How to solve $uu_{x_1}+u_{x_2}=1$ with characteristic method?

Question

$$uu_{x_1}+u_{x_2}=1\text{ with initial condition }u(x_1,x_1)=\frac{1}{2}x_1$$ I have problem to use following characteristic method to solve it.

Let $$x(s) = (x_1(s), x_2(s))$$$$z(s) = u(x(s))$$$$p(s) = (u_{x_1}(x(s)), u_{x_2}(x(s))$$$$F(x,z,p) = zp_1 + p_2 - 1 = 0$$

Assume, $$\frac{dx}{ds} = F_p = (z,1) $$$$\frac{dz}{ds} = F_p \cdot p = 1$$$$\frac{dp}{ds} = -F_x - F_z p = 0 - p_1(p_1,p_2) = - (p_1^2,p_1p_2)$$

Now, we can have, $$z(s) = s + c_0$$$$\dot x_1 = s + c_0 \Rightarrow x_1 = \frac{1}{2}s^2 + sc_0 + c_1$$$$\dot x_2 = 1 \Rightarrow x_2 = s + c_2$$

I'm stucked here. I don't know what to do next.

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. $u-x_2=F\left(x_1-x_2u+\dfrac{x_2^2}{2}\right)$

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

$F(x_1)=-\dfrac{x_1}{2}$

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

$4u-4x_2=-2x_1+2x_2u-x_2^2$

$(2x_2-4)u=2x_1+x_2^2-4x_2$

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

Answer 2

$uu_{x_{1}} + u_{x_{2}} = 1, u(x_{1}, x_{1}) = \frac{1}{2} x_{1}$

Let \begin{cases} x(s) = (x_1(s), x_2(s))\\ z(s) = u(x(s))\\ p(s) = (u_{x_1}(x(s)), u_{x_2}(s)) \end{cases}

Need to solve the equation:

\begin{cases} \frac{dx_1}{ds} = z(s) \\ \frac{dx_2}{ds} = 1 \\ \end{cases}

Observe that,

\begin{cases} \frac{dz}{ds} = 1 \end{cases}

Therefore, we have

\begin{cases} \frac{dx_1}{ds} = z(s) \\ \frac{dx_2}{ds} = 1 \Rightarrow x_2 = s + c_0 \\ \frac{dz}{ds} = 1 \Rightarrow z = s + z(0) \end{cases}

where we let $x_1(0) = x_2(0) = c_0. $ and $z(0) = \frac{1}{2} c_0$

\begin{cases} \frac{dx_1}{ds} = z(s) \\ z(0) = \frac{1}{2} c_0 \\ z = s + z(0) \\ x_1(0) = c_0 \end{cases} $\Rightarrow x_1 = \frac{1}{2} s^2 + \frac{1}{2}c_0s + c_0$

Combine above, we now have

\begin{cases} x_1 = \frac{1}{2} s^2 + \frac{1}{2}c_0s + c_0 \\ x_2 = s + c_0 \end{cases} $\Rightarrow$ \begin{cases} s = \frac{2x_1 - 2 x_2}{x_2 - 2} \\ c_0 = \frac{x_2^2 - 2 x_1}{x_2 - 2} \end{cases}

$\text{Therefore, for point } (x_1,x_2) \in \mathbb{R}^2 \text{are at the line } x_{c_0}(s) \text{ coefficient } c_0 = \frac{x_2^2 - 2 x_1}{x_2 - 2} \text{ and parameter is } s = \frac{2x_1 - 2 x_2}{x_2 - 2} \text{, that is, } x_{c_0}(s) = (x_1,x_2).$

$\text{Also, we know that } u(x_1, x_2)=z(x_{c_0}(s) = s + \frac{1}{2}c_0 \text{. Then, } u(x_1,x_2) = \frac{x_1 - 2 x_2 + \frac{1}{2} x_2^2}{x_2 - 2} $