User:Egm6322.s09.bit.la/hw1

= General non-Linear PDE's =

N independent variables

 * {xi}={x1,...,xn}

if expand the index "i"=1,...,n

Example

1D case with time variable (1D time independent problem)

(x,t)=(x1,x2)

x1=x  spatial variable

x2=t  temporal variable

3D time independent problem

(x,y,z,t)=(x1,...,x4)

Unknown Functions: u1,u2,...

e.g. Navier-Stokes in 3D

ui({xj})  i=1,2,

ui={u1,u2,u3}  j=1,2,3,4 ui={u1,u2,u3}$$\rightarrow$$ velocity field (x,y,z)

Restrict to 1 unknown function u
n independent variable {xi}  i=1,...,n

mth partial derivatie: $$\frac{\partial u^m}{\partial x_{i1} ... \partial x_{im}}$$

{i1,...,im} subset of m index among n possible index

i1,...,im=1,...,n

=Nonlinear PDE's=

n independent variables {xi} 1 Unknown function u
$$F=\left(\left\{xi \right\},\left\{\frac{\partial u}{\partial x_{i}} \right\},\left\{\frac{\partial u^2}{\partial x_{i} \partial x_{j}} \right\},... \right)=0$$

$$\left\{xi \right\}$$ - n arguments

$$\left\{\frac{\partial u}{\partial x_{i}} \right\}$$ - n arguments (components of grad. u)

$$\left\{\frac{\partial u^2}{\partial x_{i} \partial x_{j}} \right\}$$ - $$\frac{n\left(n+1 \right)}{2}$$ arguments (assuming $$\left\{\frac{\partial u^2}{\partial i \partial j} \right\}=\left\{\frac{\partial u^2}{\partial i \partial j} \right\}$$

i=1,...,n

Hij:= $$\frac{\partial u^2}{\partial x_{i} \partial x_{j}}$$

Hnxn:=[Hij]nxn and HT=H

$$\begin{vmatrix} U_{11}  U_{12} U_{13}\\ U_{21} U_{22} U_{23}\\ U_{31} U_{32} U_{33} \end{vmatrix}$$

if you just the elements above the main diagonal - $$\frac{n^{2}-n}{2}$$

if you just the elements above the main diagonal (including the diagonal) - $$\frac{n\left(n+1 \right)}{2}$$

2 Independent variables, 1 unknown function
$$F((x,y),u(u_{x},u_{y}),(u_{xx},u_{xy},u_{yy}),...)=0$$

Example

2uxx+3uyy=ax2+bx

uxx+uyy=0  Laplace Equation

div(grad u)

$$\triangledown. \triangledown u = \triangledown ^{2} u$$

5(uxx2)+7(uyy)3=0

Linearity with respect to u (unknown function)
"We must welcome the future, remembering that soon it will be the past; and we must respect the past, remembering that it was once all that was humanly possible" George Santayana

$$\mathcal{L}$$: an operator $$\mathcal{L}$$ (u) is linear with respect to u if

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

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

(x,y) $$\rightarrow$$ u(x,y)



$$\mathbb{R}$$=set of real numbers

(x,y) $$\epsilon$$ $$\Omega$$



$$\Omega$$ is domain of u, $$\mathbb{R}$$ is range of u,

$$u\left(\Omega \right)$$ is image of $$\Omega$$ under mapping $$u$$,

$$v: \Omega \rightarrow \mathbb{R}$$ just anather function like $$u$$,

$$\alpha ,\beta \epsilon \mathbb{R}$$ (2 real numbers)

Example:

Egm6322.s09.bit.sahin 21:02, 24 April 2009 (UTC)

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

$$F\left(x,y,u,u_{x},u_{y}... \right)=0$$

Therefore,

$$\mathcal{L}\left(\cdot \right)=2{\frac{\partial ^{2}\left(\cdot  \right)}{\partial x^2}}{}+3{\frac{\partial ^{2}\left(\cdot  \right)}{\partial y^2}}{}-7x^{2}-x$$ and

$$\mathcal{L}\left(u \right)=2u_{xx}+3u_{yy}-7x^{2}-x$$

$$\mathcal{L}\left(v \right)=2v_{xx}+3v_{yy}-7x^{2}-x$$

let, $$f\left(x \right)=-7x^{2}-x$$

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

and

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

so,

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

It is clear that the partial differential equation is non-linear (Figure 3).



If we define,

$$\mathcal{L}\left(\cdot \right)=2\left(\cdot  \right)_{xx}+3\left(\cdot  \right)_{yy}$$

then cleary

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

this partial differential equation is linear (Figure 3).

= Order of PDE =

Egm6322.s09.bit.sahin 21:03, 24 April 2009 (UTC)

The order of a differential equation is the order of the highest derivative that appears in the equation. Followings are both first order PDEs. First one is a linear first order PDE, whereas second one is non-linear.

$$ 5u_{x}-7u_{y}=0$$

$$6u_{x}^{3}+2u_{y}^{2}+u^{1/2}+x^{2}+siny=0$$

An example for the second order PDE is,

$$div\left(grad\ u \right)+f\left(x,y \right)=0$$

This equation can also be written in cartesian coordinates as:

$$u_{xx}+u_{yy}+f\left(x,y \right)=0$$

Another example for 2nd order example is:

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

Here $$\kappa$$ is known as conductivity tensor (2nd order tensor).

$$\kappa =\kappa _{ij}e_{i}\otimes e_{j}$$

Here $$e_{i}$$ shows the unit vectors. Cartesian coordinate system and corresponding unit vectors can be seen in Figure 4.



$$\kappa \cdot gradu=\left(\kappa _{ij} e_{i}\otimes e_{j}\right)\cdot \left(\frac{\partial u}{\partial x_{k}}e_{k} \right)=\kappa _{ij}\frac{\partial u}{\partial x_{k}}\left(e_{i}\otimes e_{j} \right)e_{k}$$

$$=\kappa _{ij}\frac{\partial u}{\partial x_{k}}e_{i}\left(e_{j}\cdot e_{k} \right)$$

Since $$e_{j}\cdot e_{k}=\delta _{jk} $$

$$\kappa \cdot grad u=\kappa _{ij}\frac{\partial u}{\partial x_{j}}e_{i}=: v_{i}e_{i}$$

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

$$divv=\frac{\partial }{\partial x_{i}}\kappa _{ij}\frac{\partial u}{\partial x_{j}}$$

The conductivity tensor is a second order tensor and obeys the usual transformation rule when the coordinate system rotated. Moreover it is usually assumed that $$\kappa$$ is symmetric.

$$\left[\kappa _{ij} \right]=\begin{bmatrix} \kappa _{11} & \kappa _{12}\\ \kappa _{21}& \kappa _{22} \end{bmatrix}$$, $$\kappa _{ij}=\kappa _{ji}$$ or $$\kappa ^{T}=\kappa $$

If $$\kappa$$ is constant, the equation is a linear 2nd order PDE.

If $$\kappa=\kappa(x,y)$$, the equation is a linear 2nd order PDE.

If $$\kappa=\kappa(x,y,u)$$, the equation is a quasi-linear 2nd order PDE.

=Home work Solutions=

Further Examples of Linear PDE
{| class="toccolours collapsible collapsed" width="60%" style="text-align:left" !Examples of Linear and Non Linear PDE


 * A simple example of a linear PDE is the 2-D Laplace Equation

$$L\left( \phi\right)=\frac{\partial^2\left( \phi\right) }{\partial^2 x}+\frac{\partial^2\left( \phi\right) }{\partial^2 y}$$

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, we have,

$$L\left( \right)=\frac{\partial^2\left( \right) }{\partial^2 x}+\frac{\partial^2\left( \right) }{\partial^2 y}$$

$$\Rightarrow L\left(\alpha \phi \right)=\frac{\partial^2\left(\alpha \phi \right) }{\partial^2 x}+\frac{\partial^2\left(\alpha \phi \right) }{\partial^2 y}$$

Where, $$\phi=\phi\left(x,y\right)$$

$$\Rightarrow L\left(\alpha \phi \right)=\alpha L\left(\phi\right)$$

$$\Rightarrow Linear$$

This also is an example to illustrate the homogeneous property of the linear PDE


 * Another example for linear PDE

$$\frac{\partial\left(\ u\right)}{\partial t}+\frac{\partial\left(\ u\right)}{\partial x}=0$$

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, we have,

$$L\left( \right)=\frac{\partial\left( \right) }{\partial t}+\frac{\partial\left( \right) }{\partial x}$$

$$L\left(\alpha u+\beta v \right)=\frac{\partial\left(\alpha u+\beta v \right) }{\partial t}+\frac{\partial\left(\alpha u+\beta v \right) }{\partial x}$$

$$\Rightarrow L\left( \alpha u+\beta v \right)=\alpha\frac{\partial\left(\ u\right) }{\partial t}+\beta\frac{\partial\left(\ v\right) }{\partial t}+\alpha\frac{\partial\left(\ u\right) }{\partial x}+\beta\frac{\partial\left(\ v\right) }{\partial t}$$

$$\Rightarrow L\left(\alpha u+\beta v \right)=\alpha\left\{\frac{\partial\left(\ u \right) }{\partial t}+\frac{\partial\left(\ u \right) }{\partial x} \right\}+\beta\left\{\frac{\partial\left(\ v\right) }{\partial t}+\frac{\partial\left(\ v\right) }{\partial x} \right\}$$

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

This also is an example to illustrate the additivity property of the linear PDE

Further Examples of non linear PDE

 * The inviscid Burgers equation

$$\frac{\partial\left(\ u\right)}{\partial t}+u\frac{\partial\left(\ u\right)}{\partial x}=0$$

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, we have,

$$L\left( \right)=\frac{\partial\left( \right) }{\partial t}+\left(\right)\frac{\partial\left( \right) }{\partial x}$$

$$L\left(\alpha u+\beta v \right)=\frac{\partial\left(\alpha u+\beta v \right) }{\partial t}+\left(\alpha u+\beta v \right)\frac{\partial\left(\alpha u+\beta v \right) }{\partial x}$$

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

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


 * Another example for non linear PDE

$$L\left( u\right)=\frac{\partial^2\left( u\right) }{\partial^2 t}+\frac{\partial^2\left( u\right) }{\partial^2 x}-u^2$$

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, we have,

$$L\left( \right)=\frac{\partial^2\left( \right) }{\partial^2 t}+\frac{\partial^2\left( \right) }{\partial^2 x}-\left(\right)^2$$

From the definition of operator $$L\left(\right)$$,

$$L\left(\alpha u+\beta v \right)=\frac{\partial^2\left(\alpha u+\beta v \right) }{\partial^2 t}+\frac{\partial^2\left(\alpha u+\beta v \right) }{\partial^2 x}-\left(\alpha u+\beta v \right)^2$$

Clearly,

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

Egm6322.s09.bit.gk 20:27, 24 April 2009 (UTC)
 * }

Problem 5
128.227.51.230 22:29, 26 January 2009 (UTC) Egm6322.s09.bit.sahin 22:31, 26 January 2009 (UTC) Egm6322.s09.bit.la 22:34, 26 January 2009 (UTC) Egm6322.s09.bit.gk 22:37, 26 January 2009 (UTC)