Solutions to the Practice Problems
Math 3150
November 25, 2003
These are only practice problems for the material in chapter 7. You might also want to make sure you know how to
do the problem on the earlier practice sheets.
1. Compute the Fourier transform of the following functions.
?1 ?1 < x < 0
(a) f (x) = : 1 0 < x < 1
0 otherwise
8
<
Z 1
Z 0
Z 1
1
1
1
?
ix!
?
ix!
^
e
dx ? p
e?ix! dx
f (! ) = p
f (x)e
dx = p
2 0
2 ?1
2 ?1
1 e?ix! x=1 + 1 e?ix! x=0 ] = p1 [? e?i! + 1 + 1 ? ei! ]
= p1 [? i!
x=0 i!
x=?1
2 i! i! i! i!
2
2
?
2
cos
!
1
i!
?
i!
p [2 ? e ? e ] = p :
=
i! 2
i! 2
(b) f (x) =
?x jxj < 1
0 jxj 1
Z 1
Z 1
Z 1
1
1
x ?ix! x=1
1
1
?
ix!
?
ix!
^
+
e?ix! dx]
f (! ) = p
e
f (x)dx = p
xe
dx = p [ ? e
x=?1 i! ?1
2 ?1
2 ?1
2 i!
i!
?i!
1 [? 2 cos ! + e?i! ? ei! ] = p1 [ 2i cos(!) + 2 sin(!) ]
p
]
=
= p1 [? ei! ? ei! + !12 e?ix! xx=1
=?1
i!
!2
!
!2
2p
2
2
2
= i!p 2 [! cos(!) + sin(!)]
1 + x ?1 x 1
(c) f (x) = : 1 ? x 0 x 1
0 otherwise
First let's do some preliminary computations:
8
<
Z 1
Z
Z 1
0
1
= ? i!
0
0
1
e?ix! dx = ?
i!
e?ix!
?1
?xe?ix! dx =
Z 0
?1
xe?ix! dx
=
e?ix! 0 =
1
1 [1 ? e?i! ]
i!
e?ix! ?1 =
1 [ei! ? 1]
0
i!
1 e?ix! 1 = e?i! + 1 ? e?i!
x ?ix! 1 1 1 ?ix!
e?i!
e
?
e
dx =
?
0
i!
i! 0
i!
!2
i!
!2
0
Z 0
0
i!
i!
? i!x e?ix! ?1 + !12 e?ix! dx = ? ei! + !12 e?ix! 0?1 = ? ei! + !12 [1 ? ei! ]:
?1
Z
Now put everything together:
f^(! )
Z 1
1 Z 1 (1 ? x)e?ix! dx + p1 Z 0 (1 + x)e?ix! dx
= p1
f (x)?ix! dx = p
2 ?1
2 0
2 ?1
Z 1
Z 0
Z 1
Z 0
1
1
1
1
?
ix!
?
ix!
?
ix!
p
p
p
p
e
dx +
e
dx +
?xe dx +
xe?ix! dx
=
2 0
2 ?1
2 0
2 ?1
?i!
i!
p1 [1 ? e?i! ] + p1 [ei! ? 1] + e p + 2 p1 (1 ? e?i! ) + ?pe + 2 p1 (1 ? ei! )
=
i! 2
i! 2
i! 2 ! 2
i! 2 ! 2
1
2
?
2
cos
! p
1
?
cos
!
i!
?
i!
= 2p (2 ? e ? e ) = 2 p
= 2= !2 :
! 2
! 2
(d) f (x) = e?jxj
f^(! )
Z 1
1 Z 0 e?ix! exdx + p1 Z 1 e?ix! e?x dx
f (x)e?ix! dx = p
= p1
2 ?1
2 ?1
2 0
Z 0
Z 1
1
1 ex(1?i!) 0 ? 1 ex(?1?i!) 1 ]
1
= p1
ex(1?i!) dx + p
ex(?1?i!) dx = p [
?1 1 + i!
0
2 ?1
2 0
2 1 ? i!
1
1
1
1
1
1
+
i! + 1 ? i! p
= p [ 1 ? i! + 1 + i! ] = p (1 ? i!)(1 + i!) = 2= 1 + !2
2
2
2. Let f = f (x) and g = g(x) be piece-wise continuous, real-valued functions such that f (x) = 0 for jxj > 1 and
g (x) = 0 for jxj > 1. In particular, this implies both f and g are bounded and achieves their maxima.
(a) Show that the Fourier transform f^ makes sense (i.e. the improper integral converges) and jf^(!)j maxx jf (x)j for all !.
The improper integral is in fact a proper integral, from x = ?1 to x = 1. This is because f (x) = 0 for
jxj > 1. We also have the bound
jf^(!)j p1
Z
1
je?ix! jjf (x)jdx = p1
Z 1
p jf j
jf (x)jdx max
Z 1
dx
2 ?1
2 ?1
2 ?1
2
= p max jf j max jf j:
2
p
The last inequality comes from 2 > 2.
(b) Is it true that f^(!) = 0 for j!j suciently large? Explain your answer.
No. For instance, if f is a square pulse (i.e. f (x) = 1 for jxj 1 and f (x) = 0 for jxj > 1), then the
Fourier transform is
p
sin(!) ;
f^(! ) = 2=
!
which is only zero at ! = ; 2; 3; : : :
R1
(c) Is it true that f g(x) = ?1
f (x ? y )g (y )dy is zero for jxj suciently large? Explain your answer.
Yes. Indeed, if jyj > 1 and jxj > 2 then both f (x ? y) = 0 and g(y) = 0, and so the integrand in the
convolution is zero.
(d) If the Fourier transform f^ is also real-valued, what can you say about f ?
Recall that
Z 1
Z 1
1
1
?
ix!
^
e
f (x)dx = p
(cos(x!) ? i sin(x!))f (x)dx
f (! ) = p
2 Z?1
2 ?1
Z
1
1
cos(x!)f (x)dx ? pi
sin(x!)f (x)dx:
= p1
2 ?1
2 ?1
The rst integral picks out the even part of f and the second integral picks out the odd part. Therefor,
if f^(!) is real valued, the second integral must be zero for all !, and so f must be an even function.
p
(e) Show that F (f g)(!) = 2f^(!)^g(!).
F (f g)(!) = p1
=
=
=
=
=
Z
1
e?ix! f g (x)dx
2 Z?1
Z 1
1
p1
e?ix!
f (x ? y )g (y )dydx
2 Z?1 Z
?1
1 1
p1
e?ix! f (x ? y )g (y )dxdy
2 Z?1 Z?1
1 1
p1
e?ix! e?iy! eiy! f (x ? y )g (y )dxdy
2 Z?1 ?1
Z 1
1
p1
e?i!(x?y) f (x ? y )d(x ? y ) e?iy! g (y )dy
2
?1
?1
p ^
2f (!) g^(!):
3. Solve the wave equation
@t2 u = @x2 u
with the initial conditions
1 :
1 + x2
You may leave your answer in terms of the inverse Fourier transform (even though it is possible to simplify it
more).
Let
1 = @ u(x; 0):
g (x) =
1 + x2 t
u(x; t) = f (x) = 0;
@t u(x; 0) =
Take the Fourier transform of each side of the equation in the x variable. Then we get
u^00 (!; t) = ?! u^(!; t);
This has the solution
u^(!; 0) = 0;
u^(!; t) =
and so
u^0 (!; 0) = g^(! ):
g^(! )
sin(!t);
!
g^(! ) sin(!t)
:
u(x; t) = F ?1
!
4. Solve the dierential equation
@t u =
for t 0, with the initial condition
1 2
1 + t @x u
u(x; 0) = f (x) =
1 jxj < 1
0 jxj 1:
You may leave your answer in terms of the inverse Fourier transform.
Again, take th Fourier transform in x. We get
u^0 = ?
!2
1+t
This is a separable, rst order ODE, so
which we can integrate to get
Thus
u^;
u^(!; 0) = f^(! ) =
du^
u^
2= sin(!!) :
p
= ?!2 1 dt
+ t;
ln u^ = ?!2 ln(1 + t) + c = ln[c(1 + t)?! ]:
2
p
2
2
= u^(!; 0)(1 + t)?! = 2= sin! ! (1 + t)?! :
Taking the inverse Fourier transform, we have
p
sin ! (1 + t)?!2 :
u(x; t) = 2= F ?1
u^(!; t) = c(1 + t)?!
2
!
5. Solve the dierential equation
for t 0, with the initial conditions
@t u = @x2 u + u
u(x; 0) = f (x) =
8
<
:
x+1
1?x
?1 x 0
0x1
0 otherwise.
You may leave your answer in terms of the inverse Fourier transform.
Taking the Fourier transform in x, we get
u^0 = ?! 2 u^ + u^ = (1 ? ! 2 )^
u;
u^(!; 0) = f^(! ) =
!
2= 1 ?!cos
:
2
p
This ODE has the solution
2
p
2 )t
e(1?! )t (1 ? cos ! )
?
!
(1?! 2 )t
^
u^(!; t) = u^(!; 0)e
= f (!)e
= 2=
:
2
(1
!
Taking the inverse Fourier transform, we have
u(x; t) =
p
!
6. Solve the heat equation
for t 0, with the initial condition
#
"
(1?! 2 )t
(1 ? cos !) :
2=F ?1 e
2
@t u = @x2 u
2
u(x; 0) = f (x) = e?x =2 :
You may leave your answer in terms of a convolution with the heat kernel (even though it is possible to simplify
it more).
The solution to the heat equation is given by
Z 1
Z 1
2
1
1
u(x; t) = ? f (x; y ) p
?(x ? y; t)f (y)dy = p
e?(x?y) =(4t) f (y )dy
2t ?1
2t ?infty
Z 1
2
2
1
e?(x?y) =(4t) e?y =2 dy
= p
2t ?1
There is another way to look at the solution. Recall that the solution to the transformed heat equation is a
Gaussian. Indeed, the transformed equation (after the Fourier transform) is
u^00 = ?! 2 u^;
u^(!; 0) = f^(! );
which has the solution
p
p
2
2
2
2
u^(!; t) = f^(! )e?! t = f^(! )e?( 2t!) =2 = e?! =2 e?( 2t!) =2 :
We have used the fact that in this case f is a Gaussian, so f^(!) = f (!), and^u is a product of Gaussians.
There is a general rule hiding here. Indeed, if g(x) = e?ax2=2 and h(x) = e?bx2 =2 for two positive numbers a
and b, then
2
1
^h(!) = p1 e?!2 =(2b) ;
g^(! ) = p e?! =(2a) ;
a
which implies that
b
p
F (g h)(!) = 2g^(!)h^ (!) =
r
2 e?!2 =2a e?!2=2b =
ab
r
2 e?!2 (2ab)=(a+b) :
ab
However, the right hand side of the equation above is the Fourier transform of
r
2 e?x2ab=(2(a+b)) ;
a+b
which means (after taking the inverse Fourier transform) that
g h(x) =
r
2 e?x2 ab=(2(a+b)) :
a+b
p
p
Now we can 2use this to nd an explicit formula for the u(x; t) = F ?1 (^u). Indeed, if ft(x) = 2tf (x= 2t), then
f^t (! ) = e?! t and
u^(!; t) = f^(! ) f^t (! ) = F (f ft )(! ):
Taking the inverse Fourier transform, we have
r
2 e?x2 =(2t+1) :
1 + 2t
Here we have used the formula we derived for the convolution of two Gaussians.
u(x; t) = f ft (x) =
7. Solve the heat equation
for t 0, with the initial condition
@t u = @x2 u
1 jxj < 1
0 jxj 1:
You may leave your answer in terms of a convolution with the heat kernel (even though it is possible to simplify
it more).
u(x; 0) = f (x) =
Z 1
Z 1
2
= ? f (x; t) = p1
?(x ? y; t)f (y)dy = p1
e?(x?y) =(4t) dy
2t ?1
2
t ?1
p
Z x+1
p Z (x+1)=(2 t) ?w2
2 =(4t)
1
?
z
e
dz = 2
= p
p e dw
2t x?1
(x?1)=(2 t)
p
p
p
= 2[erf((x + 1)=(2 t)) ? erf((x ? 1)=(2 t))]
u(x; t)
8. Let u be a solution to the heat equation
@t = @x2 u
on a line, with u(x; t) ! 0 very rapidly as jxj ! 1. For instance, a sucient decay condition is
ju(x; t)j 1 +1 x2
for all x and t 0.
(a) Dene the energy E (t) by
Z 1
1
E (t) =
(u(x; t))2 dx
2
?1
and show that E is a decreasing function of t. (Hint: use the equation and integrate by parts!)
To show E is decreasing, we will show that E 0 < 0. First observe that u is not identically zero, and it
decays for large x. This means @x u cannot be identicially zero. We have
dE
dt
Z
Z 1
Z 1
d 1 2
= dt
u (x; t)dx =
@t u2 (x; t)dx =
2u(x; t)@t u(x; t)dx
= 2
Z
?1
1
?1
?1
?1Z
=1
u(x; t)@x2 u(x; t)dx = 2 u@x ujxx=
?1 ? 2
1
?1
(@x u(x; t))2 dx = ?2
Z
1
?1
(@x u(x; t))2 dx < 0:
In the last inequality, we have used the fact that @x u is not identically zero.
(b) Use the condition on E to say what happens to u(x; t) as t ! 1. You might want to also think about
this physically.
Notice that E (t) is the integral of a non-negative function, so E (t) 0 for all t. However, E is also a
decreasing function, so we mus thave E (t) ! 0 as t ! 1. Thus
Z
1
?1
(u(x; t))2 dx ! 0;
which can only happen if u(x; t) ! 0 as t ! 1 for all x. Physically, this make sense; all the heat should
leak out to 1, and so temperature should fall to 0.