Given the Poisson equation $$\vec{\nabla} ^2 u(\vec{r}) = f(\vec{r})$$ and its solution as a convolution with the appropriate Green's function $$u(\vec{r}) = \int G(\vec{r},\vec{r}^\prime) f(\vec{r}^\prime)\, \mathrm{d}\vec{r}^\prime, $$ over an appropriate domain, my question is what happens when we take $f \to 0$?
We then obtain Laplace's equation $\vec{\nabla} ^2 u_0(\vec{r}) = 0$, which can be solved by several methods depending on the context.
In particular, I am confused by the following: what is the full solution to the Poisson equation? Is it
- $u = u_0 + \int G(\vec{r},\vec{r}^\prime) f(\vec{r}^\prime)\, \mathrm{d}\vec{r}^\prime$, where $u_0$ is harmonic, in which case $$\lim_{f \to 0} \int G(\vec{r},\vec{r}^\prime) f(\vec{r}^\prime)\, \mathrm{d}\vec{r}^\prime \stackrel{?}{=} 0 $$ making some kind of intuitive sense?
- Or, is the harmonic part $u_0$ somehow already contained in the convolution integral? If so, how does $$\lim_{f \to 0} \int G(\vec{r},\vec{r}^\prime) f(\vec{r}^\prime)\, \mathrm{d}\vec{r}^\prime \stackrel{?}{=} u_0(\vec{r}) $$ work?
Thanks!