# Explaining the method of characteristics

I am learning about solving p.d.e.s by the method of characteristics at the moment. I was given an "algorithm" to solve these problems but I want to know also what is going on, how it works and what it is doing intuitively/physically/graphically. It would be great if someone could either explain it or provide a good link To reference material. Thank you.

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The basic idea is that we look for parametric curves along which the PDE tells us how the function changes. Suppose you have a smooth parametric curve in the $x-y$ plane: $x = X(t)$, $y = Y(t)$. Consider how a smooth function $u(x,y)$ changes as you move along the curve. The chain rule says $$\frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dX}{dt} + \frac{\partial u}{\partial y} \frac{dY}{dt}$$ Now this looks rather like the left side of a first-order PDE $$a(x,y) \frac{\partial u}{\partial x} + b(x,y) \frac{\partial u}{\partial y} = c(x,y)$$ In fact, if you can find $X(t)$ and $Y(t)$ such that $\frac{dX}{dt} = a(X(t),Y(t))$ and $\frac{dY}{dt} = b(X(t),Y(t))$, this tells you that along that curve $\frac{du}{dt} = c(X(t), Y(t))$, so that if you know $u(X(0), Y(0))$ you can get $$u(X(s),Y(s)) = u(X(0), Y(0)) + \int_0^s c(X(t),Y(t))\ dt$$
Now, for any point $(x_1, y_1)$ in the plane, suppose you want to find $u(x_1, y_1)$, where $u(x,y)$ satisfies the PDE $a(x,y) \frac{\partial u}{\partial x} + b(x,y) \frac{\partial u}{\partial y} = c(x,y)$ plus some boundary condition. Then you want to find a curve $x = X(t)$, $y = Y(t)$ satisfying the system of ordinary differential equations $\frac{dX}{dt} = a(X(t),Y(t))$ and $\frac{dY}{dt} = b(X(t),Y(t))$ that passes through the given point $(x_1, y_1)$, follow that curve to some $(x_0, y_0)$ where your boundary condition tells you the value of $u$, and then you can get $u(x_1, y_1)$ by an integral as above.