Find the Euler-Lagrange equation for the functional

$$I(y) = \int_0^1(py\,'\,^2-qy^2)\mathrm dx$$

subject to the constraint

$$\int_0^1ry^2 = 1.$$

Answer: $\frac{d}{dx}(py') + (q-\lambda r)y = 0$.

Can anyone answer this question? I tried but the answer does not come correct.

  • 1
    $\begingroup$ You also need some boundary conditions on $y$. $\endgroup$ – user7530 Sep 16 '13 at 18:55
  • $\begingroup$ This is all i am given.. $\endgroup$ – manayay Sep 16 '13 at 18:57
  • $\begingroup$ @Davide Giraudo , Hi How did you edit the question? Please help me do that y self $\endgroup$ – manayay Sep 16 '13 at 19:00
  • $\begingroup$ I added the tag "calculus of variations" and pout a "\mathrm" in the integral for the $d$. $\endgroup$ – Davide Giraudo Sep 16 '13 at 19:14
  • $\begingroup$ constraints are solved with Lagrange multipliers. The new functional is $I(y)+\lambda\int_0^1ry^2dx$ (with $\lambda$ an undetermined constant), get its Euler Lagrange equations; finally, among the solutions, take only those satisfying the constraint $\endgroup$ – user8268 Sep 16 '13 at 19:17

First, stick the constraint into the objective using a Lagrange multiplier $\lambda$:

$$\min_{y,\lambda} \int_0^1 (py'^2-qy^2)\,dx + \lambda\left(\int_0^1 ry^2dx-1\right).$$

Now take variations with respect to $y$. You get

$$0 = \int_0^1 2py'\delta y' -2qy\delta y +2\lambda r y \delta y\, dx = \int_0^1 2py'\delta y'\,dx + \int_0^1 2y(-q+r\lambda)\,dx.$$

You now integrate the first term by parts:

$$0 = 2p(y'(1)\delta y(1)-y'(0)\delta y(0)) + \int_0^1 \left(-2(py')' + 2y(r\lambda - q)\right)\,dx$$ and assuming appropriate boundary conditions on $y$ (such as fixed endpoints) you recover your expected solution.

  • $\begingroup$ Thanks for your answer, but this is what i have as my question.I am not sure like can we take any assumptions to reach at the answer i have posted? $\endgroup$ – manayay Sep 16 '13 at 20:19
  • $\begingroup$ To get the answer you posted, you must have boundary constraints; typically if you're talking about Euler-Lagrange equations you are enforcing that the path has fixed endpoints, $\delta y(0) =\delta y(1) = 0$. Notice that the problem as written is extremely sloppy: there is no explicit indication of what $p, q, r$ depend on, which variable is being integrated over in the constraint, etc. It's all implicit. Possibly whoever gave you this problem forgot about the boundary conditions, or thought they were obvious. $\endgroup$ – user7530 Sep 16 '13 at 21:16
  • $\begingroup$ Thanks for thr response, really appriciate it. $\endgroup$ – manayay Sep 16 '13 at 21:19

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