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Method of characteristics

From Wikipedia, the free encyclopedia

In mathematics, the method of characteristics is a technique for solving partial differential equations. Typically, it applies to first-order equations, although more generally the method of characteristics is valid for any hyperbolic partial differential equation. The method is to reduce a partial differential equation to a family of ordinary differential equations along which the solution can be integrated from some initial data given on a suitable hypersurface.

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  • 17. Method of Characteristics

Transcription

In this video we're going to talk about the 'Method of Characteristics'. The Method of Characteristics is a technique for solving partial differential equations, and it is a technique which shares a common idea with many other techniques, and that is to reduce the PDE to (usually) a 'system' of ordinary differential equations, and then hope that you can solve those rather than trying to solve the original PDE. So let's start by defining, essentially, the position of a moving observer. So let's say that: " x(t) ".. is a position coordinate, depending on time, and call this a moving observer, (think of.. me on my 'cart'.. ..from the last video) and let's.. let: " u(x,t) ".. just be any old: " function of 'x' and 't' ". (it doesn't need to be a solution to a specific PDE at the moment) Now the question that I am going to ask is: How is 'u(x,t)' changing.. ..from the perspective of the moving observer? So: " how does 'u' change ".. " from observer's perspective "? Now.. let we draw couple of pictures just to make this is this question I'm asking clearer. So.. let's say here we have.. the 'u-axis'.. and here is the 'x-axis'.. and.. and I'm looking at a fixed point 'in time', So here's some particular point in time, and I see.. there's the graph of 'u' versus 'x', (so at some fixed point in time) and if I'm looking at this graph from my cart (here) and my cart is moving, and at each particular point in time it's located at 'x(t)'. (so here I am standing on it) This question.. is the question: What do I see? How do I see the point right in front of me.. moving? So if I look straight in front of me I see 'this' point on the curve. Now, as time passes.. my cart may move back and forth and I will see another point right in front of me.. (and the curve may also move up and down) and so the question is: How do I see this curve changing? Let me draw this in another, slightly different way: Let's say I now have 'x' versus 't'. So you can imagine what I did, I just sort of added a 'time' dimension right here, ( but drawn straight up ) like a 't' dimension. So this is the plot right here, and here's 'time'. Let's say that here is my path: okay so I start at some point ( for 't' = 0 ) and here I go.. my path is: " (x(t),t) " and we know that, at every point in this plot, everywhere.. I have a value of 'u(x,t)' And so the question: how does 'u(x,t)' change from the observer's perspective? says: how does 'u(x,t)' change when we move along this line? Let's answer this question. The way we're going to do that is by computing the derivative of 'u(x,t)'.. ..where I plug-in, for the 'position' coordinate', the ( time dependent ) position of the moving observer, (and the 'time' coordinate just as normal) Let's write this as: " d(u)/dt ". That's a sort of funny-looking thing, but we think of 'u' as being.. this function : "u(x(t),t)" (of 't' alone) and this is 'u_x'.. 'd(x)/dt'.. plus 'u_t' (..that's by the 'chain-rule') What does this tell us? Well, look at this expression right here.. ( in box ) ..we've discovered: this 'equality'. Now it looks a little bit like the Transport equation, for instance: (let's write down the Transport equation, just to compare it) the Transport equation was: " 0 = c*u_x + u_t " . So if this ( 'd(u)/dt' ) was '0'.. and 'd(x)/dt' was 'c', it would look 'just like' the Transport equation. So let's write that down too: " d(x)/dt = c " and. " d(u)/dt = 0 ".. then that expression is just like Transport equation. What does it mean for 'd(x)/dt' to be equal to 'c'? Well it means that the moving observer is moving at a constant speed of 'c'. So: " moving ".. " with speed 'c' ". (just like in the last video where I was watching the wave.. ..and moving along at constant speed, equal to the wave-speed) What does 'd(u)/dt' equals '0' mean? It means that 'u' is not changing. Notice: this is exactly the situation I was describing in the previous video: ..the wave was moving along with a constant speed of 'c' ..'I' sat on the cart and moved with that same constant speed and watched the wave 'not' change. So what this is saying is: the Transport equation is the 'same'.. as saying that these two ODEs are satisfied. In other words, along these lines.. '(x(t),t)', when the line is given by a constant slope of..in this picture, '1/c'.. ( since 'd(x)/dt' = 'c' ) ..the solution to the Transport equation.. is 'not' changing ( i.e. 'contant' ) So if this function is not changing along this line.. then it's a solution to the Transport equation ..(and vice versa) It's going to seem like we're beating a dead horse at this point, but I'm going to solve the Transport equation yet again. (and I promise that this is the last time I'm going to do it... ...we'll move on to something different in the very next slide) So: the " Transport eqn (last time) " and I'm going to do at this time using the Method of Characteristics, in the way that I have outlined it before. So what is it? its: 'u_t + c*u_x' equals '0', and the Method of Characteristics says that this is the same: as the statement: 'd(u)/dt' is equal to '0' " along ".. " curves ".. " given by ".. " 'd(x)/dt = 'c' " (in this case they are just lines) This is the technique of solution: to look at the problem like 'this' ..rather than like 'this'. So first of all let's figure what these guys are: this is not so hard (right?) This is: 'd(x)/dt' = 'c', (that's just a constant) and that implies: 'x' = 'c*t + x_0' (where 'x_0' is a constant of integration) Okay, let's draw ourselves a little picture here. So we have: 'x-axis'.. and the 't-axis'.. and the graph of this guy right here.. will be a straight line, and it will have slope '1/c', and it will intersect the 'x-axis' at the point 'x_0'. If you're confused by the '1/c' think about the fact that these two coordinate axes have 'flipped''. ('x' is a function of 't' ...but the whole thing is 'flipped') and this is the line: 'x - c*t = x_0' Now, 'u' is a constant along this line, that's what Characteristics 'say'. ('u' is constant along this line) So in other words, if I know 'u' at this point, then I know it along this entire line, and similarly, for any of these other lines: if I knew 'u' at this point.. 'here' on the axis, I would know it along the entire ( parallel ) line 'here'. So this problem is practically begging for an 'initial condition'. If I know what 'u' is ( just ) along the 'x-axis', (in other words at time '0') then I can just use the 'characteristic' lines to.. 'propagate' the solution.. throughout all of this space. So let's write that down. we: " Need initial condition ", and that's going to be (say) " u(x,0) = ".. and let's just say: " f(x) "., and this is an initial condition along the 'x-axis', ..what the value of 'u' is along this entire axis. (in other words for 't = 0') So since we have an initial condition, we know what the value of 'u' is along the 'x-axis'. If we know what the value is at a particular point, let's say 'x_0', then since 'u' is constant along the line, that value will 'propagate' along the line as being the same value. So along this red line, 'u' = 'f(x_0)'. So let's write that down: along 'characteristic' lines: 'x - c*t = x_0', 'u(x,t) = f(x_0)', which is equal to: 'f(x - c*t)' (..because 'x_0 = x - c*t'.) Now you notice here that it didn't really depend on which 'line' I picked because all I would have to say is that 'u(x,t)' is equal to 'f(x-ct)', and we know that 'x - c*t' wherever I plop down in the 'x-t plane', (here I am) 'x - c*t' is going to be equal to that point.. ..where the line will intercept the 'x-axis'. So really, the conclusion is: I don't need to reference that 'x_0' point at all, all I have to do is say 'u(x,t)' is equal to 'f(x-c*t)'. So let's write down a full conclusion. the solution to the initial-value problem: " u_t + c*u_x = 0 " and.. " u(x,0) = f(x) ".. the solution is: " u(x,t) = f(x - c*t) ". (and this is via the Method of Characteristics) and you'll notice that this is.. exactly the same solution we derived in the last video, although we didn't talk about 'initial conditions' explicitly in that video because we didn't really have to. (and here it was more 'pertinent' to do so) In the time remaining I'm going to give you an exercise: So I kind of 'lied' when I said I wasn't going to talk about the Transport equation anymore but now 'you' have to talk about it. (because you have to solve this exercise) The exercise is to solve the initial-value problem: remember an 'initial-value' problem consists of a PDE plus some initial condition, and the 'PDE' in this case is: the Transport equation with a '1' on the right hand side' (so it's no longer 'homogeneous': not a '0' but '1') and then an initial condition which says that: 'u(x,0) = sin x'. Use the Method of Characteristics to solve this equation. (and go ahead and pause the video now and I'll give you a hint after pause) [PAUSE] Alright: 'Hint #1': Remember the general equation that describes the way that 'u(x,t)' changes along the 'characteristic' lines? That's: 'd(u)/dt' = 'u_x*d(x)/dt' + 'u_t'. And so to compare it with this guy I would have essentially the 'd(u)/dt' would be a '1' and then 'u_x'.. and then the.. 'd(x)/dt' would be a 'c'. plus 'u_t'. And so my equations for the 'characteristic' lines would be: 'd(x)/dt' equals 'c',, and then I'd be left with this 'd(u)/dt' equal to 1' instead of '0'. Okay that's 'Hint #1', so go ahead and pause if you want to continue... [PAUSE] Alright here's 'Hint #2': The fact that 'd(u)/dt' is '1' means that along 'characteristic' lines 'u' changes as: 't' plus the some constant 'A'. (this is along 'characteristic' lines) So at time, 't = 0' the 't' will become '0' and the 'A' will become simply the initial condition that we have which is: 'sin(x_0)' and remember that in 'sin(x_0)', 'x_0' = 'x - c*t' wherever I am in the whole 'x-t plane'. Pause now if you don't want to see the solution I'll put the solution of in the upper right-hand corner. momentarily... [PAUSE] And the solution 'magically appears'..

Characteristics of first-order partial differential equation

For a first-order PDE (partial differential equation), the method of characteristics discovers curves (called characteristic curves or just characteristics) along which the PDE becomes an ordinary differential equation (ODE).[1] Once the ODE is found, it can be solved along the characteristic curves and transformed into a solution for the original PDE.

For the sake of simplicity, we confine our attention to the case of a function of two independent variables x and y for the moment. Consider a quasilinear PDE of the form

(1)

Suppose that a solution z is known, and consider the surface graph z = z(x,y) in R3. A normal vector to this surface is given by

As a result,[2] equation (1) is equivalent to the geometrical statement that the vector field

is tangent to the surface z = z(x,y) at every point, for the dot product of this vector field with the above normal vector is zero. In other words, the graph of the solution must be a union of integral curves of this vector field. These integral curves are called the characteristic curves of the original partial differential equation and are given by the LagrangeCharpit equations[3]

A parametrization invariant form of the Lagrange–Charpit equations[3] is:

Linear and quasilinear cases

Consider now a PDE of the form

For this PDE to be linear, the coefficients ai may be functions of the spatial variables only, and independent of u. For it to be quasilinear,[4] ai may also depend on the value of the function, but not on any derivatives. The distinction between these two cases is inessential for the discussion here.

For a linear or quasilinear PDE, the characteristic curves are given parametrically by

such that the following system of ODEs is satisfied

(2)

(3)

Equations (2) and (3) give the characteristics of the PDE.

Proof for quasilinear case

In the quasilinear case, the use of the method of characteristics is justified by Grönwall's inequality. The above equation may be written as

We must distinguish between the solutions to the ODE and the solutions to the PDE, which we do not know are equal a priori. Letting capital letters be the solutions to the ODE we find

Examining , we find, upon differentiating that

which is the same as

We cannot conclude the above is 0 as we would like, since the PDE only guarantees us that this relationship is satisfied for , , and we do not yet know that .

However, we can see that

since by the PDE, the last term is 0. This equals

By the triangle inequality, we have

Assuming are at least , we can bound this for small times. Choose a neighborhood around small enough such that are locally Lipschitz. By continuity, will remain in for small enough . Since , we also have that will be in for small enough by continuity. So, and for . Additionally, for some for by compactness. From this, we find the above is bounded as

for some . It is a straightforward application of Grönwall's Inequality to show that since we have for as long as this inequality holds. We have some interval such that in this interval. Choose the largest such that this is true. Then, by continuity, . Provided the ODE still has a solution in some interval after , we can repeat the argument above to find that in a larger interval. Thus, so long as the ODE has a solution, we have .

Fully nonlinear case

Consider the partial differential equation

(4)

where the variables pi are shorthand for the partial derivatives

Let (xi(s),u(s),pi(s)) be a curve in R2n+1. Suppose that u is any solution, and that

Along a solution, differentiating (4) with respect to s gives

The second equation follows from applying the chain rule to a solution u, and the third follows by taking an exterior derivative of the relation . Manipulating these equations gives

where λ is a constant. Writing these equations more symmetrically, one obtains the Lagrange–Charpit equations for the characteristic

Geometrically, the method of characteristics in the fully nonlinear case can be interpreted as requiring that the Monge cone of the differential equation should everywhere be tangent to the graph of the solution. The second order partial differential equation is solved with Charpit method.

Example

As an example, consider the advection equation (this example assumes familiarity with PDE notation, and solutions to basic ODEs).

where is constant and is a function of and . We want to transform this linear first-order PDE into an ODE along the appropriate curve; i.e. something of the form

where is a characteristic line. First, we find

by the chain rule. Now, if we set and we get

which is the left hand side of the PDE we started with. Thus

So, along the characteristic line , the original PDE becomes the ODE . That is to say that along the characteristics, the solution is constant. Thus, where and lie on the same characteristic. Therefore, to determine the general solution, it is enough to find the characteristics by solving the characteristic system of ODEs:

  • , letting we know ,
  • , letting we know ,
  • , letting we know .

In this case, the characteristic lines are straight lines with slope , and the value of remains constant along any characteristic line.

Characteristics of linear differential operators

Let X be a differentiable manifold and P a linear differential operator

of order k. In a local coordinate system xi,

in which α denotes a multi-index. The principal symbol of P, denoted σP, is the function on the cotangent bundle TX defined in these local coordinates by

where the ξi are the fiber coordinates on the cotangent bundle induced by the coordinate differentials dxi. Although this is defined using a particular coordinate system, the transformation law relating the ξi and the xi ensures that σP is a well-defined function on the cotangent bundle.

The function σP is homogeneous of degree k in the ξ variable. The zeros of σP, away from the zero section of TX, are the characteristics of P. A hypersurface of X defined by the equation F(x) = c is called a characteristic hypersurface at x if

Invariantly, a characteristic hypersurface is a hypersurface whose conormal bundle is in the characteristic set of P.

Qualitative analysis of characteristics

Characteristics are also a powerful tool for gaining qualitative insight into a PDE.

One can use the crossings of the characteristics to find shock waves for potential flow in a compressible fluid. Intuitively, we can think of each characteristic line implying a solution to along itself. Thus, when two characteristics cross, the function becomes multi-valued resulting in a non-physical solution. Physically, this contradiction is removed by the formation of a shock wave, a tangential discontinuity or a weak discontinuity and can result in non-potential flow, violating the initial assumptions.[5]

Characteristics may fail to cover part of the domain of the PDE. This is called a rarefaction, and indicates the solution typically exists only in a weak, i.e. integral equation, sense.

The direction of the characteristic lines indicates the flow of values through the solution, as the example above demonstrates. This kind of knowledge is useful when solving PDEs numerically as it can indicate which finite difference scheme is best for the problem.

See also

Notes

  1. ^ Zachmanoglou, E. C.; Thoe, Dale W. (1976), "Linear Partial Differential Equations : Characteristics, Classification, and Canonical Forms", Introduction to Partial Differential Equations with Applications, Baltimore: Williams & Wilkins, pp. 112–152, ISBN 0-486-65251-3
  2. ^ John, Fritz (1991), Partial differential equations (4th ed.), Springer, ISBN 978-0-387-90609-6
  3. ^ a b Delgado, Manuel (1997), "The Lagrange-Charpit Method", SIAM Review, 39 (2): 298–304, Bibcode:1997SIAMR..39..298D, doi:10.1137/S0036144595293534, JSTOR 2133111
  4. ^ "Partial Differential Equations (PDEs)—Wolfram Language Documentation".
  5. ^ Debnath, Lokenath (2005), "Conservation Laws and Shock Waves", Nonlinear Partial Differential Equations for Scientists and Engineers (2nd ed.), Boston: Birkhäuser, pp. 251–276, ISBN 0-8176-4323-0

References

External links

This page was last edited on 24 March 2024, at 04:18
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