Дефинисање exp у $R^2$:

Pretpostavimo sada da neka funkcija $F : {R}^2 \rightarrow {R}^2$ zadovoljava $1^\circ - 5^\circ$ $$1^\circ F(x,0)=(e^x, 0)=e^x (1,0)$$ $$2^\circ F(x,t)=e^x(1\cdot T_{21}(t), T_{22}(t))$$ $$3^\circ <(1\cdot T_{21}(t),T_{21}(t)), ( (1 \cdot T_{21}(t))',(T_{21}(t))')>=0$$ $$4^\circ <((1 \cdot T_{21}(t))',(T_{21}(t))'), ( (1 \cdot T_{21}(t))'',(T_{21}(t))'')>=0$$ $$5^\circ (1\cdot T'_{21}(0), T'_{22}(0))=(0,1)$$ Pokušajmo da odredimo funkciju $F$. Iz $1^\circ$ i $2^\circ$ sledi $$T_{21}(0)=1$$ $$T_{22}(0)=0\label{*}$$ Iz $3^\circ$ sledi $$T'_21(0)=0$$ $$T'_22(0)=1\label{**}$$ Iz $3^\circ$ sledi $$T_{21}(t)\cdot T'_{21}(t) + T_{22}(t)\cdot T'_{22}(t)=0 / \int$$ $$\frac{1}{2}(T_{21}(t))^2 + \frac{1}{2}(T_{22}(t))^2= C$$ kako to važi za svako $t$, neka je onda $t=0$. Dobijamo $$\frac{1}{2}1^2 + \frac{1}{2} 0^2=C$$ $$C=\frac{1}{2}.$$ pa važi \begin{equation} (T_{21}(t))^2 + (T_{22}(t))^2=1 \end{equation} Iz $4^\circ$ sledi $$T'_{21}(t)\cdot T''_{21}(t) + T'_{22}(t)\cdot T''_{22}(t)=0 / \int$$ $$\frac{1}{2}(T'_{21}(t))^2 + \frac{1}{2}(T'_{22}(t))^2= C$$ kako to važi za svako $t$, neka je onda $t=0$. Dobijamo $$\frac{1}{2}1^2 + \frac{1}{2} 0^2=C$$ $$C=\frac{1}{2}.$$ pa važi \begin{equation} (T'_{21}(t))^2 + (T'_{22}(t))^2=1 \end{equation} Iz $(T_{21}(t))^2 + (T_{22}(t))^2=1$ i $(T'_{21}(t))^2 + (T'_{22}(t))^2=1 $sledi $$T_{21}(t)=\cos \phi (t) $$ $$T_{22}(t)=\sin \phi (t)$$ $$T'_{21}(t)=-\phi'(t) \sin \phi (t) $$ $$T'_{22}(t)=\phi'(t)\cos \phi (t) $$ $$(\phi'(t))^2\sin^2\phi(t)+ (\phi'(t))^2\cos^2\phi(t)=1$$ odnosno \begin{equation} (\phi'(t))^2=1 \end{equation} $$\cos \phi (0) =1 $$ $$\sin \phi (0) =0 $$ $$-\phi(0) \sin \phi (0) =1 $$ $$\phi(0) \cos \phi (0) =1 .$$ Odnosno $\phi(0)=2k\pi$ za neko $k \in Z$ i $\phi'(0)=1$. Otuda, koristeći neprekidnost funkcije $\phi'$, sledi $$\phi'(t)=1.$$ Dalje, $$\phi(t)=t+C$$ $$\phi(0)=2k\pi, k\in \mathbb{Z}$$ $$\phi(0)= 0 + C$$ Sledi da je $C=2k\pi, k\in \mathbb{Z}$. Konačno $$\phi(t)=t+2k\pi ,$$ pa je $$T_{21}(t)=\cos (t+2k\pi)=\cos t$$ $$T_{22}(t)=\sin(t+2k\pi)=\sin t.$$ Odnosno $$F(x,t)=e^x(1\cdot \cos t, \sin t)=(e^x \cos t, e^x \sin t).$$