Partial Differential Equations of Applied Mathematics
MATH 220
MATH 220: PROBLEM SET 3 DUE THURSDAY, OCTOBER 15, 2009
so 0, (x) = 0 if x 1 and R (x) dx = 1. Let j (x) = j(jx), so j (x) = 0 if x 1/j, and j (x) dx = 1 for all j. (1) Show that j 0 in D (R) (i.e., to be pedantic, j 0 ). (Hint: this is essentially wri
Partial Differential Equations of Applied Mathematics
MATH 220
MATH 220: PROBLEM SET 2 DUE THURSDAY, OCTOBER 8, 2009
Problem 1. Show that the only solution u D (R) of u = 0 is u = c, c a constant function. Hint: u = 0 means that u( ) = 0 for all Cc (R). You need to show that there is a constant c such that u() = c dx
Partial Differential Equations of Applied Mathematics
MATH 220
MATH 220: PRACTICE MIDTERM
This is a closed book, closed notes, no calculators exam. There are 5 problems. Solve all of them. Total score: 100 points. Problem 1. (i) (13 points) Find the general C 1 solution of the PDE 3y 2 ux + uy = 0. (ii) (6 points) So
Partial Differential Equations of Applied Mathematics
MATH 220

Fall 2012
MATH 220: Problem Set 6
Solutions
Problem 1. g is obviously not a Schwartz function, but it is a tempered
distribution. Therefore we can make sense of the Fourier transform of the distribution g (and see it as a function since the distribution is going to
Math 220A
Practice Final Exam 2
Fall 2002
1. Solve the following initialvalue problem,
ut + ux + u2 = 0
u(x, 0) = x.
2. Find the unique weak solution of
ut + u2 ux = 0
u(x, 0) = (x)
where
(x) =
0 x<0
1 x>0
which satises the Oleinik entropy condition.
3.
Partial Differential Equations of Applied Mathematics
MATH 220

Fall 2012
MATH 220: Problem Set 4
Solutions
Problem 1.
Lets solve the wave equation on the line:
utt
c2 uxx = 0,
u(x, 0) = (x),
ut (x, 0) = (x),
(1)
with
and
8
> 0,
>
<
1 + x,
(x) =
> 1 x,
>
:
0,
8
< 0,
2,
(x) =
:
0,
x < 1,
1 < x < 0,
0 < x < 1,
x > 1,
x < 1,
1 < x
Partial Differential Equations of Applied Mathematics
MATH 220

Fall 2012
MATH 220: Problem Set 2
Solutions
Problem 1. Show that the only solution u 2 D0 (R) of u0 = 0 is u = c, c a
constant function.
1
u0 = 0 in D0 (R) means that u( 0 ) = 0, 8 2 Cc (R). And we want to show
that:
Z
1
9c 2 R, 8 2 Cc (R), u( ) =
c (x)dx.
(1)
R
1
Partial Differential Equations of Applied Mathematics
MATH 220

Fall 2012
MATH 220: PROBLEM SET 1, SOLUTIONS
DUE THURSDAY, OCTOBER 4, 2012
Problem 1. Classify the following PDEs by degree of nonlinearity (linear, semilinear, quasilinear, fully nonlinear):
(1) (cos x) ux + uy = u2 .
(2) u utt = uxx .
(3) ux ex uy = cos x.
(4) u
Partial Differential Equations of Applied Mathematics
MATH 220

Fall 2012
MATH 220: Problem Set 3
Solutions
2 C(R) be given by:
8
x < 1,
> 0,
>
<
1 + x,
1 < x < 0,
(x) =
> 1 x, 0 < x < 1,
>
:
0,
x > 1,
R
so that it veries
0, (x) = 0 if x 1 and R (x)dx = 1.
Consider ( j )j 1 constructed as j (x) = j (jx), so that j (x) = 0 if
Partial Differential Equations of Applied Mathematics
MATH 220

Fall 2012
MATH 220: Problem Set 7
Solutions
Problem 1. Lets solve the inhomogeneous heat equation on the halfline for
Dirichlet boundary conditions:
ut
kuxx = f,
x
0,
u(x, 0) = (x),
u(0, t) = 0.
(1)
(i) Using Duhamels principle:
We know from the class that the sol
Partial Differential Equations of Applied Mathematics
MATH 220

Fall 2012
MATH 220: Problem Set 8
Solutions
Problem 1. By proposition 0.11 in the Inner product spaces handout, we
know that kf
k is minimal if we take the generalized Fourier coe cients.
Meaning, if we call cn (x) = cos(nx) and sn (x) = sin(nx):
8
Z
> a = h , c0
Partial Differential Equations of Applied Mathematics
MATH 220

Fall 2012
MATH 220: Problem Set 5
Solutions
Problem 1. Let f L1 (Rn ) and a Rn . The whole problem is a matter of
change of variables with integrals.
(i)
(F fa ) () =
Rn
eix fa (x)dx =
Rn
eix f (x a)dx
ei(a+y) f (y)dy = eia
=
=e
Rn
ia
(F fa ) ().
Rn
eiy f (y)dy
(1)
Math 220A
Practice Midterm  Fall 2002
1. Classify the following in terms of degree of nonlinearity:
(a) ut + eu ux = x2
(b) ut + x2 ux = eu
(c) x3 ut + u2 = 1
x
(d) ut + x3 ux = sin(x2 )
(e) uxt +
u2
2
x
= cos(ux )
2. Find the unique weak solution of
ut
Math 220A
Second Practice Midterm Exam
Fall 2002
1. Classify the following in terms of degree of nonlinearity:
(a) u2 + x2 uxt = sin(u)
t
(b) ux + [u3 ]y = x2 + y 2
(c) [eu ]x + u2 uy = 0
(d) [x3 u]x + y 3 u = sin(x2 + y 2 )
(e) [u3 ]t + euxt = 0
x
2. Sol
Math 220A
Practice Final Exam I  Fall 2002
1. (a) Suppose S (t) is the solution operator associated with the homogeneous equation
ut + aux = 0
u(x, 0) = (x).
()
In particular, assume the solution of (*) is given by u(x, t) = S (t)(x). Show that
t
v (x, t
Partial Differential Equations of Applied Mathematics
MATH 220
MATH 220: PROBLEM SET 1, SOLUTIONS DUE THURSDAY, OCTOBER 1, 2009
Problem 1. Classify the following PDEs by degree of nonlinearity (linear, semilinear, quasilinear, fully nonlinear): (1) (cos x) ux + uy = u2 . (2) u utt = uxx . (3) ux  ex uy = cos x. (4)
Partial Differential Equations of Applied Mathematics
MATH 220
MATH 220: PROBLEM SET 4 DUE THURSDAY, OCTOBER 22, 2009
Problem 1. Solve the wave equation on the line: utt  c2 uxx = 0, u(x, 0) = (x), ut (x, 0) = (x), with 0, 1 + x, (x) = 1  x, 0, x < 1, 1 < x < 0, 0 < x < 1, x > 1.
and
Also describe in t > 0 where
Partial Differential Equations of Applied Mathematics
MATH 220
MATH 220: PROBLEM SET 5 DUE TUESDAY, OCTOBER 27, 2009
Problem 1. Suppose that f is (piecewise) continuous on Rn with xN f (x) bounded for some N > n (or indeed simply that f L1 (Rn ). Throughout this problem, a Rn . (i) Let fa (x) = f (x  a). Show that
Partial Differential Equations of Applied Mathematics
MATH 220
MATH 220: PRACTICE MIDTERM, SOLUTIONS
This is a closed book, closed notes, no calculators exam. There are 5 problems. Solve all of them. Total score: 100 points. Problem 1. (i) (13 points) Find the general C 1 solution of the PDE 3y 2 ux + uy = 0. (ii) (6
Partial Differential Equations of Applied Mathematics
MATH 220
MATH 220: PROBLEM SET 1 DUE THURSDAY, OCTOBER 1, 2009
Problem 1. Classify the following PDEs by degree of nonlinearity (linear, semilinear, quasilinear, fully nonlinear): (1) (cos x)ux + uy = u2 . (2) uutt = uxx . (3) ux  ex uy = cos x. (4) utt  uxx +