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# 10.7: Our Initial Problem and Bessel Functions

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We started the discussion from the problem of the temperature on a circular disk, solved in polar coordinates, Since the initial conditions do not depend on $$\phi$$, we expect the solution to be radially symmetric as well, $$u(\rho,t)$$, which satisfies the equation

\begin{aligned} \frac{\partial u}{\partial t} &= k\left[ \frac{\partial^2 u}{\partial \rho^2} + \frac{1}{\rho} \frac{\partial u}{\partial \rho} \right],\nonumber\\ u(c,t) &=0,\nonumber\\ u(\rho,0) &= f(\rho).\end{aligned}

With $$u(\rho,t)=R(\rho)T(t)$$ we found the equations

\begin{aligned} \rho^2 R''+\rho R' + \lambda \rho^2 R &= 0\;\;\;\;R(c)=0\nonumber\\ T'+\lambda k T = 0.\end{aligned}

The equation for $$R$$ is clearly self-adjoint, it can be written as

$[\rho R']' + \lambda \rho R = 0$

So how does the equation for $$R$$ relate to Bessel’s equation? Let us make the change of variables $$x= \sqrt{\lambda} \rho$$. We find

$\frac{d}{d\rho} = \sqrt{\lambda} \frac{d}{dx},$

and we can remove a common factor $$\sqrt{\lambda}$$ to obtain ($$X(x)=R(\rho)$$)

$[x X']' + x X = 0,$

which is Bessel’s equation of order $$0$$, i.e.,

$R(\rho) = J_0(\rho \sqrt{\lambda}).$

The boundary condition $$R(c)=0$$ shows that $c \sqrt{\lambda_n} = x_n,$

where $$x_n$$ are the points where $$J_0(x)=0$$. We thus conclude $R_n(\rho) = J_0(\rho \sqrt{\lambda_n}).$ the first five solutions $$R_n$$ (for $$c=1$$) are shown in Fig. $$\PageIndex{1}$$. Figure $$\PageIndex{1}$$: A graph of the first five functions $$R_n$$

From Sturm-Liouville theory we conclude that

$\int_0^\infty \rho d\rho \,R_n(\rho)R_m(\rho)=0\;\;{\rm if\ }n \neq m.$

Together with the solution for the $$T$$ equation,

$T_n(t)= \exp(-\lambda_n k t)$

we find a Fourier-Bessel series type solution

$u(\rho,t) = \sum_{n=1}^\infty A_n J_0(\rho\sqrt{\lambda_n})\exp(-\lambda_n k t),$

with $$\lambda_n= (x_n/c)^2$$.

In order to understand how to determine the coefficients $$A_n$$ from the initial condition $$u(\rho,0)=f(\rho)$$ we need to study Fourier-Bessel series in a little more detail.