The answet is "Yes", in the sense that a solution will exist in terms of a Fourier-Bessel series. However in practice solving the problem numerically is probably more efficient than trying to solve the Sturm-Liouville problem for the radial basis functions or calculating the coefficients.
The boundary condition on [itex]\Gamma_2[/itex] is not a self-adjoint condition, but you can solve that by taking [itex]\phi = \psi + \frac{K_2}{K_1}[/itex] so that [itex]\psi[/itex] satisfies Laplace's equation together with [itex]\psi = V_1 - \frac{K_2}{K_1}[/itex] on [itex]\Gamma_1[/itex], [itex]\psi = V_2 - \frac{K_2}{K_1}[/itex] on [itex]\Gamma_3[/itex], [itex]\partial \psi/\partial r = 0[/itex] on [itex]\Gamma_4[/itex] and [itex]\partial \psi /\partial r + K_1 \psi = 0[/itex] on [itex]\Gamma_2[/itex].
Since [itex]r = 0[/itex] is not in the domain, this is one of those rare instances where we will need to use the Bessel function of the second kind [itex]Y_0[/itex] as well as [itex]J_0[/itex], and the radial dependence must be [tex]\rho_n(r) = \cos \alpha_n J_0(k_nr) + \sin \alpha_nY_0(k_nr)[/tex] where the eigenvalues [itex]k_n[/itex], [itex]n = 0, 1, \cdots,[/itex] satisfy [tex]
\left| \begin{array}{cc} k_nJ_0'(k_nR_1) & k_nY_0'(k_nR_1) \\ k_nJ_0'(k_nR_2) + K_1 J_0(k_nR_2) & k_nY_0'(k_nR_2) + K_1Y_0(k_nR_2)<br />
\end{array} \right| = 0[/tex] and [tex]
\tan \alpha_n = -\frac{J_0'(k_nR_1)}{Y_0'(k_nR_1)}.[/tex] The [itex]\rho_n[/itex] are orthogonal with respect to the inner product
[tex]\langle f, g \rangle = \int_{R_1}^{R_2} f(r) g(r) r\,dr.[/tex]
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