Differential Equations - Fourier Series: Boundary Value Problems

Differential Equations - Fourier Series: Boundary Value Problems - ii

Example 3 Solve the following BVP.

y^{″} + 4 y = 0 y (0) = - 2 y (2 π) = 3

Again, we have the following general solution,

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

This time the boundary conditions give us,

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

In this case we have a set of boundary conditions each of which requires a different value of

c_{1}

in order to be satisfied. This, however, is not possible and so in this case have no solution.

So, with Examples 2 and 3 we can see that only a small change to the boundary conditions, in relation to each other and to Example 1, can completely change the nature of the solution. All three of these examples used the same differential equation and yet a different set of initial conditions yielded, no solutions, one solution, or infinitely many solutions.

Note that this kind of behavior is not always unpredictable however. If we use the conditions

y (0)

and

y (2 π)

the only way we’ll ever get a solution to the boundary value problem is if we have,

y (0) = a y (2 π) = a

for any value of

a

. Also, note that if we do have these boundary conditions we’ll in fact get infinitely many solutions.

All the examples we’ve worked to this point involved the same differential equation and the same type of boundary conditions so let’s work a couple more just to make sure that we’ve got some more examples here. Also, note that with each of these we could tweak the boundary conditions a little to get any of the possible solution behaviors to show up (i.e.zero, one or infinitely many solutions).

Example 4 Solve the following BVP.

y^{″} + 3 y = 0 y (0) = 7 y (2 π) = 0

The general solution for this differential equation is,

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

Applying the boundary conditions gives,

\begin{aligned} 7 & = y (0) = c_{1} \\ 0 & = y (2 π) = c_{1} \cos (2 \sqrt{3} π) + c_{2} \sin (2 \sqrt{3} π) \Rightarrow c_{2} = - 7 \cot (2 \sqrt{3} π) \end{aligned}

In this case we get a single solution,

y (x) = 7 \cos (\sqrt{3} x) - 7 \cot (2 \sqrt{3} π) \sin (\sqrt{3} x)

Example 5 Solve the following BVP.

y^{″} + 25 y = 0 y^{'} (0) = 6 y^{'} (π) = - 9

Here the general solution is,

y (x) = c_{1} \cos (5 x) + c_{2} \sin (5 x)

and we’ll need the derivative to apply the boundary conditions,

y^{'} (x) = - 5 c_{1} \sin (5 x) + 5 c_{2} \cos (5 x)

Applying the boundary conditions gives,

\begin{aligned} 6 & = y^{'} (0) = 5 c_{2} \Rightarrow c_{2} = \frac{6}{5} \\ - 9 & = y^{'} (π) = - 5 c_{2} \Rightarrow c_{2} = \frac{9}{5} \end{aligned}

This is not possible and so in this case have no solution.

All of the examples worked to this point have been nonhomogeneous because at least one of the boundary conditions have been non-zero. Let’s work one nonhomogeneous example where the differential equation is also nonhomogeneous before we work a couple of homogeneous examples.

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Calculus III - 3-Dimensional Space: Equations of Lines