Step 2 of 2
Hence,
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Chapter 6.2, Problem 6E
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and
So,
Thus,
By boundary condition
, we get
Hence,
Therefore, the solution can be expressed as
, where c is any constant with
coeff
Step 2 of 2
At
, we have
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Chapter 6.3, Problem 3E
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Matching the terms, we have
and all other coefficients are zero. So
.
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Step 2 of 2
(b)
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Without finding the solution we need to calculate the value of u at the origin.
Chapter 6.3, Problem 1E
We calculate the value of u at the origin using the mean value property
Adding them and simplifying, we get
.
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Chapter 6.1, Problem 11E
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where we have used the divergence theorem.
Thus, there is no solution of
in D,
on bdy D
in three dimensions, unless
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Step 2 of 3
In case o
Step 2 of 2
At
, we have
Comparing terms, we have
and all other coefficients are zero.
So,
.
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The University of
Note that u is continuous except where
Let
that is except at the point
.
be a boundary point.
Then,
.
For
;
.
Thus, u is 0 at the boundary except at
In the interior
Since,
,
is not continuous on
.
.
;
But
on
so
So, we are left with
Since each of
Hence,
constant.
But we know
That is
is nonnegative and continuous, this implies that
on
so
This implies that
Therefore the solution to the Dirichlet probl
on
and
where
are arbitrary
constants and we must show that
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Chapter 6.1, Problem 11E
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We have
Thus,
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We have
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Chapter 6.1, Problem 11E
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Using Greens Theorem, we have
Thus,
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Step 3 of 3
In the case of one dimension D is just an interval
and
specifying boundary conditions
Since right hand value of the above expression is function of
is function of
alone, so they must be equal to a constant say
alone, and left hand side value
.
Therefore,
on
,
and
on
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Therefore, eigenvalue problem for X has solutions
for
with eigenfunctions
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Step 2 of 3
For Y, we solve
The solution have the form
The condition at
implies that
as
So we must have
satisfies
Then
Chapter 6.2, Problem 6E
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Jeisel Q.
28
The University of Texas.
Victor C.
155
California Institute of T.
Katrina S.
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Coastal Carolina
Step 2 of 2
(b)
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Without finding the solution we need to calculate the value of u at the origin.
Chapter 6.3, Problem 1E
We calculate the value of u at the origin using the mean value property
Hence, the required harmonic function is
where,
Comment
Step 3 of 3
(b)
We need to determine what would go awry if we omit the condition at infinity.
If we omit the condition at infinity the problem i
Step 2 of 2
At
, we have
Comparing terms, we have
and all other coefficients are zero.
So,
.
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on
Comment
Step 2 of 3
Multiplying by w, we have
on D. Now we integrate this equation to show that
Note that,
So,
Now we integrate it as follows
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Step 3 of 3
We can integrate the first term usi
Thus, the solution of (6) is;
.
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Chapter 6.2, Problem 4E
Comment
Step 6 of 9
Substitute values of
and
in (3);
Use the initial condition to obtain;
This represents the Fourier cosine series of
,
Therefore, solution is,
Comment
Step 6 of 6
The boundary conditions,
implies that
Without loss of generality assume that
.
Therefore,
Putt together the solutions for
Now to determine the constants
C
M412 Assignment 6 Solutions, due Friday October 28
1. [10 pts] Show that for the eigenvalue problem
a x b,
(p(x)ux )x + q (x)u + (x)u = 0;
eigenvalues are related to their eigenfunctions u by the Rayl
M412 Assignment 5 Solutions
1. [10 pts] Haberman 2.5.1 (a).
Solution. Setting u(x, y ) = X (x)Y (y ), we nd that X and Y satisfy
X + X = 0
Y Y = 0,
with boundary conditions
X (0) = 0
X (L) = 0
Y (0) =
M412 Assignment 4 Solutions
Two errors in the Practice Problems for Exam 2 have been brought to my attention. In Problem 3, f (x)
should be given explicitly as x2 . Also, in the solution to Problem 3,
M412 Assignment 3 Solutions
1. [10 pts] Use the method of characteristics to solve the PDE
ux uy + 2y =0
u(x, y ) = xy on the line x + 2y = 1.
Solution. In this case, set U (t) = u(x(t), y (t) and cho
M412 Assignment 2, due Friday September 9
1. [10 pts] Use the method of diagonalization to determine a general solution for the ODE system
3
y1 = y1 + y 2
4
y2 = 5 y1 + 3 y2 .
Solution. In matrix form
Assignment 1, Solutions
1. [5 pts] We solve this problem by separation of variables. We have
dy
= 3x2 dx
+1
y2
dy
=
+1
3x2 dx tan1 y = x3 + C y (x) = tan(x3 + C ).
y2
Using the initial condition y (0
M412 Practice Problems for Exam 3
Qualititative properties of the Laplace equation
1. For the Laplace equation
0 x 1,
uxx + uyy = 0;
u(x, 0) = 10x(1 x);
u(0, y ) = 10y (1 y );
0 y 1,
u(x, 1) = 1 x
u(1
M412 Practice Problems for Exam 2
Please keep in mind that Exam 2 will be given Wednesday, October 19 7:009:00.
Equilibrium Problems
1. Haberman 1.4.1 Parts (d) and (h).
2. Haberman Problem 1.4.11.
3.