The question:

Find the minimum value of $\int_0^1 y'^2 dx$ subject to the conditions $y(0)=y(1)=0$ and $\int_0^1y^2dx=1$.

In another question, I proved that, if we have an integral of the form


With end conditions $y(a)=y(b)=0$ and subject to the constraint $\int_a^b r(x)y^2dx=1$, the Euler-Lagrange equation becomes

$$\frac{d}{dx}(p\frac{dy}{dx})+(q+\lambda r)y=0$$

Where $\lambda$ is our Lagrange multiplier. Now here, we have a similar integral, with $p=1$ and $q=0$. The above equation becomes


Which has the solution


Is this the minimum value for my integral? If it is, how can I solve for C and $\lambda$? I can use the constraint, but that would only give me one term.

Thank you in advance.

  • $\begingroup$ If $y=Ce^{\sqrt{\lambda_i}}$ and $y(0)=0$, then $y\equiv0$. $\endgroup$ – John Gowers Sep 24 '13 at 18:03

The general solution is actually $$y=C_1e^{\sqrt{\lambda}ix}+C_2e^{-\sqrt{\lambda}ix}$$ but it's more convenient to write it as $$y=A\cos \omega x+B\sin \omega x$$ where $\omega=\sqrt{\lambda}$. The condition $y(0)=0$ gives $A=0$, while $y(1)=0$ tells you that $\omega =\pi n$ for some nonzero integer $n$.

It remains to optimize $n$. From $$\frac{\int_0^1 (y')^2}{\int_0^1 y^2} = \frac{B^2\omega^2/2}{B^2/2}$$ you can conclude that $n= 1$ gives the minimizing function.


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