WEBVTT - generated by VCS
1
00:00:03.898 --> 00:00:07.167
Have a wonderful good day, ladies and gentlemen.
2
00:00:07.167 --> 00:00:10.500
I would like to extend a warm welcome to you
for another contribution
3
00:00:10.500 --> 00:00:12.667
in the electromagnetic theory series.
4
00:00:12.667 --> 00:00:16.167
My name is Hans Georg Krauthäuser, I am the
holder of the professorship
5
00:00:16.167 --> 00:00:20.800
for Electromagnetic Theory and Compatibility
at TU Dresden.
6
00:00:20.800 --> 00:00:27.700
Today we will deal with another contribution
in the block "Electrostatics".
7
00:00:27.700 --> 00:00:33.900
We will deal with the formal solution of the
basic problem of electrostatics,
8
00:00:33.900 --> 00:00:42.500
so deal with the Poisson equation in the event
that we also have boundary conditions in the
9
00:00:42.500 --> 00:00:44.333
To have at last.
10
00:00:44.333 --> 00:00:51.867
Yes, and for that we will start with a little
excursion
11
00:00:51.867 --> 00:00:58.967
actually into mathematics because the important
subject matter of Green's functions
12
00:00:58.967 --> 00:01:02.000
maybe not quite as present anymore.
13
00:01:02.000 --> 00:01:07.000
So: Green's function or fundamental solutions
are always assigned
14
00:01:07.000 --> 00:01:12.000
a linear differential operator, which we call
L here.
15
00:01:12.000 --> 00:01:18.000
This linear differential operator acts on
functions, actually better said
16
00:01:18.000 --> 00:01:26.000
on distributions which are defined on a part
of the space of the R_n.
17
00:01:26.000 --> 00:01:33.000
Such a Green function or also a fundamental
solution, we shall call it G.
18
00:01:33.000 --> 00:01:41.000
As already said, G of x and s is assigned
to an operator L and then becomes
19
00:01:41.000 --> 00:01:46.000
evaluated at a point s from omega.
20
00:01:46.000 --> 00:01:51.000
When is a Green function a function or a fundamental
solution?
21
00:01:51.000 --> 00:01:57.000
Then, if it fulfills this very special inhomogeneous
differential equation,
22
00:01:57.000 --> 00:02:07.000
that is, if L applied to Green's function
is equal to the delta of x minus s.
23
00:02:07.000 --> 00:02:12.000
This is considered to be a special inhomogeneity.
24
00:02:12.000 --> 00:02:17.000
Normally the inhomogeneity will look different,
but we'll see how
25
00:02:17.000 --> 00:02:26.000
from the solution of the equation for this
inhomogeneity to the general one
26
00:02:26.000 --> 00:02:27.000
Solution can close.
27
00:02:27.000 --> 00:02:31.000
Let's do that now, just a few small steps.
28
00:02:31.000 --> 00:02:37.000
And for that we will first of all get the
equation we have up here from
29
00:02:37.000 --> 00:02:46.000
multiply on the right by any function f of
s and then on both sides
30
00:02:46.000 --> 00:02:51.000
form the integral ds.
31
00:02:51.000 --> 00:02:54.000
And then we have what's down here.
32
00:02:54.000 --> 00:03:01.000
And the right integral - the integral on the
right side - can immediately
33
00:03:01.000 --> 00:03:03.000
be evaluated immediately.
34
00:03:03.000 --> 00:03:08.000
Because here is the integral delta x minus
sf of sd s.
35
00:03:08.000 --> 00:03:15.000
And by the definition of the delta function
, that's just f of x.
36
00:03:15.000 --> 00:03:25.000
On the other hand, the differential operator
is currently below the
37
00:03:25.000 --> 00:03:30.000
Integral, but according to the assumption that
is a linear operator and he
38
00:03:30.000 --> 00:03:35.000
also only affects x - so it was introduced.
39
00:03:35.000 --> 00:03:40.000
So it is clear that we can pull it in front
of the integral.
40
00:03:40.000 --> 00:03:47.000
That is, L applied to this integral is just
f of x.
41
00:03:47.000 --> 00:03:55.000
And f of x is an arbitrary function - that
is, it appears here as arbitrary
42
00:03:55.000 --> 00:04:02.000
Inhomogeneity. So it is already clear what
the whole thing is going to be.
43
00:04:02.000 --> 00:04:10.000
So if you are looking for the solution of an
inhomogeneously linear differential equation
of the form L.
44
00:04:10.000 --> 00:04:17.000
of y equals f of x, with any inhomogeneity
and one knows one
45
00:04:17.000 --> 00:04:25.000
Green's functions, a fundamental solution G
of x and s already, ie
46
00:04:25.000 --> 00:04:29.000
we already know the solution of this differential
equation, which we do
47
00:04:29.000 --> 00:04:31.000
have introduced up here.
48
00:04:31.000 --> 00:04:33.000
So the G is known.
49
00:04:33.000 --> 00:04:42.000
Then you can immediately see that the solution
to any inhomogeneity f
50
00:04:42.000 --> 00:04:50.000
of x is given in the form: integral G of x
and sf of sd s.
51
00:04:50.000 --> 00:04:56.000
In other words, one knows a Green's function
to be a linear one
52
00:04:56.000 --> 00:05:01.000
Differential operator, then one can use the
corresponding differential equation for
53
00:05:01.000 --> 00:05:09.000
solve any inhomogeneities - can write down
the solution in the form immediately.
54
00:05:09.000 --> 00:05:15.000
Let's take a look at some of the properties
of the Green function.
55
00:05:15.000 --> 00:05:21.000
L - our linear differential operator - that
is a mapping.
56
00:05:21.000 --> 00:05:27.000
And in general it will be the case that the
kernel of L is not trivial.
57
00:05:27.000 --> 00:05:33.000
Trivial would mean that only the zero function
is mapped to 0.
58
00:05:33.000 --> 00:05:38.000
In general, other functions are mapped to
0 by the operator.
59
00:05:38.000 --> 00:05:41.000
In that case the kernel of L is not trivial.
60
00:05:41.000 --> 00:05:50.000
Then the problem of L applied to G equals
the delta function entirely
61
00:05:50.000 --> 00:05:52.000
obviously infinite solutions.
62
00:05:52.000 --> 00:06:00.000
You could go and find a solution G of x and
s any element from the
63
00:06:00.000 --> 00:06:02.000
Add core of L, add to it.
64
00:06:02.000 --> 00:06:07.000
And because the operator is linear, the whole
thing would still be a solution.
65
00:06:07.000 --> 00:06:09.000
That will be the rule.
66
00:06:09.000 --> 00:06:13.000
In the case of the operators we are considering,
it will usually be the case that we
67
00:06:13.000 --> 00:06:15.000
actually have an infinite number of solutions.
68
00:06:15.000 --> 00:06:23.000
And that is actually something that we will
make use of for ourselves.
69
00:06:23.000 --> 00:06:27.000
That is, this ambiguity is not bad for us
, on the contrary, it is
70
00:06:27.000 --> 00:06:28.000
is very welcome.
71
00:06:28.000 --> 00:06:32.000
Because we don't just have a homogeneous differential
equation, we have
72
00:06:32.000 --> 00:06:38.000
have also given boundary values on the
edges of the solution area.
73
00:06:38.000 --> 00:06:44.000
It is only through the ambiguity of Green's
function that it will succeed
74
00:06:44.000 --> 00:06:51.000
adapt so that the boundary values are
then also fulfilled.
75
00:06:51.000 --> 00:07:00.000
Without proof, it should be mentioned that
Green's functions are adjoint
76
00:07:00.000 --> 00:07:01.000
are symmetrical, ie
77
00:07:01.000 --> 00:07:08.000
we have the identity here, so if we move on
to adjoint only if
78
00:07:08.000 --> 00:07:09.000
we also swap the arguments.
79
00:07:09.000 --> 00:07:15.000
However, it is often the case that the associated
differential operator is a self
80
00:07:15.000 --> 00:07:20.000
adjoint operator, i.e. L is equal to L *.
81
00:07:20.000 --> 00:07:25.000
And in this case, which will be the most common
for us, is Green's function,
82
00:07:25.000 --> 00:07:29.000
then actually symmetrical. That too should
be noted.
83
00:07:29.000 --> 00:07:36.000
As a rule, with us, Green's functions are
symmetrical functions.
84
00:07:36.000 --> 00:07:43.000
Let's get back to the boundary value problems
and now we just apply what we think of
85
00:07:43.000 --> 00:07:45.000
Green's functions rely on our boundary value
problems.
86
00:07:45.000 --> 00:07:51.000
The concrete boundary value problem we have
is the Poisson equation.
87
00:07:51.000 --> 00:07:57.000
I have now written the r to the Laplace operator
again with ran to
88
00:07:57.000 --> 00:08:04.000
to always make it clear to which size r or
r 'this is actually affecting .
89
00:08:04.000 --> 00:08:14.000
Delta r phi of r is equal to minus 1 epsilon_0
roh_V of r, that's ours
90
00:08:14.000 --> 00:08:17.000
Differential equation that we consider.
91
00:08:17.000 --> 00:08:23.000
And this operator - the Laplace operator -
is self-adjoint.
92
00:08:23.000 --> 00:08:29.000
With that, Green's function would actually
also be seen as symmetrical.
93
00:08:29.000 --> 00:08:34.000
The equation for the Green function - just
repeat it
94
00:08:34.000 --> 00:08:42.000
written down- would be Laplace of G is equal
to minus 1 through
95
00:08:42.000 --> 00:08:44.000
epsilon_0 delta of r minus r '.
96
00:08:44.000 --> 00:08:49.000
One might ask: yes, why have we now left this
factor here?
97
00:08:49.000 --> 00:08:55.000
That is insignificant, we could have dragged
it into the operator, but
98
00:08:55.000 --> 00:08:58.000
it is just common practice to do it this way
now.
99
00:08:58.000 --> 00:09:04.000
This is not really a different homogeneity,
so we can do that too
100
00:09:04.000 --> 00:09:08.000
still call it Green's function.
101
00:09:08.000 --> 00:09:16.000
Yes, we have already got to know a different
identity for the delta function .
102
00:09:16.000 --> 00:09:21.000
We know that delta of r minus r 'is nothing
other than minus 1
103
00:09:21.000 --> 00:09:28.000
by 4 pi Laplace of 1 by r minus r 'amount.
104
00:09:28.000 --> 00:09:33.000
And only when I compare that , the Laplace
operator is included in both cases.
105
00:09:33.000 --> 00:09:37.000
So you could use the delta function up here
and then just do it
106
00:09:37.000 --> 00:09:40.000
do a term comparison.
107
00:09:40.000 --> 00:09:47.000
In this way you can immediately have a Green
function of this
108
00:09:47.000 --> 00:09:51.000
Read the operator of the Laplace operator,
ie
109
00:09:51.000 --> 00:09:57.000
we already have a Green function immediately,
without having to continue calculating
110
00:09:57.000 --> 00:10:05.000
found and this Green's function of the Laplace
operator is 1 through 4 pi epsilon_0
111
00:10:05.000 --> 00:10:09.000
by r minus r 'amount.
112
00:10:09.000 --> 00:10:18.000
Anyone can actually do this special Green
function
113
00:10:18.000 --> 00:10:24.000
Element gamma -we call the times- the kernel
of the Laplace operator- and that is
114
00:10:24.000 --> 00:10:30.000
the Laplace equation is finally added, ie
115
00:10:30.000 --> 00:10:34.000
we have a solution here - here a solution to
Poisson's equation - and we
116
00:10:34.000 --> 00:10:38.000
can be any solution of the to this solution
of the Poisson equation
117
00:10:38.000 --> 00:10:40.000
Add Laplace equation
118
00:10:40.000 --> 00:10:42.000
and it's still a solution to Poisson's equation.
119
00:10:42.000 --> 00:10:45.000
This is quite obvious - it is due to the linearity
of the operator.
120
00:10:45.000 --> 00:10:54.000
And here, however, the gamma must actually
be a symmetrical function.
121
00:10:54.000 --> 00:10:57.000
Because the Laplace operator itself is adjoint.
122
00:10:57.000 --> 00:11:04.000
I. E. without us changing the name now - we
will
123
00:11:04.000 --> 00:11:09.000
Always call Green's function G - Green's function
would now be in the form
124
00:11:09.000 --> 00:11:14.000
write that G is just this special Green function,
1 through 4 pi
125
00:11:14.000 --> 00:11:17.000
epsilon_0 by r minus r 'amount.
126
00:11:17.000 --> 00:11:25.000
And then just plus a gamma of r and r ', which
is the solution to Laplace's equation
127
00:11:25.000 --> 00:11:27.000
and that is symmetrical.
128
00:11:27.000 --> 00:11:37.000
Yes, and with the preparations we can now
go so that we can do that too
129
00:11:37.000 --> 00:11:46.000
Calculate the scalar potential, we will actually
do it quite analogously to the species
130
00:11:46.000 --> 00:11:51.000
and way as we have already calculated the
general scalar potential, ie
131
00:11:51.000 --> 00:11:59.000
we go back to Green's second identity, which
is for any two
132
00:11:59.000 --> 00:12:02.000
scalar fields f and g had already been written
down.
133
00:12:02.000 --> 00:12:09.000
And now we replace f and g in such a way that
we take f equal to phi - the scalar potential
134
00:12:09.000 --> 00:12:14.000
-and g we are just taking the Green function.
135
00:12:14.000 --> 00:12:19.000
We are allowed to do this because Green's second
identity holds for any f and g.
136
00:12:19.000 --> 00:12:24.000
That would be one side of the equation to
begin with.
137
00:12:24.000 --> 00:12:32.000
And of course we can reform that because we
have here in the first
138
00:12:32.000 --> 00:12:36.000
Summands under which integral Laplace G stand.
139
00:12:36.000 --> 00:12:41.000
But Laplace G is minus 1 through epsilon_0
times the delta function.
140
00:12:41.000 --> 00:12:49.000
The second term, that's where we have Laplace
phi.
141
00:12:49.000 --> 00:13:01.000
And according to Poisson's equation, this is
nothing more than minus 1 through epsilon_0
times rho.
142
00:13:01.000 --> 00:13:05.000
With the minus, this minus becomes a plus here.
143
00:13:05.000 --> 00:13:15.000
The other side of the second Green intensity
is now simply written down here again.
144
00:13:15.000 --> 00:13:19.000
Yes, and just as we used last time also
145
00:13:19.000 --> 00:13:25.000
have already seen: Here the first integral,
that is of course special
146
00:13:25.000 --> 00:13:35.000
simply because here phi times the delta function
and delta of r minus r 'results
147
00:13:35.000 --> 00:13:39.000
then just put the integrand at the front here
at r.
148
00:13:39.000 --> 00:13:47.000
That is, the integral is nothing other than
phi of r and so we can do that
149
00:13:47.000 --> 00:13:51.000
Then of course dissolve the whole according
to phi of r and get one again
150
00:13:51.000 --> 00:13:55.000
Expression that looks similar to what we had
before.
151
00:13:55.000 --> 00:14:04.000
Phi of r is thus integral rho _V at the point
r ' times G of r and r', integrated
152
00:14:04.000 --> 00:14:07.000
over the entire room times epsilon_0.
153
00:14:07.000 --> 00:14:11.000
And then there is a surface integral again
with the two terms, where once the
154
00:14:11.000 --> 00:14:19.000
It contains the normal derivation of the potential
and the potential itself.
155
00:14:19.000 --> 00:14:24.000
So again our Neumann and Dirichletschen fringe
terms.
156
00:14:24.000 --> 00:14:29.000
We now have the free choice of gamma.
157
00:14:29.000 --> 00:14:31.000
The gamma is in the G here.
158
00:14:31.000 --> 00:14:38.000
And that will now allow us to adapt to the
given boundary conditions
159
00:14:38.000 --> 00:14:42.000
to undertake. That's what we'll look at next.
160
00:14:42.000 --> 00:14:46.000
We start with the Dirichletschen boundary conditions.
161
00:14:46.000 --> 00:14:53.000
That is, we assume that phi is given on the
surface of V.
162
00:14:53.000 --> 00:14:58.000
We simply wrote down the term - this is the
complete term - again.
163
00:14:58.000 --> 00:15:05.000
Well, if phi is given on the surface, then
that is it
164
00:15:05.000 --> 00:15:11.000
here is a term that we want to keep and the
other term is there
165
00:15:11.000 --> 00:15:13.000
Normal derivation in there, but not phi itself.
166
00:15:13.000 --> 00:15:18.000
That is, of course, we would like to get rid
of this term.
167
00:15:18.000 --> 00:15:25.000
In other words, the wish would be to choose
Gamma - since we are free to do so - just that
168
00:15:25.000 --> 00:15:30.000
this term, which is written down again here,
just disappears.
169
00:15:30.000 --> 00:15:34.000
Yes and how can you do that?
170
00:15:34.000 --> 00:15:38.000
There's an obvious approach to this, and it
is one too
171
00:15:38.000 --> 00:15:40.000
Approach that is often used.
172
00:15:40.000 --> 00:15:46.000
You just do it so that you now select the
gamma so that it is out of it
173
00:15:46.000 --> 00:15:53.000
resulting G on the surface of V vanishes.
174
00:15:53.000 --> 00:16:00.000
If that is the case, if the gamma could be
selected in such a way that G on the
175
00:16:00.000 --> 00:16:05.000
Surface of the volume disappears, then we
have a zero here.
176
00:16:05.000 --> 00:16:11.000
So the entire integral is 0 and this term falls
out of the
177
00:16:11.000 --> 00:16:14.000
Just look out.
178
00:16:14.000 --> 00:16:20.000
In other words, the expression is reduced accordingly
to an expression in which only
179
00:16:20.000 --> 00:16:25.000
still known quantities are in it, namely the
charge distribution in the volume, which
180
00:16:25.000 --> 00:16:33.000
Greens function that we of course need for
this and the potential in the places
181
00:16:33.000 --> 00:16:36.000
the surface, which is given according to the
assumption.
182
00:16:36.000 --> 00:16:42.000
That means that one could actually use it to
calculate the scalar potential.
183
00:16:42.000 --> 00:16:45.000
We have a specific calculation rule - everything
would be known.
184
00:16:45.000 --> 00:16:49.000
It then only has to be calculated.
185
00:16:49.000 --> 00:16:55.000
Yes, and let's look at that for Neumann's
boundary conditions.
186
00:16:55.000 --> 00:16:59.000
There is not the potential itself on the surface
, but the
187
00:16:59.000 --> 00:17:01.020
Give normal derivative of the potential.
188
00:17:01.020 --> 00:17:07.020
And we know, normal derivatives of the potential
-also gradient phi times vector n-
189
00:17:07.020 --> 00:17:12.020
is nothing else than minus E times vector n.
190
00:17:12.020 --> 00:17:14.020
n is the normal vector of the surface.
191
00:17:14.020 --> 00:17:18.020
That means, given the normal derivative of
the potential means nothing else than that
the
192
00:17:18.020 --> 00:17:24.020
Normal component, the negative normal component
of the electric field
193
00:17:24.020 --> 00:17:27.020
the place in the surface is given.
194
00:17:27.020 --> 00:17:30.020
Yes, the same procedure in principle.
195
00:17:30.020 --> 00:17:37.020
We first write down the entire expression again
and then now would be the wish
196
00:17:37.020 --> 00:17:42.020
of course, that this second summand disappears
back here.
197
00:17:42.020 --> 00:17:47.020
Speak, more precisely, the whole integral with
the second summand as the integrand
198
00:17:47.020 --> 00:17:54.020
disappears because there is the unknown potential
in there.
199
00:17:54.020 --> 00:17:59.020
We have just given the normal derivations
on the
200
00:17:59.020 --> 00:18:01.080
Surface but not the potential itself.
201
00:18:01.080 --> 00:18:02.080
So you try to get rid of this term.
202
00:18:02.080 --> 00:18:09.080
So the wish would be to get this integral
to 0 , very similar to the one we just did
.
203
00:18:09.080 --> 00:18:15.080
The obvious approach, I'll say right away it
won't work, is that
204
00:18:15.080 --> 00:18:20.080
one now demands that the normal derivatives
of the Greens function
205
00:18:20.080 --> 00:18:26.080
should disappear here on the surface.
206
00:18:26.080 --> 00:18:30.080
We can now see why that won't work.
207
00:18:30.080 --> 00:18:37.080
Because on the one hand we can look at this
integral.
208
00:18:37.080 --> 00:18:42.080
Integral over Laplace G taken over the volume.
209
00:18:42.080 --> 00:18:52.080
Yes, because G is the very special solution
to the Laplace operator
210
00:18:52.080 --> 00:18:56.080
Is inhomogeneity, you can write it down here
in this way.
211
00:18:56.080 --> 00:19:02.040
So that's just our Poisson equation with the
special one
212
00:19:02.040 --> 00:19:08.040
Inhomogeneity of the delta function. And you
can use the integral over the delta function
213
00:19:08.040 --> 00:19:11.040
evaluate of course.
214
00:19:11.040 --> 00:19:18.040
That just gives a 1 by definition of the delta
function, so here for that
215
00:19:18.040 --> 00:19:23.040
On the one hand, it can be calculated integrally
that it is minus 1 through epsilon_0.
216
00:19:23.040 --> 00:19:34.040
On the other hand, we can also go a different
way with the same integral .
217
00:19:34.040 --> 00:19:38.040
We can write down Laplace as a divergence gradient.
218
00:19:38.040 --> 00:19:44.040
Then there is a volume integral over the divergence
of something.
219
00:19:44.040 --> 00:19:50.040
And we can convert that into the corresponding
surface integral according to Gauss
220
00:19:50.040 --> 00:19:53.040
the function itself. So here the divergence
no longer stands.
221
00:19:53.040 --> 00:20:06.000
The term that appears here, gradient times
n, that's just the normal derivative
222
00:20:06.000 --> 00:20:11.000
from G, that is, it is nothing other than this
expression.
223
00:20:11.000 --> 00:20:19.000
And when I now try the approach , I try the
normal derivation
224
00:20:19.000 --> 00:20:27.000
would be equal to zero, then this entire integral
would be equal to 0.
225
00:20:27.000 --> 00:20:32.000
On the other hand, we have already shown that
epsilon_0 equals 1 and it
226
00:20:32.000 --> 00:20:36.000
cannot be 0 and minus 1 at the same time through
epsilon_0.
227
00:20:36.000 --> 00:20:40.000
Thus, this approach obviously leads to a contradiction,
228
00:20:40.000 --> 00:20:46.000
that is, it doesn't work that way, even if
it had been easy.
229
00:20:46.000 --> 00:20:50.000
So we have to start all over again.
230
00:20:50.000 --> 00:20:54.000
We write down the entire term again and think
about, well, what is it
231
00:20:54.000 --> 00:20:57.000
the next best approach we could take instead.
232
00:20:57.000 --> 00:21:04.060
The next best approach would be for the integral
not to become 0, but maybe it will
233
00:21:04.060 --> 00:21:06.060
at least constant.
234
00:21:06.060 --> 00:21:12.060
The idea behind it, if it's constant, then
we just have an additional one
235
00:21:12.060 --> 00:21:17.060
constant factor that is added to the total
potential.
236
00:21:17.060 --> 00:21:21.060
But it doesn't really bother you because it
affects the transition to field strength
237
00:21:21.060 --> 00:21:24.060
is just lost again.
238
00:21:24.060 --> 00:21:29.060
Because E is equal to minus Gradient_phi,
a constant bothers us
239
00:21:29.060 --> 00:21:34.060
not in potential. The potential is indefinite
anyway except for a constant.
240
00:21:34.060 --> 00:21:38.060
So that would be the next best approach.
241
00:21:38.060 --> 00:21:45.060
One says ok, minus epsilon_0 times this integral
should be phi_0.
242
00:21:45.060 --> 00:21:50.060
We just take the minus epsilon_0 with us so
that it comes out nicely afterwards.
243
00:21:50.060 --> 00:21:59.060
Often one now chooses a special one in order
to fulfill this so that it becomes constant
244
00:21:59.060 --> 00:22:09.020
Value for the normal derivative. One demands
that the normal derivative of G, the value
245
00:22:09.020 --> 00:22:15.020
assume minus 1 by epsilon_0 times S.
246
00:22:15.020 --> 00:22:20.020
And here S is nothing other than this total
surface of this volume.
247
00:22:20.020 --> 00:22:26.020
So this surface O of S has the value, the size
S.
248
00:22:26.020 --> 00:22:36.020
If we do that, we just need to use it accordingly.
249
00:22:36.020 --> 00:22:45.020
We can of course extract this constant value
from the integral and use it for phi_0
250
00:22:45.020 --> 00:22:52.020
then this expression remains, which we can
easily interpret.
251
00:22:52.020 --> 00:23:02.080
This is nothing more than the mean value of
the potential on the surface of V.
252
00:23:02.080 --> 00:23:12.080
If you then put all of that in, then we would
get from phi, more precisely for phi
253
00:23:12.080 --> 00:23:21.080
of r minus phi_0, i.e. minus the mean value
on the surface, just this
254
00:23:21.080 --> 00:23:25.080
Expression in which everything is now determined
again.
255
00:23:25.080 --> 00:23:29.080
The rho_v, the Green function, is known up
to now.
256
00:23:29.080 --> 00:23:37.080
And the normal derivation of phi is given according
to the assumption.
257
00:23:37.080 --> 00:23:42.080
That means, in this case too, we would now
be finished.
258
00:23:42.080 --> 00:23:46.080
We could, everything is known, everything can
be calculated clearly.
259
00:23:46.080 --> 00:23:54.080
We swept a little something under the carpet
here.
260
00:23:54.080 --> 00:24:01.040
This little thing is: how do you do that so
that you actually do that now
261
00:24:01.040 --> 00:24:08.040
gets concrete G, ie, what do concrete solution
procedures actually look like .
262
00:24:08.040 --> 00:24:15.040
We actually have to have a speaking gamma
- and with it
263
00:24:15.040 --> 00:24:19.040
finally find the G- too.
264
00:24:19.040 --> 00:24:25.040
When we find it, we're done.
265
00:24:25.040 --> 00:24:33.040
And the concrete methods for the solution,
for finding the gamma and thus the G, the
266
00:24:33.040 --> 00:24:37.040
we will actually consider now in the following.
267
00:24:37.040 --> 00:24:40.040
Not in this block, then in the next blocks.
268
00:24:40.040 --> 00:24:45.040
But I would like to point out which methods
these will be.
269
00:24:45.040 --> 00:24:52.040
We will start with the mirroring method or
also called the image loading method.
270
00:24:52.040 --> 00:25:00.000
And the idea is there, you just place extra
charges, but not now
271
00:25:00.000 --> 00:25:06.000
within V, that would be forbidden because that
would mean the first integral
272
00:25:06.000 --> 00:25:11.000
-I'm going there again- with that we would
influence this integral and that
273
00:25:11.000 --> 00:25:13.000
we are certainly not allowed to do it.
274
00:25:13.000 --> 00:25:26.000
Rather, we place charges outside of V, precisely
in such a way that the field what
275
00:25:26.000 --> 00:25:30.000
that generate, or the potential that they
generate with the
276
00:25:30.000 --> 00:25:33.000
Superposition of the potential or
277
00:25:33.000 --> 00:25:40.000
of the field of the original charge just met
the boundary conditions.
278
00:25:40.000 --> 00:25:46.000
If we can do that - that will depend a bit
on the geometry like
279
00:25:46.000 --> 00:25:51.000
that's easy - but if we can do that, then
it's a very nice one
280
00:25:51.000 --> 00:25:58.000
Possibility of defining the gamma and thus
ultimately the G.
281
00:25:58.000 --> 00:26:01.060
And then we actually found the solution.
282
00:26:01.060 --> 00:26:07.060
The second thing we will do is develop according
to orthogonal functions.
283
00:26:07.060 --> 00:26:13.060
Orthogonal functions: You know function spaces,
you know that you are in
284
00:26:13.060 --> 00:26:23.060
Function spaces also have bases and such orthogonal
functions are even,
285
00:26:23.060 --> 00:26:28.060
if they are also complete, they correspond
to bases of functional spaces.
286
00:26:28.060 --> 00:26:37.060
And we can, if we can now find a base that
fits the geometry well
287
00:26:37.060 --> 00:26:42.060
is adapted so that the basic function can
then meet the boundary conditions
288
00:26:42.060 --> 00:26:48.060
we just general solutions as superpositions
of these basis functions
289
00:26:48.060 --> 00:26:52.060
represent and then also have the boundary conditions
290
00:26:52.060 --> 00:26:56.060
met and thus solved the problem.
291
00:26:56.060 --> 00:27:03.020
And the last thing you can do is a mathematically
oriented process and
292
00:27:03.020 --> 00:27:10.020
is only possible for very specific geometries:
the separation process
293
00:27:10.020 --> 00:27:14.020
the variables you might get from other mathematics
contexts
294
00:27:14.020 --> 00:27:17.020
with ordinary differential equations, with
partial differential equations
295
00:27:17.020 --> 00:27:19.020
have already met.
296
00:27:19.020 --> 00:27:25.020
Again, we'll see how we apply it to our underlying
problem
297
00:27:25.020 --> 00:27:29.020
Be able to apply boundary value problems.
298
00:27:29.020 --> 00:27:38.020
Well, that would actually have been our program
for this session now.
299
00:27:38.020 --> 00:27:44.020
Thank you for your attention and, as always,
you will find more information
300
00:27:44.020 --> 00:27:48.020
also on the specified website. Thank you very
much.