with the velocity field of the flowing water being \(\vv(x,y)\text{.}\) We are assuming, for simplicity, that the velocity field does not depend 2
This is not such an unreasonable assumption. The flow often changes on a larger time scale.
on time \(t\text{.}\) The stick will move along with the water 3
This is also not an unreasonable approximation.
. When the stick is at \(\vr\text{,}\) its velocity will be the same as the velocity of the water at \(\vr\text{,}\) which is \(\vv(\vr)\text{.}\) Thus if the stick is at \(\vr(t)\) at time \(t\text{,}\) we will have
field line or integral curve (for general vector fields) or a
stream line or flow line (when the vector field \(\vv\) is being thought of as a velocity field) or a
line of force (when the vector field \(\vv\) is being thought of as a force field)
of the vector field \(\vv\text{.}\)
Example2.2.2.Flow Line Sketch for the Vortex of Example 2.1.4.
Consider the vortex vector field, \(\vv(x,y) = \Om\big(-y\hi +x\hj\big)\) of Example 2.1.4. Once we sketched the vector field, as in Figure 2.1.6, or even made the “skeleton sketch” of Figure 2.1.5, we can get rough idea of what the stream lines look like just by following the arrows. For example, suppose that we start a stream line (i.e. drop the stick into the stream) on the positive \(x\)-axis. Looking at Figure 2.1.5, which is repeated here,
the stick
starts by moving in the \(+y\) direction, i.e. straight upward.
As it moves farther into the first quadrant it develops a larger and larger negative \(x\)-component of velocity. So it also moves leftwards toward the \(y\)-axis.
Eventually it crosses the positive \(y\)-axis moving in the \(-x\) direction, i.e. to the left.
As it moves farther into the second quadrant it develops a larger and larger negative \(y\)-component of velocity. So it also moves downwards toward the \(x\)-axis.
Eventually it crosses the negative \(x\)-axis moving in the \(-y\) direction, i.e. straight downward.
As it moves farther into the third quadrant it develops a larger and larger positive \(x\)-component of velocity. So it also moves rightward towards the \(y\)-axis.
Eventually it crosses the negative \(y\)-axis moving in the \(+x\) direction, i.e. to the right.
As it moves farther into the fourth quadrant it develops a larger and larger positive \(y\)-component of velocity. So it also moves upwards toward the \(x\)-axis.
With this type of analysis we cannot tell if the streamline, which is the red line in the figure above, will return to the \(x\)-axis
exactly at its starting point, forming a closed curve, or
inside its starting point, spiralling inwards, or
outside its starting point, spiralling outwards.
While the above procedure is a good way to get a qualitative feel for trajectories, we can develop more precise, detailed descriptions of field lines by working analytically. As we saw above, thinking of \(\vr(t)\) as the position at time \(t\) of a stick dropped into water whose velocity at \((x,y)\) is \(\vv(x,y)\text{,}\) the velocity of the stick at time \(t\) will be the same as the velocity of the water at \(\vr(t)\text{,}\) which is \(\vv\big(\vr(t)\big)\text{.}\) Thus \(\vr(t)\) will obey the system of first order differential equations
Notice that if we reparametrize \(\vr(t)\text{,}\) say to \(\vR(u) = \vr\big(t(u)\big)\text{,}\) then \(\vR'(u) = \vr'\big(t(u)\big)\ t'(u)\) is parallel to (though not necessarily equal to) \(\vr'\big(t(u)\big)= \vv\big(\vr\big((t(u)\big)\big)
=\vv\big(\vR(u)\big)\text{.}\) So if we only care about the curve traced out by the stick, and not about when the stick is at each point of the path, then it suffices to impose the weaker condition 4
We’ll have a more careful discussion of this in the optional §2.2.1.
that, when the stick is at \(\vr(t)\text{,}\) its velocity \(\vr'(t)\) is parallel to (though not necessarily equal to) \(\vv\big(\vr(t)\big)\text{.}\) In three dimensions, \(\vr'(t)\) is parallel to \(\vv\big(\vr(t)\big)\) when the cross product is zero:
In two dimensions we can still use the cross product by the simple expedient of thinking of \(\vr'(t)\) and \(\vv\big(\vr(t)\big)\) as three component vectors whose third components are zero.
A more convenient way to implement the weaker “just parallel” condition, involves reparametrizing our streamline. Suppose that we are in two dimensions with \(\vr'(t)=\big(\diff{x}{t}(t)\,,\,\diff{y}{t}(t)\big)\) and \(\vv(\vr)=\big(v_1(\vr)\,,\,v_2(\vr)\big)\) and fix some \(t_0\text{.}\) If \(\diff{x}{t}(t_0)\) is nonzero 5
If \(\diff{x}{t}(t_0)=0\text{,}\) but \(\diff{y}{t}(t_0)\ne 0\text{,}\) we should use \(y\) rather than \(x\) as the parameter. If \(\diff{x}{t}(t_0)=\diff{y}{t}(t_0)=0\text{,}\) then \(\vr(t)=\vr(t_0)\) for all \(t\) and the streamline doesn’t move. It is just a single point.
, we can reparametrize the curve (at least near \(\vr(t_0)\)) so as to use \(x\text{,}\) rather than \(t\) as the parameter. To do so, we
solve \(x=x(t)\) for \(t\) as a function of \(x\text{.}\) Call the solution \(T(x)\text{.}\) Then
the point on the curve which has \(x\)-coordinate \(x\) is \(\vR(x)=\big(X(x)\,,\,Y(x)\big)\) with \(X(x)=x\) and \(Y(x) = y\big(T(x)\big)\text{.}\)
Then the condition that \(\vR'(x)=\big(1,Y'(x)\big)\) is parallel to \(\vv\big(\vR(x)\big)\) says that \(\vR'(x)\) is a scalar multiple of \(\vv\big(\vR(x)\big)\) so that there is a nonzero number \(c(x)\) so that \(\vR'(x)=c(x) \vv\big(\vR(x)\big)\text{.}\) That is
Of course \(\diff{y}{x}\) is not the ratio of \(\dee{y}\) and \(\dee{x}\text{.}\) However pretending that it is provides a simple way to remember the technique that is used to solve the equation. You may have used this mnemonic device before when you learned how to solve separable differential equations. Section 2.4 of the CLP-2 text contains a treatment of separable differential equations, including a justification for the mnemonic device.
that \(\diff{y}{x}\) is the ratio of \(\dee{y}\) and \(\dee{x}\) and rewrite the differential equation 7
Here is another nonrigorous, but intuitive way to come up with this equation. Suppose that our stick is at \((x,y)\) and has velocity \(\big(\diff{x} {t}(t)\,,\,\diff{y}{t}(t)\big)\text{.}\) In a tiny time interval \(\dee{t}\) the stick moves by \(\big(\diff{x}{t}(t)\,,\,\diff{y}{t}(t)\big)\dee{t}
=(\dee{x},\dee{y})\text{,}\) which is parallel to \(\big(v_1(x,y)\,,\,v_2(x,y)\big)\) if \(\frac{\dee{x}}{v_1(x,y)}=\frac{\dee{y}}{v_2(x,y)}\text{.}\)
Let us apply this to two examples, in which the stream lines of the vortex field of Example 2.1.4 are found by two different methods.
Example2.2.6.Stream lines for the vortex field using \(\vr'(t)
\!\parallel\!\vv(\vr(t))\).
In this example we will find the stream lines for the vortex field, \(\vv(x,y) = \Om\big(-y\hi +x\hj\big)\) of Example 2.1.4, by using the requirement that, on a stream line, the velocity vector \(\vr'(t)\) must be parallel to \(\vv\big(\vr(t)\big)\text{.}\) By 2.2.5 one way to express this requirement mathematically is
This is a simple separable differential equation. We can solve it by cross multiplying and integrating both sides. (Recall that \(\Om\) is a constant.)
where \(C'\) and \(C=\frac{2}{\Om}C'\) are just arbitrary constants. So the stream lines of the vortex field are exactly circles centred on the origin.
We can come to exactly the same conclusion by using the cross product formulation of 2.2.4.
Example2.2.7.Stream lines for the vortex field using \(\vr'(t)\!=\!\vv(\vr(t))\).
This time we will find the stream lines for the vortex field, \(\vv(x,y) = \Om\big(-y\hi +x\hj\big)\) of Example 2.1.4, by using 2.2.3, which is
\begin{align*}
\diff{x}{t} &= -\Om y\\
\diff{y}{t} &= \Om x
\end{align*}
We can convert this system of first order linear ordinary differential equations into a single second order linear constant coefficient differential equation 8
In Example 2.1.7 we converted a second order ordinary differential equation into a system of first order ordinary differential equations. We are now just reversing the procedure we used there.
, by differentiating the first equation, to get \(\difftwo{x}{t} = -\Om\diff{y}{t}\text{,}\) and then substituting in the second equation to get
\begin{equation*}
\difftwo{x}{t} + \Om^2 x = 0
\end{equation*}
This equation is a special case of the ordinary differential equation treated in Example A.9.3 of the Appendix A.9, entitled “Review of Linear Ordinary Differential Equations”. In fact it is exactly (A.9.6) with \(R=0\text{,}\)\(L=C=\frac{1}{\Om}\text{.}\) So the general solution is (A.9.8) with \(\rho=0\) and \(\nu=\Om\text{,}\) which is
with \(A\) and \(\theta\) being arbitrary constants 9
Even if you don’t know how \(x(t) = A\cos(\Om t-\theta)\) was arrived at, you should be able to easily verify that it really does obey \(x''+\Om^2 x=0\text{.}\)
Subsection2.2.1More about \(\vr'(t)\times\vv\big(\vr(t)\big)=\vZero\)
Here is a lemma that gives a more precise version of “if we only care about the curve traced out by the stick, and not about when the stick is at each point of the path, then it suffices to impose the weaker condition \(\vr'(t)\times\vv\big(\vr(t)\big)=\vZero\)”.
Lemma2.2.8.
Lat \(a \lt b\) and let \(\vv(\vr)\) be a vector field. Assume that, for all \(a \lt u \lt b\text{,}\)\(\vR(u)\) is defined, both \(\vR'(u)\) and \(\vv\big(\vR(u)\big)\) are continuous and nonzero and
This \(a(u)=\frac{\vR'(u)\cdot\vv(\vR(u))}{\vv(\vR(u))\cdot\vv(\vR(u))}\) is necessarily nonzero and continuous. Since \(a(u)\) is nonzero and continuous, it never changes sign. That is, either \(a(u) \gt 0\) for all \(u\text{,}\) or \(a(u) \lt 0\) for all \(u\text{.}\) Let \(T(u)\) be an antiderivative of \(a(u)\text{.}\) Then \(T(u)\) is strictly monotone (and continuous) and hence is invertible. That is, there is a continuous function \(U(t)\) that obeys \(U\big(T(u)\big)=u\) for all \(a \lt u \lt b\) and \(T\big(U(t)\big)=t\) for all \(t\) in the range of \(U\text{.}\) Differentiating \(T\big(U(t)\big)=t\) gives \(T'\big(U(t)\big)\,U'(t)=1\) and hence \(U'(t) = \frac{1}{T'(U(t))}\text{.}\) Set \(\vr(t) = \vR\big(U(t)\big)\text{.}\) Then
which forces both \(x(t)\) and \(y(t)\) to be constants. So each field line is just a single point. On the other hand every nonconstant \(\vR(u)\) obeys \(\vR'(u)\times\vv\big(\vR(u)\big)=\vZero\) but is not contained in a field line. (As \(\vR(u)\) is not constant, it covers more than one point, while each field line is just a single point.)
Now here is a more interesting example.
Example2.2.10.
Consider the vector field \(\vv(x,y)=x\,\hi\text{.}\) This vector field takes the value \(\vZero\) at each point on the \(y\)-axis, is a positive multiple of \(\hi\) at every point of the right half-plane and is a negative multiple of \(\hi\) at every point of the left half-plane. Let’s find the field lines. Any field line must obey
So \(y(t)\) must be a constant. We can solve the linear ordinary differential equation \(\diff{x}{t}(t)=x(t)\) by moving the \(x(t)\) to the left hand side, and multiplying by the (integrating factor) \(e^{-t}\text{.}\) This gives
which forces \(e^{-t}x(t)\) to be a constant. So our field lines are \(\big(Ce^t\,,\,D\big)\text{,}\) with \(C\) and \(D\) being arbitrary constants. Note that
if \(C=0\text{,}\) the field line is just the single point \((0,D)\) on the \(y\)-axis. It is illustrated by the black dot in the figure below.
If \(C \gt 0\text{,}\) then as \(t\) runs from \(-\infty\) to \(+\infty\text{,}\) the field line covers the horizontal half-line
But as \(u\) runs from \(-\infty\) to \(+\infty\text{,}\)\(\vR(u)\) runs over the full line \(\Set{(x,D)}{-\infty \lt x \lt \infty}\text{.}\) It is not contained in any single field line and, in fact, completely covers three different field lines.
Exercises2.2.2Exercises
Exercise Group.
Exercises — Stage 1
1.
Suppose that the vector field \(\vv(x,y)\) sketched below represents the velocity of moving water at the point \((x,y)\) in the first quadrant of the \(xy\)-plane.
Sketch the path followed by a rubber ducky dropped in at the point
\(\displaystyle (0,2)\)
\(\displaystyle (1,0)\)
\(\displaystyle (1,2)\)
2.
Find a vector field \(\vv(x,y)\) for which
\begin{align*}
x(t) &= e^{-t}\cos t\\
y(t) &= e^{-t}\sin t
\end{align*}
is a field line.
Exercise Group.
Exercises — Stage 2
3.(✳).
Consider the function \(f(x,y) = xy\text{.}\)
Explicitly determine the field lines (flow lines) of \(\vF(x,y) = \vnabla f\text{.}\)
Sketch the field lines of \(\vF\) and the level curves of \(f\) in the same diagram.
4.(✳).
Find the field line of the vector field \(\vF= 2y\,\hi+ \frac{x}{y^2}\,\hj+e^y\hk\) that passes through \((1,1,e)\text{.}\)
5.(✳).
Find and sketch the field lines of the vector field \(\vF= x\,\hi+ 3y\,\hj\text{.}\)
