Initial data and method of characteristic

Question

So I'm still trying my VERY best to wrap my head around the method of characteristic. It doesn't seem like a difficult topic but the concept is still fairly fuzzy in my head.

Let's suppose an example:

$xU_x + (x+y)U_y=1$

with an initial data of $U(1,y)=y$

Why is $x(0) = 1$, $y(0) = \xi$ and $u(0) = \xi$?

How do I interpret this? I need to be able to understand this concept.

Answer 1

As mentioned in this answer, the idea of the method of characteristics (for first-order equations; second-orders are rather different) is that you want to create a new basis for the domain that is better suited to the problem than just $(x,y)$: in particular, we choose one basis vector/variable, often called $s$, so that the partial differential equation is reduced to $$ \frac{dU}{ds} = 1, \tag{1} $$ in your example, which is an ODE which we can solve as usual. We need another variable: the idea of this one, normally called $\xi$, is to parametrise the initial data. In other words, changing $s$ tells us how the initial data changes along a characteristic, whereas varying $\xi$ changes what point of the initial data we're looking at.

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Now, in your case, since you're concerned with the variable $\xi$, I'll just note that we can produce characteristics $$x^C = ae^s, \quad y^C = a se^s + be^s, $$ (the upper index $C$ denoting "on the characteristic") and then along each of these, we get the evolution equation (1).

Here I shall now actually answer your question: in this case, the initial data is given on the line $x=1$. We can write this usefully with the parametric equations $$ x^I = 1, \quad y^I = \xi, $$ ($I$ being "initial") and then our initial condition $U(x^I,y^I)=y^I$ becomes $$ U(1,\xi) = \xi. $$

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Remark: sometimes the problem people have in understanding comes from the first examples of characteristics using quite simple sets for initial values, such as one of the axes. Suppose instead I had the initial data $u(x^I,y^I)=1+x^I$ for $(x^I,y^I)$ on the parabola $y^2=4ax$. You probably remember this has a simple parametrisation, and maybe even that the parametrisation is $(x,y)=(at^2,2at)$. Therefore I can parametrise the initial data on this parabola by taking $$ x^I = a\xi^2, \quad y^I = 2a\xi, $$ and the initial data is then parametrised as $$ u(a\xi^2,2a\xi) = 1+a\xi^2. $$

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Okay, back to your problem. What we haven't done yet is find the specific characteristics that come from each point on the initial data curve. This means finding $a$ and $b$ in terms of $\xi$ so that (usually) $s=0$ puts the point $(x^C,y^C)$ on the initial data curve at the point with parameter $\xi$. For us, the point where $s=0$ is just $(a,b)$, so we are immediately able to discover that $(a,b)=(1,\xi)$, and so the characteristic that begins at the point $(1,\xi)$ is $$ x^C = e^s, \quad y^C = (s + \xi)e^s. $$ Then we solve the ODE (1): $U(s,\xi) = s + c, $ where $c$ is chosen, of course, so that $U(s=0,\xi)=\xi$, the initial condition we discussed above. Clearly here we get $c=\xi$.

Therefore $U(s,\xi) = s+\xi$. Transforming back, we find that $U(x,y) =y/x$.

We'd better check this works, if only to confirm our reasoning. $$ U(1,y) = y/1=y, \\ xU_x + (x+y)U_y = x(-y/x^2) + (x+y)/x = 1, $$ as we hoped.

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To answer your questions in the comments:

1. I'd have to see examples: they look like the characteristics for the wave equation, which everyone takes as $x \pm ct = \text{constant}$. It does seem surprising if the initial data are using them. If you post another question with more details, you'd probably get a better answer.
2. Initial data is one name we use for the given values of the function (in this case, $U(1,y)=y$). You may also call it boundary data, or boundary conditions. In the Method of Characteristics, often we think of $s$ as a time variable (because it looks like the data evolves along the characteristics from where they intersect the boundary data), so we call the line given by $s=0$ the initial data, and use words associated with time evolution (like "propagates", "evolves" and so on.) We have to take our general solution and make it agree with the given data.
3. I'm not sure what you mean here. $U$ does nothing to determine the characteristics. (See 4.)
4. Solving the equations for $x^C$ and $y^C$ in terms of $s$ to construct the characteristics, because they are two first-order equations, we acquire two constants of integration, that do not depend on $s$. These can be thought of as determining "which characteristic we are on", or to look at it the other way, if we know what one point on a characteristic is, the characteristic equations tell us the characteristic curve that passes through that point (assuming no nasty degenerate cases). It's like if I have the line $y-y_0 = m(x-x_0)$, and parametrise it as $ (s+x_0,m(s+y_0)) $, I know that at $s=0$, the point is $(x_0,y_0)$. In our case, we want $(x^C,y^C)$ to lie on the initial data curve when $s=0$, which is parametrised by $\xi$, so $a$ and $b$ will be functions of $\xi$.