The PDE is $u_t = D \, u_{xx}$

**The steady state solution is constant: $u_{ss}(x)=1$.**

The initial data are the $k=1$, $k=2$, and $k=3$ modes: $u(x,0)= 1 + \sin(\pi x/L)$, $u(x,0) = 1 + \sin(2 \pi x/L)$, and $u(x,0)=1 + \sin(3 \pi x/L)$

The resultant solutions are \begin{align*} u(x,t) &= 1 + e^{-D \, (\pi/L)^2 t} \, \sin(\pi x/L)\\ u(x,t) &= 1 + e^{-D (2 \pi/L)^2 t} \, \sin(2 \pi x/L)\\ u(x,t) &= 1 + e^{-D (3 \pi/L)^2 t} \, \sin(3 \pi x/L) \end{align*}

The steady state is denoted with a dashed line.

Note that the $k=3$ mode decays faster than the $k=2$ mode which decays
faster than the $k=1$ mode.

The initial data is chosen by choosing random numbers and then
multiplying them by $\sin(k \pi x/L)$ for $k=1$ to $20$.

The steady state is denoted with a dashed line.

Note that the $k=1$ mode decays the slowest and as the solution relaxes
it looks more and more like a multiple of $\sin(\pi x/L)$.

The initial data is a discontinuous function: $$ u(x,0)= \begin{cases} 1 & \mbox{if $L/4 < x < L/2$} \\ 0 & \mbox{otherwise} \end{cases} $$

The steady state
$u_{ss}(x)=1$ is denoted with a dashed line.

Note that the solution instantaneously smooths. The regions that were
flat in the initial data appear to "take a while" to stop being flat
if you look away from the two discontinuities. (In reality, there’s
infinite speed of propagation and there’s no flatness anywhere, it’s
just that our eye can’t see the deviation at first.) The boundary
condition is driving the solution down to the steady state; note that
the $x=0$ boundary is "felt" by the solution before the $x=L$ boundary.
Again, this has to do with infinite speed of propagation and how the
"width" of the fundamental solution depends on time.

The initial data is a “witch’s hat”. The solution is provided for
your amusement; I have nothing to say that I didn’t say for the
solution with top-hat initial data.

The steady state
$u_{ss}(x)=1$ is denoted with a dashed line.

**The steady state solution is a line: $u_{ss}(x) =1 + x/L$**

The initial data are the $k=1$, $k=2$, and $k=3$ modes: $u(x,0) =
u_{ss}(x) + \sin(\pi x/L)$, $u(x,0) = u_{ss}(x) + \sin(2 \pi x/L)$,
and $u(x,0) = u_{ss}(x) + \sin(3 \pi x/L)$.

The resultant solutions are \begin{align*} u(x,t) &= u_{ss}(x) + e^{-D (\pi/L)^2 t} \sin(\pi x/L)\\ u(x,t) &= u_{ss}(x) + e^{-D (2 \pi/L)^2 t} \sin(2 \pi x/L)\\ u(x,t) &= u_{ss}(x) + e^{-D (3 \pi/L)^2 t} \sin(3 \pi x/L). \end{align*}

The steady state is denoted with a dashed line.

Note that the $k=3$ mode decays faster than the $k=2$ mode which decays
faster than the $k=1$ mode.

The initial data is chosen by choosing random numbers and then
multiplying them by $\sin(k \pi x/L)$ for $k=1$ to $20$.

The steady state is denoted with a dashed line.

Note that the $k=1$ mode decays the slowest and as the solution relaxes
it looks more and more like a multiple of $\sin(\pi x/L)$ added to the steady state $u_{ss}(x)$.

The initial data is $u_{ss}(x)$ with a discontinuous function added to it: $$ u(x,0) = \begin{cases} u_{ss}(x) + 1 & \mbox{if $L/4 < x < L/2$}\\ u_{ss}(x) & \mbox{otherwise} \end{cases} $$

The steady state
is denoted with a dashed line.

Note that the solution instantaneously smooths. The regions that were
flat in the initial data appear to “take a while” to stop being flat
if you look away from the two discontinuities. (In reality, there’s
infinite speed of propagation and there’s no flatness anywhere, it’s
just that our eye can’t see the deviation at first.) The boundary
conditions are driving the solution down to the steady state; note
that the x=0 boundary is “felt” by the solution before the x=L
boundary. Again, this has to do with infinite speed of propagation
and how the “width” of the fundamental solution depends on time.

**
In the
no-flux boundary condition case, there are infinitely many steady
states: they're all constant.
The long-time limit of an initial value problem is steady state that's determined by the
initial data: $u_{ss}(x) =$ the average of $u(x,0)$.
**

The initial data is chosen by choosing random numbers and then
multiplying them by $\cos(k \pi x/L)$ for $k=1$ to $20$.

The steady state is found by computing the average vaule of the inital data and is denoted with a dashed line.

Note that the $k=1$ mode decays the slowest and as the solution
relaxes it looks more and more like a multiple of $\cos(\pi x/L)$
added to the steady state.

The initial data is a discontinuous function: $$ u(x,0)= \begin{cases} 1 & \mbox{if $L/4 < x < L/2$} \\ 0 & \mbox{otherwise} \end{cases} $$

The steady state is found by computing the average vaule of the inital data and is denoted with a dashed line.

Note that the solution instantaneously smooths. The regions that were
flat in the initial data appear to “take a while” to stop being flat
if you look away from the two discontinuities. (In reality, there’s
infinite speed of propagation and there’s no flatness anywhere, it’s
just that our eye can’t see the deviation at first.) The boundary
conditions are driving the solution down to the steady state; note
that the x=0 boundary is “felt” by the solution before the x=L
boundary. Again, this has to do with infinite speed of propagation
and how the “width” of the fundamental solution depends on time.

Here the heat equation is $u_t = ( D(x) \, u_x )_x$ where the diffusivity $D(x)$ depends on $x$. It’s one constant on $(0,L/2)$, and a larger constant on $(L/2,L)$. As a result, in the material to the right the diffusion is “faster”.

The boundary conditions are Dirichlet boundary conditions $u(0,t) = 0$ and $u(L,t) = 0$. And so the steady
state is: $u_{ss}(x) = 0$.

The initial data is $\sin(\pi x/L)$: a single bump with its peak initially
at the interface between the two materials.

Note: If the boundary conditions had been $u(0,t) = 1$ and $u(L,t) = 2$
then the steady state would be piecewise linear. To find it, you would need to
use the fact that both $u_{ss}(x)$ and the flux $D(x) \, {u_{ss}}_x(x)$ are
continuous at $x=L/2$.

The PDE is $u_t = c^2 \, u_{xx}$ on the line. No boundary conditions are needed. For initial data, both the initial displacement $u(x,0)$ and the initial velocity $u_t(x,0)$ are given.

The initial displacement is $u(x,0)$ is the discontinuous function
$$ u(x,0)=
\begin{cases} 1 & \mbox{if $0 < x < 2$} \\
0 & \mbox{otherwise}
\end{cases}
$$
The initial
velocity is zero.

Note that there is no instantaneous smoothing. In fact the initial
data has only two jump discontinuities in it while the solution at t>0
has more than two. Note that if you wait long enough you get two
copies of the initial data running away from each other off to
infinity and -infinity; their height is half the original height.
This is exactly what DAlembert’s formula told you to expect.

The initial displacement is continuous but with jumps in the derivative (corners):
$$ u(x,0)=
\begin{cases} (2-x)(x-0) & \mbox{if $0 < x < 2$} \\
0 & \mbox{otherwise}
\end{cases}
$$
The initial velocity is zero.

Note that there is no instantaneous smoothing. In fact the initial
data has only two “corners” in it while the solution at
$t>0$ has more than two corners. Note that if you wait long enough
you get two copies of the initial data running away from each other
off to $\infty$ and $-\infty$; their height is half the original
height. This is exactly what D'Alembert’s formula told you to
expect.

The PDE is $u_t = c^2 \, u_{xx}$.

The boundary conditions are Dirichlet:
$u(0,t) = u(L,t) = 0$.

The first three standing wave solutions are
$u(x,t) = \cos( c \pi/L t) \sin(\pi x/L)$,
$u(x,t) = \cos( c 2 \pi/L t) \sin(2 \pi x/L)$, and
$u(x,t) = \cos( c 3 \pi/L t) \sin(3 \pi x/L)$.

As expected, they don’t decay in amplitude (in contrast with
the heat equation) and the larger $k$ is the faster the temporal
oscillation.

The initial displacement is chosen by choosing random numbers and then
multiplying them by $\sin(k \pi x/4)$ for $k=1$ to $20$. The initial velocity
is zero.

Note there’s no “infinity” for solutions to run off to (in contrast
to the wave equation on the line). Things that hit the boundary
get sent right back in.

I chose the speed $c$ and the run time so that the final snap-shot is
the same as the initial data. Can you figure out how I did this?

The initial displacement, $u(x,0)$, is a discontinuous function:
$$ u(x,0)=
\begin{cases} 1 & \mbox{if $L/4 < x < L/2$} \\
0 & \mbox{otherwise}
\end{cases}
$$
The initial
velocity is zero.

Note there’s no “infinity” for solutions to run off to (in contrast
to the wave equation on the line). Things that hit the boundary
get sent right back in.

I chose the speed $c$ and the run time so that the final snap-shot is
the same as the initial data. Can you figure out how I did this?