Calculus of Variations – Integrals with Additional Conditions

calculuscalculus-of-variationsmultivariable-calculussystems of equations

Find the functions $y_1$ and $y_2$ with which the functional $$J = \int_{0}^{\pi/2}(2y_1y_2 + (y_1^{'})^2+(y_2{'})^2)dx$$ has conditional extremum.
$$\begin{cases}y_1(0)=-1,\\y_2(0)=1,&\end{cases}\quad\quad\begin{cases}y_1(\pi/2)=\frac{\pi^2}4-1,\\y_2(\pi/2)=\frac{\pi^2}4+1.&\end{cases}$$
With additional condition $y_1^{\prime}+y_2^{\prime}=4x.$

Here are my ideas or hints I've been given so far:

It is a conditional functional problem (due to additional condition). We need to set up a "helping" functional such $J^{*}=\int_{a}^{b}(F+\lambda(x)g)dx$ and solve using Lagrange multipliers. Then, I am not sure I know how to proceed further. The answer should look like $y_1=$ and $y_2=$, without constants because to find them we are given the boundary conditions. I would appreciate if someone could show how to solve this problem.

Here is my work so far:

$J^{*}=\int_{a}^{b}((F+\lambda(x)g)dx=\int_{0}^{\pi/2}(2y_1y_2+ (y_1^{'})^2+(y_2{'})^2)+\lambda (y'_1+y_2^{'})) dx $

Now I was suggested to set some system of equations, but I don't know what they mean by that.

Best Answer

Too long for a comment.

You missed a term in $J^{*}$; it should be $J^{*}=\int_0^{\pi/2}F^{*}(x,y_1,y_2,y_1',y_2',\lambda)\,dx$, where $$ F^{*}(x,y_1,y_2,y_1',y_2',\lambda)=2y_1y_2+(y_1')^2+(y_2')^2+\lambda(y_1'+y_2'-4x). \tag{1} $$ Now, write the Euler-Lagrange equations for $F^{*}$: \begin{align} &\frac{\partial F^{*}}{\partial y_1}-\frac{d}{dx}\left(\frac{\partial F^{*}}{\partial y_1'}\right)=0\implies 2y_2-2y_1''-\lambda'=0, \tag{2a} \\ &\frac{\partial F^{*}}{\partial y_2}-\frac{d}{dx}\left(\frac{\partial F^{*}}{\partial y_2'}\right)=0\implies 2y_1-2y_2''-\lambda'=0, \tag{2b} \\ &\frac{\partial F^{*}}{\partial \lambda}=0\implies y_1'+y_2'-4x = 0. \tag{2c} \end{align} Can you continue from here?

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