#### Computational Physics: Truncation and Rounding Errors

Posted 15th October 2021 by

In a previous post, I talked about accuracy and precision in numerical calculations. Ideally one would like to perform calculations that are perfect in these two aspects. However, this is almost never possible in practical situations. The reduction of accuracy or precision is due to two numerical errors. These errors can be classified into two main groups, round-off errors and truncation errors.

#### Rounding Error

Round-off errors occur due to the limits of numerical precision at which numbers are stored in the computer. As I discussed here a 32-bit floating-point number for example can only store up to 7 or 8 decimal digits. Not just the final result but every intermediate result of a calculation will be rounded to this precision. In some cases, this can result in a much lower precision of the final result. One instance where round-off errors can become a problem happens when the result of a calculation is given by the difference of two large numbers.

#### Truncation Error

Truncation errors occur because of approximations the numerical scheme makes with respect to the underlying model. The name truncation error stems from the fact that in most schemes the underlying model is first expressed as an infinite series which is then truncated allowing it to be calculated on a computer.

### Example: Approximating Pi

Let’s start with a simple task. Use a series to approximate the value of \(\pi\).

#### Naive summation

One of the traditional ways of calculating \(\pi\) is by using the \(\arctan\) function together with the identity \[

\arctan(1) = \frac{\pi}{4}.

\] One can expand \(\arctan\) into its Taylor series, \[

\arctan(x)

= x – \frac{x^3}{3} +\frac{x^5}{5} – \frac{x^7}{7} + \ldots

= \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}.

\] The terms of the series become increasingly smaller and you could try to add up all the terms up to some maximum \(N\) in the hope that the remaining infinite sum is small and can be neglected. Inserting \(x=1\) into the sum will give you an approximation for \(\pi\), \[

\pi \approx 4\sum_{n=0}^N \frac{(-1)^n }{2n+1}.

\]

Here are e implementations for this approximation in C++, Python and JavaScript.

**C++**

```
double pi_summation_slow(int N) {
double sum = 0.0;
int sign = 1;
for (int i=0; i<N; ++i) {
sum += sign/(2*i + 1.0);
sign = -sign;
}
return 4*sum;
}
```

**Python**

```
def pi_summation_slow(N):
sum = 0.0
sign = 1
for i in range(0,N):
sum = sum + sign/(2*i + 1.0)
sign = -sign
return 4*sum
```

**JavaScript**

```
function pi_summation_slow(N) {
let sum = 0.0;
let sign = 1;
for (let i=0; i<N; ++i) {
sum += sign/(2*i + 1.0);
sign = -sign;
}
return 4*sum;
}
```

Let’s call this function with \(N=10\). All the results I am showing here are calculated using a Python implementation. We get a result of around 3.0418. The relative error is 0.0318 and is, of course, unacceptable. This error falls under the category of truncation errors because it is caused by not summing up enough terms of the Taylor series. Calling the function with \(N=1000\) gives us a result of 3.14059 with a relative error of \(3.183\times 10^{-4}\). The error has improved but is still far off from the possible \(10^{-14}\) to \(10^{-15}\) achievable in double-precision arithmetic. The figure below shows how the relative error decreases with the number of iterations.

From this curve, one h long m wl hat the error decreases with \(1/N\). If one extrapolates the curve, one finds that it would take \(10^{14}\) iterations to reach an error below \(10^{-14}\). Even if this was computationally feasible, the round-off errors of such a long sum would eventually prevent the error from being lowered to this limit.

#### Improvements using Machin’s formula

The technique of calculating \(\pi\) can be improved in two ways. Firstly, instead of using the Taylor series, you can use Euler’s series for the \(\arctan\) function.

\[

\arctan(x) = \sum_{n=0}^\infty \frac{2^{2n} (n!)^2}{(2n + 1)!} \frac{x^{2n + 1}}{(1 + x^2)^{n + 1}}.

\]

This series converges much more quickly than the Taylor series. The other way to improve convergence is to use trigonometric identities to come up with formulas that converge more quickly. One of the classic equations is the Machin formula for \(\pi\), first discovered by John Machin in 1706, \[

\frac{\pi}{4} = 4 \arctan \frac{1}{5} – \arctan \frac{1}{239}

\] Here are the implementations for this formula.

**C++**

```
double pi_summation_fast(int order) {
using boost::math::factorial;
double sum = 0.0;
for (unsigned int n=0; n<order; ++n) {
double f = factorial<double>(n);
double common = pow(2.0, 2*n)*f*f/factorial<double>(2*n + 1);
double A = pow(25./26., n+1)/pow(5., 2*n+1);
double B = pow(239.*239. / (239.*239. + 1.), n+1)/pow(239., 2*n+1);
sum += common*( 4*A - B );
}
return 4*sum;
}
```

**Python**

```
def pi_summation_fast(N):
sum = 0.0
for n in range(0,N):
f = factorial(n)
common = math.pow(2.0, 2*n)*f*f/factorial(2*n + 1)
A = math.pow(25/26, n+1)/math.pow(5, 2*n+1)
B = math.pow(239*239 / (239*239 + 1), n+1)/math.pow(239, 2*n+1)
sum = sum + common*( 4*A - B )
return 4*sum;
```

**JavaScript**

```
function pi_summation_fast(N) {
let sum = 0.0;
for (let n=0; n<N; ++n) {
const f = factorial(n);
const common = Math.pow(2.0, 2*n)*f*f/factorial(2*n + 1);
const A = pow(25/26, n+1)/pow(5, 2*n+1);
const B = pow(239*239 / (239*239 + 1), n+1)/pow(239, 2*n+1);
sum += common*( 4*A - B );
}
return 4*sum;
}
```

The table below shows the computed values for \(\pi\) together with the relative error. You can see that each iteration reduces the error by more than an order of magnitude and only a few iterations are necessary to achieve machine precision accuracy.

N | \(S_N\) | error |
---|---|---|

1 | 3.060186968243409 | 0.02591223443732105 |

2 | 3.139082236428362 | 0.0007990906009289966 |

3 | 3.141509789149037 | 2.6376570705797483e-05 |

4 | 3.141589818359699 | 9.024817686074192e-07 |

5 | 3.141592554401089 | 3.157274505454055e-08 |

6 | 3.141592650066872 | 1.1213806035463463e-09 |

7 | 3.1415926534632903 | 4.0267094489200705e-11 |

8 | 3.1415926535852132 | 1.4578249079970333e-12 |

9 | 3.1415926535896266 | 5.3009244691058615e-14 |

10 | 3.1415926535897873 | 1.8376538159566985e-15 |

11 | 3.141592653589793 | 0.0 |

### Example: Calculating sin(x)

Calculate the value of \(\sin(x)\) using it’s Taylor series around x=0.

The Taylor series for \(\sin(x)\) is \[

\sin x = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}.

\] this series is much more well-behaved than the Taylor series for \(\arctan\) we saw above. Because of the factorial in the denominator, the individual terms of this series will converge reasonably quickly. Here are some naive implementations of this function where the infinite sum has been replaced by a sum from zero to \(N\).

**C++**

```
double taylor_sin(double x, int order)
{
using boost::math::factorial;
double sum = 0.0;
int sign = 1;
for (unsigned int n=0; n<order; ++n)
{
sum += sign*pow(x, 2*n + 1)/factorial<double>(2*n +1);
sign = -sign;
}
return sum;
}
```

**Python**

```
def taylor_sin(x, N):
sum = 0.0
sign = 1
for n in range(0,N):
sum = sum + sign*math.pow(x, 2*n + 1)/factorial(2*n + 1)
sign = -sign
return sum
```

**JavaScript**

```
function taylor_sin(x, N) {
let sum = 0.0;
let sign = 1;
for (let n=0; n<N; n++) {
sum += sign*pow(x, 2*n + 1)/factorial(2*n +1);
sign = -sign;
}
return sum;
}
```

A good test for this function is the evaluation of \(\sin(x)\) at values \(x = k\pi\), where \(k\) is an integer. We know that \(\sin(k\pi) = 0\) and the return value from the numeric function can directly be used as the absolute error of the computation. The figure below shows results for some values of \(k\) plotted against \(N\).

For small values of \(k\), this series converges relatively quickly. But for larger \(k\) you can see that more and more terms are needed. The error even grows first before being reduced. Just like the example above, the truncation error requires large values of \(N\) to reach a good accuracy of the result. In practice, you would not calculate the \(\sin\) function this way. Instead you would make use of known properties, such as \(\sin(2k\pi + x) = \sin(x)\) for integer \(k\), to transform the argument into a range where fast convergence is guaranteed.

However, I would like to continue my analysis of this function because it shows two more interesting pitfalls when performing long sums. First, you will notice that the curves in the figure above show dashed lines for \(N>85\). This is because the implementation I showed above will actually fail with a range error. The `pow`

function and the factorial both start producing numbers that exceed the valid range of `double`

floating-point numbers. The quotient of the two, on the other hand, remains well-behaved. It is, therefore, better to write the Taylor series using a recursive definition of the terms.

\[

\sin x = \sum_{n=0}^\infty a_n(x),

\] with \[

a_0 = x

\] and \[

a_{n} = -\frac{x^2}{2n(2n+1)}a_{n-1}

\]

The implementations are given again below.

**C++**

```
double taylor_sin_opt(double x, int order)
{
double sum = x;
double an = x;
for (unsigned int n=1; n<order; ++n)
{
an = -x*x*an/(2*n*(2*n+1));
sum += an;
}
return sum;
}
```

**Python**

```
def taylor_sin_opt(x, N):
sum = x
an = x
for n in range(1,N):
an = -x*x*an/(2*n*(2*n+1))
sum = sum + an
return sum
```

**JavaScript**

```
function taylor_sin_opt(x, N) {
let sum = x;
let an = x;
for (let n=1; n<N; n++) {
an = -x*x*an/(2*n*(2*n+1));
sum += an;
}
return sum;
}
```

The other takeaway from the graphs of the errors is that they don’t always converge to machine accuracy. The reason for this originates from fact that the initial terms of the sum can be quite large but with opposite signs. They should cancel each other out exactly, but they don’t because of numerical round-off errors.

#### Spherical Blast Wave Simulation

Posted 2nd December 2020 by Holger Schmitz

Here is another animation that I made using the Vellamo fluid code. It shows two very simple simulations of spherical blast waves. The first simulation has been carried out in two dimensions. The second one shows a very similar setup but in three dimensions.

You might have seen some youtube videos on blast waves and dimensional analysis on the Sixty Symbols channel or on Numberphile. The criterion for the dimensional analysis given in those videos is true for strong blast waves. The simulation that I carried out, looks at the later stage of these waves when the energy peters out and the strong shock is replaced by a compression wave that travels at a constant velocity. You can still see some of the self-similar behaviour of the Sedov-Taylor solution during the very early stages of the explosion. But after the speed of the shock has slowed down to the sound speed, the compression wave continues to travel at the speed of sound, gradually losing its energy.

The video shows the energy density over time. The energy density includes the thermal energy as well as the kinetic energy of the gas.

For those of you who are interested in the maths and the physics, the code simulates the Euler equations of a compressible fluid. These equations are a model for an ideal adiabatic gas. For more information about the Euler equations check out my previous post.

#### The Euler Equations: Sod Shock Tube

Posted 30th September 2020 by Holger Schmitz

In the last post, I presented a simple derivation of the Euler fluid equations. These equations describe hydrodynamic flow in the form of three conservation equations. The three partial differential equations express the conservation of mass, the conservation of momentum, and the conservation of energy. These fundamental conservation equations are written in terms of fluxes of the densities of the conserved quantities.

In general, it is impossible to solve the Euler equations for an arbitrary problem. This means that in practice when we want to find the hydrodynamic flow in a particular situation, we have to use computers and numerical methods to calculate an approximation to the solution.

Numerical simulation codes approximate the continuous mathematical solution by storing the values of the functions at discrete points. The values are stored using a finite precision format. On most CPUs nowadays numbers will be stored as 64-bit floating-point, but for codes running on GPUs, this might be reduced to 32 bits. In addition to the discretisation in space, the equations are normally integrated using discrete steps in time. All of these factors mean that the computer only stores part of the information of the exact function.

These discretisations mean that the continuous differential equations have to be turned into a prescription how to update the discrete values from one time step to the next. For each differential equation, there are numerous ways to translate them into a numerical algorithm. Each algorithm will make different approximations and introduce different errors into the system. In general, it is impossible for a numerical algorithm to reproduce the solutions of the mathematical equations exactly. This will have implications for the physics that will be simulated with the code. Talking in terms of the Euler equations, some numerical algorithms will have problems capturing shocks, while others might smear out the solutions. Even others might cause the density to turn negative in places. Making sure that physical invariants, such as mass, momentum, and energy are conserved by the algorithm is often an important aspect when designing a numerical scheme.

### The Sod Shock Tube

One of the standard tests for numerical schemes simulating the Euler equations is the Sod shock tube problem. This is a simple one-dimensional setup that is initialised with a single discontinuity in the density and energy density. It then develops a left-going rarefaction wave, a right-going shock and a somewhat slower right-going contact discontinuity. The shock tube has been proposed as a test for numerical schemes by Gary A. Sod 1978 [1]. It was then picked up by others like Philip Roe [2] or Bram van Leer [3] who made it popular. Now, it is one of the standard tests for any new solver for the Euler equations. The advantage of the Sod Shock Tube is that it has an analytical solution that can be compared against. I won’t go into the details of deriving the solution in this post. You can find a sketch of the procedure on Wikipedia.

Here, I want to present an example of the shock tube test using an algorithm by Kurganov, Noelle, and Petrova [4]. The advantage of the algorithm is that it is relatively easy to implement and that is straightforward to extend to multiple dimensions. I have implemented the algorithm in my open-source fluid code Vellamo. At the current stage, the code can only simulate the Euler equations, but it can run in 1d, 2d, or 3d. It is also parallelised so that it can run on computer clusters It uses the Schnek library to manage the computational grids, their distribution onto multiple CPUs, and the communication between the individual processes.

The video shows four simulations run with different grid resolutions. At the highest resolution of 10000 grid points, the results are more or less identical to the analytic solution. The left three panels show the density $\rho$, the momentum density $\rho v$, and the energy density $E$. These are the quantities that are actually simulated by the code. The right three panels show quantities derived from the simulation, the temperature $T$, the velocity $v$, and the pressure $p$.

Starting from the initial discontinuity at $x=0.5$ you can see a rarefaction wave moving to the left. The density $\rho$ has a negative slope whereas the momentum density $\rho v$ increases in this region. In the plot of the velocity, you can see a linear slope, indicating that the fluid is being accelerated by the pressure gradient. The expansion of the fluid causes cooling and the temperature $T$ decreases inside the rarefaction wave.

While the rarefaction wave moves left, a shock wave develops on the very right moving into the undisturbed fluid. All six graphs show a discontinuity at the shock. The density jumps $\rho$ as the fluid is compressed and the velocity $v$ rises from 0 to almost 1 as the expanding fluid pushes into the undisturbed background. The compression also causes a sudden increase in pressure $p$ and temperature $T$.

To the left of the shock wave, you can observe another discontinuity in some but not all quantities. This is what is known as a contact discontinuity. To the left of the discontinuity, the density $\rho$ is high and the temperature $T$ is low whereas to the right density $\rho$ is low and the temperature $T$ is high. Both effects cancel each other when calculating the pressure which is the same on both sides of the discontinuity. And because there is no pressure difference, the fluid is not accelerated either, and the velocity $v$ is the same on both sides as well.

If you look at the plots of the velocity $v$ and the pressure $p$ you can’t make out where the contact discontinuity is. In a way, this is quite extraordinary because these two quantities are calculated from the density $\rho$, the momentum density $\rho v$ and the energy density $E$. All of these quantities have jumps at the location of the contact discontinuity. The level to which the derived quantities stay constant across the discontinuity is an indicator of the quality of the numerical scheme.

The simulations with lower grid resolutions show how the scheme degrades. At 1000 grid points, the results are still very close to the exact solution. The discontinuities are slightly smeared out. If you look closely at the temperature profile, you can see a slight overshoot on the high-temperature side of the contact discontinuity. At 100 grid points, the discontinuities are more smeared out. The overshoot in the temperature is more pronounced and now you can also see an overshoot at the right end of the rarefaction wave. This overshoot is present in most quantities.

As a rule of thumb, a numerical scheme for hyperbolic differential equations has to balance accuracy against numerical diffusion. In regions where the solution is well behaved, high order schemes will provide very good accuracy. Near discontinuities, however, a high order scheme will tend to produce artificial oscillations. These artefacts can create non-physical behaviour when, for instance, the numerical solution predicts a negative density or temperature. A good scheme detects the conditions where the high order algorithm fails and falls back to a lower order in those regions. This will naturally introduce some numerical diffusion into the system.

At the lowest resolution with 50 grid points, the discontinuities are even more spread out. Nonetheless, at later times in the simulation, all discontinuities can be seen and the speed at which these discontinuities move compares well with the exact result.

I hope you have enjoyed this article. If you have any questions, suggestions, or corrections please leave a comment below. If you are interested in contributing to Vellamo or Schnek feel free to contact me via Facebook or LinkedIn.

[1] Sod, G.A., 1978. *A survey of several finite difference methods for systems of nonlinear hyperbolic conservation laws.* Journal of computational physics, **27**(1), pp.1-31.

[2] Roe, P.L., 1981. Approximate Riemann solvers, parameter vectors, and difference schemes. Journal of computational physics, 43(2), pp.357-372.

[3] Van Leer, B., 1979. Towards the ultimate conservative difference scheme. V. A second-order sequel to Godunov’s method. Journal of Computational Physics, 32(1), pp.101-136.

[4] Kurganov, A., Noelle, S. and Petrova, G., 2001. Semidiscrete central-upwind schemes for hyperbolic conservation laws and Hamilton–Jacobi equations. SIAM Journal on Scientific Computing, 23(3), pp.707-740.

#### The Euler Equations

Posted 3rd September 2020 by Holger Schmitz

Leonhard Euler was not only brilliant but also a very productive mathematician. In this post, I want to talk about one of the many things named after the Swiss genius, the Euler equations. These should not be confused with the Euler identity, that famous relation involving $e$, $i$, and $\pi$. No, the Euler equations are a set of partial differential equations that can be used to describe fluid flow. They are related to the Navier-Stokes equations which are at the heart of one of the unsolved Millenium Problems of the Clay Mathematics Institute.

In this short series of posts, I want to present a phenomenological derivation of the equations in one dimension. This derivation is not mathematically rigorous, but will hopefully give you an understanding of the physics that the equations are describing.

### Conservation Laws

One way to understand the Euler equations is to write them in the form of conservation laws. Each equation represents a basic conservation law of physics. A central concept in these conservation equation is **flux**. In a 1d description, the flux of a quantity is a measure for the amount of that quantity that passes through a point. In 2d it’s the amount passing through a line and in 3d through a surface, but let’s stick to 1d during this article.

Take some quantity $N$. To make it concrete, imagine something like the total mass or number of atoms in a volume. I am using the word *volume* loosely. In 1d a volume is the same as an interval. Let’s call the flux of this quantity $F$. The flux can be different at different locations, so $F$ is a function of position $x$, in other words, the flux is $F(x)$.

Next, consider a small volume between $x$ and $x+h$. Here $h$ is the width of the volume and you should think of it as a small quantity. The rate of change of the quantity $N$ within this volume is determined by the difference between the flux into the volume and the flux out of the volume. If we take positive flux to mean that the quantity (mass, number of atoms, stuff) is moving to the right, then we can write down the following equation.

$$\frac{dN}{dt} = F(x) – F(x+h)$$

Here the left-hand side of the equation represents the rate of change of $N$. I have not written it here but, of course, $N$ depends on the position $x$ as well as the with of the volume $h$. The next step to turn this into a conservation equation is to divide both sides by $h$.

$$\frac{d}{dt}\left(\frac{N}{h}\right) = -\frac{F(x+h) – F(x)}{h}$$

If you remember your calculus lessons, you might already see where this is going. We now make $h$ smaller and smaller and look at the limit when $h \to 0$. The right hand side of the equation will give us the derivative of $F$,

$$\lim_{h \to 0} \frac{F(x+h) – F(x)}{h} = \frac{dF}{dx}$$

The amount of stuff $N$ in the volume will shrink and shrink as $h$ decreases, but because we are dividing by $h$ the ratio $N/h$ will converge to a sensible value. Let’s call this value $n$,

$$n = \lim_{h \to 0}\frac{N}{h}.$$

The quantity $n$ is called the density of $N$. Now the conservation law is simply written as

$$\frac{\partial n}{\partial t} = -\frac{\partial F}{\partial x}.$$

This equation expresses the change of the density of the conserved quantity through the derivative of the flux of that quantity.

### The Equations

Now that the basic form of the conservation equations is established, let’s look at some quantities that are conserved and that can be used to express fluid flow. The first conservation law to look at is the conservation of mass. Note again that the conservation equation is written in terms of the density of the conserved quantity. The mass density is the mass contained in a small volume divided by that volume. It is often abbreviated by the Greek letter $\rho$ (rho).

In order to find out what the mass flux is, consider the fluid moving with velocity $v$ through some point. During a small time interval, $\Delta t$ the fluid will move by a small distance $\Delta x$. All the mass $M$ contained in the interval of width $\Delta x$ will cross the point. This mass can be calculated to be

$$M = \rho \Delta x.$$

The flux is the amount of mass crossing the point divided by the time it took the mass to cross that point, $F_{\rho} = M / \Delta t$. In other words,

$$F_{\rho} = \rho \frac{\Delta x}{\Delta t} = \rho v.$$

This now lets us write the mass conservation equation,

$$\frac{\partial \rho}{\partial t} = -\frac{\partial }{\partial x}(\rho v).$$

What does this equation mean? Let me explain this with two examples. First, imagine a constant flow velocity $v$ but an increasing density profile $\rho(x)$. At any given point the density will decrease because the density profile will constantly move towards the right without changing shape. The lower density that was previously located at somewhat left of $x$ will move to the point $x$ thus decreasing the density there.

For the other example, imagine that the density is constant but the velocity has an increasing profile. Matter to the right will flow faster than matter to the left. This will also decrease the density over time because the existing mass is constantly thinned out.

The next conservation equation expresses the conservation of momentum. Momentum is mass times velocity and therefore momentum density is mass density times velocity, $\rho v$. In the mass conservation equation, you could see that the flux consists only of the conserved density multiplied with the velocity. This is called the convective term and this term is present in all Euler equations. However, the momentum can also change through a force. Without external forces, the only force onto each fluid element is through the pressure $p$. The momentum conservation equation can be written as

$$\frac{\partial }{\partial t}(\rho v) = -\frac{\partial }{\partial x}(\rho v^2 + p).$$

The last conservation equation is the conservation of energy, expressed by the energy density $e$,

$$\frac{\partial e}{\partial t} = -\frac{\partial }{\partial x}(e v + p v).$$

The first term is again the convective term expressing the transport of energy contained in the fluid as the fluid moves. The second term is the work done by the pressure force. Note that the energy density $e$ contains the internal (heat) energy as well as the kinetic energy of the fluid moving as a whole.

Finally, we need to close the set of equations by defining what the pressure $p$ is. This closure depends on the type of fluid you want to model. For an ideal gas, the pressure is related to the other variables through

$$p = (\gamma – 1)\left(E – \frac{\rho v^2}{2}\right).$$

Here $\gamma$ is the adiabatic gas index. The second bracket on the right-hand side is the internal energy, expressed as the total energy minus the kinetic energy of the flow.

### Putting it All Together

In summary, the Euler equations are conservation equations for the mass, momentum, and energy in the system. In 1d, they can be written as

$$\begin{eqnarray}

\frac{\partial }{\partial t}\rho &=& -\frac{\partial }{\partial x}(\rho v) \\

\frac{\partial }{\partial t}(\rho v) &=& -\frac{\partial }{\partial x}(\rho v^2 + p) \\

\frac{\partial }{\partial t}e &=& -\frac{\partial }{\partial x}(e v + p v)

\end{eqnarray}

$$

In higher dimensions, the derivative with respect to $x$ is replaced by the divergence operator $\nabla$.

$$\begin{eqnarray}

\frac{\partial }{\partial t}\rho &=& -\nabla(\rho \mathbf{v}) \\

\frac{\partial }{\partial t}(\rho v) &=& -\nabla(\rho \mathbf{v}\otimes\mathbf{v} + p\mathbf{I}) \\

\frac{\partial }{\partial t}e &=& -\nabla(e \mathbf{v} + p \mathbf{v})

\end{eqnarray}

$$

The $\otimes$ symbol denotes the outer product of the velocity vectors. It multiplies two vectors to produce a matrix. I will not go into the definition of the outer product here. You can read more on Wikipedia. The symbol $\mathbf{I}$ is the identity matrix.

### Summary

The Euler equations are a set of partial differential equations describing fluid flow. When Euler published a simpler version of these equations, they were one of the earliest examples of partial differential equations. Today, the Euler equations and their generalisations, including the Navier-Stokes equations provide the basis for understanding fluid behaviour, turbulence, and the weather.

The Euler equations are difficult to solve and only some special solutions can be written down on paper. For the general case, we need computers to give us answers. In the next post, I will show the results of some simulations I have created with my freely available fluid code Vellamo.