The unit of Green’s function

dimensional analysisgreens functionmathematical physicspartial differential equationsreal-analysis

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.

Best Answer

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.

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