Simulate waves on a string that consists of two materials with different density. The tension in the string is constant, but the density has a jump at the middle of the string. Experiment with different sizes of the jump and produce animations that visualize the effect of the jump on the wave motion.

According to the section Waves on a string,
the density enters the mathematical model as \( \varrho \) in
\( \varrho u_{tt} = Tu_{xx} \), where \( T \) is the string tension. Modify, e.g., the
`wave1D_u0v.py`

code to incorporate the tension and two density values.
Make a mesh function `rho`

with density values at each spatial mesh point.
A value for the tension may be 150 N. Corresponding density values can
be computed from the wave velocity estimations in the `guitar`

function
in the `wave1D_u0v.py`

file.

Filename: `wave1D_u0_sv_discont`

.

Formulate a mathematical model for damped waves on a string.
Use data from the section Running a case, and
tune the damping parameter so that the string is very close to
the rest state after 15 s. Make a movie of the wave motion.
Filename: `wave1D_u0_sv_damping`

.

A hammer hits the end of an elastic rod. The exercise is to simulate
the resulting wave motion using the model (2.124)
from the section Elastic waves in a rod. Let the rod have length
\( L \) and let the boundary \( x=L \) be stress free so that \( \sigma_{xx}=0 \),
implying that \( \partial u/\partial x=0 \). The left end \( x=0 \) is
subject to a strong stress pulse (the hammer), modeled as
$$ \sigma_{xx}(t) = \left\lbrace\begin{array}{ll}
S,& 0 < t \leq t_s,\\
0, & t > t_s
\end{array}\right.
$$
The corresponding condition on \( u \) becomes \( u_x= S/E \)
for \( t\leq t_s \) and zero afterwards (recall that
\( \sigma_{xx} = Eu_x \)). This is a non-homogeneous
Neumann condition, and you will need to approximate this condition
and combine it with the scheme (the ideas and manipulations follow
closely the handling of a non-zero initial condition
\( u_t=V \) in wave PDEs or the corresponding
second-order ODEs for vibrations).
Filename: `wave_rod`

.

Implement a model for spherically symmetric waves using the method described in the section Spherical waves. The boundary condition at \( r=0 \) must be \( \partial u/\partial r=0 \), while the condition at \( r=R \) can either be \( u=0 \) or a radiation condition as described in Problem 2.12: Implement open boundary conditions. The \( u=0 \) condition is sufficient if \( R \) is so large that the amplitude of the spherical wave has become insignificant. Make movie(s) of the case where the source term is located around \( r=0 \) and sends out pulses $$ f(r,t) = \left\lbrace\begin{array}{ll} Q\exp{(-\frac{r^2}{2\Delta r^2})}\sin\omega t,& \sin\omega t\geq 0\\ 0, & \sin\omega t < 0 \end{array}\right. $$ Here, \( Q \) and \( \omega \) are constants to be chosen.

Use the program `wave1D_u0v.py`

as a starting point. Let `solver`

compute the \( v \) function and then set \( u=v/r \). However,
\( u=v/r \) for \( r=0 \) requires special treatment. One possibility is
to compute `u[1:] = v[1:]/r[1:]`

and then set `u[0]=u[1]`

. The latter
makes it evident that \( \partial u/\partial r = 0 \) in a plot.

Filename: `wave1D_spherical`

.

A subsea earthquake leads to an immediate lift of the water surface, see Figure 34. The lifted water surface splits into two tsunamis, one traveling to the right and one to the left, as depicted in Figure 35. Since tsunamis are normally very long waves, compared to the depth, with a small amplitude, compared to the wave length, a standard wave equation is relevant: $$ \eta_{tt} = (gH(x)\eta_x)_x,$$ where \( \eta \) is the elevation of the water surface, \( g \) is the acceleration of gravity, and \( H(x) \) is the still water depth.

To simulate the right-going tsunami, we can impose a symmetry boundary at \( x=0 \): \( \partial\eta/\partial x =0 \). We then simulate the wave motion in \( [0,L] \). Unless the ocean ends at \( x=L \), the waves should travel undisturbed through the boundary \( x=L \). A radiation condition as explained in Problem 2.12: Implement open boundary conditions can be used for this purpose. Alternatively, one can just stop the simulations before the wave hits the boundary at \( x=L \). In that case it does not matter what kind of boundary condition we use at \( x=L \). Imposing \( \eta =0 \) and stopping the simulations when \( |\eta_i^n| > \epsilon \), \( i=N_x-1 \), is a possibility (\( \epsilon \) is a small parameter).

The shape of the initial surface can be taken as a Gaussian function, $$ \begin{equation} I(x;I_0,I_a,I_m,I_s) = I_0 + I_a\exp{\left(-\left(\frac{x-I_m}{I_s}\right)^2\right)}, \tag{2.166} \end{equation} $$ with \( I_m=0 \) reflecting the location of the peak of \( I(x) \) and \( I_s \) being a measure of the width of the function \( I(x) \) (\( I_s \) is \( \sqrt{2} \) times the standard deviation of the familiar normal distribution curve).

Now we extend the problem with a hill at the sea bottom, see Figure 36. The wave speed \( c=\sqrt{gH(x)} = \sqrt{g(H_0-B(x))} \) will then be reduced in the shallow water above the hill.

One possible form of the hill is a Gaussian function, $$ \begin{equation} B(x;B_0,B_a,B_m,B_s) = B_0 + B_a\exp{\left(-\left(\frac{x-B_m}{B_s}\right)^2\right)}, \tag{2.167} \end{equation} $$ but many other shapes are also possible, e.g., a "cosine hat" where $$ \begin{equation} B(x; B_0, B_a, B_m, B_s) = B_0 + B_a\cos{\left( \pi\frac{x-B_m}{2B_s}\right)}, \tag{2.168} \end{equation} $$ when \( x\in [B_m - B_s, B_m + B_s] \) while \( B=B_0 \) outside this interval.

Also an abrupt construction may be tried: $$ \begin{equation} B(x; B_0, B_a, B_m, B_s) = B_0 + B_a, \tag{2.169} \end{equation} $$ for \( x\in [B_m - B_s, B_m + B_s] \) while \( B=B_0 \) outside this interval.

The `wave1D_dn_vc.py`
program can be used as starting point for the implementation.
Visualize both the bottom topography and the
water surface elevation in
the same plot.
Allow for a flexible choice of bottom shape:
(2.167),
(2.168),
(2.169), or \( B(x)=B_0 \) (flat).

The purpose of this problem is to explore the quality of the numerical
solution \( \eta^n_i \) for different shapes of the bottom obstruction.
The "cosine hat" and the box-shaped hills have abrupt changes in the
derivative of \( H(x) \) and are more likely to generate numerical noise
than the smooth Gaussian shape of the hill. Investigate if this is true.
Filename: `tsunami1D_hill`

.

This problem extends Problem 2.24: Earthquake-generated tsunami over a subsea hill to a three-dimensional wave phenomenon, governed by the 2D PDE $$ \begin{equation} \eta_{tt} = (gH\eta_x)_x + (gH\eta_y)_y = \nabla\cdot (gH\nabla\eta) \tag{2.170} \tp \end{equation} $$ We assume that the earthquake arises from a fault along the line \( x=0 \) in the \( xy \)-plane so that the initial lift of the surface can be taken as \( I(x) \) in Problem 2.24: Earthquake-generated tsunami over a subsea hill. That is, a plane wave is propagating to the right, but will experience bending because of the bottom.

The bottom shape is now a function of \( x \) and \( y \). An "elliptic" Gaussian function in two dimensions, with its peak at \( (B_{mx}, B_{my}) \), generalizes (2.167): $$ \begin{equation} B = B_0 + B_a\exp{\left(-\left(\frac{x-B_{mx}}{B_s}\right)^2 -\left(\frac{y-B_{my}}{bB_s}\right)^2\right)}, \tag{2.171} \end{equation} $$ where \( b \) is a scaling parameter: \( b=1 \) gives a circular Gaussian function with circular contour lines, while \( b\neq 1 \) gives an elliptic shape with elliptic contour lines. To indicate the input parameters in the model, we may write $$ B = B(x;B_0,B_a,B_{mx}, B_{my} ,B_s, b)\tp $$

The "cosine hat" (2.168) can also be generalized to $$ \begin{equation} B = B_0 + B_a\cos{\left( \pi\frac{x-B_{mx}}{2B_s}\right)} \cos{\left( \pi\frac{y-B_{my}}{2B_s}\right)}, \tag{2.172} \end{equation} $$ when \( 0 \leq \sqrt{x^2+y^2} \leq B_s \) and \( B=B_0 \) outside this circle.

A box-shaped obstacle means that $$ \begin{equation} B(x; B_0, B_a, B_m, B_s, b) = B_0 + B_a \tag{2.173} \end{equation} $$ for \( x \) and \( y \) inside a rectangle $$ B_{mx}-B_s \leq x \leq B_{mx} + B_s,\quad B_{my}-bB_s \leq y \leq B_{my} + bB_s, $$ and \( B=B_0 \) outside this rectangle. The \( b \) parameter controls the rectangular shape of the cross section of the box.

Note that the initial condition and the listed bottom shapes are symmetric around the line \( y=B_{my} \). We therefore expect the surface elevation also to be symmetric with respect to this line. This means that we can halve the computational domain by working with \( [0,L_x]\times [0, B_{my}] \). Along the upper boundary, \( y=B_{my} \), we must impose the symmetry condition \( \partial \eta/\partial n=0 \). Such a symmetry condition (\( -\eta_x=0 \)) is also needed at the \( x=0 \) boundary because the initial condition has a symmetry here. At the lower boundary \( y=0 \) we also set a Neumann condition (which becomes \( -\eta_y=0 \)). The wave motion is to be simulated until the wave hits the reflecting boundaries where \( \partial\eta/\partial n =\eta_x =0 \) (one can also set \( \eta =0 \) - the particular condition does not matter as long as the simulation is stopped before the wave is influenced by the boundary condition).

Visualize the surface elevation. Investigate how different hill shapes,
different sizes of the water gap above the hill, and different
resolutions \( \Delta x = \Delta y = h \) and \( \Delta t \) influence
the numerical quality of the solution.
Filename: `tsunami2D_hill`

.

Play with Mayavi code for visualizing 2D
solutions of the wave equation with variable wave velocity.
See if there
are effective ways to visualize both the solution and the wave
velocity scalar field at the same time.
Filename: `tsunami2D_hill_mlab`

.

Create some fancy 3D visualization of the water waves *and* the subsea hill
in Problem 2.25: Earthquake-generated tsunami over a 3D hill.
Try to make the hill transparent. Possible visualization tools are
Mayavi,
Paraview, and
OpenDX.
Filename: `tsunami2D_hill_viz`

.

Extend the program from Problem 2.25: Earthquake-generated tsunami over a 3D hill such
that the loops over mesh points, inside the time loop, are
implemented in compiled languages.
Consider implementations in
Cython, Fortran via `f2py`

, C via Cython, C via `f2py`

, C/C++ via Instant,
and C/C++ via `scipy.weave`

.
Perform efficiency experiments to investigate the relative performance
of the various implementations. It is often advantageous to normalize
CPU times by the fastest method on a given mesh.
Filename: `tsunami2D_hill_compiled`

.

The goal of this exercise is to simulate seismic waves using the
PDE model (2.134) in a
2D \( xz \) domain with geological layers.
Introduce \( m \) horizontal layers of thickness \( h_i \), \( i=0,\ldots,m-1 \).
Inside layer number \( i \) we have a vertical wave velocity \( c_{z,i} \)
and a horizontal wave velocity \( c_{h,i} \). Make a program
for simulating such 2D waves. Test it on a case with 3 layers where
$$ c_{z,0}=c_{z,1}=c_{z,2},\quad c_{h,0}=c_{h,2},\quad c_{h,1} \ll c_{h,0}
\tp
$$
Let \( s \) be a localized point source at the middle of the Earth's surface
(the upper boundary)
and investigate how the resulting wave travels through the medium.
The source can be a localized Gaussian peak that oscillates in
time for some time interval.
Place the boundaries far enough from the expanding wave so that the
boundary conditions do not disturb the wave. Then the type of
boundary condition does not matter, except that we physically need
to have \( p=p_0 \), where \( p_0 \) is the atmospheric pressure,
at the upper boundary.
Filename: `seismic2D`

.

The equation for sound waves in air is derived in the section Sound waves in liquids and gases and reads $$ p_{tt} = c^2\nabla^2 p,$$ where \( p(x,y,z,t) \) is the pressure and \( c \) is the speed of sound, taken as 340 m/s. However, sound is absorbed in the air due to relaxation of molecules in the gas. A model for simple relaxation, valid for gases consisting only of one type of molecules, is a term \( c^2\tau_s\nabla^2 p_t \) in the PDE, where \( \tau_s \) is the relaxation time. If we generate sound from, e.g., a loudspeaker in the room, this sound source must also be added to the governing equation.

The PDE with the mentioned type of damping and source then becomes $$ \begin{equation} p_tt = c^2\nabla^p + c^2\tau_s\nabla^2 p_t + f, \tag{2.174} \end{equation} $$ where \( f(x,y,z,t) \) is the source term.

The walls can absorb some sound. A possible model is to have a "wall layer" (thicker than the physical wall) outside the room where \( c \) is changed such that some of the wave energy is reflected and some is absorbed in the wall. The absorption of energy can be taken care of by adding a damping term \( bp_t \) in the equation: $$ \begin{equation} p_tt + bp_t = c^2\nabla^p + c^2\tau_s\nabla^2 p_t + f\tp \tag{2.175} \end{equation} $$ Typically, \( b=0 \) in the room and \( b>0 \) in the wall. A discontinuity in \( b \) or \( c \) will give rise to reflections. It can be wise to use a constant \( c \) in the wall to control reflections because of the discontinuity between \( c \) in the air and in the wall, while \( b \) is gradually increased as we go into the wall to avoid reflections because of rapid changes in \( b \). At the outer boundary of the wall the condition \( p=0 \) or \( \partial p/\partial n=0 \) can be imposed. The waves should anyway be approximately dampened to \( p=0 \) this far out in the wall layer.

There are two strategies for discretizing the \( \nabla^2 p_t \) term: using a center difference between times \( n+1 \) and \( n-1 \) (if the equation is sampled at level \( n \)), or use a one-sided difference based on levels \( n \) and \( n-1 \). The latter has the advantage of not leading to any equation system, while the former is second-order accurate as the scheme for the simple wave equation \( p_tt = c^2\nabla^2 p \). To avoid an equation system, go for the one-sided difference such that the overall scheme becomes explicit and only of first order in time.

Develop a 3D solver for the specified PDE and introduce a wall layer.
Test the solver with the method of manufactured solutions. Make some
demonstrations where the wall reflects and absorbs the waves (reflection
because of discontinuity in \( b \) and absorption because of growing \( b \)).
Experiment with the impact of the \( \tau_s \) parameter.
Filename: `acoustics`

.

We shall study the wave equation
$$
\begin{equation}
u_t + cu_x = 0,\quad x\in (0,L],\ t\in (0, T],
\tag{2.176}
\end{equation}
$$
with initial condition
$$
\begin{equation}
u(x,0) = I(x),\quad x\in [0,L],
\tag{2.177}
\end{equation}
$$
and *one* periodic boundary condition
$$
\begin{equation}
u(0,t) = u(L,t)
\tp
\tag{2.178}
\end{equation}
$$
This boundary condition means that what goes out of the domain at \( x=L \)
comes in at \( x=0 \). Roughly speaking,
we need only one boundary condition because of
the spatial derivative is of first order only.

**Physical interpretation.**
The parameter \( c \) can be constant or variable, \( c=c(x) \). The
equation (2.176) arises in *transport*
problems where a quantity \( u \), which could be temperature or
concentration of some contaminant, is transported with the velocity
\( c \) of a fluid. In addition to the transport imposed by "travelling with
the fluid", \( u \) may also be transported by diffusion (such as
heat conduction or Fickian diffusion), but we have in
the model \( u_t + cu_x \) assumed that diffusion effects are negligible,
which they often are.

**a)**
Show that under the assumption of \( a=\hbox{const} \),
$$
\begin{equation}
u(x,t) = I(x - ct)
\tag{2.179}
\end{equation}
$$
fulfills the PDE as well as the initial and boundary condition
(provided \( I(0)=I(L) \)).

A widely used numerical scheme for (2.176) applies a forward difference in time and a backward difference in space when \( c>0 \): $$ \begin{equation} [D_t^+ u + cD_x^-u = 0]_i^n \tag{2.180} \tp \end{equation} $$ For \( c < 0 \) we use a forward difference in space: \( [cD_x^+u]_i^n \).

**b)**
Set up a computational algorithm and implement it in a function.
Assume \( a \) is constant and positive.

**c)**
Test the implementation by using the remarkable property that
the numerical solution is exact at the mesh points if
\( \Delta t = c^{-1}\Delta x \).

**d)**
Make a movie comparing the numerical and exact solution for the
following two choices of initial conditions:
$$
\begin{equation}
I(x) = \left\lbrack\sin\left(\pi\frac{x}{L}\right)\right\rbrack^{2n}
\tag{2.181}
\end{equation}
$$
where \( n \) is an integer, typically \( n=5 \),
and
$$
\begin{equation}
I(x) = \exp{\left( -\frac{(x-L/2)^2}{2\sigma2}\right)} \tp
\end{equation}
\tag{2.182}
$$
Choose \( \Delta t = c^{-1}\Delta x, 0.9c^{-1}\Delta x, 0.5c^{-1}\Delta x \).

**e)**
The performance of the suggested numerical scheme can be investigated
by analyzing the numerical dispersion relation.
Analytically, we have that the *Fourier component*
$$ u(x,t) = e^{i(kx-\omega t)},$$
is a solution of the PDE if \( \omega = kc \). This is the *analytical
dispersion relation*. A complete solution of the PDE can be built by
adding up such Fourier components with different amplitudes, where the
initial condition \( I \) determines the amplitudes. The solution \( u \) is
then represented by a Fourier series.

A similar discrete Fourier component at \( (x_p,t_n) \) is $$ u_p^q = e^{i(kp\Delta x -\tilde\omega n\Delta t)},$$ where in general \( \tilde\omega \) is a function of \( k \), \( \Delta t \), and \( \Delta x \), and differs from the exact \( \omega =kc \).

Insert the discrete Fourier component in the numerical scheme and derive an expression for \( \tilde\omega \), i.e., the discrete dispersion relation. Show in particular that if \( \Delta t/(c\Delta x)=1 \), the discrete solution coincides with the exact solution at the mesh points, regardless of the mesh resolution (!). Show that if the stability condition $$ \frac{\Delta t}{c\Delta x}\leq 1,$$ the discrete Fourier component cannot grow (i.e., \( \tilde\omega \) is real).

**f)**
Write a test for your implementation where you try to use information
from the numerical dispersion relation.

We shall hereafter assume that \( c(x)>0 \).

**g)**
Set up a computational algorithm for the variable
coefficient case and implement it in a function.
Make a test that the function works for constant \( a \).

**h)**
It can be shown that
for an observer moving with velocity \( c(x) \), \( u \) is constant.
This can be used to derive an exact solution when \( a \) varies
with \( x \). Show first that
$$
\begin{equation}
u(x,t) = f(C(x) - t),
\tag{2.183}
\end{equation}
$$
where
$$ C'(x) = \frac{1}{c(x)},$$
is a solution of (2.176) for any
differentiable function \( f \).

**i)**
Use the initial condition to show that an exact solution is
$$ u(x,t) = I(C^{-1}(C(x)-t)),$$
with \( C^{-1} \) being the inverse function of \( C = \int c^{1}dx \).
Since \( C(x) \) is an integral \( \int_0^x (1/c)dx \),
\( C(x) \) is monotonically increasing and there exists hence an
inverse function \( C^{-1} \) with values in \( [0,L] \).

To compute (2.183) we need to integrate \( 1/c \) to obtain \( C \) and then compute the inverse of \( C \).

The inverse function computation can be easily done if we first think discretely. Say we have some function \( y=g(x) \) and seek its inverse. Plotting \( (x_i,y_i) \), where \( y_i=g(x_i) \) for some mesh points \( x_i \), displays \( g \) as a function of \( x \). The inverse function is simply \( x \) as a function of \( g \), i.e., the curve with points \( (y_i,x_i) \). We can therefore quickly compute points at the curve of the inverse function. One way of extending these points to a continuous function is to assume a linear variation (known as linear interpolation) between the points (which actually means to draw straight lines between the points, exactly as done by a plotting program).

The function `wrap2callable`

in `scitools.std`

can take a set of points
and return a continuous function that corresponds to linear
variation between the points. The computation of the inverse of a
function \( g \) on \( [0,L] \) can then be done by

```
def inverse(g, domain, resolution=101):
x = linspace(domain[0], domain[L], resolution)
y = g(x)
from scitools.std import wrap2callable
g_inverse = wrap2callable((y, x))
return g_inverse
```

To compute \( C(x) \) we need to integrate \( 1/c \), which can be done by
a Trapezoidal rule. Suppose we have computed \( C(x_i) \) and need
to compute \( C(x_{i+1}) \). Using the Trapezoidal rule with \( m \) subintervals
over the integration domain \( [x_i,x_{i+1}] \) gives
$$
\begin{equation}
C(x_{i+1}) = C(x_i) + \int_{x_i}^{x_{i+1}} \frac{dx}{c}
\approx h\left(
\half\frac{1}{c(x_i)} + \half\frac{1}{c(x_{i+1})}
+ \sum_{j=1}^{m-1} \frac{1}{c(x_i + jh)}\right),
\tag{2.184}
\end{equation}
$$
where \( h=(x_{i+1}-x_i)/m \) is the length of the subintervals used
for the integral over \( [x_i,x_{i+1}] \).
We observe that (2.184) is a
*difference equation* which we can solve by repeatedly
applying (2.184) for
\( i=0,1,\ldots,N_x-1 \) if a mesh \( x_0,x_,\ldots,x_{N_x} \) is prescribed.
Note that \( C(0)=0 \).

**j)**
Implement a function for computing \( C(x_i) \) and one for computing
\( C^{-1}(x) \) for any \( x \). Use these two functions for computing
the exact solution \( I(C^{-1}(C(x)-t)) \).
End up with a function `u_exact_variable_c(x, n, c, I)`

that returns
the value of \( I(C^{-1}(C(x)-t_n)) \).

**k)**
Make movies showing a comparison of the numerical and exact solutions
for the two initial conditions
(2.181) and (2.182).
Choose \( \Delta t = \Delta x /\max_{0,L} c(x) \)
and the velocity of the medium as

- \( c(x) = 1 + \epsilon\sin(k\pi x/L) \), \( \epsilon < 1 \),
- \( c(x) = 1 + I(x) \), where \( I \) is given by (2.181) or (2.182).

Filename: `advec1D`

.

We consider an initial-boundary value problem for the damped wave equation: $$ \begin{alignat*}{2} u_{tt} +bu_t &= c^2 u_{xx}, \quad &x\in (0,L),\ t\in (0,T]\\ u(0,t) &= 0,\\ u(L,t) &=0,\\ u(x, 0) &= I(x),\\ u_t(x, 0) &= V(x)\tp \end{alignat*} $$ Here, \( b\geq 0 \) and \( c \) are given constants. The aim is to derive a general analytical solution of this problem. Familiarity with the method of separation of variables for solving PDEs will be assumed.

**a)**
Seek a solution on the form \( u(x,t)=X(x)T(t) \). Insert this solution in
the PDE and show that it leads to two differential equations for \( X \) and \( T \):
$$ T'' + bT' + \lambda T = 0,\quad c^2 X'' +\lambda X = 0, $$
with \( X(0)=X(L)=0 \) as boundary conditions, and \( \lambda \) as a constant
to be determined.

**b)**
Show that \( X(x) \) is on the form
$$ X_n(x) = C_n\sin kx,\quad k = \frac{n\pi}{L},\quad n=1,2,\ldots$$
where \( C_n \) is an arbitrary constant.

**c)**
Under the assumption that \( (b/2)^2 < k^2 \), show that \( T(t) \) is on
the form
$$ T_n(t) = e^{-{\half}bt}(a_n\cos\omega t + b_n\sin\omega t),
\quad\omega = \sqrt{k^2 - \frac{1}{4}b^2},\quad n=1,2,\ldots $$
The complete solution is then
$$ u(x,t) = \sum_{n=1}^\infty \sin kx e^{-{\half}bt}(
A_n\cos\omega t + B_n\sin\omega t),$$
where the constants \( A_n \) and \( B_n \) must be computed from the initial
conditions.

**d)**
Derive a formula for \( A_n \) from \( u(x,0)=I(x) \) and developing \( I(x) \) as a sine
Fourier series on \( [0,L] \).

**e)**
Derive a formula for \( B_n \) from \( u_t(x,0)=V(x) \) and developing \( V(x) \) as
a sine Fourier series on \( [0,L] \).

**f)**
Calculate \( A_n \) and \( B_n \) from vibrations of a string where \( V(x)=0 \) and
$$
\begin{equation}
I(x) = \left\lbrace
\begin{array}{ll}
ax/x_0, & x < x_0,\\
a(L-x)/(L-x_0), & \hbox{otherwise}
\end{array}\right.
\tag{2.185}
\end{equation}
$$

**g)**
Implement a function `u_series(x, t, tol=1E-10)`

for
the series for \( u(x,t) \),
where `tol`

is a tolerance for truncating the series. Simply sum
the terms until \( |a_n| \) and \( |b_b| \) both are less than `tol`

.

**h)**
What will change in the derivation of the analytical solution if we
have \( u_x(0,t)=u_x(L,t)=0 \) as boundary conditions?
And how will you solve the problem with \( u(0,t)=0 \) and \( u_x(L,t)=0 \)?

Filename: `damped_wave1D`

.

Carry out Problem 2.32: General analytical solution of a 1D damped wave equation in the 2D
case: \( u_{tt}+bu_t = c^2(u_{xx}+u_{yy}) \), where \( (x,y)\in (0,L_x)\times
(0,L_y) \). Assume a solution on the form \( u(x,y,t)=X(x)Y(y)T(t) \).
Filename: `damped_wave2D`

.