I was just introduced to method of characteristics for solving PDE's. We solved the wave equation that is inifinitely long using this method. However I am very confused about this method. Here is a question from the book I am trying to do. It says to solve the first order equations for $u(x,t)$ using this method. No initial conditions are given. $$ 2 \frac{\partial u}{\partial t} + 3 \frac{\partial u}{\partial x} = 0$$

There is an example in the book (Haberman) that solves PDE in this form $$ \frac{\partial u}{\partial t} + c \frac{\partial u}{\partial x} = 0$$

no naturally, for my solution I divide everything by $2$ to get $$ \frac{\partial u}{\partial t} + \frac{3}{2} \frac{\partial u}{\partial x} = 0$$ Then I let $u(x,t) = u(x(t),t)$ and so if I take the derivative of this I get $$\frac{du}{dt}(x(t),t) = \frac{\partial u}{\partial t} + \frac{\partial u}{\partial x} \frac{dx}{dt}$$ so it must mean that $\frac{dx}{dt} = \frac{3}{2}$ so this is easy enough so we get a solution $$ x = \frac{3}{2}t + x_0$$

okay but I need this in terms uf $u(x,t)$. I am lost at this point on what to do since the book continues using intial conditions.

Similarly I have a second question that is bothering me even more $$ 2 \frac{\partial u}{\partial t} + 3 \frac{\partial u}{\partial x} = 1$$

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    $\begingroup$ Your second question seems to indicate that you haven't actually looked at the Wikipedia article yet? $\endgroup$ – joriki Dec 5 '11 at 4:44
  • $\begingroup$ I just read the wikipedia article and have made adjustments to my first question. The fact is that I shouldn't really divide by 2. Infact I can just find two characteristic equations? $\endgroup$ – Tyler Hilton Dec 5 '11 at 4:54
  • $\begingroup$ It's hard to tell what the problem is because we don't have your book. We do have the Wikipedia article, however, so it would help if you describe the problems you run into if you try to apply the method as described there. (In fact, the example in that article is almost exactly your problem.) $\endgroup$ – joriki Dec 5 '11 at 5:05

You're on the right track, but so far you haven't used the most important fact about characteristics: that the solution is constant along them. It is correct that for your first question, the characteristics are given by $x(s) = 3/2(s-t_0)+ x_0$. But how do we use this?

Suppose we are given the initial condition $u(x, t_0) = f(x)$. To find the value of the solution at a point $(x,t)$, we have to find a characteristic that passes through $(x,t)$. But this is easy: $$ x(s) = \frac{3}{2} (s-t_0) + x_0 $$ When $s=t$ this is supposed to take the value $x$, so we find $$ x_0 = x - \frac{3}{2}(t-t_0) $$ Plugging it back in, we have $$ x(s) = \frac{3}{2}(s-t_0) + x - \frac{3}{2}(t-t_0) $$ Since $u(x,t)$ is constant along characteristics, we have $$u(x,t) = u(x(s), s) = u(x(t_0), t_0) = f(x(t_0))$$ where we used the initial condition $u(x, t_0) = f(x)$. Using the equation for $x(s)$, we find $$ x(t_0) = x - \frac{3}{2}(t-t_0) $$ Hence the solution is $$ u(x,t) = f(x - \frac{3}{2}(t-t_0)).$$

For your second question, consider the function $u(x(s),s) - s$ and you'll find that it reduces to the first question.


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