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diff --git a/ClassJuly5.page b/ClassJuly5.page index 632ecec..19b3d47 100644 --- a/ClassJuly5.page +++ b/ClassJuly5.page @@ -7,28 +7,45 @@ Below are answers to some questions that came up during the lecture. If anyone Suppose instead of having a metal ring we had a metal rod of length $L$, and we kept the ends of the rod at constant temperatures $u_0$ and $u_L$. How might we solve the heat equation in this context? There is one obvious solution that satisfies these boundary conditions, namely the time-independent or steady state solution + $$ g(x,t) = u_0 + \frac{u_L- u_0}{L}x $$ + 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 + $$ u(x,t) = g(x,t) - h(x,t)$$ -also satisfies the heat equation, but it is zero at both endpoints. + +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 + $$u(x,t) = a(t)b(x)$$ + 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: + $$ \frac{da}{dt}/a = \frac{d^2b}{dx^2}/b $$ + 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 $-\lambda^2$. First we solve for $a$. It satisfies the equation + $$ \frac{da}{dt} = - \lambda^2 a$$ + and therefore (up to a constant multiple) it is given by: + $$ a(t) = e^{-\lambda^2 t} $$ + Next we solve for $b$. It satisfies the equation $$ \frac{d^2b}{dx^2} = -\lambda b $$ However we have to be a bit more careful in picking our solutions because $b$ is supposed to satisfy the boundary conditions |