Let $S = \{s \in S: s'(a) = s'(b) = 0 \}$ be the spline space that holds all cubic splines with derivate at startpoint (a) and endpoint (b) =0. I want to find a basis for this vector space. I looked at bsplines, but I think that is the wrong approach
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2 Answers
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Let's just solve the problem first for $a = 0$ and $b = 1$. Then I'll write the generic spline as \begin{align} s(t) &= at^3 + bt^2(t-1) + ct(t-1)^2 + d (t-1)^3 & \text{ so that} \\ s'(t) &= 3at^2 + 2bt(t-1) + bt^2 + c(t-1)^2 + 2ct(t-1) + 3d (t-1)^2 \\ s'(t) &= (3a+b) t^2 + 2(b+c) t(t-1) + (3d+c) (t-1)^2 \\ \end{align} [OMG...I'm sorry. The "a" and "b" here have nothing to do with the "a" and "b" in the problem as stated; they're just two coefficients of an unknown polynomial.]
Now saying that $s'(0) = 0$ is the same as saying that \begin{align} 3d + c = 0 \end{align} And saying that $s'(1) = 0$ is the same as saying that \begin{align} 3a + b &= 0 \\ \end{align} So we have two equations in the four unknowns $a,b,c,d$. Let's find two independent solutions. We can do this by picking $a = 1, d = 0$, yielding $(a,b,c,d) = (1, -3, 0, 0)$, and $a = 0, d=1$, yielding $(a,b,c,d) = (0,0,1, -3)$, or writing out the polynomials, $$ b_1(t) = t^3 -3t^2(t-1) \\ b_2(t) = -3t(t-1)^2 + (t-1)^3 $$
Those have the property we want ... but only on an interval between $0$ and $1$. If we write $$ c_1(t) = b_1\left(\frac{t-a}{b-a}\right) \\ c_2(t) = b_2\left(\frac{t-a}{b-a}\right) $$ where now "a" and "b" are the interval endpoints from your origonal problem, you can check that $c_1$ and $c_2$ are in fact polynomials with the desired properties, and any linear combination of them will have the property that $s'(a) = s'(b) = 0$ (and indeed, that all such cubics are covered by linear combinations of $c_1$ and $c_2$).
If you want to do quartic, or quintic, or higher order splines, I think you can mimic the approach I've taken here, which is more or less "use the Bézier (or is it Bernstein?) basis and follow your nose."
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It sounds like you might be interested in cubic polynomials, rather than cubic splines. If so, read on.
I’ll talk about the interval $[0,1]$; rescaling to an arbitrary interval $[a,b]$ is straightforward.
A useful basis for cubic polynomials is the Hermite basis. The four basis functions are: $$ 𝜙_0(𝑢) = 2𝑢^3 −3𝑢^2 +1 \quad ; \quad 𝜙_1(𝑢) = 3𝑢^2 −2𝑢^3 \\ 𝜓_0(𝑢) = 𝑢−2𝑢^2 +𝑢^3 \quad ; \quad 𝜓_1(𝑢) = 𝑢^3 −𝑢^2 $$ Then, given two values $p_0$ and $p_1$ and two derivatives $d_0$ and $d_1$, we construct the curve $$ f(u) = \phi_0(u)p_0 + \phi_1(u)p_1 + \psi_0(u)d_0 + \psi_1(u)d_1 $$ It is easy to check that $$ f(0) = p_0 \quad ; \quad f(1) = p_1 \\ f’(0) = d_0 \quad ; \quad f’(1) = d_1 \\ $$ So, any curve of the form $$ f(u) = \phi_0(u)p_0 + \phi_1(u)p_1 $$ Satisfies your requirements. The two functions $\phi_0$ and $\phi_1$ are the basis functions you want.
