
Hermite polynomials.
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(x1)²(2x+1) + b(x1)²x + cx²(32x) + dx²(x1). But how do I get the cubic which substitutes u = a(x1)²(2x+1) + b(x1)²x + cx²(32x) + dx²(x1) 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..

Hi
Using the following base
x(xh)²
(xh)^3
x²(xh)
x²
instead of the canonical base x^3, x², x, 1
I can find
$\displaystyle u(x) = \frac{a}{h^3} (xh)^2(2x+h) + \frac{b}{h^2}x(xh)^2 + \frac{c}{h^3}x^2(3h2x) + \frac{d}{h^2}x^2(xh)$

.
Hi runninggag,
Thanks for taking a look at my question, but I couldn't quite figure out how to get from the canonical base to the new one. Could you show me how?

You are looking for a third degree polynomial
To find it you can use the canonical base $\displaystyle x^3$, $\displaystyle x^2$, $\displaystyle x^3$, $\displaystyle 1$,
set $\displaystyle u(x) = \alpha x^3 + \beta x^2 + \gamma x + \delta$
and use the conditions $\displaystyle u(0) = a$, $\displaystyle \frac{du}{dx}(0) = b$, $\displaystyle u(h) = c$, $\displaystyle \frac{du}{dx}(h) = d$
But you can use another base in order to simplify the calculations
I used $\displaystyle x(xh)^2$, $\displaystyle (xh)^3$, $\displaystyle x^2(xh)$, $\displaystyle x^2$ also because I wanted to find a result with a form close to the one you already had for h=1
To show that $\displaystyle x(xh)^2$, $\displaystyle (xh)^3$, $\displaystyle x^2(xh)$, $\displaystyle x^2$ is a base you set
$\displaystyle \alpha x(xh)^2 + \beta(xh)^3 + \gamma x^2(xh) + \delta x^2 = 0$
and you show that this leads to $\displaystyle \alpha= 0$, $\displaystyle \beta = 0 $, $\displaystyle \gamma = 0 $ and $\displaystyle \delta = 0$
To find u(x) you set
$\displaystyle u(x) = \alpha x(xh)^2 + \beta (xh)^3 +\gamma x^2(xh) + \delta x^2$
and you use the conditions $\displaystyle u(0) = a$, $\displaystyle \frac{du}{dx}(0) = b$, $\displaystyle u(h) = c$, $\displaystyle \frac{du}{dx}(h) = d$

Thanks runninggag, 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...