# A question on the existence and uniqueness of a cubic Hermite interpolant

I have been trying to solve a particular problem that establishes both the existence and uniqueness of a cubic hermite interpolant on some generic interval $$[a,b]$$. Briefly, for a function $$f$$ we require a cubic polynomial $$p$$ that satisfies

$$p(a) = f(a), \ p^{\prime}(a) = f^{\prime}(a),$$ $$p(b) = f(b), \ p^{\prime}(b) = f^{\prime}(b).$$

I am aware that there are several approaches for this but the approach that my book recommends is to use the fact (which I have successfully proven) that the differential equation $$u^{(4)}(x) = f(x)$$ with $$u(a) = u(b) = u^{\prime}(a) = u^{\prime}(b) = 0$$ has a unique solution. Using this one can then easily show that the cubic hermite interpolant is unique but I am baffled as to how this proves existence and therefore I would appreciate some help.

I should mention that this is an exercise from Prenter's "Splines and Variational Methods".

The problem of existence and uniqueness of a cubic polynomial passing through two given points with two given slopes reduces to a linear system of four equations in four unknowns, as follows. Take the most general cubic, $$P(x)=\alpha x^3+\beta x^2 + \gamma x + \delta$$. Then $$P'(x)=3 \alpha x^2 + 2\beta x + \gamma$$, and you want: \begin{aligned} \alpha a^3 + \beta a^2 + \gamma a + \delta &= f(a)\\ \alpha b^3 + \beta b^2 + \gamma b + \delta &= f(b)\\ 3\alpha a^2 + 2\beta a + \gamma &= f'(a)\\ 2\alpha b^2 + 2\beta b + \gamma &= f'(b)\\ \end{aligned} The corresponding matrix of coefficients is $$A= \begin{pmatrix} a^3 & a^2 & a & 1\\ b^3 & b^2 & b & 1\\ 3a^2 & 2a & 1 & 0 \\ 3b^2 & 2b & 1 & 0 \end{pmatrix}$$
A straighforward calculation shows that the determinant of this matrix is $$-(a-b)^4$$, which is nonzero if we assume that $$a\neq b$$, and of course we assume that. From Linear Algebra we know that the above linear system of equations will have a unique solution. That is, there exists a solution, and it is unique. Moreover, using standard techniques it is possible to express the coefficients of the interpolating polynomial as functions of the input data $$(f(a),f(b),f'(a),f'(b))$$.
• I don't see how existence of an interpolating cubic follows from the uniqueness of the solution to the IVP $u^{(4)}(x)=f(x)$. The way things are usually done, is that if you know that this differential equation has a unique solution you look at $f-P$ where you already have proven that $P$ exists. Maybe you are just reading too much into the exercise. – uniquesolution Apr 11 '19 at 10:41