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!