Question

(Link to document inside)
I do not understand the intermediate steps that were taken to solve for a constant of an integral.

Answer 1

http://people.whitman.edu/~hundledr/courses/M367/HWSect01_4.pdf

Problem 1.4.2 (B)

Answer 2

Where they solve the ODE:
\[u _{xx} = -\frac{ Q }{ K_{0} } = -x \rightarrow u_{x} = -\frac{1}{2}x ^{2}+\frac{1}{6}L^{2}\]

I do not understand where they got:
\[+\frac{1}{6}L^{2}\]

Because if you solve the integral:
\[\int\limits_{0}^{L}u_{xx} dx =\int\limits_{0}^{L}(-x) dx\]

Don't you just get

\[-\frac{x^{2}}{2}+C_{1}\]

Right?

Answer 3

I mean, if you solve for
\[C_{1}\]
You just get:

\[C_{1} = \frac{L^{2}}{2}\]

and
\[u_{x} = -\frac{1}{2}x ^{2}+\frac{1}{2}L^{2}\]
?

Answer 4

Hmm I don't remember PDE's very well :\

How are you solving for the constant?
They gave us information about the boundary conditions for u, but not for u', so I'm a little confused.

\(\large\rm u_x(L)=?\)
\(\large\rm u_x(0)=?\)

Answer 5

I am also confused. I was just using x = 0, x = L as the upper and lower bounds since it said to integrate the entire rod

So,

\[\int\limits_{0}^{L}u_{xx} dx =\int\limits_{0}^{L}(-x) dx \]

Then
\[u_{x} = -\frac{x^{2}}{2}+C_{1}|_{0}^{L}\]

\[=-\frac{(L)^{2}}{2} + C_{1} - (-\frac{0^{2}}{2})\]

\[C_{1}=\frac{L^{2}}{2}\]

I think this might be the better way

\[u_{x} =-\frac{x^{2}}{2}+C_{1}\]
for
\[ x = 0\]
\[u_{0} =-\frac{(0)^{2}}{2}+C_{1}\]

Then
\[C_{1} = 0\]

\[u_{x} =-\frac{x^{2}}{2}\]
for
\[ x = L\]
\[u_{x} =-\frac{(L)^{2}}{2}+C_{1}\]

Then
\[C_{1} = \frac{L^{2}}{2}\]
\[u_{x} =-\frac{x^{2}}{2} + \frac{L^{2}}{2}\]

Answer 6

UPDATE: Solved
at x = 0 and x = L
\[\phi(x,t) = -K_{0}u_{x}(x,t)\]
We have:
ODE:
\[u_{x}(x,t) = K_{0} x\]
BC:
\[u(0) = 0 \]
and
\[u(L) = 0\]

Solving the ODE:
\[\phi(x,t) = -K_{0}u_{x}(x,t)\]

\[\int\limits_{}^{}u_{xx}dx = \int\limits_{}^{}K_{0}xdx\]
\[u_{x} = \frac{K_{0}x^{2}}{2} + C_{1}\]
\[\int\limits_{}^{}u_{x}dx = \int\limits_{}^{}\frac{K_{0}x^{2}}{2} + C_{1}dx\]
\[ = \int\limits_{}^{}\frac{K_{0}x^{2}}{2}dx + \int\limits_{}^{} C_{1}dx\]
\[= \frac{K_{0}}{2}\int\limits_{}^{}x^{2}dx + C_{1}x\]
\[= \frac{K_{0}}{2}(\frac{x^3}{3}) + C_{1}x\]
\[= \frac{K_{0}x^3}{6} + C_{1}x\]
\[u(x,t) = \frac{K_{0}x^3}{6} + C_{1}x + C_{2}\]

Apply boundary conditions (BC) to
\[u(x,t)\]
and solve for C1 and C2, and then take the derivative of u(x,t) so that you can use it in:
\[\phi(x,t) = -K_{0}u_{x}(x,t)\]
Where x = 0, and x = L