PDE:Method of characteristics

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solution $u_t + a u_x = 0, u(x,0)=f(x)\,$

solution $u_t + u u_x = 0, u(x,0)=x\,$

solution $y^{-1} u_x + u_y = 0, u(x,1)=x^2\,$

PDE BOOKS

solution $u_x + 2u_y = u^2, u(x,0)=h(x)\,$

solution $u_x + xu_y = u^2\,$

solution $u_x + xu_y -u_z = u\,$ , $u(x,y,1)=x+y\,$

solution $xu_x + u_y = y, u(x,0)=x^2\,$

solution $xu_x + yu_y + u_z = u, u(x,y,0) = h(x,y)\,$

solution $u_x + u_y + u = e^{x + 2y}\,$ , $u(x,0) = 0\,$

solution Show that if $z=u(x,y)\,$ is an integral surface of $V=<a,b,c>\,$ containing a point $P\,$ , then the surface contains the characteristic curve $\chi\,$ passing through $P\,$ . (Assume the vector field $V\,$ is $C^1\,$ ).

solution If $S_1\,$ and $S_2\,$ are two graphs $\left[ S_i = u_i(x,y), i=1,2\right]\,$ that are integral surfaces of $V=<a,b,c>\,$ and intersect in a curve $\chi\,$ , show that $\chi\,$ is a characteristic curve.

solution $(x+u)u_x + (y+u)u_y = 0\,$

solution $u_t+uu_x=0, u(x,0)=\begin{cases} 1, & x\le 0 \\ 1-x, & 0<x\le 1\\ 0, & x>1 \end{cases}\,$

solution Solve the initial value problem $a(u)u_x+u_y=0\,$ with $u(x,0)=h(x)\,$ and show the solution becomes singular for some $y>0\,$ unless $a(h(s))\,$ is a nondecreasing function of $s\,$ .