# Solving a PDE: basic first-order hyperbolic equation $u_t = -u_x$

So I have to solve the first-order hyperbolic equation $u_t = -u_x$. It is a PDE, since there is the time and spatial variable, but I'm overwhelmed by the maths given in books of how this works. Could anyone help me throught the steps?

The conditions are $0\leq x\leq 1$ and $u(0,x)=0$ and $u(t,0)=0, t \geq 0$.

So I figure that I have to get some ODE. But am not sure how..

Best Regards

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This is a form of a transport equation, standard first semester PDE problem. Probably in a chapter about characteristics?

First, the "divinely inspired" solution; consider a solution of the form $u(t,x) = f(x-t)$. Then, $u_t=-f'(x-t)$ and $u_x=f'(x-t)$, implying $u_t=-u_x$. With that said, your initial conditions yield $f(x-0)=f(x)=0$. In other words, your solution is identically zero (at least on the domain of interest).

Why is this? Well, you write a linear PDE with 0 assigned to the side conditions. This sort of means that your solution must be 0 -- its a homogeneous equation, so 0 is definitely a solution (existence), and 0 also happens to satisfy your side conditions. Since we have uniqueness (think back to ODEs, similar sort of idea), 0 is the solution.

Second, more interestingly, how did I know so "guess" a form of $u(t,x)=f(x-t)$? Recall from... high school physics (?) that sometimes problems have a "natural frame of reference". (Back in AP Physics, that was often rotating your axes 45 degrees). In first order PDEs, the same idea (but different in the details, of course) is often at play -- our transport equation has a "natural" frame of reference -- that is, moving along the curve $z=x-t$. You'll learn in your class how to identify such natural frames of reference; it really is a powerful technique.

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So the solution is zero for all t and x? I have to make a 3D plot from t=0 to t=2. And I have to solve it using a ODE solver; it doesn't look like there is any ODE whatsoever? –  user61001 Feb 28 '14 at 7:59
Correct; as I hinted at, you actually don't need any fancy math -- if we can find one solution that works, existence and uniqueness gives us the rest. Clearly, $u(t,x)=0$ satisfies the PDE, as well as the initial and boundary conditions. Done. If u(0,x) were something other than 0 , then it gets more interesting. –  Jason Feb 28 '14 at 8:50

One can use separation of variables to solve it. Let $u=X(x)T(t)$ then $u_x=X'T,u_t=XT'$ and so $$X'T=-XT'\implies \frac{X'}{-X}=\frac{T'}{T}=c$$ Now we have two ordinary differential equations $$X'=-cX\implies \ln X=-cx+a\implies X=Ae^{-cx}$$ and $$T'=cT\implies \ln T=ct+b\implies X=Be^{ct}$$ The solution is then $$u(x,t)=Ce^{c(t-x)}$$Using the given conditions we get $$0=Ce^{c(-x)}\ \ \ \ \ \text{ and } \ \ \ \ 0=Ce^{c(t)}$$ These conditions implies that $u(x,t)\equiv 0.$

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