Math 678. Homework 3 Solutions.

Math 678. Homework 3 Solutions.
http://math.gmu.edu/~memelian/teaching/Fall11/math678/hw/hw3sol.pdf
#1
We need to derive the formula for the solution of the IVP
ut 􀀀 u + cu = f: in Rn (0;1)
u = g; on Rn ft = 0g
Some observations: (1) the solution to the non-homogeneous problem can be
obtained from the solution to the homogeneous problem via Duhamel's principle;
(2) if the term cu is not present, we know the exact solution to this IVP, (3)
ut + cu = 0 is equivalent to ectut + cectu = 0 which converts to (ectu)t = 0 and
leads to u = Ce􀀀ct as a solution (should remind you of an integrating factor
technique).
From the above observations, we see that a good way to proceed is to mul-
tiply both sides by the function ect. This gives us:
ectut + cectu 􀀀 ect u = ectf
Since (ectu) = ect u, the above can be written as
(ectu)t 􀀀 (ectu) = ectf
and hence this converts into a regular heat equation formulation in terms of ectu,
with non-homogeneous right hand side. The initial condition is unchanged, since
(ectu)(x; 0) = u(x; 0) = g(x). So now we can use Duhamel's principle and write
the exact solution for this modi ed IVP as:
ectu(x; t) =
Z
Rn
(x 􀀀 y; t)g(y)dy +
Z t
0
Z
Rn
(x 􀀀 y; t 􀀀 s)ecsf(y; s)dyds
The solution to the original IVP then follows:
u(x; t) = e􀀀ct
h Z
Rn
(x 􀀀 y; t)g(y)dy +
Z t
0
Z
Rn
(x 􀀀 y; t 􀀀 s)ecsf(y; s)dyds
i
#2
Now let us consider the usual heat equation ut = uxx in 1d with u(x; t) being a
solution.
(a) Take v(x; t) = u(x􀀀y; t). How does one show this is also a solution? The
easiest way is to plug it into the equation. First we need to use chain rule and
compute partial derivatives: vt = (u(x􀀀y; t))t = ut(x􀀀y; t), vx = ux(x􀀀y; t),
relying on the fact that (x 􀀀 y)x = 1. Since ut(x; t) = uxx(x; t) for any x 2 R,
we have vt = vxx. Notice that if we had an IVP originally with u(x; 0) = g(x),
the initial condition for v would be hy(x) = g(x 􀀀 y), and the formula for the
solution using fundamental function (x; t) would give us:
u(x 􀀀 y; t) =
Z
R
(x 􀀀 y 􀀀 s; t)hy(s)ds
1
where s0 = y + s
(b) For any derivative D , where is a multiindex, and the derivative is
taken either in t or in x, we have (D u)t = D ut = D uxx = (D u)xx since
derivatives commute. it follows that any derivative of the solution of the heat
equation is again a solution.
(c) (au + bv)t = aut + bvt = auxx + bvxx = (au + bv)xx, hence au + bv is
again a solution, if u; v are solutions.
(d) This follows from the fact that the operations of integration and di er-
entiation are commutative:
@
@t
Z x
0
u(y; t)dy

=
Z x
0
ut(y; t)dy =
Z x
0
uxx(y; t)dy = @2
@x2
Z x
0
u(y; t)dy

(e) Let v(x; t) = u(
p
ax; at). Then vt(x; t) = av(x; t); vx(x; t) =
p
av(x; t); vxx =
av(x; t). It follows that vt = vxx.
#3
The easiest example of a Dirichlet problem with no solution can be constructed
as follows. Let UT = U (0; T) and = U
T 􀀀 UT be the boundary of this
cylinder including the top, bottom and the sides, while denoting 􀀀T to be the
parabolic boundary comprised of the bottom and vertical sides only. Consider
ut 􀀀 u = 0; in UT
u = f; on
Suppose u 2 C2
1 (UT ) \ C( U
T ) solves this IVP. Then it should satisfy the
weak maximum principle, namely
max
U
T
u(x; t) = max
􀀀T
u(x; t)
Since u(x; t) = f(x) on 􀀀T , we need the solution to satisfy max
U
T
u = max
􀀀T
f.
In particular, f(x; t = T) < max
􀀀T
f, which does not hold for all continuous
functions.
If we allow t = T to be part of the cylinder UT = U (0; T], we need to
construct a more elaborate example of a Dirichlet problem with no solution.
One such example is a ball in R3 with deformable surface. We can push in a
sharp spike at some point on this surface and assume that near the tip of the
spike the surface takes the form of a conical surface obtained by rotating the
curve
y =

e􀀀1=x: for x > 0
0: for x = 0
about the x-axis. Then we can consider heat conduction on the interior of the
deformed ball de ned this way, called
. If the temperature distribution on @

is given by a continuous function f which is equal to zero at points of the spike
2
and is equal to a large positive constant temperature T at points away from the
spike, the seady state temperature u(x) should be close to T for all x in
. But
this is impossible, since u(x) won't be able to approach the zero temperature
as x approaches the spike from within of
. Basically, the spike doe not have
enough surface area to keep the temperature at surrounding points close to zero,
hence the solution fails to be continuous in the closure
. More details about
the subject are available in Helms "Introduction to potential theory" (1975).
#4
Consider v(x; t) = k(x; t)u(x=t;􀀀1=t); t > 0. To see that this solved the heat
equation, let us compute its partial derivatives, using chain rule and product
rule:
vt(x; t) = kt(x; t)u(x
t
;􀀀
1
t
) + k(x; t)(
1
t2 )ut(x
t
;􀀀
1
t
) 􀀀 k(x; t)( x
t2 )ux(x
t
;􀀀
1
t
)
vx(x; t) = kx(x; t)u(x
t
;􀀀
1
t
) + k(x; t)(
1
t
)ux(x
t
;􀀀
1
t
)
vxx(x; t) = kxx(x; t)u(x
t
;􀀀
1
t
) + 2kx(x; t)(
1
t
)ux(x
t
;􀀀
1
t
) + k(x; t)(
1
t2 )uxx(x
t
;􀀀
1
t
)
Now using the fact that kt = kxx; ut = uxx, we can simplify this to
vt 􀀀 vxx = 􀀀
1
t
xk(x; t)
t
+ 2kx(x; t))

ux(x
t
;􀀀
1
t
)
Using the de nition of k(x; t), we can easily see that kx(x; t) = 􀀀(x=2t)k(x; t),
so that xk(x; t) + 2tkx(x; t) = 0. This con rms the claim that v(x; t) solves
the heat equation. Since u(x; t) was de ned for all t < 0, s = 􀀀1=t covers the
domain (0;1). In other words, v(x; t) is a solution for all t > 0.
3