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

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

$$\Rightarrow Linear$$


 * 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,

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

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

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

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

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

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,

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

Problem 5
Egm6322.s09.bit.gk 22:37, 26 January 2009 (UTC)