5.2.2 Polar coordinates

5.2.2.1 [404] no $$\theta$$ dependency, ﬁxed boundary, general case

problem number 404

Circular disk. ﬁxed edge of disk, no $$\theta$$ dependency, with initial position and velocity given

Solve for $$u(r,t)$$ with $$0<r<a$$ and $$t>0$$. $u_{tt}= c^2 \left ( u_{rr} + \frac {1}{r} u_r \right )$ With boundary conditions \begin {align*} u(a,t) &=0 \end {align*}

With initial conditions \begin {align*} u(r,0) &=f(r) \\ \frac {\partial u}{\partial t}(r,0) &= g(r) \end {align*}

Mathematica

$\left \{\left \{u(r,t)\to \underset {n=1}{\overset {\infty }{\sum }}\fbox {\frac {2 J_0\left (\frac {r j_{0,n}}{a}\right ) \left (\sqrt {c^2} j_{0,n} \cos \left (\frac {\sqrt {c^2} t j_{0,n}}{a}\right ) \int _0^a r J_0\left (\frac {r j_{0,n}}{a}\right ) f(r) \, dr+a \left (\int _0^a r J_0\left (\frac {r j_{0,n}}{a}\right ) g(r) \, dr\right ) \sin \left (\frac {\sqrt {c^2} t j_{0,n}}{a}\right )\right )}{a^2 \sqrt {c^2} \left (J_0\left (j_{0,n}\right ){}^2+J_1\left (j_{0,n}\right ){}^2\right ) j_{0,n}}\text { if }j_{0,n}\geq 0}\right \}\right \}$

Maple

$u \left (r , t\right ) = \frac {-\mathcal {L}^{-1}\left (\BesselK \left (0, \frac {r s}{c}\right ) \left (\int \left (s f (a )+g (a )\right ) a \BesselI \left (0, \frac {a s}{c}\right )d a \right ), s , t\right )+\mathcal {L}^{-1}\left (\BesselK \left (0, \frac {r s}{c}\right ) \left (\int \left (s f (r )+g (r )\right ) r \BesselI \left (0, \frac {r s}{c}\right )d r \right ), s , t\right )-\mathcal {L}^{-1}\left (\BesselI \left (0, \frac {r s}{c}\right ) \left (\int \left (s f (r )+g (r )\right ) r \BesselK \left (0, \frac {r s}{c}\right )d r \right ), s , t\right )+\mathcal {L}^{-1}\left (\frac {\BesselI \left (0, \frac {a s}{c}\right ) \BesselK \left (0, \frac {r s}{c}\right ) \left (\int \left (s f (a )+g (a )\right ) a \BesselK \left (0, \frac {a s}{c}\right )d a \right )}{\BesselK \left (0, \frac {a s}{c}\right )}, s , t\right )}{c^{2}}$ Has unresolved Invlaplace calls

Hand solution

Assuming $$u=T\left ( t\right ) R\left ( r\right )$$. Substituting in the PDE gives$\frac {1}{c^{2}}T^{\prime \prime }R=R^{\prime \prime }T+\frac {1}{r}R^{\prime }T$ Dividing by $$RT$$$\frac {1}{c^{2}}\frac {T^{\prime \prime }}{T}=\frac {R^{\prime \prime }}{R}+\frac {1}{r}\frac {R^{\prime }}{R}$ Hence\begin {align*} \frac {1}{c^{2}}\frac {T^{\prime \prime }}{T} & =-\lambda \\ \frac {R^{\prime \prime }}{R}+\frac {1}{r}\frac {R^{\prime }}{R} & =-\lambda \end {align*}

The time ODE is$T^{\prime \prime }+c^{2}\lambda T=0$ And the $$r$$ ODE is (Sturm-Liouville)

$rR^{\prime \prime }+R^{\prime }+\lambda rR=0$ Where $$p=r,q=0,\sigma =r$$. This is singular SL.  The solution turns out to be  $R_{n}\left ( r\right ) =A_{n}J_{0}\left ( \sqrt {\lambda _{n}}r\right ) \qquad n=1,2,3,\cdots$ Where $$\lambda _{n}$$ is found from roots of $$0=J_{n}\left ( \sqrt {\lambda _{n}}a\right )$$ giving the eigenvalues. Now the time ODE is solved\begin {align*} T_{n}^{\prime \prime }+c^{2}\lambda _{n}T_{n} & =0\\ T_{n} & =B_{n}\cos \left ( c\sqrt {\lambda _{n}}t\right ) +C_{n}\sin \left ( c\sqrt {\lambda _{n}}t\right ) \qquad n=1,2,3,\ldots , \end {align*}

Hence the solution is\begin {align} u\left ( r,t\right ) & =\sum _{n=1}^{\infty }T_{n}R_{n}\nonumber \\ & =\sum _{n=1}^{\infty }A_{n}\cos \left ( c\sqrt {\lambda _{n}}t\right ) J_{0}\left ( \sqrt {\lambda _{n}}r\right ) +B_{n}\sin \left ( c\sqrt {\lambda _{n}}t\right ) J_{0}\left ( \sqrt {\lambda _{n}}r\right ) \tag {1} \end {align}

Now initial conditions $$u\left ( r,0\right ) =f\left ( r\right )$$ is used to ﬁnd $$A_{n}\,$$using orthogonality. At $$t=0$$ the solution simpliﬁes to $u\left ( r,0\right ) =\sum _{n=1}^{\infty }A_{n}J_{0}\left ( \sqrt {\lambda _{n}}r\right )$ Hence\begin {align*} f\left ( r\right ) & =\sum _{n=1}^{\infty }A_{n}J_{0}\left ( \sqrt {\lambda _{n}}r\right ) \\ \int _{0}^{a}f\left ( r\right ) J_{0}\left ( \sqrt {\lambda _{n}}r\right ) rdr & =A_{n}\int _{0}^{a}J_{0}^{2}\left ( \sqrt {\lambda _{n}}r\right ) rdr\\ A_{n} & =\frac {\int _{0}^{a}f\left ( r\right ) J_{0}\left ( \sqrt {\lambda _{n}}r\right ) rdr}{\int _{0}^{a}J_{0}^{2}\left ( \sqrt {\lambda _{n}}r\right ) rdr} \end {align*}

Now we will look at the second initial conditions $$\frac {\partial u}{\partial t}\left ( r,0\right ) =g\left ( r\right ) .$$ Taking derivative w.r.t. time $$t$$ of the solution in (1) gives$\frac {\partial u}{\partial t}\left ( r,t\right ) =\sum _{n=1}^{\infty }-c\sqrt {\lambda _{n}}A_{n}\sin \left ( c\sqrt {\lambda _{n}}t\right ) J_{0}\left ( \sqrt {\lambda _{n}}r\right ) +B_{n}c\sqrt {\lambda _{n}}\cos \left ( c\sqrt {\lambda _{n}}t\right ) J_{0}\left ( \sqrt {\lambda _{n}}r\right )$ At time $$t=0$$ the above becomes $g\left ( r\right ) =\sum _{n=1}^{\infty }B_{n}c\sqrt {\lambda _{n}}J_{0}\left ( \sqrt {\lambda _{n}}r\right )$ Now orthogonality is used. The above becomes$B_{n}=\frac {\int _{0}^{a}g\left ( r\right ) J_{0}\left ( \sqrt {\lambda _{n}}r\right ) rdr}{c\sqrt {\lambda _{n}}\int _{0}^{a}J_{0}^{2}\left ( \sqrt {\lambda _{n}}r\right ) rdr}$ Summary of solution$u\left ( r,t\right ) =\sum _{n=1}^{\infty }A_{n}\cos \left ( c\sqrt {\lambda _{n}}t\right ) J_{0}\left ( \sqrt {\lambda _{n}}r\right ) +B_{n}\sin \left ( c\sqrt {\lambda _{n}}t\right ) J_{0}\left ( \sqrt {\lambda _{n}}r\right )$$A_{n}=\frac {\int _{0}^{a}f\left ( r\right ) J_{0}\left ( \sqrt {\lambda _{n}}r\right ) rdr}{\int _{0}^{a}J_{0}^{2}\left ( \sqrt {\lambda _{n}}r\right ) rdr}$$B_{n}=\frac {\int _{0}^{a}g\left ( r\right ) J_{0}\left ( \sqrt {\lambda _{n}}r\right ) rdr}{c\sqrt {\lambda _{n}}\int _{0}^{a}J_{0}^{2}\left ( \sqrt {\lambda _{n}}r\right ) rdr}$ With $$\lambda _{n}$$ being the solutions for $$0=J_{0}\left ( \sqrt {\lambda _{n}}a\right )$$. We have inﬁnite number of zeros. This generates all the needed $$\lambda _{n}$$. Hence $$\sqrt {\lambda _{n}}a=BesselJZero\left ( 0,n\right )$$, therefore $$\sqrt {\lambda _{n}}=\frac {a}{BesselJZero\left ( 0,n\right ) }$$

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5.2.2.2 [405] no $$\theta$$ dependency. Speciﬁc example. Both initial conditions not zero

problem number 405

Taken from Mathematica helps pages on DSolve

In circular disk. ﬁxed edge of disk, no $$\theta$$ dependency, with initial position and velocity given

Solve for $$u(r,t)$$ with $$0<r<1$$ and $$t>0$$. $u_{tt}= c^2 \left ( u_{rr} + \frac {1}{r} u_r \right )$ With boundary conditions \begin {align*} u(1,t) &=0 \end {align*}

With initial conditions \begin {align*} u(r,0) &=1 \\ \frac {\partial u}{\partial t}(r,0) &= \frac {r}{3} \end {align*}

Mathematica

$\left \{\left \{u(r,t)\to \underset {n=1}{\overset {\infty }{\sum }}\frac {2 J_0\left (r j_{0,n}\right ) \left (9 \sqrt {c^2} J_1\left (j_{0,n}\right ) \cos \left (c t j_{0,n}\right )+\, _1F_2\left (\frac {3}{2};1,\frac {5}{2};-\frac {1}{4} \left (j_{0,n}\right ){}^2\right ) \sin \left (\sqrt {c^2} t j_{0,n}\right )\right )}{9 \sqrt {c^2} \left (J_0\left (j_{0,n}\right ){}^2+J_1\left (j_{0,n}\right ){}^2\right ) j_{0,n}}\right \}\right \}$

Maple

$u \left (r , t\right ) = \frac {r t}{3}+\frac {\pi \left (-\mathcal {L}^{-1}\left (\frac {\StruveL \left (0, \frac {r s}{c}\right )}{s^{3}}, s , t\right )+\mathcal {L}^{-1}\left (\frac {\BesselI \left (0, \frac {r s}{c}\right ) \StruveL \left (0, \frac {s}{c}\right )}{s^{3} \BesselI \left (0, \frac {s}{c}\right )}, s , t\right )\right ) c}{6}-\frac {\mathcal {L}^{-1}\left (\frac {\BesselI \left (0, \frac {r s}{c}\right )}{s^{2} \BesselI \left (0, \frac {s}{c}\right )}, s , t\right )}{3}-\mathcal {L}^{-1}\left (\frac {\BesselI \left (0, \frac {r s}{c}\right )}{s \BesselI \left (0, \frac {s}{c}\right )}, s , t\right )+1$ Has unresolved Invlaplace calls

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5.2.2.3 [406] no $$\theta$$ dependency. Speciﬁc example. Both initial conditions not zero

problem number 406

In circular disk. ﬁxed edge of disk, no $$\theta$$ dependency, with initial position and velocity given

Solve for $$u(r,t)$$ with $$0<r<a$$ and $$t>0$$. $u_{tt}= c^2 \left ( u_{rr} + \frac {1}{r} u_r \right )$ With boundary conditions \begin {align*} u(a,t) &=0 \end {align*}

With initial conditions \begin {align*} u(r,0) &=f(r) \\ \frac {\partial u}{\partial t}(r,0) &= g(r) \end {align*}

Using $$a=1,c=\frac {2}{10},g(r)=0,f(r)=r$$.

Mathematica

$\left \{\left \{u(r,t)\to \underset {n=1}{\overset {\infty }{\sum }}\frac {2 J_0\left (r j_{0,n}\right ) \cos \left (\frac {t j_{0,n}}{5}\right ) \, _1F_2\left (\frac {3}{2};1,\frac {5}{2};-\frac {1}{4} \left (j_{0,n}\right ){}^2\right )}{3 \left (J_0\left (j_{0,n}\right ){}^2+J_1\left (j_{0,n}\right ){}^2\right )}\right \}\right \}$

Maple

$u \left (r , t\right ) = r +\int _{0}^{r -1}\mathcal {L}^{-1}\left (\frac {\BesselI \left (0, 5 \left (r -\tau \right ) s \right ) \mathcal {L}\left (-\frac {1}{\tau +1}, t , s\right )}{\BesselI \left (0, 5 s \right )}, s , t\right )d \tau +\mathcal {L}^{-1}\left (\frac {\BesselI \left (0, 5 r s \right ) \mathcal {L}\left (-1, t , s\right )}{\BesselI \left (0, 5 s \right )}, s , t\right )$ Has unresolved Invlaplace calls. How to get series solution?

Hand solution

The basic solution for this type of PDE was already given in problem 5.2.2.1 on page 1886 as

$u\left ( r,t\right ) =\sum _{n=1}^{\infty }A_{n}\cos \left ( c\sqrt {\lambda _{n}}t\right ) J_{0}\left ( \sqrt {\lambda _{n}}r\right ) +B_{n}\sin \left ( c\sqrt {\lambda _{n}}t\right ) J_{0}\left ( \sqrt {\lambda _{n}}r\right )$$A_{n}=\frac {\int _{0}^{a}f\left ( r\right ) J_{0}\left ( \sqrt {\lambda _{n}}r\right ) rdr}{\int _{0}^{a}J_{0}^{2}\left ( \sqrt {\lambda _{n}}r\right ) rdr}$$B_{n}=\frac {\int _{0}^{a}g\left ( r\right ) J_{0}\left ( \sqrt {\lambda _{n}}r\right ) rdr}{c\sqrt {\lambda _{n}}\int _{0}^{a}J_{0}^{2}\left ( \sqrt {\lambda _{n}}r\right ) rdr}$ With $$\lambda _{n}$$ being the solutions for $$0=J_{0}\left ( \sqrt {\lambda _{n}}a\right )$$. We have inﬁnite number of zeros. This generates all the needed $$\lambda _{n}$$. Hence $$\sqrt {\lambda _{n}}a=BesselJZero\left ( 0,n\right )$$, therefore $$\sqrt {\lambda _{n}}=\frac {a}{BesselJZero\left ( 0,n\right ) }$$.

In this problem $$c=\frac {2}{10},a=1,g\left ( r\right ) =0$$ and $$f\left ( r\right ) =r$$, hence the solution becomes

$u\left ( r,t\right ) =\sum _{n=1}^{\infty }A_{n}\cos \left ( \frac {2}{10}\sqrt {\lambda _{n}}t\right ) J_{0}\left ( \sqrt {\lambda _{n}}r\right )$

Where $$\sqrt {\lambda _{n}}=\frac {1}{BesselJZero\left ( 0,n\right ) }$$.

This animation runs for 40 seconds.

Source code for all the above animation

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5.2.2.4 [407] no $$\theta$$ dependency. Using integral transforms. Source present. Speciﬁc example

problem number 407

Solve ${\frac {\partial ^{2}}{\partial {r}^{2}}}u \left ( r,t \right ) +{\frac {{\frac {\partial }{\partial r}}u \left ( r,t \right ) }{r}}+{\frac { \partial ^{2}}{\partial {t}^{2}}}u \left ( r,t \right ) =-Q_{0}\,q \left ( r \right )$ With initial conditions \begin {align*} u(r, 0) &= 0 \end {align*}

Mathematica

$\left \{\left \{u(r,t)\to \text {Q0} \left (\int _0^{\infty } \frac {J_0(r K[1]) \int _0^{\infty } r q(r) J_0(r K[1]) \, dr}{K[1]} \, dK[1]-\int _0^{\infty } \frac {e^{-t K[1]} J_0(r K[1]) \int _0^{\infty } r q(r) J_0(r K[1]) \, dr}{K[1]} \, dK[1]\right )\right \}\right \}$

Maple

$u \left (r , t\right ) = \left (\int _{0}^{\infty }\frac {\BesselJ \left (0, r s \right ) \left (\int _{0}^{\infty }r \BesselJ \left (0, r s \right ) q (r )d r \right )}{s}d s -\left (\int _{0}^{\infty }\frac {\BesselJ \left (0, r s \right ) \left (\int _{0}^{\infty }r \BesselJ \left (0, r s \right ) q (r )d r \right ) {\mathrm e}^{-s t}}{s}d s \right )\right ) Q_{0}$

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5.2.2.5 [408] no $$\theta$$ dependency. Using integral transforms. Source present. Speciﬁc example

problem number 408

Solve ${c}^{2} \left ( {\frac {\partial ^{2}}{\partial {r}^{2}}}u \left ( r,t\right ) +{\frac {{\frac {\partial }{\partial r}}u \left ( r,t \right ) }{r}} \right ) ={\frac {\partial ^{2}}{\partial {t}^{2}}}u \left ( r,t\right )$ With initial conditions \begin {align*} u(r, 0) &= \frac {A a}{\sqrt {a^2+r^2}}\\ \frac {\partial u(r,0)}{\partial t} &=0 \end {align*}

Mathematica

Failed

Maple

$u \left (r , t\right ) = \frac {\left (\sqrt {2 i a c t -c^{2} t^{2}+a^{2}+r^{2}}+\sqrt {-2 i a c t -c^{2} t^{2}+a^{2}+r^{2}}\right ) A a}{2 \sqrt {-2 i a c t -c^{2} t^{2}+a^{2}+r^{2}}\, \sqrt {2 i a c t -c^{2} t^{2}+a^{2}+r^{2}}}$

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5.2.2.6 [409] $$\theta$$ dependency, ﬁxed on edges, general solution

problem number 409

Solve for $$u(r,\theta ,t)$$ with $$0<r<a$$ and $$t>0$$ and $$-\pi <\theta <\pi$$ $\frac {\partial ^2 u}{\partial t^2} = c^2 \left ( \frac {\partial ^2 u}{\partial r^2} + \frac {1}{r} \frac {\partial u}{\partial r} +\frac {1}{r^2} \frac {\partial ^2 u}{\partial \theta ^2} \right )$ With boundary conditions \begin {align*} u(a,\theta ,t) &=0 \\ |u(0,\theta ,t)| < \infty \\ u(r,-\pi ,t) &= u(r,\pi ,t) \\ \frac {\partial u}{\partial \theta }(r,-\pi ,t) &= \frac {\partial u}{\partial \theta }(r,\pi ,t)\\ \end {align*}

With initial conditions \begin {align*} u(r,\theta ,0) &= f(r,\theta ) \\ \frac {\partial u}{\partial t}(r,\theta ,0) &= g(r,\theta ) \end {align*}

Mathematica

$\left \{\left \{u(r,\theta ,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[3]=1}{\overset {\infty }{\sum }}-\frac {\sqrt {\frac {2}{\pi }} J_0\left (\frac {r j_{0,K[3]}}{a}\right ) \left (-\frac {\sqrt {\frac {2}{\pi }} \cos \left (\frac {\sqrt {c^2} t j_{0,K[3]}}{a}\right ) \int _0^a\int _{-\pi }^{\pi }r J_0\left (\frac {r j_{0,K[3]}}{a}\right ) f(r,\theta )d\theta dr}{a J_1\left (j_{0,K[3]}\right )}-\frac {\sqrt {\frac {2}{\pi }} \left (\int _0^a\int _{-\pi }^{\pi }r J_0\left (\frac {r j_{0,K[3]}}{a}\right ) g(r,\theta )d\theta dr\right ) \sin \left (\frac {\sqrt {c^2} t j_{0,K[3]}}{a}\right )}{| c| J_1\left (j_{0,K[3]}\right ) j_{0,K[3]}}\right )}{a J_1\left (j_{0,K[3]}\right )}+\underset {K[3]=1}{\overset {\infty }{\sum }}\left (\underset {K[1]=1}{\overset {\infty }{\sum }}\left (\frac {\sqrt {\frac {2}{\pi }} J_{K[1]}\left (\frac {r j_{K[1],K[3]}}{a}\right ) \cos (\theta K[1]) \left (\frac {\sqrt {\frac {2}{\pi }} \cos \left (\frac {\sqrt {c^2} t j_{K[1],K[3]}}{a}\right ) \int _0^a\int _{-\pi }^{\pi }r J_{K[1]}\left (\frac {r j_{K[1],K[3]}}{a}\right ) \cos (\theta K[1]) f(r,\theta )d\theta dr}{a J_{K[1]-1}\left (j_{K[1],K[3]}\right )}+\frac {\sqrt {\frac {2}{\pi }} \left (\int _0^a\int _{-\pi }^{\pi }r J_{K[1]}\left (\frac {r j_{K[1],K[3]}}{a}\right ) \cos (\theta K[1]) g(r,\theta )d\theta dr\right ) \sin \left (\frac {\sqrt {c^2} t j_{K[1],K[3]}}{a}\right )}{| c| J_{K[1]-1}\left (j_{K[1],K[3]}\right ) j_{K[1],K[3]}}\right )}{a J_{K[1]-1}\left (j_{K[1],K[3]}\right )}+\frac {\sqrt {\frac {2}{\pi }} J_{K[1]}\left (\frac {r j_{K[1],K[3]}}{a}\right ) \left (\frac {\sqrt {\frac {2}{\pi }} \cos \left (\frac {\sqrt {c^2} t j_{K[1],K[3]}}{a}\right ) \int _0^a\int _{-\pi }^{\pi }r J_{K[1]}\left (\frac {r j_{K[1],K[3]}}{a}\right ) f(r,\theta ) \sin (\theta K[1])d\theta dr}{a J_{K[1]-1}\left (j_{K[1],K[3]}\right )}+\frac {\sqrt {\frac {2}{\pi }} \left (\int _0^a\int _{-\pi }^{\pi }r J_{K[1]}\left (\frac {r j_{K[1],K[3]}}{a}\right ) g(r,\theta ) \sin (\theta K[1])d\theta dr\right ) \sin \left (\frac {\sqrt {c^2} t j_{K[1],K[3]}}{a}\right )}{| c| J_{K[1]-1}\left (j_{K[1],K[3]}\right ) j_{K[1],K[3]}}\right ) \sin (\theta K[1])}{a J_{K[1]-1}\left (j_{K[1],K[3]}\right )}\right )\right ) & (K[1]|K[3])\in \mathbb {Z}\land K[1]\geq 1\land K[3]\geq 1\land c^2 \left (j_{K[1],K[3]}\right ){}^2>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}$

Maple

sol=()

Hand solution

Assuming $$u=T\left ( t\right ) R\left ( r\right ) \Theta \left ( \theta \right )$$ and substituting in the PDE gives$\frac {1}{c^{2}}T^{\prime \prime }R\Theta =R^{\prime \prime }T\Theta +\frac {1}{r}R^{\prime }T\Theta +\frac {1}{r^{2}}\Theta ^{\prime \prime }RT$ Dividing by $$RT\Theta$$$\frac {1}{c^{2}}\frac {T^{\prime \prime }}{T}=\frac {R^{\prime \prime }}{R}+\frac {1}{r}\frac {R^{\prime }}{R}+\frac {1}{r^{2}}\frac {\Theta ^{\prime \prime }}{\Theta }$ Hence\begin {align*} \frac {1}{c^{2}}\frac {T^{\prime \prime }}{T} & =-\lambda \\ \frac {R^{\prime \prime }}{R}+\frac {1}{r}\frac {R^{\prime }}{R}+\frac {1}{r^{2}}\frac {\Theta ^{\prime \prime }}{\Theta } & =-\lambda \end {align*}

The time ODE is$T^{\prime \prime }+c^{2}\lambda T=0$ Now we separate again the space ODE’s (remember to move the $$\lambda$$ with the $$R$$ and not the $$\Theta$$)\begin {align*} \frac {R^{\prime \prime }}{R}+\frac {1}{r}\frac {R^{\prime }}{R}+\lambda & =-\frac {1}{r^{2}}\frac {\Theta ^{\prime \prime }}{\Theta }\\ r^{2}\frac {R^{\prime \prime }}{R}+r\frac {R^{\prime }}{R}+r^{2}\lambda & =-\frac {\Theta ^{\prime \prime }}{\Theta } \end {align*}

Let the new separation constant be $$\mu$$, therefore\begin {align*} -\frac {\Theta ^{\prime \prime }}{\Theta } & =\mu \\ \Theta ^{\prime \prime }+\mu \Theta & =0 \end {align*}

With periodic boundary conditions and\begin {align*} r^{2}\frac {R^{\prime \prime }}{R}+r\frac {R^{\prime }}{R}+r^{2}\lambda & =\mu \\ r^{2}R^{\prime \prime }+rR^{\prime }+\lambda r^{2}R-\mu R & =0\\ rR^{\prime \prime }+R^{\prime }-\frac {\mu }{r}R & =-\lambda rR \end {align*}

Now it is in Sturm Liouville form, where $$p=r,q=-\frac {\mu }{r},\sigma =r$$. This is singular SL. Can be written as$R^{\prime \prime }+\frac {1}{r}R^{\prime }+\left ( \lambda -\frac {\mu }{r^{2}}\right ) R=0$ Before we solve the above $$R$$ ODE, we solve the $$\Theta ^{\prime \prime }+\mu \Theta =0$$ to ﬁnd $$\mu$$ Eigenvalues. The solution is$\Theta =A\cos \left ( \sqrt {\mu }\theta \right ) +B\sin \left ( \sqrt {\mu }\theta \right )$ With B.C $$\Theta \left ( -\pi \right ) =\Theta \left ( \pi \right )$$ and $$\Theta ^{\prime }\left ( -\pi \right ) =\Theta ^{\prime }\left ( \pi \right )$$. From ﬁrst B.C. we obtain\begin {align} A\cos \left ( \sqrt {\mu }\pi \right ) -B\sin \left ( \sqrt {\mu }\pi \right ) & =A\cos \left ( \sqrt {\mu }\pi \right ) +B\sin \left ( \sqrt {\mu }\pi \right ) \nonumber \\ 2B\sin \left ( \sqrt {\mu }\pi \right ) & =0\tag {1} \end {align}

Looking at second B.C. $$\Theta ^{\prime }\left ( -\pi \right ) =\Theta ^{\prime }\left ( \pi \right )$$$\Theta ^{\prime }\left ( \theta \right ) =-A\sqrt {\mu }\sin \left ( \sqrt {\mu }\theta \right ) +\sqrt {\mu }B\cos \left ( \sqrt {\mu }\theta \right )$ Hence\begin {align} A\sqrt {\mu }\sin \left ( \sqrt {\mu }\pi \right ) +\sqrt {\mu }B\cos \left ( \sqrt {\mu }\pi \right ) & =-A\sqrt {\mu }\sin \left ( \sqrt {\mu }\pi \right ) +\sqrt {\mu }B\cos \left ( \sqrt {\mu }\pi \right ) \nonumber \\ A\sqrt {\mu }\sin \left ( \sqrt {\mu }\pi \right ) & =-A\sqrt {\mu }\sin \left ( \sqrt {\mu }\pi \right ) \nonumber \\ 2A\sin \left ( \sqrt {\mu }\pi \right ) & =0\tag {2} \end {align}

From (1,2), we see that both are satisﬁed if \begin {align*} \sqrt {\mu }\pi & =n\pi \qquad n=1,2,3,\ldots \\ \mu & =n^{2} \end {align*}

Hence $\Theta _{n}=A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right )$ There is another solution for $$\mu =0$$ which is constant (that is why one of the sums below starts from $$n=0$$). We can combine the zero eigenvalue with the above and write$\Theta _{n}=A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \qquad n=0,1,2,3,\ldots$ Since at $$n=0$$ the above reduces to constant $$A_{0}$$.

Now that we know $$\mu _{n}=n^{2}$$, from solving the $$\theta$$ part, we go and solve the $$r$$ ODE. For each $$n$$, the solution to the $$r$$ (Bessel) ode

$R^{\prime \prime }+\frac {1}{r}R^{\prime }+\left ( \lambda -\frac {n^{2}}{r^{2}}\right ) R=0$ The solution turns out to be  $R_{nm}\left ( r\right ) =J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \qquad m=1,2,3,\cdots$ Where $$\lambda _{nm}$$ is found from roots of $$0=J_{n}\left ( \sqrt {\lambda _{nm}}a\right )$$ giving the eigenvalues. Now the time ODE is solved\begin {align*} T_{nm}^{\prime \prime }+c^{2}\lambda _{nm}T_{nm} & =0\\ T_{nm} & =C_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) +D_{nm}\sin \left ( c\sqrt {\lambda _{nm}}t\right ) \qquad n=0,1,2,3,\ldots ,m=1,2,3,\cdots \end {align*}

Hence the solution is\begin {align*} u\left ( r,\theta ,t\right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }T_{nm}R_{nm}\Theta _{n}\\ & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }\left ( C_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) +D_{nm}\sin \left ( c\sqrt {\lambda _{nm}}t\right ) \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \left ( A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \right ) \end {align*}

We now break this sum as follows\begin {align*} u\left ( r,\theta ,t\right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }\left ( C_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) +D_{nm}\sin \left ( c\sqrt {\lambda _{nm}}t\right ) \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) A_{n}\cos \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }\left ( C_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) +D_{nm}\sin \left ( c\sqrt {\lambda _{nm}}t\right ) \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) B_{n}\sin \left ( n\theta \right ) \end {align*}

Or\begin {align*} u\left ( r,\theta ,t\right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }C_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) A_{n}\cos \left ( n\theta \right ) +D_{nm}\sin \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) A_{n}\cos \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }C_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) B_{n}\sin \left ( n\theta \right ) +D_{nm}\sin \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) B_{n}\sin \left ( n\theta \right ) \end {align*}

Then we break the above into 4 sums\begin {align*} u\left ( r,\theta ,t\right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }C_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) A_{n}\cos \left ( n\theta \right ) \\ & +\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }D_{nm}\sin \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) A_{n}\cos \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }C_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) B_{n}\sin \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }D_{nm}\sin \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) B_{n}\sin \left ( n\theta \right ) \end {align*}

Finally, we merge constants in the above as follows\begin {align*} A_{n}C_{nm} & \equiv A_{nm}\\ A_{n}D_{nm} & \equiv B_{nm}\\ B_{n}C_{nm} & \equiv C_{nm}\\ B_{n}D_{nm} & \equiv D_{nm} \end {align*}

Hence the ﬁnal solution now becomes\begin {align} u\left ( r,\theta ,t\right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }A_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) \nonumber \\ & +\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }B_{nm}\sin \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) \nonumber \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }C_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \nonumber \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }D_{nm}\sin \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \tag {3} \end {align}

Now initial conditions $$u\left ( r,\theta ,0\right ) =f\left ( r,\theta \right )$$ is used to ﬁnd $$A_{nm},C_{nm}\,$$using orthogonality. At $$t=0$$ the solution simpliﬁes to (all terms with $$\sin \left ( c\sqrt {\lambda _{nm}}t\right )$$ vanish giving\begin {align*} u\left ( r,\theta ,t\right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }A_{nm}J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }C_{nm}J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \end {align*}

Hence\begin {equation} f\left ( r,\theta \right ) =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }A_{nm}J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }C_{nm}J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \tag {4} \end {equation} When iterating over $$m$$ index, the terms $$\cos \left ( n\theta \right )$$ and $$\sin \left ( n\theta \right )$$ will be constant. So for each $$n$$, we have $$\sum _{m=1}^{\infty }A_{nm}J_{n}\left ( \sqrt {\lambda _{nm}}r\right )$$ and $$\sum _{m=1}^{\infty }C_{nm}J_{n}\left ( \sqrt {\lambda _{nm}}r\right )$$. So orthogonality is carried out on the $$m$$ index on the Bessel functions. Multiplying (4) by $$J_{n}\left ( \sqrt {\lambda _{nk}}r\right )$$ and integrating\begin {align*} \int _{0}^{a}f\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nk}}r\right ) rdr & =\sum _{n=0}^{\infty }\left ( \int _{0}^{a}\sum _{m=1}^{\infty }A_{nm}J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) J_{n}\left ( \sqrt {\lambda _{nk}}r\right ) rdr\right ) \cos \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\left ( \int _{0}^{a}\sum _{m=1}^{\infty }C_{nm}J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) J_{n}\left ( \sqrt {\lambda _{nk}}r\right ) rdr\right ) \sin \left ( n\theta \right ) \end {align*}

Or\begin {align*} \int _{0}^{a}f\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nk}}r\right ) rdr & =\sum _{n=0}^{\infty }A_{nk}\left ( \int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nk}}r\right ) rdr\right ) \cos \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }C_{nk}\left ( \int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nk}}r\right ) rdr\right ) \sin \left ( n\theta \right ) \end {align*}

Replacing $$k$$ back with $$m$$, the above becomes\begin {align} \int _{0}^{a}f\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) rdr & =\sum _{n=0}^{\infty }A_{nm}\left ( \int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rdr\right ) \cos \left ( n\theta \right ) \nonumber \\ & +\sum _{n=1}^{\infty }C_{nm}\left ( \int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rdr\right ) \sin \left ( n\theta \right ) \tag {5} \end {align}

We now apply orthogonality on $$n$$ using the $$\cos \left ( n\theta \right )$$ results in$\int _{-\pi }^{\pi }\left ( \int _{0}^{a}f\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) r\ dr\right ) \cos \left ( n\theta \right ) d\theta =A_{nm}\int _{-\pi }^{\pi }\left ( \int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rdr\right ) \cos ^{2}\left ( n\theta \right ) d\theta$ But $$\int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rdr$$ does not depend on $$\theta$$, therefore the above becomes\begin {align*} \int _{-\pi }^{\pi }\left ( \int _{0}^{a}f\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) r\ dr\right ) \cos \left ( n\theta \right ) d\theta & =A_{nm}\left ( \int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rdr\right ) \int _{-\pi }^{\pi }\cos ^{2}\left ( n\theta \right ) d\theta \\ & =A_{nm}\pi \int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rdr \end {align*}

Therefore$A_{nm}=\frac {\int _{-\pi }^{\pi }\left ( \int _{0}^{a}f\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) r\ dr\right ) \cos \left ( n\theta \right ) d\theta }{\pi \int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rdr}$ Similarly for $$\sin \left ( n\theta \right )$$, which gives$C_{nm}=\frac {\int _{-\pi }^{\pi }\left ( \int _{0}^{a}f\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) r\ dr\right ) \sin \left ( n\theta \right ) d\theta }{\pi \int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rdr}$ Now we will look at the second initial conditions $$\frac {\partial u}{\partial t}\left ( r,\theta ,0\right ) =g\left ( r,\theta \right ) .$$ Taking derivative w.r.t. time $$t$$ of the solution in (3) gives\begin {align*} \frac {\partial u}{\partial t}\left ( r,\theta ,t\right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }-c\sqrt {\lambda _{nm}}A_{nm}\sin \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) \\ & +\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }c\sqrt {\lambda _{nm}}B_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }-c\sqrt {\lambda _{nm}}C_{nm}\sin \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }c\sqrt {\lambda _{nm}}D_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \end {align*}

At time $$t=0$$ the above becomes (all terms with $$\sin \left ( c\sqrt {\lambda _{nm}}t\right )$$ vanish).\begin {align*} g\left ( r,\theta \right ) = & \sum _{n=0}^{\infty }\sum _{m=1}^{\infty }c\sqrt {\lambda _{nm}}B_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }c\sqrt {\lambda _{nm}}D_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \end {align*}

Now orthogonality is used. At $$t=0$$ the above becomes\begin {align*} g\left ( r,\theta \right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }c\sqrt {\lambda _{nm}}B_{nm}J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) \\ & +\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }c\sqrt {\lambda _{nm}}D_{nm}J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \end {align*}

Similarly to the above we now ﬁnd $$B_{nm}$$ and $$D_{nm}$$. The only diﬀerence, is that now we have extra $$c\sqrt {\lambda _{nm}}$$ terms that show up. The ﬁnal result will be$B_{nm}=\frac {\int _{-\pi }^{\pi }\int _{0}^{a}g\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) r\ d\theta dr}{c\pi \sqrt {\lambda _{nm}}\int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) r\ dr}$ And$D_{nm}=\frac {\int _{-\pi }^{\pi }\int _{0}^{a}g\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) r\ d\theta dr}{c\pi \sqrt {\lambda _{nm}}\int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) r\ dr}$ Summary of solution\begin {align*} u\left ( r,\theta ,t\right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }A_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) \\ & +\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }B_{nm}\sin \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }C_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }D_{nm}\sin \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \end {align*}

$A_{nm}=\frac {\int _{-\pi }^{\pi }\left ( \int _{0}^{a}f\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) r\ dr\right ) \cos \left ( n\theta \right ) d\theta }{\pi \int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rdr}$$C_{nm}=\frac {\int _{-\pi }^{\pi }\left ( \int _{0}^{a}f\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) r\ dr\right ) \sin \left ( n\theta \right ) d\theta }{\pi \int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rdr}$$B_{nm}=\frac {\int _{-\pi }^{\pi }\int _{0}^{a}g\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) r\ d\theta dr}{c\pi \sqrt {\lambda _{nm}}\int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) r\ dr}$$D_{nm}=\frac {\int _{-\pi }^{\pi }\int _{0}^{a}g\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) r\ d\theta dr}{c\pi \sqrt {\lambda _{nm}}\int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) r\ dr}$ With $$\lambda _{nm}$$ being the solutions for $$0=J_{n}\left ( \sqrt {\lambda _{nm}}a\right )$$. For each $$n$$, we ﬁnd $$\lambda _{n,1},\lambda _{n,2},\lambda _{n,3},\cdots$$, which are the zeros of the Bessel $$J_{n}\left ( x\right )$$ function. So for each $$n$$, there are inﬁnite number of zeros. This generates all eigenvalues $$\lambda _{nm}$$. Hence $$\sqrt {\lambda _{nm}}a=BesselJZero\left ( n,m\right )$$, therefore $$\sqrt {\lambda _{nm}}=\frac {a}{BesselJZero\left ( n,m\right ) }$$

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5.2.2.7 [410] $$\theta$$ dependency, ﬁxed on edges, zero initial velocity, general solution

problem number 410

Solve for $$u(r,\theta ,t)$$ with $$0<r<a$$ and $$t>0$$ and $$-\pi <\theta <\pi$$ $\frac {\partial ^2 u}{\partial t^2} = c^2 \left ( \frac {\partial ^2 u}{\partial r^2} + \frac {1}{r} \frac {\partial u}{\partial r} +\frac {1}{r^2} \frac {\partial ^2 u}{\partial \theta ^2} \right )$ With boundary conditions \begin {align*} u(a,\theta ,t) &=0 \\ |u(0,\theta ,t)| < \infty \\ u(r,-\pi ,t) &= u(r,\pi ,t) \\ \frac {\partial u}{\partial \theta }(r,-\pi ,t) &= \frac {\partial u}{\partial \theta }(r,\pi ,t)\\ \end {align*}

With initial conditions \begin {align*} u(r,\theta ,0) &= f(r,\theta ) \\ \frac {\partial u}{\partial t}(r,\theta ,0) &= 0 \end {align*}

Mathematica

$\left \{\left \{u(r,\theta ,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[3]=1}{\overset {\infty }{\sum }}\frac {2 J_0\left (\frac {r j_{0,K[3]}}{a}\right ) \cos \left (\frac {\sqrt {c^2} t j_{0,K[3]}}{a}\right ) \int _0^a\int _{-\pi }^{\pi }r J_0\left (\frac {r j_{0,K[3]}}{a}\right ) f(r,\theta )d\theta dr}{a^2 \pi J_1\left (j_{0,K[3]}\right ){}^2}+\underset {K[3]=1}{\overset {\infty }{\sum }}\left (\underset {K[1]=1}{\overset {\infty }{\sum }}\left (\frac {2 J_{K[1]}\left (\frac {r j_{K[1],K[3]}}{a}\right ) \cos \left (\frac {\sqrt {c^2} t j_{K[1],K[3]}}{a}\right ) \cos (\theta K[1]) \int _0^a\int _{-\pi }^{\pi }r J_{K[1]}\left (\frac {r j_{K[1],K[3]}}{a}\right ) \cos (\theta K[1]) f(r,\theta )d\theta dr}{a^2 \pi J_{K[1]-1}\left (j_{K[1],K[3]}\right ){}^2}+\frac {2 J_{K[1]}\left (\frac {r j_{K[1],K[3]}}{a}\right ) \cos \left (\frac {\sqrt {c^2} t j_{K[1],K[3]}}{a}\right ) \left (\int _0^a\int _{-\pi }^{\pi }r J_{K[1]}\left (\frac {r j_{K[1],K[3]}}{a}\right ) f(r,\theta ) \sin (\theta K[1])d\theta dr\right ) \sin (\theta K[1])}{a^2 \pi J_{K[1]-1}\left (j_{K[1],K[3]}\right ){}^2}\right )\right ) & (K[1]|K[3])\in \mathbb {Z}\land K[1]\geq 1\land K[3]\geq 1\land c^2 \left (j_{K[1],K[3]}\right ){}^2>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}$

Maple

sol=()

Hand solution

The basic solution for this type of PDE was already given in problem 5.2.2.6 on page 1898 as\begin {align*} u\left ( r,\theta ,t\right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }A_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) \\ & +\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }B_{nm}\sin \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }C_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }D_{nm}\sin \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \end {align*}

$A_{nm}=\frac {\int _{-\pi }^{\pi }\left ( \int _{0}^{a}f\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) r\ dr\right ) \cos \left ( n\theta \right ) d\theta }{\pi \int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rdr}$$C_{nm}=\frac {\int _{-\pi }^{\pi }\left ( \int _{0}^{a}f\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) r\ dr\right ) \sin \left ( n\theta \right ) d\theta }{\pi \int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rdr}$$B_{nm}=\frac {\int _{-\pi }^{\pi }\int _{0}^{a}g\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) r\ d\theta dr}{c\pi \sqrt {\lambda _{nm}}\int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) r\ dr}$$D_{nm}=\frac {\int _{-\pi }^{\pi }\int _{0}^{a}g\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) r\ d\theta dr}{c\pi \sqrt {\lambda _{nm}}\int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) r\ dr}$ With $$\lambda _{nm}$$ being the solutions for $$0=J_{n}\left ( \sqrt {\lambda _{nm}}a\right )$$. For each $$n$$, we ﬁnd $$\lambda _{n,1},\lambda _{n,2},\lambda _{n,3},\cdots$$, which are the zeros of the Bessel $$J_{n}\left ( x\right )$$ function. So for each $$n$$, there are inﬁnite number of zeros. This generates all eigenvalues $$\lambda _{nm}$$. Hence $$\sqrt {\lambda _{nm}}a=BesselJZero\left ( n,m\right )$$, therefore $$\sqrt {\lambda _{nm}}=\frac {a}{BesselJZero\left ( n,m\right ) }$$. Since $$g\left ( r,\theta \right ) =0$$ in this case, then $$B_{nm}=0,D_{nm}=0$$ and the solution simpliﬁes to

$u\left ( r,\theta ,t\right ) =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }A_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }C_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right )$

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5.2.2.8 [411] $$\theta$$ dependency, ﬁxed on edges, zero initial velocity, speciﬁc example

problem number 411

Solve for $$u(r,\theta ,t)$$ with $$0<r<a$$ and $$t>0$$ and $$-\pi <\theta <\pi$$ $\frac {\partial ^2 u}{\partial t^2} = c^2 \left ( \frac {\partial ^2 u}{\partial r^2} + \frac {1}{r} \frac {\partial u}{\partial r} +\frac {1}{r^2} \frac {\partial ^2 u}{\partial \theta ^2} \right )$ With boundary conditions \begin {align*} u(a,\theta ,t) &=0 \\ |u(0,\theta ,t)| < \infty \\ u(r,-\pi ,t) &= u(r,\pi ,t) \\ \frac {\partial u}{\partial \theta }(r,-\pi ,t) &= \frac {\partial u}{\partial \theta }(r,\pi ,t)\\ \end {align*}

With initial conditions \begin {align*} u(r,\theta ,0) &= f(r,\theta ) \\ \frac {\partial u}{\partial t}(r,\theta ,0) &= 0 \end {align*}

Using $$a=1,c=0.2,f(r,\theta )=r \theta$$.

Mathematica

$\left \{\left \{u(r,\theta ,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[3]=1}{\overset {\infty }{\sum }}\left (\underset {K[1]=1}{\overset {\infty }{\sum }}\frac {(-1)^{K[1]+1} J_{K[1]}\left (r j_{K[1],K[3]}\right ) j_{K[1],K[3]} \cos \left (\frac {1}{5} t j_{K[1],K[3]}\right ) \operatorname {Gamma}\left (\frac {1}{2} (K[1]+3)\right ) \, _1\tilde {F}_2\left (\frac {1}{2} (K[1]+3);\frac {1}{2} (K[1]+5),K[1]+1;-\frac {1}{4} \left (j_{K[1],K[3]}\right ){}^2\right ) \sin (\theta K[1])}{J_{K[1]-1}\left (j_{K[1],K[3]}\right ) \, _0\tilde {F}_1\left (;K[1];-\frac {1}{4} \left (j_{K[1],K[3]}\right ){}^2\right ) K[1]}\right ) & (K[1]|K[3])\in \mathbb {Z}\land K[1]\geq 1\land K[3]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}$

Maple

sol=()

Hand solution

The basic solution for this type of PDE was already given in problem 5.2.2.6 on page 1898 as\begin {align*} u\left ( r,\theta ,t\right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }A_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) \\ & +\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }B_{nm}\sin \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }C_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }D_{nm}\sin \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \end {align*}

$A_{nm}=\frac {\int _{-\pi }^{\pi }\left ( \int _{0}^{a}f\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) r\ dr\right ) \cos \left ( n\theta \right ) d\theta }{\pi \int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rdr}$$C_{nm}=\frac {\int _{-\pi }^{\pi }\left ( \int _{0}^{a}f\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) r\ dr\right ) \sin \left ( n\theta \right ) d\theta }{\pi \int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rdr}$$B_{nm}=\frac {\int _{-\pi }^{\pi }\int _{0}^{a}g\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) r\ d\theta dr}{c\pi \sqrt {\lambda _{nm}}\int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) r\ dr}$$D_{nm}=\frac {\int _{-\pi }^{\pi }\int _{0}^{a}g\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) r\ d\theta dr}{c\pi \sqrt {\lambda _{nm}}\int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) r\ dr}$ With $$\lambda _{nm}$$ being the solutions for $$0=J_{n}\left ( \sqrt {\lambda _{nm}}a\right )$$. For each $$n$$, we ﬁnd $$\lambda _{n,1},\lambda _{n,2},\lambda _{n,3},\cdots$$, which are the zeros of the Bessel $$J_{n}\left ( x\right )$$ function. So for each $$n$$, there are inﬁnite number of zeros. This generates all eigenvalues $$\lambda _{nm}$$. Hence $$\sqrt {\lambda _{nm}}a=BesselJZero\left ( n,m\right )$$, therefore $$\sqrt {\lambda _{nm}}=\frac {a}{BesselJZero\left ( n,m\right ) }$$.

In this problem $$g\left ( r,\theta \right ) =0,f\left ( r,\theta \right ) =r\theta ,a=1,c=\frac {2}{10}$$, then $$B_{nm}=0,D_{nm}=0$$ and the solution simpliﬁes to

$u\left ( r,\theta ,t\right ) =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }A_{nm}\cos \left ( \frac {2}{10}\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }C_{nm}\cos \left ( \frac {2}{10}\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right )$

Where

\begin {align*} A_{nm} & =\frac {\int _{-\pi }^{\pi }\left ( \int _{0}^{a}r\theta J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) r\ dr\right ) \cos \left ( n\theta \right ) d\theta }{\pi \int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rdr}\\ C_{nm} & =\frac {\int _{-\pi }^{\pi }\left ( \int _{0}^{a}r\theta J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) r\ dr\right ) \sin \left ( n\theta \right ) d\theta }{\pi \int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rdr} \end {align*}

The following animations run for 80 seconds. They are for diﬀerent $$n,m\,\$$modes. (This only show in the HTML version)

Cases for $$n=0$$

Cases for $$n=1$$

Cases for $$n=2$$

Cases for $$n=3$$

Source code for all the above animations

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5.2.2.9 [412] $$\theta$$ dependency, ﬁxed on edges, zero initial position, speciﬁc example

problem number 412

Math 322 UW exam problem. 2018.

Solve for $$u(r,\theta ,t)$$ with $$0<r<a$$ and $$t>0$$ and $$-\pi <\theta <\pi$$ $\frac {\partial ^2 u}{\partial t^2} = c^2 \left ( \frac {\partial ^2 u}{\partial r^2} + \frac {1}{r} \frac {\partial u}{\partial r} +\frac {1}{r^2} \frac {\partial ^2 u}{\partial \theta ^2} \right )$ With boundary conditions \begin {align*} u(a,\theta ,t) &=0 \\ |u(0,\theta ,t)| < \infty \\ u(r,-\pi ,t) &= u(r,\pi ,t) \\ \frac {\partial u}{\partial \theta }(r,-\pi ,t) &= \frac {\partial u}{\partial \theta }(r,\pi ,t)\\ \end {align*}

With initial conditions \begin {align*} u\left ( r,\theta ,0\right ) & =0\\ u_{t}\left ( r,\theta ,0\right ) & =\left \{ \begin {array} [c]{ccc}\frac {1}{\pi \epsilon ^{2}} & & \text {if }r\leq \epsilon \\ 0 & & \text {otherwise}\end {array} \right . \end {align*}

Where $$0<\epsilon <1$$

Using $$a=1,c=1,\epsilon =\frac {1}{2}$$.

Mathematica

$\left \{\left \{u(r,\theta ,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[3]=1}{\overset {\infty }{\sum }}\frac {2 J_0\left (r j_{0,K[3]}\right ) \, _0\tilde {F}_1\left (;2;-\frac {1}{16} \left (j_{0,K[3]}\right ){}^2\right ) \sin \left (t j_{0,K[3]}\right )}{\pi J_1\left (j_{0,K[3]}\right ){}^2 j_{0,K[3]}} & (K[1]|K[3])\in \mathbb {Z}\land K[1]\geq 1\land K[3]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}$

Maple

sol=()

Hand solution

The basic solution for this type of PDE was already given in problem 5.2.2.6 on page 1898 as

\begin {align*} u\left ( r,\theta ,t\right ) & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }A_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) \\ & +\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }B_{nm}\sin \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }C_{nm}\cos \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }D_{nm}\sin \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) \end {align*}

$A_{nm}=\frac {\int _{-\pi }^{\pi }\left ( \int _{0}^{a}f\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) r\ dr\right ) \cos \left ( n\theta \right ) d\theta }{\pi \int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rdr}$$C_{nm}=\frac {\int _{-\pi }^{\pi }\left ( \int _{0}^{a}f\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) r\ dr\right ) \sin \left ( n\theta \right ) d\theta }{\pi \int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rdr}$$B_{nm}=\frac {\int _{-\pi }^{\pi }\int _{0}^{a}g\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) r\ d\theta dr}{c\pi \sqrt {\lambda _{nm}}\int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) r\ dr}$$D_{nm}=\frac {\int _{-\pi }^{\pi }\int _{0}^{a}g\left ( r,\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right ) r\ d\theta dr}{c\pi \sqrt {\lambda _{nm}}\int _{0}^{a}J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) r\ dr}$ With $$\lambda _{nm}$$ being the solutions for $$0=J_{n}\left ( \sqrt {\lambda _{nm}}a\right )$$. For each $$n$$, we ﬁnd $$\lambda _{n,1},\lambda _{n,2},\lambda _{n,3},\cdots$$, which are the zeros of the Bessel $$J_{n}\left ( x\right )$$ function. So for each $$n$$, there are inﬁnite number of zeros. This generates all eigenvalues $$\lambda _{nm}$$. Hence $$\sqrt {\lambda _{nm}}a=BesselJZero\left ( n,m\right )$$, therefore $$\sqrt {\lambda _{nm}}=\frac {a}{BesselJZero\left ( n,m\right ) }$$.

In this problem $$f\left ( r,\theta \right ) =0,a=1,c=1$$, then $$A_{nm}=0,C_{nm}=0$$ and the solution simpliﬁes to

$u\left ( r,\theta ,t\right ) =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }B_{nm}\sin \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }D_{nm}\sin \left ( c\sqrt {\lambda _{nm}}t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right )$ Taking time derivative gives$u_{t}\left ( r,\theta ,t\right ) =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }B_{nm}\cos \left ( n\theta \right ) \lambda _{nm}\cos \left ( \lambda _{nm}t\right ) J_{n}\left ( \lambda _{nm}r\right ) +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }D_{nm}\sin \left ( n\theta \right ) \lambda _{nm}\cos \left ( \lambda _{nm}t\right ) J_{n}\left ( \lambda _{nm}r\right )$ Applying the second initial condition at $$t=0$$ gives\begin {equation} \sum _{n=0}^{\infty }\sum _{m=1}^{\infty }B_{nm}\cos \left ( n\theta \right ) \lambda _{nm}J_{n}\left ( \lambda _{nm}r\right ) +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }D_{nm}\sin \left ( n\theta \right ) \lambda _{nm}J_{n}\left ( \lambda _{nm}r\right ) =\left \{ \begin {array} [c]{ccc}\frac {1}{\pi \epsilon ^{2}} & & \text {if }r\leq \epsilon \\ 0 & & \text {otherwise}\end {array} \right . \tag {9} \end {equation} Case $$n=0$$ (9) becomes$\sum _{m=1}^{\infty }B_{0m}\lambda _{0m}J_{0}\left ( \lambda _{0m}r\right ) =\left \{ \begin {array} [c]{ccc}\frac {1}{\pi \epsilon ^{2}} & & \text {if }r\leq \epsilon \\ 0 & & \text {otherwise}\end {array} \right .$ Applying orthogonality on $$J_{0}\left ( \lambda _{0m}r\right )$$ results in\begin {align} B_{0m}\lambda _{0m}\int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{0m}r\right ) dr & =\frac {1}{\pi \epsilon ^{2}}\int _{0}^{\epsilon }rJ_{0}\left ( \lambda _{0m}r\right ) dr\nonumber \\ B_{0m} & =\frac {1}{\pi \epsilon ^{2}\lambda _{0m}}\frac {\int _{0}^{\epsilon }rJ_{0}\left ( \lambda _{0m}r\right ) dr}{\int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{0m}r\right ) dr} \tag {9A} \end {align}

Case $$n>1$$ Applying orthogonality on $$\cos \left ( n\theta \right ) ,$$ equation (9) becomes\begin {align*} \sum _{m=1}^{\infty }B_{nm}\left ( \int _{-\pi }^{\pi }\cos ^{2}\left ( n\theta \right ) d\theta \right ) \lambda _{nm}J_{n}\left ( \lambda _{nm}r\right ) & =\left \{ \begin {array} [c]{ccc}\frac {1}{\pi \epsilon ^{2}}\int _{-\pi }^{\pi }\cos \left ( n\theta \right ) d\theta & & \text {if }r^{2}\leq \epsilon \\ 0 & & \text {otherwise}\end {array} \right . \\ \sum _{m=1}^{\infty }\pi B_{nm}\lambda _{nm}J_{n}\left ( \lambda _{nm}r\right ) & =\left \{ \begin {array} [c]{ccc}0 & & \text {if }r^{2}\leq \epsilon \\ 0 & & \text {otherwise}\end {array} \right . \end {align*}

Hence $$B_{nm}=0$$ for all $$n>0$$.

The same is now done to ﬁnd $$\bar {D}_{nm}$$. Applying orthogonality on $$\sin \left ( n\theta \right ) ,$$ equation (9) becomes\begin {align*} \sum _{m=1}^{\infty }D_{nm}\left ( \int _{-\pi }^{\pi }\sin ^{2}\left ( n\theta \right ) d\theta \right ) \lambda _{nm}J_{n}\left ( \lambda _{nm}r\right ) & =\left \{ \begin {array} [c]{ccc}\frac {1}{\pi \epsilon ^{2}}\int _{-\pi }^{\pi }\sin \left ( n\theta \right ) d\theta & & \text {if }r^{2}\leq \epsilon \\ 0 & & \text {otherwise}\end {array} \right . \\ \sum _{m=1}^{\infty }D_{nm}\left ( \int _{-\pi }^{\pi }\sin ^{2}\left ( n\theta \right ) d\theta \right ) \lambda _{nm}J_{n}\left ( \lambda _{nm}r\right ) & =\left \{ \begin {array} [c]{ccc}0 & & \text {if }r^{2}\leq \epsilon \\ 0 & & \text {otherwise}\end {array} \right . \end {align*}

Hence all $$D_{nm}=0$$ for all $$n>0$$.

Therefore the solution (8) reduces to only using $$n=0,m=1,2,3,\cdots$$. The solution can now be written as\begin {equation} u\left ( r,\theta ,t\right ) =\sum _{m=1}^{\infty }B_{0m}\sin \left ( \lambda _{0m}t\right ) J_{0}\left ( \lambda _{0m}r\right ) \tag {10} \end {equation} Where $$B_{0m}=\frac {1}{\pi \epsilon ^{2}\lambda _{0m}}\frac {\int _{0}^{\epsilon }rJ_{0}\left ( \lambda _{0m}r\right ) dr}{\int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{0m}r\right ) dr}$$ And $$\lambda _{0m}$$ are all the positive zeros of $$J_{0}\left ( z\right )$$, $$m=1,2,3,\cdots$$.

$$B_{0m}$$ is now simpliﬁed more. Considering ﬁrst the numerator of $$B_{0m}$$ which is $$\int _{0}^{\epsilon }rJ_{0}\left ( \lambda _{0m}r\right ) dr$$. The hint given says that$\frac {d}{dr}\left ( rJ_{1}\left ( r\right ) \right ) =rJ_{0}\left ( r\right )$ This is the same as saying\begin {equation} rJ_{1}\left ( r\right ) =\int rJ_{0}\left ( r\right ) dr \tag {10A} \end {equation} However the integral in $$B_{0m}$$ is $$\int rJ_{0}\left ( \lambda _{0m}r\right ) dr$$ and not $$\int rJ_{0}\left ( r\right ) dr$$. To transform it so that the hint can be used, let $$\lambda _{0m}r=\bar {r}$$, then $$\frac {dr}{d\bar {r}}=\frac {1}{\lambda _{0m}}$$ or $$dr=\frac {d\bar {r}}{\lambda _{0m}}$$. Now $$\int rJ_{0}\left ( \lambda _{0m}r\right ) dr$$ becomes $$\int \frac {\bar {r}}{\lambda _{0m}}J_{0}\left ( \bar {r}\right ) \frac {d\bar {r}}{\lambda _{0m}}$$ or $$\frac {1}{\lambda _{0m}^{2}}\int \bar {r}J_{0}\left ( \bar {r}\right ) d\bar {r}$$ and now the hint (10A) can be used on this integral giving$\frac {1}{\lambda _{0m}^{2}}\left ( \int \bar {r}J_{0}\left ( \bar {r}\right ) d\bar {r}\right ) =\frac {1}{\lambda _{0m}^{2}}\left ( \bar {r}J_{1}\left ( \bar {r}\right ) \right )$ Replacing $$\bar {r}$$ back by $$\lambda _{0m}r$$, gives the result needed\begin {align*} \frac {1}{\lambda _{0m}^{2}}\left ( \bar {r}J_{1}\left ( \bar {r}\right ) \right ) & =\frac {1}{\lambda _{0m}^{2}}\left ( \lambda _{0m}rJ_{1}\left ( \lambda _{0m}r\right ) \right ) \\ & =\frac {1}{\lambda _{0m}}rJ_{1}\left ( \lambda _{0m}r\right ) \end {align*}

Now the limits are applied, using the fundamental theory of calculus\begin {align} \int _{0}^{\epsilon }rJ_{0}\left ( \lambda _{0m}r\right ) dr & =\frac {1}{\lambda _{0m}}\left [ rJ_{1}\left ( \lambda _{0m}r\right ) \right ] _{0}^{\epsilon }\nonumber \\ & =\frac {\epsilon }{\lambda _{0m}}J_{1}\left ( \lambda _{0m}\epsilon \right ) \tag {10B} \end {align}

This completes ﬁnding the numerator integral in $$B_{0m}$$. The denominator integral in $$B_{0m}$$ is $$\int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{0m}r\right ) dr$$. This was found before which is $\int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{0m}r\right ) dr=\frac {1}{2}\left [ J_{0}^{\prime }\left ( \lambda _{0m}\right ) \right ] ^{2}$ But $$J_{0}^{\prime }\left ( \lambda _{0m}\right ) =-J_{1}\left ( \lambda _{0m}\right )$$, hence the above becomes\begin {equation} \int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{0m}r\right ) dr=\frac {1}{2}J_{1}^{2}\left ( \lambda _{0m}\right ) \tag {10C} \end {equation} Applying (10B) and (10C), $$B_{0m}$$ simpliﬁes to the following expression\begin {align*} B_{0m} & =\frac {1}{\pi \epsilon ^{2}\lambda _{0m}}\frac {\frac {\epsilon }{\lambda _{0m}}J_{1}\left ( \lambda _{0m}\epsilon \right ) }{\frac {1}{2}J_{1}^{2}\left ( \lambda _{0m}\right ) }\\ & =\frac {2}{\pi \epsilon \lambda _{0m}^{2}}\frac {J_{1}\left ( \lambda _{0m}\epsilon \right ) }{J_{1}^{2}\left ( \lambda _{0m}\right ) } \end {align*}

Therefore the ﬁnal solution becomes\begin {align} u\left ( r,\theta ,t\right ) & =\sum _{m=1}^{\infty }B_{0m}\sin \left ( \lambda _{0m}t\right ) J_{0}\left ( \lambda _{0m}r\right ) \nonumber \\ u\left ( r,\theta ,t\right ) & =\frac {2}{\pi \epsilon }\sum _{m=1}^{\infty }\frac {1}{\lambda _{0m}^{2}}\frac {J_{1}\left ( \lambda _{0m}\epsilon \right ) }{J_{1}^{2}\left ( \lambda _{0m}\right ) }J_{0}\left ( \lambda _{0m}r\right ) \sin \left ( \lambda _{0m}t\right ) \tag {11} \end {align}

When $$\epsilon =\frac {1}{2}$$, the above solution (11) becomes\begin {equation} u\left ( r,\theta ,t\right ) =\frac {4}{\pi }\sum _{m=1}^{\infty }\frac {1}{\lambda _{0m}^{2}}\frac {J_{1}\left ( \frac {1}{2}\lambda _{0m}\right ) }{J_{1}^{2}\left ( \lambda _{0m}\right ) }J_{0}\left ( \lambda _{0m}r\right ) \sin \left ( \lambda _{0m}t\right ) \tag {11A} \end {equation} Here is animation for 5 seconds made in Mathematica

Mathematica Source code for all the above animations

Here is the same animation made in Maple 2018

Maple source code for all the above animations

Here is the same animation made in Matlab 2016a

Matlab source code for all the above animations

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5.2.2.10 [413] $$\theta$$ dependency, ﬁxed on edges, zero initial position with internal source (Haberman 8.5.5. (b)

problem number 413

Problem 8.5.5. (b) Richard Haberman applied partial di fb00erential equations book, 5th edition

Solve wave PDE inside circular membrane for $$u(r,\theta ,t)$$ with $$0<r<a$$ and $$t>0$$ and $$-\pi <\theta <\pi$$ $u_{tt} = c^2 \nabla ^2 u(r,\theta ) + Q(r,\theta ,t)$ With boundary conditions \begin {align*} u(a,\theta ,t) &=0 \\ |u(0,\theta ,t)| < \infty \\ u(r,-\pi ,t) &= u(r,\pi ,t) \\ \frac {\partial u}{\partial \theta }(r,-\pi ,t) &= \frac {\partial u}{\partial \theta }(r,\pi ,t)\\ \end {align*}

With initial conditions \begin {align*} u\left ( r,\theta ,0\right ) & =f(r,\theta )\\ u_{t}\left ( r,\theta ,0\right ) & =0 \end {align*}

Mathematica

Failed

Maple

sol=()

Hand solution

The solution to the corresponding homogeneous wave PDE

$\frac {\partial ^{2}u}{\partial t^{2}}=c^{2}\nabla ^{2}$ Is known to be $u\left ( r,\theta ,t\right ) =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }a_{n}\left ( t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \cos \left ( n\theta \right ) +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }a_{n}\left ( t\right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \sin \left ( n\theta \right )$ Where $$\lambda _{nm}$$ are found by solving roots of $$J_{n}\left ( \sqrt {\lambda _{nm}}a\right ) =0$$. To make things simpler, we will write $u\left ( r,\theta ,t\right ) =\sum _{i}a_{i}\left ( t\right ) \Phi _{i}\left ( r,\theta \right )$ Where the above means the double sum of all eigenvalues $$\lambda _{i}$$. So $$\Phi _{i}\left ( r,\theta \right )$$ represents $$J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \left \{ \cos \left ( n\theta \right ) ,\sin \left ( \theta \right ) \right \}$$ combined. So double sum is implied everywhere. Given this, we now expand the source term$Q\left ( r,\theta ,t\right ) =\sum _{i}q_{i}\left ( t\right ) \Phi _{i}\left ( r,\theta \right )$ And the original PDE becomes\begin {equation} \sum _{i}a_{i}^{\prime \prime }\left ( t\right ) \Phi \left ( \lambda _{i}\right ) =c^{2}\sum _{i}a_{i}\left ( t\right ) \nabla ^{2}\left ( \Phi _{i}\left ( r,\theta \right ) \right ) +\sum _{i}q_{i}\left ( t\right ) \Phi _{i}\left ( r,\theta \right ) \tag {1} \end {equation} But $\nabla ^{2}\left ( \Phi _{i}\left ( r,\theta \right ) \right ) =-\lambda _{i}\Phi _{i}\left ( r,\theta \right )$ Hence (1) becomes\begin {align*} \sum _{i}a_{i}^{\prime \prime }\left ( t\right ) \Phi _{i}\left ( r,\theta \right ) +c^{2}\lambda _{i}a_{i}\left ( t\right ) \Phi _{i}\left ( r,\theta \right ) & =\sum _{i}q_{i}\left ( t\right ) \Phi _{i}\left ( r,\theta \right ) \\ \sum _{i}\left ( a_{i}^{\prime \prime }\left ( t\right ) +c^{2}\lambda _{i}a_{i}\left ( t\right ) \right ) \Phi _{i}\left ( r,\theta \right ) & =\sum _{i}q_{i}\left ( t\right ) \Phi _{i}\left ( r,\theta \right ) \end {align*}

Applying orthogonality gives$a_{i}^{\prime \prime }\left ( t\right ) +c^{2}\lambda _{i}a_{i}\left ( t\right ) =q_{i}\left ( t\right )$ Where $q_{i}\left ( t\right ) =\frac {\int _{0}^{a}\int _{-\pi }^{\pi }Q\left ( r,\theta ,t\right ) \Phi _{i}\left ( r,\theta \right ) rdrd\theta }{\int _{0}^{a}\int _{-\pi }^{\pi }\Phi _{i}^{2}\left ( r,\theta \right ) rdrd\theta }$ The solution to the homogenous ODE is$a_{i}^{h}\left ( t\right ) =A_{i}\cos \left ( c\sqrt {\lambda _{i}}t\right ) +B_{i}\sin \left ( c\sqrt {\lambda _{i}}t\right )$ And the particular solution is found if we know what $$Q\left ( r,\theta ,t\right )$$ and hence $$q_{i}\left ( t\right )$$. For now, lets call the particular solution as $$a_{i}^{p}\left ( t\right )$$. Hence the solution for $$a_{i}\left ( t\right )$$ is$a_{i}\left ( t\right ) =A_{i}\cos \left ( c\sqrt {\lambda _{i}}t\right ) +B_{i}\sin \left ( c\sqrt {\lambda _{i}}t\right ) +a_{i}^{p}\left ( t\right )$ Plugging the above into the $$u\left ( r,\theta ,t\right ) =\sum _{i}a_{i}\left ( t\right ) \Phi _{i}\left ( r,\theta \right )$$, gives\begin {equation} u\left ( r,\theta ,t\right ) =\sum _{i}\left ( A_{i}\cos \left ( c\sqrt {\lambda _{i}}t\right ) +B_{i}\sin \left ( c\sqrt {\lambda _{i}}t\right ) +a_{i}^{p}\left ( t\right ) \right ) \Phi _{i}\left ( r,\theta \right ) \tag {2} \end {equation} We now ﬁnd $$A_{i},B_{i}$$ from initial conditions. At $$t=0$$$f\left ( r,\theta \right ) =\sum _{i}\left ( A_{i}+a_{i}^{p}\left ( 0\right ) \right ) \Phi _{i}\left ( r,\theta \right )$ Applying orthogonality\begin {align*} \int _{0}^{a}\int _{-\pi }^{\pi }f\left ( r,\theta \right ) \Phi _{j}\left ( r,\theta \right ) rdrd\theta & =\int _{0}^{a}\int _{-\pi }^{\pi }\sum _{i}\left ( A_{i}+a_{i}^{p}\left ( 0\right ) \right ) \Phi _{i}\left ( r,\theta \right ) \Phi _{j}\left ( r,\theta \right ) rdrd\theta \\ \int _{0}^{a}\int _{-\pi }^{\pi }f\left ( r,\theta \right ) \Phi _{j}\left ( r,\theta \right ) rdrd\theta & =\left ( A_{j}+a_{j}^{p}\left ( 0\right ) \right ) \int _{0}^{a}\int _{-\pi }^{\pi }\Phi _{j}^{2}\left ( r,\theta \right ) rdrd\theta \\ \left ( A_{i}+a_{i}^{p}\left ( 0\right ) \right ) & =\frac {\int _{0}^{a}\int _{-\pi }^{\pi }f\left ( r,\theta \right ) \Phi _{i}\left ( r,\theta \right ) rdrd\theta }{\int _{0}^{a}\int _{-\pi }^{\pi }\Phi _{i}^{2}\left ( r,\theta \right ) rdrd\theta } \end {align*}

Taking time derivative of  (2)$\frac {\partial u\left ( r,\theta ,t\right ) }{\partial t}=\sum _{i}\left ( -A_{i}c\sqrt {\lambda _{i}}\sin \left ( c\sqrt {\lambda _{i}}t\right ) +c\sqrt {\lambda _{i}}B_{i}\cos \left ( c\sqrt {\lambda _{i}}t\right ) +\frac {da_{i}^{p}\left ( t\right ) }{dt}\right ) \Phi _{i}\left ( r,\theta \right )$ At $$t=0$$$0=\sum _{i}\left ( c\sqrt {\lambda _{i}}B_{i}+\frac {da_{i}^{p}\left ( 0\right ) }{dt}\right ) \Phi _{i}\left ( r,\theta \right )$ Hence $$B_{i}=0$$. Therefore the ﬁnal solution is$u\left ( r,\theta ,t\right ) =\sum _{i}\left ( A_{i}\cos \left ( c\sqrt {\lambda _{i}}t\right ) +a_{i}^{p}\left ( t\right ) \right ) \Phi _{i}\left ( r,\theta \right )$ Where$\left ( A_{i}+a_{i}^{p}\left ( 0\right ) \right ) =\frac {\int _{0}^{a}\int _{-\pi }^{\pi }f\left ( r,\theta \right ) \Phi _{i}\left ( r,\theta \right ) rdrd\theta }{\int _{0}^{a}\int _{-\pi }^{\pi }\Phi _{i}^{2}\left ( r,\theta \right ) rdrd\theta }$ This complete the solution.