Mixed boundary condition for the heat equation












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$begingroup$


Would someone help me understand the way the solution obtained in this question:
Heat Equation Mixed Boundaries Case: Fourier Coefficients



I did not understand why in the final solution, he took $b_m$ and not $b_{2m-1}$. I am confused not just because of the rank of the coefficients, this is because according to this rank $$b_m= frac{1}{L} int_{0}^{2L} f(x) sinBig(frac{m pi x}{2L}Big)$$ or
$$b_{2m-1}= frac{1}{L} int_{0}^{2L} f(x) sinBig(frac{(2m-1) pi x}{2L}Big)$$



Sorry if I appear much confused, but I really want to understand it.
Thanks in adavnce.










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  • 1




    $begingroup$
    Use $sin$ for $sin$ (instead of $sin$ for $sin$).
    $endgroup$
    – Shaun
    Dec 8 '18 at 13:22










  • $begingroup$
    Both formulas are equivalent. The first one restricts the subscript $m$ to odd integers; the second one doesn't.
    $endgroup$
    – Dylan
    Dec 13 '18 at 4:28
















0












$begingroup$


Would someone help me understand the way the solution obtained in this question:
Heat Equation Mixed Boundaries Case: Fourier Coefficients



I did not understand why in the final solution, he took $b_m$ and not $b_{2m-1}$. I am confused not just because of the rank of the coefficients, this is because according to this rank $$b_m= frac{1}{L} int_{0}^{2L} f(x) sinBig(frac{m pi x}{2L}Big)$$ or
$$b_{2m-1}= frac{1}{L} int_{0}^{2L} f(x) sinBig(frac{(2m-1) pi x}{2L}Big)$$



Sorry if I appear much confused, but I really want to understand it.
Thanks in adavnce.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Use $sin$ for $sin$ (instead of $sin$ for $sin$).
    $endgroup$
    – Shaun
    Dec 8 '18 at 13:22










  • $begingroup$
    Both formulas are equivalent. The first one restricts the subscript $m$ to odd integers; the second one doesn't.
    $endgroup$
    – Dylan
    Dec 13 '18 at 4:28














0












0








0





$begingroup$


Would someone help me understand the way the solution obtained in this question:
Heat Equation Mixed Boundaries Case: Fourier Coefficients



I did not understand why in the final solution, he took $b_m$ and not $b_{2m-1}$. I am confused not just because of the rank of the coefficients, this is because according to this rank $$b_m= frac{1}{L} int_{0}^{2L} f(x) sinBig(frac{m pi x}{2L}Big)$$ or
$$b_{2m-1}= frac{1}{L} int_{0}^{2L} f(x) sinBig(frac{(2m-1) pi x}{2L}Big)$$



Sorry if I appear much confused, but I really want to understand it.
Thanks in adavnce.










share|cite|improve this question











$endgroup$




Would someone help me understand the way the solution obtained in this question:
Heat Equation Mixed Boundaries Case: Fourier Coefficients



I did not understand why in the final solution, he took $b_m$ and not $b_{2m-1}$. I am confused not just because of the rank of the coefficients, this is because according to this rank $$b_m= frac{1}{L} int_{0}^{2L} f(x) sinBig(frac{m pi x}{2L}Big)$$ or
$$b_{2m-1}= frac{1}{L} int_{0}^{2L} f(x) sinBig(frac{(2m-1) pi x}{2L}Big)$$



Sorry if I appear much confused, but I really want to understand it.
Thanks in adavnce.







pde boundary-value-problem heat-equation linear-pde






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 14:11







Nizar

















asked Dec 8 '18 at 13:10









NizarNizar

2,39921023




2,39921023








  • 1




    $begingroup$
    Use $sin$ for $sin$ (instead of $sin$ for $sin$).
    $endgroup$
    – Shaun
    Dec 8 '18 at 13:22










  • $begingroup$
    Both formulas are equivalent. The first one restricts the subscript $m$ to odd integers; the second one doesn't.
    $endgroup$
    – Dylan
    Dec 13 '18 at 4:28














  • 1




    $begingroup$
    Use $sin$ for $sin$ (instead of $sin$ for $sin$).
    $endgroup$
    – Shaun
    Dec 8 '18 at 13:22










  • $begingroup$
    Both formulas are equivalent. The first one restricts the subscript $m$ to odd integers; the second one doesn't.
    $endgroup$
    – Dylan
    Dec 13 '18 at 4:28








1




1




$begingroup$
Use $sin$ for $sin$ (instead of $sin$ for $sin$).
$endgroup$
– Shaun
Dec 8 '18 at 13:22




$begingroup$
Use $sin$ for $sin$ (instead of $sin$ for $sin$).
$endgroup$
– Shaun
Dec 8 '18 at 13:22












$begingroup$
Both formulas are equivalent. The first one restricts the subscript $m$ to odd integers; the second one doesn't.
$endgroup$
– Dylan
Dec 13 '18 at 4:28




$begingroup$
Both formulas are equivalent. The first one restricts the subscript $m$ to odd integers; the second one doesn't.
$endgroup$
– Dylan
Dec 13 '18 at 4:28










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