User:Egm6322.mafia

=General Non-Linear PDEs=

Defining Variables
For these notes, we will assume we have n independent variables:

$$  \left \{ {x}_{i} \right \}=   \left \{ {x}_{1},...,{x}_{n}   \right \} $$.

An example of this would be single dimensional case with a time variable.

' $$ (  x,t )=( {x}_{1}, { x}_{2 } )  $$'

where $$ {x}_{1} = {x}  $$, a spatial variable

and $$ {x}_{2} = {t}  $$, a temporal variable.

A more complex situation would be a three dimensional, time dependent problem:

$$(x,y,z,t)=({x}_{1},...,{x}_{4})$$.

In this case, $$ {x}_{1} = {x}  $$, $$  {x}_{2} = {y}  $$, $$  {x}_{3} = {z}  $$, and $$  {x}_{4} = {t}  $$.

Definition of Functions
The unknown function $$ u $$ maps numbers from the domain $$ \Omega $$ to the real numbers $$ \mathbb{R} $$ In two dimensions, this means:

$$ \left( x,y \right) \in \Omega $$

$$ u \left( x,y \right) \in \mathbb{R} $$

and

$$ \Omega \rightarrow \mathbb{R}$$

$$ \left( x,y \right) \mapsto u \left( x,y \right) $$

which means that $$\Omega $$ is the domain of u, $$ \mathbb{R}$$ is the range of u, and $$u \left ( \Omega \right )$$ is the image of $$\Omega $$ under the mapping of $$ u $$.

Example Functions
For these notes, we will assume we have unknown functions: $$ {u}_{1},  {u}_{2} ,...  $$.

The Navier Stokes Equation
An example of this is the Navier-Stokes Equations in 3-D:

$$ {u}_{i} ( \left \{  {x}_{j} \right \} ) $$

where $$ {u}_{i} $$ is the velocity field in x, y, z; $$ {u}_{1}, {u}_{2}, {u}_{3} $$, dependent upon variables $$ {x}_{1}, ... {x}_{4}$$, where $$ {x}_{1} = {x}  $$, $$  {x}_{2} = {y}  $$, $$  {x}_{3} = {z}  $$, and $$  {x}_{4} = {t}  $$.

One Unknown Function
For our present discussion, this can be restricted to one unknown function $$ u $$, containing $$ n $$ independent variables $$ {{x}_{i}} i=1,...,n $$

The mth partial derivative of this function can be expressed as:

$$ \frac{\partial^m u}{\partial x_i\ ,...,\partial x_m} $$ where

$$ {{i}_{1},...,{i}_{m}} $$ is the subset of m indices among n possible indices

$$ {i}_{i}, ..., {i}_{m}=1,...,n $$.

Another Example
Another example of a non-linear PDE is:

$$ F \left ( \left \{ {x}_{i} \right \}, \left \{ \frac{\partial u}{\partial x_i} \right \}, \left \{ \frac{\partial^2 u}{\partial x_i \partial x_j} \right \}, ... \right ) =0$$

where $$ \left \{ {x}_{i} \right \}$$ contains $$ n $$ arguments,

$$ \left \{ \frac{\partial u}{\partial x_i} \right \}$$ are the components of the $$ \nabla u$$, and

$$ \left \{ \frac{\partial^2 u}{\partial x_i \partial x_j} \right \}$$ are the components of the Hessian of $$ u $$.

The Hessian
The Hessian is a symmetric $$ n \times n $$ matrix, thus H = HT

and $${H}_{n \times n}$$ := $$ {\left [ {H}_{ij}\right ]}_{n \times n}$$.

The Hessian is defined as:

$$ {H}_{ij} := \left \{ \frac{\partial^2 u}{\partial x_i \partial x_j} \right \}$$.

This has many interesting properties.



The Laplace Equation
The Laplace Equation is a single function of two (or more) variables:

$$ F \left ( \left (x,y \right) ,u,\left ({u}_{x},{u}_{y} \right), \left ({u}_{xx},{u}_{xy},{u}_{yy} \right ), ...   \right )=0 $$

Example of the Laplace Equation include:

$$2{u}_{xx}+3{u}_{yy}=a{x}^{2}+bx$$

$${u}_{xx} +{u}_{yy}=0$$

The Definition of Linearity
If $$ u $$ is an unknown function, the operator $$ L $$ is considered a linear operator with respect to $$ u $$ if:

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

Example of Checking for Linearity
Let's take the function:

$$2{u}_{xx}+3{u}_{yy}-7{x}^{2}+x=0$$

The operator is :

$$ L= 2 \frac{\partial^2 }{\partial x^2} + 3 \frac{\partial^2 }{\partial y^2} -7 x^2-x$$

where $$-7 x^2-x:= f(x)$$

To check the linearity, we assume $$\alpha$$ and $$\beta$$ are real number constants, and evaluate: $$ L(\alpha u+ \beta v)= 2 \frac{\partial^2 (\alpha u+ \beta v)}{\partial x^2} + 3 \frac{\partial^2 (\alpha u+ \beta v)}{\partial y^2} + f(x)$$

which is equivalent to:

$$ L(\alpha u+ \beta v)=2 {\left (\alpha u + \beta v \right )}_{xx}+3 {\left (\alpha u +\beta v\right )}_{yy}+f(x) $$

If we distribute the integers, as well as the second derivative operators, the equation becomes:

$$ L(\alpha u+ \beta v)=2 \alpha {u}_{xx}+2 \beta {v}_{xx}+3 \alpha {u}_{yy} + 3 \beta {v}_{yy}+f(x) $$.

Grouping the terms by the constants $$\alpha$$ and $$\beta$$ concludes:

$$ L(\alpha u+ \beta v)= \alpha \left [2 {u}_{xx}+3{u}_{yy} \right ] +\beta \left [ 2 {v}_{xx} +3 {v}_{yy} \right ] +f(x) $$.

The operator $$ L $$ is nonlinear because

$$ \alpha L(u)+ \beta L(v)= \alpha \left [2 {u}_{xx} +3 {u}_{yy} +f(x) \right ] +\beta \left [2 {v}_{xx}+3 {v}_{yy} + f(x) \right ] $$

and

$$ f(x) \ne \alpha f(x) +\beta f(x) $$


 * $$J=\begin{bmatrix} \dfrac{\partial y_1}{\partial x_1} & \cdots & \dfrac{\partial y_1}{\partial x_n} \\ \vdots & \ddots & \vdots \\ \dfrac{\partial y_m}{\partial x_1} & \cdots & \dfrac{\partial y_m}{\partial x_n} \end{bmatrix}. $$

Definition of the order of PDE
Order of highest derivative in the PDE.

Example of first order PDE
linear: $$5u_x-7u_y=0$$

non-linear: $$6(u_x)^3+2(u_y)^2+(u)^{1/2}+x^2+sin xy=0$$

we can assume that $$D(u)=6(u_x)^3+2(u_y)^2+(u)^{1/2}$$

Homework
Show D(u) is not linear.

(Solution)

Again, $$D \left( u \right) = 6 \left( u_x \right)^3 + 2 \left( u_y \right)^2 + \left( u \right)^{1/2}$$.

D(u) is linear if the following formula holds:

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

OR,

$$6 \left( \alpha\cdot u + \beta\cdot v \right)^3 + 2 \left( \alpha\cdot u + \beta\cdot v  \right)^2 +  \left( \alpha\cdot u + \beta\cdot v  \right)^{1/2} = \alpha\cdot \left [ 6 \left( u_x  \right)^3 + 2 \left( u_y  \right)^2 +  \left( u  \right)^{1/2} \right ] + \beta\cdot \left [ 6 \left( v_x  \right)^3 + 2 \left( v_y  \right)^2 +  \left( v  \right)^{1/2} \right ]$$

The expressions on either side of the equality sign in the above equation are NOT identical. Therefore, D(u) is NOT linear.

Example of second order PDE
$$div(gradu)+f(x,y)=0$$

Notice: tensorial notation = coord free notation.

Cartesian Coordinate (x,y)
Let's take the function:

$$u_xx+u_yy+f(x,y)=0$$

We can also write it as div( $$\kappa$$ $$\cdot$$$$grad u$$$$+f(x,y)=0$$

Here, $$\kappa$$ is a continuity tensor (2nd order tensor).

$$\kappa$$ = $$\kappa$$ $$e_i$$ $$\otimes$$ $$e_j$$ 2nd-order tensor

$$v$$ = $$v_i$$ $$e_i$$ vector,1st-order tensor

Operators
$$grad$$ $$u$$ $$=$$ $$\frac{\partial u}{\partial x_i}$$ $$e_i$$

$$div$$ $$v$$ $$=$$ $$\frac{\partial v_i}{\partial x_i}$$ $$e_i$$

$$\kappa$$ $$\cdot$$ $$grad$$ $$u$$ =( $$\kappa_{ij}$$ $$e_i$$ $$\otimes$$ $$e_j$$ ) $$\cdot$$ ($$\frac{\partial u_k}{\partial x_k}$$ $$e_k$$ )

=$$\kappa_{ij}$$ $$\frac{\partial u}{\partial x_k}$$ ( $$e_i$$ $$\otimes$$ $$e_j$$ ) $$\cdot$$ $$e_k$$

Here, $$e_i$$ $$\otimes$$ $$e_j$$ $$\cdot$$ $$e_k$$ = $$e_i$$ ( $$e_j$$ $$\cdot$$ $$e_k$$ )

Kronecker delta
Definition: If $$e_j$$ $$\otimes$$ $$e_k$$ = $$\delta_{jk}$$, $$\delta_{jk}$$ is called Kronecker delta.

$$ \delta_{jk} = \begin{cases} 1 & for \ j=k \\ 0 & for \ j \neq k \end{cases} $$ = $$\kappa_{jk} \frac{\partial u}{\partial x_i} $$ $$e_i$$

$$u(x,y)$$ is a scaler function $$\Rightarrow$$ 0th order tensor $$gradu$$ is a vector field $$\Rightarrow$$ 1st order tensor

$$grad$$  ($$\cdot$$) increases tensor order by 1

$$div$$  ($$\cdot$$) decreases tensor order by 1

We can let $$v$$ : = $$\kappa$$ $$\cdot$$ $$gradu$$

$$div$$ $$v$$ = $$\frac{\partial }{\partial x_i}$$ $$\kappa_{jk}$$ $$\frac{\partial u}{\partial x_j}$$

Homework
Expand the function above.

What's more
$$\kappa_ij$$ = $$\begin{bmatrix} \kappa_{11} & \kappa_{12} \\ \kappa_{21} & \kappa_{22} \end{bmatrix} $$

Notice: in PDE, $$\kappa_{ij}=\kappa_{ji}$$ or $$\kappa^T=$$ $$\kappa$$

$$v$$ = $$v_i$$ $$e_i$$ $$\rightarrow$$ $$\begin{Bmatrix} v_i\end{Bmatrix} _{3 \times 1 or 2 \times 1} $$

And if we change $$e_i$$ coordinate into different system $$\underline{\overline{e_i}}$$

$$\overline{v_i}$$ $$\underline{\overline{e_i}}$$ $$\to$$ $$\overline{\begin{Bmatrix} v_i\end{Bmatrix}}$$

Linearity
if $$\kappa$$ $$= const$$ $$\Rightarrow$$ is a linear 2nd order PDE

$$\kappa$$ $$= \underline{\kappa} (x,y)$$ $$\Rightarrow$$ is also a linear 2nd order PDE

$$\kappa$$ $$= \underline{\kappa} (x,y,u)$$ $$\Rightarrow$$ is a quasilinear 2nd order PDE

Homework
Show that $$\kappa$$ $$= \underline{\kappa} (x,y)$$ $$\Rightarrow$$ is also a linear 2nd order PDE