= Solution
The <chain rule> along a smooth curve is
$$
\boxed{\frac{d}{dt}f(\mathbf X(t))=\sum_{i=1}^n\frac{\partial f}{\partial x_i}(\mathbf X(t))X_i'(t)=\nabla f(\mathbf X(t))\cdot\mathbf X'(t).}
$$
Differentiating the circular parametrization gives the <tangent vector>
$$
\mathbf x'(t)=(-a\sin t,a\cos t)=(-y(t),x(t)).
$$
Thus the <chain rule> turns the angular derivative of $u$ into $-yu_x+xu_y$. The given <partial differential equation> makes this equal to $u$, so along every portion of the circle inside the upper half-plane, $U(t)=u(\mathbf x(t))$ satisfies the <ordinary differential equation> $U'=U$. Its solution is $U(t)=U(t_0)e^{t-t_0}$.
For the two specified points take $a=\sqrt2$. The initial point corresponds to $t_0=\pi/4$ and the terminal point to $t_1=3\pi/4$. The entire intervening arc lies in $y>0$, so no continuation outside the domain is used. \b[The requested value is]
$$
\boxed{u(-1,1)=10e^{\pi/2}.}
$$
These circular <characteristic curves> also show locally that the general solution has the polar-coordinate form $u(r,\theta)=C(r)e^\theta$, with $0<\theta<\pi$; there is no requirement of angular periodicity on the upper half-plane.
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