User:Egm6322.s09.bit.gk/r2

Quasilinear Partial differential equation
For a PDE of order n,i.e,highest order derivative are of order n ,coefficients of the nth

order derivative are functions of $$(x,y,u,u_x,u_y,...,\frac{\partial^m u }{\partial x^p\partial x^q},...)$$

Where,

$$(x,y,u,u_x,u_y,...,\frac{\partial^m u }{\partial x^p\partial x^q},...)$$

is the mth order derivative

and

$$p+q=m$$

The general solution of a Quasilinear partial differential equation is given by

$$A(x,y,u)\frac{\partial u }{\partial x}+B(x,y,u)\frac{\partial u }{\partial y}=C(x,y,u) $$

The quasilinear PDE is non linear PDE but linear in $$u_{x}$$ and $$ u_{y}$$

For Example a PDE linear wrt 2nd order derivative but still non linear in general can be given as

$$Div\left(\kappa.grad u\right)+f(x,y,u_x,u_y)=0$$

where $$\kappa=\kappa \left(x,y\right)$$

For example

$$Div\left(\kappa.grad u\right)+ax^2+by+\sqrt u_x+(u_x)^4+2(u_y)^2=0$$

Proof of linearity of a particular differential operator
We Have,

$$Div V :=\kappa.Grad u$$

Therefore exapnding it,assuming 2-D

$$Div V=\frac{\partial }{\partial x}\left(\kappa_{11}\frac{\partial u}{\partial x}\right)+\frac{\partial }{\partial x}\left(\kappa_{12}\frac{\partial u}{\partial y}\right)+\frac{\partial }{\partial y}\left(\kappa_{21}\frac{\partial u}{\partial x}\right)+\frac{\partial }{\partial y}\left(\kappa_{22}\frac{\partial u}{\partial y}\right)$$

To prove that the above Divergence is linear ,we have to show,

Let,

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

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

Therefore, in the present problem, we have,

$$D=\frac{\partial }{\partial x}\left(\kappa_{11}\frac{\partial }{\partial x}\right)+\frac{\partial }{\partial x}\left(\kappa_{12}\frac{\partial }{\partial y}\right)+\frac{\partial }{\partial y}\left(\kappa_{21}\frac{\partial }{\partial x}\right)+\frac{\partial }{\partial y}\left(\kappa_{22}\frac{\partial}{\partial y}\right)$$

Let $$\kappa=\kappa(x,y)$$

$$\Rightarrow D= \left(\kappa_{11x}\frac{\partial }{\partial x}+\kappa_{11}\frac{\partial^2  }{\partial x^2}\right)

+\left(\kappa_{12x}\frac{\partial }{\partial y}+\kappa_{12}\frac{\partial^2  }{\partial x\partial y}\right)$$

$$+\left(\kappa_{21y}\frac{\partial }{\partial x}+\kappa_{21}\frac{\partial^2  }{\partial x\partial y}\right)

+\left(\kappa_{22y}\frac{\partial }{\partial y}+\kappa_{22}\frac{\partial^2  }{\partial y^2}\right)$$

$$\therefore$$

$$\Rightarrow D(\alpha u+\beta v)= \left(\kappa_{11x}\frac{\partial (\alpha u+\beta v) }{\partial x}+\kappa_{11}\frac{\partial^2 (\alpha u+\beta v) }{\partial x^2}\right)

+\left(\kappa_{12x}\frac{\partial (\alpha u+\beta v) }{\partial y}+\kappa_{12}\frac{\partial^2 (\alpha u+\beta v) }{\partial x\partial y}\right)$$


 * $$+\left(\kappa_{21y}\frac{\partial (\alpha u+\beta v) }{\partial x}+\kappa_{21}\frac{\partial^2 (\alpha u+\beta v) }{\partial x\partial y}\right)

+\left(\kappa_{22y}\frac{\partial (\alpha u+\beta v) }{\partial y}+\kappa_{22}\frac{\partial^2 (\alpha u+\beta v) }{\partial y^2}\right)$$

$$\Rightarrow D(\alpha u+\beta v)= \left(\kappa_{11x}\left[\alpha\frac{\partial ( u) }{\partial x}+\beta\frac{\partial ( v)}{\partial x}\right]+\kappa_{11}\left[\alpha\frac{\partial^2 ( u)}{\partial x^2}+\beta\frac{\partial^2 (v) }{\partial x^2}\right]\right)$$

$$+ \left(\kappa_{12x}\left[\alpha\frac{\partial ( u) }{\partial y}+\beta\frac{\partial ( v)}{\partial y}\right]+\kappa_{12}\left[\alpha\frac{\partial^2 ( u)}{\partial x\partial y}+\beta\frac{\partial^2 (v) }{\partial x\partial y}\right]\right)$$

$$+ \left(\kappa_{21x}\left[\alpha\frac{\partial ( u) }{\partial x}+\beta\frac{\partial ( v)}{\partial x}\right]+\kappa_{21}\left[\alpha\frac{\partial^2 ( u)}{\partial x\partial y}+\beta\frac{\partial^2 (v) }{\partial x\partial y}\right]\right)$$

$$+ \left(\kappa_{22y}\left[\alpha\frac{\partial ( u) }{\partial y}+\beta\frac{\partial ( v)}{\partial y}\right]+\kappa_{22}\left[\alpha\frac{\partial^2 ( u)}{\partial y^2}+\beta\frac{\partial^2 (v) }{\partial y^2}\right]\right)$$

Therefore by rearranging we have,

$$\Rightarrow D(\alpha u +\beta v)= \alpha\left[\left(\kappa_{11x}\frac{\partial (u) }{\partial x}+\kappa_{11}\frac{\partial^2 (u) }{\partial x^2}\right) +\left(\kappa_{12x}\frac{\partial (u) }{\partial y}+\kappa_{12}\frac{\partial^2 (u) }{\partial x\partial y}\right)\right]$$

$$+\alpha\left[\left(\kappa_{21y}\frac{\partial (u) }{\partial x}+\kappa_{21}\frac{\partial^2 (u) }{\partial x\partial y}\right) +\left(\kappa_{22y}\frac{\partial (u) }{\partial y}+\kappa_{22}\frac{\partial^2 (u) }{\partial y^2}\right)\right] $$

$$+\beta\left[\left(\kappa_{11x}\frac{\partial (v) }{\partial x}+\kappa_{11}\frac{\partial^2 (v) }{\partial x^2}\right) +\left(\kappa_{12x}\frac{\partial (v) }{\partial y}+\kappa_{12}\frac{\partial^2 (v) }{\partial x\partial y}\right)\right]$$

$$+\beta\left[\left(\kappa_{21y}\frac{\partial (v) }{\partial x}+\kappa_{21}\frac{\partial^2 (v) }{\partial x\partial y}\right) +\left(\kappa_{22y}\frac{\partial (v) }{\partial y}+\kappa_{22}\frac{\partial^2 (v) }{\partial y^2}\right)\right] $$

$$\therefore$$

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

$$\Rightarrow$$ the operator $$ D\left(\right)$$ is Linear

Additive and Homogeneity of PDE


$$\mathfrak{L}(0)=0$$

Image of 0 under $$\mathfrak{L}(.)$$ is zero "0" if $$\mathfrak{L}(.)$$ is linear


 * Additive: The additivity of an operator is defined as follows,

$$\forall u ,v:\omega\rightarrow \R$$

In the present case if $$\alpha$$=1 and $$\beta$$=1, we can clearly check the additive property of the operator

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


 * Homogeneous:The Homogeneity of an operator is defined as follows,

$$\forall u ,v:\omega\rightarrow \R$$

$$ L\left( \alpha u\right)=\alpha L\left(\ u\right)$$

If either $$\alpha$$=0 or $$\beta$$ =0, we notice the operator is homogeneous.

The adjoining figure in essence demonstrates that a zero intercept indicates homogeneity.

Proof of Equivalency of linearity
Note: Equivalency $$A\equiv B$$

"if and only of" $$\equiv$$ "equivalent to" $$\equiv$$ "necessary and sufficient condition

$$A\Rightarrow B$$ A is a sufficient condition for B

$$A\Leftarrow B$$ B is a sufficient condition for A

A is a necessary condition for B

$$\begin{bmatrix}

A\Rightarrow B\end{bmatrix}\Leftrightarrow \begin{bmatrix}

\mathbf{B}\Rightarrow \mathbf{A}\end{bmatrix}$$

Where: B is non B or negation of B

and A is non A or negation of A

1) $$(\Rightarrow) $$ (sufficient condition)

2) $$(\Leftarrow) $$ (necessary condition)

$$\R^m$$,$$\R^n$$ are species of vectors (tensors,column matrix). Div maps, vector fields

vector fields (vector-value function) into a scalar function.

In other words, domain and range of div(.) are function spaces

General Form
The General form of a Second-order linear PDE in two independant variable is given as ,

$$A(x,y)\frac{\partial^2u }{\partial x^2}+2B(x,y)\frac{\partial^2u }{\partial x\partial y}+C(x,y)\frac{\partial^2u }{\partial y^2}+D(x,y)\frac{\partial u }{\partial x}+E(x,y)\frac{\partial u }{\partial y}+F(x,y)u+G(x,y)=0 $$

The coefficients $$A$$,$$B$$,$$C$$ are constant
With the assumption that in the general form of the Second order PDE, the coefficients $$A$$, $$B$$,$$C$$

are constants and the remaining coefficients $$D$$, $$E$$, $$F$$,

$$G$$ are still functions of $$(x,y)$$, we can write the general equation  in

the matrix form as follows

$$\left[{\begin{matrix} \frac{\partial }{\partial x} &  \frac{\partial }{\partial x}  \\ \end{matrix}}\right]\left[{\begin{matrix} A & B \\ B & C \\ \end{matrix}}\right]\left[{\begin{matrix}

\frac{\partial u }{\partial x}   \\

\frac{\partial u}{\partial y}    \\ \end{matrix}}\right] +\left[{\begin{matrix} D &  E  \\ \end{matrix}}\right]\left[{\begin{matrix}

\frac{\partial u }{\partial x}   \\

\frac{\partial u}{\partial y}    \\ \end{matrix}}\right]+F(x,y)u+G(x,y)=0 $$

Let,

$$\alpha=\left[{\begin{matrix} \frac{\partial }{\partial x} &  \frac{\partial }{\partial x}  \\ \end{matrix}}\right]\left[{\begin{matrix} A & B \\ B & C \\ \end{matrix}}\right]\left[{\begin{matrix}

\frac{\partial u }{\partial x}   \\

\frac{\partial u}{\partial y}    \\ \end{matrix}}\right]$$

and let,

$$\beta=\left[{\begin{matrix} D &  E  \\ \end{matrix}}\right]\left[{\begin{matrix}

\frac{\partial u}{\partial x}   \\

\frac{\partial u}{\partial y}    \\ \end{matrix}}\right]+F(x,y)u+G(x,y)$$

Expanding $$\alpha$$,

$$\alpha=\left[{\begin{matrix} \frac{\partial }{\partial x} &  \frac{\partial }{\partial x}  \\ \end{matrix}}\right]

\left[{\begin{matrix} A\frac{\partial u }{\partial x}+B\frac{\partial u}{\partial y}   \\

B\frac{\partial u }{\partial x}+A\frac{\partial u}{\partial y}   \\ \end{matrix}}\right]$$

$$\Rightarrow \alpha=(A\frac{\partial^2 u }{\partial x^2}+A_x\frac{\partial u}{\partial x})+(B\frac{\partial^2 u }{\partial x\partial y}+B_x\frac{\partial u}{\partial y})+(B\frac{\partial^2 u }{\partial x\partial y}+B_y\frac{\partial u}{\partial x})+(C\frac{\partial^2 u }{\partial^2 y}+C_y\frac{\partial u}{\partial y})$$

Expanding $$\beta$$,

$$\Rightarrow \beta=D(x,y)\frac{\partial u }{\partial x}+E(x,y)\frac{\partial u }{\partial y}+F(x,y)u+G(x,y)$$

If $$A, B, C$$ where to be a constant and we make the assumption that$$\frac{\partial }{\partial x\partial y}=\frac{\partial }{\partial y\partial x}$$ then in $$\alpha $$ the $$A_x$$,    $$B_y$$ and $$C_y$$ terms would be zero.

$$\therefore$$

$$\alpha=(A\frac{\partial^2 u }{\partial x^2})+(B\frac{\partial^2 u }{\partial x\partial y})+(B\frac{\partial^2 u }{\partial x\partial y})+(C\frac{\partial^2 u }{\partial^2 y})$$

Adding $$\alpha$$ and $$\beta$$, we have the gen general form of the Second order PDE,

$$\Rightarrow \alpha+\beta=A(x,y)\frac{\partial^2u }{\partial x^2}+2B(x,y)\frac{\partial^2u }{\partial x\partial y}+C(x,y)\frac{\partial^2u }{\partial y^2}+D(x,y)\frac{\partial u }{\partial x}+E(x,y)\frac{\partial u }{\partial y}+F(x,y)u+G(x,y)=0$$

The coefficients $$A$$,$$B$$,$$C$$ are constant are functions of $$(x,y)$$
If we have have to incorporate the coefficients into the partial differential equation,the general form of the Second order PDE would be

$$\frac{\partial^2A(x,y)u }{\partial x^2}+\frac{\partial^2B(x,y)u }{\partial x\partial y}+\frac{\partial^2C(x,y)u }{\partial y^2}+\frac{\partial D(x,y)u }{\partial x}+\frac{\partial E(x,y)u }{\partial y}+F(x,y)u+G(x,y)=0$$

Expanding this we have ,

$$[A_x\frac{\partial^2u }{\partial x^2}+A(x,y)\frac{\partial u}{\partial x}]+[2B(x,y)\frac{\partial^2u }{\partial x\partial y}+B_y\frac{\partial u}{\partial x}+B_x\frac{\partial u}{\partial y}]+[C(x,y)\frac{\partial^2u }{\partial y^2}+C_y\frac{\partial u}{\partial y}]$$

$$+D(x,y)\frac{\partial u }{\partial x}+E(x,y)\frac{\partial u }{\partial y}+F(x,y)u+G(x,y)=0 $$

Rearranging the terms ,we have

$$A(x,y)\frac{\partial^2u }{\partial x^2}+2B(x,y)\frac{\partial^2u }{\partial x\partial y}+C(x,y)\frac{\partial^2u }{\partial y^2}+(D+A_x+B_y)\frac{\partial u }{\partial x}+(E+B_x+C_y)\frac{\partial u }{\partial y}+F(x,y)u+G(x,y)=0$$

Let,

$$\overline D=D+A_x+B_y$$

and

$$\overline E=E+B_x+C_y$$

Rotation of Coordinate axes


Rotation of coordinate axes is an example of linear coordinate translation.

Following is general representation of linear coordinate translation.

$$\begin{pmatrix} x\\

y\end{pmatrix} = V_{2\times2}\begin{pmatrix} \overline x\\

\overline y\end{pmatrix} $$

Example of linear map for $$\R^n$$ (in a n dimension case)

$$A:\R^n \to \R^n$$

$$x \mapsto y=Ax$$

$$A \in R^{nxn} $$ nxn matrix



In this case:

$$\mathfrak{D}(Domain)=\R^n$$

$$\mathfrak{R}(Range)=\R^n$$

Analyzing a more general where m≠n:

$$A:\R^n \to \R^n$$

$$x \mapsto y=Ax$$

$$A \in R^{nxn}$$

where: n=row and m=columns

y=Ax

Now consider:

y=Ax

where: y=nx1 matrix

Ax=nx1

b=nx1

$$M: \R^m \mapsto \R^n$$

$$x \mapsto y=Ax+b$$

Clearly:

$$M(0)=b\neq 0\Rightarrow M(.)$$

The $$\neq$$ signal means that $$M(.)$$ is not homogenic

$$M(.)$$ is not a linear map, affine map

Example: Rotation followed by translation

$$y=Rx+b$$

m=n=2

$$\begin{Bmatrix} y_1\\y_2

\end{Bmatrix}=\begin{bmatrix} R_{11} &R_{12} \\ R_{21} &R_{22} \end{bmatrix}\begin{Bmatrix} x_1\\x_2

\end{Bmatrix}+\begin{Bmatrix} b_1\\b_2

\end{Bmatrix}$$

Coordinates:

$$x=\phi (\bar{x},\bar{y})$$

$$y=\psi (\bar{x},\bar{y})$$

Linear coordinate transformation

Let consider:

$$u=(x,y)= u(\phi (\bar{x},\bar{y}),y(\psi (\bar{x},\bar{y}))$$

$$u=(x,y)= u(\bar{x},\bar{y})$$ * abuse of notation by using "u"

$$u=(x,y)= \bar{u}(\bar{x},\bar{y})$$ * this is more rigorous notation

Example:

Let: $$u(x)=ax+b$$

Consider: $$x=x(\bar{x})=sin\bar{x}$$ where x $$\to $$ general function or $$(\phi (\bar{x}))$$

$$u(x)=u(\phi (\bar{x}))=a sin \bar{x}+b$$

$$u(x)=u(\bar{x})=\bar{u}$$

where:

$$u(\bar{x})$$ is an abuse of notation

and $$\bar{u}(\bar{x})$$ is a more rigorous notation

$$u_x(x,y)=\frac{\partial u}{\partial x\,}(x,y)=u_x(\phi (\bar{x},\bar{y}),y(\psi (\bar{x},\bar{y}))$$

$$=\frac{\partial \bar{u}}{\partial x\,}(\bar{x},\bar{y})=\frac{\partial \bar{u}}{\partial \bar{x}\,}\frac{\partial \bar{x}}{\partial x\,}+\frac{\partial \bar{u}}{\partial \bar{y}\,}\frac{\partial \bar{y}}{\partial x\,}$$

where:

$$\frac{\partial \bar{u}}{\partial \bar{x}\,}\frac{\partial \bar{x}}{\partial x\,}=\bar{u}_\bar{x}$$

$$\frac{\partial \bar{u}}{\partial \bar{y}\,}\frac{\partial \bar{y}}{\partial x\,}=\bar{u}_\bar{y}$$

Define:

$$\bar{x}=\bar{x}(x,y)=\phi (x,y)$$

$$\bar{y}=\bar{y}(x,y)=\psi (x,y)$$

$$u_y(x,y)=\frac{\partial \bar{u}}{\partial y\,}(\bar{x},\bar{y})=\bar{u}_{\bar{x}}\frac{\partial \bar{x}}{\partial y\,}+\bar{u}_{\bar{y}}\frac{\partial \bar{y}}{\partial y\,}$$

In a matrix form:

$$=\partial \bar{x}(u)=\begin{bmatrix}\frac{\partial \bar{x}}{\partial x\,} & \frac{\partial \bar{y}}{\partial x\,}\end{bmatrix} \begin{bmatrix}\partial \bar{x} \\ \partial \bar{y}\end{bmatrix}(\bar{u})$$

$$\begin{Bmatrix}\partial x \\\partial y\end{Bmatrix}(u)=\begin{bmatrix}\frac{\partial \bar{x}}{\partial x} & \frac{\partial \bar{y}}{\partial x}\\ \frac{\partial \bar{x}}{\partial y} & \frac{\partial \bar{y}}{\partial y}\end{bmatrix}\begin{Bmatrix}\partial \bar{x} \\\partial \bar{y}\end{Bmatrix}(\bar {u})$$

$$\begin{bmatrix}\frac{\partial \bar{x}}{\partial x} & \frac{\partial \bar{y}}{\partial x}\\ \frac{\partial \bar{x}}{\partial y} & \frac{\partial \bar{y}}{\partial y}\end{bmatrix}$$ This matrix is known as the transpose of Jacobian matrix: $$\underline{J}^{T}$$ (sometimes defined as Jacobian matrix)

Note: Easier and more general

$$(x_1,...,x_n)\to (\bar{x}_1,...,\bar{x}_n)$$

Ind. notation: $$\bar{x}_i=\bar{x}_i(x_1,...,x_n)$$

$$\mathbf{J}_{nxn}=\begin{bmatrix}\partial \bar{x}_i\\ \partial \bar{x}_j\end{bmatrix}_{nxn}$$

where:

i= row index

j=column index

= References = Egm6322.s09.bit.sahin 20:50, 6 February 2009 (UTC) Egm6322.s09.bit.gk 20:51, 6 February 2009 (UTC) Egm6322.s09.bit.la 20:52, 6 February 2009 (UTC)