I am given a first order quasilinear PDE $uu_{x_1} + u_{x_2} = 1$ which has the Cauchy condition $u(x_1,x_1) = \frac 12 x_1$. And I am asked to find $u=(x_1,x_2)$ which satisfies this PDE.

My attempt is:

Using the Cauchy problem formula $a_1(x_1,x_2)u_{x_1} + a_2(x_1,x_2)u_{x_2}=b(x_1,x_2,u(x_1,x_2))$ I get that the coefficients of $a_1 = u$, $a_2 = 1$, $b=1$

I then go to paramaterise the cauchy curve $\gamma (s) = (x_{1_0}(s), x_{2_0}(s)) = (s,0)$

This leads to $z_0(s) = u_0(x_{1_0}(s), x_{2_0}(s)) = \frac s2$

$\tilde x_{1_\tau} = a_1(\tilde x_1,\tilde x_2, \tilde z)=\tilde z$, $\tilde x_{2_\tau} = a_2(\tilde x_1,\tilde x_2, \tilde z)=1$, $\tilde z_\tau = b(\tilde x_1,\tilde x_2, \tilde z)=1$

From these 3 equations I get that $\tilde z = \tau + A(s)$ where $A(s)$ is a constant. Then I find $\tilde x_2 = \tau + B(s)$ where $B(s)$ is a constant. Finally I get $\tilde x_1 = \frac {\tau^2}2 + \frac {s\tau}2 +C(s)$ where $C(s)$ is a constant.

Using the initial conditions earlier specified I get $\tilde z = \tau + \frac s2$, $\tilde x_2 = \tau$ and $\tilde x_1 = \frac {\tau^2}2 + \frac {s\tau}2 +s$

I have gotten this far, but I am unsure whether what I have done up to this stage is correct, or what the next step is to take.

Any hints or help would be great. Thank you.

I know this question has been answered in another place, however the method the used was different to the one I have attempted to use. I am struggling to follow the other method, so would still be gracious if someone could help me out


1 Answer 1


Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx_2}{dt}=1$ , letting $x_2(0)=0$ , we have $x_2=t$

$\dfrac{du}{dt}=1$ , letting $u(0)=u_0$ , we have $u=t+u_0=x_2+u_0$

$\dfrac{dx_1}{dt}=u=t+u_0$ , letting $x_1(0)=f(u_0)$ , we have $x_1=\dfrac{t^2}{2}+u_0t+f(u_0)=\dfrac{x_2^2}{2}+(u-x_2)x_2+f(u-x_2)=x_2u-\dfrac{x_2^2}{2}+f(u-x_2)$ , i.e. $u-x_2=F\left(x_1-x_2u+\dfrac{x_2^2}{2}\right)$

$u(x_1,x_2=x_1)=\dfrac{x_1}{2}$ :


$\therefore u-x_2=-\dfrac{x_1}{2}+\dfrac{x_2u}{2}-\dfrac{x_2^2}{4}$





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