WEBVTT

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The method of characteristics
turns a partial differential equation

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into an ordinary differential
equation along carefully chosen paths.

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We will build that idea
from a moving shape,

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then follow it until those paths collide.

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Let's start with water.

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A long, narrow tank
of it, seen from the side,

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and something has disturbed
the surface: a swell, sitting here.

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Now pick a spot along the tank (any
spot) and ask one very small question.

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How high is the water, right there?

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Down here it is barely raised.

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Walk toward the swell and it is higher.

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Keep going, up to the top, higher still,
and past that it drops away again.

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So every position hands
you back one number.

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That is all a function is,
and this one has a name:

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u.

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Except water does not hold
still, and that ruins it.

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Watch.

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And I owe you an explanation
for what you are about to see,

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because the swell is going
to keep its shape exactly.

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That is not a law of water:

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it is the simplest case,
and the one to understand first.

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Picture the whole tank drifting
downstream at a steady speed,

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carrying whatever sits on it
the way a river carries a leaf.

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Nothing spreads, nothing dies.

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It just goes.

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Now stop moving, and let
it come to you instead.

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Stand at this one place.

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The water climbs past you,
and then it sinks away again,

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and you never took a step.

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So knowing where you are is no longer
enough to know how high the water is.

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You have to say when as well.

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u of x and t.

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Two inputs means two ways to change it,
so let us change them one at a time.

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First, freeze the clock.

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Nothing moves;

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this is one instant, held.

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Stand here, and walk
this far along the tank.

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You changed x by that much, and look:

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u changed too, by that much.

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Same instant, different
place, different height.

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Take a shorter walk and u changes less.

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A longer one, and it changes more.

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There is a rate hiding

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in there (height gained per step of x)
and that rate is what u sub x means.

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Now the other one. Put x back where it was
and leave it there: one place, this stake,

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and you do not move again.

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On the right I will keep a record of the
height at that one stake, against time.

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Not against position;
the position is fixed now.

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Against time.

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Watch both dots.

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The swell arrives, the water climbs
the stake, and the record climbs with it.

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The crest passes, and the two
of them come back down together.

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One number, drawn twice.

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You never moved, and u changed anyway.

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So there is a second rate
(height gained per tick

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of the clock) and that one is u sub t.

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Two rates, then.

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And they are not independent: here is why.

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Here is the surface at one instant,
and here it is a moment later.

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Nothing about the shape changed.

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Every part of it simply moved to the
right, by the speed c times that moment.

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So the height at your feet
now is not a new number at all.

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It is the height that used

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to sit a little way upstream (exactly c
delta t upstream) and has since arrived.

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Now read that as a rate.

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The water rises under you at whatever
rate the shape slopes downward, multiplied

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by how fast the shape is coming.

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Steeper slope, or faster water,
and the height changes quicker.

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Tidy it up, and there
is the transport equation.

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Every symbol in it
is something we pointed at first.

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And now the reason anyone
would want to solve it.

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This is a rule about rates, here and now.

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A solution is the height at every
place at every time, including times

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that have not happened yet.

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Solving it means predicting the water.

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And there is more than one way to do that.

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You could guess that the shape
merely travels and verify it;

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we have very nearly done that already.

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You could break the profile into sine
waves and march each one forward.

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Or you could hunt

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for paths through the x t plane along
which u is not allowed to change at all.

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That last one is the method
of characteristics.

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It is one method among several, but it is
the one that turns a partial differential

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equation into an ordinary one,
and that is a trade worth understanding.

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Let's try something
that looks like a trick,

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and turns out to be the whole method.

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Instead of asking what u does everywhere
at once, we are going to choose a curve

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with x prime of t equal to c, then walk
across the x t plane and watch u as we go.

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Here is that curve,

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and here is where we stand on it at time
t: at the position x of t, at height t.

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Now give the value of u
at that moving point a name.

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Call it z of t.

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And z is a function of one variable.

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One variable means we
can simply differentiate it,

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and the chain rule tells
us exactly what comes out.

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u sub x times x prime of t,

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coming through the x slot, plus u
sub t, coming through the t slot.

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Nothing clever has happened yet.

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But look hard at what is sitting there:

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u sub x, u sub t, and a factor x prime of
t that nobody has told us how to choose.

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So let's choose it.

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Set x prime of t equal to c,
the speed out of the equation.

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And now read the right hand side.

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u sub t plus c u sub x.

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That is precisely the left
hand side of our equation,

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and our equation says it is zero.

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So z prime is zero.

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The value of u does not change
at all as we walk along that curve.

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And which curve is it?

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x prime equals c says
that x is x naught plus c t.

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A straight line
of slope c in the x t plane.

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That line has a name:
it is called a characteristic.

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Every equation of this kind
has a whole family of them,

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one through each starting point.

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And the partial differential equation,

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which was a statement about
two derivatives at once,

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has become a statement about
one derivative along a line.

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We already have the family
of parallel characteristics.

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Keep the initial profile
f in view on the right.

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Every point of the x t plane sits
on exactly one of those paths,

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so one path is enough
to recover its value.

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Now pick any point x, t, up here.

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Follow its characteristic
back down to time zero.

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It lands on the horizontal
axis at x minus c t.

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And u is constant all
the way along that line.

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So the value up at x,
t is the value down at the foot,

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which is nothing but the initial
data f, evaluated at x minus c t.

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And there is our solution.

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That is the whole answer.

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No integrals, no infinite series,
no separation of variables.

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Whatever shape you start with, you get
that same shape back, only shifted.

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Watch what that means.

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Here is f, the blob
sitting there at time zero.

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A moment later, the whole graph has picked
itself up and moved to the right.

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Not one thing about its shape has changed.

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Later still, further along,
and it is still exactly the same blob.

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That is the only thing
this equation ever does.

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And here is one particle of dye.

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On the left it climbs its own
characteristic, from that foot up

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to the point we asked about.

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On the right it rides
the crest along at speed c.

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Solving this equation was never
more than following the flow.

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One last remark,
and it matters in a minute.

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The data is carried along untouched,
so if f has a sharp corner in it,

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that corner rides along
forever and never smooths out.

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Transport has no memory and no
diffusion. It only translates.

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Now let's change exactly one thing.

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Instead of a constant speed c,
let the speed depend on time.

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u sub t plus t u sub x equals zero.

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Look back at what
the derivation actually used.

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Nothing in it needed c to be constant.

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Along a curve where u does not change,

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the chain rule asks x prime of t to equal
whatever sits in front of u sub x.

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Here, that is t.

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So the characteristics
solve x prime equals t.

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That is an ordinary differential
equation, and an easy one.

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Integrate, and x of t is x
naught plus t squared over two.

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Parabolas.

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Each one leaves the axis
at its own x naught and then takes off.

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At time zero the slope is zero,
so everything starts out standing still.

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Later they lean over more and more.

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The dye is accelerating.

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Now invert it.

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If I am standing at position x
at time t, which parabola am I on?

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Solve for x naught, and it
is x minus t squared over two.

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And u is constant along
each parabola, so u of x,

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t is f of x minus t squared over two.

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Same recipe, two lines of work, and the
answer is once again the starting profile

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with a rewritten argument.

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It is worth checking. Differentiate:

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u sub t is minus t times f
prime, and u sub x is f prime.

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So u sub t plus t u sub
x is minus t f prime,

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plus t f prime, which is zero.

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It works.

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One last equation, and it is the one
that made this method famous.

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u sub t plus u times u sub x equals zero.

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The coefficient in front
of u sub x is now u itself:

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the value being carried
sets its own speed.

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Along any curve, the chain rule gives u
sub t plus x prime of t times u sub x.

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Choose x prime of t equal to u,

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and that derivative becomes the
equation's left hand side, which is zero.

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So u stays constant
along the characteristic.

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A characteristic leaving x naught
therefore carries the fixed value f

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of x naught.

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Its slope x prime is that same fixed
number, so each path is a straight line,

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but different starting
values produce different slopes.

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Take decreasing initial data.

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The piece starting at x naught equals

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zero point five carries the large value
one point seven five, so it travels fast.

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The piece starting at three point
five carries only zero point two five,

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so it barely crawls.

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The two middle values lie between them.

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The quick characteristics
chase the slow ones ahead.

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All four paths meet at x
equals four, t equals two.

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They arrive carrying four different values
of u, but u is supposed to be a function.

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At that point it cannot
remain smooth and single-valued.

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The same collision appears
in the wave profile.

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At first the front is smooth.

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As faster values catch slower ones,

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it steepens, and steepens again,
until the front is nearly vertical.

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That breaking point is a shock.

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The smooth solution stops exactly
when the first characteristics meet.

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The method is four steps long,
and their order matters.

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First read the transport
speed from the equation.

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First, write the equation
with the two derivatives lined up,

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and read off the coefficient
sitting in front of u sub x.

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Second, solve x prime
of t equals that coefficient.

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Its solutions are the characteristics,
one through each starting point x naught.

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Third, use the equation to show
that u does not change along those curves.

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Thus u equals f of x
naught all the way along.

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Fourth, invert.

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Express x naught in terms of x
and t, then substitute it into f.

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That final expression is the solution.

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Each step is ordinary calculus:
read the speed, find its curves, carry

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the initial value along them, and trace
the requested point back to its start.

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The partial differential equation becomes
ordinary along exactly those paths.