Value of k and boundary conditions for "Heat on $\mathbb R$" and "Heat on $[-1,1]$"
@mark For the final part of both questions where we are supposed to compute estimates of $u(0,0.1)$, should we keep the default value of $k=1$?
Also, for the "Heat on $[-1,1]$, problem, is the boundary condition $u(2,0) = 0$ correct, or should it be $u(2,t) = 0$, analogous to the other boundary condition $u(0,t) = 0$ in the same problem?
On a related note, don't the boundary conditions for the second problem mean that $u(0,0.1)=0$, since $u(0,t)=0$ for any $t$?
Comments
@mark Also, I just realized: The ObservableHQ page on unbounded heat flow (and our notes) says that the solution to an unbounded heat flow problem is $$u(x,t) = \int_{-\infty}^{\infty} f(y)G(y-x,t)dy,$$
while the book writes the same equation as $$u(x,t) = \int_{-\infty}^{\infty} f(y)G(x-y,t)dy.$$
Which is correct?
For the final part of both questions where we are supposed to compute estimates of $u(0,0.1)$, should we keep the default value of $k=1$?
$k$ is a parameter; any positive number generates a reasonable problem.
Also, for the "Heat on $[-1,1]$, problem, is the boundary condition $u(2,0) = 0$ correct, or should it be $u(2,t) = 0$, analogous to the other boundary condition $u(0,t) = 0$ in the same problem?
On a related note, don't the boundary conditions for the second problem mean that $u(0,0.1)=0$, since $u(0,t)=0$ for any $t$?
Sure.
(On unbounded heat flow) Which is correct?
You could try both and then report back on why the results are identical.
I see that for this problem, the two are identical because $(x-y)$/$(y-x)$ is squared, so you get the same result. I was just curious about the general formula for convolution. I would imagine there are some functions where the convolution would yield different results depending on which definition you used.