This course is an introduction to the finite element method as applicable to a range of problems in physics and engineering sciences. The treatment is mathematical, but only for the purpose of clarifying the formulation. The emphasis is on coding up the formulations in a modern, open-source environment that can be expanded to other applications, subsequently. The course includes about 45 hours of lectures covering the material I normally teach in an introductory graduate class at University of Michigan. The treatment is mathematical, which is natural for a topic whose roots lie deep in functional analysis and variational calculus. It is not formal, however, because the main goal of these lectures is to turn the viewer into a competent developer of finite element code. We do spend time in rudimentary functional analysis, and variational calculus, but this is only to highlight the mathematical basis for the methods, which in turn explains why they work so well. Much of the success of the Finite Element Method as a computational framework lies in the rigor of its mathematical foundation, and this needs to be appreciated, even if only in the elementary manner presented here. A background in PDEs and, more importantly, linear algebra, is assumed, although the viewer will find that we develop all the relevant ideas that are needed. The development itself focuses on the classical forms of partial differential equations (PDEs): elliptic, parabolic and hyperbolic. At each stage, however, we make numerous connections to the physical phenomena represented by the PDEs. For clarity we begin with elliptic PDEs in one dimension (linearized elasticity, steady state heat conduction and mass diffusion). We then move on to three dimensional elliptic PDEs in scalar unknowns (heat conduction and mass diffusion), before ending the treatment of elliptic PDEs with three dimensional problems in vector unknowns (linearized elasticity). Parabolic PDEs in three dimensions come next (unsteady heat conduction and mass diffusion), and the lectures end with hyperbolic PDEs in three dimensions (linear elastodynamics). Interspersed among the lectures are responses to questions that arose from a small group of graduate students and post-doctoral scholars who followed the lectures live. At suitable points in the lectures, we interrupt the mathematical development to lay out the code framework, which is entirely open source, and C++ based. Books: There are many books on finite element methods. This class does not have a required textbook. However, we do recommend the following books for more detailed and broader treatments than can be provided in any form of class: The Finite Element Method: Linear Static and Dynamic Finite Element Analysis, T.J.R. Hughes, Dover Publications, 2000. The Finite Element Method: Its Basis and Fundamentals, O.C. Zienkiewicz, R.L. Taylor and J.Z. Zhu, Butterworth-Heinemann, 2005. A First Course in Finite Elements, J. Fish and T. Belytschko, Wiley, 2007. Resources: You can download the deal.ii library at dealii.org. The lectures include coding tutorials where we list other resources that you can use if you are unable to install deal.ii on your own computer. You will need cmake to run deal.ii. It is available at cmake.org.
01.05. Strong form of the partial differential equation. Analytic solution
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From the course by University of Michigan
The Finite Element Method for Problems in Physics
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1
This unit is an introduction to a simple one-dimensional problem that can be solved by the finite element method.
Meet the Instructors
Krishna Garikipati, Ph.D.Professor of Mechanical Engineering, College of Engineering - Professor of Mathematics, College of Literature, Science and the Arts
Welcome back, we continue with our preamble into the finite element method.
And you will recall that in the previous segment,
we had started working with this linear, one dimensional,
partial differential equation, which was of elliptic type.
Okay? And, we,
we, we'd written out this differential equation.
We had, also motivated it in the context of elasticity.
And also looked at how essentially the same
partial differential equation arises in other problems.
Right? We'd looked at heat transport
as well as mass diffusion.
Okay?
We'll pick up now, but we'll focus on the PDE itself.
All right?
So, the topic of this segment is what we
call the strong form of the p, of the partial differential equation.
Okay?
So.
So the strong form of
a linear PDE, or
partial differential equation,
of elliptic type.
Right?
In one dimension.
Okay? Now let me write the, the PDE and
this is the same PDE that we were working with in
the previous segment, but let me put it down now.
Okay?
So what you want to do is the following.
Right?
We've set ourselves the task of
finding U, function of X,
given certain data.
Right?
We're given Ug, U not.
We're given Ug, or we're given t, right?
And remember how we qualify the difference between Ug and
tUg would be the Dirichlet boundary data, and t would be the Neumann boundary data.
Okay? We're given these.
We're also given the function f, function of x.
And the constitutive relation.
And that constitutive relation is sigma equals E.
U, X.
Right?
Given all of this, what we want to do is sorry,
we already said that we wanted to find U.
So we want to find U given all of this information, such that
the spatial gradient of sigma plus
f equals 0 in the open interval 0, L.
Okay?
With boundary conditions,
Boundary conditions being u(0) equals U not.
And, either
U at L equals U
given or sigma
at L equals T.
Okay?
This is our specification of of the linear partial differential equation,
of elliptic type in 1D.
Now, we call this a strong form of the equation, right?
That's new terminology that I will introduce.
We call it the strong form for two reasons, one is that if you look at
the PDE itself, you will observe that the PDE has been specified to
hold essentially at every point in the interval of interest, right?
When we say that d segment dx plus f equals 0 in the open interval 0,L,
we're specifying that that PDE has to hold at every point.
Right? So, in a sense,
that is a strong condition, we've seen that at every point we want it to hold.
Additionally, you observe that when we account for this,
constitutive relation, right?
What this implies is that, if we substitute
that considerative relationship for sigma in the partial differential equation,
what we get is the following form.
Right?
We get d, d, x of sigma.
But sigma itself is E du,
dx plus f equals 0 in
the open integral.
Now, if you observe that term, what we see is
that there are two derivatives effectively on the field of interest, right?
There are two derivatives implied here on the prime of field U.
And that is a slightly strong condition, because really for this equation to
make sense, classically, we do require that U should be
a function of a type that allows us to take two derivatives okay.
As you may imagine if U is not a terribly smooth function,
if it has discontinuities and so on.
If it, if, if even worse, right, if it has delta functions, right?
If there are delta functions in it.
There are certain difficulties we are going to encounter with taking two
derivatives of it.
Okay?
All right?
And so, what, what we're doing here is w,
w, we are, we, we require
so called strong conditions
of what I will loosely call smoothness.
Okay?
On U.
Right?
And why is this?
Because the strong form,
and I will capitalize it here to emphasize that,
for now, it's something special to us, right?
We require strong conditions of smoothness on U effects,
because the strong form has to Spatial derivatives.
'Kay, and there's also this idea that we require the whole point wise, 'kay,
so we require The partial
differential equation to hold pointwise
In the interval of interest.
Okay, these are the reasons for
which we call this form of the equation the strong form.
All right, and
this is of course is the classical form in which you have seen partial differential
equations posed up to this point, assuming of course that you do not already know
something about the mathematical background to finite element methods.
Okay, all right, so, so this is the strong form with a PDE of interest and, and,
of course, we observed in the previous segment that it, it models elasticity,
it models heat transport, mass diffusion and so on.
What I'm going to do now in the next few minutes is sketch out for
us the approach to an analytic solution, 'kay?
And I say just sketch out because I'm not actually going to
produce an analytical an, an analytic solution.
But I will point us in the direction of of how we would proceed, okay?
Of course I'll be using the fact that this is all in 1D, so
it makes life a little easier for me, okay?
So so let's like now look at
an analytic, analytic solution.
'Kay?
The approach is very straightforward and
you probably already know how I'm going to approach it.
It is to simply integrate them, okay?
So, we have d/dx of sigma equals minus f,
all right, and I did that by just moving f over to
the other side, it's just integrate, okay?
So we integrate this now.
I'm going to integrate this from 0 not all the way up to L, 'kay,
but I'm going to integrate this from 0 to say y, okay?
And here the the dx, of course.
And here I have integral of f.
Also over the same limits, right?
0 to y, dx, right?
Where y belongs to the open integral,
sorry, it belongs now to the closed integral.
Right, y belongs to the closed interval, 0 to L.
Importantly, we do not necessarily, we do not actually in so, in solving this PDE,
we do not want to integrate all the way up to L, right,
because we want to maintain a general solution applicable at any point, right,
and that's why we don't take the upper limit all the way up to L.
Okay this again is completely straight forward,
right, so what we, what we know we get from here is
we get sigma at y minus sigma at 0 equals
minus integral of fdx over the limit 0 to y.
F in general is at this point it's, it's not quite unknown.
It's data, which we would know if we set out to actually solve the problem.
But I want to leave it general.
I do not want to assume a particular form of f or anything, right?
All right.
So we have this.
But then we realize that what this
implies is, of course this,
right, it is that E u,
x, evaluated at y minus E,
u comma x evaluated at 0 equals
minus integral 0 to y, fdx, okay?
because that's what it implies for us, okay?
And you, and you see how I get this last line by simply making the substitution for
the constitutive relation for sigma, all right, at the two limit points.
Now in in sa,
in what is actually some what
of an abuse of notation,
let me simplify this
to write as let me simplify and
write this as E du/dy
equals minus integral 0
to y fdx plus E du/dx
evaluated at x equals 0, okay?
Right? Okay, so and now you know what,
what the next step is.
Well, we need to do one more integration, right?
So what that does for us then is the forage now
we'll integrate this again now let's integrate from 0 to z, right?
Or 0 to zed depending on which part of the world you come from.
Okay, so that becomes E that
the, the integral here is E,
du/dy dy equals minus integral 0 to y of fdx,
and then we're doing another integration,
right, 0 to z dy.
Okay, and just to make it clear, let's do this, okay?
And then we have another term here,
which is plus E du/dx evaluated at 0, okay?
But this term also gets
integrated from 0 to z,
dy, all right?
Okay again, straightforward enough what, what we get here,
now, now note that of course E could be a function of position.
Right, so, so, you're, you're allowed to make E a function of position.
So E could also depend upon y.
All right, so now when you do that integral we know what we get,
we, we essentially get to make things simpler,
let me forget about that particular dependent, okay?
Let me, let me leave things as they are.
Okay, so let's suppose E's a constant and
we go ahead with the integration what we get here is now E.
U at z minus E.
Again, u at zero equals everything that we have here on the right hand side, correct?
So we have integral zero two z,
integral zero two y f d x,
d y plus integral zero to z.
Well actually this last integral can be made very simple now, right?
Right, because it is just E.
D u, d x evaluated at zero, right?
Times z.
Okay.
So this is the final form that we get, and
since we want to write this all as a, as a result for u of z.
We write it as one over e.
Of integral zero two z.
Integral zero two y, f, d x,
d y, plus e, d u, d x.
Evaluated at zero times z +
E multiplying u at zero and
close the brackets, ok?
So this is the final, the general form of our equation.
Of course I made things even a little simpler for
us by saying that lets just assume that E is not a function of position.
Okay, and if E were a function of position, of course,
would involve a little more.
But just for the point of fixing ideas, I want to take this approach.
Okay.
Now, you observed that we have some unknowns here.
Would, we know how to apply them, them, right.
We would apply the boundary conditioning,
we would apply the differential boundary condition here for u of zero.
That is equal to u not, okay?
That leaves us however with one more unknown, which is this okay?
And this would be obtained by applying the boundary condition that we have at l.
Okay?
This term which is determined.
By applying the given boundary condition.
BC for boundary condition at X.
Equals L.
Right?
X equals l or in this case z equals l because we
are finally expressing this as, a function of z.
Okay?
Alright.
Now.
If the bound recondition that we have z equals l is a displacement boundary
condition, if its a directional boundary condition it could be straight forward to
have to apply it.
What if it were a Neuman boundary condition?
Think about it.
Straight forward of course its strictly almost if it were a Neuman boundary
condition we would simply carry out a derivative.
A few right and thereby allow us.
That would allow us to apply the boundary condition right.
So if boundary
condition at x
equals l is a Neuman
boundary condition,
we first calculate,
calculate or
go to u comma z.
Using that form of the function, okay?
And that will then allow us to apply the Neuman boundary condition, right?
And then apply
The boundary condition,
e u comma z at l.
Okay?
That will then allow us to determine this unknown.
Okay?
I also note now, that I have just 1 sign off in this, slide.
I have a minus sign here, which I forgot to carry over.
So let me just put it in here.
Okay?
That takes care of things.
So, this would be the.
General approach to getting an analytic solution of the strong form
of the pde that we are working with.
Okay.
The trouble with this of course is that we can do this only for
very special forms, or for the forcing function f.
All right?
And furthermore, we did observe during the process that I
simplified things by saying that E is a constant.
Okay?
I'm assuming that E is a constant.
It does not always have to be the case.
I assumed E were a constant to go through the process.
If E is not a constant it makes the process a little more complicated right?
And likewise if F were some sort of complicated function.
Furthermore this sort of simple minded approach or
simplistic approach works only in one dimension right?
As we go up to,
to higher dimensions integrating up the equations of, of this form.
This, this particular linear elliptic PDE.
It is not at all a trivial matter, alright?
Exact solutions or analytic solutions exist only for very special cases.
And what we are trying develop here is really an approach to solve general
boundary value problems.
All right? With general data,
general data in the form of F and E.
So that's what we're going to go on to.
Shortly, but we'll do that in the next segment, and we'll end this segment here.
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