I've been working on solving an ODE equation in trying to find the pressure gradient along a hydrostatic bearing, as far as literature goes,
the given ode $\frac{\partial }{\partial r}\left(\frac{h^3}{12 \eta }\text{}r\text{ }\frac{\partial p}{\partial r}\right)=0$ with a simply given solution of;
$p(r)\text{:=}\frac{\text{pt} (\log (r)-\log (R))}{\log (\delta )-\log (R)}$ With $p(\delta )=\text{pt},\text{ }p(R)=0$ as the Initial values.
For me, it seems to look like a simple double integrated ode $\frac{\partial ^2p}{\partial r^2}=\text{ }0$ , with a solution of:
$p(r)\to \frac{\text{pt} R-\text{pt} r}{R-\delta }$
and infact, when evaluating with mathematica, I get the same answer as by hand.
My solution is a very simply linear one...and since I'm dealing with typically non-linear systems I can't imagine it's as simple as I have solved it.
So, to ask the question, how did they get their answer?
Thanks for the help!
$$\frac{\partial }{\partial r}\left(\frac{h^3}{12 \eta }\text{}r\text{ }\frac{\partial p}{\partial r}\right)=0$$ Integrating once with respect to $r$ we get $$\frac{h^3}{12\eta}\frac{\partial{p}}{\partial{r}}=\frac{c_{1}}{r}$$ for some constant $c_{1}$. Integrating the second time, $$\frac{h^3}{12\eta}p(r)=c_{1}\ln{(r)}+c_{2}$$ or, using $A=\frac{12c_{1}\eta}{h^{3}}$ and $B=\frac{12c_{2}\eta}{h^{3}}$ $$p(r)=A\ln{r}+B$$ using the boundary conditions we have $$B=-A\ln{R}$$ $$A=\frac{pt}{\ln{\delta}-\ln{R}}$$