John C. Chien

December 18, 1998 13:12 
Re: Stability Analysis
The equation can be written as Y'' =(B/A) * Y  (C/A). If you use finitedifference for Y'' , you get (Y(i+1) 2Y(i) +Y(i1)) / ((x(i+1)x(i1))/2)**2. Also Y can be written as Y(i). If we let dx2=((x(i+1)x(i1))/2)**2, then the final equation for Y(i) is Y(i)=(Y(i+1)+Y(i1))/(2 + (B/A)*dx2) + (dx2*(C/A))/(2 + (B/A)*dx2 ). When the mesh size is small, dx2 is small. Y(i) is approximately equal to (Y(i+1)+Y(i1))/2. So, as long as Y(i+1) and Y(i1) are finite, the solution Y(i) ( the value between x(i+1) and x(i1)) will be finite and bounded by Y(i+1) and Y(i1). In this case is the average value of Y(i+1) and Y(i1). For finite value of mesh size, dx2 will be finite ( not a very small value ). And for large value of B, the denominator (2+(B/A)*dx2) can become a large number. For example, for finite Y(i+1) and Y(i1), the solution Y(i) can become very small or approach zero. In this case, Y(i1) > Y(i) < Y(i+1). So a Vshaped profile is created. This result is artificial because the real solution should always close to the average value of the Y(i+1) and Y(i1) by making mesh size very small, and thus very small dx2. So you should try to keep (B/A)*dx2 << 2, by making mesh size small. Or dx2 << (2/ (B/A)). I think, the finite element works in a similar way( with additional assumptions).
