# Hermite polynomials.

• Apr 23rd 2010, 02:14 PM
francoiskalff
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(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..
• Apr 24th 2010, 12:53 AM
running-gag
Hi

Using the following base
x(x-h)²
(x-h)^3
x²(x-h)

instead of the canonical base x^3, x², x, 1

I can find

$u(x) = \frac{a}{h^3} (x-h)^2(2x+h) + \frac{b}{h^2}x(x-h)^2 + \frac{c}{h^3}x^2(3h-2x) + \frac{d}{h^2}x^2(x-h)$
• Apr 24th 2010, 03:13 AM
francoiskalff
.
Hi running-gag,
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?
• Apr 24th 2010, 08:56 AM
running-gag
You are looking for a third degree polynomial

To find it you can use the canonical base $x^3$, $x^2$, $x^3$, $1$,
set $u(x) = \alpha x^3 + \beta x^2 + \gamma x + \delta$
and use the conditions $u(0) = a$, $\frac{du}{dx}(0) = b$, $u(h) = c$, $\frac{du}{dx}(h) = d$

But you can use another base in order to simplify the calculations

I used $x(x-h)^2$, $(x-h)^3$, $x^2(x-h)$, $x^2$ also because I wanted to find a result with a form close to the one you already had for h=1

To show that $x(x-h)^2$, $(x-h)^3$, $x^2(x-h)$, $x^2$ is a base you set
$\alpha x(x-h)^2 + \beta(x-h)^3 + \gamma x^2(x-h) + \delta x^2 = 0$
and you show that this leads to $\alpha= 0$, $\beta = 0$, $\gamma = 0$ and $\delta = 0$

To find u(x) you set
$u(x) = \alpha x(x-h)^2 + \beta (x-h)^3 +\gamma x^2(x-h) + \delta x^2$
and you use the conditions $u(0) = a$, $\frac{du}{dx}(0) = b$, $u(h) = c$, $\frac{du}{dx}(h) = d$
• Apr 24th 2010, 12:27 PM
francoiskalff
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...