# Boundary condition in FEM method

Categories: Math
Following the previous post, I'm going to discuss about the boundary conditions in FEM method. For FEM method, there are two types of boundary conditions: 1. Natural boundary condition 2. Essential boundary condition The difference between these two kinds of boundary conditions are well summarized in the following two references: - https://caendkoelsch.wordpress.com/2018/06/09/what-is-the-difference-between-essential-and-natural-boundary-conditions-in-fem/ - https://www.researchgate.net/post/What_is_the_difference_between_essential_boundary_conditions_and_natural_boundary_conditions - https://math.stackexchange.com/questions/2089253/how-to-identify-natural-and-essential-boundary-conditions-of-this-differential-e - http://textofvideo.nptel.ac.in/105106051/lec3.pdf To clearly show the difference, I combined the two references: Essential boundary conditions Natural boundary conditions They are imposed explicitly during solving These conditions are added during the formulation of FEM problem and are automatically, “naturally” satisfied without any external conditions They are satisfied exactly by the construction of additional trial function or modification of the linear system to solve They are satisfied up to the order of the shape function Some examples would be Displacement in stress analysis Bending moment or shear forces in stress analysis Temperature in thermal analysis Adiabatic boundary in heat conduction analysis The part that I'm mainly interested in is the natural boundary condition.

# Uniqueness of Poisson Equation

Categories: Math
Recently, I'm working on the numerical simulation of Nano-second pulsed plasma. The equation to solve is the drift diffusion equation, which has a similar form to 🍔's equation. If it's a steady problem, then the equation becomes elliptical one. However, there are several challenges for me including: 1. Finite element method with weak form 2. New solver: COMSOL 3. The settings of boundary condition So, to solve the problem, I decide to start with the simple one.

The problem is how to solve the following ODE: $$\ddot{x} = A\exp\left(-Bx\right)$$ with the following initial condition: $$x(0)=0,\quad x'(0)=0$$ The actual method is quite traditional. We first make $v(t)=\dot{x}(t)$, then the ODE becomes: $$v\frac{dv}{dx}=A\exp(-Bx)$$ The general solution of it is: $$\frac{v^2}{2}=-\frac{A}{B}\exp(-Bx)+C$$ Using the initial condition: $$\dot{x}=\sqrt{\frac{2A}{B}\left[1-\exp(-Bx)\right]}$$ The tricky part now is we make a another substitution now and make $y=\exp(-Bx)$, then we get: $$-\frac{\dot{y}}{By}=\sqrt{\frac{2A}{B}\left[1-y\right]}$$ Then the following ODE can be easily solved: $$\dfrac{dy}{y\sqrt{1-y}}=-\sqrt{2AB}dt$$ The result: $$\log\left(\dfrac{1-\sqrt{1-y}}{1+\sqrt{1-y}}\right)=-\sqrt{2AB}t$$