Solving an equation with boundary conditions

[Mad_MechE]

Hey all,

I finally figured out how to solve the integral:

\int{dp} = \int{6U\eta(\frac{h-\overline{h}}{h^{3}})}{dx} + C

using maple and have it export to MATLAB where:

h=R+h0-\sqrt{R+x}\sqrt{R-x}
\overline{h}=R+h0-\sqrt{R+\overline{x}}\sqrt{R-\overline{x}}

how do i find the boundary conditions to satisfy the constants \overline{x} and C?

my boundary conditions are:

p = 0 \ @ \ x = R
and
p = 0 \ @ \ x = -\overline{x} \mbox{ where } \overline{x} \mbox{ is where } \frac{dp}{dx} = 0 \mbox{ (maximum pressure)}

i don't know if there is an easy way to do it or not! Thanks for your help!

MT

 

[HallsofIvy]

Hey all,

I finally figured out how to solve the integral:

\int{dp} = \int{6U\eta(\frac{h-\overline{h}}{h^{3}})}{dx} + C

using maple and have it export to MATLAB where:

h=R+h0-\sqrt{R+x}\sqrt{R-x}
\overline{h}=R+h0-\sqrt{R+\overline{x}}\sqrt{R-\overline{x}}

how do i find the boundary conditions to satisfy the constants \overline{x} and C?

my boundary conditions are:

p = 0 \ @ \ x = R
and
p = 0 \ @ \ x = -\overline{x} \mbox{ where } \overline{x} \mbox{ is where } \frac{dp}{dx} = 0 \mbox{ (maximum pressure)}

i don't know if there is an easy way to do it or not! Thanks for your help!

MT

Well, the obvious thing to do would be to put those conditions into your equation, giving you two equations for \overline{x} and C- except that the conditions say "p= 0" and there is NO p in your equation!

Since your original equation, in terms of integrals, has \int dp on the left side, you should get an equation of the form "p= the integral on the right". I have no idea what the equations you give for h and \overline{h} have to do with that equation!