User:Egm6322.s09.lapetina

=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 $$v$$ : = $$\kappa$$ $$\cdot$$ $$gradu$$

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

We can expand it as: $$div$$ $$v$$ = $$\frac{\partial \kappa_{ij} }{\partial x_i}$$ $$\frac{\partial u }{\partial x_j}$$ + $$\kappa_{ij}$$ $$\frac{\partial u^2 }{\partial x_i x_j}$$

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

Affine map
In geometry, an affine transformation or affine map or an affinity (from the Latin, affinis, "connected with") between two vector spaces (strictly speaking, two affine spaces) consists of a linear transformation followed by a translation:

To find more:Affine_transformation

Linear map
Definition:

In mathematics, a linear map (also called a linear transformation, or linear operator) is a function between two vector spaces that preserves the operations of vector addition and scalar multiplication. The expression "linear transformation" is in particularly common use, especially for linear maps from a vector space to itself (endomorphisms). In advanced mathematics, the definition of linear function coincides with the definition of linear map.

To find more:Linear_transformation

=HW2=

Constant Coefficients
This can be rewritten as:

$$ \left \lfloor \partial_x \; \partial_y \right \rfloor \begin{bmatrix} a & b\\ b & c \end{bmatrix}

\begin{Bmatrix} \partial_x u \\ \partial_y u \end{Bmatrix} +

\left \lfloor d \; e \right \rfloor

\begin{Bmatrix} \partial_x u \\ \partial_y u \end{Bmatrix}

+ fu + g = 0

$$

$$ \alpha =

\left \lfloor \partial_x \; \partial_y \right \rfloor

\begin{Bmatrix} au_x+bu_y \\ bu_x+cu_y \end{Bmatrix}

=

(au_x)_x+(bu_y)_x+(bu_x)_y+(cu_y)_y

$$, if we assume $$ {u}_{xy}={u}_{yx}$$ and $$a, b, c $$ are real constants.

Two Independent Variables
$$(au_x)_x+ (bu_x)_y +(bu_y)_x +(cu_y)_y + du_x +eu_y +fu+g=0$$

$$\left [ a {u}_{xx}+a_x u_x \right ] + \left [ 2b {u}_{xy} +b_y u_x+ b_x u_y \right ] + \left [c {u}_{yy} + c_y u_y \right ] +d u_x + e u_y + fu+ g=0$$

$$ a {u}_{xx}+2 b {u}_{xy}+c {u}_{yy}+\bar d\ u_x+\bar e\ u_y+f u+g=0 $$

$$\bar d\ := d+a_x+b_y$$

$$\bar e\ := e+b_x+c_y$$

One case
$$\kappa$$ $$= \underline{\kappa} (x,y,u,u_x,u_y)$$ $$\Rightarrow$$

Definition of quasilinear: For a PDE of order n (i.e. highest derivative terms are of order n),coeffients of nth order derivative are functions $$= (x,y,u,u_x,u_y,\frac{\partial^m u}{\partial {x^p y^q}})$$

Here, we assume there are two independent variables, mth order derivative, and $$p+q=m,m<n$$

Another Case
PDEs linear with respect to 2nd derivative, but still non-linear in general:

$$div($$ $$\kappa$$   $$\cdot grad u)+$$ $$f(x,y,u,u_x,u_y)=0$$

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

$$f(x,y,u,u_x,u_y)=0$$ is nonlinear with respect to argues in general, e.g.

$$div($$ $$\kappa$$   $$\cdot grad u)+$$ $$ax^2+by+\sqrt{u}+(u_x)^4+2(u_y)^2=0$$

Homework
Show it linear to 2nd derivative, but non-linear in general.

To make it clear,

$$D_1(\cdot):=$$ $$div[$$  $$\kappa(x,y)$$   $$\cdot grad (\cdot)]$$

We can do this:

$$D_2(\cdot):=$$ $$D_1(\cdot)+$$ $$(\cdot)^{1/2}+$$ $$[(\cdot)_x]^4+$$ $$[2(\cdot)_y]^2$$

$$D_3(\cdot):=$$ $$D_2(\cdot)+$$ $$ax^2+by$$

=Homework 4=

The Power Law
The Power Law is a very common relationship in the universe. It is defined as:

$$y=b x^a $$.

The Power Law is observed in classical physics, biology, economics, and many other natural and social sciences. The exponent $$a$$ dominates the nature of the equation. In electrostatics and gravitation, $$a=2$$, while in Stefan-Boltzmann equations, $$a=4$$. More can be found on the Power Law here.

The inverse of the exponent function is the logarithm.

Application of the Power Law
The thermal conductivity of solids is summarized in the following graph.

Plotting here is on a log-log scale. As a result, the slopes appear linear, rather than exponential.

The exponential varying thermal conductivity of solids is very important for solving Fourier's Law:

$$q=- \kappa \; grad \; T$$

where $$q$$ is the heat flux.

From this equation, we can find the units of $$\kappa$$ in the following fashion:

$$q \equiv \left [ \frac{Power}{Unit \; Area} \right ]=\left [ \frac{W}{m^2} \right ]$$ while $$grad \; T= \frac {dT}{dx} = \left [ \frac {K}{m} \right ]$$

Therefore: $$\left [ \kappa \right ]= \left [ \frac {\frac {W}{m^2}}{\frac{K}{m}}\right ] = \frac {W}{mK}$$

If we consider the thermal conductivity $$\kappa$$ as $$\kappa \left ( T \right )$$ where $$T$$ is temperature, we can find the heat flux at any given temperature using the power law by the following equation:

$$\kappa \left ( T \right )=b T^a $$.

This equation can be solved over a given domain if $$a$$ and $$b$$ are known.

Using data from the aforementioned graph, we see that for diamond, $$T \in \left [ 1K, 10.7 K \right ]$$ :

$$a=\frac {log \kappa_2 - log \kappa_1}{log T_2 - log T_1}=\frac {log (1000)-log (0.4)}{log (10.7)-log (1)} \cong 3.39 $$

while $$b \cong 0.4 \frac {W}{mK}$$

so $$\kappa (T) = (0.4) T^{3.39} \frac {W}{mK}$$.

For $$T \in \left [ 100 K, 1000 K \right ]$$, we want to find $$\kappa \left ( T \right ) $$. We can estimate $$a$$ for diamond using the slope of graphite parallel to layers:

$$a=\frac {log \kappa_2 - log \kappa_1}{log T_2 - log T_1}=\frac {log (3)-log (80)}{log (1000)-log (100)} \cong -1.43 $$.

Extrapolating $$b$$ backwards shows its value is $$10^7$$.

Therefore, for all $$T \in \left [ 100 K, 1000 K \right ]$$,

$$\kappa (T) = (10^7) T^{-1.43} \frac {W}{mK}$$

The Wave Equation and String Vibration
The Wave Equation can be studied by examining the physics of string vibration in one spatial dimension. An excellent book on this topic is The Science of Sound by Rossing, Moore and Wheeler.

In the accompanying free-body diagram, forces in the $$x$$ direction are :

$$\sum F_x = -\tau \left ( x,t \right ) cos \; \theta ( x,t )+ \tau \left ( x+dx,t \right ) cos \; \theta ( x+dx,t )$$

$$\theta$$ is very small here, therefore $$cos \theta \cong 1-\frac{\theta^2}{2}$$.

Inserting this into the $$x$$ equation results in:

$$\sum F_x = -\tau \left ( x,t \right ) (1-\frac{{\theta(x,t)}^2}{2})+ \tau \left ( x+dx,t \right ) (1-\frac{{\theta (x+dx,t)}^2}{2})$$.

We can neglect second order terms here, since they are very small and nearly cancel, leaving:

$$\tau (x+dx,t)=\tau (x,t)=\tau$$, suggesting $$\tau$$ is constant throughout the string.

In the $$y$$ direction:

$$\sum F_y = \tau sin \; \theta ( x,t )+ \tau sin \; \theta ( x+dx,t )+ P (x,t) dx -m(dx) \ddot w$$

where $$\ddot w={w}_{tt}$$.

Where $$\theta$$ is small, we can simplify the first two terms as:

$$\theta (x+dx,t) -\theta (x,t) \cong \frac{d\theta}{dx} dx= {(w_x)}_{x} dx= {w}_{xx}dx$$

multiplied by the constant $$\tau$$.

Therefore, for an infinitely small length of string $$dx$$,

$$\tau {w}_{xx} +P=m {w}_{tt}$$

with a Jacobian of the form

$$

\mathbf{J_\beta}=

\begin{bmatrix} 0 & d_y \\ d_x & 0 \end{bmatrix}

$$

where $$d_x = \frac{\sqrt{|\lambda_2|}}{\sqrt{g}}$$

and

$$d_y = \frac{\sqrt{|\lambda_1|}}{\sqrt{g}}$$

The inverse of this is:

$$

\mathbf{J_\beta^-1}=

\begin{bmatrix} 0 & \frac{1}{d_y} \\ \frac{1}{d_x} & 0 \end{bmatrix}

$$

By substitution, the matrix operator form of the conic then becomes:

$$

\left \lfloor \xi \ \eta \right \rfloor

\begin{bmatrix} 0 & \frac{1}{d_y} \\ \frac{1}{d_x} & 0 \end{bmatrix}

\begin{bmatrix} \lambda_{1} \ 0 \\ 0 \ \lambda_2 \end{bmatrix}

\begin{bmatrix} 0 & \frac{1}{d_y} \\ \frac{1}{d_x} & 0 \end{bmatrix}

\begin{Bmatrix}

\xi \\ \eta

\end{Bmatrix}

=g $$

This leaves us with the polynomial equation:

$$ \lambda_2 \frac{\xi^2}{d_x^2}+ \lambda_1 \frac{\eta^2}{d_y^2}=g$$.

Upon substitution, this becomes

$$(\frac{\sqrt{g}}{\sqrt{|\lambda_2|}})^2 \lambda_2 (\xi)^2- (\frac{\sqrt{g}}{\sqrt{|\lambda_1|}})^2 \lambda_1 (\eta)^2=g$$.

Constant values cancel, leaving:

$$\xi ^2- \eta^2= 1$$ for the case where:

$$\lambda_1<0$$, $$\lambda_2>0$$, and $$g<0$$.

For the case where