I am working on this problem: Solve the following IVP using characteristic method. $$u_t + (u_x)^2 = t,\quad (x,t) \in \mathbb{R} \times (0, \infty)$$ $$u(x,0) = x, \quad x \in \mathbb{R}, t = 0$$
I rewrote it as $F(p,z,x) = (u_x ,1)p - t = 0$ and had some difficulty in selecting $s$ and $x^0$. Can anyone help me? Thanks in advance!
$$F(x,t,p,q,u)=q+p^2-t=0\qquad;\quad p=u_x\qquad;\quad q=u_t$$ $$F_x=0\quad;\quad F_t=-1\quad;\quad F_p=2p\quad F_q=1\quad F_u=0$$ $$\frac{dx}{F_p}=\frac{dt}{F_q}=\frac{dp}{-F_x-pF_u}=\frac{dq}{-F_t-qF_u}=\frac{du}{pF_p+qF_q}$$ $$\frac{dx}{2p}=\frac{dt}{1}=\frac{dp}{0-p*0}=\frac{dq}{-(-1)-q*0}=\frac{du}{p(2p)+q*1}$$ $$\frac{dx}{2p}=\frac{dt}{1}=\frac{dp}{0}=\frac{dq}{1}=\frac{du}{2p^2+q}$$ A first characteristic equation comes from $dp=0$ $$p=c_1$$ There's not even any need for further search of other characteristic equation because $$u_x=c_1\quad\implies\quad u=c_1x+f(t)$$ $$u_t=f'(t)\quad;\quad f'(t)+(c_1)^2=t\quad\implies f(t)=\frac12 t^2-(c_1)^2t+c_2$$ $$u=c_1x+\frac12 t^2-(c_1)^2t+c_2$$ Condition : $$u(x,0)=x=c_1x+c_2\quad\implies\quad c_1=1\quad\text{and}\quad c_2=0$$ $$u(x,t)=x+\frac12 t^2-t$$
