- #1
AnotherParadox
- 35
- 3
- Homework Statement
- Solve the Boundary value problem
[itex]u_{t}=u_{xx}[/itex]
[itex]u(0, t) = u(\pi ,t)=0[/itex]
[itex]u(x, 0) = f(x)[/itex]
[tex]f(x)=\left\{\begin{matrix}
x; 0 < x < \frac{\pi}{2}\\
\pi-x; \frac{\pi}{2} < x < \pi
\end{matrix}\right.[/tex]
- Relevant Equations
- [tex]u(x, t) = \sum_{n=1}^{\infty} A_{n} \exp(- \lambda t)\sin(\sqrt{\lambda}\cdot x))[/tex]
[tex]A_{n}=\frac{2}{L}\int_{0}^{L}f(x)sin(\sqrt{\lambda}\cdot x)dx[/tex]
Solve the boundary value problem
Given
[itex]u_{t}=u_{xx}[/itex]
[itex]u(0, t) = u(\pi ,t)=0[/itex]
[itex]u(x, 0) = f(x)[/itex]
[tex]f(x)=\left\{\begin{matrix}
x; 0 < x < \frac{\pi}{2}\\
\pi-x; \frac{\pi}{2} < x < \pi
\end{matrix}\right.[/tex]
L is π - 0=π
λ = α2 since 0 and -α lead to trivial solutions
Let
[itex]u = XT[/itex]
[itex]X{T}'={X}''T[/itex]
[itex]\frac{{X}''}{X}=\frac{{T}'}{T}=-\lambda [/itex]
[itex]{X}''+\lambda =0[/itex]
[itex]X(0)=X(\pi)=0[/itex]
[itex]{T}'+\lambda T=0[/itex]
[itex]X(x)=C_{1}\cos (\sqrt{\lambda }\cdot x)+C_{2}\sin(\sqrt{\lambda }\cdot x)[/itex]
[itex]\sqrt{\lambda } \cdot \pi = n\pi[/itex]
[itex]\lambda =n ^{2}; n=1,2,3,..[/itex]
[itex]{T}'+n^{2}T=0[/itex]
(I'm not entirely sure how t got into this next step, I know T is actually T(t) but still not sure what it's doing in the next step with the exponential)
[itex]T(t)=C_{3}\exp (-n ^{2}t)[/itex]
[itex]u_{n}(x,t)=C_{2}C_{3} \exp(-n^{2}t)\sin(nx)=A_{n} \exp(-n^{2}t)\sin(nx)); n = 1, 2, 3, ..[/itex]
[tex]u(x, t) = \sum_{n=1}^{\infty} A_{n} \exp(-n^{2}t)\sin(nx))[/tex]
[tex]u(x,0)=f(x)[/tex]
[tex]A_{n}=\frac{2}{\pi}\left [\int_{0}^{\pi/2}x\sin(nx)dx + \int_{\pi/2}^{\pi}(\pi-x)\sin(nx)dx \right ][/tex]
[tex]=\frac{2}{\pi}\left [\left [ \frac{x(-\cos(nx))}{n} \right ]_{0}^{\pi/2} + \int_{0}^{\pi/2} \frac{\cos(nx)}{n}dx + \left [\frac{(\pi-x)(-\cos(nx))}{n} \right ] _{\pi/2}^{\pi} - \int_{\pi}^{\pi/2}\frac{\cos(nx)}{n}dx \right ][/tex]
[tex]=\frac{-1}{n}\cos (\frac{n \pi}{2}) + \frac{2}{\pi n^{2}}\sin(\frac{n \pi}{2}) + \frac{1}{n}\cos (\frac{n \pi}{2}) + \frac{2}{\pi n^{2}}\sin(\frac{n \pi}{2})[/tex]
[tex]=\frac{4}{\pi n^{2}}\sin(\frac{n \pi}{2})[/tex]
[tex]u(x, t) = \sum_{n=1}^{\infty} \frac{4}{\pi n^{2}}\sin(\frac{n \pi}{2}) \exp(-n^{2}t)\sin(nx))[/tex]
Ok so far so good. An was found and put back into the general form of the solution.
Here's where it gets tricky
Apparently there is further simplification of this, now these next steps were provided to me by my instructor (as were the previous too) during lecture and I cannot make sense of them. I see what is supposed to happen, cos and sin follow a pattern on intervals of π or nπ where they evaluate to predictable values like 1,0,-1 etc. and we can ignore some values because they lead to all 0s or the trivial solution.
Apparently he continues by the following steps
[tex]
u(x, t) = \sum_{k=0}^{\infty} \frac{4}{\pi (4k+1)^{2}}\exp(-(4k+1)^{2}t)\sin((4k+1)x)) -\sum_{k=0}^{\infty} \frac{4}{\pi (4k+3)^{2}}\exp(-(4k+3)^{2}t)\sin((4k+3)x))[/tex]
Separated into two series and replaced n with 4k+1 and 4k+3, eliminated the first trig function with some version of previously stated pattern recognition and started the series at 0 instead of 1
Now I can't even verify if this is equivalent to the other series let alone figure out the pattern or logic to this to figure it out for other scenarios.
And someone stated that it can be reduced further into
[tex]
u(x, t) = \sum_{k=0}^{\infty} (-1)^{k}\exp(-(2k+1)^{2}t)\sin((2k+1)x))
[/tex]
If someone can please break it down and explain how this is done I will be very grateful as the exam is soon and I cannot seem to figure out how to follow the process to come up with these series reductions
Thank you!
Also it would be nice to know how t ends up in this step[itex]{T}'+n^{2}T=0[/itex]
[itex]T(t)=C_{3}\exp (-n ^{2}t)[/itex]
but not nearly as important.
Thanks again
Given
[itex]u_{t}=u_{xx}[/itex]
[itex]u(0, t) = u(\pi ,t)=0[/itex]
[itex]u(x, 0) = f(x)[/itex]
[tex]f(x)=\left\{\begin{matrix}
x; 0 < x < \frac{\pi}{2}\\
\pi-x; \frac{\pi}{2} < x < \pi
\end{matrix}\right.[/tex]
L is π - 0=π
λ = α2 since 0 and -α lead to trivial solutions
Let
[itex]u = XT[/itex]
[itex]X{T}'={X}''T[/itex]
[itex]\frac{{X}''}{X}=\frac{{T}'}{T}=-\lambda [/itex]
[itex]{X}''+\lambda =0[/itex]
[itex]X(0)=X(\pi)=0[/itex]
[itex]{T}'+\lambda T=0[/itex]
[itex]X(x)=C_{1}\cos (\sqrt{\lambda }\cdot x)+C_{2}\sin(\sqrt{\lambda }\cdot x)[/itex]
[itex]\sqrt{\lambda } \cdot \pi = n\pi[/itex]
[itex]\lambda =n ^{2}; n=1,2,3,..[/itex]
[itex]{T}'+n^{2}T=0[/itex]
(I'm not entirely sure how t got into this next step, I know T is actually T(t) but still not sure what it's doing in the next step with the exponential)
[itex]T(t)=C_{3}\exp (-n ^{2}t)[/itex]
[itex]u_{n}(x,t)=C_{2}C_{3} \exp(-n^{2}t)\sin(nx)=A_{n} \exp(-n^{2}t)\sin(nx)); n = 1, 2, 3, ..[/itex]
[tex]u(x, t) = \sum_{n=1}^{\infty} A_{n} \exp(-n^{2}t)\sin(nx))[/tex]
[tex]u(x,0)=f(x)[/tex]
[tex]A_{n}=\frac{2}{\pi}\left [\int_{0}^{\pi/2}x\sin(nx)dx + \int_{\pi/2}^{\pi}(\pi-x)\sin(nx)dx \right ][/tex]
[tex]=\frac{2}{\pi}\left [\left [ \frac{x(-\cos(nx))}{n} \right ]_{0}^{\pi/2} + \int_{0}^{\pi/2} \frac{\cos(nx)}{n}dx + \left [\frac{(\pi-x)(-\cos(nx))}{n} \right ] _{\pi/2}^{\pi} - \int_{\pi}^{\pi/2}\frac{\cos(nx)}{n}dx \right ][/tex]
[tex]=\frac{-1}{n}\cos (\frac{n \pi}{2}) + \frac{2}{\pi n^{2}}\sin(\frac{n \pi}{2}) + \frac{1}{n}\cos (\frac{n \pi}{2}) + \frac{2}{\pi n^{2}}\sin(\frac{n \pi}{2})[/tex]
[tex]=\frac{4}{\pi n^{2}}\sin(\frac{n \pi}{2})[/tex]
[tex]u(x, t) = \sum_{n=1}^{\infty} \frac{4}{\pi n^{2}}\sin(\frac{n \pi}{2}) \exp(-n^{2}t)\sin(nx))[/tex]
Ok so far so good. An was found and put back into the general form of the solution.
Here's where it gets tricky
Apparently there is further simplification of this, now these next steps were provided to me by my instructor (as were the previous too) during lecture and I cannot make sense of them. I see what is supposed to happen, cos and sin follow a pattern on intervals of π or nπ where they evaluate to predictable values like 1,0,-1 etc. and we can ignore some values because they lead to all 0s or the trivial solution.
Apparently he continues by the following steps
[tex]
u(x, t) = \sum_{k=0}^{\infty} \frac{4}{\pi (4k+1)^{2}}\exp(-(4k+1)^{2}t)\sin((4k+1)x)) -\sum_{k=0}^{\infty} \frac{4}{\pi (4k+3)^{2}}\exp(-(4k+3)^{2}t)\sin((4k+3)x))[/tex]
Separated into two series and replaced n with 4k+1 and 4k+3, eliminated the first trig function with some version of previously stated pattern recognition and started the series at 0 instead of 1
Now I can't even verify if this is equivalent to the other series let alone figure out the pattern or logic to this to figure it out for other scenarios.
And someone stated that it can be reduced further into
[tex]
u(x, t) = \sum_{k=0}^{\infty} (-1)^{k}\exp(-(2k+1)^{2}t)\sin((2k+1)x))
[/tex]
If someone can please break it down and explain how this is done I will be very grateful as the exam is soon and I cannot seem to figure out how to follow the process to come up with these series reductions
Thank you!
Also it would be nice to know how t ends up in this step[itex]{T}'+n^{2}T=0[/itex]
[itex]T(t)=C_{3}\exp (-n ^{2}t)[/itex]
but not nearly as important.
Thanks again
Last edited: