Some Notes on the Solutions of non Homogeneous Differential Equations

Where µ is the Moebius function. See also and [Ap] chapter 2 for the Moebius as also for other multiplicative functions.

Examples of such identities are 1) Let

Where φ(n) is the Euler Totient function. Proposition 2. If A(n) is arbitrary sequence of numbers we have for x > 0

Proof.

See also [B] We will use Proposition 1 to find the solution of the N -th degree linear differential equation

Lemma 1. Set

then the solution of (3) is

Proof 1. From Proposition 2, it is clear that if

Eq.6 shows clearly that

Also one can see that we have

then setting into (3) the above expansion we get the same result in a more easy way.

Theorem 1.

, where

Proof. As in Lemma 1.

Next we proceed with the 2-dimension problem with a similar way.

We set

and

Theorem 2. The equation

have solution

where

and

Then differentiating with respect to x we get

then with respect to y we get

combining the above we get the result.

Note. The polynomials P apparently must have no solutions in natural numbers.

Then the equation

where

) Observe that in this example we are not able to split the solution into two parts in x and y. Note. For no confusion the form of the equation is defined by

where the an and bn are respectively that of Px and Py.

If F is an operator such that

We will try to solve the equation

Hence the operator exp t ∂ ∂x produces the solution. From Eq.( 3) and Lemma one can take the limit N → ∞ then

thus according to Lemma the solution of the above equation must be

Using the parameter λ which can take and complex values one can arrive to the conclusion that

One can see that, a solution of Schrodingers equation

where

This is the general case in which the potential depends only in time.

Let again

Consider now the equation

the solution is

Where ρ k is the roots of P (x) = 0 The same equation have solution according to the Theorems of section 1:

An interesting question is how one can extract from ( 21) and ( 22) the roots ρ k .

Anyway when if we let N → ∞, then Theorem 3. i)

ii)

iii)

and also

in this example the diferential eq. is

The series in a first view can not become more fast convergent. If we consider for example

But for the diferential equation holds

If we try with F (x) = cos(πx) + x + 1 then we have

and for the diferential equation holds

which is the same. Thus we can say that even the equations are not easy to solve numericaly, the solutions itself under certain conditions may behave very good i.e yM (x) is very fast convergent.

3. Let us consider now a curius case. The L2(R) function h(x) = e x-e x then if F (x) = h(x) + x + 1 (we dont need the roots), we have

is the solution indeed.

We proceed with the following Lemma 2. Let

fnx n be analytic function in R such that for every a, b > 0 there exist constant depending from f , M f :

Let also φ(x) real valued function with values in C, such that for every c > 0 there exist costant M φ :

hence we can write

ii) The solution of

This theorem shows clearly that we can find solutions in integral-closedform, of the general not homogeneous equation (if existing the Laplace transforms).

Examples. 1) If hapens g(x) = 1/x 2 , then (L (-1) g)(x) = x and thus the solution of

where the form of the equations ( 36) and ( 37) is that of ( 34) and ( 35). In the same way as in the above example we can set other values for g(x)

2. Set

Then the equation

have solution

This method is like solving (38) with Fourier or Laplace transforms but we avoid some restrictions of y and g to be in L2(R). Note also that it is solved with Laplace theory.

3. We try now to evaluate T = 1

we use (39) and get

solving with respect to g we have

In general holds

under certain conditions of convergence. For example if y(x) =polynomial in x. Also for h(x) = log(1 + x), then

Thus for example if we consider the equation

with f (x) = x + log(1 + x), then the differential equation (e) is actualy the dy(x) dx

and have solution

Relation ( 40) is very useful if one can set a one to one relation between h and a function of y. For example if one take h(x) = e x then for all the functions Q(x) + e x with Q(x) = N k=0 a k x k , the equation will be

will have the same solution type

the solution is

Another related equation is

which have a solution

From the above examples and ( 39) and ( 40) one can see that the inversion with respect to some g(x) is Theorem 5. (Inversion)

then

Where

Where ψ is the Polygamma function i.e

(see and Mathematica notes). The above example is trivial and can be solved with Laplace theory. Now we will find a way to solve the equation (a1x + b1)f ′′ (x) + (a2x + b2)f ′ (x) + (a3x + b3)f (x) = g(x)

where f , g ∈ L2(R).

Let the Fourier Transform of a function of L2(R) is

the Inverse Fourier Transform is Proof.

The proof of ( 45) and ( 46) can be found in [Pa]. The relations (47) and ( 48) are obtained with integr

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