User:Egm6322.s09.bit.sahin/r2

Homework 1
$$div \left [\underline{\kappa}\left (x,y \right )\cdot grad u \right ]+ax_{2}+by+\sqrt{u}+\left (u_{x} \right )^{4}+2\left (u_{y} \right )^{2}=0$$

Linear operator for second order derivatives,

$$\mathcal{L}_{1}\left (\cdot \right )=div \left [\underline{\kappa}\left (x,y \right )\cdot grad\left (\cdot  \right )\right ]$$

To show the linearity,

$$\mathcal{L}_{1}\left (\alpha u+\beta v \right )=div \left [\underline{\kappa}\cdot grad\left (\alpha u+\beta v \right )  \right ]$$

$$grad\left (\cdot \right )$$  and matrix multiplication are linear operators,

$$\underline{\kappa }\cdot grad\left (\alpha u+\beta v \right )=\alpha \underline{\kappa }\cdot gradu+\beta \underline{\kappa }\cdot grad v$$

Since $$div\left (\cdot \right )$$ is also a linear operator we can write that

$$\mathcal{L}_{1}\left (\alpha u+\beta v \right )=\alpha div\left (\underline{\kappa }\cdot gradu \right )+\beta div\left (\underline{\kappa }\cdot gradv \right )$$

and

$$\alpha \mathcal{L}_{1}\left (u \right )+\beta \mathcal{L}_{1}\left (v \right )=\alpha div\left (\underline{\kappa }\cdot gradu \right )+\beta div\left (\underline{\kappa }\cdot gradv \right )$$

$$\mathcal{L}_{1}\left (\alpha u+\beta v \right )=\alpha \mathcal{L}_{1}\left (u \right )+\beta \mathcal{L}_{1}\left (v \right )$$

Therefore PDE is linear with respect to second order derivatives.

For the whole equation, linear operator is,

$$\mathcal{L}_{2}\left (\cdot \right )=\mathcal{L}_{1}\left (\cdot  \right )+\sqrt{\left (\cdot  \right )}+\left [\left (\cdot  \right )_{x} \right ]^{4}+2\left [\left (\cdot  \right )_{y} \right ]^{2}+ax^{2}+by$$

$$\mathcal{L}_{2}\left (\alpha u+\beta v \right )=\mathcal{L}_{1}\left (\alpha u+\beta v   \right )+\sqrt{\left (\alpha u+\beta v   \right )}+\left [\left (\alpha u+\beta v   \right )_{x} \right ]^{4}+2\left [\left (\alpha u+\beta v   \right )_{y} \right ]^{2}+ax^{2}+by$$

$$\mathcal{L}_{2}\left (\alpha u+\beta v \right )=\alpha \mathcal{L}_{1}\left (u \right )+\beta \mathcal{L}_{1}\left (v \right )+\sqrt{\left (\alpha u+\beta v   \right )}+\left [\left (\alpha u+\beta v   \right )_{x} \right ]^{4}+2\left [\left (\alpha u+\beta v   \right )_{y} \right ]^{2}+ax^{2}+by$$

On the other hand,

$$\alpha \mathcal{L}_{2}\left (u \right )+\beta \mathcal{L}_{2}\left (v \right )=\alpha \mathcal{L}_{1}\left (u \right )+\alpha u^{1/2}+\alpha \left (u_{x} \right )^{4}+2\alpha \left (u_{y} \right )^{2}+\alpha \left (ax^{2}+by \right )+\beta \mathcal{L}_{1}\left (v \right )+\beta v^{1/2}+\beta \left (v_{x} \right )^{4}+2\beta \left (v_{y} \right )^{2}+\beta \left (ax^{2}+by \right )$$

$$\alpha \mathcal{L}_{2}\left (u \right )+\beta \mathcal{L}_{2}\left (v \right )=\alpha \mathcal{L}_{1}\left (u \right )+\beta \mathcal{L}_{1}\left (v \right )+\alpha u^{1/2}+\beta v^{1/2}+\alpha \left (u_{x} \right )^{4}+\beta \left (v_{x} \right )^{4}+2\alpha \left (u_{y} \right )^{2}+2\beta \left (v_{y} \right )^{2}+\alpha ax^{2}+\alpha by+\beta ax^{2}+\beta by$$

Eventually,

$$\mathcal{L}_{2}\left (\alpha u+\beta v \right )\neq \alpha \mathcal{L}_{2}\left (u \right )+\beta \mathcal{L}_{2}\left (v \right )$$

PDE is non-linear as a whole.

Homework 2
$$ \mathcal{L}\left (\cdot \right )=a\left (\cdot  \right )_{xx}+2b\left (\cdot  \right )_{xy}+c\left (\cdot  \right )_{yy}+d\left (\cdot  \right )_{x}+e\left (\cdot  \right )_{y}+f\left (\cdot  \right )$$

If it is a linear PDE, it has to satisfy that

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

$$\alpha \mathcal{L}\left (u \right )=\alpha au_{xx}+2\alpha bu_{xy}+\alpha cu_{yy}+\alpha du_{x}+\alpha eu_{y}+\alpha fu$$

$$\beta \mathcal{L}\left (v \right )=\beta av_{xx}+2\beta bv_{xy}+\beta cv_{yy}+\beta dv_{x}+\beta ev_{y}+\alpha fv$$

$$\alpha \mathcal{L}\left (u \right )+\beta \mathcal{L}\left (v \right )=\alpha au_{xx}+\beta av_{xx}+2\alpha bu_{xy}+2\beta bv_{xy}+\alpha cu_{yy}+\beta cv_{yy}+\alpha du_{x}+\beta dv_{x}+\alpha eu_{y}+\beta ev_{y}+fu+fv$$

and

$$\alpha \mathcal{L}\left (u \right )+\beta \mathcal{L}\left (v \right )=a\left (\alpha u+\beta v \right )_{xx}+2b\left (\alpha u+\beta v \right )_{xy}+c\left (\alpha u+\beta v \right )_{yy}+d\left (\alpha u+\beta v \right )_{x}+e\left (\alpha u+\beta v \right )_{y}+f\left (\alpha u+\beta v \right )$$

$$\alpha \mathcal{L}\left (u \right )+\beta \mathcal{L}\left (v \right )=a\left (\alpha u_{xx}+\beta v_{xx} \right )+2b\left (\alpha u_{xy}+\beta v_{xy} \right )+c\left (\alpha u_{yy}+\beta v_{yy} \right )+d\left (\alpha u_{x}+\beta v_{x} \right )+e\left (\alpha u_{y}+\beta v_{y} \right )+f\left (\alpha u+\beta v \right )$$

$$\alpha \mathcal{L}\left (u \right )+\beta \mathcal{L}\left (v \right )=a\alpha u_{xx}+a\beta v_{xx} +2b\alpha u_{xy}+2b\beta v_{xy}+c\alpha u_{yy}+c\beta v_{yy}+d\alpha u_{x}+d\beta v_{x}+e\alpha u_{y}+e\beta v_{y}+f\alpha u+f\beta v$$

Clearly it satisfies the linearity condition. Since the highest derivative is second, PDE is a second order and linear.

Homework 3
 Equality of mixed partials  Theorem: If $$f\left (x,y \right )$$ is of class $$ C^{2}$$ (is twice continuously differentiable), then the mixed partial derivatives are equal; that is

$$\frac{\partial ^{2}f}{\partial x\partial y}=\frac{\partial ^{2}f}{\partial y\partial x}$$

Proof: Consider the following expression (Figure 1)

$$S\left (\bigtriangleup x,\bigtriangleup y \right )=f\left (x_{0}+\bigtriangleup x,y_{0}+\bigtriangleup y \right )-f\left (x_{0}+\bigtriangleup x,y_{0} \right )-f\left (x_{0},y_{0}+\bigtriangleup y \right )+f\left (x_{0},y_{0} \right )$$



Holding $$y_{0}$$ and $$\bigtriangleup y$$  fixed, define

$$g\left (x \right )=f\left (x,y_{0}+\bigtriangleup y \right )-f\left (x,y_{0} \right )$$

so that $$S\left (\bigtriangleup x,\bigtriangleup y \right )=g\left (x_{0}+\bigtriangleup x \right )-g\left (x_{0} \right )$$, which expresses S as a difference of differences. By the mean-value theorem for functions of one variable,$$g\left (x_{0}+\bigtriangleup x \right )-g\left (x_{0} \right )$$ equals $$g'\left (\overline{x }\right )\bigtriangleup x $$ for some $$\overline{x }$$ between $$x_{0}$$ and $$x_{0}+\bigtriangleup x$$. Hence,

$$S\left (\bigtriangleup _{x},\bigtriangleup _{y} \right )=\left [\frac{d f}{d x}\left (\overline{x},y_{0}+\bigtriangleup y \right )-\frac{d f}{d x}\left (\overline{x},y_{0} \right ) \right ]\bigtriangleup x$$

Applying the mean-value theorem again, there is a $$\overline{y }$$ between $$y_{0}$$ and $$y_{0}+\bigtriangleup y$$ such that

$$S\left (\bigtriangleup _{x},\bigtriangleup _{y} \right )=\frac{\partial ^{2}f}{\partial y\partial x}\left (\overline{x},\overline{y} \right )\Delta x\Delta y$$

Because $$\frac{\partial ^{2}f}{\partial y\partial x}$$ is continuous, it follows that

$$\frac{\partial ^{2}f}{\partial y\partial x}\left (x_{0},y_{0} \right )=\lim_{\left (\Delta x,\Delta y \right )\to \left (0,0 \right )}\frac{1}{\Delta x\Delta y}\left [S\left (\Delta x,\Delta y \right ) \right ].$$

Noting that S is symmetric in $$\Delta x$$ and $$\Delta y$$, one shows in a similar way that $$\frac{\partial ^{2}f}{\partial x\partial y}$$ is given by the same limit formula, which proves the result.

Homework 4
$$\left (au_{x} \right )_{x}+\left (bu_{x} \right )_{y}+\left (bu_{y} \right )_{x}+\left (cu_{y} \right )_{y}+du_{x}+eu_{y}+fu+g=0$$

Here a,b and c are the functions of (x,y). This equation can be rewritten as

$$\left (a_{x}u_{x} +au_{xx}\right )+\left (b_{y}u_{x}+u_{xy}b \right )+\left (b_{x}u_{y}+bu_{xy} \right )+\left (c_{y}u_{y}+cu_{yy} \right )+eu_{y}+fu+g=0$$

$$au_{xx}+cu_{yy}+2bu_{xy}+\left (a_{x}u_{x}+b_{y}u_{x}+du_{x} \right )+\left (b_{x}u_{y}+c_{y}u_{y}+eu_{y} \right )+fu+g=0$$

In order to show that it is a linear PDE, we define the linearity operator for the second order derivatives as

$$\mathcal{L}_{1}\left (\cdot \right )=a\left (\cdot  \right )_{xx}+c\left (\cdot  \right )_{yy}+2b\left (\cdot  \right )_{xy}$$

$$\alpha \mathcal{L}_{1}\left (u \right )=\alpha a\left (u  \right )_{xx}+\alpha c\left (u  \right )_{yy}+2\alpha b\left (u  \right )_{xy}$$

$$\beta \mathcal{L}_{1}\left (v \right )=\beta a\left (v  \right )_{xx}+\beta c\left (v  \right )_{yy}+2\beta b\left (v  \right )_{xy}$$

$$\alpha \mathcal{L}_{1}\left (u \right )+\beta \mathcal{L}_{1}\left (v  \right )=\alpha a\left (u  \right )_{xx}+\beta a\left (v  \right )_{xx}+\alpha c\left (u  \right )_{yy}+\beta c\left (v  \right )_{yy}+2\alpha b\left (u  \right )_{xy}+2\beta b\left (v  \right )_{xy}$$

and

$$ \mathcal{L}_{1}\left (\alpha u+\beta v \right )= a\left (\alpha u+\beta v  \right )_{xx}+c\left (\alpha u+\beta v  \right )_{yy}+2 b\left (\alpha u+\beta v  \right )_{xy}$$

$$ \mathcal{L}_{1}\left (\alpha u+\beta v \right )=\alpha a\left (u  \right )_{xx}+\beta a\left (v  \right )_{xx}+\alpha c\left (u  \right )_{yy}+\beta c\left (v  \right )_{yy}+2\alpha b\left (u  \right )_{xy}+2\beta b\left (v  \right )_{xy}$$

Because $$ \mathcal{L}_{1}\left (\alpha u+\beta v \right )=\alpha \mathcal{L}_{1}\left (u  \right )+\beta \mathcal{L}_{1}\left (v  \right )$$  this PDE is linear wuth respect to second order derivatives.

For the first order derivatives,

$$\mathcal{L}_{2}\left (\cdot \right )=\mathcal{L}_{1}\left (\cdot  \right )+a_{x}\left (\cdot  \right )_{x}+b_{y}\left (\cdot  \right )_{x}+d\left (\cdot  \right )_{x}+b_{x}\left (\cdot  \right )_{y}+c_{y}\left (\cdot  \right )_{y}+e\left (\cdot  \right )_{y}$$

$$\alpha \mathcal{L}_{2}\left (u \right )=\alpha \mathcal{L}_{1}\left (u  \right )+\alpha a_{x} u_{x}+\alpha b_{y}u_{x}+\alpha du_{x}+\alpha b_{x}u_{y}+\alpha c_{y}u_{y}+\alpha eu_{y}$$

$$\beta \mathcal{L}_{2}\left (v \right )=\beta \mathcal{L}_{1}\left (v  \right )+\beta a_{x} v_{x}+\beta b_{y}v_{x}+\beta dv_{x}+\beta b_{x}v_{y}+\beta c_{y}v_{y}+\beta ev_{y}$$

$$\alpha \mathcal{L}_{2}\left (u \right )+\beta \mathcal{L}_{2}\left (v  \right )=\alpha \mathcal{L}_{1}\left (u  \right )+\beta \mathcal{L}_{1}\left (v  \right )+\alpha a_{x} u_{x}+\beta a_{x} v_{x}+\alpha b_{y}u_{x}+\beta b_{y}v_{x}+\alpha du_{x}+\beta dv_{x}+\alpha b_{x}u_{y}+\beta b_{x}v_{y}+\alpha c_{y}u_{y}+\beta c_{y}v_{y}+\alpha eu_{y}+\beta ev_{y}$$

On the other hand,

$$\mathcal{L}_{2}\left (\alpha u+\beta v \right )=\mathcal{L}_{1}\left (\alpha u+\beta v  \right )+a_{x}\left (\alpha u+\beta v \right )_{x}+b_{y}\left (\alpha u+\beta v  \right )_{x}+d\left (\alpha u+\beta v \right )_{x}+b_{x}\left (\alpha u+\beta v  \right )_{y}+c_{y}\left (\alpha u+\beta v  \right )_{y}+e\left (\alpha u+\beta v  \right )_{y}$$

$$\mathcal{L}_{2}\left (\alpha u+\beta v \right )=\alpha \mathcal{L}_{1}\left (u  \right )+\beta \mathcal{L}_{1}\left (v  \right )+\alpha a_{x} u_{x}+\beta a_{x} v_{x}+\alpha b_{y}u_{x}+\beta b_{y}v_{x}+\alpha du_{x}+\beta dv_{x}+\alpha b_{x}u_{y}+\beta b_{x}v_{y}+\alpha c_{y}u_{y}+\beta c_{y}v_{y}+\alpha eu_{y}+\beta ev_{y}$$

It is obvious that $$\mathcal{L}_{2}\left (\alpha u+\beta v \right )=\alpha \mathcal{L}_{2}\left (u  \right )+\beta \mathcal{L}_{2}\left (v  \right )$$ so, it is linear with respect to second order derivatives. As a result, even if a,b and c are functions of (x,y), PDE is still second order and linear with respect to all derivatives.

Homework 5
Second order, linear PDE:

$$au_{xx}+2bu_{xy}+cu_{yy}+du_{x}+eu_{y}+fu+g=0$$

here coefficients {a,b,....,g} are functions of (x,y). We can write this general form in matrix form as

$$\begin{bmatrix} \partial _{x} & \partial _{y} \end{bmatrix}\begin{bmatrix} a & b\\ b & c \end{bmatrix}\begin{bmatrix} \partial _{x}u\\\partial _{y}u

\end{bmatrix}+\begin{bmatrix} \left (d-a_{x}-b_{y} \right )&\left (e-b_{x}-c_{y} \right ) \end{bmatrix}\begin{bmatrix} \partial _{x}u\\\partial _{y}u

\end{bmatrix}+fu+g=0 $$

Homework 6
If system is rotated through a clockwise angle θ relative to the xy-coordinate system, this relationship may be written:

$$x=\overline{x}cos\theta +\overline{y}sin\theta$$

$$y=-\overline{x}sin\theta +\overline{y}cos\theta $$

In the matrix form,

$$\begin{bmatrix} x\\y

\end{bmatrix}=\mathbf{V}\begin{bmatrix} \overline{x} \\ \overline{y} \end{bmatrix}$$

where $$\mathbf{V}$$ is rotation matrix:

$$\mathbf{V}=\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}$$

Homework 7
Partial operator: $$\frac{\partial }{\partial x_{i}}\left (\cdot \right )$$

$$\frac{\partial }{\partial x_{i}}\left (\alpha u+\beta v \right )=\frac{\partial }{\partial x_{i}}\left (\alpha u \right )+\frac{\partial }{\partial x_{i}}\left (\beta v \right )=\alpha \frac{\partial u}{\partial  x_{i}}+u \frac{\partial \alpha}{\partial  x_{i}}+\beta \frac{\partial v}{\partial  x_{i}}+v \frac{\partial \beta}{\partial  x_{i}}$$

Since $$\alpha$$ and $$\beta$$ are real numbers

$$\frac{\partial \alpha }{\partial x_{i}}=0, \frac{\partial \beta}{\partial  x_{i}}=0$$

$$\frac{\partial }{\partial x_{i}}\left (\alpha u+\beta v \right )=\alpha \frac{\partial u}{\partial x_{i}}+\beta \frac{\partial v}{\partial x_{i}}$$

Therefore partial operator is a linear operator.

Homework 8
Proof of equivalence of definitions of linearity

Definition of linear operator in class,

$$\mathcal{L}\left (\alpha u+\beta v \right )=\alpha \mathcal{L}\left ( u\right )+\beta \mathcal{L}\left ( v\right )$$ ....(1)

Definition in Kolmogorov & Fomin, 1975 p.123,

1- Additivity: $$\forall u,v \epsilon \mathbb{R}:\Omega \to \mathbb{R}$$

$$ \mathcal{L}\left ( u+ v \right )= \mathcal{L}\left ( u\right )+\mathcal{L}\left ( v\right )$$ ...(2)

2- Homogeneity: $$\forall \alpha \epsilon \mathbb{R},\forall u :\Omega \to \mathbb{R}$$

$$\mathcal{L}\left ( \alpha u \right )=\alpha  \mathcal{L}\left ( u\right )$$ ...(3)

Part 2) Necessary condition ($$\Leftarrow $$): [Eq. 1]$$\Leftarrow $$[Eq. 2 & Eq. 3]

Lets define $$ \alpha $$ and $$\beta$$ as arbitrary numbers.Using the additivity property we can write that,

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

Using the homogeneity property,

$$\mathcal{L}\left ( \alpha u\right )=\alpha \mathcal{L}\left ( u\right )$$

$$\mathcal{L}\left ( \beta v\right )=\beta \mathcal{L}\left ( v\right )$$

Hence,

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

Eventually,

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

It proves that definition of linear operators as shown in class is equivalent to the definition in Kolmogorov & Fomin, 1975 p.123.

Homework 9
$$u_{y}\left (x,y \right )=\frac{\partial }{\partial y}\overline{u}\left (\overline{x},\overline{y} \right )=\overline{u}_{\overline{x}}\frac{\partial \overline{x}}{\partial y}+\overline{u}_{\overline{y}}\frac{\partial \overline{y}}{\partial y}$$

In matrix form

$$\frac{\partial u}{\partial y}=\begin{bmatrix} \frac{\partial \overline{x}}{\partial y} & \frac{\partial \overline{y}}{\partial y} \end{bmatrix}\begin{bmatrix} \partial _{\overline{x}}\\ \partial \overline{y}

\end{bmatrix}\left (\overline{u} \right )$$

References: