Minimization of a Functional, exercize
Exercize 1
Find stationary point of the functional
\mathcal{F}(x) = \int_0^2 \left( x(t)^2 + x(t)(x'(t))^2 + t x(t) \right) \, \mathrm{d}t
with constaint
x(0)=0,\qquad x(1) = 1
Applying the First Variation
We start by applying the first variation to the functional: \mathcal{F}(x). To compute the variation, we first take the variation of the integrand:
\begin{aligned} \delta\mathcal{F}(x) & = \delta \int_0^2 \left( x(t)^2 + x(t)(x'(t))^2 + t x(t) \right) \, \mathrm{d}t \\ & = \int_0^2 \delta \left( x(t)^2 + x(t)(x'(t))^2 + t x(t) \right) \, \mathrm{d}t \\ & = \int_0^2 \left( 2x(t) \delta x(t) + \delta x(t) (x'(t))^2 + 2x(t) x'(t) \delta x'(t) + t \delta x(t) \right) \, \mathrm{d}t \\ & = \int_0^2 \left( \left( 2x(t) + (x'(t))^2 + t \right) \delta x(t) + \textcolor{blue}{2x(t)x'(t) \delta x'(t)} \right) \, \mathrm{d}t \end{aligned}
Here, we expanded the variation term by term and grouped the expressions. The blue term \textcolor{blue}{2x(t)x'(t) \delta x'(t)} is crucial for the integration by parts.
Integration by Parts
To eliminate the \delta x'(t) term, we apply integration by parts to the expression \textcolor{blue}{2x(t)x'(t) \delta x'(t)}:
\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} \left( 2x(t) x'(t) \delta x(t) \right) & = \textcolor{blue}{2x(t) x'(t) \delta x'(t)} + \frac{\mathrm{d}}{\mathrm{d}t} \left( 2x(t) x'(t) \right) \delta x(t) \\ & = \textcolor{blue}{2x(t) x'(t) \delta x'(t)} + 2 \left( (x'(t))^2 + x(t)x''(t) \right) \delta x(t) \end{aligned}
This allows us to rewrite the variation of \mathcal{F}(x) as:
\begin{aligned} \delta\mathcal{F}(x) & = \int_0^2 \left( 2x(t) + (x'(t))^2 + t \right) \delta x(t) + \textcolor{blue}{2x(t)x'(t) \delta x'(t)} \, \mathrm{d}t \\ & = \int_0^2 \left( 2x(t) + (x'(t))^2 + t \right) \delta x(t) \, \mathrm{d}t \\ & \quad + \textcolor{red}{\left[ \frac{\mathrm{d}}{\mathrm{d}t} \left( 2x(t)x'(t)\delta x(t) \right) - 2 \left( (x'(t))^2 + x(t) x''(t) \right) \delta x(t) \right]} \, \mathrm{d}t \end{aligned}
Now, after applying the integration by parts, we can group terms again:
\begin{aligned} \delta\mathcal{F}(x) & = \int_0^2 \left( 2x(t) - (x'(t))^2 + t - 2x(t) x''(t) \right) \delta x(t) \, \mathrm{d}t \\ & \quad + \Big[ 2x(t)x'(t)\delta x(t) \Big]_0^2 \end{aligned}
Since \delta x(0) = \delta x(2) = 0 (from the boundary conditions), the boundary term vanishes. Therefore, we are left with:
\delta\mathcal{F}(x) = \int_0^2 \left( 2x(t) - (x'(t))^2 + t - 2x(t)x''(t) \right) \delta x(t) \, \mathrm{d}t
Applying the Fundamental Lemma of Calculus of Variations
For the variation \delta\mathcal{F}(x) to be zero for all variations \delta x(t), the integrand must vanish. Therefore, we obtain the Euler-Lagrange equation:
2x(t) - (x'(t))^2 + t - 2x(t)x''(t) = 0
Boundary Conditions
The problem also specifies boundary conditions:
x(0) = 0, \quad x(2) = 1
Final Problem
Thus, the full problem is to solve the following boundary value problem:
\left\{ \begin{aligned} & 2x(t) - (x'(t))^2 + t - 2x(t)x''(t) = 0, \\ & x(0) = 0, \\ & x(2) = 1. \end{aligned} \right.
Exercize 2
Find stationary point of the functional
\mathcal{F}(x)=\int_a^b x^2 + 4xx^\prime+2(x^\prime)^2 \,\mathrm{d}t
with constaint
x(a) = x_a
Applying the First Variation
\mathcal{F}(x+\alpha\delta x)=\int_a^b (x+\alpha\delta x)^2 + 4(x+\alpha\delta x)(x^\prime+\alpha\delta x^\prime) +2(x^\prime+\alpha\delta x^\prime)^2 \,\mathrm{d}t
perform variation
\begin{aligned} \dfrac{\mathrm{d}}{\mathrm{d}\alpha} \mathcal{F}(x+\alpha\delta x) &= \underbrace{\dfrac{\mathrm{d}}{\mathrm{d}\alpha}\int_a^b}_{\longleftrightarrow} \cdots \mathrm{d}t \\ &= \underbrace{\int_a^b\dfrac{\mathrm{d}}{\mathrm{d}\alpha}}_{\longleftrightarrow} \cdots \mathrm{d}t \\ &= \int_a^b (x+\alpha\delta x)(2\delta x+4\delta x^\prime)+ 4(x^\prime+\alpha\delta x^\prime)(\delta x+\delta x^\prime) \,\mathrm{d}t \\ \end{aligned}
evaluating at \alpha=0:
\begin{aligned} \delta\mathcal{F}(x) &= \dfrac{\mathrm{d}}{\mathrm{d}\alpha} \mathcal{F}(x+\alpha\delta x)\bigg|_{\alpha=0} \\ &= \int_a^b x(2\delta x+4\delta x^\prime)+ 4x^\prime(\delta x+\delta x^\prime) \,\mathrm{d}t \\ &= \int_a^b (2x+4x^\prime)\delta x+ 4(x+x^\prime)\delta x^\prime \,\mathrm{d}t \end{aligned}
Integration by Parts
For the term 4(x+x^\prime)\delta x^\prime
\begin{aligned} \dfrac{\mathrm{d}}{\mathrm{d}t} \left( (x+x^\prime)\delta x \right) &= (x+x^\prime)\delta x^\prime+ \dfrac{\mathrm{d}}{\mathrm{d}t} \left( (x+x^\prime) \right)\delta x \\ &= (x+x^\prime)\delta x^\prime+ (x^\prime+x^{\prime\prime}) \delta x \end{aligned}
and
\begin{aligned} \int_a^b (x+x^\prime)\delta x^\prime \,\mathrm{d}t &= \int_a^b \dfrac{\mathrm{d}}{\mathrm{d}t} \left( (x+x^\prime)\delta x \right) \,\mathrm{d}t - \int_a^b (x^\prime+x^{\prime\prime}) \delta x \,\mathrm{d}t \\ &= (x+x^\prime)\delta x\bigg|_a^b - \int_a^b (x^\prime+x^{\prime\prime}) \delta x \,\mathrm{d}t \end{aligned}
so that
\begin{aligned} \delta\mathcal{F}(x) &= \int_a^b (2x+\cancel{4x^\prime}-\cancel{4x^{\prime}}-4x^{\prime\prime})\delta x \,\mathrm{d}t+\\ &\qquad (x(b)+x^\prime(b))\delta x(b)- \underbrace{\cancel{(x(a)+x^\prime(a))\delta x(a)}}_{\delta x(a)=0} \end{aligned}
Applying the Fundamental Lemma of Calculus of Variations
\begin{aligned} \delta x(b) & \implies x(b)+x^\prime(b) = 0 \\ \delta x & \implies 2x(t)-4x^{\prime\prime}(t) = 0 \\ \end{aligned}
Final Problem
\left\{\begin{aligned} x-2x^{\prime\prime} & = 0 \\ x(a) & = x_a \\ x(b)+x^\prime(b) & = 0 \end{aligned}\right.
The same result can be obtained using Euler-Lagrange equation
L(x,x',t) = x^2 + 4 x x^\prime + 2(x^\prime)^2
so that
\begin{aligned} \dfrac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L(x,x',t)}{\partial x^\prime} - \frac{\partial L(x,x',t)}{\partial x} &= 0 \\ \dfrac{\mathrm{d}}{\mathrm{d}t} (4x+4x^{\prime}) - (2x+4x^{\prime}) &= 0 \\ 4x^{\prime\prime} -2x &= 0 \\ -2(x-2x^{\prime\prime})&= 0 \end{aligned}
and for the free boundary
0= \frac{\partial L(x,x',t)}{\partial x^\prime} \bigg|_{x=b} = (4x+4x^{\prime})\Big|_{x=a} = 4(x(b)+x^\prime(b)) = 0
Exercize 3
Find stationary point of the functional
\mathcal{F}(x) = \int_0^1 x^\prime (\log x^\prime)^2 \mathrm{d}t
with no constaint
Applying the First Variation
\begin{aligned} \delta\mathcal{F}(x) &= \delta\int_0^1 x^\prime (\log x^\prime)^2 \mathrm{d}t\\ &= \int_0^1 \delta\left( x^\prime (\log x^\prime)^2 \right) \mathrm{d}t\\ &= \int_0^1 \delta x^\prime (\log x^\prime)^2 +2x^\prime\log x^\prime\dfrac{\delta x^\prime}{x^\prime} \mathrm{d}t\\ &= \int_0^1 \left( (\log x^\prime)^2 +2\log x^\prime\right) \delta x^\prime \mathrm{d}t \end{aligned}
Remark 1. Do not do this wrong argument: \delta x is arbitrary and so that using foundamental lemma of calculus o f variation it follows
\color{red} (\log x^\prime)^2 +2\log x^\prime = 0
the lemma is true for \delta x not \delta x^\prime.
Integration by Parts
\begin{aligned} \dfrac{\mathrm{d}}{\mathrm{d}t}\left[ \left((\log x^\prime)^2 +2\log x^\prime\right) \delta x \right] & = \dfrac{\mathrm{d}}{\mathrm{d}t}\left[ (\log x^\prime)^2 +2\log x^\prime \right] \delta x \\ & + \left((\log x^\prime)^2 +2\log x^\prime\right) \delta x^\prime \end{aligned}
so that
\delta\mathcal{F}(x) = \left((\log x^\prime)^2 +2\log x^\prime\right) \delta x\Big|_0^1 - \int_0^1 \dfrac{\mathrm{d}}{\mathrm{d}t}\left[ (\log x^\prime)^2 +2\log x^\prime \right] \delta x \,\mathrm{d}t
Applying the Fundamental Lemma of Calculus of Variations
\begin{aligned} \delta x(a) & \implies (\log x^\prime(a))^2 +2\log x^\prime(a) = 0 \\ \delta x(b) & \implies (\log x^\prime(b))^2 +2\log x^\prime(b) = 0 \\ \delta x(t) & \implies \dfrac{\mathrm{d}}{\mathrm{d}t}\left[ (\log x^\prime)^2 +2\log x^\prime \right] = 0, \end{aligned}
Final Problem
\left\{ \begin{aligned} (\log x^\prime)^2 +2\log x^\prime & = C, \\ (\log x^\prime(a))^2 +2\log x^\prime(a) & = 0 \\ (\log x^\prime(b))^2 +2\log x^\prime(b) & = 0 \end{aligned} \right.
notice that the ODE part is now first order and the solution depend on a constant C (determined by the boundary conditions).
Exercize 4
Find stationary point of the functional
with constaint