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9.3 (continued)

Boundary-Value Problems (BVP)

\[x'' + ax = f(t)\]
\(x(0) = 0, x(L) = 0\)

values at the ends and not at initial \(t\)

Dirichlet condition

(values specified)

\(x'(0) = 0, x'(L) = 0\)

Neumann condition

(rates specified)

let's look at \(x'' + ax = 0\) with boundary conditions \(x(0) = x(L) = 0\)

\[x(t) = A \cos(\sqrt{a} t) + B \sin(\sqrt{a} t)\]
  • \(x(0) = 0 \rightarrow A = 0\)
  • \(x(L) = 0 \rightarrow 0 = B \sin(\sqrt{a} L)\)\(B \neq 0\) (otherwise \(x = 0\) for ALL \(t\) \(\rightarrow\) trivial solution)
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then \(\sin(\sqrt{a} L) = 0 \implies \sqrt{a} L = n\pi \quad n = 1, 2, 3, \dots\)

for each \(n\), the fundamental solution is

\[x_n = \sin\left(\frac{n\pi}{L}t\right) \quad \text{(B is scaling constant, not important)}\]

then general solution is a linear combination of ALL

\[x(t) = b_1 \sin\left(\frac{\pi}{L}t\right) + b_2 \sin\left(\frac{2\pi}{L}t\right) + b_3 \sin\left(\frac{3\pi}{L}t\right) + \dots\]

this is solution to \(x'' + ax = 0, \quad x(0) = x(L) = 0\)

Sine series w/ half period L

if we had used the Neumann condition \(x'(0) = x'(L) = 0\)

we would get a Cosine series.

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Non-Homogeneous Boundary Value Problems

Now back to:

\[x'' + ax = f(t) \quad x(0) = x(L) = 0\]

Sine series if \(f(t) = 0\):

\[\sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi t}{L}\right) \quad \text{each } b_n \text{ is dependent on } f(t)\]

We need to make \(f(t)\) periodic (because solution \(x\) is periodic) by adding either even or odd extensions.

→ Pick the one that would make \(f(t)\) satisfy the boundary conditions.

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Example

\[x'' + 2x = 1 \quad x(0) = 0, \quad x(\pi) = 0\]

Note: \(L = \pi\)

\(f(t) = 1\). Let's look at both extensions to see which is appropriate (satisfies BC's).

Odd Extension

With odd extensions, its Fourier series (in red) satisfies the BC's: \(f(0) = f(\pi) = 0\).

Graph of an odd square wave extension of f(t)=1 from -pi to pi, showing zero values at boundaries.

Even Extension

With even extensions, its Fourier series (in red) does NOT satisfy BC's.

But, notice it would satisfy \(x'(0) = x'(\pi) = 0\).

Graph of an even extension of f(t)=1, showing a constant value of 1 across the interval.
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For this example, we must pick odd extensions.

\[ f(t) = \begin{cases} -1 & -\pi < t < 0 \\ 1 & 0 < t < \pi \end{cases} \quad \text{period } 2\pi \quad (L = \pi) \]

Its Fourier (sine) series is

\[ f(t) \sim \sum_{n=1}^{\infty} \frac{2}{n\pi} \left[ 1 - (-1)^n \right] \sin(nt) \]

Back to

\[ x'' + 2x = 1 \quad x(0) = x(\pi) = 0 \]

has sines series solution if \( f(t) = 0 \) w/ unknown coefficients

\[ x(t) = \sum_{n=1}^{\infty} B_n \sin(nt) \]

Sub into the diff. eq. and replace right side w/ its own series

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\[ x'(t) = \sum_{n=1}^{\infty} n B_n \cos(nt) \]\[ x''(t) = \sum_{n=1}^{\infty} -n^2 B_n \sin(nt) \]

\( x'' + 2x = 1 \) becomes

\[ \sum_{n=1}^{\infty} -n^2 B_n \sin(nt) + 2 \sum_{n=1}^{\infty} B_n \sin(nt) = \sum_{n=1}^{\infty} \frac{2}{n\pi} \left[ 1 - (-1)^n \right] \sin(nt) \]

for each \( n = 1, 2, 3, \dots \) equate coefficients of \( \sin(nt) \)

\[ -n^2 B_n + 2 B_n = \frac{2}{n\pi} \left[ 1 - (-1)^n \right] \]\[ B_n = \frac{2 \left[ 1 - (-1)^n \right]}{n\pi (2 - n^2)} \quad n = 1, 2, 3, \dots \]

Solution:

\[ x(t) = \sum_{n=1}^{\infty} \frac{2 \left[ 1 - (-1)^n \right]}{n\pi (2 - n^2)} \sin(nt) \]\[ = \frac{4}{\pi} \sin(t) - \frac{4}{21\pi} \sin(3t) - \frac{4}{105\pi} \sin(5t) - \dots \]