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Step 2 of 2
At
, we have
Comparing terms, we have
and all other coefficients are zero.
So,
.
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Chapter 6.3, Problem 1E
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Partial Dierential Equations (2nd
We have
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Using Greens Theorem, we have
Thus,
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In the case of one dimension D is just an interval
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specifying boundary conditions means we are specifying
is just two points a and b. So
on
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Chapter 6.1, Problem 10E
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Step 2 of 2
At
, we have
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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
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.
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home / study / math / dierential equations / solutions manual / partial dierential equations / 2nd edition / chapter 6.2 / problem 7e
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Hence, the required harmonic function is
where,
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(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 is solve in the same may except we no longer
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Without finding the solution we need to calculate the value of u at the origin.
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.
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Chapter 6.3, Problem
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Step 2 of 2
Hence,
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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
coefficients
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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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Step 2 of 6
Use the initial boundary conditions of
to determine th
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Therefore, eigenvalue problem for X has solutions
for
with eigenfunctions
Comment
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 the function
and three of the four boundary conditions
Chapter 6.2, Problem 6E
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