Dirichlet condition (values at ends)
not at same place like initial-value problem
Neumann condition (rates at ends)
if \(f(t) = 0\),
\[x(t) = A \cos(\sqrt{a} t) + B \sin(\sqrt{a} t)\]
\[x(0) = 0 \rightarrow 0 = A\]
trivial solution
\(B \neq 0\) (otherwise \(x = 0\) for all \(t\))
this means, for each \(n = 1, 2, 3, \dots\) there is one fundamental solution:
Note: \(\sqrt{a} = \frac{n\pi}{L}\)
there are infinitely-many \(n\)'s, so the general solution is a linear combination of ALL:
Sine series w/ half period L
The equation above is solved by the sine series. Each term \(\sin\left(\frac{n\pi}{L}t\right)\) satisfies the boundary conditions.
If we had \(x'(0) = x'(L) = 0\), we would get a cosine series.
Now let's look at \(f(t) \neq 0\):
Dirichlet condition
(HW: Neumann)
This leads to a sine series if \(f(t) = 0\).
We need to add the appropriate extensions (even or odd) to \(f(t)\) to make it periodic such that it satisfies the boundary conditions.
The boundary conditions \(x(0)=0\) and \(x(\pi)=0\) tell us that \(L = \pi\).
For \(f(t) = 1\), we need to add even or odd extensions so it satisfies BC's and is periodic.
With odd extensions, its Fourier series (indicated in red on the graph) satisfies the boundary conditions:
Notice it satisfies \(f(0) = f(\pi) = 0\) (BC's).
With even extensions, its Fourier series (indicated in red) does NOT satisfy the BC's.
BUT, it DOES satisfy if \(x'(0) = x'(\pi) = 0\).
So, we want odd extensions in this example for \( f(t) = 1 \), with \( L = \pi \).
We write out a Fourier series for this:
Back to \( x'' + 2x = 1 \) with boundary conditions \( x(0) = x(\pi) = 0 \).
If right side is 0: Sine series
Express \( x(t) \) as a sine series with unknown coefficients with \( L = \pi \):
Sub into \( x'' + 2x = 1 \).
Sub into \( x'' + 2x = 1 \) with 1 replaced by its Fourier Series:
For each \( n = 1, 2, 3, \dots \) equate coefficients of \( \sin(nt) \):