#### 4.5.1 Spherical coordinates

4.5.1.1 [332] Chain reaction PDE

##### 4.5.1.1 [332] Chain reaction PDE

problem number 332

Assume $$\phi$$ independence. Solve for $$u(r,\theta ,t)$$

\begin {align*} u_t = k ( \lambda u + \nabla ^2(u) ) \end {align*}

Where $$\nabla ^2(u)= \frac {1}{r^2} \frac {\partial }{\partial r}( r^2 u_r)+ \frac {1}{r^2 \sin \theta } \frac {\partial }{\partial \theta }(\sin \theta u_\theta )$$ with $$k>0$$.

Boundary conditions $$u(R,\theta ,t)=0$$.

Mathematica

Failed

Maple

$u(r,theta,t) = 0$ trivial solution

Hand solution

Solve for $$u\left ( r,\theta ,t\right )$$ in spherical coordinates (assuming $$\phi$$ independence) the chain reaction equation $$\frac {1}{k}u_{t}=\lambda u+\nabla ^{2}u$$ with boundary conditions $$u\left ( R,\theta ,t\right ) =0$$.\begin {align*} \frac {1}{k}u_{t} & =\lambda u+\nabla ^{2}u\\ & =\lambda u+\left ( \frac {1}{r^{2}}\frac {\partial }{\partial r}\left ( r^{2}u_{r}\right ) +\frac {1}{r^{2}\sin \theta }\frac {\partial }{\partial \theta }\left ( \sin \theta u_{\theta }\right ) \right ) \\ & =\lambda u+\frac {1}{r^{2}}\left ( 2ru_{r}+r^{2}u_{rr}\right ) +\frac {1}{r^{2}\sin \theta }\frac {\partial }{\partial \theta }\left ( \sin \theta u_{\theta }\right ) \end {align*}

Let $$u=R\left ( r\right ) \Theta \left ( \theta \right ) T\left ( t\right )$$. Substituting into the above gives\begin {align*} \frac {1}{k}T^{\prime }R\Theta & =\lambda TR\Theta +\frac {1}{r^{2}}\left ( 2rR^{\prime }\Theta T+r^{2}R^{\prime \prime }\Theta T\right ) +\frac {1}{r^{2}\sin \theta }\frac {\partial }{\partial \theta }\left ( \sin \theta \left ( \Theta ^{\prime }RT\right ) \right ) \\ \frac {1}{k}T^{\prime }R\Theta & =\lambda TR\Theta +\frac {2}{r}R^{\prime }\Theta T+R^{\prime \prime }\Theta T+\frac {RT}{r^{2}\sin \theta }\frac {\partial }{\partial \theta }\left ( \Theta ^{\prime }\sin \theta \right ) \end {align*}

Dividing by $$TR\Theta$$ gives$\frac {1}{k}\frac {T^{\prime }}{T}=\lambda +\frac {2}{r}\frac {R^{\prime }}{R}+\frac {R^{\prime \prime }}{R}+\frac {1}{\Theta r^{2}\sin \theta }\frac {\partial }{\partial \theta }\left ( \Theta ^{\prime }\sin \theta \right )$ The left side depends on $$t$$ only and the right depends on $$r,\theta$$ only. Let the separation variable be $$-n$$. This gives the following 2 equations\begin {align} \frac {1}{k}\frac {T^{\prime }}{T} & =-n\tag {1}\\ \lambda +\frac {2}{r}\frac {R^{\prime }}{R}+\frac {R^{\prime \prime }}{R}+\frac {1}{\Theta r^{2}\sin \theta }\frac {\partial }{\partial \theta }\left ( \Theta ^{\prime }\sin \theta \right ) & =-n\tag {2} \end {align}

Now we consider (2). Multiplying both sides of (2) by $$r^{2}$$ gives\begin {align*} \lambda r^{2}+2r\frac {R^{\prime }}{R}+r^{2}\frac {R^{\prime \prime }}{R}+\frac {1}{\Theta \sin \theta }\frac {\partial }{\partial \theta }\left ( \Theta ^{\prime }\sin \theta \right ) & =-nr^{2}\\ 2r\frac {R^{\prime }}{R}+r^{2}\frac {R^{\prime \prime }}{R}+r^{2}\left ( \lambda +n\right ) & =-\frac {1}{\Theta \sin \theta }\frac {\partial }{\partial \theta }\left ( \Theta ^{\prime }\sin \theta \right ) \end {align*}

The left side depends on $$r$$ and the right side depends on $$\theta$$. Let the separation variable be $$l\left ( l+1\right )$$ where $$l$$ is integer. Hence we obtain the following two equations\begin {align} -\frac {1}{\Theta \sin \theta }\frac {\partial }{\partial \theta }\left ( \Theta ^{\prime }\sin \theta \right ) & =l\left ( l+1\right ) \tag {4}\\ 2r\frac {R^{\prime }}{R}+r^{2}\frac {R^{\prime \prime }}{R}+r^{2}\left ( \lambda +n\right ) & =l\left ( l+1\right ) \tag {5} \end {align}

Starting with (4)\begin {align*} \frac {\partial }{\partial \theta }\left ( \Theta ^{\prime }\sin \theta \right ) +l\left ( l+1\right ) \Theta \sin \theta & =0\\ \Theta ^{\prime \prime }\sin \theta +\Theta ^{\prime }\cos \theta +l\left ( l+1\right ) \Theta \sin \theta & =0 \end {align*}

Using the substitution $$z=\cos \theta$$ the above becomes$\left ( 1-z^{2}\right ) \Theta ^{\prime \prime }-2z\Theta ^{\prime }+l\left ( l+1\right ) \Theta =0$ This Legendre ODE. Solution is $$P_{l}\left ( \theta \right )$$. The other solution to the above ODE is ignored as not bounded. Now back to solving (5). Writing it as\begin {align} 2rR^{\prime }+r^{2}R^{\prime \prime }+r^{2}\left ( \lambda +n\right ) R & =l\left ( l+1\right ) R\nonumber \\ r^{2}R^{\prime \prime }+2rR^{\prime }+\left ( r^{2}\left ( \lambda +n\right ) -l\left ( l+1\right ) \right ) R & =0\tag {6} \end {align}

This can be converted to Bessel ODE using substitution. First let $$v=r\sqrt {\left ( \lambda +n\right ) }$$. Then $$R^{\prime }\left ( r\right ) =\sqrt {\left ( \lambda +n\right ) }R^{\prime }\left ( v\right ) ,R^{\prime \prime }\left ( r\right ) =\left ( \lambda +n\right ) R^{\prime \prime }\left ( v\right )$$ and (6) becomes\begin {align} \left ( \lambda +n\right ) r^{2}R^{\prime \prime }\left ( v\right ) +2r\sqrt {\left ( \lambda +n\right ) }R^{\prime }\left ( v\right ) +\left ( r^{2}\left ( \lambda +n\right ) -l\left ( l+1\right ) \right ) R & =0\nonumber \\ v^{2}R^{\prime \prime }\left ( v\right ) +2vR^{\prime }\left ( v\right ) +\left ( v^{2}-l\left ( l+1\right ) \right ) R & =0\tag {7} \end {align}

Now, we apply second transformation $$R\left ( v\right ) =\frac {Z\left ( v\right ) }{\sqrt {v}}$$ Then\begin {align*} R^{\prime }\left ( v\right ) & =\frac {Z^{\prime }\left ( v\right ) }{\sqrt {v}}-\frac {1}{2}Z\left ( v\right ) \frac {1}{v^{\frac {3}{2}}}\\ R^{\prime \prime }\left ( v\right ) & =\frac {Z^{\prime \prime }\left ( v\right ) }{\sqrt {v}}-\frac {1}{2}Z^{\prime }\left ( v\right ) \frac {1}{v^{\frac {3}{2}}}-\frac {1}{2}Z^{\prime }\left ( v\right ) \frac {1}{v^{\frac {3}{2}}}-\frac {1}{2}\left ( -\frac {3}{2}\right ) Z\left ( v\right ) \frac {1}{v^{\frac {5}{2}}}\\ & =\frac {Z^{\prime \prime }\left ( v\right ) }{\sqrt {v}}-Z^{\prime }\left ( v\right ) \frac {1}{r^{\frac {3}{2}}}+\frac {3}{4}Z\left ( v\right ) \frac {1}{v^{\frac {5}{2}}} \end {align*}

Hence (7) becomes$v^{2}\left ( \frac {Z^{\prime \prime }\left ( v\right ) }{\sqrt {v}}-Z^{\prime }\left ( v\right ) \frac {1}{r^{\frac {3}{2}}}+\frac {3}{4}Z\left ( v\right ) \frac {1}{v^{\frac {5}{2}}}\right ) +2v\left ( \frac {Z^{\prime }\left ( v\right ) }{\sqrt {v}}-\frac {1}{2}Z\left ( v\right ) \frac {1}{v^{\frac {3}{2}}}\right ) +\left ( v^{2}-l\left ( l+1\right ) \right ) \frac {Z\left ( v\right ) }{\sqrt {v}}=0$ Multiplying by $$\sqrt {v}$$ gives\begin {align*} v^{2}\left ( Z^{\prime \prime }\left ( v\right ) -Z^{\prime }\left ( v\right ) \frac {1}{v}+\frac {3}{4}Z\left ( v\right ) \frac {1}{v^{2}}\right ) +2v\left ( Z^{\prime }\left ( v\right ) -\frac {1}{2}Z\left ( v\right ) \frac {1}{v}\right ) +\left ( v^{2}-l\left ( l+1\right ) \right ) Z\left ( v\right ) & =0\\ \left ( v^{2}Z^{\prime \prime }\left ( v\right ) -vZ^{\prime }\left ( v\right ) +\frac {3}{4}Z\left ( v\right ) \right ) +\left ( 2vZ^{\prime }\left ( v\right ) -Z\left ( v\right ) \right ) +\left ( v^{2}-l\left ( l+1\right ) \right ) Z\left ( v\right ) & =0\\ v^{2}Z^{\prime \prime }\left ( v\right ) +vZ^{\prime }\left ( v\right ) +\frac {1}{4}Z\left ( v\right ) +\left ( v^{2}-l\left ( l+1\right ) \right ) Z\left ( v\right ) & =0\\ v^{2}Z^{\prime \prime }\left ( v\right ) +vZ^{\prime }\left ( v\right ) +\left ( v^{2}-l\left ( l+1\right ) +\frac {1}{4}\right ) Z\left ( v\right ) & =0\\ v^{2}Z^{\prime \prime }\left ( v\right ) +vZ^{\prime }\left ( v\right ) +\left ( v^{2}-l^{2}+l+\frac {1}{4}\right ) Z\left ( v\right ) & =0\\ v^{2}Z^{\prime \prime }\left ( v\right ) +vZ^{\prime }\left ( v\right ) +\left ( v^{2}-\left ( l+\frac {1}{2}\right ) ^{2}\right ) Z\left ( v\right ) & =0 \end {align*}

This is now in standard Bessel ODE form. Comparing it to $$v^{2}Z^{\prime \prime }\left ( v\right ) +vZ^{\prime }\left ( v\right ) +\left ( v^{2}-d^{2}\right ) Z\left ( v\right ) =0$$ shows the order is $$d=l+\frac {1}{2}$$. The solutions are $Z\left ( v\right ) =c_{1}J_{l+\frac {1}{2}}\left ( v\right ) +c_{2}Y_{l+\frac {1}{2}}\left ( v\right )$ But $$R\left ( v\right ) =\frac {Z\left ( v\right ) }{\sqrt {v}}$$, hence$R\left ( v\right ) =c_{1}\frac {J_{l+\frac {1}{2}}\left ( v\right ) }{\sqrt {v}}+c_{2}\frac {Y_{l+\frac {1}{2}}\left ( v\right ) }{\sqrt {v}}$ But $$v=r\sqrt {\lambda +n}$$ then above becomes$R\left ( r\right ) =c_{1}\frac {J_{l+\frac {1}{2}}\left ( r\sqrt {\left ( \lambda +n\right ) }\right ) }{\sqrt {r\sqrt {\left ( \lambda +n\right ) }}}+c_{2}\frac {Y_{l+\frac {1}{2}}\left ( r\sqrt {\left ( \lambda +n\right ) }\right ) }{\sqrt {r\sqrt {\left ( \lambda +n\right ) }}}$ The solution assumed bounded at $$r=0$$ hence $$c_{2}=0$$ and the above becomes$R\left ( r\right ) =c_{1}\frac {J_{l+\frac {1}{2}}\left ( r\sqrt {\left ( \lambda +n\right ) }\right ) }{\sqrt {r\sqrt {\left ( \lambda +n\right ) }}}$ Let $$m^{2}=\left ( \lambda +n\right )$$, hence\begin {align*} R\left ( r\right ) & =c_{1}\frac {J_{l+\frac {1}{2}}\left ( mr\right ) }{\sqrt {mr}}\\ & =j_{l}\left ( mr\right ) \end {align*}

where $$j_{l}\left ( mr\right )$$ are the spherical Bessel functions. Boundary conditions at $$r=R$$ gives $j_{l}\left ( mR\right ) =0$ Hence $$mR$$ or $$\sqrt {\lambda +n}R$$ are the zeros of spherical Bessel functions $$j_{l}\left ( mR\right )$$. There are inﬁnite zeros for each $$l$$. Let the $$v^{th}$$ zero of $$j_{l}$$ be called $$Z_{l,v}$$. Then $$mR_{l,v}=Z_{l,v}$$ or $$\sqrt {\lambda +n}_{l,v}=\frac {Z_{l,v}}{R}$$. or $n=\left ( \frac {Z_{l,v}}{R}\right ) ^{2}-\lambda$ The solution to the time ODE is therefore\begin {align*} T_{l,v} & =A_{l,v}e^{-nkt}\\ & =A_{l,v}e^{-\left ( \left ( \frac {Z_{l,v}}{R}\right ) ^{2}-\lambda \right ) kt} \end {align*}

Hence the complete solution is\begin {align*} u\left ( r,\theta ,t\right ) & =e^{-i\alpha kt}P_{l}\left ( \theta \right ) j_{l}\left ( mr\right ) \\ & =\sum _{l=1}^{\infty }\sum _{v=0}^{\infty }A_{l,v}e^{-nkt}P_{l}\left ( \theta \right ) j_{l}\left ( \frac {Z_{l,v}}{R}r\right ) \end {align*}

$$A_{l.v}$$ constants still need to be found from initial conditions. For each $$l$$, we have inﬁnite sum over all $$v^{\prime }s$$ zeros of $$j_{l}$$.