- #1

Math100

- 792

- 220

- Homework Statement
- For each of the following functionals, find the Gateaux differential and use it to derive the associated Euler-Lagrange equation. In each case, be sure to specify all boundary conditions that the stationary path must satisfy.

a) ## S[y]=\int_{1}^{2}(3y'^2-2y^2)dx, y(1)=1, y(2)=3 ##.

b) ## S[y]=y(0)+\frac{1}{2}\int_{0}^{1}(2y'^2+x^2y^2)dx, y(1)=2 ##.

- Relevant Equations
- Gateaux differential: ## \bigtriangleup S[y, h]=\displaystyle \lim_{\epsilon \to 0}\frac{d}{d\epsilon}S[y+\epsilon h] ## is defined on admissible paths ## y+\epsilon h ## for all ## \epsilon ## in the neighborhood of zero.

A path ## y(x) ## is a stationary path of a functional if the Gateaux differential ## \bigtriangleup S[y, h] ## is zero for all admissible paths ## y+\epsilon h ##.

For the functional ## S[y]=\int_{a}^{b}F(x, y, y')dx, y(a)=A, y(b)=B ##, 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 ##.

a) We have ## S[y+\epsilon h]=\int_{1}^{2}[3(y'+\epsilon h')^2-2(y+\epsilon h)^2]dx ##.

Note that ## \frac{d}{d\epsilon}S[y+\epsilon h]=\int_{1}^{2}[6(y'+\epsilon h')h'-4(y+\epsilon h)h]dx=2\int_{1}^{2}[3(y'+\epsilon h')h'-2(y+\epsilon h)]dx ##.

Then ## \bigtriangleup S[y, h]=\frac{d}{d\epsilon}S[y+\epsilon h]\rvert_{\epsilon=0}=2\int_{1}^{2}(3y'h'-2yh)dx ##.

Thus, the Gateaux differential is ## \bigtriangleup S[y, h]=2\int_{1}^{2}(3y'h'-2yh)dx ##.

Consider the functional ## S[y]=\int_{1}^{2}(3y'^2-2y^2)dx, y(1)=1, y(2)=3 ##.

Let ## F(x, y, y')=3y'^2-2y^2 ##.

Then ## \frac{\partial F}{\partial y'}=6y' ## and ## \frac{\partial F}{\partial y}=-4y ##.

Observe that ## \frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0\implies 6\frac{d^2y}{dx^2}+4y=0 ##.

Therefore, the Euler-Lagrange equation is ## 6\frac{d^2y}{dx^2}+4y=0, y(1)=1, y(2)=3 ##.

b) We have ## S[y+\epsilon h]=y(0)+\frac{1}{2}\int_{0}^{1}(2(y'+\epsilon h')^2+x^2(y+\epsilon h)^2)dx ##.

Note that ## \frac{d}{d\epsilon}S[y+\epsilon h]=y(0)+\frac{1}{2}\int_{0}^{1}[4(y'+\epsilon h')h'+2x^2(y+\epsilon h)h]dx=y(0)+\int_{0}^{1}[2(y'+\epsilon h')h'+x^2(y+\epsilon h)h]dx ##.

Thus, the Gateaux differential is ## \bigtriangleup S[y, h]=\int_{0}^{1}(2y'h'+x^2yh)dx ##.

Consider the functional ## S[y]=y(0)+\frac{1}{2}\int_{0}^{1}(2y'^2+x^2y^2)dx, y(1)=2 ##.

Let ## F(x, y, y')=2y'^2+x^2y^2 ##.

Then ## \frac{\partial F}{\partial y'}=4y' ## and ## \frac{\partial F}{\partial y}=2x^2y ##.

Observe that ## \frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0\implies 4\frac{d^2y}{dx^2}-2x^2y=0 ##.

Therefore, the Euler-Lagrange equation is ## 4\frac{d^2y}{dx^2}-2x^2y=0, y(1)=2 ##.

Note that ## \frac{d}{d\epsilon}S[y+\epsilon h]=\int_{1}^{2}[6(y'+\epsilon h')h'-4(y+\epsilon h)h]dx=2\int_{1}^{2}[3(y'+\epsilon h')h'-2(y+\epsilon h)]dx ##.

Then ## \bigtriangleup S[y, h]=\frac{d}{d\epsilon}S[y+\epsilon h]\rvert_{\epsilon=0}=2\int_{1}^{2}(3y'h'-2yh)dx ##.

Thus, the Gateaux differential is ## \bigtriangleup S[y, h]=2\int_{1}^{2}(3y'h'-2yh)dx ##.

Consider the functional ## S[y]=\int_{1}^{2}(3y'^2-2y^2)dx, y(1)=1, y(2)=3 ##.

Let ## F(x, y, y')=3y'^2-2y^2 ##.

Then ## \frac{\partial F}{\partial y'}=6y' ## and ## \frac{\partial F}{\partial y}=-4y ##.

Observe that ## \frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0\implies 6\frac{d^2y}{dx^2}+4y=0 ##.

Therefore, the Euler-Lagrange equation is ## 6\frac{d^2y}{dx^2}+4y=0, y(1)=1, y(2)=3 ##.

b) We have ## S[y+\epsilon h]=y(0)+\frac{1}{2}\int_{0}^{1}(2(y'+\epsilon h')^2+x^2(y+\epsilon h)^2)dx ##.

Note that ## \frac{d}{d\epsilon}S[y+\epsilon h]=y(0)+\frac{1}{2}\int_{0}^{1}[4(y'+\epsilon h')h'+2x^2(y+\epsilon h)h]dx=y(0)+\int_{0}^{1}[2(y'+\epsilon h')h'+x^2(y+\epsilon h)h]dx ##.

Thus, the Gateaux differential is ## \bigtriangleup S[y, h]=\int_{0}^{1}(2y'h'+x^2yh)dx ##.

Consider the functional ## S[y]=y(0)+\frac{1}{2}\int_{0}^{1}(2y'^2+x^2y^2)dx, y(1)=2 ##.

Let ## F(x, y, y')=2y'^2+x^2y^2 ##.

Then ## \frac{\partial F}{\partial y'}=4y' ## and ## \frac{\partial F}{\partial y}=2x^2y ##.

Observe that ## \frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0\implies 4\frac{d^2y}{dx^2}-2x^2y=0 ##.

Therefore, the Euler-Lagrange equation is ## 4\frac{d^2y}{dx^2}-2x^2y=0, y(1)=2 ##.