solve $uu_{x_1}+u_{x_2} = 1$ using Method of Characteristics

Question

I am solving this PDE: $$uu_{x_1}+u_{x_2} = 1; u(x_1,x_1)= \frac {x_1}2$$

For the characteristic equations I have:

$$(x^1)' = z$$

$$(x^2)' = 1$$

$$z' = 1$$

Attempting to solve these ODEs I get: $$x^2(s) = x^1_o + s$$ but I am confused on how to solve the first and third characteristic ODEs.

Also, how do I know what z(s) is?

Sorry, I am new to PDEs and this method but am trying to learn.

Answer 1

$$ \begin{cases} \dfrac {dx_1}{ds}=u \\ \dfrac {dx_2}{ds}=1 \\ \dfrac {du}{ds}=1 \end{cases} $$ $$ \implies \begin{cases} \dfrac {dx_1}{ds}=u \\ x_2=s+x_2(0) \\ u=s+u(0) \end{cases} $$

First equation is easy to integrate $$\frac {dx_1}{ds}=u $$ $${dx_1}=\int u ds $$ $$\implies x_1=\int s+u(0) ds \\$$ $$ x_1=\frac { s^2}2+u(0)s+x_1(0) \\$$

Apply intitial conditions now $$ \begin{cases} x_1(0)=r \\ x_2(0)=r \\ u(0)=\frac {r}2 \end{cases} $$ Therefore: $$ \begin{cases} x_1=\dfrac { s^2}2+s\dfrac r2+r \\ x_2=s+r \\ u=s+\dfrac {r}2 \end{cases} $$

$$ \begin{cases} x_1=\dfrac { s^2}2+s\dfrac {r}2+r \\ x_2=s+r \\ \end{cases} \implies \begin{cases} s=2\dfrac { x_1-x_2}{x_2-2} \\ r=x_2- s\\ \end{cases} $$ From the third equation we have: $$u(x_1,x_2)=s+\dfrac r2$$ $$u(x_1,x_2)=\frac s2+\dfrac {x_2}2$$ $$u(x_1,x_2)=\dfrac { x_1-x_2}{x_2-2}+\dfrac {x_2}2$$ Finally, $$\boxed {u(x_1,x_2)=\dfrac { x_1-2}{x_2-2}+\dfrac {x_2}2-1}$$

Answer 2

We are given

$$uu_{x_1}+u_{x_2} = 1$$ $$u(x_1,x_1)= \frac{x_1}2$$

The initial data curve is $\Gamma: \langle s,s,s/2 \rangle$. Therefore, the Jacobian evaluated with the boundary data is

$$J\Bigg|_\Gamma=\begin{vmatrix} {x_1}_s & {x_1}_t \\ {x_2}_s & {x_2}_t \end{vmatrix}\Bigg|_\Gamma=\begin{vmatrix} 1 & \dfrac{s}{2} \\ 1 & 1 \end{vmatrix}=1-\frac{s}{2}$$

So, there is a unique solution in the neighborhood of $\Gamma$ provided $s\neq 2$ . The characteristic equations are

\begin{cases} \dfrac{dx_1}{dt}=u, & x_1(s,0)=s \\[1em] \dfrac{dx_2}{dt}=1, & x_2(s,0)=s \\[1em] \dfrac{du}{dt}=1, & u(s,0)=\dfrac{s}{2} \end{cases}

For the third equation,

$$\frac{du}{dt}=1 \implies du=dt\implies u = t+c_1(s)$$

Plugging in the initial condition of $ u(s,0)=\dfrac{s}{2} \implies c_1(s)=\dfrac{s}{2}$. Hence,

$$\boxed{u=t+\frac{s}{2}}\tag{1}$$

For the second equation,

$$\frac{dx_2}{dt}=1 \implies dx_2 = dt \implies x_2=t+c_2(s)$$

Plugging in the initial condition of $ x_2(s,0)=0 \implies c_2(s)=s$. Hence,

$$\boxed{x_2=t+s}\tag{2}$$

For the first equation,

$$\frac{dx_1}{dt}=u \implies dx_1 =u\,dt \implies x_1=\frac{t^2}{2}+\frac{st}{2}+c_3(s)$$

Plugging in the initial condition of $ x_1(s,0)=s \implies c_3(s)=s$. Hence,

$$\boxed{x_1=\frac{t^2}{2}+\frac{st}{2}+s}\tag{3}$$

Combining $(1),(2),(3),$ we see that

$$s=\frac{x_2^2-2x_1}{x_2-2}$$ $$t=\frac{2(x_1-x_2)}{x_2-2}$$

and upon solving we find

$$\boxed{u=u(x_1,x_2)=t+\frac{s}{2}=\frac{2(x_1-x_2)}{x_2-2}+\frac{x_2^2-2x_1}{2(x_2-2)}=\frac{x_1-2}{x_2-2}+\frac{x_2}{2}-1}\tag{4}$$

in which the solution does not exist for $x_2=2$ as we found from our initial parametrization.