User:Egm6322.s09.bit.gk/r4



We can see from the free body diagram ,equating the horizontal forces ,we have

$$ -T_0cos(\alpha)dy+T_0cos(\alpha+d\alpha)dy-T_0cos(\beta)dx+T_0cos(\beta+d\beta)dx=0 $$

and equating the vertical forces, we have,

$$ -T_0sin(\alpha)dy+T_0sin(\alpha+d\alpha)dy-T_0sin(\beta)dx+T_0sin(\beta+d\beta)dx=P(x,y)dxdy $$

where $$P(x,y)$$ is the transverse load acting on the membrane

$$\Rightarrow T_0\left[\left (sin(\alpha+d\alpha) -sin(\alpha)\right )dy+\left (sin(\beta+d\beta)-sin(\beta)\right )dx\right]+P(x,y)dxdy=0$$

When $$\theta$$ is small ,we can assume, $$ sin(\theta)=Tan(\theta)$$

$$\therefore$$ the above equation becomes,

$$T_0\left[\left (tan(\alpha+d\alpha)-tan(\alpha)\right )dy+\left (tan(\beta+d\beta)-tan(\beta)\right )dx\right]+P(x,y)dxdy=0 $$

Let this equation be $$(1)$$

but,$$tan(\alpha)$$ is the slope the membrane with respect to the $$x$$ and $$y$$ axes

where the displacement is given as $$w=w(x,y)$$

$$\therefore$$

$$tan(\alpha)=\left[\frac{\partial w}{\partial x}\right]_{x,y}$$ and $$ tan(\alpha+d\alpha)=\left[\frac{\partial w}{\partial x}\right]_{x+dx,y}$$

$$tan(\beta)=\left[\frac{\partial w}{\partial y}\right]_{x,y}$$ and $$ tan(\beta+d\beta)=\left[\frac{\partial w}{\partial y}\right]_{x,y+dy}$$

Substituting the slopes in $$(1)$$,we have ,

$$T_0\left[\left(\left[\frac{\partial w}{\partial x}\right]_{x+dx,y}-\left[\frac{\partial w}{\partial x}\right]_{x,y}\right)dy+\left(\left[\frac{\partial w}{\partial y}\right]_{x,y+dy}-\left[\frac{\partial w}{\partial y}\right]_{x,y}\right)dx\right]+P(x,y)dxdy=0 $$

Let this be equation $$(2)$$

We have Taylor series expansion as ,

$$f(x+dx,y)=f(x,y)+\frac{\partial f}{\partial x}dx+...$$

substituting taylor series expansion in equation $$(2)$$,we have

$$T_0\left(\frac{\partial^2 w}{\partial x^2}+\frac{\partial^2 w}{\partial y^2}\right)+P(x,y)=0$$

$$ \Rightarrow T_0\triangledown^{2}w(x,y)+P(x,y)=0 $$

hw2
To show that $$\kappa (u)grad(u)+f(x,y)=0$$ is non linear

Let,

$$L\left( \right):$$ be an operator,such that $$L\left(\ u\right)$$ is linear with respect to $$u$$ if,

$$ L\left(\alpha u+\beta v\right)=\alpha L\left(\ u\right)+\beta L\left(\ v\right)$$

Therefore, in the present problem ,assuming 2D, we have,

$$L\left( \right)=\kappa \left(\overline i\frac{\partial\left( \right) }{\partial x}+\overline j\frac{\partial\left( \right) }{\partial y}\right)+f(x,y)$$

$$L\left(\alpha u+\beta v \right)=\kappa (\alpha u+\beta v)\left(\overline i\frac{\partial\left(\alpha u+\beta v \right) }{\partial x}+\overline j\frac{\partial\left(\alpha u+\beta v \right) }{\partial y}\right)+f(x,y)$$

$$\Rightarrow L\left( \alpha u+\beta v \right)=\kappa (\alpha u+\beta v)\left(\overline i\alpha\frac{\partial\left(\ u\right) }{\partial x}+\overline i\beta\frac{\partial\left(\ v\right) }{\partial x}+\overline j\alpha\frac{\partial\left(\ u\right) }{\partial y}+\overline j\beta\frac{\partial\left(\ v\right) }{\partial y}\right)+f(x,y)$$

$$\Rightarrow L\left(\alpha u+\beta v \right)=\kappa (\alpha u+\beta v)\left[\alpha\left\{\overline i\frac{\partial\left(\ u \right) }{\partial x}+\overline j\frac{\partial\left(\ u \right) }{\partial y} \right\}+\beta\left\{\overline i\frac{\partial\left(\ v\right) }{\partial x}+\overline j\frac{\partial\left(\ v\right) }{\partial y} \right\}\right]+f(x,y)$$

\therefore,we can see that

$$\Rightarrow L\left(\alpha u+\beta v\right)\neq\alpha L\left(\ u\right)+\beta L\left(\ v\right)$$

$$ \therefore \kappa (u)grad (u)+f(x,y)$$ is non linear