Solve the one-dimensional wave equation ∂
2y
∂x2
=
1
c
2
∂
2y
∂t2
subject to the boundary con-
ditions y(0, t) = y(L, t) = 0 and initial conditions y(x, 0) = f(x),
∂y
∂t (x, 0) = g(x) where f(x) is the
initial deflection and g(x) is the initial velocity. Here y = y(x, t). Also find the solution when f(x) =
2k
L
x if 0 < x < L/2
2k
L
(L − x) if L/2 < x < L and the initial velocity is 0.
Answers
Answer:
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Step-by-step explanation:
Notes
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I apologize for the outage on the site yesterday and today. Lamar University is in Beaumont Texas and Hurricane Laura came through here and caused a brief power outage at Lamar. Things should be up and running at this point and (hopefully) will stay that way, at least until the next hurricane comes through here which seems to happen about once every 10-15 years. Note that I wouldn't be too suprised if there are brief outages over the next couple of days as they work to get everything back up and running properly. I apologize for the inconvienence.
Paul
August 27, 2020
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Section 9-5 : Solving The Heat Equation
Okay, it is finally time to completely solve a partial differential equation. In the previous section we applied separation of variables to several partial differential equations and reduced the problem down to needing to solve two ordinary differential equations. In this section we will now solve those ordinary differential equations and use the results to get a solution to the partial differential equation. We will be concentrating on the heat equation in this section and will do the wave equation and Laplace’s equation in later sections.
The first problem that we’re going to look at will be the temperature distribution in a bar with zero temperature boundaries. We are going to do the work in a couple of steps so we can take our time and see how everything works.
The first thing that we need to do is find a solution that will satisfy the partial differential equation and the boundary conditions. At this point we will not worry about the initial condition. The solution we’ll get first will not satisfy the vast majority of initial conditions but as we’ll see it can be used to find a solution that will satisfy a sufficiently nice initial condition.
Example 1 Find a solution to the following partial differential equation that will also satisfy the boundary conditions.
∂
u
∂
t
=
k
∂
2
u
∂
x
2
u
(
x
,
0
)
=
f
(
x
)
u
(
0
,
t
)
=
0
u
(
L
,
t
)
=
0
Show Solution
So, there we have it. The function above will satisfy the heat equation and the boundary condition of zero temperature on the ends of the bar.
The problem with this solution is that it simply will not satisfy almost every possible initial condition we could possibly want to use. That does not mean however, that there aren’t at least a few that it will satisfy as the next example illustrates.
Example 2 Solve the following heat problem for the given initial conditions.
∂
u
∂
t
=
k
∂
2
u
∂
x
2
u
(
x
,
0
)
=
f
(
x
)
u
(
0
,
t
)
=
0
u
(
L
,
t
)
=
0
f
(
x
)
=
6
sin
(
π
x
L
)
f
(
x
)
=
12
sin
(
9
π
x
L
)
−
7
sin
(
4
π
x
L
)
Show All Solutions Hide All Solutions
a
f
(
x
)
=
6
sin
(
π
x
L
)
Show Solution
b
f
(
x
)
=
12
sin
(
9
π
x
L
)
−
7
sin
(
4
π
x
L
)
Show Solution
So, we’ve seen that our solution from the first example will satisfy at least a small number of highly specific initial conditions.
Now, let’s extend the idea out that we used in the second part of the previous example a little to see how we can get a solution that will satisfy any sufficiently nice initial condition. The Principle of Superposition is, of course, not restricted to only two solutions. For instance, the following is also a solution to the partial differential equation.
u
(
x
,
t
)
=
M
∑
n
=
1
B
n
sin
(
n
π
x
L
)
e
−
k
(
n
π
L
)
2
t
and notice that this solution will not only satisfy the boundary conditions but it will also satisfy the initial condition,
u
(
x
,
0
)
=
M
∑
n
=
1
B
n
sin
(
n
π
x
L
)
Let’s extend this out even further and take the limit as
M
→
∞
. Doing this our solution now becomes,
u
(
x
,
t
)
=
∞
∑
n
=
1
B
n
sin
(
n
π
x
L
)
e
−
k
(
n
π
L
)
2
t
This solution will satisfy any initial condition that can be written in the form,
u
(
x
,
0
)
=
f
(
x
)
=
∞
∑
n
=
1
B
n
sin
(
n
π
x
L
)
This may still seem to be very restrictive, but the series on the right should look awful familiar to you after the previous chapter. The series on the left is exactly the Fourier sine series we looked at in that chapter. Also recall that when we can write down the Fourier sine series for any piecewise smooth function on
0
≤
x
≤
L
.
So, provided our initial condition is piecewise smooth after applying the initial condition to our solution we can determine the
B
n
as if we were finding the Fourier sine series of initial condition. So we can either proceed as we did in that section and use the orthogonality of the sines to derive them or we can acknowledge that we’ve already done that work and know that coefficients are given by,
B
n
=
2
L
∫
L
0
f
(
x
)
sin
(
n
π
x
L
)
d
x
n
=
1
,
2
,
3
,
…
So, we finally can completely solve a partial differential equation.
Example 3 Solve the following BVP.
∂
u
∂
t
=
k
∂
2
u
∂
x
2
u
(
x
,
0
)
=
20
u
(
0
,
t
)
=
0
u
(
L
,
t
)
=
0