the realm of elements such as the Argyris
element, the Clough-Tocher element and others, all developed in the
late 1960s. From a twenty-first century perspective, they can only be
described as bizarre in their construction. They are also exceedingly
cumbersome to implement if one wants to use general meshes. As a
consequence, they have largely fallen out of favor and deal.II currently
does not contain implementations of these shape functions.


<h3> What to do instead? </h3>

So how does one approach solving such problems then? That depends a
bit on the boundary conditions. If one has the first set of boundary
conditions, i.e., if the equation is
@f{align*}{
  \Delta^2 u(\mathbf x) &= f(\mathbf x)
  \qquad \qquad &&\forall \mathbf x \in \Omega, \\
  u(\mathbf x) &= g(\mathbf x) \qquad \qquad
  &&\forall \mathbf x \in \partial\Omega, \\
  \Delta u(\mathbf x) &= h(\mathbf x) \qquad \qquad
  &&\forall \mathbf x \in \partial\Omega,
@f}
then the following trick works (at least if the domain is convex, see
below): In the same way as we obtained the
mixed Laplace equation of step-20 from the regular Laplace equation by
introducing a second variable, we can here introduce a variable
$v=\Delta u$ and can then replace the equations above by the
following, "mixed" system:
@f{align*}{
  -\Delta u(\mathbf x) +v(\mathbf x) &= 0
  \qquad \qquad &&\forall \mathbf x \in \Omega, \\
  -\Delta v(\mathbf x) &= -f(\mathbf x)
  \qquad \qquad &&\forall \mathbf x \in \Omega, \\
  u(\mathbf x) &= g(\mathbf x) \qquad \qquad
  &&\forall \mathbf x \in \partial\Omega, \\
  v(\mathbf x) &= h(\mathbf x) \qquad \qquad
  &&\forall \mathbf x \in \partial\Omega.
@f}
In other words, we end up with what is in essence a system of two
coupled Laplace equations for $u,v$, each with Dirichlet-type boundary
conditions. We know how to solve such problems, and it should not be
very difficult to construct good solvers and preconditioners for this
system either using the techniques of step-20 or step-22. So this
case is pretty simple to deal with.

@note It is worth pointing out that this only works for domains whose
  boundary has corners if the domain is also convex -- in other words,
  if there are no re-entrant corners.
  This sounds like a rather random condition, but it makes
  sense in view of the following two facts: The solution of the
  original biharmonic equation must satisfy $u\in H^2(\Omega)$. On the
  other hand, the mixed system reformulation above suggests that both
  $u$ and $v$ satisfy $u,v\in H^1(\Omega)$ because both variables only
  solve a Poisson equation. In other words, if we want to ensure that
  the solution $u$ of the mixed problem is also a solution of the
  original biharmonic equation, then we need to be able to somehow
  guarantee that the solution of $-\Delta u=v$ is in fact more smooth
  than just $H^1(\Omega)$. This can be argued as follows: For convex
  domains,
  <a href="https://en.wikipedia.org/wiki/Elliptic_operator#Elliptic_regularity_theorem">"elliptic
  regularity"</a> implies that if the right hand side $v\in H^s$, then
  $u\in H^{s+2}$ if the domain is convex and the boundary is smooth
  enough. (This could also be guaranteed if the domain boundary is
  sufficiently smooth -- but domains whose boundaries have no corners
  are not very practical in real life.)
  We know that $v\in H^1$ because it solves the equation
  $-\Delta v=f$, but we are still left with the condition on convexity
  of the boundary; one can show that polygonal, convex domains are
  good enough to guarantee that $u\in H^2$ in this case (smoothly
  bounded, convex domains would result in $u\in H^3$, but we don't
  need this much regularity). On the other hand, if the domain is not
  convex, we can not guarantee that the solution of the mixed system
  is in $H^2$, and consequently may obtain a solution that can't be
  equal to the solution of the original biharmonic equation.

The more complicated situation is if we have the "clamped" boundary
conditions, i.e., if the equation looks like this:
@f{align*}{
  \Delta^2 u(\mathbf x) &= f(\mathbf x)
  \qquad \qquad &&\forall \mathbf x \in \Omega, \\
  u(\mathbf x) &= g(\mathbf x) \qquad \qquad
  &&\forall \mathbf x \in \partial\Omega, \\
  \frac{\partial u(\mathbf x)}{\partial \mathbf n} &= j(\mathbf x) \qquad \qquad
  &&\forall \mathbf x \in \partial\Omega.
@f}
The same trick with the mixed system does not work here, because we
would end up with <i>both</i> Dirichlet and Neumann boundary conditions for
$u$, but none for $v$.


The solution to this conundrum arrived with the Discontinuous Galerkin
method wave in the 1990s and early 2000s: In much the same way as one
can use <i>discontinuous</i> shape functions for the Laplace equation
by penalizing the size of the discontinuity to obtain a scheme for an
equation that has one derivative on each shape function, we can use a
scheme that uses <i>continuous</i> (but not $C^1$ continuous) shape
functions and penalize the jump in the derivative to obtain a scheme
for an equation that has two derivatives on each shape function. In
analogy to the Interior Penalty (IP) method for the Laplace equation,
this scheme for the biharmonic equation is typically called the $C^0$ IP
(or C0IP) method, since it uses $C^0$ (continuous but not continuously
differentiable) shape functions with an interior penalty formulation.

It is worth noting that the C0IP method is not the only one that has
been developed for the biharmonic equation. step-82 shows an
alternative method.


<h3> Derivation of the C0IP method </h3>

We base this program on the $C^0$ IP method presented by Susanne
Brenner and Li-Yeng Sung in the paper "C$^0$ Interior Penalty Method
for Linear Fourth Order Boundary Value Problems on polygonal
domains"
@cite Brenner2005
where the method is
derived for the biharmonic equation with "clamped" boundary
conditions.

As mentioned, this method relies on the use of $C^0$ Lagrange finite
elements where the $C^1$ continuity requirement is relaxed and has
been replaced with interior penalty techniques. To derive this method,
we consider a $C^0$ shape function $v_h$ which vanishes on
$\partial\Omega$. We introduce notation $ \mathbb{F} $ as the set of
all faces of $\mathbb{T}$, $ \mathbb{F}^b $ as the set of boundary faces,
and $ \mathbb{F}^i $ as the set of interior faces for use further down below.
Since the higher order derivatives of $v_h$ have two
values on each interface $e\in \mathbb{F}$ (shared by the two cells
$K_{+},K_{-} \in \mathbb{T}$), we cope with this discontinuity by
defining the following single-valued functions on $e$:
@f{align*}{
  \jump{\frac{\partial^k v_h}{\partial \mathbf n^k}}
  &=
  \frac{\partial^k v_h|_{K_+}}{\partial \mathbf n^k} \bigg |_e
  - \frac{\partial^k v_h|_{K_-}}{\partial \mathbf n^k} \bigg |_e,
  \\
  \average{\frac{\partial^k v_h}{\partial \mathbf n^k}}
  &=
  \frac{1}{2}
  \bigg( \frac{\partial^k v_h|_{K_+}}{\partial \mathbf n^k} \bigg |_e
  + \frac{\partial^k v_h|_{K_-}}{\partial \mathbf n^k} \bigg |_e \bigg )
@f}
for $k =1,2$ (i.e., for the gradient and the matrix of second
derivatives), and where $\mathbf n$ denotes a unit vector normal to
$e$ pointing from $K_+$ to $K_-$. In the
literature, these functions are referred to as the "jump" and
"average" operations, respectively.

To obtain the $C^0$ IP approximation $u_h$, we left multiply the
biharmonic equation by $v_h$, and then integrate over $\Omega$. As
explained above, we can't do the integration by parts on all of
$\Omega$ with these shape functions, but we can do it on each cell
individually since the shape functions are just polynomials on each
cell. Consequently, we start by using the following
integration-by-parts formula on each mesh cell $K \in {\mathbb{T}}$:
@f{align*}{
  \int_K v_h (\Delta^2 w_h)
  &= \int_K v_h (\nabla\cdot\nabla) (\Delta w_h)
  \\
  &= -\int_K \nabla v_h \cdot (\nabla \Delta w_h)
     +\int_{\partial K} v_h (\nabla \Delta w_h \cdot \mathbf n).
@f}
At this point, we have two options: We can integrate the domain term's
$\nabla\Delta w_h$ one more time to obtain
@f{align*}{
  \int_K v_h (\Delta^2 w_h)
  &= \int_K (\Delta v_h) (\Delta w_h)
     +\int_{\partial K} v_h (\nabla \Delta w_h \cdot \mathbf n)
     -\int_{\partial K} (\nabla v_h \cdot \mathbf n) \Delta w_h.
@f}
For a variety of reasons, this turns out to be a variation that is not
useful for our purposes.

Instead, what we do is recognize that
$\nabla\Delta w_h = \text{grad}\,(\text{div}\,\text{grad}\, w_h)$, and we
can re-sort these operations as
$\nabla\Delta w_h = \text{div}\,(\text{grad}\,\text{grad}\, w_h)$ where we
typically write $\text{grad}\,\text{grad}\, w_h = D^2 w_h$ to indicate
that this is the "Hessian" matrix of second derivatives. With this
re-ordering, we can now integrate the divergence, rather than the
gradient operator, and we get the following instead:
@f{align*}{
  \int_K v_h (\Delta^2 w_h)
  &= \int_K (\nabla \nabla v_h) : (\nabla \nabla w_h)
     +\int_{\partial K} v_h (\nabla \Delta w_h \cdot \mathbf n)
     -\int_{\partial K} (\nabla v_h \otimes \mathbf n) : (\nabla\nabla w_h)
  \\
  &= \int_K (D^2 v_h) : (D^2 w_h)
     +\int_{\partial K} v_h (\nabla \Delta w_h \cdot \mathbf n)
     -\int_{\partial K} (\nabla v_h) \cdot (D^2 w_h \mathbf n).
@f}
Here, the colon indicates a double-contraction over the indices of the
matrices to its left and right, i.e., the scalar product between two
tensors. The outer product of two vectors $a \otimes b$ yields the
matrix $(a \otimes b)_{ij} = a_i b_j$.

Then, we sum over all cells $K \in  \mathbb{T}$, and take into account
that this means that every interior face appears twice in the
sum. If we therefore split everything into a sum of integrals over
cell interiors and a separate sum over cell interfaces, we can use
the jump and average operators defined above. There are two steps
left: First, because our shape functions are continuous, the gradients
of the shape functions may be discontinuous, but the continuity
guarantees that really only the normal component of the gradient is
discontinuous across faces whereas the tangential component(s) are
continuous. Second, the discrete formulation that results is not
stable as the mesh size goes to zero, and to obtain a stable
formulation that converges to the correct solution, we need to add
the following terms:
@f{align*}{
-\sum_{e \in \mathbb{F}} \int_{e}
  \average{\frac{\partial^2 v_h}{\partial \mathbf n^2}}
  \jump{\frac{\partial u_h}{\partial \mathbf n}}
+ \sum_{e \in \mathbb{F}}
  \frac{\gamma}{h_e}\int_e
  \jump{\frac{\partial v_h}{\partial \mathbf n}}
  \jump{\frac{\partial u_h}{\partial \mathbf n}}.
@f}
Then, after making cancellations that arise, we arrive at the following
C0IP formulation of the biharmonic equation: find $u_h$ such that $u_h =
g$ on $\partial \Omega$ and
@f{align*}{
\mathcal{A}(v_h,u_h)&=\mathcal{F}(v_h) \quad \text{holds for all test functions } v_h,
@f}
where
@f{align*}{
\mathcal{A}(v_h,u_h):=&\sum_{K \in \mathbb{T}}\int_K D^2v_h:D^2u_h \ dx
\\
&
 -\sum_{e \in \mathbb{F}} \int_{e}
  \jump{\frac{\partial v_h}{\partial \mathbf n}}
  \average{\frac{\partial^2 u_h}{\partial \mathbf n^2}} \ ds
 -\sum_{e \in \mathbb{F}} \int_{e}
 \average{\frac{\partial^2 v_h}{\partial \mathbf n^2}}
 \jump{\frac{\partial u_h}{\partial \mathbf n}} \ ds
\\
&+ \sum_{e \in \mathbb{F}}
 \frac{\gamma}{h_e}
 \int_e
 \jump{\frac{\partial v_h}{\partial \mathbf n}}
 \jump{\frac{\partial u_h}{\partial \mathbf n}} \ ds,
@f}
and
@f{align*}{
\mathcal{F}(v_h)&:=\sum_{K \in \mathbb{T}}\int_{K} v_h f \ dx
-
\sum_{e \in \mathbb{F}, e\subset\partial\Omega}
\int_e \average{\frac{\partial^2 v_h}{\partial \mathbf n^2}} j \ ds
+
\sum_{e \in \mathbb{F}, e\subset\partial\Omega}
\frac{\gamma}{h_e}
\int_e \jump{\frac{\partial v_h}{\partial \mathbf n}} j \ ds.
@f}
Here, $\gamma$ is the penalty parameter which both weakly enforces the
boundary condition
@f{align*}{
\frac{\partial u(\mathbf x)}{\partial \mathbf n} = j(\mathbf x)
@f}
on the boundary interfaces $e \in \mathbb{F}^b$, and also ensures that
in the limit $h\rightarrow 0$, $u_h$ converges to a $C^1$ continuous
function. $\gamma$ is chosen to be large enough to guarantee the
stability of the method. We will discuss our choice in the program below.


<h4>Convergence Rates </h4>
On polygonal domains, the weak solution $u$ to the biharmonic equation
lives in $H^{2 +\alpha}(\Omega)$ where $\alpha \in(1/2, 2]$ is
determined by the interior angles at the corners of $\Omega$. For
instance, whenever $\Omega$ is convex, $\alpha=1$; $\alpha$ may be less
than one if the domain has re-entrant corners but
$\alpha$ is close to $1$ if one of all interior angles is close to
$\pi$.

Now suppose that the $C^0$ IP solution $u_h$ is approximated by $C^0$
shape functions with polynomial degree $p \ge 2$. Then the
discretization outlined above yields the convergence rates as
discussed below.


<b>Convergence in the $C^0$ IP-norm</b>

Ideally, we would like to measure convergence in the "energy norm"
$\|D^2(u-u_h)\|$. However, this does not work because, again, the
discrete solution $u_h$ does not have two (weak) derivatives. Instead,
one can define a discrete ($C^0$ IP) seminorm that is "equivalent" to the
energy norm, as follows:
@f{align*}{
 |u_h|_{h}^2 :=
 \sum\limits_{K \in \mathbb{T}} \big|u_h\big|_{H^2(K)}^2
 +
 \sum\limits_{e \in \mathbb{F} }
 \frac{\gamma }{h_e} \left\|
 \jump{\frac{\partial u_h}{\partial \mathbf n}} \right\|_{L^2(e)}^2.
@f}

In this seminorm, the theory in the paper mentioned above yields that we
can expect
@f{align*}{
 |u-u_h|_{h}^2 = {\cal O}(h^{p-1}),
@f}
much as one would expect given the convergence rates we know are true
for the usual discretizations of the Laplace equation.

Of course, this is true only if the exact solution is sufficiently
smooth. Indeed, if $f \in H^m(\Omega)$ with $m \ge 0$,
$u \in H^{2+\alpha}(\Omega)$ where $ 2 < 2+\alpha  \le m+4$,
then the convergence rate of the $C^0$ IP method is
$\mathcal{O}(h^{\min\{p-1, \alpha\}})$. In other words, the optimal
convergence rate can only be expected if the solution is so smooth
that $\alpha\ge p-1$; this can
only happen if (i) the domain is convex with a sufficiently smooth
boundary, and (ii) $m\ge p-3$. In practice, of course, the solution is
what it is (independent of the polynomial degree we choose), and the
last condition can then equivalently be read as saying that there is
definitely no point in choosing $p$ large if $m$ is not also
large. In other words, the only reasonably choices for $p$ are $p\le
m+3$ because larger polynomial degrees do not result in higher
convergence orders.

For the purposes of this program, we're a bit too lazy to actually
implement this equivalent seminorm -- though it's not very difficult and
would make for a good exercise. Instead, we'll simply check in the
program what the "broken" $H^2$ seminorm
@f{align*}{
 \left(|u_h|^\circ_{H^2}\right)^2
 :=
 \sum\limits_{K \in \mathbb{T}} \big|u_h\big|_{H^2(K)}^2
 =
 \sum\limits_{K \in \mathbb{T}} \big|D^2 u_h\big|_{L_2}^2
@f}
yields. The convergence rate in this norm can, from a theoretical
perspective, of course not be <i>worse</i> than the one for
$|\cdot|_h$ because it contains only a subset of the necessary terms,
but it could at least conceivably be better. It could also be the case that
we get the optimal convergence rate even though there is a bug in the
program, and that that bug would only show up in sub-optimal rates for
the additional terms present in $|\cdot|_h$. But, one might hope
that if we get the optimal rate in the broken norm and the norms
discussed below, then the program is indeed correct. The results
section will demonstrate that we obtain optimal rates in all norms
shown.


<b>Convergence in the $L_2$-norm</b>

The optimal convergence rate in the $L_2$-norm is $\mathcal{O}(h^{p+1})$
provided $p \ge 3$. More details can be found in Theorem 4.6 of
@cite Engel2002 .

The default in the program below is to choose $p=2$. In that case, the
theorem does not apply, and indeed one only gets $\mathcal{O}(h^2)$
instead of $\mathcal{O}(h^3)$ as we will show in the results section.


<b>Convergence in the $H^1$-seminorm</b>

Given that we expect
$\mathcal{O}(h^{p-1})$ in the best of cases for a norm equivalent to
the $H^2$ seminorm, and $\mathcal{O}(h^{p+1})$ for the $L_2$ norm, one
may ask about what happens in the $H^1$ seminorm that is intermediate
to the two others. A reasonable guess is that one should expect
$\mathcal{O}(h^{p})$. There is probably a paper somewhere that proves
this, but we also verify that this conjecture is experimentally true
below.



<h3>Other Boundary Conditions</h3>

We remark that the derivation of the $C^0$ IP method for the
biharmonic equation with other boundary conditions -- for instance,
for the first set of boundary conditions namely $u(\mathbf x) =
g(\mathbf x)$ and $\Delta u(\mathbf x)= h(\mathbf x)$ on
$\partial\Omega$ -- can be obtained with suitable modifications to
$\mathcal{A}(\cdot,\cdot)$ and $\mathcal{F}(\cdot)$ described in
the book chapter @cite Brenner2011 .


<h3>The testcase</h3>

The last step that remains to describe is what this program solves
for. As always, a trigonometric function is both a good and a bad
choice because it does not lie in any polynomial space in which we may
seek the solution while at the same time being smoother than real
solutions typically are (here, it is in $C^\infty$ while real
solutions are typically only in $H^3$ or so on convex polygonal
domains, or somewhere between $H^2$ and $H^3$ if the domain is not
convex). But, since we don't have the means to describe solutions of
realistic problems in terms of relatively simple formulas, we just go
with the following, on the unit square for the domain $\Omega$:
@f{align*}{
  u = \sin(\pi x) \sin(\pi y).
@f}
As a consequence, we then need choose as boundary conditions the following:
@f{align*}{
  g &= u|_{\partial\Omega} = \sin(\pi x) \sin(\pi y)|_{\partial\Omega},
  \\
  j &= \frac{\partial u}{\partial\mathbf n}|_{\partial\Omega}
  \\
    &= \left.\begin{pmatrix}
                \pi\cos(\pi x) \sin(\pi y) \\
                \pi\sin(\pi x) \cos(\pi y)
             \end{pmatrix}\right|_{\partial\Omega} \cdot \mathbf n.
@f}
The right hand side is easily computed as
@f{align*}{
  f = \Delta^2 u = 4 \pi^4 \sin(\pi x) \sin(\pi y).
@f}
The program has classes `ExactSolution::Solution` and
`ExactSolution::RightHandSide` that encode this information.
