Let f be defined on [a, b] and let the nodes $a = x_{0} < x_{1} < x_{2} = b$ be given. A quadratic spline interpolating function S consists of the quadratic polynomial $S_{0} = a_{0} + b_{0}(x - x_{0}) + c_{0}(x - x_{0})^2 on [x_{0}, x_{1}]$ and $S_{1} = a_{1} + b_{1}(x - x_{1}) + c_{1}(x - x_{1})^2 on [x_{1}, x_{2}]$ such that

$
(i) S(x_{0}) = f(x_{0}), S(x_{1}) = f(x_{1}), S(x_{2}) = f(x_{2})$

$(ii) S \in C^1[x_{0}, x_{2}]$

Show that conditions (i) and (ii) lead to five equations in the six unknowns $a_{0}, b_{0}, c_{0}, a_{1}, b_{1}, c_{1}$. What additional condition can we impose to make the solution unique?

How do I find the five equations and make it unique?

2. Originally Posted by Makall
Let f be defined on [a, b] and let the nodes $a = x_{0} < x_{1} < x_{2} = b$ be given. A quadratic spline interpolating function S consists of the quadratic polynomial $S_{0} = a_{0} + b_{0}(x - x_{0}) + c_{0}(x - x_{0})^2 on [x_{0}, x_{1}]$ and $S_{1} = a_{1} + b_{1}(x - x_{1}) + c_{1}(x - x_{1})^2 on [x_{1}, x_{2}]$ such that

$
(i) S(x_{0}) = f(x_{0}), S(x_{1}) = f(x_{1}), S(x_{2}) = f(x_{2})$

$(ii) S \in C^1[x_{0}, x_{2}]$

Show that conditions (i) and (ii) lead to five equations in the six unknowns $a_{0}, b_{0}, c_{0}, a_{1}, b_{1}, c_{1}$. What additional condition can we impose to make the solution unique?

How do I find the five equations and make it unique?
Just as the problem says, use conditions (i) and (ii).
With $S_{0} = a_{0} + b_{0}(x - x_{0}) + c_{0}(x - x_{0})^2 on [x_{0}, x_{1}]$, " $S(x_0)= f(x_0)$" becomes $a_0= f(x_0)$ and " $S(x_1)= f(x_1)$" becomes $a_0+ b_0(x_1-x_0)+ c_0(x_1- x_0)^2= f(x_1)$ and $a_1= f(x_1)$. " $S(x_2)= f(x_2)$" becomes $a_1+ b_1(x_2- x_1)+ b_2(x2- x_1)^2= f(x_2)$. Saying that " $S \in C^1[x_{0}, x_{2}]$ means the derivative must be continuous at $x_1$: $b_0+ 2c_0(x_1- x_0)= b_1$.

There are a number of additional conditions you could add- you could require a specific derivative at either endpoint. Another condition is the "not a knot" condition- requiring that the second derivative be continuous at $x_1$ as well- which would be equivalent to having a single quadratic function through the three points.