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

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.

• It's not true in your case that $u$ is constant along characteristics – after all, it's governed by the ODE $\partial z/\partial s = z$, not $\partial z/\partial s=0$. Sep 1, 2021 at 11:59

$$u_x+u_t=u$$ 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 $$F(x-t)=e^{-(x-t)}\cos(x-t)$$ : $$u(x,t)=e^xe^{-(x-t)}\cos(x-t)$$ $$u(x,t)=e^t\cos(x-t)$$

• The solutions that you and @EditPiAf game me are quite simpler (although I'm not sure if it is ok to use them in my exams). Is it possible to mimic them for first order quasilinear PDE's? Just a yes or no answer. Sep 3, 2021 at 10:23
• Yes. No need for mimic, just apply : en.wikipedia.org/wiki/Method_of_characteristics Sep 3, 2021 at 10:41

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).

• Why $t(0)=0$ and not $t_0$? And how do I get so quickly that $z(0)=\cos (x_0)$? Sep 2, 2021 at 15:49
• The explanation that I gave is right? I am still not sure. I understand yours. @EditPiAf Sep 3, 2021 at 8:32
• @ΝικολέταΣεβαστού We can do the same thing by setting the boundary at an arbitrary value $s=s_0$ along which $z(s_0) = \cos x_0$ and $(x(s_0), t(s_0)) = (x_0, 0)$. Doing so, we have $$x = s-s_0+x_0, \qquad t = s-s_0, \qquad z = \cos(x_0) e^{s-s0}$$ which yields the exact same final expression. Sep 3, 2021 at 10:27
• I was talking about the answer below. Sorry I wasn't clear. The one that I wrote. Sep 3, 2021 at 10:29

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.

• The fact that $t-x = t_0 - x_0$ is constant along any characteristic curve follows from $\frac{d (t-x)}{d s} = 0$. That being said, this condition is not sufficient to complete the resolution, and the boundary condition at $t=0$ needs to be enforced. This can be done by noting that $u e^{-t}$ is constant too. Sep 3, 2021 at 10:33