#finite element methods

Ok, so suppose we have a two-point boundary value problem of the form

-d2u/dx2 = f in (0,1)

u(0) = 0

u’(1) = 0

where f is a fixed continuous function.

Turns out this is soluble by variational methods.

Throughout, assume that (f,g) refers to the standard integral-based inner product on [0,1], defined by Int(f(x)g(x),x,0,1) in pseudocode.

Define V = {v in L2(0,1): (v’,v’) < infty and v(0) = 0}.

The claim here is that u in V solves the boundary value problem above iff (u’,v’)=(f,v) for all v in V.

(=>)

If we have a solution to the boundary value problem, then in particular notice that for any v in V, we have

(f,v) = Int(f(x)v(x),x,0,1) = Int(-u’’(x)v(x),0,1)

Now as long as u’ and v are continuously differentiable (which is implicit (? I think) in the fact that u’’ shows up in the boundary-value problem, and that (v’,v’) < infty)), it must be that we can apply the Integration by Parts theorem, which states that

u’v]_0^1 = Int(u’’(x)v(x),x,0,1) + Int(u’(x)v’(x),0,1)

But since u’(1) = 0 and v(0) = 0, the right-hand side goes away.

In particular, Int(-u’’(x)v(x),x,0,1) = Int(u’(x),v’(x),0,1), which implies that the variational formulation given above holds.

(<=)

In the other direction, suppose that u is a continuously twice-differentiable function satisfying the variational formulation.

Then in particular, u’ is continuously differentiable and so is any v in V by definition. Thus the above reasoning gives (f,v) = (-u’’(x)v(x),x,0,1) + u’(1)v(1) (since v(0) = 0).

In particular, (f-(-u’’),v) = 0 for all v in V such that v(1) = 0 as well. Let w = f+u’’, which is at least continuous even if it’s not continuously twice-differentiable. If w is *not* identically zero (read: u doesn’t solve the DE), then w(x) is nonzero and either strictly positive or strictly negative on some closed interval [x_0-e,x_0+e] by continuity. Choose v(x) so that it is a twice-differentiable function that is zero off of this interval and twice differentiable and positive on it. This v(x) will in particular be such that v(1)=0 since our closed interval [x_0-e,x_0+e] could be chosen not to include this endpoint. But, for this v, we will contradict the assumption that (w,v) = 0. Thus it must have been the case that -u’’=f on the interval. 

Notice that plugging in v(x) = x to the equation (f-(-u’’),v) = u’(1)v(1) will imply in particular that 0 = (0,x) = u’(1), so u also satisfies the right-hand condition on its derivative.