Integral Curves

Section 9.2 Integral Curves

Subsection 9.2.1 Vector Fields and Parametric Curve

I said in the previous section that fields of force and fluid flows would be my primary motivating examples. Let me start by talking about these two fields.

If I have a field of force, it acts on objects in it. (Gravitation force acts on object with mass, electromagnetic force acts on object with charge). A natural question is: if I put an object in a field, what will it do? How will the force act upon it? Presumably, the object will move due to the force. How will it move? What path will it take?

I can ask a similar question about a fluid flow. If I put a bouyant object (not affected by other forces such as gravity) in a fluid, what will happen to it? Presumably, it will move with the movement of the fluid. What path will it take?

In both of these contexts, I ended up asking about a path. A path is a parametric curve. So, so answer this question, there should be a connection between vector field and parametric curves. Let me define this connection.

Definition 9.2.1.

Let \(F: S \rightarrow \RR^n\) be a vector field on a region \(S \subset \RR^n\text{.}\) A parametric curve \(\gamma: [a,b] \rightarrow S \) is called a integral curves of the vector field if, at all points on the curve, \(\gamma^\prime(t) = F(\gamma(t))\text{.}\) That is, the tangents of the parametric curves are the same as the vector field along the whole curve.

The parametric curves are the paths I wanted for my application. If an boyant object starts at rest in a fluid, or a charged object starts in an area with electromagnetic force, the integral curve going through the starting point will be the path the object will move along. (This is not entirely true, I admit: there is a subtlety about the starting velocity of the object and how that interacts with the curve, but I’ll leave that aside for now.)

Subsection 9.2.2 Examples of Integral Curves

Example 9.2.2.

Figure 9.2.3. The Vector Field \(F(x,y) = (-y,x)\)

Figure 9.2.4. The Integral Curves for \(F(x,y) = (-y,x)\)

Consider the vector field \(F(x,y) = (-y,x)\text{,}\) as shown in Figure 9.2.3, with its integral curves in Figure 9.2.4. These integral curves of \(F(x,y) = (-y,x)\) aren’t surprising. Looking just at the vector fields, I can clearly tell that the fluid is travelling in a circular paths. Mathematically, the circles are curves \(\gamma(t) = (a \cos t, a \sin t)\) for a parameter \(a > 0\text{.}\) I can calculate \(\gamma^\prime(t) = (-a\sin t, a \cos t)\text{.}\) Now, I can notice that the first component of the derivative, \(-a \sin t\text{,}\) is the negative of the \(y\) component of the curve. Likewise, the second component of the derivative is exactly the first component of the curve. Therefore, I can write \(\gamma^(t) = (-y,x)\text{,}\) where \(x\) and \(y\) are the component of the curve. In this way, the derivative of the curve (expressed in the original component of the curve) and the vector field have the same expression. The tangents to the curve are the vectors in the vector field.

Example 9.2.5.

Figure 9.2.6. The Vector Field \(F(x,y) = (x,y)\)

Figure 9.2.7. Integral Curves for \(F(x,y) = (x,y)\)

Let \(F(x,y) = (x,y)\text{,}\) as in Figure 9.2.6. The curves \(\gamma(t) = (ae^t,be^t)\) where \(a,b \in \RR\) are integral curves, shown in Figure 9.2.7. They have \(x = ae^t\) and \(y = be^t\text{,}\) and I can calculate tangents: \(\gamma^\prime(t) = (ae^t, be^t)\text{.}\) If I compare with the component of the original curve, this is exactly \((x,y)\text{.}\)

Subsection 9.2.3 Calculating Integral Curves

Integral curves are a very powerful conceptual tool. However, they are usually very hard to calculate and most examples are beyond the scope of this course. However, there will be a few I can attempt. I’ll describe the general process in \(\RR^3\text{.}\)

If \(F = (F_1,F_2,F_3)\) is a vector field and \(\gamma(t) = (\gamma_1, \gamma_2,\gamma_3)\) is a parametric curve, then \(\gamma\) is an integral curve if it satisfies the following system.

\begin{align*} \gamma_1^\prime(t) \amp = F_1(\gamma_1, \gamma_2, \gamma_3)\\ \gamma_2^\prime(t) \amp = F_2(\gamma_1, \gamma_2, \gamma_3)\\ \gamma_3^\prime(t) \amp = F_3(\gamma_1, \gamma_2, \gamma_3) \end{align*}

This is a system (generally non-linear) of three differential equations in three functions, the \(\gamma_i\text{.}\) Even if it is linear, this is still a difficult system to solve, if not impossible. Even a simple field such as \(F(x,y,z) = (y,z,x)\) leads to a difficult system.

\begin{align*} \gamma_1^\prime \amp = \gamma_2\\ \gamma_2^\prime \amp = \gamma_3\\ \gamma_3^\prime \amp = \gamma_1 \end{align*}

Currently, I have no method for approaching this system. (For those who have taken the differential equations course, you may recognize that this is a linear system and that course did cover solving linear systems of differential equation). However, there are some examples with reasonable systems of equations which I will be able to solve with integration.

Example 9.2.8.

Let \(F = (x,y,z)\text{.}\) The system that defined the integral curve here is pleasant since each component only related to itself. Let me write the setup.

\begin{align*} \gamma_1^\prime \amp = \gamma_1 \\ \gamma_2^\prime \amp = \gamma_2 \\ \gamma_3^\prime \amp = \gamma_3 \end{align*}

Each differential equation can be solved directly. These are seperable differential equation, but I can also simply notice that these are the basic exponential equation. Therefore, the solutions must be exponential function. Each exponential function will have an unknown starting value. Here are the solutions.

\begin{align*} \gamma_1^\prime \amp = \gamma_1 \implies \gamma_1(t) = ae^t\\ \gamma_2^\prime \amp = \gamma_2 \implies \gamma_2(t) = be^t\\ \gamma_3^\prime \amp = \gamma_3 \implies \gamma_3(t) = ce^t \end{align*}

The values of \(a\text{,}\) \(b\) and \(c\) will be determined by the starting point of the parametric curve. As with all differential equation, initial conditions determine the unknown constant in the solution. This also gives the entire family of integral curves by making all the possible independent choices for these three constants.

Example 9.2.9.

Let \(F = (1,2x,3y)\text{.}\) Here is the resulting system of differential equation for the integral curves.

\begin{align*} \gamma_1^\prime \amp = 1\\ \gamma_2^\prime \amp = 2\gamma_1\\ \gamma_3^\prime \amp = 3\gamma_2 \end{align*}

This system can be solved by iteration. I can solve the first equation by integration. Then the second equation relies on the first component, so I can use the first solution to solve the second by integration again. Finally, the third relies on the second, so I can use the second function and integration to calculate the third.

\begin{align*} \gamma_1 \amp = \int 1 dt = t + a\\ \gamma_2 \amp = 2 \int \gamma_1 dt = \int 2t + 2a dt = t^2 + 2at + 2b\\ \gamma_3 \amp = 3 \int \gamma_2 dt = 3\int t^2 + 2at + 2b dt = t^3 + 3at^2 + 6bt + 3c\\ \gamma(t) \amp = (t+a, t^2 + 2ta + b, t^3 + 3at^2 + 6bt + 3c) \end{align*}

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