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1. Linear Partial Differential Equations:
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2. Problem 1: Distributions
This problem concerns distributions as defined in the notes: continuous linear
functionals f{φ} from test functions φ ∈ D, where D is the set of infinitely
differentiable functions with compact support (i.e. φ = 0 outside some region with
finite diameter [differing for different φ], i.e. outside some finite interval [a, b] in
1d).
In this part, you will consider the function and its (weak)
derivative, which is connected to something called the Cauchy Principal Value.
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3. (i) Show that f(x) defines a regular distribution, by showing that f(x) is locally
integrable for
all intervals [a, b].
(ii) Consider the 18.01 derivative of f(x), which gives
Suppose we
just set “f'(0) = 0” at the origin to define Show that this g(x) is not
locally
integrable, and hence does not define a distribution.
But the weak derivative f'{φ} must exist, so this means that we have to do something
different from the 18.01 derivative, and moreover f'{φ} is not a regular distribution.
What is it?
(iii) Write where
since this limit exists and equals f{φ} for all φ from your proof in the previous part.
Compute the distributional derivative and show that f'{φ} is
precisely the Cauchy Principal Value (google the definition, e.g. on Wikipedia) of the
integral of g(x)φ(x).
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4. (iv) Alternatively, show that (which is a well-
defined integral for all φ ∈ D).
(b) In class, we only looked explicitly at 1d distributions, but a distribution in d
dimensions Rd can obviously be defined similarly, as maps f{φ} from smooth localized
functions φ(x) to numbers. Analogous to class, define the distributional gradient 'f by
'f{φ} = f{−'φ}.
Consider some finite volume V with a surface ∂V , and assume ∂V is differentiable so
that at each point it has an outward-pointing unit normal vector n, as shown in figure
1. Define a “surface delta function” δ(∂V ){φ} = φ(x)dd−1x to give the surface integral of
the test function.
Suppose we have a regular distribution f{φ} defined by the function
where we may have a discontinuity f2 − f1 =0 at ∂V .
(i) Show that the distributional gradient of f is
where the second term is a regular distribution given by the ordinary 18.02 gradient of
f1 and f2 (assumed to be differentiable), while the first term is the singular
distribution
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5. You can use the integral identity that to help you integrate by
parts.
(ii) Defining '2f{φ} = f{'2φ}, derive a similar expression to the above for '2f. Note that you
should have one term from the discontinuity f1 − f2, and another term from the
discontinuity 'f1 −'f2. (Recall how we integrated '2 by parts in class some time ago.)
Problem 2: Green’s functions
Consider Green’s functions of the self-adjoint indefinite operator Aˆ= −'2 − ω2 (κ> 0)
over all space (Ω= R3 in 3d), with solutions that → 0 at infinity. (This is the
multidimensional version of problem 2 from pset 5.) As in class, thanks to the
translational and rotational invariance of this problem, we can find G(x, x')= g(|x − x'|)
for some g(r) in spherical coordinates.
(a) Solve for g(r) in 3d, similar to the procedure in class.
(i) Similar to the case of Aˆ= −'2 in class, first solve for g(r) for r> 0, and write g(r) =
limf→0+ ff(r) where ff(r)=0 for r ≤ E. [Hint: although Wikipedia writes the spherical
'2g(r) as 1/r2 (rg')', it may be more convenient to write it equivalently as '2g =1/2 (rg)'',
as in class, and to solve for h(r)= rg(r) first.
Hint: if you get sines and cosines from this differential equation, it will probably be
easier to use complex exponentials, e.g. eiωr, instead.]
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6. (ii) In the previous part, you should find two solutions, both of which go to zero at
infinity. To choose between them, remember that this operator arose from a e−iωt time
dependence. Plug in this time dependence and impose an “outgoing wave” boundary
condition (also called a Sommerfield or radiation boundary condition): require that
waves be traveling outward far away, not inward.
(iii) Then, evaluate Agˆ= Ag){q} = g{ ˆ= q(0) for an arbitrary (smooth,
δ(x) in the distributional sense: (ˆAq}localized) test function q(x) to solve for the
unknown constants in g(r). [Hint: when evaluating g{ Aq}, you may need to integrate
by parts on the radial-derivative term of '2q; don’t forget the boundary term(s).]
(b) Check that the ω → 0+ limit gives the answer from class.
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7. Problem 1:
(a) Solutions:
f(x) is bounded in every interval except intervals containing x =0, so local integrability
is trivial except for intervals containing x =0. It is sufficient to consider integrals
because any interval [a, b] containing 0 can be broken up into and f(−x)= f(x)
so we only need to show that the latter is finite. But we can now just do the integral
explicitly:
since
as can easily be seen e.g. by L’Hôpital’s rule applied to
(ii) g(x) is not locally integrable for intervals containing the origin. For example
Therefore, f'{φ} = f{−φ'} is a singular distribution.
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8. (iii) We write f'{φ} = liml→0+ fl'{φ}, and integrate by parts in fl'{φ} = fl{−φ'}:
In the last line, the first limit is precisely the Cauchy Principal Value of g(x)φ(x)dx
(CPV = remove a ball of radius E around the singularity, do the integral, and then take
the E → 0+ limit). The second term vanishes because, since φ(x) is continuous and
infinitely differentiable, φ(E)−φ(−E) vanishes at least as fast as E as E → 0, so its
product with ln E vanishes in the limit as in part (i)
(iv) Use f'{φ} = f{−φ'} = f{−[φ − φ(0)]'} = f'{φ − φ(0)}. Then substitute φ − φ(0) into the
previous part, and note that since the integrand is now finite as x → 0 [since
φ(x) − φ(0) goes to zero as x → 0, at least proportionally to x or faster as in the previous
part], we can now just take the limit to write without using
the CPV.
(b) Solutions:
(i) Let V c denote the complement of V (the exterior region outside V , i.e. V c = RdV ),
and note that ∂V c = ∂V (but with the outward-normal vector reversed in sign). We
write:
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9. as desired.
(ii) In this case, we will need to integrate by parts twice, but we can just quote the
results from the “notes on elliptic operators” from class (where we integrated by parts
twice with −V2 already), albeit keeping the boundary terms from ∂V that were zero in
the notes:
But the first term is δ(∂V )[f1(x) − f2(x)] {n·Vφ} =(n·V)δ(∂V )[f2(x) − f1(x)] {φ} by the
definition (note the sign change) of the distributional derivative n·V (note that this is
a scalar derivative in the n direction, not a gradient vector). The second term is a
surface delta function weighted by (n ·Vf1 − n ·Vf2) ,the discontinuity in the normal
derivative. And the last terms are just a regular distribution. So, we have
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10. As noted in class, n ·V of a delta function is a “dipole” oriented in the n direction, so
the first term is a “dipole layer”.
Problem 2:
(a) We solve for g(r) in 3d as follows:
(i) For r> 0, −V2g − ω2g =0 and hence −ω2g = V2g =1/r(rg)'' =⇒ = −ω2h where h(r)=
rg(r)the solution to this is h(r)= cefor some constants c and d, or
It is a little more tricky to determine whether we should use the c or the d term than
in class, since both decay at the same rate. The ratio c/d will be determined by some
kind of boundary condition at infinity, but what might this be? It is acceptable for
you to just punt on this here; since e±iωr are complex conjugates of each other, your
analysis will apply equally well to either one, and you can arbitrarily pick one, say
ceiωr/r, to analyze.
However, to see why there should be a sensible choice, recall that this operator arose
in pset 5 by assuming a time dependence eiωtmultiplying the solution, in which case
we are looking at wave solutions
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11. where the c term describes waves moving out towards r →∞, while the d term
describes
waves moving in from infinity. In wave problems, we typically impose a boundary
condition of outgoing waves at infinity, in which case we must set d =0. (However, the
choice would have been reversed if we picked the opposite sign convention, e+iωt, for the
time dependence.)
(ii) Let’s focus on g(r)= ceiωr/r. As in class, the 1/r singularity is no problem in 3d (it is
cancelled by the Jacobian factor r2dr), so g is a regular distribution. Given an arbitrary
test function q(x), we now evaluate
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12. and hence c =1/4π . Thus,
assuming boundary conditions such that d =0. More generally, since exactly the same
result applies to the de−iωr/r term, we obtain,
for c + d =1/4π , with the ratio c/d being set by the boundary conditions at ∞. The value
at x = x being irrelevant in the distribution sense, e.g. we can assign it to zero, since
this is a regular distribution with a finite integral, similar to class.)
(b) The ω → 0 limit gives 1/4π|x − x'| as in class, by inspection.
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