A Deriving the derivative boundary conditions from natural formulation

maistral
Messages
235
Reaction score
17
TL;DR Summary
How to derive the finite difference derivative formulation from the natural boundary formulation?
PS: This is not an assignment, this is more of a brain exercise.

I intend to apply a general derivative boundary condition f(x,y). While I know that the boxed formulation is correct, I have no idea how to acquire the same formulation if I come from the general natural boundary condition formulation. I honestly do not know what am I doing wrong. Can someone check where am I incorrect?
67846231_357745925149080_6309915412955922432_n.jpg
 
  • Like
Likes exln
Physics news on Phys.org
Looks OK. So what is the problem?
 
Chestermiller said:
Looks OK. So what is the problem?

The left side and the right side have different results.
 
If ##q_x## is the heat flux in the x direction, then ##q_x## is always given by $$q_x=-k\frac{dT}{dx}$$irrespective of whether it is the left- or the right boundary. Of course, the sign of the flux in the x direction can be negative. So, at the fictitious point at the left boundary, you have:
$$T(-\Delta x)=T(+\Delta x)+q_x(0)\Delta x$$and, at the fictitious point at the right boundary, you have:
$$T(L+\Delta x)=T(L-\Delta x)-q_x(L)\Delta x$$
 
that is a good answer
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top