# Method of Characteristics or How I Started Worrying and Loathe Quasi-linear PDEs: Part 1

I emailed my adviser…at least I think she’s my adviser. I had an adviser, but was later given her as an adviser, then she went on sabbatical and was reassigned to my original adviser, but now he’s on sabbatical, so I’m not 100% sure. Regardless, she’s someone of mathematical authority who tells me what I do and I subsequently obey.

After failing my qual on financial mathematics I decided to change some things in my life. One of them is actually being proactive and begin preparing for my qual on Partial Differntial Equations (PDEs) in four months.Her advice was “be sure to solve any problem in Applied Partial Differential Equations” by Ockendon, Howison, Lacey, and Movchan. Which is doable if I can solve two a day. Well, that was one piece, the others were take different quals.

Quasi-linear PDEs are the second worst type of PDEs after and nonlinear  PDEs. Or to phrase it positively, the third best type of PDEs after linear PDEs and semi-linear PDEs.

Life is already complex, so we restrict ourselves to a function $u$ of two variables $x,y$. Anymore variables doesn’t provide additional insight, any less and it is not a PDE and this isn’t a post about how I loathe Ordinary Differential Equations (ODEs).

Suppose our function $u(x,y)$ satisfies the PDE $a(x,y,u)\frac{\partial u}{\partial x}+b(x,y,u)\frac{\partial u}{\partial y}=c(x,y,u)$. We can graph $u$ and get a surface laying in three dimensional space. We can rewrite our PDE as an inner product of two vectors $(a,b,c) \cdot (\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},-1) =0$.

$(\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},-1)$ is the normal vector to the surface $u(x,y)$, which means $(a,b,c)$ is in the tangent plane of $u(x,y)$ at any specific point.

Our plan of attack is introducing a “time” parameter to construct a curve $(x(t),y(t),u(t))$ that will satisfy our PDE and consequently trace a path on the solution surface of $u$. This approach allows us to convert our PDE into a system of ODEs, which makes life simpler.

Thus we need to solve $\frac{dx}{dt}=a(x(t),y(t),u(t)), \frac{dy}{dt}=b(x(t),y(t),u(t)), \frac{du}{dt}=c(x(t),y(t),u(t))$. We can’t fully solve a system of ODEs without specifying initial conditions or starting point. As we change the initial conditions we produce different curves. We add a second parameter to vary our starting points so we have a family of curves that when collected produces the two-dimensional surface of $u$. So at time $t=0$ we let $x(0)=x_0(s),y(0)=y_0(s),u(0)=u_0(s)$. Our choice of $x_0(s),y_0(s),u_0(s)$ will be determined by the boundary/initial condition of our original PDE.

Ockendon, Howison, Lacey, and Movchan’s give a cute problem. They try to model the process of proof-reading a textbook. They construct a family of functions $p_n(t)$ which represent the probability after time $t$ spent proof-reading, there are still $n$ errors in the draft. Then they construct the function $p(x,t) = \sum_{n=0}^\infty p_n(t)x^n$, and the resulting generating function satisfies the PDE $\frac{\partial p}{\partial t}+\mu(x-1)\frac{\partial p}{\partial x}=0$. We now need to set an initial or boundary condition. If the book contains $N$ errors, they set the initial condition $p(x,0)=x^N$ where $x$ is between 0 and 1.

If you’re having difficulties interpretation is confusing, it’s okay. I actually emailed Lacey trying to understand the intuition of $x$ and he replied the important part are the family of functions $p_n(t)$, and the generating function $p(x,t)$ is actually obscure. Besides, this isn’t a post about quantifying the editorial process, but about how to solve a certain class of PDEs.

We’ve already introduced a lot, and now have used $t$ in two completely different ways. So let’s rewrite their example PDE as $\mu(x-1)\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y} = 0$ to be consistent with the notation in the exposition. This is actually a pleasant introduction to the method of characteristics. This PDE is actually not just semi-linear but entirely linear (it’d be strictly semi-linear if the right-hand side was a function $c(x,y,u)$  instead of 0 (and then in the case $c(x,y,u)=g(x,y)u+h(x,y)$ we would then again get a linear PDE)), which is a nicer class to deal with than quasi-linear PDEs, and nonlinear PDEs are just big “fuck you”s. Our system of ODEs are $\frac{dx}{dt}=\mu(x-1),\frac{dy}{dt}=1,\frac{du}{dt}=0$. Our initial condition $u(x,0)=x^N$ hints that are initial conditions should probably be $x(0)=x_0(s)=s,y(0)=y_0(s)=0,u(0)=u_0(s)=u(x_0(s),y_0(s))=s^N$. I’m not going to do all the algebra for you.

Let’s solve. $\frac{dy}{dt}=1$, so $y(t)=t+d_y(s)$ (our constant of integration is actually a function of $s$ since our curve depends on the value of starting point parameter $s$), and $y(0)=d_y(s)$, since $y(0)=y_0(s)=0$ we have $y(t)=t$. $\frac{du}{dt}=0$, $u(t)=d_u(s)$, since $u(0)=u_0(s)=s^N$ we have $u(t)=s^N$. Now for the non-trivial ODE. $\frac{dx}{dt}=\mu(x-1)$, or $\frac{dx}{x-1}=\mu dt$. Thus $\ln |x-1| = \mu t+ d_x(s)$. By assumption $x$ is between 0 and 1, so $|x-1| = 1-x$ (I originally skipped this nuance and got the completely wrong answer), hence $1-x(t) = e^{d_x(s)}e^{\mu t}$. Plugging in $t=0$ we have $1-s=e^{d_x(s)}$. Therefore $1-x(t)=(1-s)e^{\mu t}$. I’m not going to do all the algebra for you.

Sometimes life isn’t simple and we can’t solve for $s$ and $t$ explicitly in terms of $x$ and $y$, and we just have to plot the characteristic lines but thankfully we can arrive at an explicit solution. We have $t=y$ and $s=1+(x-1)e^{-\mu t}$, and thus $u(x,y)=u(x(t),y(t))=u(t)=(1+(x-1)e^{-\mu y})^N$. I’m not going to do all the algebra for you.

Part 2 will cover a quasi-linear PDE with an explicit solution, while Part 3 will cover a quasi-linear PDE with a non-explicit solution expressed parametrically.

As always, all corrections and suggestions are welcomed.

EDITED: Rephrased a paragraph at the suggestion of Andrew Lacey.