# What is the unit of Green's function

The solution of heat equation using Green's function is given by \begin{aligned} T(\hat{r}, t)=&\left.\int_{R^{\prime}} G\left(\hat{r}, t \mid \hat{r}^{\prime}, \tau\right)\right|_{\tau=0} F\left(\hat{r}^{\prime}\right) d V^{\prime} \\ &+\frac{\alpha}{k} \int_{\tau=0}^{t} \int_{R^{\prime}} G\left(\hat{r}, t \mid \hat{r}^{\prime}, \tau\right) g\left(\hat{r}^{\prime}, \tau\right) d V^{\prime} d \tau \\ &+\alpha \sum_{i=1}^{N}\left[\left.\int_{\tau=0}^{t} \int_{S_{i}^{\prime}} G\left(\hat{r}, t \mid \hat{r}^{\prime}, \tau\right)\right|_{r^{\prime}=r_{i}} \frac{1}{k} f_{i}\left(\hat{r}^{\prime}, \tau\right) d A_{i}^{\prime} d \tau\right] \end{aligned}

Where $$F$$ is initial condition, $$g$$ is heat source and $$f_i$$ is boundary conditions. From this solution the unit of Green's function should be $$\frac{1}{\text{meter}^n}$$ where $$n$$ is the dimension of the domain. But Green's function is also heat response to a impulse located at $$(\hat{r}^{\prime}, \tau)$$ so intuitively $$G$$ should have unit $$K$$. Why the unit of $$G$$ here is simply just in space?

Also from auxiliary problem of the Non-homogeneous heat equation $$\nabla^{2} G\left(\hat{r}, t \mid \hat{r}^{\prime}, \tau\right)+\frac{1}{k} \delta\left(\hat{r}-\hat{r}^{\prime}\right) \delta(t-\tau)=\frac{1}{\alpha} \frac{\partial G}{\partial t}$$

This is a heat equation but the unit of solution is not in $$K$$, why?

Any help is appreciated.

Looking at the Wikipedia article, the Green's function reads $$G(x,t) = \Theta(t) \left(\frac{1}{4\pi kt}\right)^{n/2} e^{-\|x\|^2/4kt}$$ for the diffusion equation $$\text{L}G = \partial_t G - k \nabla^2 G = 0, \qquad G|_{t=0} = \delta(x)$$ with $$x$$ in $${\Bbb R}^n$$. Since the Dirac delta in the initial condition has dimension $$\text{m}^{-n}$$ and the linear PDE $$\text{L}G = 0$$ is homogeneous (i.e., independent on units), the Green's function must be expressed in $$\text{m}^{-n}$$ too. This can be viewed in the above expression where $$G$$ has same unit as $$\left[\left(kt\right)^{-n/2}\right] = (\text{m}^{2})^{-n/2} = \text{m}^{-n} .$$
The same fundamental solution holds for the forced (non-homogeneous) heat equation $$\text{L}G = \delta(t)\delta(x) \, .$$ From the above analysis, we already know that $$G$$ is expressed in $$\text{m}^{-n}$$. This is consistent with the non-homogeneous PDE, where $$\text{L}$$, $$\delta(t)$$, $$\delta(x)$$ have physical units $$\text{s}^{-1}$$, $$\text{s}^{-1}$$ and $$\text{m}^{-n}$$, respectively.