All right, welcome back.
We continue and we will get down to business now.
And we so by writing out the weak form [COUGH].
The weak form for linearized elasticity.
Right so.
I'm going to take the approach we, adopted when we worked with
linear elliptic PDs with scalar variables in 3D.
Okay, which is, and I'm going to put down the weak form.
And then, I will demonstrate how we obtain it from the strong form.
I won't go back the other way to just avoid just,
because those, those arguments are, identical to what we did in 1-B.
But they are, you know, they're more subtle and, and so on.
Okay, but however we know how they, how they operate and, and they hold.
Okay, weak form of linearized elasticity.
And, and, and from here on,
what I'm going to use is called notation only, okay?
Because there are details, as you can see,
which become most transparent when we use coordinate notation.
Okay, so given
the usual data.
Given ug sub
i t-bar: i,
f, i,
the constitutive
relation.
Sigma: i,
j Equals cijkl,
epsilon kl, and
the kinematic relation,
Epsilon of kl,
equals one half.
Partial of u k with respect to x
l plus partial of u l with respect to x k.
Okay?
Given all of this,
find ui, OK?
Which, belongs to
some space S, OK?
Which consists of all u i such that u
i equals the Dirichlet boundary data
on the corresponding Dirichlet boundary reserved for
that particular component.
Right?
Of course what's implied here is i equals 1, 2, 3.
Okay?
Find u i belonging to s in this, of, of this type.
Such that.
For all wi belonging to v which now consists of all functions,
which satisfy the homogeneous boundary condition on
the corresponding Dirichlet boundary,
and corresponding Dirichlet boundary, right?
Here, too, we imply that i equals 1, 2, 3, okay?
Okay. So let me read this all over.
Given the data u, t-bar, u g, t-bar, f and constitutive relation.
The constitutive relation and
the kinematic relation, epsilon equals that turtle's derivatives of u.
Find u belonging to s such that for all w belonging to v,
the following holds, right?
Alright, so now let's write out what holds.
Integral over omega,
wi,j, sigma,
ij, dv equals
integral over
omega w i f i d v.
Plus, now, here comes the rough.
Because of the fact that our traction boundary condition needs to be
specified individually for each component, we need to straight
away here have a sum i going from 1 to number of spatial dimensions.
And of course, we're doing this in three dimensions, right?
Of integral over the corresponding traction boundary.
Okay?
Of wi t bar i ds.
I want to emphasize here that,
do we imply a sum in that
integrand over i?
Think about it.
Look at the last sum.
There is in fact no sum.
Ok, a sum is not implied there.
A sum is explicitly being carried out here.
But importantly we don't do that over the integrant, because,
why do we not do it over the integrand?
It's because that domain of integration could be different for each component i.
Okay.
So we can't quite do it as an implied sum.
What we need to do instead is compute that product for each value of i w i t-bar i.
Integrated over the corresponding Neumann boundary, right?
Which could be different for each component, right?
We get a scaler, right?
And then, we sum up those scales, right?
Over to three special dimensions.
All over the special dimensions.
Okay? That's important.
So let me state here.
No sum implied for,
for just that term,
wi t bar i, okay?
Instead, we have that explicit sum.
All right, this is our weak form.
What I will do know is derive it from the strong form.
OK?
And I, I'm going to take a bit of a short cut.
I'm not going to re, well, OK, maybe I will restart all the data very quickly.
So, what we, what am, I'm going to state here first that the strong form
Implies the weak form.
And the weak form also implies the strong form, but like we did for
the case of the scalar 3D elliptic linear equations,
We'll do only the rightward implication here, right?
The rightward equivalence just, just to see a demonstration.
All right, and that is the following, right, once again.
Given u g i,
t bar i, f i
the constitutive
relation sigma
ij = Cijkl Epsilon k l.
The kinematic relation.
Epsilon i, epsilon, let me write this as epsilon k l equals This.
Okay.
What we want to do is
find ui such that sigma
ij comma j plus fi
equals 0 in omega and
boundary conditions.
U i equals u g i on the relevant
tertiary boundary.
And sigma i j nj equals.
T bar i on.
Okay, so this has wrong form.
Alright, yet again.
Alright, we start out from here and
now we introduce as we did before, the waiting function, right?
So what we start off by saying is now let us consider.
W i, belonging to v which
has the usual property right?
Which is the v consists of all
wi such that wi vanishes on
that Dirichlet boundary.
Okay.
What we will do now, and this is something you may recall from
our previous repeated development into weak form.
Actually, think about, what do we do?
How do we proceed now?
Alright we multiply that
pde with wi and integrate.
Okay, so,
what we do is,
multiply the pde by wi,
and integrate over
our domain omicron.
Right? Which is an open domain in r, r3.
All right, fine.
When we do this, here's what we get.
We get integral over omega
wi sigma ij, comma j, db.
Plus integral over omega,
Wi Fi d V equals zero.
Okay, here sums are implied over i.
Okay, so, I'm not going to state it which means sum is implied, right.
Okay. So, do you recall what comes next?
Right, we observe that this, divergence,
right, we get rid of by, invoking integration by parts.
Right, and integration by parts, you may recall, is nothing but a combination of,
the product rule of differentiation and Gauss's Theorem of Divergence.
All right, apply to tensors now.
So, what we do here is
integrate by parts.
Right? And we integrate by
parts in that term, okay?
And in integrating that parts, remember that this is basically product rule of
differentiation.
Plus the divergence theorem.
Okay?
And the way we do that is the following, right?
We observe first that that derivative on sigma can be rewritten
if we just consider it to be the following.
Integral over omega,
a derivative of the whole product.
Okay, minus an integral over omega,
wi comma j sigma ij dv.
And what I've done here is applied the first of my
to two techniques here, right?
I've applied the product rule here.
Okay.
And of course, along for the ride is, W, I, F-I, D-V, integrated over omega.
The whole thing is equal to zero.
Okay.
The next step, we invoke off the divergence theorem on the first integral.
And in doing so,
we observe that since the i index has been contracted out here, sum being applied,
that term in parentheses is essentially a vector, right?
A vector with free index j.
We recognize that we have properly here the divergence of that vector, right.
That vector for our purposes is something that you
may probably write as w dot sigma, okay.
It means we just have the divergence of that vector.
Right, so now we apply the divergence theorem on.
That term,
right?
And in doing so, what we get is integral over the boundary, okay?
Wi sigma ij mj ds minus,
the other terms don't
get don't change.
Minus w i comma j sigma
i j d V plus integral over
omega w i f i d V equals 0.
Okay.
We have this.
And, what we are going to do now is, is two things.
I'm going to rearrange things a little.
And the way I'm going to rearrange this is by observing that
this term has a negative sign.
So I'm going to move it to the other side.
But I'm going to change my left and right hand sides.