Solve Delta u=0 for R_1≤ r≤ R_2 with u_r|_(r=R_1)=(Q)/(2πλ…

Solve $\Delta u=0$ for $R_1\le r\le R_2$ with $u_r|_{r=R_1}=\dfrac{Q}{2\pi\lambda R_1}$, $u|_{r=R_2}=u_2$; interpret physically.

Step 1 · Reduce and integrate twice

\[ \frac1r\frac{d}{dr}\!\left(r\frac{du}{dr}\right)=0 \implies r\frac{du}{dr}=C_1 \implies u(r)=C_1\ln r+C_2. \]

Step 2 · Apply the boundary conditions

$u_r(R_1)=C_1/R_1=\dfrac{Q}{2\pi\lambda R_1}\implies C_1=\dfrac{Q}{2\pi\lambda}$; $u(R_2)=u_2\implies C_2=u_2-\dfrac{Q}{2\pi\lambda}\ln R_2$.

Verification

$u''+u'/r = -Q/(2\pi\lambda r^2)+Q/(2\pi\lambda r^2)=0$. \checkmark Both boundary conditions hold by construction.

Physical interpretation

Heat conduction: a cylindrical shell with prescribed heat flux $Q$ per unit length through the inner surface and fixed outer temperature $u_2$; thermal resistance $R_{\rm th}=\ln(R_2/R_1)/(2\pi\lambda)$.
Potential flow: identifying $u$ with a velocity potential, this is a line source of strength $Q/\lambda$ at the axis between coaxial cylindrical walls, with constant volume flux $\dot V=Q/\lambda$ through every circle $r={\rm const}$ (mass conservation). The same solution also describes an infinite line charge between coaxial conductors, or flow to a well in a confined aquifer.