Last modified on 27 November 2010, at 05:37

Ordinary Differential Equations/Higher 1

Initial value problemsEdit

Initial value problems still exist in higher orders. However, since we have to integrate twice or more, there will be more than one constant of integration, and it will not always be a single arbitrary. It will be shown in a later section by the existence theorems that an equation of order n in an initial value problem satisfying suitable conditions will a solution given n arbitrary contents, and therefore will contain, in general, n arbitrary constants.

Second order initial value problemsEdit

Let's take a simple equation:

\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=2.

To solve this, we integrate twice:

\frac{\mathrm{d}y}{\mathrm{d}x}=2x+C
y=x^2+Cx+D. \,

Plug it in and check, this is the solution. Notice the 2 unknowns. In order to solve that, we need two initial conditions. For normal equations, this would do it - two equations, two unknowns. The problem here is that the two constants depend on each other. Suppose we had the initial conditions

y(x_0)=b_0 \, and y(x_1)=b_1.\,

There are multiple curves for y that could hit those two points.

How can we get around that? We can't do it with 2 initial conditions for y. Instead, we need an initial condition for y and y'. Even more - they need to be for the same point. We need

y'(a)=b_0 \, and y(a)=b_1 \,.

With these you can plug in y' and solve for C, then solve for D.

nth order initial value problemsEdit

So, generalizing, for the particular solution to an nth-order DE, we need n initial conditions. The initial conditions must take the form

y(a)=b_0,\ y'(a)=b_1,\ y''(a)=b_2,\ldots,\ y^{(n)}(a)=b_n

Separable EquationsEdit

Bad news - separable equations don't really exist in higher orders. You end up with only a small subset of problems that can be solved like that.

Second order separablesEdit

There are no real second order or higher separable equations. Mainly because we have 3 variables. However, suppose we had an equation of the form

y''=f(y')g(x). \,

While that isn't a separable equation we can solve for y - it is a first order separable for y'. We can replace y' with v (and y'' with v') and solve for v, then integrate with respect to x to solve for y.

Example needed

nth order separablesEdit

We can do the same thing in nth order equations. If we have an equation of the form

y^{(n)}=f(y^{(n-1)})g(x), \,

we can use separable techniques to solve for y(n-1) and integrate n-1 times to solve for y.

Linear EquationsEdit

Linear equations do exist in higher orders. They aren't as easy to solve though - there is no method like we demonstrated for first order equations. We'll spend the other lessons in this chapter dealing with the one easily solved subset - constant coefficients.

Existence and Uniqueness of SolutionsEdit

In first order equations, we were promised a unique solution if all terms of the function were continuous. In n dimensions, its similar.

Given an initial value problem

y^{(n)}+p_1(x)y^{(n-1)}+...+p_n(x)y=f(x)\,

with initial conditions at x=a, we are promised a unique solution if

p_1(x),\ p_2(x),\ldots,\ p_n(x) and f(x) \,

are all continuous over some interval I.

Superposition principleEdit

In many higher order equations, there are more than one solution to a problem. For example, the second-order DE

\frac{\mathrm{d}^2y}{\mathrm{d}x^2}-y=0

is satisfied by both

y=C_1e^x \, and y=C_2e^{-x} \,.

Plug them in - they both work. So how do we get that unique solution we were promised?

We add them. This is the superposition principle:

If y_1=f(x) \, and y_2=g(x) \, are both solutions to a linear differential equation, then y=f(x)+g(x) \, is also a solution.

Let's illustrate this. Suppose that u and v are solutions, where y=u and y=v are distinct solutions, comprised of functions of x, of the DE

(1)a\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+b\frac{\mathrm{d}y}{\mathrm{d}x}+cy=0.

Then,

(2)a\frac{\mathrm{d}^2u}{\mathrm{d}x^2}+b\frac{\mathrm{d}u}{\mathrm{d}x}+cu=0

and

(3)a\frac{\mathrm{d}^2v}{\mathrm{d}x^2}+b\frac{\mathrm{d}v}{\mathrm{d}x}+cv=0

A second order DE needs two arbitrary constants to properly define it. So, assume that the general solution is

(4)y=Au+Bv, \,

where A and B are our constants. Then,

(5)\frac{\mathrm{d}y}{\mathrm{d}x}=A\frac{\mathrm{d}u}{\mathrm{d}x}+B\frac{\mathrm{d}v}{\mathrm{d}x}

and

(6)\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=A\frac{\mathrm{d}^2u}{\mathrm{d}x^2}+B\frac{\mathrm{d}^2v}{\mathrm{d}x^2}

Substituting back into our original DE, (1), we get

(7)a \left( A\frac{\mathrm{d}^2u}{\mathrm{d}x^2}+B\frac{\mathrm{d}^2v}{\mathrm{d}x^2} \right)
+b \left( A\frac{\mathrm{d}u}{\mathrm{d}x}+B\frac{\mathrm{d}v}{\mathrm{d}x} \right)
+c \left(Au +Bv \right)
=0

Now, regrouping to get all functions of u together, and functions of v together, we have

(8)A \left( a \frac{\mathrm{d}^2u}{\mathrm{d}x^2}+b\frac{\mathrm{d}u}{\mathrm{d}x}+cu \right)
+B \left( a \frac{\mathrm{d}^2v}{\mathrm{d}x^2}+b\frac{\mathrm{d}v}{\mathrm{d}x}+cv \right)
=0

From our definitions in (2) and (3), we have

(9)A\times0+B\times0=0 \,

Now, since A and B are just numbers, this is true, therefore our original assumption must be true, so

y=Au+Bv \,

is the general solution, and y=u and y=v are distinct solutions. Now to solve the equation, we need to find u and v. This is done in the next lesson.