How should I calculate the stationary value of ## S[y] ##?

[Math100]

Homework Statement

Use the method of Lagrange multipliers to find the function $y(x)$ that makes the functional $S[y]=\int_{1}^{2}x^2y^2dx$ stationary subject to the two constraints $\int_{1}^{2}xydx=1$ and $\int_{1}^{2}x^2ydx=2$. Calculate the stationary value of $S[y]$.

Relevant Equations

None.

Consider the functional $S[y]=\int_{1}^{2}x^2y^2dx$ stationary subject to the two constraints $\int_{1}^{2}xydx=1$ and $\int_{1}^{2}x^2ydx=2$.
Then the auxiliary functional is $\overline{S}[y]=\int_{1}^{2}(x^2y^2+\lambda_{1}xy+\lambda_{2}x^2y)dx, y(1)=y(2)=0$ where $\lambda_{1}$ and $\lambda_{2}$ are the Lagrange multipliers.
By definition, the Euler-Lagrange equation is $\frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0, y(a)=A, y(b)=B$ for the functional $S[y]=\int_{a}^{b}F(x, y, y')dx, y(a)=A, y(b)=B$.
Let $F(x, y, y')=x^2y^2+\lambda_{1}xy+\lambda_{2}x^2y$.
This gives $\frac{\partial F}{\partial y}=2x^2y+\lambda_{1}x+\lambda_{2}x^2$.
Thus, the Euler-Lagrange equation is $2x^2y+\lambda_{1}x+\lambda_{2}x^2=0\implies 2x^2y=-\lambda_{1}x-\lambda_{2}x^2\implies y=-\frac{\lambda_{1}}{2x}-\frac{\lambda_{2}}{2}$.
The first constraint $\int_{1}^{2}xydx=1$ gives $1=\int_{1}^{2}x[-\frac{\lambda_{1}}{2x}-\frac{\lambda_{2}}{2}]dx\implies 1=-\frac{\lambda_{1}}{2}-\frac{3\lambda_{2}}{4}$ and the second constraint $\int_{1}^{2}x^2ydx=2$ gives $2=\int_{1}^{2}x^2[-\frac{\lambda_{1}}{2x}-\frac{\lambda_{2}}{2}]dx\implies 2=-\frac{3\lambda_{1}}{4}-\frac{7\lambda_{2}}{6}$, so $\lambda_{1}=16$ and $\lambda_{2}=-12$.
Hence $y(x)=-\frac{\lambda_{1}}{2x}-\frac{\lambda_{2}}{2}\implies y(x)=-\frac{8}{x}+6$.
Observe that $S[y]=\int_{1}^{2}x^2y^2dx\implies S[y]=\int_{1}^{2}x^2(-\frac{8}{x}+6)^2dx\implies S[y]=\int_{1}^{2}x^2(\frac{64}{x^2}-\frac{96}{x}+36)dx\implies S[y]=4$.
Therefore, the function $y(x)$ that makes the functional $S[y]=\int_{1}^{2}x^2y^2dx$ stationary subject to the two constraints $\int_{1}^{2}xydx=1$ and $\int_{1}^{2}x^2ydx=2$ is $y(x)=-\frac{8}{x}+6$ and the stationary value of $S[y]$ is $4$.

I just want to know if the work and answer shown above is correct or not. Please check/verify/confirm to see if this is correct or not.

 

[pasmith]

This is the correct method; I hve not checked your arithmetic.

Note that $y'$ does not appear in $F$, so you do not end up with an ODE and you do not have boundary conditions.

 

[Math100]

This is the correct method; I hve not checked your arithmetic.

Note that $y'$ does not appear in $F$, so you do not end up with an ODE and you do not have boundary conditions.

Thank you so much for confirming/verifying!