So the point of this exercise was basically to define suitably this
function u and of course since u of x c t should be equal to v of x t the
definition of u is that we put it equal to v of x t over c. Therefore of course u is
c2 over Rn times 0 plus infinity because v is c2 and this regularity is clearly
preserved by the composition with the map which sends x and t in x and t over
c, c is not zeros or positive so no problem. And now we have to check that
actually this u in this way defined is actually the solution of this initial
value problem. Of course if I take the second derivative of u I get 1 over c
squared the second derivative of t when instead the Laplacian of u is equal to
the Laplacian of v. Therefore u to t minus Laplacian u is equal to 1 over c
squared v to t of x evaluated in t over c minus Laplacian v of x Laplacian of v in x t over c.
Collecting the term 1 over c squared we get v to t minus c squared Laplacian v but this is 0 for
every x and t in Rn 0 plus infinity. And dividing by a positive factor c of course doesn't change
anything in the range of t. Then if we set t equal to 0 of course u x 0 is equal to v x 0 and
therefore it's equal to g by the assumption of v and instead u t of x 0 will be I mean u t of x t
will be 1 over c v t of x t over c. Therefore setting t equal to 0 we get the factor 1 over c
here then v t at time 0 is h so we get h over c and we are done. Now the second exercise well
consistent in having you to derive again Dalambert's formula in another way, if you want in a more
classical way in some sense. That is a vi change of variables. So in one dimension we have our
initial value problem for the wave equation so u to t minus c squared u x x equal to 0, u equal to g
at time 0 and u t equal to h at time 0, g c 2 in R h c 1. Point a was to show that if we have the
pd v x y for x and y both one dimensional variables equal to 0 in R 2 then the general solution is
given by f of x plus g of y for arbitrary function f and g c 1 of R. And this is interesting because
one would say that I need c 2 to make sense of this expression but notice that if I take a
derivative in say x the term g vanishes and I get f prime so I need f to be c 1. Then if I take a
derivative in y, f does not depend on y so vanish and this function is twice differentiable once in
x and once in y. It's not c 2. That's kind of interesting. Then part b was actually let's
exploit this fact to actually get to the solution of the wave equation that is consider the change
of variables psi equal to x plus c t and eta equal to x minus c t in order to show that u t t minus c
squared u x x is equal to 0 and r times 0 plus infinity if and only if u psi eta is equal to 0
in R 2. Then combine points a and b to get d'Alembert's formula. And finally a little more
interesting question. Under what conditions on the initial data g and h is the solution u of 1
a right moving or left moving wave? What does it mean basically? What does it mean? Right moving
wave means that seeing the graph as a function of x and t as a translation parameter the
translation is moving from left to right. That is basically that you are subtracting t and
vice versa for the left moving wave. So point a assume that v is c 1 over 2 such that v x and v y
are again differentiable in y and x respectively. These are the minimal assumptions to make the
equation make sense. Now define w to be v x then v x y equal to 0 implies that w of y equal to 0.
Well clearly and this makes sense because I'm assuming that v x is differentiable in y.
But then if w y is equal to 0 it means that w of x y is actually a function f small f let's
call it of x because it doesn't it's constant in y. For some f it should be at least 0 because
w x is a continuous function in x differentiable in y. I mean now we see it's constant.
Very well therefore v x is equal to this f of x but this is possible if and only if v x y
is equal to some capital f of f plus g of y where f prime of capital f prime is equal to f I'm
integrating and g is a constant in x and therefore depending only on y function.
Why I write g c prime c 1 sorry I mean one can say it should be only continuous
but the regulator g follows from the fact that v the function small small v should be c 1. Or one
can argue the argument is clearly specular in x and y so arguing the opposite way I should find
that g must be c 1 whereas f can be even c 0. So the only way that these two assumptions are
coherent is that both capital f and capital g are c 1. Now if we define xi and eta as before
it's clear that x is xi plus eta divided by 2 and t is xi minus eta divided by 2 c.
So that u x t can be seen as u of xi plus eta over 2 xi minus eta over 2 c. Therefore if I
take the derivative of u inside this is a chain rule I get 1 half u x evaluated in xi plus eta over
2 xi minus eta over 2 c plus 1 over 2 c u t evaluated etc. And then obviously if I take the
derivative of u in eta I will get 1 over 2 u x minus 1 over 2 c u t. Therefore if I consider u xi eta
that is that I derive this term in eta this time I will get 1 over 4 u x x minus 1 over 4 c u x t
plus 1 over 4 c u x t. So first I derive the first term and there you get the minus. Then if I derive
this term in the in x in the xi which is in the in the argument of x or should be x I get another
factor 1 over 2 with a plus. And then I take the last term of the chain rule so the derivative of
u again in t times dt in the xi that is 1 over 2 no in the eta sorry that is minus 1 over 2 c. So you
get a term minus 1 over 4 c squared u t t. So the mixed term simplify I collect 1 over 4 c squared
so that I get c squared u x x minus u t t. But this is the wave equation I mean change the sign but
therefore u xi eta equal to 0 if and only if u t t is equal to c squared u x x.
Now point c thanks to b it's enough to solve what we say xi eta equal to 0 in our term. Thanks to a
we know that this means that big c eta can be written as f of xi plus g of eta for some g and
f in c1 of r. But then u of x t is equal to u of xi plus eta over 2 times m. xi minus eta over 2 c.
This is clearly our function big xi eta therefore this is f of xi plus g of eta therefore this is
f of x plus ct plus g of x minus ct so the the true traveling waves. Now we exploit the initial
conditions so u of x0 should be small g but this is equal therefore to f of g f of x plus g of x.
u t x0 should be h and this must be therefore equal to c f prime of x minus c g prime of x
minus because there is a minus in front of t. But this is true if and only if well
the first equation remains the same in the second I integrate by parts and divide by c
no sorry I integrate by parts sorry I integrate from 0 to x on both sides all of this makes
sense because f capital f and capital g are c1 and I divide by c for simplicity. I get the term
here that is f of 0 minus g of 0 and this is a constant is in general non-zero don't worry it
will go away and now I'm done because if I say sum up these two equation I get the expression
for f that is one alpha g of x plus integral from 0 to x h over c plus this constant k and if instead
I subtract the second from the first I get an expression for g which is the same just with a
minus sign here and minus k therefore u x t is the sum of these two functions in particular
plus k and minus k simplify as it should f gives me g of x plus c t small g and g gives it small g
of x minus c t divided by 2 and then 1 over 2 c integral from 0 to x plus c t of h minus
integral from 0 to x minus c t of h of course x minus c t is less than x plus c t because
c and t are both positive and therefore we get the usual expression of d'Alembert formula and
finally f of x plus c t is a left moving wave because I mean the the in the geometrical intuition
is this if at time 0 this is the profile of the curve y equal to capital f of x when I subtract
sorry if I go to a point x minus c t in the graph of the function f of x plus c t I get again f of x
so given a translation c t inside the function I have to go to move to the left to be again
at the same level I mean to get again the same value for the function so for fixed I mean for
sorry the the profile of this function is moving towards the left as t positive increases
and the opposite g of x minus c t is moving to the right therefore u is the right moving
if and only if a capital f of x is equal to 0 but this means remember the expression f of x is equal
to this and therefore g of x small g of x plus 1 over c t integral from 0 to x h of y should be
equal to minus this constant for every x in R but you may say I don't know what is this constant
but this means that the map which sends x into small g of x plus 1 over c integral from 0 to x
h of y must be constant or you could express again this by saying that g prime plus 1 over c h
h of x should be equal to 0 for every x in R and analogously you will be a right moving wave if
and only if the oh I'm so sorry I should write left here yeah g is a left moving wave and only if
the right moving wave capital G satisfies the same condition therefore basically
this term is constant