• <blockquote id="a6oys"></blockquote>
  • <tt id="a6oys"></tt>
  • Paul's Online Notes
    Paul's Online Notes
    Home / Differential Equations / Partial Differential Equations / Vibrating String
    Show General Notice Show Mobile Notice Show All Notes Hide All Notes
    General Notice

    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

    Mobile Notice
    You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

    Section 9-8 : Vibrating String

    This will be the final partial differential equation that we’ll be solving in this chapter. In this section we’ll be solving the 1-D wave equation to determine the displacement of a vibrating string. There really isn’t much in the way of introduction to do here so let’s just jump straight into the example.

    Example 1 Find a solution to the following partial differential equation.

    \[\begin{align*}& \frac{{{\partial ^2}u}}{{\partial {t^2}}} = {c^2}\frac{{{\partial ^2}u}}{{\partial {x^2}}}\\ & u\left( {x,0} \right) = f\left( x \right)\hspace{0.25in}\frac{{\partial u}}{{\partial t}}\left( {x,0} \right) = g\left( x \right)\\ & u\left( {0,t} \right) = 0\hspace{0.25in}u\left( {L,t} \right) = 0\end{align*}\]
    Show Solution

    One of the main differences here that we’re going to have to deal with is the fact that we’ve now got two initial conditions. That is not something we’ve seen to this point but will not be all that difficult to deal with when the time rolls around.

    We’ve already done the separation of variables for this problem, but let’s go ahead and redo it here so we can say we’ve got another problem almost completely worked out.

    So, let’s start off with the product solution.

    \[u\left( {x,t} \right) = \varphi \left( x \right)h\left( t \right)\]

    Plugging this into the two boundary conditions gives,

    \[\varphi \left( 0 \right) = 0\hspace{0.25in}\varphi \left( L \right) = 0\]

    Plugging the product solution into the differential equation, separating and introducing a separation constant gives,

    \[\begin{align*}\frac{{{\partial ^2}}}{{\partial {t^2}}}\left( {\varphi \left( x \right)h\left( t \right)} \right) & = {c^2}\frac{{{\partial ^2}}}{{\partial {x^2}}}\left( {\varphi \left( x \right)h\left( t \right)} \right)\\ \varphi \left( x \right)\frac{{{d^2}h}}{{d{t^2}}} & = {c^2}h\left( t \right)\frac{{{d^2}\varphi }}{{d{x^2}}}\\ \frac{1}{{{c^2}h}}\frac{{{d^2}h}}{{d{t^2}}} & = \frac{1}{\varphi }\frac{{{d^2}\varphi }}{{d{x^2}}} = - \lambda \end{align*}\]

    We moved the \({c^2}\) to the left side for convenience and chose \( - \lambda \) for the separation constant so the differential equation for \(\varphi \) would match a known (and solved) case.

    The two ordinary differential equations we get from separation of variables are then,

    \[\begin{align*}\frac{{{d^2}h}}{{d{t^2}}} + {c^2}\lambda h = 0\hspace{0.25in} & \frac{{{d^2}\varphi }}{{d{x^2}}} + \lambda \varphi=0 \\ & \varphi \left( 0 \right) = 0\hspace{0.25in}\varphi \left( L \right) = 0\end{align*}\]

    We solved the boundary value problem above in Example 1 of the Solving the Heat Equation section of this chapter and so the eigenvalues and eigenfunctions for this problem are,

    \[{\lambda _{\,n}} = {\left( {\frac{{n\pi }}{L}} \right)^2}\hspace{0.25in}{\varphi _n}\left( x \right) = \sin \left( {\frac{{n\,\pi \,x}}{L}} \right)\hspace{0.25in}n = 1,2,3, \ldots \]

    The first ordinary differential equation is now,

    \[\frac{{{d^2}h}}{{d{t^2}}} + {\left( {\frac{{n\pi c}}{L}} \right)^2}h = 0\]

    and because the coefficient of the \(h\) is clearly positive the solution to this is,

    \[h\left( t \right) = {c_1}\cos \left( {\frac{{n\pi c\,t}}{L}} \right) + {c_2}\sin \left( {\frac{{n\pi c\,t}}{L}} \right)\]

    Because there is no reason to think that either of the coefficients above are zero we then get two product solutions,

    \[\begin{array}{*{20}{c}}{{u_n}\left( {x,t} \right) = {A_n}\cos \left( {\frac{{n\pi c\,t}}{L}} \right)\sin \left( {\frac{{n\,\pi \,x}}{L}} \right)}\\{{u_n}\left( {x,t} \right) = {B_n}\sin \left( {\frac{{n\pi c\,t}}{L}} \right)\sin \left( {\frac{{n\,\pi \,x}}{L}} \right)}\end{array}\hspace{0.25in}n = 1,2,3, \ldots \]

    The solution is then,

    \[u\left( {x,t} \right) = \sum\limits_{n = 1}^\infty {\left[ {{A_n}\cos \left( {\frac{{n\pi c\,t}}{L}} \right)\sin \left( {\frac{{n\,\pi \,x}}{L}} \right) + {B_n}\sin \left( {\frac{{n\pi c\,t}}{L}} \right)\sin \left( {\frac{{n\,\pi \,x}}{L}} \right)} \right]} \]

    Now, in order to apply the second initial condition we’ll need to differentiate this with respect to \(t\) so,

    \[\frac{{\partial u}}{{\partial t}} = \sum\limits_{n = 1}^\infty {\left[ { - \frac{{n\pi c}}{L}{A_n}\sin \left( {\frac{{n\pi c\,t}}{L}} \right)\sin \left( {\frac{{n\,\pi \,x}}{L}} \right) + \frac{{n\pi c}}{L}{B_n}\cos \left( {\frac{{n\pi c\,t}}{L}} \right)\sin \left( {\frac{{n\,\pi \,x}}{L}} \right)} \right]} \]

    If we now apply the initial conditions we get,

    \[\begin{align*}& u\left( {x,0} \right) = f\left( x \right) = \sum\limits_{n = 1}^\infty {\left[ {{A_n}\cos \left( 0 \right)\sin \left( {\frac{{n\,\pi \,x}}{L}} \right) + {B_n}\sin \left( 0 \right)\sin \left( {\frac{{n\,\pi \,x}}{L}} \right)} \right]} = \sum\limits_{n = 1}^\infty {{A_n}\sin \left( {\frac{{n\,\pi \,x}}{L}} \right)} \\ & \frac{{\partial u}}{{\partial t}}\left( {x,0} \right) = g\left( x \right) = \sum\limits_{n = 1}^\infty {\frac{{n\pi c}}{L}{B_n}\sin \left( {\frac{{n\,\pi \,x}}{L}} \right)} \end{align*}\]

    Both of these are Fourier sine series. The first is for \(f\left( x \right)\) on \(0 \le x \le L\) while the second is for \(g\left( x \right)\) on \(0 \le x \le L\) with a slightly messy coefficient. As in the last few sections we’re faced with the choice of either using the orthogonality of the sines to derive formulas for \({A_n}\) and \({B_n}\) or we could reuse formula from previous work.

    It’s easier to reuse formulas so using the formulas form the Fourier sine series section we get,

    \[\begin{align*}{A_{\,n}} & = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\,\,\,\,\,\,\,n = 1,2,3, \ldots \\ \frac{{n\pi c}}{L}{B_{\,n}} & = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{g\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\,\,\,\,\,\,\,n = 1,2,3, \ldots \end{align*}\]

    Upon solving the second one we get,

    \[\begin{align*}{A_{\,n}} & = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\,\,\,\,\,\,\,n = 1,2,3, \ldots \\ {B_{\,n}} & = \frac{2}{{n\pi c}}\int_{{\,0}}^{{\,L}}{{g\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\,\,\,\,\,\,\,n = 1,2,3, \ldots \end{align*}\]

    So, there is the solution to the 1-D wave equation and with that we’ve solved the final partial differential equation in this chapter.

    亚洲欧美中文日韩视频 - 视频 - 在线观看 - 影视资讯 - av网 <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <文本链> <文本链> <文本链> <文本链> <文本链> <文本链>