A proof of ordinary differential equation about the comparative theorem and the extension theorem OD

Using the extension theorem of the solution, let y=u(x) be the initial value of the problem (e'):y'=f(x,y),y(x1)=y1 (definitely exists), considering the region in the rectangular region r bounded by y=w(x) and y=z(x) and the boundary and the point (x0, y0) (in this region), apply the extension theorem of the solution, y=u(x) extends to the right to cross the boundary of this region, you may wish to intersect y=w(x), then y=u can be constructed'(x), take u(x) before intersection, take w(x) after intersection to (x0, y0), smoothness can be guaranteed, u"(x) the conditions are satisfied, and other circumstances can be proved accordingly. I don't understand, then pm me, I also use this textbook ==, the answer at the back of the book is a few words "the extension theorem using the solution".

If that point falls on the largest or minimum solution, then the maximum or minimum solution can be used directly on that small interval; And if that point falls between the largest or smallest solution, then we try to write the solution to be constructed as an interpolation of the largest and minimum solutions.

u(x)=(1-a(x))w(x)+a(x)z(x), and then construct a(x) through ode.

It's also my homework for the week. Together.

The two solutions are represented by h(x), g(x), h(x 0)x0, such that h(x 1)>=h(x 1).

So, since h(x), g(x) are continuous, there must be x 1>=x 2>x 0 such that h(x 2)=g(x 2), which shows that there are two different integration curves for the crossing point (x 2,h(x 2)), which contradicts any point on r 2 and has only one integration curve.

Where does this begin? Your handwriting is beautiful.

What's the good explanation for this??? It's that when those conditions are met, there is a unique solution...

There are a large number of differential equations that cannot be solved by the elementary integral method, and in practice, it is necessary to know whether the equation has a solution and whether the solution is unique, so mathematicians began to study the existence uniqueness of the solution.

Refer to the first section of chapter 6 of the book in the link.

Let f(x) = inequality left - right.

Derivative of f(x).

f'(x)=0+ln(x+(1+x 2) ln(x+(1+x 2) Note: The last two items are removed.

It also compares the size of x+(1+x 2) with the size of 1.

x+(1+x^2)^

The square of the minus sign gives 1+x 2 and 1+x 2-2x, because x<0 and the right side is larger, so x+(1+x 2) is less than 1, and ln(x+(1+x 2) is less than 0

x<0, f(x).

f(x,y)=x-y^2

f(x,y1)-f(x,y2)| y1^2-y2^2|And |y1+y2|<=1, hence |f(x,y1)-f(x,y2)| = |(y1-y2)|

Lipschitz conditions are met.

So there is a unique solution.

Note: The <= above means less than or equal to.

Have you studied functional analysis? The Banach image compression principle can be used to prove the uniqueness of the solution of ordinary differential equations.