Instability of cubic splines

Question

I'm supposed to show, that the calculation of cubic splines with tridiagonal matrices is unstable. To show this I'm supposed to consider the function $s_1: [x_0, x_n] \to \mathbb{R}$ which is a spline that "corrects" the first derivative by interpolating $s_1(x_j)=0$ for $0 \le j \le n$ with additional conditions $s_1'(x_0)=1$ and $s_1''(x_0)=0$. Furthermore our Intervall is split up in equidistant points with $x_{j+1}-x_j=1$. With that being said I'm supposed to calculate the coefficients and show that with an increasing amount of knots these coefficients are unbounded. Here is where my troubles begin: I'm pretty much a beginner when it comes to spline Interpolation, which makes me unsure how to calculate the coefficients. Since this is neither going to be a natural spline nor a clamped one, but rather some half-product of both, what algorithm do I use to calculate these coefficients? Thanks in advance.

Answer 1

Consider $p(x)=s_1(x_j+x)=s_1(j+x)$ on the interval $[0,1]$. By assumption $p(0)=p(1)=0$, and $p'(0)$, $p''(0)$ are fixed from the prior computations. As $p$ is a cubic polynomial, it has the form $$ p(x)=x(1-x)(ax+b)\\ \implies p'(x)=(1-2x)(ax+b)+x(1-x)a,\\ p''(x)=-2(ax+b)+2(1-2x)a $$ so that $p'(0)=b$ and $p''(0)=-2b+2a$, that is, $a=p'(0)+\tfrac12p''(0)$ and $b=p'(0)$. Then one has for the derivatives at $x=1$ $$ p'(1)=-(a+b)=-2p'(0)-\tfrac12p''(0),\\ p''(1)=-4a-2b=-6p'(0)-2p''(0) $$ or $$ \pmatrix{p'(1)\\p''(1)} = -\pmatrix{2&\frac12\\6&2}\pmatrix{p'(0)\\p''(0)} $$ The matrix without the sign has characteristic polynomial $\lambda^2-4\lambda+1$ with roots $λ=2\pm\sqrt3$, so one is outside the unit circle, larger than $3+\frac23$. Thus the progression of the derivative pairs has a rapidly growing component.