so this second exercise actually has a little funny story I don't know if you
found it on Evans book because it comes from there it's exercise 20 at the end
of the section of problems of chapter 2 and if you check there there is
written at the end of the text the Marley Siam reviews 27 1985 page 69 71
because this exercise is actually the content of this little paper was title
is quite interesting the title is a simple proof that the world is
three-dimensional indeed the offer this Tom Marley argues that you can prove
that I mean it's also funny of course it's not completely serious it says
that Huyen principle says that distortion less wave propagation so
wave propagation of a certain form is possible only in all dimension actually
radially symmetric wave propagation he says it's possible only in dimension 1
and 3 which is the content of this exercise and therefore since we are not
on the line we are in dimension 3 okay it's only for fun the deeper meaning of
this exercise is I mean the idea the reason why I gave you is that it's not
unusual in many applications in in physics but also in other fields to
look for solutions of some PD with special forms and then there is of
course a basic question is it possible and when is it possible to have
solutions of this given form let's even assume that the functions are
smooth enough so that everything makes sense in a classical sense but still can
we have it or not so in this precise more precisely you were asked to check
her under which conditions we can write our solution to the wave equation in our
end excluding the origin times are so minus infinity plus infinity for you
given by some alpha of the norm of X where alpha is a c2 function 0 plus
infinity times Phi where Phi is a c2 function of our evaluated in t minus B
of the norm of X where B beta sorry is a c2 of a 0 plus infinity continues
up to 0 non-negative and equal to 0 in the origin why then why am I taking away
the origin here of course because the norm of X is not differentiable and let
alone c2 what is the point of this I want to say that the propagation on my
wave follows a certain profile Phi I mean may follow any profile Phi to
given any Phi I can always have this solution which can be written in this
way there is some function in front which depends only on the norm of X so
how far Phi I am from the origin in the space and the dependence in time is given
by this profile function Phi evaluated in the time minus another function which
depends only on the length on the norm of the vector X and is giving me
basically a delay because it's non-negative and I'm subtracting it
from the time so when can I can write my solution in this way for any choice of
Phi the answer is for non-trivial cases so for alpha different from zero yes I
mean of course alpha should not be zero otherwise everything is trivial even
only if n is equal to 1 and 3 and in that case you have to show which which
is the form of the functions alpha and beta because it turns out they have a
unique expression also now no it's not difficult to find the the second
derivative of this kind of u in time it's simply alpha of the norm of X 5
second of T minus beta of X the gradient of u will be alpha prime
norm of X times X over the norm of X Phi of T minus beta of X minus alpha
norm of X Phi prime of T minus beta norm of X beta prime of norm of X X over
norm of X the minus comes from here of course now notice that the divergence of
X over the norm of X is equal to the sum over J from 1 to n d index J XJ over
norm of X and this is 1 over norm of X minus XJ squared over norm of X to the
power 3 but therefore this is n over norm of X minus norm of X squared over
norm of X to power 3 so it's n minus 1 over the norm of X for all X nonzero and
if n equal 1 therefore it is 0 some of you may now may notice that if n equal
1 here there is written the sign of X and so if n equal 1 we get actually twice
the Dirac delta centered in the origin but but we don't care X should not be
0 so outside of the origin we get this function everything is differential so
we get this function so that we can write the Laplacian of u now these are a
lot of boring calculations but everything is doable notice that
whenever we take a second derivative of alpha prime of alpha or beta we get
another factor X divided by its norm so that multiplying scalarly with X divided
by norm by its norm becomes 1 because again we are not in the origin so you
can check all of these in the notes I'm not going to read everything we can
reorder this more interesting all these terms let the Laplacian of u will be
written as phi t minus beta norm of X multiplied by alpha second norm of X
plus alpha prime norm of X times n minus 1 over norm of X plus phi prime t
minus times minus 2 alpha prime norm of X beta prime of norm of X minus alpha
beta second minus alpha beta prime and minus 1 over the norm of X and then phi
second of t minus beta norm of X alpha beta prime squared therefore utt minus
Laplacian u well it will be the same thing is the only difference is that in
phi second there is also an additional term alpha of X here and all of this
should be 0 this equation should be z I mean this this function here should be
equal to 0 for all phi's C2 over R and this means that each coefficient of phi
prime 5 second and phi itself should be 0 for all X in Rn minus the origin how
to see this when one can first select phi to be constant so that this then
vanish here there is a constant so this should be 0 alpha second plus alpha
prime over n minus 1 over norm of X then if you take phi of Z equal to Z
well this first term is already vanished this one vanishes and this gives
that also this coefficient to be 0 and finally phi of Z equal to Z square over 2
that's a standard way to justify this fact now if we call norm of X R because
yeah these functions are actually regularly symmetric so they depend
actually on only one variable all this system here can be rewritten in this way
it becomes a system of all these so alpha times 1 minus beta prime squared
equal to 0 2 alpha prime beta prime plus alpha beta second plus alpha beta
prime times n minus 1 over R equal to 0 and alpha second plus alpha prime n
minus 1 over R equal to 0 notice we have two unknown functions and three
differential equations that's bad it means that there are more conditions
that then unknowns so there will be there is the point is that there is
actually another unknown to be that is going to be to be found and it is the
dimension of the space and over here and here now since alpha is not
identically 0 otherwise everything is trivial the first equation give us beta
prime equal squared equal to 1 therefore beta prime equal to 1 or beta
prime equal to minus 1 remember that beta is C2 so it can beta prime in
particular is continuous is actually C1 but it is continuous so given that it
can have only two values it must be either identically equal to 1 or
either identically equal to minus 1 but since beta is 0 0 and is no negative
then beta prime must be equal to 1 and beta bar should be equal to R why so
because if we choose beta prime equal to minus 1 then integrate from 0 to R we
get the beta of R beta of 0 0 beta of R equal to minus R so it will be negative
which is against the assumptions wonderful so beta of R is equal to R but
this means that beta second is 0 so this term over here and in the second
equation vanish so we get this system two way alpha prime plus alpha n minus
1 over R equal to 0 and alpha second plus n minus 1 over R alpha prime equal
to 0 so the first equation is very simple it's an Euler type equation I
would say alpha of R is equal to some constant say alpha 1 R to the n minus 1
over 2 alpha 1 because if R is equal to 1 and now we plug in this solution in
the second equation notice that alpha prime is equal to minus n minus 1 over 2
alpha 1 R to the minus n minus 1 over 2 minus 1 and alpha second give us as n
minus 1 over 2 n plus 1 over 2 alpha prime alpha 1 sorry R to the n minus 3 m
plus 3 sorry over 2 therefore the second equation can be reduced to this becomes
this alpha of 1 so here there is n squared minus 1 over 4 n minus 1 times
well there is a factor minus so 1 minus n over 2 R to the power minus n plus 3
over 2 because there is a factor R here that gives a minus 2 here but this should hold
alpha 1 is different from 0 of course this should hold for all R positive and
therefore this equation should be 0 but this is a second order this is a
quadratic equation and so it's even it's actually trivial the solutions are n
equal 1 or n equal 3 if n equal 1 the exponent of R is 0 so alpha is actually
a constant not 0 if n equal 3 the exponent becomes minus 1 so alpha of R is a
constant divided by R