I have to minimize the functional

$$J[y] =\int_0^\pi y' ^2 - ky^2 dx$$

subject to $y(0)=y(\pi)=0$. The parameter $k$ is positive. Writing down the Euler-Lagrange equation, I have:

$$y'' +ky =0,$$

which implies $y=A\cos\sqrt{k}x + B\sin\sqrt{k}x $. However, imposing the boundary conditions gives first of all $A=0$, and then $B$ arbitrary, given that $\sqrt{k} \in \mathbb{Z}$, otherwise no solution. Thus $k=m^2$ for $m \in \mathbb{N}$.

Ultimately, the solutions are

$$y_m = B\sin mx$$

Since I want to find the minimal value of the functional, I substitute into the integral:

$$J[y_m]=\int_0^\pi B^2 m^2 \cos (2mx) \ dx = 0.$$

Is this solution correct?


If your method were correct, the minimum would be achieved by the solution corresponding to $B=0$, and the minimum would be equal to $0$.

But, for $y=B\sin x$, we get $J[y]=\frac{\pi}{2}B^2(1-k)$, which for $k>1$, $$\lim_{B\to\infty}J[y]=-\infty.$$ Hence, in order to have a minimum either $k$ should be $\le 1$, or something else is missing from the formulation.

  • $\begingroup$ So I guess my error is in the final substitution. I'll take another look at it soon. The range of $k$ is actually not given in the exercise, I assumed it was positive, with no good reason actually. $\endgroup$ – Spine Feast Jan 13 '14 at 21:33
  • $\begingroup$ @Yiorgos S. Smyrlis: since $k = m^2$, and the solution $y = B \sin x$ corresponds to $m = 1$, we have $k = 1$, whence $J[y] = (\pi / 2)B^2 (1 - k) = 0$ for all $B$. Or do I err? Am I missing something; if so, please set me straight! $\endgroup$ – Robert Lewis Jan 14 '14 at 7:49
  • $\begingroup$ If $k\le 1$, then $J$ is bounded from below. (Wirtinger inequality). $\endgroup$ – Yiorgos S. Smyrlis Jan 14 '14 at 7:55
  • $\begingroup$ The solutions of the E-L equation are $y_m = B\sin(mx)$ where $m^2 = k $ is integral. So the functional becomes $\int y_m ' ^2 - m^2 y_m ^2 dx$, which evaluates to zero. Is this right or wrong, I'm getting confused here. $\endgroup$ – Spine Feast Jan 14 '14 at 11:51
  • $\begingroup$ Try $u=\sum a_n\sin nx$, and observe that $J=\frac{\pi}{2}\sum(n^2-k)a_n^2$. $\endgroup$ – Yiorgos S. Smyrlis Jan 14 '14 at 12:36

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