Remember the following result for the Loewner equation:
If $\lambda:[0,\infty)\to\mathbb R$ is continuous, then for all $z\in\mathbb C\setminus\{\lambda(0)\}$ there is a uniqe $\zeta(z)\in(0,\infty]$ and a unique continuous $g(\;\cdot\;,z):[0,\zeta(t))\to\mathbb C$ with $$g(t,z)\ne\lambda(t)\tag1$$ and $$g(t,z)=z+\int_0^t\frac2{g(s,z)-\lambda(s)}\:{\rm d}s\tag2$$ for all $t\in[0,\zeta(z))$.
Now assume $$\lambda(t)=2\sqrt{\kappa t}\;\;\;\text{for all }t\ge0$$ for some $\kappa\ge0$. How can we determine $g$ here and how can we determine $\zeta(z)$?
Let $y_\pm:=\sqrt\kappa\pm\sqrt{\kappa+4}$ and $$H(w):=\frac{2y_+\ln(w-y_-)-2y_-\ln(w-y_+)}{y_+-y_-}.$$ Using this, I was able to find the relation $$H\left(\frac{g(t,z)}{\sqrt t}\right)=2\ln\frac z{\sqrt t}\tag3,$$ but how do we solve this for $g(t,z)$?
EDIT: From $(3)$, I was only able to obtain $$\gamma(t):=g_t^{-1}(\lambda(t))=c\sqrt t,$$ where $$c:=\exp\left(\frac12H(2\sqrt\kappa)\right).$$
If we cannot find an explicit expression for $g$, can we at least determien $\zeta(z)$? My guess is that $\zeta(z)=\infty$ for all $z\in\mathbb C\setminus\{\lambda(0)\}$, but how do we prove this?