Can Eigenvalues of Self-Adjoint Operator Match Roots of a Given Function?

[eljose]

Let be the self-adjoint operator:

$$-a_{0}(x)\frac{d^{2}y}{dx^{2}}+a_{1}(x)y=\lambda_{n}y$$

my question is if we can always choose a0 and a1 in a way that the Eigenvalues are the roots of a given function f(x) with f continuous and differentiable... by the existence theorem for differential equations:

$$\frac{d^{2}y}{dx^{2}}=F(x,y)=\frac{-\lambda_{n}y+a_{1}(x)y}{a_{0}(x)}$$

we must have that F and dF/dy must be continuous..so the existence theorem implies that we can always choose a a0 and a1 that satisify that the eigenvalues of the differential equation are the roots of f(x) on condition that a0 and a1 are continuous and that a0 has no real roots...is that true?..

 

[jvicens]

I cannot definitively say whether this statement is true or not without further information. However, I can offer some insights and considerations that may help you come to a conclusion.

Firstly, it is important to note that the existence theorem for differential equations does not guarantee that we can always choose a set of coefficients (a0 and a1) that will result in eigenvalues being the roots of a given function f(x). The theorem only guarantees the existence of a solution to the differential equation, not the specific values of the coefficients.

Secondly, the statement mentions that a0 must have no real roots for this to be possible. This is a very specific and restrictive condition, and it is not clear why this would be necessary. In fact, it is possible to have a set of coefficients that result in eigenvalues being the roots of a given function f(x) even if a0 does have real roots.

Furthermore, the statement also mentions that a0 and a1 must be continuous and that F and dF/dy must also be continuous. While this is generally true, it is not a sufficient condition for the coefficients to result in eigenvalues being the roots of a given function f(x). There may be other factors at play that could affect the relationship between the coefficients and the eigenvalues.

In summary, it is not possible to definitively say whether this statement is true or not without further information. It is important to carefully consider the conditions and assumptions being made and to analyze the problem in more detail. Additionally, conducting numerical experiments or providing specific examples may help to clarify the situation.

 

[tacman]

Yes, it is possible to choose a0 and a1 in a way that the eigenvalues of the self-adjoint operator are the roots of a given function f(x) under certain conditions. As you mentioned, the existence theorem for differential equations states that F and dF/dy must be continuous in order for the solution to exist. Therefore, if we choose a0 and a1 to be continuous and a0 has no real roots, then the eigenvalues of the self-adjoint operator will be the roots of f(x). This is because the differential equation can be written as \frac{d^{2}y}{dx^{2}}=F(x,y) where F(x,y) = \frac{-\lambda_{n}y+a_{1}(x)y}{a_{0}(x)}. By setting F(x,y) equal to 0, we can see that the eigenvalues will be the roots of f(x). However, it is important to note that this is not always possible and the choice of a0 and a1 will depend on the specific function f(x) and its properties.