Hi
Using the following base
x(x-h)²
(x-h)^3
x²(x-h)
x²
instead of the canonical base x^3, x², x, 1
I can find
I’m practicing some problems for applied mathematics, and couldn’t quite solve the following:
In differential equations of equilibrium, from u(0) = a, du/dx(0) = b, u(1) = c, du/dx(1) = d we get the Hermite polynomial u = a(x-1)²(2x+1) + b(x-1)²x + cx²(3-2x) + dx²(x-1). But how do I get the cubic which substitutes u = a(x-1)²(2x+1) + b(x-1)²x + cx²(3-2x) + dx²(x-1) when we move the right end of the interval from x=1 to x=h?
And how do I get the new form of 2(s0) + 1(s1) = 3(u1) – 3(u0) for the case described above, and given that d²u/dx² = 0 at x = 0 (in which u are the heights and s the slopes)
If you know the solution, or could guide me in the right direction, I would appreciate it a lot! Thanks in advance..
You are looking for a third degree polynomial
To find it you can use the canonical base , , , ,
set
and use the conditions , , ,
But you can use another base in order to simplify the calculations
I used , , , also because I wanted to find a result with a form close to the one you already had for h=1
To show that , , , is a base you set
and you show that this leads to , , and
To find u(x) you set
and you use the conditions , , ,
Thanks running-gag, its clear to me that you know your math quite well! Thanks for showing me the way..
Do you by any chance know the second part of my question? I think I should end up with 2s(0)+s(1)=(3u(1)-3u(0))/h but I'm not quite sure...