Return to Question

My self-adjount differential operator $L$ is defined by $$L f(x) \equiv u(x) \frac{\partial^2}{\partial x^2} \left( u(x) f(x) \right)$$ where $u(x)$ is a known but arbitrary smooth function that satisfies $u(x) > 0$ $\forall x$.

I seek a basis of eigenfunctions of $L$, and their eigenvalues.

Are there any results known that help with this problem?

Are there any specific forms for $u(x)$ (other than $u\equiv 1$) that make the problem easier?

Edit: I am looking for eigenfunctions that do not explode, and form a useful basis in which I could in principle expand other "well behaved" functions. For example, if $u(x)\equiv 1$ I would select $e^{i\lambda x}$ as eigenfunctions, with eigenvalues $-\lambda^2$.

Edit 2: In practice, $u(x)$ is a function that is close to $1$. Its shape might be roughly $1$ at $-\infty$ smoothly reducing to $0.5$ at $0$ and back to $1$ at $\infty$.

Edit 3: As I said it is self-adjoint, I have a specific inner product in mind, namely $<f,g> = \int dx f(x) g(x)$. In that case, $L = - D^T D$ which shows, I think, that the eigenvalues of $L$ are negative, as long as $u(x) f(x)$ is square integrable for every square integral $f$. This would be a very helpful result for my particular application if it is correct

My self-adjount differential operator $L$ is defined by $$L f(x) \equiv u(x) \frac{\partial^2}{\partial x^2} \left( u(x) f(x) \right)$$ where $u(x)$ is a known but arbitrary smooth function that satisfies $u(x) > 0$ $\forall x$.

I seek a basis of eigenfunctions of $L$, and their eigenvalues.

Are there any results known that help with this problem?

Are there any specific forms for $u(x)$ (other than $u\equiv 1$) that make the problem easier?

Edit: I am looking for eigenfunctions that do not explode, and form a useful basis in which I could in principle expand other "well behaved" functions. For example, if $u(x)\equiv 1$ I would select $e^{i\lambda x}$ as eigenfunctions, with eigenvalues $-\lambda^2$.

Edit 2: In practice, $u(x)$ is a function that is close to $1$. Its shape might be roughly $1$ at $-\infty$ smoothly reducing to $0.5$ at $0$ and back to $1$ at $\infty$.

Edit 3: As I said it is self-adjoint, I have a specific inner product in mind, namely $<f,g> = \int dx f(x) g(x)$. In that case, I think we can write $L = - D^T D$. Does this help?

My self-adjount differential operator $L$ is defined by $$L f(x) \equiv u(x) \frac{\partial^2}{\partial x^2} \left( u(x) f(x) \right)$$ where $u(x)$ is a known but arbitrary smooth function that satisfies $u(x) > 0$ $\forall x$.

I seek a basis of eigenfunctions of $L$, and their eigenvalues.

Are there any results known that help with this problem?

Are there any specific forms for $u(x)$ (other than $u\equiv 1$) that make the problem easier?

Edit: I am looking for eigenfunctions that do not explode, and form a useful basis in which I could in principle expand other "well behaved" functions. For example, if $u(x)\equiv 1$ I would select $e^{i\lambda x}$ as eigenfunctions, with eigenvalues $-\lambda^2$.

Edit 2: In practice, $u(x)$ is a function that is close to $1$. Its shape might be roughly $1$ at $-\infty$ smoothly reducing to $0.5$ at $0$ and back to $1$ at $\infty$.

My self-adjount differential operator $L$ is defined by $$L f(x) \equiv u(x) \frac{\partial^2}{\partial x^2} \left( u(x) f(x) \right)$$ where $u(x)$ is a known but arbitrary smooth function that satisfies $u(x) > 0$ $\forall x$.

I seek a basis of eigenfunctions of $L$, and their eigenvalues.

Are there any results known that help with this problem?

Are there any specific forms for $u(x)$ (other than $u\equiv 1$) that make the problem easier?

Edit: I am looking for eigenfunctions that do not explode, and form a useful basis in which I could in principle expand other "well behaved" functions. For example, if $u(x)\equiv 1$ I would select $e^{i\lambda x}$ as eigenfunctions, with eigenvalues $-\lambda^2$.

Edit 2: In practice, $u(x)$ is a function that is close to $1$. Its shape might be roughly $1$ at $-\infty$ smoothly reducing to $0.5$ at $0$ and back to $1$ at $\infty$.

Edit 3: As I said it is self-adjoint, I have a specific inner product in mind, namely $<f,g> = \int dx f(x) g(x)$. In that case, $L = - D^T D$ which shows, I think, that the eigenvalues of $L$ are negative, as long as $u(x) f(x)$ is square integrable for every square integral $f$. This would be a very helpful result for my particular application if it is correct

My self-adjount differential operator $L$ is defined by $$L f(x) \equiv u(x) \frac{\partial^2}{\partial x^2} u(x) f(x)$$ where $u(x)$ is a known but arbitrary smooth function that satisfies $u(x) > 0$ $\forall x$.

I seek a basis of eigenfunctions of $L$, and their eigenvalues.

Are there any results known that help with this problem?

Are there any specific forms for $u(x)$ (other than $u\equiv 1$) that make the problem easier?

Edit: I am looking for eigenfunctions that do not explode, and form a useful basis in which I could in principle expand other "well behaved" functions. For example, if $u(x)\equiv 1$ I would select $e^{i\lambda x}$ as eigenfunctions, with eigenvalues $-\lambda^2$.

Edit 2: In practice, $u(x)$ is a function that is close to $1$. Its shape might be roughly $1$ at $-\infty$ smoothly reducing to $0.5$ at $0$ and back to $1$ at $\infty$.

My self-adjount differential operator $L$ is defined by $$L f(x) \equiv u(x) \frac{\partial^2}{\partial x^2} \left( u(x) f(x) \right)$$ where $u(x)$ is a known but arbitrary smooth function that satisfies $u(x) > 0$ $\forall x$.

I seek a basis of eigenfunctions of $L$, and their eigenvalues.

Are there any results known that help with this problem?

Are there any specific forms for $u(x)$ (other than $u\equiv 1$) that make the problem easier?

Edit: I am looking for eigenfunctions that do not explode, and form a useful basis in which I could in principle expand other "well behaved" functions. For example, if $u(x)\equiv 1$ I would select $e^{i\lambda x}$ as eigenfunctions, with eigenvalues $-\lambda^2$.

Edit 2: In practice, $u(x)$ is a function that is close to $1$. Its shape might be roughly $1$ at $-\infty$ smoothly reducing to $0.5$ at $0$ and back to $1$ at $\infty$.