The 1-D semi-infinite cartesian coordinate Green's function is given by
$$G\left(x, t \mid x^{\prime}, \tau\right)=\frac{1}{[4 \pi \alpha(t-\tau)]^{1 / 2}}\left\{\exp \left[-\frac{\left(x-x^{\prime}\right)^{2}}{4 \alpha(t-\tau)}\right]-\exp \left[-\frac{\left(x+x^{\prime}\right)^{2}}{4 \alpha(t-\tau)}\right]\right\}$$
where $x^{\prime}$ denotes the location of the impulse. And the solution using Green's function is
$$\begin{aligned} T(x, t)=&\left.\int_{L^{\prime}} G\left(x, t \mid x^{\prime}, \tau\right)\right|_{\tau=0} F\left(x^{\prime}\right) x^{\prime P} d x^{\prime} \\ &+\frac{\alpha}{k} \int_{\tau=0}^{t} \int_{L^{\prime}} G\left(x, t \mid x^{\prime}, \tau\right) g\left(x^{\prime}, \tau\right) x^{\prime P} d x^{\prime} d \tau \\ &+\alpha \sum_{i=1}^{N}\left\{\left.\int_{\tau=0}^{t}\left[x^{\prime P} G\left(x, t \mid x^{\prime}, \tau\right)\right]\right|_{x^{\prime}=x_{i}} \frac{1}{k} f_{i}(\tau) d \tau\right\} \end{aligned}$$
I have one question that if $x^{\prime}$ here is $0$, then the Green's function is $0$ as well, which means that to apply type $1$ boundary condition in Green's function will give us $0$ temperature profile. Why is that?