Find (in implicit form) the solution $u(x, y)$ of the equation: $$u_x+uu_y=x$$
when $$u(0, y) = 1 + \sin y \qquad \text{for} \qquad y > 0 $$ Show that your solution is uniquely determined in the region $( −1 < x < 1,\, y_L(x)\leqslant y < ∞)$ where $y_L(x)$ is a function you should find. Show that $u_y$ becomes unbounded as x → ±1 for certain values of y.
Working: I got the implicit form: $$(t, \frac{t^3}{6}+(\sin{s}+1)t +s, \frac{t^2}{2}+\sin{s}+1)$$ Then I calculated the Jacobian, $J\frac{(x,y)}{(s,t)}$ and got: $$J= -t\cos{s}-1$$ So J=0 when $t=-1/\cos{s}$ So then I got that, since $x=t$, for $$-1\leqslant x\leqslant1$$ we get $$ \qquad J\ne0$$
But now I'm not sure how to find $y_L(x)$, and how to answer the part about $u_y$. Any help would be much appreciated!
You got correctly that $$ dx=\frac{dy}z=\frac{dz}{x} $$ so that $z=b+\frac12x^2$ and $y=a+bx+\frac16x^3=a+xz-\frac13x^3$. As $a,b$ are constants of the characteristic, and in the surface of the function $z=u(x,y)$ the characteristic curves form a one-parameter family, the constants are dependent on each other, for instance via $b=\phi(a)$. The solution is then implicitly given via $$ z-\frac12x^2=\phi(y-xz+\frac13x^3)\implies 1+\sin(y)=\phi(y) $$ so that in total the solution is given via the implicit equation $$ u(x,y)=z=1+\frac12x^2+\sin(y-xz+\frac13x^3). $$