Minimizing a functional subject to boundary conditions

Question

I want to find a smooth function $y : [0,1] \to \mathbb{R}$ that minimizes

$$ S = \int_0^1 (y(x)y'''(x) + 3 y'(x) y''(x))^2 dx $$ subject to the constraints / boundary conditions: (i) $y(0)=1$, (ii) $y(1)=0$, (iii) $y'(0)=0$, (iv) $y'(1)=0$.

What is the right approach for this? If I solve the unconstrained Euler-Poisson equation I get the square root of a fifth order polynomial, but this solution cannot satisfy all four constraints.

I understand how to implement constraints via Lagrange multipliers if the constraints are themselves differential equations, but is there a similar way to implement these boundary conditions?

(This is probably a duplicate of some sort, but I couldn't find the answer, mainly, I assume, because I don't know the right keyword to search for.)

Edit: As mentioned in the comments, I might be happy knowing just the minimal value for $S$, even if I don't find the explicit form of the minimizing $y(x)$.

With a mix of partial integration, Cauchy-Schwarz, and Taylor's theorem, I managed to show that $S \geq 1$ must hold, which is something already.

Answer 1

According to WolframAlpha, the explicit expression of the Euler-Poisson equation associated with $S$ is $$ -20y(y''')^2-30yy''y^{(4)}-12yy'y^{(5)}-2y^2y^{(6)}=0, \tag{1} $$ which can be simplified to $y(y^2)^{(6)}=0$. Therefore, the solutions to $(1)$ are $y(x)=0$ or $y(x)=\pm\sqrt{p(x)}$, where $p(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5$ is an arbitrary$^{(*)}$ fifth-degree polynomial. The boundary conditions $y(0)=1$ and $y'(0)=0$ further imply that $y(x)=+\sqrt{p(x)}$, with $a_0=1$ and $a_1=0$. In addition, $y(1)=0$ implies $$ 1+a_2+a_3+a_4+a_5=0. \tag{2} $$ To enforce the fourth boundary condition, $y'(1)=0$, is less straightforward: since the denominator of $y'(x)=\frac{p'(x)}{2\sqrt{p(x)}}$ vanishes at $x=1$, we require that not only $p'(1)=0$, i.e. $$ 2a_2+3a_3+4a_4+5a_5=0, \tag{3} $$ but also that $p''(1)=0$: $$ 2a_2+6a_3+12a_4+20a_5=0. \tag{4} $$ In other words, $x=1$ must be a triple root of $p(x)$.

The system of equations $(2)$-$(4)$ is underdetermined; it has the one-parameter family of solutions $$ (a_2,a_3,a_4,a_5)=(-t-6,3t+8,-3t-3,t). \tag{5} $$ To determine $t$, let's plug $$ y_t(x):=\sqrt{1-(t+6)x^2+(3t+8)x^3-(3t+3)x^4+tx^5} \tag{6} $$ into $S[y]$ and minimize the result with respect to $t$. Letting WolframAlpha do the hard work, we obtain the surprisingly simple result $$ S[y_t(x)]=9t^2+48t+144. \tag{7} $$ It's straightforward to show that the minimum of $S[y_t(x)]$ occurs at $t=-\frac{8}{3}$, where it attains the value $S_{\min}=80$. One can also check here that the function that minimizes $S$, $$ y_{-8/3}(x)=\sqrt{1-\frac{10}{3}x^2+5x^4-\frac{8}{3}x^5} =\sqrt{\frac{1}{3}(1-x)^3(8x^2+9x+3)}\,, \tag{8} $$ is real in the interval $[0,1]$ and satisfies all the boundary conditions.

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$^{(*)}$ To be precise, not completely arbitrary: $p(x)$ must be nonnegative in the interval $[0,1]$.

Answer 2

TL;DR: It is potentially problematic to assume a 6th-order Euler-Poisson (EP) equation in a higher-order variational problem without adequate boundary conditions (BCs).

1. OP's functional can be rewritten as $$S[z]~:=~\frac{1}{4}\int_0^1 \!dx~ (z^{\prime\prime\prime}(x))^2, \tag{1}$$ where we have introduced a new variable $$z~=~y^2\tag{2}$$ which must satisfy an inequality $$ z(x)~\geq~0,\tag{3}$$ which one should check in the end.

2. OP's BCs for $y$ $$ y(0)~=~1, \quad y^{\prime}(0)~=~0, \quad y(1)~=~0, \quad y^{\prime}(1)~=~0, \tag{4}$$ clearly imply the corresponding BCs for $z$ $$ z(0)~=~1, \quad z^{\prime}(0)~=~0, \quad z(1)~=~0, \quad z^{\prime}(1)~=~0. \tag{5}$$ Conversely, to ensure the last BC in eq. (4) [for a sufficiently nice function $y$] we need to impose a 5th BC $$ z^{\prime\prime}(1)~=~0, \tag{6}$$ as explained in Gonçalo's answer.

3. If we vary infinitesimally OP's functional, we calculate $$ \begin{align}2\delta S ~\stackrel{(1)}{=}~~&\int_0^1 \!dx~ z^{\prime\prime\prime}(x)~\delta z^{\prime\prime\prime}(x)\cr ~=~~&\int_0^1 \!dx\frac{d}{dx}\left(z^{\prime\prime\prime}(x)~\delta z^{\prime\prime}(x)\right)-\int_0^1 \!dx~ z^{(4)}(x)~\delta z^{\prime\prime} \cr ~\stackrel{(5)+(6)}{=}&-z^{\prime\prime\prime}(0)~\delta z^{\prime\prime}(0)-\int_0^1 \!dx~ z^{(6)}(x)~\delta z. \end{align}\tag{7} $$

4. Now for the 6th-order EP equation $$ z^{(6)}(x)~=~0. \tag{8}$$ to be a stationary solution, we need a 6th BC $$z^{\prime\prime\prime}(0)~=~0.\tag{9} $$

5. One may check that the minimizing 5th-order polynomial found in Gonçalo's answer happens to satisfy the 6th BC (9) as well, so it is indeed a stationary solution!

   This is not a coincidence: When we minimized the functional (1) for 5th-order polynomials satisfying EP eq. (8) and BCs (5) and (6), then eq. (7) conversely implies the 6th BC (9).