Uniqueness of solution for linear first-order partial differential equations

In Polyanin's Handbook of First Order Partial Differential Equations (2002), in Section 10.1.2, it is stated that the non-homogeneous linear, first-order partial differential equation: $$\sum_{i=1}^nf_i(x_1,\dots ,x_n)\frac{\partial w}{\partial x_i}=g(x_1,\dots ,x_n)w+h(x_1,\dots ,x_n)$$ with initial condition $$w (0,x_2,\dots ,x_n)=\psi (x_2,\dots ,x_n)$$ where $$\psi (x_2,\dots ,x_n)$$ is a given function, has unique, continuously differentiable solution in domain $$G=\{0, if $$f_1=1$$, $$f_2,\dots f_n,g,h,\psi$$ are continuously differentiable functions of $$x_1,\dots,x_n$$ in $$G$$ and the following inequality holds in $$G$$: $$\sqrt{f_2^2(x_1,\dots ,x_n)+\dots+f^2_n(x_1,\dots ,x_n)}\leq k\Big(1+\sqrt{x_2^2+\dots+x_n^2}\Big),k=const.$$ However, I couldn't find a proof for this uniqueness theorem. Can anyone give a hint?

• I would try the method of characteristics. The inequality you quote is precisely a bound on the speed of the characteristics, which should be required for existence of a smooth solution. – Jeff Mar 6 at 18:36
• Thank you Jeff for your comment. Indeed, I have solved this equation by using the method of characteristics, but my problem now is to show that this solution is unique. So, does exceeding this bound on speed means that we lose solvability or just loss of smoothness of solution? Also if you know a relevant reference, it would be most helpful for me. – KIM Mar 7 at 12:30
• The method of characteristics will give uniqueness too. You integrate any solution along the characteristics to show that it must be given by the solution you obtained with method of characteristics. – Jeff Mar 7 at 15:19