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Let $$F[y,z]=\int_a^b \left[y'(t)z(t)-\frac{1}{2}y(t)^2-\frac{1}{2}z(t)^2 \, \right]dt$$ and let $$L[x]=\int_a^b f(x(t),x'(t)) \, dt$$

I'm told that if $\tilde{x}$ is a stationary point of $L$ then $(\tilde{x},\tilde{x}')$ is a stationary point of $F$. I'm now asked to find $f$.

I have considered the Euler-Lagrange equations for $F$ and found that $y'=z$ and $z'=-y$. Using the Beltrami equation I have also found that $\frac{1}{2}y^2+\frac{1}{2}z^2=c \text{ (constant)}$. Putting this all together gives me the differential equation $(y')^2+y^2=c$.

I'm not sure if this helps me in any way but I don't know what else I can do here.

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    $\int{\mathrm{d}y \over \sqrt{c - y^{2}}} = \pm\int\mathrm{d}x$. – 2017-02-11
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    @FelixMarin Thank you. How does this help me to find $f$ though? – 2017-02-11
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    You know that $y'' = z'$ so $y'' = -y$ so $y= C_1 cos(t) + C_2 sin(t)$. If you check this, it does satisfy the thing you got from the Beltrami equation. So you have a general form for $y$ and $z$. I'm not sure if it helps though. I think you would need to know that $(x,x')$ being a stationary point of $F$ means that $x$ is a stationary point of $L$ to use this though. – 2017-02-11

2 Answers 2

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It is the Lagrangian for an unidimensional harmonic oscillator of a particle of unit mass. $$f(x,x')=\frac{1}{2}(x')^2-\frac{1}{2}x^2$$

The ODE you've found is the conservation of mechanical energy for this system.

I've found the answer with a simple substitution at F.

Added

$$F[x,x']=\int_a^b x'(t)x'(t)-\frac{1}{2}x^2-\frac{1}{2}(x')^2 \, dt=\int_a^b \frac{1}{2}(x')^2-\frac{1}{2}(x^2) \, dt$$

Now, if being stationary for $L$ must imply being stationary for $F$, the integrands must differ in the total derivative with respect to time of any function of coordinate and time. I chose this to be zero, but it can be done well:

$$f(x,x')=\frac{1}{2}(x')^2-\frac{1}{2}x^2+\frac{\mathbb d}{\mathbb dt}g(x,t)$$

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    Could explain how you've done that? – 2017-02-18
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    You simply build $L$ for $x,x'$ and that's exactly the form $F$ has to have. – 2017-02-18
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    Could you explain why the integrands must differ in the total derivative with respect to time of any function of $(x,t)$? – 2017-02-22
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    Integrating a total derivative produces the original function evaluated at $a$ and $b$. When varied to find the stationary path, the variation vanishes precisely at the integral limits, so, this part vanishes. You can check the derivation of Euler-Lagrange equations and you will see this easily. – 2017-02-22
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    http://mathworld.wolfram.com/Euler-LagrangeDifferentialEquation.html Equation (7). You can add to this equation the variation of the result of integration (the original function $g$ evaluated at $a$ and $b$), but the variation is zero in $a$ and $b$. – 2017-02-22
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    And, if all previously said is difficult to understand, use the Euler-Lagrange equation applied to $L=\frac{1}{2}(x')^2-\frac{1}{2}x^2$ and $L'=\frac{1}{2}(x')^2-\frac{1}{2}x^2+\frac{\mathbb d}{\mathbb dt}g(x,t)$ and you will see you obtain the same equation $x''-x=0$, so, the total derivative is of no influence at all! http://physics.stackexchange.com/questions/174137/adding-a-total-time-derivative-term-to-the-lagrangian – 2017-02-22
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    Sorry to come back to this question again, I thought I had it. So when you claim that the integrands differ by the derivative of $g$ w.r.t $t$ are you saying that the difference in the derivatives of the integrals is the derivative of $g$? That is, are you saying $$\frac{d}{dt}\left(\int_a^b h(y(t),z(t),y'(t),z'(t)) \, dt\right) - \frac{d}{dt}\left( \int_a^b f(x(t),x'(t) \, dt\right)=\frac{d}{dt}g(x,t)$$? – 2017-04-03
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    No. Get the original question. We, after putting $x$ and $x'$ in in $L$, have that $f(x,x')-(\frac{1}{2}(x')^2-\frac{1}{2}(x^2))=\frac{\mathbb d}{\mathbb dt}g(x,t)$, so is, the arguments differ in a total derivative wrt time of some function. The functionals are different, of course $F[x,x']-L[x]=\left.g(x,t)\right|_a^b$, but not the function $\tilde x$ that makes the functional stationary. It's the version for functionals of the constants in the function to optimize in elemental problems of optimization. – 2017-04-03
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    Better get some $g$ as example, say, $g(x)=xt\to dg/dt=x't+x$ and calculate the E-L equations for $L=\frac{1}{2}(x')^2-\frac{1}{2}(x^2)$ and for $\bar L=\frac{1}{2}(\bar x')^2-\frac{1}{2}(\bar x^2)+(x't+x)$ – 2017-04-03
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    $\partial L/\partial x'=x'\;;d/dt(\partial L/\partial x')=x'';\;\partial L/\partial x=-x\implies x''+x=0$ $\partial L/\partial \bar x'=x'+t\;;d/dt(\partial L/\partial x')=x''+1;\;\partial L/\partial x=-x+1\implies x''+x=0$ the same equation! – 2017-04-03
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Hint: One idea is to "integrate out the $z$-variable", i.e. to first find the stationary point wrt. the $z$ variable. Completing the square wrt. the $z$ variable, OP's functional becomes

$$ F[y,z]~=~\int_a^b \! \mathrm{d}t\left(y^{\prime} z -\frac{1}{2}y^2-\frac{1}{2}z^2 \right)~=~\frac{1}{2}\int_a^b \! \mathrm{d}t\left(y^{\prime 2} -y^2 \right) - \frac{1}{2}\int_a^b \! \mathrm{d}t\left(z-y^{\prime} \right)^2. \tag{1}$$

Hence $$ \max_{z}F[y,z]~=~ \frac{1}{2}\int_a^b \! \mathrm{d}t\left(y^{\prime 2} -y^2 \right) .\tag{2} $$

From eq. (2) we can directly read off a possible $f$-function:

$$f(x,x^{\prime})~=~\frac{1}{2}\left(x^{\prime 2} -x^2 \right) .\tag{3} $$

Note that $f$ is far from unique: For starters, we may scale $f$ with an overall constant or add a total derivative term. But more exotic modifications are also possible, cf. e.g. this Math.SE post.