Solving a PDE raised to the 2nd Power

[fullaclips]

Any help on this problem would be greatly appreciated.

I've never dealt with a PDE that was raised to the second power.
From what I can gather from this powerpoint http://www2.latech.edu/~sajones/Biotransport/Lecture 7 on Stokes Flow.ppt (starting on slide 56), I am supposed to take the partial derivative of psi according to the equation as if the squared term was not there, then take the partial derivate of the resulting equation once more. This is represented in the transition from slide 58 to 59. I am able to take the partial derivative the first time and get the result shown, but I do not understand what is being done on slide 59 after the 1st equation.Namely (partial/partial r^2) does not seem correct to me, since the r^2 in the denominator of the second term seems to be treated as a constant when taking the partial derivative with respect to r. I also do not understand why they are taking (partial/partial theta) of what seems to be a different function.

I am trying to explain as clearly as possible, but I realize my description may be difficult to understand. So the tl;dr version of my question is:
how do you deal with the 2nd power in equation 2.5.12

 

[HallsofIvy]

It is not "raised to the second power". Since the quantity inside the parentheses is a differential operator rather than a number, the "square" indicates that the operator is used twice.

"I am supposed to take the partial derivative of psi according to the equation as if the squared term was not there, then take the partial derivate of the resulting equation once more"
Yes, you apply the operator twice.

As far as "Namely (partial/partial r^2) does not seem correct to me, since the r^2 in the denominator of the second term seems to be treated as a constant when taking the partial derivative with respect to r." is concerned, when you are applying the differential operator a second time, you will have \(\displaystyle \frac{1}{r^2}\) times a function of r. Using the product rule you will have a term in which only the "function of r" is differentiated time \(\displaystyle \frac{1}{r^2}\) plus the function of r times the derivative of \(\displaystyle \frac{1}{r^2}\). Perhaps you are missing the second part of that.

 

[fullaclips]

Thanks

Thanks for the help. I was able to arrive at the same solution as the one presented in the slides, but my work did not look the same. This was probably preferable, since I was solving this PDE for class. Your help is much appreciated.