Second-order first-degree autonomous differential equation

From Calculus
Jump to: navigation, search

Definition

Following the convention for autonomous differential equations, we denote the dependent variable by x and the independent variable by t.

Form of the differential equation

A (one-dimensional and degree one) second-order autonomous differential equation is a differential equation of the form:

\frac{d^2x}{dt^2} = F(x, \frac{dx}{dt})

Solution method and formula

We set a variable v = dx/dt Then, we can rewrite dt = (dx)/v. In particular, d^2x/dt^2 = (d/dt)(dx/dt) = dv/dt = dv/((dx)/v) = vdv/dx. Plug this in:

v \frac{dv}{dx} = F(x, v)

Solve this to obtain the general solution for v in terms of x. Plug this expression in:

\frac{dx}{dt} = v(x)

and solve this first-order differential equation. Note that if v is not expressible as an explicit function of x, but we instead have a relational solution R(x,v) = 0, then solve the first-order differential equation:

R(x, \frac{dx}{dt}) = 0

Particular cases

Case where the function on the right depends only on x and not on dx/dt

Consider a situation of the form:

\frac{d^2x}{dt^2} = g(x)

We do the same substitution v = dx/dt and obtain:

\frac{v \, dv}{ \, dx} = g(x)

This is now a separable differential equation relating x and v. Integrate and obtain:

\int v \, dv = \int g(x) \, dx

We thus get:

\frac{v^2}{2} = \int g(x) \, dx

In particular, if G is an antiderivative for g, then we get:

v = \pm \sqrt{2G(x) + C_1}

where C_1 \in \R is a parameter. Each choice of C_1 gives a different solution.

Plug this back in and get:

\frac{dx}{dt} = \pm \sqrt{2G(x) + C_1}

(The \pm indicates that there are in fact two differential equations and we need to take the union of their solution sets).

This is a first-order autonomous differential equation, and in particular a separable differential equation. Rearrange and get:

\pm \int \frac{dx}{\sqrt{2G(x) + C_1}} = \int dt

An additional constant, C_2, arises from this indefinite integration. The upshot is that the general solution relates x to t and has two parameters C_1,C_2, as we might expect from the degree of the equation.

Case where the function on the right is multiplicatively separable

Consider a situation of the form:

\frac{d^2x}{dt^2} = g(x)h\left(\frac{dx}{dt}\right)

We do the same substitution v = dx/dt and obtain:

\frac{v \, dv}{ \, dx} = g(x)h(v)

This is a separable differential equation and we can rearrange it to obtain:

\int \frac{v \, dv}{h(v)} = \int g(x) \, dx

We now perform the integration both sides. Suppose H(v) = \int \frac{v \, dv}{h(v)} and G(x) =\int g(x) \, dx. We get:

H(v) = G(x) + C_1

If H can locally be inverted, we can write v as an explicit function of x We now plug this into the original differential equation and get:

H(dx/dt) = G(x) + C_1

This is now a first-order differential equation. If H can locally be inverted, we can write dx/dt as an explicit function of x (locally) and then solve the resultant separable differential equation. Otherwise, there may be some other method available.

Example

Separable example

Consider the following differential equation:

\frac{d^2x}{dt^2} = 2x\frac{dx}{dt}

We describe how to solve this using the technique discussed here. First, set v = dx/dt. We get:

v \frac{dv}{dx} = 2xv

Thus, we get:

v = 0 \qquad \mbox{ or } \qquad \frac{dv}{dx} = 2x

The case v = 0 solves to gives x = C_0, C_0 \in \R

The case:

\frac{dv}{dx} = 2x

solves to give:

v = x^2 + C_1

Now, we plug back the original substitution to get:

\frac{dx}{dt} = x^2 + C_1

We now make cases based on the sign of C_1.

Case C_1 = 0

In this case, the differential equation becomes:

\frac{dx}{dt} = x^2

This has a stationary solution x = 0 (which, incidentally, is already included in the prior solution family of all constant solutions) and the general solution can be obtained as:

\int \frac{dx}{x^2} = \int dt

This gives:

\frac{-1}{x} = t + C_2

This solves to give:

x = \frac{-1}{t + C_2}

To verify that this is a solution, we compute, using the chain rule for differentiation and differentiation rule for power functions:

\frac{dx}{dt} = \frac{1}{(t + C_2)^2}

and:

\frac{d^2x}{dt^2} = \frac{-2}{(t + C_2)^3}

We can now plug these into the original differential equation and verify.

Case C_1 > 0

Let k = \sqrt{C_1}. The differential equation becomes:

\frac{dx}{dt} = x^2 + k^2

There are no stationary solutions. The general solution is given by:

\int \frac{dx}{x^2 + k^2} = \int dt

This solves to give:

\frac{1}{k} \arctan\left(\frac{x}{k}\right) = t + C_2

Rearranging, we get:

x = k \tan(k(t + C_2)), \qquad k > 0, C_2 \in \R

Note that a priori, applying tan to both sides causes us to lose the information that k(t + C_2) is in the range of \arctan. However, this constraint is "artificial" anyway and can be got rid of by readjusting the value of C_2. Note that arc tangent constraints are artificial, but arc sine constraints are genuine, because the sine function repeats itself within a period but the tangent function does not.

To verify that this is a solution, we can compute:

\frac{dx}{dt} = k^2 \sec^2(k(t + C_2))

\frac{d^2x}{dt^2} = 2k^3 \sec^2(k(t + C_2)) \tan(k(t + C_2))

We can now plug in and verify the original differential equation.

Case C_1 < 0

Let l = \sqrt{-C_1}. The differential equation becomes:

\frac{dx}{dt} = x^2 - l^2

There are stationary solutions corresponding to x = l and x = -l. However, all stationary solutions are already included in another solution family, so we can ignore these. We continue solving:

\int \frac{dx}{x^2 - l^2} = \int dt

We get:

\frac{1}{2l} \ln \left|\frac{x - l}{x + l}\right| = t + C_2

This rearranges to gives:

\ln \left|\frac{x - l}{x + l}\right| = 2l(t + C_2)

Exponentiate both sides to get:

\frac{x - l}{x + l} = \exp(2l(t + C_2)) \qquad \mbox{ or } \frac{x - l}{x + l} = -\exp(2l(t + C_2))

The solutions are respectively:

x = l\left(\frac{1 + \exp(2l(t + C_2))}{1 - \exp(2l(t + C_2))}\right) \qquad \mbox{ or } x = l\left(\frac{1 - \exp(2l(t + C_2))}{1 + \exp(2l(t + C_2))}\right)

All cases

The following are the solution families:

  • x = C_0, C_0 \in \R (this is a singular solution family)
  • x = \frac{-1}{t + C_2}, C_2 \in \R
  • x = k \tan(k(t + C_2)), \qquad k > 0, C_2 \in \R
  • x = l\left(\frac{1 + \exp(2l(t + C_2))}{1 - \exp(2l(t + C_2))}\right), \qquad l > 0, C_2 \in \R
  • x = l\left(\frac{1 - \exp(2l(t + C_2))}{1 + \exp(2l(t + C_2))}\right), \qquad l > 0, C_2 \in \R