What's Happening Between Two Points?

If a load f(x) is applied to a beam with bending stiffness c(x), then the deflection (displacement) satisfies
d^2/dx^2(c d^2u/dx^2)=f(x) [that is, (cu'')''=f].
To model spline interpolation, we let c(x)=1 and f(x)=0 (no external force).
With inhomogemeous boundary conditions, we could have
u(0)=a, u'(0)=b, u(1)=c, u'(1)=d.
Here the ends of the beam are fixed, but not in a line, and the beam bends in order to satisfy the boundary conditions.

The only (uniqueness of solutions to DE's with proper boundary conditions) solution is called the Hermite cubic and is given by

In[31]:=

  Clear[a,b,c,d,x]
  hermite[x_,a_,b_,c_,d_]:=a(x-1)^2(2 x + 1) + b x(x-1)^2 + c x^2(3 - 2 x) + d x^2(x-1);
  hermite[x,a,b,c,d]

Out[31]=

            2                  2               2
  b (-1 + x)  x + c (3 - 2 x) x  + d (-1 + x) x  + 
   
              2
    a (-1 + x)  (1 + 2 x)

If you are unconvinced, perhaps Mathematica can help.

In[32]:=

  Clear[u,x]
  DSolve[{u''''[x]==0,u[0]==a,u'[0]==b,u[1]==c,u'[1]==d},
          u[x],x];
   u=Evaluate[u[x]/.Flatten[%][[1]]];
   u

Out[32]=

                                    2
  a + b x + (-3 a - 2 b + 3 c - d) x  + 
   
                         3
    (2 a + b - 2 c + d) x

In[33]:=

  Collect[u,{a,b,c,d}]

Out[33]=

        2      3              2    3         2    3
  c (3 x  - 2 x ) + b (x - 2 x  + x ) + d (-x  + x ) + 
   
              2      3
    a (1 - 3 x  + 2 x )

This looks like hermite[x] to me.

In[34]:=

  Simplify[hermite[x,a,b,c,d]-u]

Out[34]=

  0

Up to The Theory (Equally Spaced Points)