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I have the following functional

$ J(x)=\int\limits_{a}^{b}[{y(x)^2 + \dot{y}(x)^2 + 2 y(x) e^x}] dx $

and I want to find the extremum of it.

So I compute the Euler-Lagrange Equation and I get the result:

$ y^*(x) = c_1e^x+c_2e^{-x}-{e^x\over 4}+{xe^x\over 2} $

What can be done from here to find the $c_1$ and $c_2$ since I don't have any initial conditions ? Or at least the form that they should have ?

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Hints:

  1. Up to constant factors we may write OP's functional as $$ S[y]~:=~\int_a^b \!dx ~L , \qquad L ~:=~ \frac{1}{2}\dot{y}^2 + \frac{1}{2}( y+e^x)^2 ~\geq~0 .\tag{1}$$

  2. EL equation is $$ \ddot{y}-y~=~e^x.\tag{2}$$

  3. OP has already found the complete solution $$ y^{\ast}(x)~=~\left(\frac{x}{2}-\frac{3}{4}+c_+\right)e^x + c_-e^{-x},\tag{3}$$ so that $$\begin{align} \dot{y}^{\ast}(x)&~=~\left(\frac{x}{2}-\frac{1}{4}+c_+\right)e^x - c_-e^{-x}, \cr y^{\ast}(x)+e^x&~=~\left(\frac{x}{2}+\frac{1}{4}+c_+\right)e^x + c_-e^{-x}.\end{align}\tag{4}$$

  4. Therefore the Lagrangian becomes on-shell$^1$ equal to $$\begin{align}L^{\ast}(x)&~=~L^{\ast}_+(x)+L^{\ast}_-(x), \cr L^{\ast}_+(x)&~:=~ \left(\frac{x}{2}+c_+\right)^2e^{2x}~\geq~0, \cr L^{\ast}_-(x)&~:=~\left(\frac{1}{4}e^x+c_-e^{-x}\right)^2~\geq~0 ,\end{align}\tag{5}$$ while the functional on-shell is $$S[y^{\ast}]~=~S_+(c_+)+S_-(c_-), \qquad S_{\pm}(c_{\pm})~:=~\int_a^b \!dx ~L^{\ast}_{\pm}(x).\tag{6}$$

  5. Clearly it now just remains to minimize $S_{\pm}(c_{\pm})$ independently, which is a straightforward exercise.

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$^1$ The word on-shell means that the EL eq. is satisfied.

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  • $\begingroup$ "On shell" means that the argument of the Lagrangian solves the EL equations? $\endgroup$ – Giuseppe Negro Apr 10 at 11:20
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$ – Qmechanic Apr 10 at 11:38

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