What is the meaning of a characteristic curve when solving PDE?
For example, in solving $u_x + yu_y = 0 $, we get $dy/dx = y/1$ and so solving this ODE, we obtain $y=Ce^x $.
Here, what does $y=Ce^x $ really mean?
Does it mean that $x$ and $y$ have to be in the relation $y=Ce^x $?
Drawing the characteristic curves $y=Ce^x $, how does it help us determine the solution $u$?
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$\begingroup$So now if you set$$ u(x,y)=v(x,\tilde y)=v(x,e^{-x}y) $$the PDE transforms to$$ u_x=v_x-v_{\tilde y}ye^{-x}\\ u_y=v_{\tilde y}e^{-x}\\ \implies 0=u_x+yu_y=v_x $$so that $v$ is constant in its first argument and only variable in its second argument, $v(x,\tilde y)=\phi(\tilde y)$. In consequence$$ u(x,y)=\phi(e^{-x}y). $$What the function $\phi$ is and if it exists and is unique now depends on the initial conditions.
The theory of characteristic curves formalizes this procedure, it finds curves along which the first order PDE essentially behaves like a scalar ODE system. Along any curve $x(t),y(t)$ one can define $z(t)=u(x(t),y(t))$ and then compare its time derivative$$ \dot z=u_x\dot x+u_y\dot y $$with the linear PDE$$ a(x,y,u)u_x+b(x,y,u)u_y=c(x,y,u). $$One sees that if one ensures that $\dot x=a(x,y,z)$ and $\dot y=b(x,y,z)$, then also $\dot z=c(x,y,z)$. So if the initial condition $z_0(r)=u(x_0(r),y_0(r))$ is unique for the solution of the resulting IVP, then its solution gives the value of the PDE solution along this curve.
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