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Linear Partial Differential Equations:
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Consider the operator Aˆ= −c(x)Y2 in some 2d region Ω ⊆ R2 with Dirichle
tboundaries (u|∂Ω =0), where c(x) > 0. Suppose the eigen functions of Aˆ are un(x)
with eigen values λn [that is, =Aun λnun] for n =1, 2,..., numbered in order λ1 <λ2
<λ3 < ···. Let G(x, x') be the Green’s function of Aˆ.
(a) If f (x) = n αnun(x) for some coeﬃcients αn = (expression in terms of f
and un), the G(x, x')f (x')d2x' = (in terms of αn and
un).
(b) The maximum possible value of
for any possible u(x), is _____________________ (in terms of quantities
mentioned above). [Hint: min–max. Use the fact, from the handout, that if Aˆis
self-adjoint then Aˆ−1 is also self-adjoint.]
Problem 2:
In this problem, we will solve the Laplacian eigen problem −Y2u = λu in a 2d
radius-1 cylinder r ≤ 1 with Dirichlet boundary conditions u|r=1Ω =0 by “brute
force” in Julia with a 2d finite-difference discretization, and compare to the
analytical Bessel solutions.
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Problem

You will find the IJulia notebooks posted on the 18.303 website for Lecture 9 and
Lecture 11 extremely useful! (Note: when you open the notebook, you can choose
“Run All” from the Cell menu to load all the commands in it.)
(a) Using the notebook for a 100 × 100 grid, compute the 6 smallest-magnitude
eigen values and eigen functions of A with λi, Ui=eigs(Ai,nev=6,which=”SM”). The
eigen values are given by λi. The notebook also shows how to compute the exact
eigen value from the square of the root of the Bessel function. Compared with the
high-accuracy λ1 value, compute the error Δλ1 in the corresponding finite-difference
eigen value from the previous part. Also compute Δλ1 for Nx = Ny = 200 and 400.
How fast is the convergence rate with Δx? Can you explain your results, in light of
the fact that the center-difference approximation we are using has an error that is
supposed to be ∼ Δx2? (Hint: think about how accurately the boundary condition on
∂Ω is described in this finite-difference approximation.)
(b) Modify the above code to instead discretize Y·cY, by writing A0 as −GTCgG for
some G matrix that implements Y and for some Cg matrix that multiplies the gradient
by c(r)= r2 +1. Draw a sketch of the grid points at which the components of Y are
discretized—these will not be the same as the (nx,ny) where u is discretized,
because of the centered differences. Be careful that you need to evaluate c at the Y
grid points now! Hint: you can make the matrix (M1/M2)in Julia by the syntax
[M1;M2].

Hint: Notice in the IJulia notebook from Lecture 11 how a matrix r is created from a
column-vector of x values and a row-vector of y values. You will need to modify
these x and/or y values to evaluate r on a new grid(s). Given the r matrix rc on this
new grid, you can evaluate c(r) on the grid by c = rc.^2+1, and then make a
diagonal sparse matrix of these values by spdiagm(reshape(c, prod(size(c)))).
(c) Using this A ≈ Y· cY, compute the smallest-|λ| eigensolution and plot it. Given the
eigenfunction converted to a 2d Nx × Ny array u, as in the Lecture 11 notebook, plot
u(r) as a function of r, along with a plot of the exact Bessel eigenfunction J0(k0r)
from the c =1 case for comparison.
plot(r[Nx/2:end,Ny/2], u[Nx/2:end,Ny/2]) k0 = so.newton(x -> besselj(0,x), 2.0)
plot(0:0.01:1, besselj(0, k0 * (0:0.01:1))/50)
Here, I scaled J0(k0r) by 1/50, but you should change this scale factor as needed to
make the plots of comparable magnitudes. Note also that the r array here is the
radius evaluated on the original u grid, as in the Lecture 11 notebook.
Can you qualitatively explain the differences?
Problem 3:
Recall that the displacement u(x, t) of a stretched string [with fixed ends: u(0,t)= u(L,
t)=0]

satisfies the wave equation where f(x, t) is an external force
density (pressure) on the string.
(a) Suppose that f(x, t)= �[g(x)e−iωt], an oscillating force with a frequency ω. Show
that, instead of solving the wave equation with this f(x, t), we can instead use a
complex force f˜(x, t)= g(x)e−iωt, solve for a complex u˜(x, t), and then take u =
�u˜to obtain the solution for the original f(x, t).
(b) Suppose that f(x, t)= g(x)e−iωt, and we want to find a steady-state solution u(x,
t)=
−iωt
v(x)ethat is oscillating everywhere at the same frequency as the input force. (This
will be the solution after a long time if there is any dissipation in the system to
allow the initial transients to die away.) Write an equation ˆ= A self-adjoint? Av g
that v solves. Is ˆPositive/negative definite/semidefinite?
(c) Solve for the Green’s function G(x, x') of this ˆ A, assuming that ω= nπ/L for any
integer n (i.e. assume ω is not an eigenfrequency [why?]). [Write down the
continuity conditions that G must satisfy at x = x', solve for x= x', and then use the
continuity conditions to eliminate unknowns.]

(d) Form a finite-difference approximation A of your Aˆ. Compute an approximate
G(x, x') in Matlab by A dk, where dk is the unit vector of all 0’s except for one 1/Δx
at index k = x'/Δx, and compare (by plotting both) to your analytical solution from
the previous part for a couple values of x and a couple of different frequencies ω
(one < π/L and one > π/L) with L =1.
(e) Show the limit ω → 0 of your G relates in some expected way to the Green’s
function of –d2/dx2from class.

Problem 1:
(a) is self-ad joint under the inner product from class (and the
1d version in homework), and hence the eigen functions are orthogonal for
distinct eigen values (but not necessarily normalized to IunI =1), hence
The expression u(x)= is just the solution to i.e. it is u = A−1f =
(b) This is just (u, Aˆ−1u)/(u, u), which is a Rayleigh quotient for the self-adjoint
operator Aˆ−1, and hence (thanks to the min–max theorem) it is bounded above
by the largest eigen value of Aˆ−1, which is 1/λ1 (the inverse of the smallest
eigen value of Aˆ). (Note that by positive- definiteness 0 < λ1 < λ2 < · · · .)
Problem 2:
In this problem, we will solve the Laplacian eigen problem in a 2d
radius-1 cylinder r <= 1 with Dirichlet boundary conditions by “brute
force” in Julia with a 2d finite- difference discretization, and compare to the
analytical Bessel solutions. You will find the IJulia notebooks posted on the 18.303
website for Lecture 9 and Lecture 11 extremely useful! (Note: when you open the
notebook, you can choose “Run All” from the Cell menu to load all the commands
in it.)
Solutions

(a) Using the notebook for a 100 X 100 grid, compute the 6 smallest-magnitude
eigen values and eigen functions of A with λi, Ui= eigs(Ai, nev=6,which=”SM”).
The eigen values are given by λi. The notebook also shows how to compute the
exact eigen value from the square of the root of the Bessel function. Compared
with the high-accuracy λ1 value, compute the error Δλ1 in the corresponding
finite-difference eigen value from the previous part. Also compute Δλ1 for Nx = Ny
= 200 and 400. How fast is the convergence rate with Δx? Can you explain your
results, in light of the fact that the center-difference approximation we are using
has an error that is supposed to be Δx2? (Hint: think about how accurately the
boundary condition on ∂Ω is described in this finite-difference approximation.)
(b) Modify the above code to instead discretize by writing A0 as −GT CgG for
some G matrix that implements and for some Cg matrix that multiplies the
gradient by c(r)= r2 +1. Draw a sketch of the grid points at which the components
of are discretized—these will not be the same as the (nx,ny) where u is
discretized, because of the centered differences. Be careful that you need to
evaluate c at the grid points now! Hint: you can make the matrix (M1 M2 ) in
Julia by the syntax [M1;M2].
Hint: how a matrix r is created from a column-vector of x values and a row-vector of
y values. You will need to modify these x and/or y values to evaluate r on a new
grid(s). Given the r matrix rc on this new grid, you can evaluate c(r) on the grid by c
= rc.^2+1, and then make a diagonal sparse matrix of these values by
spdiagm(reshape(c, prod(size(c)))).

Using this A ≈' · c', compute the smallest-|λ| eigensolution and plot it. Given the
eigenfunction converted to a 2d Nx × Ny array u, as in the Lecture 11 notebook,
plot u(r) as a function of r, along with a plot of the exact Bessel eigenfunction
J0(k0r) from the c =1 case for comparison.
plot(r[Nx/2:end,Ny/2], u[Nx/2:end,Ny/2]) k0 = so.newton(x -> besselj(0,x), 2.0)
plot(0:0.01:1, besselj(0, k0 * (0:0.01:1))/50)
1 Here, I scaled J0(k0r) by 1/50, but you should change this scale factor as needed
to make the plots of comparable magnitudes. Note also that the r array here is the
radius evaluated on the original u grid, as in the Lecture 11 notebook.
Can you qualitatively explain the differences?
Problem 3:
Recall that the displacement u(x, t) of a stretched string [with fixed ends: u(0,t)=
u(L, t)=0]
uu
satisfies the wave equation ∂2 +f(x, t)= ∂2 , where f(x, t) is an external force density
(pressure)
∂x2 ∂t2
on the string.
∂2
u˜u˜

(a) Suppose that u˜solves ∂2 + f˜(x, t)= and satisfies u˜(0,t)= u˜(L, t)=0. Now,
consider
∂x2 ∂t2
u˜+u˜¯
u = �u˜= . Clearly, u(0,t)= u(L, t)=0, so u satisfies the same boundary conditions.
It
2
also satisfies the PDE:
since f = ˜˜¯= �f˜. The key factors that allowed us to do this are (i) linearity, and (ii)
the 2 real-ness of the PDE (the PDE itself contains no i factors or other complex
coefficients).
(b) Plugging u(x, t)= v(x)e−iωt and f(x, t)= g(x)e−iωt into the PDE, we obtain
and The boundary conditions are v(0) = v(L)=0, from the boundary
conditions on u

Since ω2 is real, this is in the general Sturm-Liouville form that we showed in
class is selfadjoint.
Subtracting a constant from an operator just shifts all of the eigenvalues by that
constant, keeping the eigenfunctions the same. Thus Aˆis still positive-definite if
ω2 is < the smallest eigenvalue of −∂2/∂x2, and positive semidefinite if ω2 = the
smallest eigenvalue. In this case, we know analytically that the eigenvalues of
−∂2/∂x2 with these boundary conditions are (nπ/L)2 for n =1, 2,.... So Aˆis positive-
definite if ω2 < (π/L)2, it is positive-semidefinite if ω2 =(π/L)2, and it is indefinite
otherwise.
(c) We know that ˆ')=0 for x x'. Also, just as in class and as in the notes, the fact
AG(x, x = that ˆ')= ') meas that G must be continuous (otherwise there would be a
δ'
AG(x,x δ(x − x factor) and ∂G/∂x must have a jump discontinuity:
for −∂2G/∂x2 to give δ(x − x'). We could also show this more explicitly by
integrating:
which gives the same result.

Now, similar to class, we will solve it separately for x<x and for x>x', and then
impose the continuity requirements at x = x to find the unknown coefficients.
For x<x , AG =0 means that ∂2 = −ω2G, hence G(x, x') is some sine or cosine of
ωx. But since G(0,x')=0, we must therefore have G(x, x')= α sin(ωx) for some
coefficient α.
Similarly, for x > x', we also have a sine or cosine of ωx. To get G(L, x') = 0, the
simplest way to do this is to use a sine with a phase shift: G(x, x') = β sin(ω[L x])
for some coeﬃcient β.
Continuity now gives two equations in the two unknowns α and β:
which has the solution
This simplifies a bit from the identity sinAcosB + cosB sinA = sin(A + B), and
hence
which obviously obeys reciprocity.

Note that if ω is an eigen frequency, i.e. ω = nπ/L for some n, then this G blows
up.
The reason is that Ain that case is singular (the nth eigen value was shifted to
zero), and defining A -1 ˆis more problematical. (Physically, this corresponds to
driving the oscillating string at a resonance frequency, which generally leads to a
diverging solution unless there is dissipation in the system.)
(d) It is critical to get the signs right. Recall that we approximated d/dx2 by −DTD
for a 1st derivative matrix D. Therefore, you want to make a matrix A = DTD − ω2I.
In Matlab, A = D’*D -omega^2 * eye(N) where N is the size of the matrix D =
diff1(N) / dx with dx = 1/(N+1). We make dk by dk = zeros(N,1); dk(k) = 1/dx. I
used N = 100.
The resulting plots are shown in figure 1, for ω =0.4π/L (left) and ω =5.4π/L (right)
and x’ =0.5 and 0.75. As expected, for ω < π/L where it is positive-definite, the
Green’s function is positive and qualitatively similar to that for ω =0, except that it
is slightly curved. For larger ω, though, it becomes oscillatory and much more
interesting in shape. The exact G and the finite-difference G match very well, as
they should, although the match becomes worse at higher frequencies—the
difference approximations become less and less accurate as the function
becomes more and more oscillatory relative to the grid Δx, just as was the case

for the eigenfunctions as discussed in class.
[If you try to plug in an eigenfrequency (e.g. 4π/L) for ω, you may notice that the
exact G blows up but the finite-difference G is finite. The reason for this is simple:
the eigenvalues of the finite-difference matrix DTD are not exactly (nπ/L)2, so the
matrix is not exactly singular, nor is it realistically possible to make it exactly
singular thanks to rounding errors. If you plug in something very close to an
eigenfrequency, then G becomes very close to an eigenfunction multipled by a
large amplitude. It is easy to see why this is if you look at the expansion of the
solution in terms of the eigenfunctions.]

(e) For small ω, sin(ωy) ≈ ωy for any y, and hence
which matches the Green’s function of –d2/dx2 from class. (Equivalently, you
get 0/0 as ω → 0 and hence must use L’Hôpital’s rule or similar to take the
limit.)

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