# How might I find, in parametric form, the solution to this (first order, quasilinear) PDE?

Consider the (first order quasilinear) PDE $$u \frac{\partial u}{\partial x} + (y+1) \frac{\partial u}{\partial y} = u \hspace{10mm} x \in \mathbb{R}, \hspace{4mm} y \in \left( 0, \hspace{2mm} \frac{1}{3} \right)$$ subject to $u(x,0) = -3x$ for $x \in \mathbb{R}$.

How might I find the solution to this PDE in parametric form? I have attempted to do this as follows:

The characteristic and compatibility equations associated with this PDE are $$\frac{dx}{ds} = u \hspace{10mm} \frac{dy}{ds} = y+1 \hspace{10mm} \frac{du}{ds} = u$$ We may directly solve the second of these equations as $$\frac{dy}{ds} = y+1 \hspace{5mm} \Rightarrow \hspace{5mm} y = y(s) = \frac{e^s}{c_1} - 1$$ and for the third of these equations $$\frac{du}{ds} = u \hspace{5mm} \Rightarrow \hspace{5mm} u = u(s) = \frac{e^s}{c_2}$$ and then, using the above, for the first of these equations we have $$\frac{du}{ds} = u = \frac{e^s}{c_2} \hspace{5mm} \Rightarrow \hspace{5mm} x = x(s) = \frac{e^s}{c_2} + c_3$$

However, I am unsure of where to go from here. Any advice for this would be appreciated.

$\dfrac{dy}{dt}=y+1$ , letting $y(0)=0$ , we have $y=e^t-1$

$\dfrac{du}{dt}=u$ , letting $u(0)=u_0$ , we have $u=u_0e^t=u_0(y+1)$

$\dfrac{dx}{dt}=u=u_0e^t$ , we have $x=f(u_0)+u_0e^t=f\left(\dfrac{u}{y+1}\right)+u$ , i.e. $u=(y+1)F(x-u)$

$u(x,0)=-3x$ :

$F(4x)=-3x$

$F(x)=-\dfrac{3x}{4}$

$\therefore u=-\dfrac{3(y+1)(x-u)}{4}$

$-4u=3x(y+1)-3(y+1)u$

$(3y-1)u=3x(y+1)$

$u=\dfrac{3x(y+1)}{3y-1}$