Linear first order PDE $u_x+u_t=u$ with the method of characteristics

Question

I miss something so that I can understand how my teacher finds the solution. I will write the exercise as it is written first.

$$u_x+u_t=u, x\in \mathbb{R}, t>0$$ $$ u(x,0)=\cos x$$ Then I have: $$\frac{\partial x(s)}{\partial s}=1, x(0)=x_0 $$ $$\frac{\partial t(s)}{\partial s}=1, t(0)=t_0 $$ $$\frac{\partial z(s)}{\partial s}=z, z(0)=u(x_0,t_0) $$ I get $$\frac{\partial t}{\partial x}=\frac{\partial t}{\partial s}\frac{\partial s}{\partial x}=1 \Rightarrow t-x=c, t(x_0)=t_0$$ $$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial s}\frac{\partial s}{\partial x}=z\Rightarrow z(x)c e^x$$ We conclude that $t-x$ is constant for all $s$, so $t-x=t_0-x_0$ and from initial conditions

$ z(x_0-t_0)=u(x_0-t_0,0)=\cos(x_0-t_0)$ and $z(x_0-t_0)=ce^{x_0-t_0}$ (I don't understand how I got the first equation)

So $z(x_0)=ce^{x_0}=e^{x_0-t_0}\cos(x_0-t_0)e^{x_0}=\cos(x_0-t_0)e^{t_0} \Rightarrow u(x_0,t_0)=\cos(x_0-t_0)e^{t_0}$ which means that $u(x,t)=\cos(x-t)e^{t}, x \in \mathbb{R}, t>0$

This is my teacher's solution. There is one other example, before this, that explains the method of characteristics which I think I have unsterstood (it's the linear transport equation). From research on the internet and some books I found, I got the idea that $u$ (any book I found doesn't explain why) is constant for $(x,t)$ belonging to the characteristic curve. Well I haven't yet understood why. My teacher says later in her notes that with the characteristics we create the surface of the solution (little-by-little for every value). I found a video and some posts here like this that explain the method but in the examples $\frac{\partial z}{\partial s}=0$ and I can't see how to manage different and more complicates cases. It seems to me that some steps of the solution are missing and I have nowhere found the ideas. In the link above the steps are more clear but I don't understand why we supposed that $s=t$. I just need some explanation around the steps and the thinking I have to do every time (for linear first order PDE generally)and how we thought of using $ x_0-t_0$ in $z(x_0-t_0)$. I think that everything else has been understood. Any explanation is very welcome because I am trying to understand this exercise since yesterday.

Answer 1

$$u_x+u_t=u$$ Charpit-Lagrange characteristic ODEs : $$\frac{dx}{1}=\frac{dt}{1}=\frac{du}{u}=ds$$ This is equivalent to the three equations that you correctly wrote.

A first characteristic equation comes from solving $\frac{dx}{1}=\frac{dt}{1}$ : $$x-t=c_1$$ A second characteristic equation comes from solving $\frac{dx}{1}=\frac{du}{u}$ : $$u\:e^{-x}=c_2$$ The general solution of the PDE expressed on implicit form $c_2=F(c_1)$ is : $$u\:e^{-x}=F(x-t)$$ $F$ is an arbitrary function (to be determined later according to a specified condition). $$\boxed{u(x,t)=e^xF(x-t)}$$ Condition : $u(x,0)=\cos(x)=e^xF(x)$ $$F(x)=e^{-x}\cos(x)$$ So the function $F(x)$ is determined. We put it into the above general solution where the argument is not $x$ but is $(x-t)$. Thus $F(x-t)=e^{-(x-t)}\cos(x-t)$ : $$u(x,t)=e^xe^{-(x-t)}\cos(x-t)$$ $$u(x,t)=e^t\cos(x-t)$$

Answer 2

I think that your teacher's solution is a bit confusing. So let us apply the method of characteristics in parametric form (see the example in §2 of the Wikipedia article):

• $\frac{d x}{d s} = 1$, letting $x(0) = x_0$ we know $x(s) = s + x_0$
• $\frac{d t}{d s} = 1$, letting $t(0) = 0$ we know $t(s) = s$
• $\frac{d z}{d s} = z$, letting $z(0) = \cos x_0$ we know $z(s) = \cos(x_0) e^s$

This system simply expresses the Cauchy problem for the PDE $u_x + u_t = u$ in differential form $$ \frac{d z}{d s} = u_x \underbrace{\frac{d x}{d s}}_{\equiv 1} + u_t \underbrace{\frac{d t}{d s}}_{\equiv 1} = \underbrace{u}_{\equiv z} \, , $$ where we have introduced the parametrization $z(s) = u(x(s), t(s))$ of the solution (${d z}/{d s}$ denotes a total derivative). The boundary condition $u(x_0, 0) = \cos x_0$ is enforced at $s=0$ by setting $$ x(0) = x_0 ,\qquad t(0) = 0 ,\qquad z(0) = \cos x_0 . $$ This condition represents the set of points where $u$ is already known, i.e. where characteristic curves $s \mapsto (x(s), t(s))$ actually 'start'. Here we note that $z = \cos(x_0) e^s$ is not constant along the characteristic curves: it increases exponentially with the parameter $s$. To express $u$ in terms of the coordinates $x, t$, we then use the first two equations above to eliminate $x_0=x-t$ and $s=t$: $$ u(x,t) = \cos(x-t)\, e^t . $$ See the answer by @JJacquelin for the parametrization invariant form (Lagrange-Charpit system).

Answer 3

Well I think that's what is missing.

We want the charactreristic to "meet" the boundary of the set $\{(x,t):x \in \mathbb{R}, t>0\}$ which is the $ x$- axis. So when we find the characteristic $t-x=t_0-x_0$ (here) we enforce $t=0$ and we find the point where they meet, which is $(x_0-t_0,0)$. This point is used in the initial condition $u(x,0)=\cos x$, because $z(s)=u(x(s),t(s)) \Rightarrow z(x)=u(x,t(x))$ (we usually omit the symbol that makes it possible to distinguish the new function that has $x$). We have $$ z(x_0-t_0)=u(x_0-t_0,0)(=\cos (x_0-t_0))$$ We always want to find a quantity that has our variables (here $t,x$) and doesn't have $s$. So when we solve the ODEs we try to make $s$ to vanish from our equations. The way is up to us.

EDIT: Just for any future reader that struggles. On the internet there are some very good lecture notes for first order PDE's from Stanford university here.