Q:发件人： "ray" <ray@vip.139sc.com> 发送时间： 2003-12-02 21:04:32 优先级： 普通

Techer Zhang:
though receiving your hint last week ,I got confidence to  read  Enlish recently
. here is my proof:

step 1:suppose u is a integral factor of equation ,U satisfies uMdx+uNdy=dU ,and
g is an arbitrary continuous function
g(U)u(Mdx+Ndy)=g(U)dU
then int g(U)dU=G(U)
therefore, g(U)u  is  an  integral factor.

step 2:suppose u1 is another integral,  U1 satisfies u1Mdx+u1Ndy=dU1,and
D=dy/dx.
Then  dU/dx=uM+uND, dU/dy=uMD^{-1}+uN
dU1/dx=u1M+u1ND, dU1/dy=u1MD^{-1}+u1N.
Thus the Jacobian determinant
det J(U, U_1)= D^{-1}u u_1 det [M+ND, M+ND; M+ND, M+ND]=0 .
So functions U and U1 are dependent, that is, U_1=h(U) for a continuuously
differentiable function h. By the fact dU/u=dU_1/u_1 and dU_1=h'(U)dU, we see
that u_1/u=h'(U).Therefore, let g=h' and u1=g(U)u .

step 3:for  a  continuous  function  g
D(g(U))=g’(U)dU=g’(U)(uMdx+uNdy)=0
from the above, we can see if u and u1 is two completely different integrals u
and u1, u1/u=g(U) and g(U)=c in a general solution
so u1/u=g(U) =c is a general solution.

but I still can’t under stand why h’(U) is contiuous?

Thanks for spending your time giving me the detailed hint.

Xie Qian

A:谢倩同学：你想问的问题是：若M,N,u_1,u_2 in C^1, u_1, u_2为方程Mdx+Ndy=0的二个积分因子，
u_1/u_2

Step 1:Prove that ug(U) is an integral factor, where g is an arbitrary continuous function and
U satisfies uMdx+uNdy=dU. This can be verified easily by showing d[\int g(U) du].

Step 2: Further prove that if u_1 is another integral factor then u_1=ug(U) for a continuously
differentiable function g.

In fact, let u_1Mdx+u_1Ndy=dU_1 and D=dy/dx. Then
dU/dx=uM+uND,           dU/dy=uMD^{-1}+uN
dU_1/dx=u_1M+u_1ND,     dU_1/dy=u_1MD^{-1}+u_1N.

Thus the Jacobian determinant
det J(U, U_1)= D^{-1}u u_1 det [M+ND, M+ND; M+ND, M+ND]=0 .
So functions U and U_1 are dependent, that is, U_1=h(U)

for a continuuously differentiable function h. By the fact
dU/u=dU_1/u_1 and dU_1=h'(U)dU, we see that u_1/u=h'(U).

Therefore, let g=h' and ....

Step 3:Finally, apply results in Steps 1 and 2to show that u_1/u_2=g(U_2)=c is a general
solution.
Consider d(g(U_2))=g'(U_2)dU_2=g'(U_2)[u_2Mdx+u_2Ndy]=0 because U_2 is a first integral.
So g(U_2)=c.
This is an outline of the proof. Hope you could understand and write the proof rigorously.

Q：发件人： 发送时间： 2003-12-19 11:58:46 优先级： 普通

A
：发件人： "zwn" <matzwn@163.net> 发送时间： 2003-12-19 12:15:12 优先级： 普通

Once we got
dx/dt_0= -sin(t_0 x(t_0))+\int_{t_0}^t s cos(...) dx/dt_0 ds,

we estimate

|dx/dt_0|\le |sin(t_0 x(t_0))|+ \int_{t_0}^t s |dx/dt_0| ds,
where \le means < or =. By Gronwall's inequality,
|dx/dt_0|\le |sin(t_0 x(t_0))|exp(\int_{t_0}^t s ds=0
when (t_0, x_0)=(0,0), because x(t_0)=x_0 and sin(0)=0.

ZWN

Q:发件人： 发送时间： 2004-01-18 11:20:13 优先级： 普通

S为一个集合，DS上的一个距离。
1 a,b,c,d,

2

3

A:

All answers can be found as I tell.

ZWN

A:廖奇峰同学

p.89  11.  Consider
(NH)_p               dx/dt=A(t)x+f(t),
where A(t), f(t) are both continuous and w-periodic.Let x^{(1)}(t),...,x^{(n)}(t)are the
basic group of solutions of equation
(LH)_p                   dx/dt=A(t)x
such that
x^{(1)}(0)=col(1,0,0,...,0),
x^{(2)}(0)=col(0,1,0,...,0),...,
x^{(n)}(t)=col(0,...,0,1).
Prove that
(a) a solution \phi(t) of (NH)_p is w-periodic if and only if \phi(0)=\phi(w).
(b) for any continuous and w-periodic function f(t),(NH)_phas a unique w-periodic
solution if and only if the matrix X(w):=(x^{(1)}(w),...,x^{(n)}(w)) has no
eigenvalues equal to 1, which is equivalent to say,(LH)_p has no periodic
solutions which are not identical to 0.
Proof:
For (a), ---> Obvious. <---- \phi(t+w)-\phi(t) satisfies (LH)_p and equals 0
at t=0. By uniqueness, \phi(t+w)-\phi(t)=0, that is,....
For (b), by result (a) and the formula of variate coefficient, we see
(*)        (X(w)-X(0))C= -\int_0^w X(w)X^{-1}(s)f(s)ds.
The periodic solution is unique <---> the algebtaic equation (*) has a unique
solution C <---> the matrix X(w)-X(0) is full of rank,i.e., det (X(w)-I)not =0 ,which implies
that 1 is not a eigenvalue of X(w).
Moreover, we refer the solution of (LH)_p which are not identical to 0 to as thetrivial
solution. If (LH)_p has a non-trivial w-periodic solution, then adding
a w-periodic solution of (NH)_p it gives another (different) w-periodic solution
of (NH)_p. This deduces a contradiction to the uniqueness of result (b).
On the other hand, if (NH)_p has two distinct w-periodic solutions,then their
difference is not zero but a  w-pewriodic solution of (LH)_p.This shows a
contradiction that (LH)_p has a trivial w-periodic solution.
The proof is completed.

Q:发件人：panlimeng2003@eyou.com
Re: Re:Thanks

过年好！

三十那天本来想向您问声好的，但是前几天在家过年，效率不高，也没什么进展，不

我和陈雷前几天已到了成都．到目前为止，我们主要是在看书，其中主要有您的那本
，和张景中与熊金城写的＜函数迭代与一维动力系统＞，我感觉求ax+b的整数次迭代根有

另外，请您再推荐一些资料，和找它们的途径如网址，有时我们想看比较老的论文，

我刚刚给您打了几次电话都没打通，可能您很忙，也不知能不能帮的上．您也可以选

潘利蒙

２００４..３０

A:2004-01-29 22:25:35
"zwn" <matzwn@tom.com>
panlimeng2003@eyou.com
Re: Re:Thanks
Dear Pan Limeng,

Tanks for your message. I am off for vacation.

I did not consider the problem deeply but I think you can try
to use a tranformation with which ax+b is conjugate to x.Then
the result of f^n(x)=x can be applied. Chen Xi-Xin gave a good
idea to investigate the same problem. You can discuss with him.

You can find some papers on iterative roots and iteration theory
in the journals of
Publicationes Mathemeticae
and
Ann. Polon. Math.,
which may be found in our university library. Of course, you have to

There are a few papers published in Chinese. Some of them are cited
in my books.

Weinian Zhang

A:To: 45424326@163.com
I only give you hints to the Problem 7 (p.102).

The solution of this Cauchy problem can be given by the constant-variation formula.
We nned to prove the formula by constructing the Picard sequence

{phi_n(t)}.

We choose phi_0(t)=x_0 and construct phi_1(t) as in (5.6) in our textbook.
Then we prove by induction that

phi_n(t)

=\sum_{k=1}^n (x_0/k!)(\int_{t_0}^{t} P(s)ds)^k

+ \int_{t_0}^{t}Q(s)\sum_{k=1}^{n-1}(1/k!)

(\int_{s}^{t}P(r)dr)^k ds,

where integration by part will be used.

Then we show that the sequence is convergent uniformly. Taking the

limit, we obtain the same expression as the constant-variation formula.

Weinian Zhang

Q:  : 2005-12-02 12:45:29( 1 , 14 小时, 43 分钟前)

: question

Dear Prof. Zhang,

I am sorry I am some late.

6. f(x) is continuous on R,please prove 初值问题
(x)'=f(x)*f(x)+exp(-t)
x(t0)=x0

have on solution.

In my book (new version) the Problem  6 on P. 102 is different.
The equation is dx/dt=x+t+1. I don't know what is wrong.

Those problems in Exercise 5.1 were given by my co-authors.
The old version (which you might have and be using) was used
by Dr. Z-D Du, one of my co-author, last year when I collaborated
at abroad. So I don't know the detail of change in exercises.

I suggest you giving up this Problem because it is put here
inappropriately as an exercise of the section on Picard Approximation.

If you really want to know its answer, I can hint you to apply

========

I worry that your problem is not formulated well since I remember
it contains a uniqueness when you asked me the first time. Maybe

Suppose that f(x) is continuous on R. Prove that the Cauchy problem

x'=(f(x))^2+exp(-t),  x(t0)=x0

has a unique solution.

Since it requires both existence and uniqueness, Peano's theorem
does not work. However, noting that x'>0, we assure the existence
of the inverse of x. Let T be the inverse, i.e., t=T(x). Then, we can equivalently consider
the equation

dT/dx=1/{(f(x))^2+ exp(-T)},  T(x_0)=t_0.

Let G(x,T) denote the function on the right hand side. Obviously,
G is continuous in (x,T) and C^1 in T. So G is Lipschitzian in T
in any bounded closed region. Thus the Picard's Theorem can be applied.

A:sherryxlee2005@sina.com
Dear Sherry,

I am sorry to answer you so late.
Concerning your questions 1(3),(4) and 2(2), (5) on p.49,

I only need to give some hints to you as follows.

1(3): The equation can be written as
y^2(1-p)=(2-p)^2
by the notation p=dy/dx. Let 2-p=yt. Then we get a parametrized form
p=1-t^2, y=1+1/t.

For p=0, we easily obtain ......
For p\not= 0, we see that dx=dy/p=d(1+1/t)/(1-t^2). Thus,...
1(4): Rewrite the equation as
x=p^2/(4y) + 2y/p.
Differentiating the both sides, we get
[(p^3-4y^2)/2yp^2]dp/dy=[(p^3-4y^2)/4y^2p].
By factorization, it is equivalent to
Either  p^3-4y^2=0
Or      dp/dy=p/2y.
From them we separately solve .......

2(2):  The equation is equivalent to
xp^2-yp +1=0,
where p=dy/dx. Clearly, the equation does not allow p=0.
From the equation we can solve
y=xp +1/p.
Differentiating the both sides, we get
(x-1/p^2)dp/dx=0.
Thus we solve the following two equations separately:
Either (x-1/p^2)=0
Or dp/dx=0.
Then .....

2(5): Let z=y-x. The equation is equivalent to
dy/dx= -z^{1/2}+1,
that is, dz/dx=dy/dx-1=- z^{1/2}.
If z\not=0, we solve the reduced equation by
variable separation and get ....
If z=0, we easily get ....