ODEs vs DAE vs ADE? - Computational Science Stack Exchange most recent 30 from scicomp.stackexchange.com 2019-10-20T03:05:07Z https://scicomp.stackexchange.com/feeds/question/23800 https://creativecommons.org/licenses/by-sa/4.0/rdf https://scicomp.stackexchange.com/q/23800 8 ODEs vs DAE vs ADE? MBM https://scicomp.stackexchange.com/users/20025 2016-04-24T13:51:57Z 2016-04-25T08:32:00Z <p>I am totally confused between ODEs which I am familiar with, and differential algebraic equations (DAE) and Algebraic Differential Equations (ADE). Are they the same but just different names or what is the key difference between them (their nature and solution methods). Thanks and best regards</p> https://scicomp.stackexchange.com/questions/23800/-/23801#23801 12 Answer by Bill Barth for ODEs vs DAE vs ADE? Bill Barth https://scicomp.stackexchange.com/users/27 2016-04-24T14:59:07Z 2016-04-25T05:58:25Z <p>At least one difference is that in a system of ODEs, all the equations are differential, e.g.: $$\dot{x}=f(x,y)\\ \dot{y}=g(x,y)$$ whereas the definition of DAEs that I'm familiar with includes some non-differential (i.e. algebraic) equations in the set, e.g.: $$\dot{x}=h(x,y)\\ y=l(x,y)$$ where $l$ is non-trival, and its solution can't be easily substituted into the first equation to simplify. These get more complex when there are more algebraic terms. </p> <p>DAEs are more challenging numerically; the challenges they entail are similar to but sometimes more severe than those attending stiff problems. A very thorough explanation of DAEs and how to solve them numerically can be found in <a href="http://www.springer.com/us/book/9783540604525">volume II of the text by Hairer and Wanner</a>.</p> https://scicomp.stackexchange.com/questions/23800/-/23802#23802 7 Answer by adhalanay for ODEs vs DAE vs ADE? adhalanay https://scicomp.stackexchange.com/users/1302 2016-04-24T17:22:49Z 2016-04-24T18:30:13Z <p>Differential-algebraic equations (DAE) are equations of the form $F(t,x,x')=0$, with the unknown function being $x(t)$. So in a way are generalizations of ODEs. A nice place to start is <a href="http://www.scholarpedia.org/article/Differential-algebraic_equations">here</a>. On the other hand an algebraic differential equation is a totally different thing. The wikipedia page gives an overview, but basically is an equation involving differential operators on a differential algebra.</p> https://scicomp.stackexchange.com/questions/23800/-/23811#23811 3 Answer by Thomas Klimpel for ODEs vs DAE vs ADE? Thomas Klimpel https://scicomp.stackexchange.com/users/534 2016-04-25T08:32:00Z 2016-04-25T08:32:00Z <p><a href="https://mathoverflow.net/questions/225840/index-reduction-of-differential-algebraic-equations-by-hand/226517#226517">Here is an identical copy of an answer on MO</a>:</p> <p>One intuitive way to understand a <a href="http://www.scholarpedia.org/article/Differential-algebraic_equations" rel="nofollow noreferrer">DAE</a> is to interpret it as a dynamical system which can be controlled by some input signals, whose output signals have to satisfy some (equational) constraints. For a typical multibody system, the input signals are the forces perpendicular to the constraints, the output signals are the positions of the bodies, and the (equational) constraints on the output signals are fixed distances between the bodies.</p> <p>The input signals must now control the dynamical system in such a way that the output signals always satisfy the constraints. This is difficult for a multibody system, because the forces only control the rate of change of the velocities, and the velocities only control the rate of change of the positions, while only the positions must satisfy the constraints.</p> <p>Reducing the index is easy in theory, because if we assume that the positions satisfy the constraints at the current time instance, then we can just replace the constraints on the positions by constraints on the velocities ensuring that the positions will continue to satisfy their constraints. In practice however, we don't want to throw away the constraint on the positions after we determined the constraints on the velocities, but we do have to throw away some of the initial (differential) equations, if we don't want to end with an overdetermined system.</p> <p>Determining the constraints on the velocities from the constraints on the positions might be tedious in practice, but at least it is straightforward (and canonical) once you understood the principle. The constraint $c(y,t)=0$ implies $\frac{d}{dt}c(y(t),t)=0=\frac{\partial c}{\partial y}*\frac{d}{dt}y+\frac{\partial c}{\partial y}$. This is not an (equational) constraint yet, because $\frac{d}{dt}y$ is not a variable but only the derivative of a variable. But the other differential equations allow us to express $\frac{d}{dt}y$ as a function of the variables, in our case $\frac{d}{dt}y=v$ for $v=\dot{y}$, so we get the equational constraint $0=\frac{\partial c}{\partial y}*v+\frac{\partial c}{\partial y}$ (or rather $0=\frac{\partial c}{\partial y}*\dot{y}+\frac{\partial c}{\partial y}$ if you manage to not get confused by using $\dot{y}$ as a variable instead of the derivative of a variable).</p> <p>Throwing away some of the initial (differential) equations is less straight forward (or canonical). If we can use a constraint equation like $y_1^2+y_2^2=1$ to determine $y_1$ as a function of the other variables (i.e. $y_1(t)=\sqrt{1-(y_2(t))^2}$ in this case), then we can throw away the differential equation for $y_1$, i.e. a differential equation of the form $\frac{d}{dt}y_1=\dots$. But we might have also decided to throw away the differential equation for $y_2$ instead, because the constraint also allow to determine $y_2$ as a function of the other variables. But no matter how easy it is to throw something away, this can easily destroy some symmetry of the system we didn't want to destroy, or we might be forced to switch which equation we throw away during the numerical simulation and thereby introduce undesired artifacts. So this part makes index reduction really challenging in practice.</p>