Gradient Tensor of Vector in Cylindrical Coordinates

Cylindrical coordinate system is not instinct as rectangular one, therefore vector gradient is derived as below.
Each gradient tensor of vector v in rectangular coordinates is expressed like following. Note ii is the component ii of del operator.

v=ui+vj+wkv=ui+vj+wk

vxx=xvi=vxi=x(ui+vj+wk)i=ux
vxy=yvi=vyi=y(ui+vj+wk)i=uy


v=[vxxvxyvxzvyxvyyvyzvzxvzyvzz]=[uxuyuzvxvyvzwxwywz]

Vector in cylindrical coordinates is described as :

v=vrˆr+vθˆθ+vzˆz

Derivatives of vector v with r,θ,z are :

vr=vrrˆr+vθrˆθ+vzrˆz
vθ=(vrˆr)θ+(vθˆθ)θ+(vzˆz)θ=vrθˆr+vrˆθ+vθθˆθvθˆr+vzθˆz

Where each term is needed product rule. Refer to this page for unit vector variation.

(vrˆr)θ=vrθˆr+vrˆrθ=vrθˆr+vrˆθ
(vrˆθ)θ=vθθˆθ+vθˆθθ=vθθˆθvθˆr
(vzˆz)θ=vzθˆz+vzˆzθ=vzθˆz

Lastly, the partial derivative of z direction is :

vz=vrzˆr+vθzˆθ+vzzˆz

Now, each tensor term is derived as following.

vrr=rvˆr=vrˆr=(vrrˆr+vθrˆθ+vzrˆz)ˆr=vrr
vrθ=θvˆr=vrθˆr=1r(vrθˆr+vrˆθ+vθθvθˆr+vzθˆz)ˆr=vrrθvθr
vrz=zvˆr=vzˆr=(vrzˆr+vθzˆθ+vzzˆz)ˆr=vrz
vθr=rvˆθ=vrˆθ=(vrrˆr+vθrˆθ+vzrˆz)ˆθ=vθr
vθθ=θvˆθ=vrθˆθ=1r(vrθˆr+vrˆθ+vθθˆθvθˆr+vzθˆz)ˆθ=vθrθ+vrr
vθz=zvˆθ=vzˆθ=(vrzˆr+vθzˆθ+vzzˆz)ˆθ=vθz
vzr=rvˆz=vrˆz=(vrrˆr+vθrˆθ+vzrˆz)ˆz=vzr
vzθ=θvˆz=vrθˆz=1r(vrθˆr+vrˆθ+vθθˆθvθˆr+vzθˆz)ˆz=vzrθ
vzz=zvˆz=vzˆz=(vrzˆr+vθzˆθ+vzzˆz)ˆz=vzz

Finally, we've got the all slots of the gradient of vector in cylindrical coordinates.

v=[vrrvrθvrzvθrvθθvθzvzrvzθvzz]=[vrrvrrθvθrvrzvθrvθrθ+vrrvθzvzrvzrθvzz]

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