Let $\frac{\partial^2}{\partial x^2}$ be the Laplacian operator on $\mathbb R$ and denoted by $R(\lambda,t)$ its resolvent kernel. I would like to know the exact expression of $R(\lambda,t)$.
Is $R(\lambda,t)= -\frac{1}{2i\lambda} e^{i t \lambda}$ or $R(\lambda,t)= \frac{1}{2i\lambda} e^{i t \lambda}$ or $R(\lambda,t) = \frac{1}{2\lambda} e^{-t \lambda}$ ?
Thank you in advance
Assuming the underlying space in $L^2(\mathbb{R})$, the resolvent is $R(\lambda)f=(\frac{d^2}{dx^2}-\lambda I)^{-1}$, which means that, given $f\in L^2$, the function $g=R(\lambda)f$ is the unique solution of $$ g''-\lambda g = f,\;\;\; g,g'\in L^2. $$ You can construct the resolvent by solving for $h\in L^2$ such that $$ h''(t)-\lambda h(t) = \delta_{x}(t). $$ That means $h''-\lambda h=0$ for $t < x$ and for $t > x$. The function $h$ must be continuous at $t=x$, but a jump discontinuity in the derivative of $h$ of $+1$ at $t=x$ is required for the $\delta_{x}$ behavior. And $h$ must be chosen to decay as $t\rightarrow\pm\infty$. Using a branch cut for $\sqrt{\lambda}$ on the negative real axis, gives $\Re\lambda > 0$ for $\lambda\notin(-\infty,0]$. Then the square integrability condition gives $$ h_x(t) =\left\{\begin{array}{ll} Ce^{\sqrt{\lambda}(t-x)}, & t < x \\ Ce^{-\sqrt{\lambda}(t-x)}, & t > x. \end{array}\right. $$ The constant $C$ must be chosen so that $$ 1=h_{x}'(x+0)-h_{x}'(x-0)=-C\sqrt{\lambda}-C\sqrt{\lambda} \\ C = -2/\sqrt{\lambda} $$ The solution $g=R(\lambda)f$ is $$ R(\lambda)f = \int_{-\infty}^{\infty}h_{x}(t)f(t)dt \\ (R(\lambda)f)(x) = -\frac{2e^{-\sqrt{\lambda}x}}{\sqrt{\lambda}}\int_{-\infty}^{x}e^{\sqrt{\lambda}t}f(t)dt-\frac{2e^{\sqrt{\lambda}x}}{\sqrt{\lambda}}\int_{x}^{\infty}e^{\sqrt{-\lambda}t}f(t)dt. $$ Unless I messed up a sign, you can directly verify that $$ (R(\lambda)f)''-\lambda R(\lambda)f = f. $$ The singularities of the resolvent lie on the negative real axis, where the branch cut of $\sqrt{\lambda}$ is found. This is expected because $\frac{d^2}{dx^2}$ is a negative operator, while $-\frac{d^2}{dx^2}$ is positive.