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Redo Exercise 4.4.3(b) by the eigenfunction expansion method
Chapter 4, Problem 4.4.4(choose chapter or problem)
Redo Exercise 4.4.3(b) by the eigenfunction expansion method.
Questions & Answers
QUESTION:
Redo Exercise 4.4.3(b) by the eigenfunction expansion method.
ANSWER:Step 1 of 4
To use the eigenfunction expansion method, first assume that the solution \(u\left( {x,t} \right)\), as well as its partial derivatives \(\frac{{\delta u}}{{\delta t}},\frac{{\delta u}}{{\delta x}},\frac{{{\delta ^2}u}}{{\delta {x^2}}}\) are continuous. The known eigenfunctions for the x-dependent boundary value problem are \(\sin \frac{{n\pi x}}{L}\), n = 1, 2, 3…. as discussed for the standard vibrating string with fixed ends. We expand the solution \(u\left( {x,t} \right)\) in terms of these eigenfunctions:
\(u\left( {x,t} \right) = \sum\limits_{n = 1}^\infty {{B_n}\left( t \right)\sin \frac{{n\pi x}}{L}}\)
Step 2 of 4
We will now plug into the differential equation \({\rho _0}\frac{{{\partial ^2}u}}{{\partial {t^2}}} = {T_0}\frac{{{\partial ^2}u}}{{\partial {x^2}}} - \beta \frac{{\delta u}}{{\delta t}}\) to determine the \({B_n}\left( t \right)\). First, we need to determine \(\frac{{\delta u}}{{\delta t}},\frac{{{\partial ^2}u}}{{\partial {t^2}}},\frac{{{\partial ^2}u}}{{\partial {x^2}}}\). We assume that \(\frac{{\delta u}}{{\delta t}},\frac{{{\partial ^2}u}}{{\partial {t^2}}}\) are piecewise smooth so that term-by-term differentiation in t is valid:
\(\frac{{\delta u}}{{\delta t}} = \sum\limits_{n = 1}^\infty {\frac{{d{B_n}\left( t \right)}}{{dt}}\sin \frac{{n\pi x}}{L}} \)
\(\frac{{{\delta ^2}u}}{{\delta {t^2}}} = \sum\limits_{n = 1}^\infty {\frac{{{d^2}{B_n}\left( t \right)}}{{d{t^2}}}\sin \frac{{n\pi x}}{L}} \)
The boundary conditions \(u\left( {0,t} \right) = 0,u\left( {L,t} \right) = 0\) allow us to differentiate term-by-term in x as well, assuming again that the derivatives are piecewise smooth:
\(\frac{{\delta u}}{{\delta x}} = \sum\limits_{n = 1}^\infty {\frac{{n\pi }}{L}{B_n}\left( t \right)\cos \frac{{n\pi x}}{L}} \)
\(\frac{{{\delta ^2}u}}{{\delta {x^2}}} = - \sum\limits_{n = 1}^\infty {{{\left( {\frac