User:Egm6322.s09.bit.sahin

$$\Omega$$ 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:

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

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

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 1).



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 1).

Order of PDE
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 2.



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

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

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.