diff options
| -rw-r--r-- | ClassJuly5.page | 3 |
1 files changed, 2 insertions, 1 deletions
diff --git a/ClassJuly5.page b/ClassJuly5.page index 19b3d47..fa72182 100644 --- a/ClassJuly5.page +++ b/ClassJuly5.page @@ -35,7 +35,7 @@ Rearranging, we get: $$ \frac{da}{dt}/a = \frac{d^2b}{dx^2}/b $$ -The left hand side only depends on $t$ and the right hand side depends only on $x$. Therefore, neither side depends on either $x$ or $t$. We conclude that both sides are equal to a constant, which for convenience we'll write as $-\lambda^2$. +The left hand side only depends on $t$ and the right hand side depends only on $x$. Therefore, neither side depends on either $x$ or $t$. We conclude that both sides are equal to a constant, which for convenience we'll write as $-\lambda^2$. First we solve for $a$. It satisfies the equation @@ -62,6 +62,7 @@ We would like to assert that any solution takes this form. One way to prove thi $$ f(x) = \sum_{n = 1}^{\infty} c_n \sin(\frac{\pi n x}{L}) $$ One could then allow the coefficients $c_n$ to vary with $t$ and apply the same method of solution that we used in the case of periodic boundary conditions. + In fact, every function of the kind described above does have a Fourier sine expansion. The link above contains a hint of how to do it. First you extend the function $f(x)$ to a certain odd, periodic function $\tilde{f}(x)$ defined on the interval $[-L,L]$. Then you can use convergence of the usual Fourier series for $\tilde{f}(x)$. ## Convergence for not-so-nice Fourier series. |