hi all...

i have 2 Q:

1-what is the diffrent between Quasi-Linear pde and Linear pde ?

2- How we can solve Quasi-Linear ??

any body can help !!!:confused:

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- Aug 18th 2007, 06:22 PMsweetQuasi-Linear
hi all...

i have 2 Q:

1-what is the diffrent between Quasi-Linear pde and Linear pde ?

2- How we can solve Quasi-Linear ??

any body can help !!!:confused: - Aug 18th 2007, 06:31 PMThePerfectHacker
Let $\displaystyle u_1 \mbox{ and }u_2$ solve a PDE, if it happens that $\displaystyle c_1u_1+c_2u_2$, i.e. any linear combination is a solution, then we say the PDE is

*linear*.

For example, consider the wave equation,

$\displaystyle \frac{\partial ^2 u}{\partial t^2} - \frac{\partial ^2 u}{\partial x^2} = 0$.

Confirm that if $\displaystyle u_1(x,t)\mbox{ and }u_2(x,t)$ are solutions then $\displaystyle c_1u_1(x,t)+c_2u_2(x,t)$ is a solution as well.

Another way to think of linear is that $\displaystyle u$, the solution, is just like what linear means in an ordinary differencial equation. So $\displaystyle u_{xx}+txu_x=0$ is linear. But $\displaystyle u_{xx}^2+txu_x=0$ is not, because by analogy we never have $\displaystyle (y'')^2+x^2y'=0$ being considered linear ODE.

Quasi-linear means (this is how I understand it) that the**highest order terms**are linear.

For example,

$\displaystyle u_{xx}+\cos(x+y)u_{yy}+u_{xy}=e^{x+y} u_x\cdot u_y^2$

Is not linear but it is quasi-linear. Because the highest order terms $\displaystyle u_{xx},u_{yy},u_{xy}$ are linear. - Aug 18th 2007, 06:35 PMThePerfectHackerQuote:

2- How we can solve Quasi-Linear ??

$\displaystyle A(x,y)\frac{\partial ^2 u}{\partial x^2}+B(x,y)\frac{\partial ^2 u}{\partial x \partial y} +C(x,y)\frac{\partial ^2 u}{\partial y^2} = F(x,y,u,u_x,u_y)$

Depending on the sign of $\displaystyle B^2-4AC$: positive, negative, zero. We have 3 cases: hyperbolic, elliptic, parabolic. Each one can be simplified by a method knows as**charachteristics curves**. When properly reduced by appropriate substitutions we get the "canonical form". From there we use certain PDE methods to solve that. - Aug 18th 2007, 06:51 PMsweetQuote:

Quasi-linear means (this is how I understand it) that the highest order terms are linear.

For example,

u_{xx}+\cos(x+y)u_{yy}+u_{xy}=e^{x+y} u_x\cdot u_y^2

Is not linear but it is quasi-linear. Because the highest order terms u_{xx},u_{yy},u_{xy} are linear.

but how can i solv quasi-linear of the first order..? - Aug 18th 2007, 06:58 PMThePerfectHacker
Here is Lagrange's method from 1772 (older than America):eek:

Consider the equation,

$\displaystyle P(x,y,z)\frac{\partial z}{\partial x}+Q(x,y,z)\frac{\partial z}{\partial y}=R(x,y,z)$.

Note, this is not a linear equation because $\displaystyle P,Q,R$ contain $\displaystyle z$, the solution, in them but still we can solve it.

We do this by solving, (I do not like using differencials but here they play a nice role),

$\displaystyle \frac{dx}{P} = \frac{dy}{Q} = \frac{dz}{R}$.

Let $\displaystyle f(x,y,z)=c_1 \mbox{ and }g(x,y,z)=c_2$ solve the differencial above (not the actual equation). Then the solution to the actual PDE is given by $\displaystyle F(f,g)=0$. - Aug 18th 2007, 07:11 PMsweet
thank u :):)

do u have a paper of the Lagrange's method - Aug 18th 2007, 07:37 PMThePerfectHacker