Differential Equations - Fourier Series: Boundary Value Problems

Differential Equations - Fourier Series: Boundary Value Problems - iii

Example 6 Solve the following BVP.

y^{″} + 9 y = \cos x y^{'} (0) = 5 y (\frac{π}{2}) = - \frac{5}{3}

The complementary solution for this differential equation is,

y_{c} (x) = c_{1} \cos (3 x) + c_{2} \sin (3 x)

Using Undetermined Coefficients or Variation of Parameters it is easy to show (we’ll leave the details to you to verify) that a particular solution is,

Y_{P} (x) = \frac{1}{8} \cos x

The general solution and its derivative (since we’ll need that for the boundary conditions) are,

\begin{aligned} y (x) & = c_{1} \cos (3 x) + c_{2} \sin (3 x) + \frac{1}{8} \cos x \\ y^{'} (x) & = - 3 c_{1} \sin (3 x) + 3 c_{2} \cos (3 x) - \frac{1}{8} \sin x \end{aligned}

Applying the boundary conditions gives,

\begin{aligned} 5 & = y^{'} (0) = 3 c_{2} \Rightarrow c_{2} = \frac{5}{3} \\ - \frac{5}{3} & = y (\frac{π}{2}) = - c_{2} \Rightarrow c_{2} = \frac{5}{3} \end{aligned}

The boundary conditions then tell us that we must have

c_{2} = \frac{5}{3}

and they don’t tell us anything about

c_{1}

and so it is can be arbitrarily chosen. The solution is then,

y (x) = c_{1} \cos (3 x) + \frac{5}{3} \sin (3 x) + \frac{1}{8} \cos x

and there will be infinitely many solutions to the BVP.

Let’s now work a couple of homogeneous examples that will also be helpful to have worked once we get to the next section.

Example 7 Solve the following BVP.

y^{″} + 4 y = 0 y (0) = 0 y (2 π) = 0

Here the general solution is,

y (x) = c_{1} \cos (2 x) + c_{2} \sin (2 x)

Applying the boundary conditions gives,

\begin{aligned} 0 & = y (0) = c_{1} \\ 0 & = y (2 π) = c_{1} \end{aligned}

c_{2}

is arbitrary and the solution is,

y (x) = c_{2} \sin (2 x)

and in this case we’ll get infinitely many solutions.

Example 8 Solve the following BVP.

y^{″} + 3 y = 0 y (0) = 0 y (2 π) = 0

The general solution in this case is,

y (x) = c_{1} \cos (\sqrt{3} x) + c_{2} \sin (\sqrt{3} x)

Applying the boundary conditions gives,

\begin{aligned} 0 & = y (0) = c_{1} \\ 0 & = y (2 π) = c_{2} \sin (2 \sqrt{3} π) \Rightarrow c_{2} = 0 \end{aligned}

In this case we found both constants to be zero and so the solution is,

y (x) = 0

In the previous example the solution was

y (x) = 0

. Notice however, that this will always be a solution to any homogeneous system given by

(5)

and any of the (homogeneous) boundary conditions given by

(1)

–

(4)

. Because of this we usually call this solution the trivial solution. Sometimes, as in the case of the last example the trivial solution is the only solution however we generally prefer solutions to be non-trivial. This will be a major idea in the next section.

Before we leave this section an important point needs to be made. In each of the examples, with one exception, the differential equation that we solved was in the form,

y^{″} + λ y = 0

The one exception to this still solved this differential equation except it was not a homogeneous differential equation and so we were still solving this basic differential equation in some manner.

So, there are probably several natural questions that can arise at this point. Do all BVP’s involve this differential equation and if not why did we spend so much time solving this one to the exclusion of all the other possible differential equations?

The answers to these questions are fairly simple. First, this differential equation is most definitely not the only one used in boundary value problems. It does however exhibit all of the behavior that we wanted to talk about here and has the added bonus of being very easy to solve. So, by using this differential equation almost exclusively we can see and discuss the important behavior that we need to discuss and frees us up from lots of potentially messy solution details and or messy solutions. We will, on occasion, look at other differential equations in the rest of this chapter, but we will still be working almost exclusively with this one.

There is another important reason for looking at this differential equation. When we get to the next chapter and take a brief look at solving partial differential equations we will see that almost every one of the examples that we’ll work there come down to exactly this differential equation. Also, in those problems we will be working some “real” problems that are actually solved in places and so are not just “made up” problems for the purposes of examples. Admittedly they will have some simplifications in them, but they do come close to realistic problem in some cases.

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Calculus III - 3-Dimensional Space: Equations of Lines