- #1

- 492

- 0

## Main Question or Discussion Point

Let be the self-adjoint operator:

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

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 continous and differentiable... by the existence theorem for differential equations:

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

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 continous and that a0 has no real roots...is that true?..

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

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 continous and differentiable... by the existence theorem for differential equations:

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

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 continous and that a0 has no real roots...is that true?..

Last edited: