Simplest two-point boundary value problem

  

 Previous blog discussed Sobolev space definition and example

Here we focus on Simplest two-point boundary value problem.

Simplest two-point boundary value problem
\begin{equation}
  -u^{''}(x) = f(x), -1 < x < 1, u(-1)=u(1)=0
\end{equation}
Energy functional
An energy functional is a mapping from a function space (often a Sobolev space) to the real numbers, which assigns a "total energy" value to each function in the space. The energy assigned typically depends on the function and its derivatives, reflecting physical or geometrical properties like potential energy, kinetic energy, or strain energy in various physical contexts.

\begin{equation}
  J(u) = \frac{1}{2} \int_{-1}^1 |u'(x)|^2 dx - \int_{-1}^1f(x)u(x)dx
\end{equation}
over \(\mathring{H^1}(-1,1)\) where this Sobolev space is for functions that go to $0$ at boundary points for weak formulation.

Find \(u \in \mathring{H^1}(-1,1)\)such that
\begin{equation}
   \int_{-1}^1u'(x)v'(x) =  \int_{-1}^1 f(x)v(x), v(x) \in \mathring{H^1}(-1,1)
\end{equation}

 Example

Imagine a fluid flowing between two long, parallel plates separated by a distance, typically normalized for simplicity. The flow is assumed to be steady (unchanging with time) and laminar (smooth with no turbulence).
The fluid is driven by a pressure gradient or external force such as gravity.
 
 Navier-Stokes Equations: For an incompressible, steady flow with no change in flow direction, the Navier-Stokes equations can simplify significantly. For a flow driven by a pressure gradient in the x-direction, or an external force such as gravity, and assuming no dependence on the y coordinate (2D flow), the equation simplifies to:
 \begin{equation}
   -\mu \frac{d^2u}{dx^2} = \frac{dp}{dx}+f
 \end{equation}
 where \(\mu)\) is viscosity of the fluid, \(\frac{dp}{dx}\) pressure gradient, \(f\) additional force such as gravity.

 Assuming constant pressure gradient and assuming \(f\) is constant, we can rewrite above equation as
 \begin{equation}
   -u^{''}(x) = g(x)
 \end{equation}
 where \(g(x) = \frac{1}{\mu}( \frac{dp}{dx}+f) \).
 Assume walls are at $x=-1,x=1$ and the flow is zero here, that is $u(-1)=u(0)=0$