# Viscous stress in the Slender-Jet Axial Equation > The origin of the $4v_2$ term used in [[slender-jets-VE-order-0|Slender jet viscous stress divergence]] Use the long-wave expansion $ v_z(r,z,t)=v_0(z,t)+\varepsilon^2 r^2 v_2(z,t)+O(\varepsilon^4 r^4). $ In the nondimensional axial equation, the viscous operator is $ \frac{1}{\varepsilon^2}\left(\partial_{rr}v_z+\frac{1}{r}\partial_r v_z\right)+\partial_{zz}v_z. $ Given (see: [[continuity-slender-jet|Continuity equation in slender jet]]) $ v_z(r,z,t)=v_0(z,t)+\varepsilon^2 r^2 v_2(z,t)+\cdots, $ $ v_r(r,z,t)= -\frac{1}{2}v_0'(z,t)\,r-\frac{1}{4}\varepsilon^2 v_2'(z,t)\,r^3+\cdots, $ Compute radial derivatives of $v_z$: $ \partial_r v_z=2\varepsilon^2 r v_2+O(\varepsilon^4 r^3), $ $ \partial_{rr}v_z=2\varepsilon^2 v_2+O(\varepsilon^4 r^2), $ $ \frac{1}{r}\partial_r v_z=2\varepsilon^2 v_2+O(\varepsilon^4 r^2). $ Therefore $ \frac{1}{\varepsilon^2}\left(\partial_{rr}v_z+\frac{1}{r}\partial_r v_z\right) =\frac{1}{\varepsilon^2}\left(2\varepsilon^2 v_2+2\varepsilon^2 v_2\right)+O(\varepsilon^2) =4v_2+O(\varepsilon^2). $ Now compute the axial derivative: $ \partial_{zz}v_z=v_0''+\varepsilon^2 r^2 v_2''+O(\varepsilon^4 r^4) =v_0''+O(\varepsilon^2). $ So the full viscous contribution at leading order is $ \frac{1}{\varepsilon^2}\left(\partial_{rr}v_z+\frac{1}{r}\partial_r v_z\right)+\partial_{zz}v_z =4v_2+v_0''+O(\varepsilon^2). $ Hence the $O(1)$ axial momentum balance contains $ \partial_t v_0+v_0 v_0'=-p_0'+\left(4v_2+v_0''\right)+O(\varepsilon^2), $ or in dimensional form $ \rho\left(\partial_t v_0+v_0 v_0'\right)=-p_0'+\eta\left(4v_2+v_0''\right)+O(\varepsilon^2). $ > [!important] Essentially: > $ f(r)=f_0+f_2 r^2+O(r^4) \quad\Longrightarrow\quad \left(\partial_{rr}+\frac{1}{r}\partial_r\right)f=4f_2+O(r^2). $