# Quasi-Linear PDE

1. Sep 16, 2009

### EAAL

1. The problem statement, all variables and given/known data

Solve
$$(Z+e^x)Z_x + (Z+e^y)Z_y = Z^2 - e^{x+y}$$
Where $$Z = Z(x,y)$$

2. Relevant equations

Equations of the form

$$PZ_x + QZ_y = R$$
Where $$P = P(x,y,z)$$ , $$Q=Q(x,y,z)$$ , $$R=R(x,y,z)$$
Are solved with the Lagrange method.
It is possible to write this in the form:
$$\frac{dx}{P} = \frac{dy}{Q} = \frac{dz}{R}$$
Where from here we get two equations. We Solve these equations like ODE's, and the Final Solution would be $$F(c_1,c_2)=0$$ And $$C_1, C_2$$ are the integrating constants of the ODE's so $$C_1=C_1(x,y,z,c_2)$$ and $$C_2=C_2(x,y,z,c_1)$$
It is possible to use the multiplier method, and try to find $$\alpha , \beta , \gamma$$ (which can be functions)
To satisfy: $$\alpha P + \beta Q + \gamma R =0$$
Which would imply
$$(\alpha)dx + (\beta)dy + (\gamma)dz = 0$$
And to make this simple we look for $$\alpha = \alpha(x) , \beta=\beta(y) , \gamma=\gamma(z)$$

3. The attempt at a solution

$$(Z+e^x)Z_x + (Z+e^y)Z_y = Z^2 - e^{x+y}$$

$$\frac{dx}{z+e^x} = \frac{dy}{z+e^y} = \frac{dz}{z^2 - e^{x+y}}$$
But we can't get 2 solvable equations from here, because it would always involve the third variable, and we couldn't treat it like an ODE.
So we go to The Multiplier Mehod:
$$\alpha(z+e^x) + \beta(z+e^y) + \gamma(z^2 - e^{x+y}) =0$$
Which would imply:
$$\alpha + \beta + \gamma z = 0$$
and
$$\alpha e^x + \beta e^y - \gamma e^{x+y} = 0$$

But I haven't been able to find any $$\alpha , \beta , \gamma$$ that do that...
I found $$\alpha = (z+e^y), \beta = (-z-e^x) , \gamma = 0$$
Fullfills, but then the equation
$$(z+e^y)dx - (z+e^x)dy = 0$$
Cannot be solved (we can't treat the z as a constant, because it is a function of x,y ...

Thank you

Last edited: Sep 16, 2009
2. Sep 18, 2009

(*Bump*)