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| -rw-r--r-- | ClassJuly5.page | 8 |
1 files changed, 4 insertions, 4 deletions
diff --git a/ClassJuly5.page b/ClassJuly5.page index dde84c7..632ecec 100644 --- a/ClassJuly5.page +++ b/ClassJuly5.page @@ -11,13 +11,13 @@ $$ g(x,t) = u_0 + \frac{u_L- u_0}{L}x $$ This satisfies the heat equation for trivial reasons since it is time-independent and its second spatial derivative is zero, hence both sides of the heat equation are zero independently of one another. Now suppose that $h(x,t)$ is another solution satisfying the same boundary conditions. Then the function -$$ u(x,t) = g(x,t) - h(x,t)$ +$$ u(x,t) = g(x,t) - h(x,t)$$ also satisfies the heat equation, but it is zero at both endpoints. $$ u(x,0) = u(x,L) = 0 $$ To solve for $h$, it is clear that we only need to solve for $u$. First we'll use what's called the ``separation of variables'' trick to generate a lot of nice solutions, then hope and pray that any other solution can be expressed as a linear combination of these. Here's how the separation of variables trick works. We seek a solution of the form -$$u(x,t) = a(t)b(x)$ +$$u(x,t) = a(t)b(x)$$ for some functions $a$ and $b$. Then the heat equation tells us: $$ \frac{da}{dt}b = a \frac{d^2b}{dx^2} $$ Rearranging, we get: @@ -42,10 +42,10 @@ So, the most general solution we can generate in this manner is: $$ u(x,t) = \sum_{n = 1}^{\infty} c_n e^{-(\frac{\pi n}{L})^2t} \sin(\frac{\pi n x}{L}) $$ We would like to assert that any solution takes this form. One way to prove this assertion would be to show that any function $f:[0,L] \to \R$ satisfying $f(0) = f(L) = 0$ has a unique ``[Fourier sine expansion](http://mathworld.wolfram.com/FourierSineSeries.html)'': -$$ f(x) = \sum_{n = 1}^{\infty} c_n \sin(\frac{\pi n x}{L}) +$$ 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 an odd, periodic function $\tilde{f}(x)$ defined on the interval $[-L,L]$. Then you can use convergence of the usual Fourier series for $\tilde{f}(x)$. +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. |