I am teaching a numerical analysis class this semester and we just started the section on differential equations. I want them to understand when we can expect to have a solution and when a solution satisfying a given initial condition is going to be unique.

We had gone over the “existence” theorem, which basically says: given and initial condition where where is some rectangle in the plane, if is a continuous function over , then we are guaranteed to have at least one solution to the differential equation which is guaranteed to be valid so long as stays in .

I might post a proof of this theorem later; however an outline of how a proof goes will be useful here. With no loss of generality, assume that and the rectangle has the lines as vertical boundaries. Let , the line of slope . Now partition the interval into and create a polygonal path as follows: use slope at , slope at and so on to the right; reverse this process going left. The idea: we are using Euler’s differential equation approximation method to obtain an initial piecewise approximation. Then do this again for step size

In this way, we obtain an infinite family of continuous approximation curves. Because is continuous over , it is also bounded, hence the curves have slopes whose magnitude are bounded by some . Hence this family is equicontinuous (for any given one can use in continuity arguments, no matter which curve in the family we are talking about. Of course, these curves are uniformly bounded, hence by the Arzela-Ascoli Theorem (not difficult) we can extract a subsequence of these curves which converges to a limit function.

Seeing that this limit function satisfies the differential equation isn’t that hard; if one chooses close enough, one shows that where then the differential equation has exactly one solution where which is valid so long as the graph remains in .

Here is the proof: where . This is clear but perhaps a strange step.

But now suppose that there are two solutions, say and where . So set and note the following: and . A Mean Value Theorem argument applied to means that we can assume that we can select our so that on that interval (since ).

So, on this selected interval about we have (we can remove the absolute value signs.).

Now we set up the differential equation: which has a unique solution whose graph is always positive; . Note that the graphs of meet at . But where .

But since on that interval.

So, no such point can exist.

Note that we used the fact that the solution to is always positive. Though this is an easy differential equation to solve, note the key fact that if we tried to separate the variables, we’d calculate and find that this is an improper integral which diverges to positive hence its primitive cannot change sign nor reach zero. So, if we had where is an infinite improper integral and , we would get exactly the same result for exactly the same reason.

Hence we can recover Osgood’s Uniqueness Theorem which states:

If is continuous on and for all we have a where where is a positive function and diverges to at then the differential equation has exactly one solution where which is valid so long as the graph remains in .

sure we see as say prof dr mircea orasanu and prof horia orasanu as followed

OSGOOD THEOREM AND LAGRANGIAN

Author Horia Orasanu

ABSTRACT

It has often been repeated that a Hamiltonian whose kinetic term depends on the position variable has in principle many different operator versions. For an arbitrary model Hamiltonian, this is of course, true

Printer rezultatele noi obtinute si relatiile dintre noile teoreme cerute, avem pe acelea care

Se refera la functiile mentionate mai sus cu o raza data in planul in care un patrat se afla

In acest plan ,si este perpendicular pe planul cercului si care se afla in miscare supuse la

Legaturi neolonome

Aici mentionam ca se poate construi un boiler inafara unui cilindru marginit de doua emis

Fere si cu pereti de o anumita grosime

Comment by perringu — February 6, 2018 @ 4:09 pm

also the Osgood theorem can be used to other as say prof dr mircea orasanu as for followed

LAGRANGIAN AND PARTIAL DERIVATIVES

Author Horia Orasanu

ABSTRACT

Okay, we’ve worked two of the four cases that would need to be solved in order to completely solve (1). As we’ve seen each case was very similar and yet also had some differences. We saw the use of both separation constants and that sometimes we need to use a “shifted” solution in order to deal with one of the boundary conditions.

Before moving on let’s note that we used prescribed temperature boundary conditions here, but we could just have easily used prescribed flux boundary conditions or a mix of the two. No matter what kind of boundary conditions we have they will work the same.

Comment by perringu — February 6, 2018 @ 7:28 pm