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I'm trying to solve the following problem:

Given that $K_a(x - y)$ and $K_b(x - y)$ are the kernels for the operators $(\Delta - aI)$ and $(\Delta - bI)$ on $L^2(\mathbb{R}^n)$, where $0 < a < b$. Show that $(\Delta - aI)(\Delta - bI)$ has a fundamental solution of the form $c_1K_a + c_2K_b$.

Use the preceding to find a fundamental solution for $\Delta^2 -\Delta$, when $n = 3$.

Edit: Originally, it it stated that $K_a(x - y)$ is a kernel for $(\Delta - aI)^{-1}$, but I believe this should be $(\Delta - aI)$ instead.

Math_Day
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1 Answers1

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A fundamental solution is one such that $(\Delta - aI)(\Delta - bI)f = 0$. We have \begin{align} (\Delta - aI)(\Delta - bI)(c_1K_a + c_2K_b) &= c_1(\Delta - aI)(\Delta - bI)K_a + c_2(\Delta - aI)(\Delta - bI)K_b\\ &= c_1(\Delta - aI)(\Delta - bI)K_a \end{align} where in the last equality we used the fact that $(\Delta - bI)K_b = 0$ since $K_b$ is a kernel of $(\Delta - bI)$. On the other hand, we have
\begin{align} c_1(\Delta - aI)(\Delta - bI)K_a&= c_1 (\Delta^2 - \Delta bI - aI\Delta + aI bI)K_a\\ &= c_1(\Delta^2 - \Delta a - b\Delta + bIaI)K_a\\ &= c_1(\Delta(\Delta - aI) - bI(\Delta - aI))K_a\\ &= c_1(\Delta(\Delta - aI) - bI(\Delta - aI))K_a\\ &= c_1(\Delta - bI)(\Delta - aI)K_a\\ &= 0 \end{align} where in the last equality we used the fact that $(\Delta - aI)K_b = 0$ since $K_a$ is a kernel of $(\Delta - aI)$.

In the case that $a = b = 1$, we have our solutions for $\Delta^2 - \Delta$ are of the form $c_1 K_1 + c_2K_2$.

Math_Day
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