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I am having difficulties to understand that how the method of characteristics works. I will clarify my question with an example. Consider a real valued function of two real variables $u(x,y)$ which satisfies the following first order partial differential equation (PDE)

$$\frac{{\partial u}}{{\partial x}}(x,y) + a\frac{{\partial u}}{{\partial y}}(x,y) = 0\tag{1}$$

where $a$ is some real constant. We know that

$$\left\{ \begin{array}{l} x = x\\ y = y\\ z = u(x,y) \end{array} \right.\tag{2}$$

is a surface with its parameters being $x$ and $y$. The PDE is telling us that the normal to the surface,$\left( {{{\partial u} \over {\partial x}},{{\partial u} \over {\partial y}}, - 1} \right)$, is perpendicular to the vector $\left( {1,a,0} \right)$. In other language, the vector $\left( {1,a,0} \right)$ is tangent to the surface.

Next, we define a curve on the surface as follows

$$\left\{ \matrix{ x = x(s) \hfill \cr y = y(s) \hfill \cr z = z(s) = u(x(s),y(s)) \hfill \cr} \right.,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \matrix{ {{dx} \over {ds}} = 1,\,\,\,\,\,\,\,x({s_0}) = {x_0} \hfill \cr {{dy} \over {ds}} = a,\,\,\,\,\,\,\,y({s_0}) = {y_0} \hfill \cr} \right.\tag{3}$$

then the PDE tells us that

$${{dz} \over {ds}} = 0,\,\,\,\,\,\,\,z({s_0}) = u({x_0},{y_0})\tag{4}$$

Now, I solve the first order ordinary differential equations (ODEs) in $(3)$ and $(4)$ to get

$$\left\{ \matrix{ x = (s - {s_0}) + {x_0} \hfill \cr y = a(s - {s_0}) + {y_0} \hfill \cr z = u({x_0},{y_0}) \hfill \cr} \right.\tag{5}$$

hence, I found my special curve on the surface $u(x,y)$. Here is where I got stuck! :)


Question

How can I proceed to get the surface $u(x,y)$ itself? I need some explanation that describes a systematic procedure to achieve this in conjunction with its geometric interpretation. Any helps will be appreciated! :)

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  • $\begingroup$ I think you need some boundary data; otherwise the solution is not at all unique. $\endgroup$ – Joey Zou Nov 3 '15 at 15:30
  • $\begingroup$ @JoeyZou: I am looking for all functions satisfying the PDE, regardless of boundary conditions! :) Are you familiar with the method of characteristics? $\endgroup$ – H. R. Nov 3 '15 at 16:12
  • $\begingroup$ In that case I think you can specify an artificial boundary condition (e.g. let $u_0(x) = u(x,0)$ be arbitrary) and solve from there. Because you're almost done based on the work you have. $\endgroup$ – Joey Zou Nov 3 '15 at 16:18
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Let $u_0(x) = u(x,0)$. Now, if we started with some $(x_0,y_0)$ such that $y_0=0$, then by what you have above we have $$ u(x,y) = u(x(s),y(s)) = z(s) = u(x_0,y_0) = u_0(x_0) $$ where $x = x_0 + s$ and $y = y_0 + as = as$ (take $s_0 = 0$ for convenience). This means that $s = \frac{y}{a}$, and hence $x_0 = x - \frac{y}{a}$, so $$ u(x,y) = u_0\left(x-\frac{y}{a}\right).$$


The above is how you would solve it provided boundary conditions, which are necessary to get a unique solution (very much like how, in solving first-order ODEs, you need some initial condition to specify a unique solution). But what if you wanted all solutions? Well, just notice that in the above case, we didn't put any conditions on the "boundary values" at $y=0$ (aside from it being differentiable so that we can have a differentiable solution). In other words, the boundary value $u_0$ is arbitrary, i.e. any differentiable $f$ can serve as the boundary value, and we would get a solution. So the solution to this PDE is $$ u(x,y) = f\left(x - \frac{y}{a}\right) $$ where $f$ is an arbitrary differentiable function (akin to the arbitrary constant in 1-dimensional indefinite integration).


Of course, in solving the boundary value problem, we didn't have to specify the boundary values at $y=0$. What if we looked at the values at $x=0$ instead, and set $u_0(y) = u(0,y)$? Then we would want to solve the equations $x = x_0 + s = s$ and $y = y_0 + as$, with $u(x,y) = u_0(y_0)$. We then have $y_0 = y-ax$, and hence $$ u(x,y) = u_0(y-ax) $$ for a specified initial data $u_0$, so the general solution is $$ u(x,y) = f(y-ax) $$ for an arbitrary differentiable $f$. I'll leave it to you to check that the two answers I've given here are actually the same. :)

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  • $\begingroup$ Thanks for the answer! :) Voted Up! I have seen all these things in the notes and books on the subject! but, what are we really doing? I don't get it! First, we find a curve on the surface for which the $u(x,y)$ is constant on it! Then what? I think there is more to understand behind these algebraic manipulations! :) $\endgroup$ – H. R. Nov 3 '15 at 17:20
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    $\begingroup$ The fact that there's a curve on which $u$ remains constant is pretty significant, since this is what allows us to solve the equation. This curve is called the "characteristic curve" or just "characteristic" (hence the name "method of characteristics"). In this case, the characteristics are just straight lines (specifically of the form $ax-y = c$ for some constant $c$), which makes solving the equation easy. $\endgroup$ – Joey Zou Nov 3 '15 at 17:36
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    $\begingroup$ Now, one feature of the characteristics for this equation (known as a transport equation) is that they don't intersect--which is good, since if two of them intersected at a point, then we don't know which one to refer to in order to find the value of $u$! But for some equations, the characteristics do eventually intersect--and then we run into problems. Finding ways to resolve these problems is a key aspect in the study of nonlinear PDEs. $\endgroup$ – Joey Zou Nov 3 '15 at 17:41
  • $\begingroup$ Can you turn your algebraic manipulation to some geometric interpretation? :) $\endgroup$ – H. R. Nov 3 '15 at 18:07
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    $\begingroup$ @H.R. I think the notion of characteristic curves is pretty geometric. Your equation, which again is called a transport equation, can be thought of as the value of your function being "transported" along these characteristic curves. In this case the characteristics are lines because the coefficient $a$ is constant, but if it weren't constant you could have characteristics that are actually curves and not just straight lines, along which the value of your function is still being transported. That's as geometric as I can think for now. Sorry if it doesn't answer your question! $\endgroup$ – Joey Zou Nov 3 '15 at 23:09
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Here is a slightly different way of looking at the method of characteristics that tries to avoid tangents and normals to surfaces which is in the standard formulation. The aim here is not to be general, but to present the method in a simpler way to make it more clear what is going on.


We are going to find curves $(x(s),y(s))$ in the plane with the property that $u(x(s),y(s))$ stays constant as we move along these curves. Then given any point $(x_1,y_1)$ in the plane then we will follow such a curve backwards until we hit a point $(x_0,y_0)$ in the initial/boundary set. Since $u$ stays constant along these curves we know that $u(x_1,y_1) = u(x_0,y_0)$ and since $u(x_0,y_0)$ is known by the initial/boundary conditions we will then know what $u(x_1,y_1)$ is.

Finding such curves is not hard. The condition $u(x(s),y(s)) = $ constant is just the statement that

$$0 = \frac{du(x(s),y(s))}{ds} = u_x \frac{dx}{ds} + u_y\frac{dy}{ds}$$

and since the PDE reads $u_x + u_ya = 0$ we see that taking $\frac{dx}{ds}=1$ and $\frac{dy}{ds}=a$ works. The curve, starting at $(x_0,y_0)$ when $s=0$, defined by these equations is the straight line

$$(x(s),~y(s)) = (x_0 + s,~y_0 + as) \implies y(s) = y_0 + a (x(s)-x_0)$$

Now lets assume the inital conditions are set at the $x$-axis, i.e. $u(x,0) = f(x)$. The characteristic curve through the point $(x_1,y_1)$, namely $y(x) = y_1 + a(x-x_1)$, intersects the $x$-axis at $(x_0,y_0) = (x_1 - \frac{y_1}{a},0)$. Since $u$ is constant along the curve we have $u(x_1,y_1) = u(x_0,y_0)$ and the general solution to the PDE can therefore be written

$$u(x_1, y_1) = u\left(x_1 - \frac{y_1}{a}, 0\right) = f\left(x_1-\frac{y_1}{a}\right)$$

Note that $(x_1,y_1)$ was an arbitrary point in the plane so the formula above holds for all $x_1,y_1$.

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  • $\begingroup$ Thanks for the contribution. :) I really don't understand what happened in the last line, i.e., $u(x,y) = u(x - \frac{y}{a},0)$! How did you concluded this? :) Also, in the first lines, $u(x,y)=u(x_0,y_0)$ means that $u(x,y)$ is constant over its domain! I think this is not a good notation as you mean that it is constant only on characteristics! :) $\endgroup$ – H. R. Nov 5 '15 at 10:17
  • $\begingroup$ @H.R. I changed the notation a bit to make it more clear and I also updated the last explanation. $\endgroup$ – Winther Nov 5 '15 at 13:19
  • $\begingroup$ It looks good now! :) Voted Up! $\endgroup$ – H. R. Nov 5 '15 at 14:03
  • $\begingroup$ If $a=0$ then you are in trouble right? :) since $(x_1,y_1)$ cannot be on the same line as $(x_0,y_0)$! :) $\endgroup$ – H. R. Nov 5 '15 at 14:08
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    $\begingroup$ @H.R. Yes the char. lines are vertical then when $a=0$. The PDE in this case is just $u_x = 0 \implies u = F(y)$ for some function $F$. The formula I put up can take this into account by defining $g$ via $f(x) = g(ax)$ then we can write $u(x,y) = g(ax - y)$ so if $a=0$ we get do the correct solution that $u(x,y)$ is just a function of $y$. $\endgroup$ – Winther Nov 5 '15 at 15:35
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I think a reconsideration should be made from Eq.$(3)$ above and what is after that. Our surface can be parameterized by any other two parameters (I don't know that why notes and books on the subject insist on writing a parametrized curve instead of a paratmetrized surface even though it causes lots of ambiguities)

$$\left\{ \matrix{ x = x(s,t) \hfill \cr y = y(s,t) \hfill \cr z = z(s,t) = u(x(s,t),y(s,t)) \hfill \cr} \right.\tag{1}$$

In fact, this is a change of variables from $(x,y)$ to $(s,t)$. If we want our change of variable to be useful then we may require that

$$\left\{ \matrix{ {{\partial x} \over {\partial s}} = 1 \hfill \cr {{\partial y} \over {\partial s}} = a \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\left\{ \matrix{ x(s,t) = s + p(t) \hfill \cr y(s,t) = as + q(t) \hfill \cr} \right.\,\tag{2}$$

because, this will turn our PDE into a simple one by just using chain-rule

$$\eqalign{ & {{\partial z} \over {\partial s}}\left( {s,t} \right) = {{\partial u} \over {\partial x}}\left( {x(s,t),y(s,t)} \right){{\partial x} \over {\partial s}}\left( {s,t} \right) + {{\partial u} \over {\partial y}}\left( {x(s,t),y(s,t)} \right){{\partial y} \over {\partial s}}\left( {s,t} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {{\partial u} \over {\partial x}}\left( {x(s,t),y(s,t)} \right) + a{{\partial u} \over {\partial y}}\left( {x(s,t),y(s,t)} \right) = 0 \cr} \tag{3}$$

and hence we can say that

$$z(s,t) = h(t)\tag{4}$$

where $h(t)$ is some arbitrary function of $t$. According to $(1)$, we can say that

$$u\left( {x(s,t),y(s,t)} \right) = h(t)\tag{5}$$

Now, if we could write $h(t)$ in terms of $x(s,t)$ and $y(s,t)$ then we were able to derive $u(x,y)$ from $(5)$. Hence, by using $(2)$ we can conclude that

$$\eqalign{ & ax(s,t) - y(s,t) = ap(t) - q(t) \cr & f\left( {ax(s,t) - y(s,t)} \right) = f\left( {ap(t) - q(t)} \right) \cr & h(t) = f\left( {ap(t) - q(t)} \right) \cr & h(t) = f\left( {ax(s,t) - y(s,t)} \right) \cr}\tag{6}$$

and then combining $(5)$ with $(6)$ we can get

$$\eqalign{ & u\left( {x(s,t),y(s,t)} \right) = f\left( {ax(s,t) - y(s,t)} \right) \cr & u(x,y) = f(ax - y) \cr} \tag{7}$$

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  • $\begingroup$ @Joey Zou: What is your opinion about this solution? Isn't it clearer? :) $\endgroup$ – H. R. Nov 4 '15 at 9:56
  • $\begingroup$ To each their own. If it makes more sense for you, then great! To me though, you're just adding another variable $t$ whose purpose I'm not really understanding, aside from changing the parametrization from a curve to a surface. I like curves better, since they depend only on one variable, and I don't really know what the "lots of ambiguities" that you mention regarding curves involve. $\endgroup$ – Joey Zou Nov 4 '15 at 20:59
  • $\begingroup$ In particular, I don't understand how you got the second half of (6). I'm guessing that you're claiming we can find $f$ such that $f(ap(t)-q(t)) = h(t)$. Why is that? I guess you are free to choose $p$ and $q$--why not make them constant? Then we'd get back the method of characteristics :) Also, how would you generalize this to more variables (i.e. $a\cdot\nabla u = 0$ for $u:\mathbb{R}^n\rightarrow\mathbb{R}$, $n\ge 3$), or when $a$ is not constant, or the right-hand side is not zero? The method of characteristics can generalize pretty well to those situations. $\endgroup$ – Joey Zou Nov 4 '15 at 21:00
  • $\begingroup$ @JoeyZou: What I have written is the method of characteristics but in the write notation! :) The second line of $(6)$ is in fact $f(g(t))=h(t)$ and nothing more! :) This can easily be generalized to higher dimension with the definition of manifold. :) $\endgroup$ – H. R. Nov 4 '15 at 21:56
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    $\begingroup$ So in (6) are you just setting $h(t)$ to be $f(ap(t)-q(t))$, where $f$ is arbitrary? That's fine then--this ends up showing that $u(x,y) = f(ax-y)$ is an acceptable solution (which you could also show by just plugging it in). How do you show that, for arbitrary $h$, you can find such an $f$? That is, how do you show all solutions must be of this form? $\endgroup$ – Joey Zou Nov 4 '15 at 22:18
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I have read just your solution of your question which always is the best chosen solution by yourself :) I think first argument (mentioned in question) is good and tells u is a "cylindrical surface". Actually it is a complete proof :) it tells solution is any cylindrical surface in a certain direction. Your solution make sense too but I cant understand what the hell is f?! Use this change of variables: $$\frac{{\partial u}}{{\partial x}}(x,y) + a\frac{{\partial u}}{{\partial y}}(x,y) = 0\tag{1}\\x(s,t)=s \,\,,\,\,y(s,t)=as+t \,\, \\ \,\,(s(x,y)=x\,\,,\,\,t(x,y)=y-ax)\\u=u(s(x,y),t(x,y))\\ {{\partial u}\over{\partial x}}={{\partial u}\over{\partial s}}-a{{\partial u}\over{\partial t}}\,\,,\,\,{{\partial u}\over{\partial y}}={{\partial u}\over{\partial t}}\\ \Rightarrow (1) \,\, \text{holds}\,\,\text{iff} \,\,{{\partial u}\over{\partial s}}=0\\ \text{or}\,\,u(s,t)=h(t)\,\,\text{for some differentiable} \,\,h\\ \text{or}\,\,u(x,y)=h(y-ax)$$

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  • $\begingroup$ When I choose the best answer, it does mean that it just works for me as the best! But others may have different tastes! :) I also Voted up the other two solutions by Winther and Joey Zou. $\endgroup$ – H. R. Nov 10 '15 at 15:32
  • $\begingroup$ The best solution is mentioned in question. (The books you have read this) $\endgroup$ – A.F.23 Nov 10 '15 at 15:35
  • $\begingroup$ I don't think that what is written in the question is really the best! It just obtains some curves on the surface! I think you have used the chain rule with a bad notation in your solution! :) Also, you have just used special cases for $p(t)$ and $q(t)$ in my solution. I will be so thankful if you take a look at this post too. $\endgroup$ – H. R. Nov 10 '15 at 16:01
  • $\begingroup$ But I think so. Which bad notation? I think you used multiple names for one thing in your solution and also in that link because of understanding chain rule badly :) maybe you want to rename your name that It's non of my biusnes :) change of variable was special in whole history of human kind :) we want answering problem. $\endgroup$ – A.F.23 Nov 10 '15 at 16:57
  • $\begingroup$ You are naming two different functions with the same letter! :/ Does the equation $u(s(x,y),t(x,y))=u(x,y)$ make sense to you? Or simply in one-variable calculus: $f(g(x))=f(x)$!? $\endgroup$ – H. R. Nov 10 '15 at 17:22

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