Domain of definition for $u_x + uu_y = 1$

Question

How do i find the domain of definition for $u_x + uu_y = 1$ with $u = x/2$ on $y=x$ , $0 \leq x \leq 1$

I parametrise by letting $y=s$ , $x=s$ , $u=s/2$ , $0 \leq s \leq 1$ at $t=0$

The characteristic equations are: $dx/dt = 1$, $dy/dt = u$, $du/dt = 1$

Solving $dx/dt = 1$ gives $x=t +s$

Solving $du/dt = 1$ gives $u=t+ s/2$

Solving $dy/dt = u$ gives $y = (1/2)t^2 + st/2 + s$

This is where I get stuck,

$t=x-s$ so $y = (1/2)(x-s)^2 + (1/2)s(x-s) + s$

and $0 \leq s \leq 1$ so am i correct in saying that the domain of definition is the region between $y = (1/2)x^2 $ and $y = (1/2)(x-1)^2 + (1/2)(x-1) + 1$

Answer 1

$$u_x+uu_y= \tag 1$$ Your three equations written on a equivalent form: $$\frac{dx}{1}=\frac{dy}{u}=\frac{du}{1}=dt$$ A first family of characteristic equations comes from $\frac{dx}{1}=\frac{du}{1}$ $$u-x=c_1$$ A second family of characteristic equations comes from $\frac{dy}{u}=\frac{du}{1}$ $$\frac{u^2}{2}-y=c_2$$ The general solution of the PDE expressed on the form of implicite equation is : $$\frac{u^2}{2}-y=F(u-x) \tag 2$$ where $F$ is an arbitrary equation, to be determined according to boundary condition : $$u(x,x)=\frac{x}{2}\quad\implies\quad\frac{x^2}{8}-x=F\left(\frac{x}{2}-x\right)$$ $X=-\frac{x}{2}\quad;\quad x=-2X$ $$F(X)=\frac{(-2X)^2}{8}-(-2X)=\frac{X^2}{2}+2X$$ So, the function $F(X)$ is determined. we put it into the general solution Eq.$(2)$ where $X=u-x$ $$\frac{u^2}{2}-y=\frac{(u-x)^2}{2}+2(u-x)$$ After simplification : $$u(x,y)=\frac{y+\frac{x^2}{2}-2x}{x-2}$$ You can find the domain of definition.

Answer 2

This PDE is a non-homogeneous inviscid Burgers' equation, for which the same question has been asked here (see previous link for details). The method of characteristics introduces the characteristic curves $y = x(x-s)/2 + s$ for $0<s<1$, along which $u$ is given by (cf. answer by JJacquelin) $$ u(x,y) = \frac{x^2/2 - 2x + y}{x-2} \, . $$ This expression is not defined at $x=2$, where the denominator vanishes. On the one hand, if we start following a characteristic curve at some point $(s,s)$ with $0<s<1$, we can decrease $x$ without restriction. On the other hand, we cannot increase $x$ further than $x=2^-$. At the point $(x,y) = (2,2)$, all characteristics coming from $\lbrace(s,s), 0<s<1\rbrace$ intersect: the classical solution collapses. Finally, the solution is only valid over the domain in yellow $$ (x,y) \in \big\lbrace \big(x, x(x-s)/2+s\big),\; 0<s<1,\; x<2\big\rbrace $$