Help solving this Heat Equation please

[Karl Karlsson]

Homework Statement

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Relevant Equations

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I want to solve the heat equation below:

I don't understand where the expression for $2/R\cdot\int_0^R q\cdot sin(k_nr)\cdot r \, dr$ came from. The r dependent function is calculated as $sin(k_nr)/r$ not $sin(k_nr)\cdot r$. I don't even know if $sin(k_nr)/r$ are orthogonal for different $k_n$ values. Why is $q_n = 2/R\cdot\int_0^R q\cdot sin(k_nr)\cdot r \, dr$ ?

 

[Paul Colby]

I think you're expanding a step function $q(r) = q$ for $r = [0,R]$ and $q(r)=0$ for $r > R$ in terms of the $\sin(k_nr)/r$ functions. You're using separation of variables. The functions $\sin(k_nr)/r$ are the spherical Bessels.

 

[pasmith]

As is indicated by the $\equiv$ sign, this is a definition of $q_n$. From the line immediately above, we can infer that $$q(r) = \sum_{m=1}^\infty \frac{q_m}{r} \sin(k_mr).$$ Now multply by $r \sin(k_n r)$ and integrate between 0 and R: $$\int_0^R q(r) \sin (k_nr) r\,dr = \sum_{m=1}^\infty q_m \int_0^R \sin(k_mr) \sin(k_nr)\,dr.$$ You should recognise that the integral on the right is zero unless $k_n = \frac{n\pi}{R} = \frac{m\pi}{R} = k_m$ (ie. $n = m$), in which case it is $R/2$ as you are integrating $\sin^2 (k_n r) = \frac12(1 - \cos (2k_n r))$ over a whole number of periods.

 

[Karl Karlsson]

As is indicated by the $\equiv$ sign, this is a definition of $q_n$. From the line immediately above, we can infer that $$q(r) = \sum_{m=1}^\infty \frac{q_m}{r} \sin(k_mr).$$ Now multply by $r \sin(k_n r)$ and integrate between 0 and R: $$\int_0^R q(r) \sin k_nr\,dr = \sum_{m=1}^\infty q_m \int_0^R \sin(k_mr) \sin(k_nr)\,dr.$$ You should recognise that the integral on the right is zero unless $k_n = \frac{n\pi}{R} = \frac{m\pi}{R} = k_m$ (ie. $n = m$), in which case it is $R/2$ as you are integrating $\sin^2 (k_n r) = \frac12(1 - \cos (2k_n r))$ over a whole number of periods.

Thanks!