- #1
Math100
- 813
- 229
- 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.
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.
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