# Minimize $\frac{\int_0^1 f(x)\phi''(x)\phi''(x)\,\text{d}x} {\int_0^1 g(x) \phi'(x)\phi'(x)\,\text{d}x}$

Below is a situation I came across during solving problems using energy method in mechanics. I have formulated it in question form, omitting details I think are unnecessary.

## Question:

Given positive continuous diffenertiable functions $$f(x)$$ and $$g(x)$$, find non-trivial $$\phi(x)$$ that minimizes the functional $$F$$:

$$F[\phi,\phi',\phi'';x]= \frac{\displaystyle\int_0^1 f(x)\phi''(x)\phi''(x)\,\text{d}x} {\displaystyle\int_0^1 g(x) \phi'(x)\phi'(x)\,\text{d}x}$$

with boundary conditions $$\phi(0)=\phi''(0)=\phi(1)=\phi''(1)=0$$.

1. For $$f(x)=1$$ and $$g(x)=1$$, prove that $$\phi(x)=C\sin(\pi x)$$, where $$C$$ is arbitrary constant, minimizes $$F$$.
2. For $$f(x)=1$$ and $$g(x) = x$$, find $$\phi(x)$$.
3. For general $$f(x)$$ and $$g(x)$$, find $$\phi(x)$$.

## Comments

One source claims that the functional $$F$$ is called the Rayleigh-Ritz quotient. Some special cases of the original problem can be solved exactly from differential equation, so there is some information regarding the expected solution:

1. For $$f(x)=1$$ and $$g(x)=1$$, the function $$\phi(x)=C\sin(\pi x)$$ corresponds to the exact solution, but not proven in terms of minimizing the functional $$F$$.
2. For $$f(x)=1$$ and $$g(x)=x$$, a similar case with different boundary condition has the exact solution $$\phi(x)$$ in the form of Bessel function of the first kind $$J_{-1/3}$$. Physically I believe Bessel function should also be the solution form in this case, but mathematically I am not sure.
3. For general $$f(x)$$ and $$g(x)$$, we normally resort to numerical methods such as finite element, because we believe that the closed-form solution is not possible. Here I am asking merely for curiosity. Would it be possible to express $$\phi(x)$$ in some form, such as using infinite series or special functions? May be imposing certain conditions for $$f(x)$$ and $$g(x)$$? Sorry for being vague, I can add more information if required.

## Edited: 2019-07-07

I would like to add the numerical results for case 2 where $$f(x)=1$$ and $$g(x)=x$$.

The mode shape $$\phi(x)$$ is expressed in power series:

$$\phi(x)=a_0+a_1 x+a_2 x^2+\cdots + a_n x^n + \cdots$$ where $$a_0=0,\quad a_1=1,\quad a_2=0,\quad a_3\approx -0.7272$$ $$a_{3n+1}=\frac{(-1)^n\xi^n}{(3n+1)!}\prod_{k=1}^n (3k-2),\quad n\ge 1$$ $$a_{3n+2}=0,\quad n\ge 1$$ $$a_{3n+3}=-6a_3\frac{(-1)^n\xi^n}{(3n+3)!}\prod_{k=1}^n (3k),\quad n\ge 1$$ $$\xi \approx 18.5687$$ The functional $$F$$ is calculated as $$F=\xi\approx 18.5687$$ It seems that my physical belief is wrong: the shape of $$\phi(x)$$ is not like a Bessel function, and not even symmetrical on domain $$[0,1].$$

For comparison, if $$\phi(x)=\sin(\pi x)$$, then $$F=2\pi^2\approx1 9.7392.$$

## 2 Answers

You can unpack this to get a "normal" Lagrange functional in $$\text{minimize}\int_0^1f(x)ϕ''(x)^2dx \text{ such that } \int_0^1g(x)ϕ'(x)^2dx=1$$ Then the Lagrange functional is $$L(ϕ,λ)=\frac12\int_0^1f(x)ϕ''(x)^2dx+\fracλ2\left(1-\int_0^1g(x)ϕ'(x)^2dx\right).$$ The saddle point condition resp. Euler-Lagrange equations then lead to \begin{align} 0=δL(ϕ,λ)&=\int_0^1f(x)ϕ''(x)δϕ''(x)dx-λ\int_0^1g(x)ϕ'(x)δϕ'(x)dx \\ &=[f(x)ϕ''(x)δϕ'(x)]_0^1-\int_0^1[f(x)ϕ''(x)]'δϕ'(x)dx-λ[g(x)ϕ'(x)δϕ(x)]_0^1+λ\int_0^1[g(x)ϕ'(x)]'δϕ(x)dx \\ &=-[f(x)ϕ''(x)]'δϕ(x)]_0^1+\int_0^1[f(x)ϕ''(x)]''δϕ(x)dx+λ\int_0^1[g(x)ϕ'(x)]'δϕ(x)dx \end{align} using the boundary conditions and that the variations have to leave the boundary conditions fixed, thus being zero at the same places in the same derivatives.

The resulting equation is $$0=[f(x)ϕ''(x)]''+λ[g(x)ϕ'(x)]'\implies C=[f(x)ϕ''(x)]'+λ[g(x)ϕ'(x)].$$

For $$f=g=1$$ this gives $$0=ϕ^{(4)}+λϕ''$$ which has non-trivial solutions $$A\cos(ωx)+B\sin(ωx)+Cx+D$$, leading to $$A=C=D=0$$ and $$ω=k\pi$$, $$λ=ω^2$$, minimum with non-trivial solution for $$k=1$$.

For the second example $$f=1$$, $$g(x)=x$$, one gets $$0=ϕ^{(4)}+λxϕ''+λϕ'\text{ or }C=ϕ'''+λxϕ'$$ The homogeneous part of the second form has a solution in terms of Airy functions for $$ϕ'$$, then apply variation of constants and integrate to get $$ϕ$$, this looks like a long unintuitive formula.

• Splendid answer! The resulting differential equations are of the same form as in original physical problem. May I ask, is it possible to calculate the functional $F$ for the second example ($f=1, g=x$), without explicitly writing down $\phi(x)$? Jul 7, 2019 at 9:34
• No, I do not see a way. You can not do this in the matrix case, so it is even less possible in the operator case. Jul 7, 2019 at 10:40

I think I have a partial answer, hopefully someone can expand on it.

Let $$\eta$$ be a function satisfying the same boundary conditions. Then the first variation of the functional is

$$\delta F = \lim_{\epsilon \to 0} \frac{F[\phi + \epsilon \eta] - F[\phi]}{\epsilon}$$

If $$\phi$$ minimizes the functional, this first variation has to be zero for all $$\eta$$.

Working this out for the given functional we get

$$\frac{\delta F}{2} = \left( \int_0^1 f \phi'' \eta'' dx \right)\left( \int_0^1 g \phi'\phi' dx \right) - \left( \int_0^1 f \phi'' \phi'' dx \right) \left( \int_0^1 g \phi' \eta' dx \right)$$

1. Now let $$\phi(x) = \sin(\pi x)$$ and $$f(x) = g(x) = 1$$. Then the first variation turns into

$$\frac{\delta F}{2} = -\left(\int_0^1 \pi^2 \sin(\pi x)\eta''\right)\frac{\pi^2}{2} - \frac{\pi^4}{2} \left( \int_0^1 \pi \cos(\pi x) \eta' dx \right)$$

Partial integration then shows that this is zero, proving $$\sin(\pi x)$$ minimizes $$F$$.