8 Leonhard Euler Quotes (Calculus of variations - The Euler–Lagrange equation).
8 Leonhard Euler Quotes (Calculus of variations – The Euler–Lagrange equation).
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The Calculus of variations is a field of mathematical analysis that uses variations , which are small changes in functions and functional to find maxima and minima of functional; mappings from a set of functions to the real numbers. The Euler-Lagrange differential equation is the fundamental equation of calculus of variation.
Leonhard Euler (15 April 1707 – 18…
Say What? Colin MacLaurin on Evidence, Reason, and Knowledge ‘It is not therefore the business of philosophy, in our present situation in the universe, to attempt to take in at once, in one view, the whole scheme of nature; but to extend, with great care and circumspection, our knowledge, by just steps, from sensible things, as far as our observations or reasonings from them will carry us, in our enquiries concerning either the greater motions and operations of nature, or her more subtle and hidden works.'
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
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.
Thus the variational formulation agrees with the DE formulation under fairly mild continuity assumptions.
