Solutions to the Practice Problems
Math 3150
November 3, 2003
1. Consider the wave equation
@t2 u = @x2 u;
u(0; t) = 0 = u(1; t);
where the intial velocity @t u(x; 0) is zero and the initial position is given by
8
< 0 0 x < 1=4
u(x; 0) = : 1 1=4 x 3=4
0 3=4 < x 1:
What is the smallest positive time t such that u(7=8; t) 6= 0? (Hint: use d'Alembert's solution.)
In this case, d'Alembert's solution is given by
u(x; t) = 21 [f (x + t) + f (x ? t)];
which is the sum of a left moving pulse (the f (x + t) term) and a right moving pulse (the f (x ? t) term).
Therefore, u(7=8; t) is nonzero as soon as the right edge of the right moving pulse hits x = 7=8. Since the pulse
moves with speed 1 and the right edge of the initial pulse is x = 3=4, this happens when
t = 7=8 ? 3=4 = 1=8:
2. Solve the following heat equations.
(a) @t u = @x2 u, u(0; t) = 0 = u(; t), u(x; 0) = 2 sin(2x) ? sin(3x).
The general solution to this type of a heat equation is
X
u(x; t) = an e?(cn)2 t=L2 sin((nx)=L);
where an are the Fourier sine coecients for the initial data f . In this case, f is already in Fourier series,
so we can read o that
a2 = 2; a3 = ?1; an = 0 for n 6= 2; 3:
Also, we have L = and c = 1. Putting this all together, we have
u(x; t) = 2e?4t sin(2x) ? e?9t sin(3x):
(b) @t u = @x2 u, u(0; t) = 0 = u(1; t), u(x; 0) = 2 sin(4x) ? sin(x).
Same general form as before. This time, L = 1 = c and
a1 = ?1; a4 = 2; a + n = 0 otherwise:
So
u(x; t) = ?e?2 t sin(x) + 2e?162t sin(4x):
(c) @t u = @x2 u, u(0; t) = 0 = u(1; t), u(x; 0) = x(1 ? x).
This time we have to do some work to nd the Fourier series of f . We have that
an = 2
Z 1
0
f (x) sin(nx)dx = 2
Z 1
0
x(1 ? x) sin(nx)dx = 2
Z 1
0
x sin(nx)dx ? 2
Z 1
0
x2 sin(nx)dx
Z 1
Z 1
2
1
2
x
x
1
1
= 2[? n cos(nx)j0 + n cos(nx)dx] ? 2[? n cos(nx)j0 + n x cos(nx)dx]
0
0
Z 1
Z 1
Z 1
1
x
1
4
4
4
= n x cos(nx)dx = n [ n sin(nx) ? n sin(nx)dx] = ? n2 2 sin(nx)dx
0
= n343 cos(nx)j10 = n343 ((?1)n ? 1):
0
0
0
This last quantity is 0 if n is even and ?8=(n33 ) if n is odd. So the Fourier series for f is
X
1 sin(nx):
f (x) = x(1 ? x) = ? 8
3
Therefore, we have
u(x; t) = ? 83
X
n odd
n odd
n3
e?2 n2 t sin(nx):
(d) @t u = @x2 u, u(0; t) = 0 = u(1; t), u(x; 0) = 1=2 + jx ? 1=2j.
In this problem, we actually want to work with f (x) = 1=2 ? jx ? 1=2j. Then the analysis below carries
through.
Again, we need to do some work to nd the Fourier series of f . This time it will help to use the symmetry
of the function, namely that f (x) = f (1 ? x). Because sin(nx) has the same symmetry, this means
Z 1
0
f (x) sin(nx)dx = 2
=
Z 1 2
0
f (x) sin(nx)dx:
The Fourier coecient is given by
an = 2
Z 1
0
f (x) sin(nx)dx = 4
=
Z 1 2
0
Z 1 2
f (x) sin(nx)dx = 4
=
Z 1 2
0
x sin(nx)dx
4 [? x cos(nx)j1=2 + = cos(nx)dx] = 4 [? 1 cos(n=2) + 1 sin(nx)j1=2 ]
= n
0
0
n 2
n
0
4 [ 1 sin(n=2) ? 1 cos(n=2)]:
= n
n
2
Thus the solution to the heat equation is given by
X
u(x; t) = 4 [ n12 sin(n=2) ? 21n cos(n=2)]e?n2 2 t sin(nx):
(e) @t u = @x2 u, u(0; t) = 0, u(; t) = , u(x; 0) = 2 sin(2x) ? sin(3x) + x.
This time the boundary data is not zero, but it is constant. So we need to subtractact o the steady state
solution u0 (x), which is the linear function interpolating the two boundary data points. We have
u0 (x) = x:
Then our modied heat equation is
@t u~ = @x2 u~; u~(0; t) = 0 = u~(; t); u~(x; 0) = f (x) ? u0(x) = 2 sin(2x) ? sin(3x):
We can read o the Fourier coecients for this last equation, and they are
a2 = 2;
Therefore, we must have
Putting this all together, we have
a3 = ?1;
an = 0 otherwise:
u~(x; t) = 2e?4t sin(2x) ? e?9t sin(3x):
u(x; t) = u~(x; t) + u0(x; t) = x + 2e?4t sin(2x) ? e?9t sin(3x):
(f) @t u = @x2 u, u(0; t) = 2,u(1; t) = 0, u(x; 0) = sin(4x) ? sin(x) + 2 ? 2x.
Again, we have to nd the steady state solution u0 (x), which is the linear function interpolating between
the boundary points. We have
u0 (x) = 2 ? 2x:
The modied problem is then
@t u~ = @x2 u~; u~(0; t) = 0 = u~(1; t); u~(x; 0) = f (x) ? u0 (x) = sin(4x) ? sin(x):
We can read o the Fourier coecients for this last equation, and they are
a1 = ?1;
Therefore, we must have
a4 = 1;
an = 0 otherwise.
u~(x; t) = ?e?2 t sin(x) + e?162 t sin(4x):
Putting this all together, we have
u(x; t) = u~(x; t) + u0 (x) = 2 ? 2x ? e?2 t sin(x) + e?162 t sin(4x):
(g) @t u = @x2 u, @x u(0; t) = 0 = @x u(; t), u(x; 0) = 3 sin(x) + 2 sin(4x).
Replace sin with cos above. The general solution for a heat equation of this type is
u(x; t) = a0 +
X
where
an e?n2 t cos(nx)
X
u(x; 0) = a0 + an cos(nx):
The initial data u(x; 0) = 3 cos(x)+2 cos(4x) is already in Fourier series, so we can read o the coecients
an :
a1 = 3; a4 = 2; an = 0 otherwise:
Thus the solution is
u(x; t) = 3e?t cos(x) + 2e?16t cos(4x):
(h) @t u = @x2 u, @x u(0; t) = 0 = @x u(1; t), u(x; 0) = ? sin(2x) + 2 sin(x).
Again, replace sin with cos above. This time the general solution is
u(x; t) = a0 +
X
an e?n2 2 t cos(nx):
We can read o the Fourier coecients an , and they are
a2 = ?1;
a1 = 2;
an = 0 otherwise:
Thus the solution is
u(x; t) = 2e?2t cos(x) ? e?42 t cos(2x):
(i) @t u = @x2 u, @x u(0; t) = 0 = @x u(1; t), u(x; 0) = 1=2 + jx ? 1=2j
Again, the general solution has the form
u(x; t) = a0 +
X
an e?n2 2 t cos(nx):
This time, we have to do some work to nd the Fourier coecients an . First, we have that
a0 =
Z 1
0
f (x)dx = 2
=
Z 1 2
0
xdx = 41 :
3. (a) Consider the heat equation
@t u = @x2 u;
and dene the energy at time t as
u(0; t) = 0 = u(L; t);
E (t) :=
L
Z
0
u(x; 0) = f (x)
u2 (x; t)dx:
Show that the energy is a decreasing function in time. Does this make sense physically? (Hint: how
can you characterize decreasing functions of one variable? You might want to use the equation and the
boundary conditions, and integrate by parts at some point.)
We compute the derivative of E :
dE = d Z L u2(x; t)dx = Z L @ u2 (x; t)dx = 2 Z L u(x; t)@ u(x; t)dx
t
t
dt
dt 0
0
0
= 2
L
Z
0
= ?2
Z
0
=L
u(x; t)@x2 u(x; t)dx = 2u(x; t)@x u(x; t)jxx=0
?2
L
Z
0
L
(@x u(x; t))2 dx
(@x u(x; t))2 dx 0:
Here we have used the equation for u, the boundary conditions @x u(0; t) = 0 = @x (L; t), and integration
by parts.
This makes sense physically, because the magnitude of temperature distribution is always decreasing in
time.
(b) Consider the same heat equation as in the previous part, but this time with the boundary conditions
@x u(0; t) = 0 = @x (L; t):
Is the energy still decreasing? Explain your answer.
Yes, the energy is still decreasing, because the boundary term in the integration by parts is still zero.
4. Solve the following dierential equations on the square f(x; y) j 0 x 1; 0 y 1g.
(a) @t2 u = @x2 u + @y2 u, u(0; y; t) = u(1; y; t) = u(x; 0; t) = u(x; 1; t) = 0,
u(x; y; 0) = sin(x) sin(2y), @t u(x; y; 0) = 0
We have that the general solution is of the form
u(x; y; t) =
X
p
p
sin(nx) sin(my)[an;m cos( n2 + m2 t) + bn;m sin( m2 + n2 t)]
where
u(x; y; 0) =
X
an;m sin(nx) sin(my);
@t u(x; y; 0) =
X
p
n2 + m2 bn;m sin(nx) sin(my):
Here the initial velocity is 0, so we have bn;m = 0 for all n and m. Also, the initial position is already in
Fourier series, so
a1;2 = 1; an;m = 0 otherwise.
Thus the solution is
p
u(x; y; t) = sin(x) sin(2y) cos( 5t):
(b) @t2 u = @x2 u + @y2 u, u(0; y; t) = u(1; y; t) = u(x; 0; t) = u(x; 1; t) = 0,
u(x; y; 0) = sin(x) sin(2y), @t u(x; y; 0) = sin(3x) sin(2y)
We have the same general solution as above. Both the initial position and the initial velocity are already
in Fourier series, so we can read o the coecients an;m and bn;m:
a1;2 = 1;
and
an;m = 0 otherwise
b3;2 = p1 ;
bn;m = 0 otherwise:
13
Thus the solution is
p
p
u(x; y; t) = sin(x) sin(2y) cos( 5t) + p1 sin(3x) sin(2y) sin( 13t):
13
(c) @t2 u = @x2 u + @y2 u, u(0; y; t) = u(1; y; t) = u(x; 0; t) = u(x; 1; t) = 0,
u(x; y; 0) = sin(x), @t u(x; y; 0) = 0
Because the initial velocity is zero, we have bn;m = 0 for all n and m. To compute the coecients an;m ,
we have
an;m = 4
= 4
Z 1Z 1
0
Z 1
0
0
f (x; y) sin(nx) sin(my)dxdy = 4
sin(my)dy Z 1
0
Z 1Z 1
0
0
sin(x) sin(nx) sin(my)dxdy =
Z 1
2
1
sin(x) sin(nx)dx = [? m cos(my)j0 ] 2 sin(x) sin(nx)dx
Z 1
2
m
= m (1 ? (?1) )2 sin(x) sin(nx)dx
0
0
This last quantity is 4=(m) for m odd and n = 1, and zero otherwise. We have used the orthogonality
properties of the functions sin(nx) here. Thus the solution is
X
4 sin(x) sin(my) cos(p1 + m2 t):
u(x; y; t) =
m odd
m
(d) @t2 u = @x2 u + @y2 u, u(0; y; t) = u(1; y; t) = u(x; 0; t) = u(x; 1; t) = 0,
u(x; y; 0) = xy(1 ? x)(1 ? y), @t u(x; y; 0) = 0
The general solution has the form
u(x; y; t) =
X
p
p
sin(nx) sin(my)[an;m cos( n2 + m2 t) + bn;m sin( n2 + m2 t)]:
The initial velocity is zero, so bn;m = 0 for all n and m. However, we need to do a little work to nd the
Fourier coecients an;m :
an;m = 4
Z 1Z 1
= (2
0
Z 1
0
0
f (x; y) sin(nx) sin(my)dxdy = 4
x(1 ? x) sin(nx)dx) (2
Z 1
0
Z 1Z 1
0
0
xy(1 ? x)(1 ? y) sin(nx) sin(my)dxdy
y(1 ? y) sin(my)dy) = [ n343 ((?1)n ? 1)] [ m343 :((?1)m ? 1)]
This last quantity is 0 if either n or m is even. If both m and n are odd, it is 16=(n3m3 6 ). Thus the
solution is
X
16 sin(nx) sin(my) cos(pn2 + m2 t):
u(x; y) =
n3 m3 6
n;m odd
(e) @t u = @x2 u + @y2 u, u(0; y; t) = u(1; y; t) = u(x; 0; t) = u(x; 1; t) = 0, u(x; y; 0) = sin(3x) sin(2y)
Here the general solution of this heat equation is
u(x; y; t) =
where
X
an;m e?(n2 +m2 )2 t sin(nx) sin(my)
u(x; y; 0) =
X
an;m sin(nx) sin(my):
Here the initial temperature is already in Fourier series, so we have
a3;2 = 1;
an;m = 0 otherwise:
Thus the solution is
u(x; y; t) = e?132 t sin(3x) sin(2y):
(f) @t u = @x2 u + @y2 u, u(0; y; t) = u(1; y; t) = u(x; 0; t) = u(x; 1; t) = 0, u(x; y; 0) = sin(2y)
This time we have to do a little work to nd the Fourier coecients. We have
an;m = 4
Z 1Z 1
0
0
sin(2y) sin(nx) sin(my)dxdy = (2
Z 1
Z 1
0
sin(nx)dx) (2
Z 1
0
sin(2y) sin(my)dy)
2 (1 ? (?1)n ) (2 sin(2y) sin(my)dy):
= n
0
This last quantity is 4=(n) for n odd and m = 2 and 0 otherwise. Thus the solution is
X
4 e?2(4+n2 )t sin(nx) sin(2y):
u(x; y; t) =
n odd
n
(g) @t u = @x2 u + @y2 u, u(0; y; t) = u(1; y; t) = u(x; 0; t) = u(x; 1; t) = 0, u(x; y; 0) = 1=2 + jx ? 1=2j
(h) @t u = @x2 u + @y2 u, @x u(0; y; t) = @x u(1; y; t) = @y u(x; 0; t) = @y u(x; 1; t) = 0, u(x; y; 0) = cos(2x) cos(y)
This time the general solution is
u(x; y; t) = a0 +
where
X
u(x; y; 0) = a0 +
e?2 (n2 +m2 )t cos(nx) cos(my)
X
an;m cos(nx) cos(my):
Fortunately, our initial temperature is already in Fourier series, we we can read o the Fourier coecients:
a2;1 = 1;
Thus we have
an;m = 0 otherwise:
u(x; y; t) = e?52 t cos(2x) cos(y):
(i) @t u = @x2 u + @y2 u, u(0; y; t) = 0, u(1; y; t) = 1, u(x; 0; t) = x = u(x; 1; t), u(x; y; 0) = x + sin(4x) sin(3y)
This time we have to subtract o the steady state solution u0 (x; y). It turns out that u0 = x (see below),
so our modied problem is
@t u1 = @x2 u1 + @y2 u1 ; u1 (0; y; t) = u1 (1; y; t) = u1 (x; 0; t) = u1(x; 1; t); u1(x; y; 0) = sin(4x) sin(3y):
The initial temperature is already in Fourier series, so we can read o that the Fourier coecients are
a4;3 = 1;
Therefore,
an;m = 0 otherwise:
u1 (x; y; t) = e?252 t sin(4x) sin(3y)
and the whole solution is
u(x; y; t) = u0 + u1 = x + e?252 t sin(4x) sin(3y):
(j) 0 = @x2 u + @y2 u, u(0; y; t) = u(1; y; t) = u(x; 0; t) = 0, u(x; 1; t) = 1.
We will separate variables. Let u(x; y) = X (x)Y (y). Then the equation becomes
X 00 = ? Y 00 = ?2 ;
X
Y
with the boundary conditions
X (0) = 0 = X (1);
The boundary conditions on X force
Y (0) = 0;
X (x) = sin(nx);
Y (1) = constant:
= n;
n = 1; 2; 3; ; : : :
Then the equation for Y becomes
Y 00 = n2 2 Y;
Y (0) = 0;
Y (1) = an :
Therefore we have Y (y) = an sinh(ny), so
u(x; y) =
where
X
an sin(nx) sinh(ny);
X
1 = an sinh(n) sin(nx):
It remains to nd the Fourier series for the function 1. These Fourier coecients are
Z 1
c = 2 sin(nx)dx = ? 2 cos(nx)j1 = 2 (1 ? (?1)n ):
n
0
n
n
This last quantity is 0 for n even and 4=(n) for n odd. Thus we have
so
0
2 (1 ? (?1)n );
an sinh(n) = n
u(x; y) =
X
n odd
4 sinh(n) sinh(ny) sin(nx):
n
(k) 0 = @x2 u + @y2 u, u(0; y; t) = 1, u(1; y; t) = 2, u(x; 0; t) = 0 = u(x; 1; t).
Again, we separate variables and write u(x; y) = X (x)Y (y). Then
with the boundary conditions
X (0) = 1;
X 00 = ? Y 00 = 2 ;
X
Y
X (1) = 2;
Y (0) = 0 = Y (1):
Solving for the equation in Y rst, we see
Y (y) = sin(ny); = n:
Thus the equation for X is X 00 = n2 2 X and
X (x) = an cosh(nx) + bn sinh(nx):
Our solution is
u(x; y) =
X
sin(ny)[an cosh(nx) + bn sinh(nx)]:
We can determine the coecients by matching the boundary data. Evaluating at x = 0 we have
X
X
4 sin(ny);
1 = u(0; y) = a sin(ny) =
n
n odd
n
so an = 0 for n even and an = 4=(n) for n odd. Evaluating at x = 1, we have
X
X
8
2 = u(1; y) = sin(ny)[an cosh(n) + bn sinh(n)] =
n sin(ny):
n odd
So we have bn = 0 for n odd and, for n even, we have
4 2 ? cosh(n) :
bn = n
sinh(n)
Thus we have
X
4 sin(ny)[cosh(nx) + (2 ? cosh(n)) sinh(nx)]:
u(x; y) =
n
sinh(n)
n odd
(l) 0 = @x2u + @y2 u, u(0; y; t) = y = u(1; y; t), u(x; 0; t) = 0, u(x; 1; t) = 1. (Hint: you might be able to guess
the solution to this one.)
This is one of those problems where it helps to draw a picture. The boundary data for this problem all
lies in the plane z = x, which is the graph of the linear function u(x; y) = x. So the solution is just
u(x; y) = x:
5. Consider the dierential equation
@t u = @x2 u + @y2 u;
u(x; y; 0) = f (x; y)
with the boundary conditions
u(x; 0; t) = 0 = u(x; 1; t); @xu(0; y; t) = 0 = @x u(1; y; t):
(a) Using the form u(x; y; t) = X (x)Y (y)T (t), separate variables and write down the dierential equations X ,
Y and T must satisfy.
Let u(x; y; t) = X (x)y(y)T (t). Then the dierential equation reads
XY T 0 = X 00Y T + XY 00 T;
which we can rearrange to read
T 0 (t) = X 00 (x) + Y 00 (y):
T
X
Y
Since one side of the equation depends on t alone, and the other side depends on x and y, both sides of
the equation must be equal to a constant, which we will call ?2 . Thus we have
T 0 = ?2 ; X 0 + Y 0 = ?2 :
T
X Y
We need to separate variables again in the second of these two equations, which we can rearrange to read
X 00 (x) = ?2 ? Y 00 (y):
X
Y
This time, one side of the equation depends on x alone while the other side depends on y alone, so both
sides of the equation must be equal to a constant which we will call ?2 . The two equations we have now
are
X 00 = ?2 ;
Y 00 = ?(2 ? 2 ) = ? 2 :
X
Y
The solution to these last two equations are
X (x) = c1 cos(x) + c2 sin(x);
Y (y) = c1 cos(y) + c2 sin(y):
We also have
T (t) = c3 e?2 t :
Here the relation between the separation constants , and is given by
2 = 2 + 2 ;
so the formula for T should really read
T (t) = c3 e?(2 + 2 )t :
(b) What are the boundary conditions for X and Y ?
The boundary conditions on the equation for x come from the boundary conditions on the vertical sides,
where we have @x u(0; y; t) = 0 = @x u(1; y; t). Thus we have
X 0 (0) = 0 = X 0 (1):
Similarly, the boundary conditions on the equation for y come from the boundary conditions on the
horizontal sides, where we have u(x; 0; t) = 0 = u(x; 1; t). Thus we have
Y (0) = 0 = Y (1):
(c) What are the initial conditions for T ?
The initial condition for T in nding T (0) is given by
f (x; y) = u(x; y; 0) = X (x)Y (y)T (0):
We'll return to this later.
(d) Write down what X and Y are from the boundary conditions.
From these boundary condition at x = 0 and the form of X above, we see that
X (x) = c1 cos(x):
From the boundary condition at x = 1 we have = n for n = 0; 1; 2; 3; : : : Thus we may as well take
X (x) = cos(nx); n = 0; 1; 2; 3; : : :
From the boundary condition at y = 0 and the form of Y above, we see that
Y (y) = c2 sin(y):
From the boundary condition at y = 1 we have = m for m = 0; 1; 2; 3; : : : Thus we may as well take
Y (y) = sin(my); m = 0; 1; 2; 3; : : :
(e) What is the form of the general solution to this equation?
So far we have
Y (y) = sin(my); T (t) = cn;me?(n2 +m2 )t :
To nd the general solution we have to summ over all n = 0; 1; 2; 3; : : : and m = 0; 1; 2; 3; : : : Thus the
X (x) = cos(nx);
general form is
u(x; y; t) = a0 +
1
X
n;m=1
an;me?(n2 +m2 )t cos(nx) sin(my);
where the coecients an;m are given by
a0 =
Z 1Z 1
0
0
f (x; y)dxdy;
an;m = 4
Z 1Z 1
0
0
f (x; y) cos(nx) sin(my)dxdy:
Alternatively, we could say
f (x; y) = a0 +
X
n;m
an;m cos(nx) sin(my):
6. Consider the dierential equation
@t u + u = @x2 u;
u(0; t) = 0 = u(L; t);
u(x; 0) = f (x):
Separate variables and write down the general solution to this equation.
We will try a solution of the form u(x; t) = X (x)T (t). Then the equation reads
XT 0 + XT = X 00T;
which we can rearrange to read
T 0 + T = X 00 :
T
X
The left hand side is a function of t alone, while the right hand side is a function of x alone, so both sides of
the equation must be equal to a constant which we will call ?2 . Now the equations are
T 0 + T = ?2 ;
X 00 = ?2 :
T
X
Let's look at the equation for x rst. We have the boundary conditions u(0; t) = 0 = u(L; t), so
X (0) = 0 = X (L):
On the other hand, the general solution to the ODE for X is
X (x) = c1 cos(x) + c2 sin(x):
Plugging in the boundary condition X (0) = 0 we see c1 = 0. Also,
0 = X (L) = c2 sin(L);
which implies L = n for some n = 1; 2; 3; : : :, or
We may as well take
The dierential equation for T is now
which we can rearrange to read
The general solution to this equation is
= n
L:
X (x) = sin((nx)=L):
T 0 + T = ? n2 2 ;
T
L2
T 0 = ?((n2 =L2 + 1)T:
T (t) = cn e?(n2 2 =L2 +1)t :
Putting this all together, we get that the general solution has the form
u(x; t) =
1
X
cn e?(n2 2 =L2 +1)t sin((nx)=L);
1
where cn are the Fourier coecients of the initial data f (x) = u(x; 0).
7. For each of the following functions of two variables, compute the Laplacian u = @x2 u + @y2 u. You may want
to use a dierent coordinate system for some of these.
(a) u = x2 ? y2
u = @x2 (x2 ? y2 ) + @y2 (x2 ? y2 ) = 2 ? 2 = 0.
(b) u = x3 ? 3x2 y + 3xy2 ? y3
u = @x2 (x3 ? 3x2 y + 3xy2 ? y3 ) + @y2 (x3 ? 3x2 y + 3xy2 ? y3 ) = 6x + 6y + 6x ? 6y = 12x
(c) u = (x2 + y2 )?3=2
This time it's a good idea to use polar coordinates. Recall that
u = @ 2 u + @ 2 u = @ 2u + 1 @ u + 1 @ 2 u:
y
x
r
Our function is just u = r?3=2 , so
r r
r2 u = @r2 (r?3=2 ) + 1r @r (r?3=2 ) = 94 r?7=2 = 49 (x2 + y2 )?7=2 :
(d) u = arctan(y=x)
Again, we should use polar coordinates to see that u(r; ) = . Thus u = 0.
(e) u = (x2 + y2 ) arctan(y=x)
This time U = r2 , so
u = @ 2 (r2 ) + 1 @ (r2 ) + 1 @ 2(r2 ) = [@ 2 (r2 ) + 1 @ (r2 )] = 4 = 4 arctan(y=x):
r
r r
r2 r r
8. For each of the give functions f = f () on the unit circle in the plane, solve the boundary value problem
u(r; ) = 0 for r 1;
u(1; ) = f ():
r
The general form of a harmonic function on the unit disc is
u(r; ) = a0 +
X
rn [an cos(n) + bn sin(n)]
where the boundary data is given by
u(1; ) = f () = a0 +
X
[an cos(n) + bn sin(n)]:
(a) f () = cos() + sin(2)
Here the boundary is already in Fourier series, so we can read o the coecients:
a1 = 1;
b2 = 1;
and all the other coecients are zero. Thus we have
u(r; ) = r cos() + r2 sin(2):
(b) f () = cos2 ? sin2 This time we have to use a trig identity to get the boundary data in Fourier series. We have
f () = cos2 ? sin2 = cos(2);
so a2 = 1 and all the other coecients are zero. Therefore,
u(r; ) = r2 cos(2):
(c) f () = 0 ? 0 < 2
This time we have to do some work to nd the Fourier series for f . First we have
Z 2
Z 1
1
a0 = 2
f ()d = 2 ( ? )d = 2 ? 12 2 0 = 4 :
0
0
For the an terms, we have
Z 2
an = 1
Z
=
0
1
= n
0
f () cos(n)d = 1
Z
0
( ? ) cos(n)d
Z Z cos(n)d ? 1 cos(n)d = ? 1 [ n sin(n) ? n1 sin(n)d]
Z
0
Z 0
0
0
sin(n)d = ? n12 cos(n)j0 = n12 [1 ? (?1)n ]:
0
This last quantity is o for n even and 2=(n2) for n odd. Finally, we have
Z 2
Z
Z
f () sin(n)d = sin(n) ? 1 sin(n)d
0
0
0
Z n
n
cos(
n
)
1
1
= ? n ? [ ? n cos(n) + n cos(n)d] = n1 ? (?n1) ? 1 (? (?n1) ) = n1 :
0
0
0
bn = 1
Thus we can write the solution as
u(r; ) = 4 +
rn n1 2 (1 ? (?1)n ) cos(n) + n1 sin(n) :
n
X