- #1

Math100

- 756

- 204

- Homework Statement
- Let ## a, b ## and ## w ## be constants, with ## w\neq 0 ##.

a) Show that the Euler-Lagrange equation for the functional ## S[y]=\int_{1}^{2}(y'^2+w^2y^2+2y(a \sin(wx)+b \sinh(wx)))dx, y(1)=0, y(2)=1 ##, is ## y''-w^2y=a \sin(wx)+b \sinh(wx), y(1)=0, y(2)=1 ##.

b) Solve this equation to show that the stationary path is ## y=\alpha\sinh(w(x-1))+\beta\sinh(w(x-2))-\frac{a \sin(wx)}{2w^2}+\frac{bx \cosh(wx)}{2w} ##, where ## \alpha=\frac{1}{\sinh(w)}[1+\frac{a \sin(2w)}{2w^2}-\frac{b \cosh(2w)}{2}], \beta=\frac{1}{\sinh(w)}[\frac{b\cosh(w)}{2w}-\frac{a \sin(w)}{2w^2}] ##.

c) Determine whether ## S[y] ## has a maximum or a minimum on this stationary path, and whether it is global or local.

- Relevant Equations
- Euler-Lagrange equation:

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) The Euler-Lagrange equation is of the form ## \frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0, y(a)=A, y(b)=B ##.

Let ## F(x, y, y')=(y'^2+w^2y^2+2y(a \sin(wx)+b \sinh(wx))) ##.

Then ## \frac{\partial F}{\partial y'}=2y' ## and ## \frac{\partial F}{\partial y}=2w^2y+2(a \sin(wx)+b \sinh(wx)) ##.

Note that ## \frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0\implies 2\frac{d^2y}{dx^2}-2w^2y-2(a \sin(wx)+b \sinh(wx))=0 ##, so we have ## \frac{d^2y}{dx^2}-w^2y=a \sin(wx)+b \sinh(wx) ##.

Therefore, the Euler-Lagrange equation is ## y''-w^2y=a \sin(wx)+b \sinh(wx), y(1)=0, y(2)=1 ##.

b) Consider the differential equation ## y''-w^2y=a \sin(wx)+b \sinh(wx), y(1)=0, y(2)=1 ##.

The characteristic equation is ## r^2-w^2=0\implies y_{h}=c_{1}e^{x}+c_{2}e^{-x} ##.

Let the particular solution be of the form ## y_{p}=A \sin(wx)+Bx \cosh(wx) ##.

Then ## y_{p}'=Aw \cos(wx)+Bwx \sinh(wx)+B \cosh(wx) ## and ## y_{p}''=-Aw^2 \sin(wx)+Bw^2x \cosh(wx)+Bw \sinh(wx)+Bw \sinh(wx) ##.

Substituting ## y_{p}'', y_{p} ## into our original differential equation produces:

## -Aw^2 \sin(wx)+Bw^2x \cosh(wx)+Bw \sinh(wx)+Bw \sinh(wx)-w^2[A \sin(wx)+Bx \cosh(wx)]=a \sin(wx)+b \sinh(wx) ##.

Now we have ## -2Aw^2 \sin(wx)=a \sin(wx) ## and ## 2Bw \sinh(wx)=b \sinh(wx) ##, so ## A=\frac{a}{-2w^2}, B=\frac{b}{2w} ##.

Thus ## y=c_{1}e^{x}+c_{2}e^{-x}-\frac{a \sin(wx)}{2w^2}+\frac{bx \cosh(wx)}{2w} ##.

By our boundary conditions, we have ## y(1)=0\implies 0=c_{1}e+c_{2}e\implies c_{1}=-c_{2} ## and ## y(2)=1\implies 1=c_{1}e^{2}+c_{2}e^{-2}-\frac{a \sin(2w)}{2w^2}+\frac{2b \cosh(2w)}{2w} ##.

Observe that ## c_{2}(-e^{2}+e^{-2})-\frac{a \sin(2w)}{2w^2}+\frac{b(e^{2w}+e^{-2w})}{2w}=1 ##.

From here, I cannot see why does the stationary path is ## y=\alpha\sinh(w(x-1))+\beta\sinh(w(x-2))-\frac{a \sin(wx)}{2w^2}+\frac{bx\cosh(wx)}{2w} ##, even with the given forms of both ## \alpha ## and ## \beta ##. I know that the hyperbolic functions of ## \sinh(x), \cosh(x) ## are ## \sinh(x)=\frac{e^{x}-e^{-x}}{2} ## and ## \cosh(x)=\frac{e^{x}+e^{-x}}{2} ## but I still don't see how this gives our desired answer.

c) Does anyone know how to find the second variation? If the second variation is positive, then the given functional ## S[y] ## has a minimum on the stationary path and if the second variation is negative, then the given functional ## S[y] ## has a maximum on the stationary path.

Let ## F(x, y, y')=(y'^2+w^2y^2+2y(a \sin(wx)+b \sinh(wx))) ##.

Then ## \frac{\partial F}{\partial y'}=2y' ## and ## \frac{\partial F}{\partial y}=2w^2y+2(a \sin(wx)+b \sinh(wx)) ##.

Note that ## \frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0\implies 2\frac{d^2y}{dx^2}-2w^2y-2(a \sin(wx)+b \sinh(wx))=0 ##, so we have ## \frac{d^2y}{dx^2}-w^2y=a \sin(wx)+b \sinh(wx) ##.

Therefore, the Euler-Lagrange equation is ## y''-w^2y=a \sin(wx)+b \sinh(wx), y(1)=0, y(2)=1 ##.

b) Consider the differential equation ## y''-w^2y=a \sin(wx)+b \sinh(wx), y(1)=0, y(2)=1 ##.

The characteristic equation is ## r^2-w^2=0\implies y_{h}=c_{1}e^{x}+c_{2}e^{-x} ##.

Let the particular solution be of the form ## y_{p}=A \sin(wx)+Bx \cosh(wx) ##.

Then ## y_{p}'=Aw \cos(wx)+Bwx \sinh(wx)+B \cosh(wx) ## and ## y_{p}''=-Aw^2 \sin(wx)+Bw^2x \cosh(wx)+Bw \sinh(wx)+Bw \sinh(wx) ##.

Substituting ## y_{p}'', y_{p} ## into our original differential equation produces:

## -Aw^2 \sin(wx)+Bw^2x \cosh(wx)+Bw \sinh(wx)+Bw \sinh(wx)-w^2[A \sin(wx)+Bx \cosh(wx)]=a \sin(wx)+b \sinh(wx) ##.

Now we have ## -2Aw^2 \sin(wx)=a \sin(wx) ## and ## 2Bw \sinh(wx)=b \sinh(wx) ##, so ## A=\frac{a}{-2w^2}, B=\frac{b}{2w} ##.

Thus ## y=c_{1}e^{x}+c_{2}e^{-x}-\frac{a \sin(wx)}{2w^2}+\frac{bx \cosh(wx)}{2w} ##.

By our boundary conditions, we have ## y(1)=0\implies 0=c_{1}e+c_{2}e\implies c_{1}=-c_{2} ## and ## y(2)=1\implies 1=c_{1}e^{2}+c_{2}e^{-2}-\frac{a \sin(2w)}{2w^2}+\frac{2b \cosh(2w)}{2w} ##.

Observe that ## c_{2}(-e^{2}+e^{-2})-\frac{a \sin(2w)}{2w^2}+\frac{b(e^{2w}+e^{-2w})}{2w}=1 ##.

From here, I cannot see why does the stationary path is ## y=\alpha\sinh(w(x-1))+\beta\sinh(w(x-2))-\frac{a \sin(wx)}{2w^2}+\frac{bx\cosh(wx)}{2w} ##, even with the given forms of both ## \alpha ## and ## \beta ##. I know that the hyperbolic functions of ## \sinh(x), \cosh(x) ## are ## \sinh(x)=\frac{e^{x}-e^{-x}}{2} ## and ## \cosh(x)=\frac{e^{x}+e^{-x}}{2} ## but I still don't see how this gives our desired answer.

c) Does anyone know how to find the second variation? If the second variation is positive, then the given functional ## S[y] ## has a minimum on the stationary path and if the second variation is negative, then the given functional ## S[y] ## has a maximum on the stationary path.

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