# Quasi-Linear

• Aug 18th 2007, 06:22 PM
sweet
Quasi-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 PM
ThePerfectHacker
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 PM
ThePerfectHacker
Quote:

2- How we can solve Quasi-Linear ??
A general 2nd order Quasilinear equation is,
$\displaystyle A(x,y)\frac{\partial ^2 u}{\partial x^2}+B(x,y)\frac{\partial ^2 u}{\partial x \partial y} +C(x,y)\frac{\partial ^2 u}{\partial y^2} = F(x,y,u,u_x,u_y)$

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 PM
sweet
Quote:

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.

really iwas need it , thank u

but how can i solv quasi-linear of the first order..?
• Aug 18th 2007, 06:58 PM
ThePerfectHacker
Quote:

Originally Posted by sweet
but how can i solv quasi-linear of the first order..?

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 PM
sweet
thank u :):)

do u have a paper of the Lagrange's method
• Aug 18th 2007, 07:37 PM
ThePerfectHacker
Quote:

Originally Posted by sweet
do u have a paper of the Lagrange's method

Yes, I have the orginal paper. I am a direct descendent of Lagrange.