I.1 If y(x) is a solution for x > 0, and z(x)=y(-x)
for x negative. One has:
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I.2 jl(x) = lj(x/l) follow the pre-requisite of question 2 and is a solution of (E) iff j is also one.
I.3 If j is a solution, then -j is also a solution. One therefore obtains the solutions of each quarter-of-plane from the ones defined on (x > 0,y > 0) by some axial symmetries, horizontal or vertical or both.
II.1.a (x2j ¢)¢ = 2xj¢ +x2j" = -x2/j < 0
II.1.b from the previous question x2j ¢ is a decreasing function, thus can vanish at most only once. If so, then from (E) we get that j¢¢ = -1/j < 0 at this point, thus it's a local maximum. Therefore, one cannot have a minimum.
II.1.c If x2j¢ stay with a constant sign, then it is also the case for j¢ and consequently j is a monotonous function. If there is a maximum, then j is increasing before and decreasing after.
II.2.a On the considered interval, y¢¢ is a function of x, y and y¢ that is of class C2 (at least) relatively to these 3 parameters. One has thus the local uniqueness if the value of the function and the value of its derivative are given at one point. Thus the equality of the derivatives at x0 would mean that the solutions are exactly equal, which is contradictory with the hypothesis.
II.2.b Using II.1.a one gets in a straightfoward way x2j2¢(x) -x2j1 ¢(x) = x02(j2¢(x0)-j 1¢(x0)) -òx0x t2(1/(j2) -1/(j1)) between x0 and x1 one has j2 > j1 because it's true in a neighborhood of x0 (by comparison of the derivatives at this point), and the two functions cannot be equal at a point closer than x1 by definition of that point. Using this in the previous calculation one gets that x2j 2¢(x) -x2 j1¢(x) > 0 ,thus j2¢(x) > j1¢(x)+ C with c > 0. By integration one gets j2(x1) > j1(x1) +c(x1-x0) which is incompatible with hypothesis H1.
II.3.a x2j ¢ is decreasing (II.1.b), so j ¢(x) < g2 j¢(g)/x2. After integration between g and x one gets that j>(x) < j(g) +g2 j¢ (g)(1/g-1/x), which implies that j is bounded.
II.3.b x2j ¢(x)- g2j ¢(g) = -ògx t2/j(t), but j < M from II.3.a, so x2j¢(x) < g2j¢ (g) = -ò gx t2/M and the right part of the inequality tends to -¥ when x tends to +¥.
II.3.c Since j ¢ tends to - ¥, after integration we necessarily get that j , which also tends to - ¥ and this is incompatible with j > 0. The H2 hypothesis is therefore not acceptable.
II.4.a From II.3.c we got that H2 is false, thus the definition interval of j is necessarily bounded. j being monotonous around b (see II.1.c), j has a finite limit denoted by l. Using x2j¢(x) -g2 j¢ (g) = - òg x t2/j(t) we deduce that j¢ also admit a finite limit if l>0. In this case the solution could be prolongated, which is incompatible with the maximal nature of this solution. Thus, l=0.
II.4.b j¢ tends to l at x=b. Notice that l £ 0 from II.4.a and j > 0. Therefore, for all e > 0 one can find a such that if b-a < x < b then l- e < j¢(x) < l+e. Then, since the hypothesis allow us to integrate, we have j(x) < -(l- e)(b-x). But if a=b-a we have òax (t2j ¢)¢=-òax t2/j < [(b2)/(l-e)] òax [ 1/(b-t)] and the integral on the right diverges when x is close to b. So, j¢ tends to - ¥, which is incompatible with the hypothesis. j¢ is therefore not bounded around b and since j y is locally monotonous, j¢ tends to -¥.
II.5.a Setting x=0 inside (E) one has y¢(0)=0. Since x2y¢ is a decreasing function, psi¢(c) < 0.
II.5.b We have shown that H1 is false
therefore, under the requirements of this section y1 >
y for x between a1 and c.
y1 is piecewise-monotonous (II.1.c).
In order to show that y1
has a finite limit in a1, it is sufficient to show that
y1 has an upper bound.
c2y1¢(c) <
x2y1¢(x) therefore
c2y1¢(c)/x2
< y1¢(x),
and if we integrate from x to c we get that y1
is bounded in a neighborhood of a1. Thus, y1
can be continuously prolongated and since c2y
1¢(c)-
x2y1¢(x)
= -òxc
t2/y1
and that y1 > y,
y1¢
has also a continuous prolongation at a1. This contradicts the maximal nature of the
solution. Thus a1=0.
x2y1¢(x)
-x2y¢(x)
= c2(y1¢(c)
-y
¢(c)) +ò
xc t2(1/(y1)
- 1/(y)) < 0
because y1¢(c) <
y¢(c) and
y1 > y.
Since y has a finite, non-vanishing limit at 0,
y1 has a lower bound that is strictly non-negative
around 0, which makes the integral convergent. Thus the limit exists and it has a negative value.
Therefore x2y1
¢ has a finite, negative limit at 0, and thus
y1¢ tends to
-¥.
We can therefore bound y1¢
between (l-e)/x2 and
(l+e)/x2 in a neighborhood of 0.
By integrating we get that xy1(x) >
-(l+e)
(1-a/c)
for x < a, thus y1
tends to +¥.
II.5.c (xy1)¢¢ =-x/y1 < 0. xy1 has a decreasing derivatique (it therefore concave). In case it vanishes at x=a, this point corresponds to a global maximum, which gives an upper bound of xy1. Otherwise, we have either xy1 increasing, therefore upper bounded by b1y1(b1) or decreasing and in that case the graph of the function is under its tangent at b1, which intersects the vertical axis x=0 at a point, which y coordinates gives the upper bound we are looking for. Since the function is concave, it is piecewise monotonous, and since it is also a bounded function between 0 and the previous upper bound, it has necessarily a limit at x=0.
II.5.d x2y2¢(x) -x2y¢(x) = c2y2¢(c) -c2y¢(c) -òcx t2(1/y2 -1/y) > 0 on ]a2,c[ (the arguments are similar to II.5.b). Therefore x2y2¢(x) > x2y¢(x) +c2(y2¢(c) -y¢(c)). Assume now that a2=0. y¢ tends to 0 at x=0 so we have x2y2¢(x) > c2(y2¢(c) -y¢(c)) -e > 0 in a neighborhood of a2. Thus y2¢(x) > M/x2 tends to +¥ and after integration one gets that y2 tends to -¥, which cantradicts the fact that y2 > 0. We must therefore have a2 > 0.
II.6 You can do this part by yourself I think ....
III.1.a x2y
¢-y
¢(1) = ò1x
(t2y¢)
¢ = òx1 t2/y
(xy¢(x)
+y(x))¢
= xy"+2y
¢ = -x/y and therefore
xy¢(x)
+y(x)-y¢(1)-h = òx1
t/y(t)
Applying the first equality with x=0 we get y¢(1)
= -ò01
t2/y(i)
III.1.b T(1/y)(0)
= ò01 t/y
-ò01
t2/y = y(0)
-y¢(1)
-h
+y¢(1)=y(0)
-h
and
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III.2.a T(f)¢(x)
= -1/(x2)
ò0x t2f(t) and
T(f)¢¢(x) = 2/(x3)ò0x t2f(t) dt
-f(x)
T(f)¢(0)=0 can be obtained by bounding f that is continuous at x=0.
Consequently T(f)¢¢(0) = lim -1/(x3)
ò0x t2f(t)
= -f(0)/3
This values are respectively equal to the limits of T(f)¢ and T(f)¢¢
when x tends to 0 that are easily computed, and T(f) is therefore of class C2 sur [0,1].
III.2.b T(f) > 0 if f ³ 0
and cannot be identiquely 0 because f > 0 at least on an interval that has a strictly positive length (due to its
continuity), and the integral kernels are positive and vanish only at x=0 or x=1.
Therefore, if f is non-negative, T(f) can be identically 0 only if f is also identically 0.
III.2.c Linearity of T is straightforward.
Using also the positivity propety we have just seen, we will further be able to argue
that f > g Þ T(f) > T(g).
||T(f)|| £
||f||
[ (1/x-1)ò0x
t2 +òx1 (t-
t2) ] and the expression inside the brackets is
(1-x2)/6 which is less than 1/6.
III.3.a g0 is in F and since gn+1 ³ h , 1/gn is continuous, so T(1/gn) is of class C2. The recursion is therefore checked.
III.3.b We have that g1 ³
g0, and if g2p-1
³ g2p-2 then
T(1/g2p-1)
£ T(1/g2p-2)
so g2p £ g2p-1
and repeating the same argument, g2p+1 ³ g2p.
We could add that the inequalities are strict for x < 1 but this is not asked.
Since g2 ³ h
we have g2 ³ g0 and if g2p
³ g2p-2 then
g2p+1 £ g2p-1
and finally g2p+2 ³
g2p. (g2p) is therefore an increasing sequence.
This last inequality implies also similarly that g2p+3 £
g2p+1 and thus (g2p+1) is a decreasing sequence of functions.
III.3.c (g2p) is increasing, upper-bounded by g2p+1 , itself being upper bounded by g1 because (g2p+1) is a decreasing sequence, so (g2p) is increasing and upper-bounded independently of p, so this sequence must converge to a function g. (g2p+1) is decreasing and lower bounded by h so it converges also towards a function G.
III.4.a gn+1¢ = T(1/gn)¢ = -1/(x2) ò0x t2/gn(t) and gn ³ h so |gn+1¢ | £ ||gn+1¢ || £ x/(3h) £ 1/(3h)
III.4.b
|gn¢|
£ M, thus |gn(x)
-gn(y)|
£ M|x
-y| (mean value theorem).
If we cut the [0,1] interval in small intervals of length less than
e/M we then have that gn(x) lies in an interval
of length e for x moving inside each sub-interval of the partition.
III.4.c For a given e
we cut the [0,1] interval as described in the III.4.b, the edges of the subintervals being
xk = k/N where k is an integer between 0 and N.
The inequality |gn(x)
-gn(y)| £
M|x-y|
remains valid when taking the limit and
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III.4.d Since the convergence
is uniform, the limit being strictly non-negative, and the two integral kernels in the definition of T
being continuous, the limit of T(1/g2n) is equal to T(1/g) and the limit of
T(1/g2n+1) is equal to T(1/G).
III.5.a u(0)=0, u(1)=G(1) -g(1) and since T(f)(1)=0 we also have that u(1)=0. u¢(0)=G(0)-g(0)
III.5.b u¢
=G-g+x(G¢
-g¢)
=G-g +x(T(1/g)¢
-T(1/G)¢)
= G-g+[ 1/x]ò0x
t2(1/G-1/g)
and u¢¢=G¢
-g¢-
1/(x2)ò0x t2(1/G-1/g)
+x(1/G-1/g)
= x(1/G-1/g) = - u/gG
III.5.c The answer to question seems to be u=0.
It is actually straightforward to show that u=0 if
h > 1/Ö6.
In this case we have 1/(gG) < 6 and
||G-g||
= ||T((G-g)/gG)||
< ||G-g||
if G-g is not identically 0. It is indeed quite easy to make a numerical
simulation, and it shows that gn seems to converge (thus g=G) quite rapidly even for
a small value of h (of order 0.01 for my tests).
The limit g therefore obeys the relation g=h
+T(1/g), and under derivation of this equality one easily notice that g is a solution
of (E) that follows the requirements of this part III.
The solution that was proposed in the "Revue de Math spé" in 1987
follows the partial answer I have given above. The author of the article
notices that maximal solutions defined at least on [0,1] are all related
through some scale transformation from the origin (see I.2).
And for h as small as we want we will be able to find
such a scale transformation that drives a solution containing the point x=1,y=1
onto a solution containing the point x=1,y=h.
To get this, it is straightforward to show that y=hx
intersects the solution that containts (x=1,y=1) at a point "I".
The scale transformations that carries this point to the point
x=1,y=h is the transformation we need.
One can this way build all the maximal solutions obeying the requirements of part III.
Going back to the problem of proving that u=0, the differential equation obtained in III.5.b is
Sturm-Liouville-like, and the solution of this equation must admit 0 and 1
as zeros, with no other zero in between.
May be we could find some bounds for u that would be incompatible with these zeros ...???
For that purpose one usually uses a wronskian. If for instance we set
v=sin(px) solution of v¢¢ +
p2 v=0. Let w be the wronskian of u and v,
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If someone has the solution ....send me an email: eric.chopin@wanadoo.fr
III.6 I think you can do this yourself now ...