Added January 12, 2020
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 ✓
ClearAll["Global`*"]; pde = D[u[r, t], {t, 2}] == c^2*(D[u[r, t], {r, 2}] + 1/r*D[u[r, t], r]); ic = {u[r, 0] == f[r], Derivative[0, 1][u][r, 0] == g[r]}; bc = u[a, t] == 0; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[r, t], {r, t},Assumptions->{t>0,r>0,r<a}], 60*10]]; sol = sol /. K[1] -> n;
\[\left \{\left \{u(r,t)\to \underset {n=1}{\overset {\infty }{\sum }}\fbox {$\frac {2 J_0\left (\frac {r j_{0,n}}{a}\right ) \left (\sqrt {c^2} j_{0,n} \cos \left (\frac {\sqrt {c^2} t j_{0,n}}{a}\right ) \int _0^a r J_0\left (\frac {r j_{0,n}}{a}\right ) f(r) \, dr+a \left (\int _0^a r J_0\left (\frac {r j_{0,n}}{a}\right ) g(r) \, dr\right ) \sin \left (\frac {\sqrt {c^2} t j_{0,n}}{a}\right )\right )}{a^2 \sqrt {c^2} \left (J_0\left (j_{0,n}\right ){}^2+J_1\left (j_{0,n}\right ){}^2\right ) j_{0,n}}\text { if }j_{0,n}\geq 0$}\right \}\right \}\]
Maple ✓
restart; pde := diff(u(r, t), t$2) = c^2*( diff(u(r,t), r$2)+ (1/r)* diff(u(r,t),r) ); ic := u(r,0)=f(r), D[2](u)(r,0)=g(r); bc := u(a,t)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic,bc], u(r, t),HINT = boundedseries(r=0)) assuming t>0,r>0,r<a),output='realtime'));
\[u \left ( r,t \right ) ={\frac {1}{{c}^{2}} \left ( {\it invlaplace} \left ( {\BesselK \left ( 0,{\frac {sr}{c}} \right ) \int \!\BesselK \left ( 0,{\frac {sa}{c}} \right ) a \left ( f \left ( a \right ) s+g \left ( a \right ) \right ) \,{\rm d}a\BesselI \left ( 0,{\frac {sa}{c}} \right ) \left ( \BesselK \left ( 0,{\frac {sa}{c}} \right ) \right ) ^{-1}},s,t \right ) -{\it invlaplace} \left ( \BesselK \left ( 0,{\frac {sr}{c}} \right ) \int \!\BesselI \left ( 0,{\frac {sa}{c}} \right ) a \left ( f \left ( a \right ) s+g \left ( a \right ) \right ) \,{\rm d}a,s,t \right ) -{\it invlaplace} \left ( \int \!\BesselK \left ( 0,{\frac {sr}{c}} \right ) \left ( g \left ( r \right ) +sf \left ( r \right ) \right ) r\,{\rm d}r\BesselI \left ( 0,{\frac {sr}{c}} \right ) ,s,t \right ) +{\it invlaplace} \left ( \int \!\BesselI \left ( 0,{\frac {sr}{c}} \right ) \left ( g \left ( r \right ) +sf \left ( r \right ) \right ) r\,{\rm d}r\BesselK \left ( 0,{\frac {sr}{c}} \right ) ,s,t \right ) \right ) }\] 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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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 ✓
ClearAll["Global`*"]; pde = D[u[r, t], {t, 2}] == c^2*(D[u[r, t], {r, 2}] + 1/r*D[u[r, t], r]); ic = {u[r, 0] == 1, Derivative[0, 1][u][r, 0] == r/3}; bc = u[1, t] == 0; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[r, t], {r, t}], 60*10]]; sol = sol /. K[1] -> n; sol = FullSimplify[sol];
\[\left \{\left \{u(r,t)\to \underset {n=1}{\overset {\infty }{\sum }}\frac {2 J_0\left (r j_{0,n}\right ) \left (9 \sqrt {c^2} J_1\left (j_{0,n}\right ) \cos \left (c t j_{0,n}\right )+\, _1F_2\left (\frac {3}{2};1,\frac {5}{2};-\frac {1}{4} \left (j_{0,n}\right ){}^2\right ) \sin \left (\sqrt {c^2} t j_{0,n}\right )\right )}{9 \sqrt {c^2} \left (J_0\left (j_{0,n}\right ){}^2+J_1\left (j_{0,n}\right ){}^2\right ) j_{0,n}}\right \}\right \}\]
Maple ✓
restart; pde := diff(u(r, t), t$2) = c^2*( diff(u(r,t), r$2)+ (1/r)* diff(u(r,t),r) ); ic := u(r,0)=1, eval( diff(u(r,t),t),t=0)=r/3; bc := u(1,t)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic,bc], u(r, t)) assuming t>0,r>0,r<1),output='realtime'));
\[u \left ( r,t \right ) =-{\it invlaplace} \left ( {\frac {1}{s}\BesselI \left ( 0,{\frac {sr}{c}} \right ) \left ( \BesselI \left ( 0,{\frac {s}{c}} \right ) \right ) ^{-1}},s,t \right ) -1/3\,{\it invlaplace} \left ( {\frac {1}{{s}^{2}}\BesselI \left ( 0,{\frac {sr}{c}} \right ) \left ( \BesselI \left ( 0,{\frac {s}{c}} \right ) \right ) ^{-1}},s,t \right ) +1/6\,\pi \, \left ( {\it invlaplace} \left ( {\frac {1}{{s}^{3}}\BesselI \left ( 0,{\frac {sr}{c}} \right ) \StruveL \left ( 0,{\frac {s}{c}} \right ) \left ( \BesselI \left ( 0,{\frac {s}{c}} \right ) \right ) ^{-1}},s,t \right ) -{\it invlaplace} \left ( {\frac {1}{{s}^{3}}\StruveL \left ( 0,{\frac {sr}{c}} \right ) },s,t \right ) \right ) c+1+1/3\,tr\] Has unresolved Invlaplace calls
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Added January 12, 2020.
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 ✓
ClearAll["Global`*"]; c=2/10; a=1; g[r_]:=0; f[r_]:=r; pde = D[u[r, t], {t, 2}] == c^2*(D[u[r, t], {r, 2}] + 1/r*D[u[r, t], r]); ic = {u[r, 0] == f[r], Derivative[0, 1][u][r, 0] == g[r]}; bc = u[a, t] == 0; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[r, t], {r, t},Assumptions->{t>0,r>0}], 60*10]]; sol = sol /. K[1] -> n;
\[\left \{\left \{u(r,t)\to \underset {n=1}{\overset {\infty }{\sum }}\frac {2 J_0\left (r j_{0,n}\right ) \cos \left (\frac {t j_{0,n}}{5}\right ) \, _1F_2\left (\frac {3}{2};1,\frac {5}{2};-\frac {1}{4} \left (j_{0,n}\right ){}^2\right )}{3 \left (J_0\left (j_{0,n}\right ){}^2+J_1\left (j_{0,n}\right ){}^2\right )}\right \}\right \}\]
Maple ✓
restart; c:=2/10; a:=1; g:=r->0; f:=r->r; pde := diff(u(r, t), t$2) = c^2*( diff(u(r,t), r$2)+ (1/r)* diff(u(r,t),r) ); ic := u(r,0)=f(r), D[2](u)(r,0)=g(r); bc := u(a,t)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic,bc], u(r, t)) assuming t>0,r>0),output='realtime'));
\[u \left ( r,t \right ) ={\it invlaplace} \left ( {\frac {{\it laplace} \left ( -1,t,s \right ) \BesselI \left ( 0,5\,sr \right ) }{\BesselI \left ( 0,5\,s \right ) }},s,t \right ) +\int _{0}^{r-1}\!{\it invlaplace} \left ( {\frac {\BesselI \left ( 0,5\,s \left ( r-\tau \right ) \right ) }{\BesselI \left ( 0,5\,s \right ) }{\it laplace} \left ( - \left ( \tau +1 \right ) ^{-1},t,s \right ) },s,t \right ) \,{\rm d}\tau +r\] 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 1545 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
(*By Nasser M Abbasi*) SetDirectory[NotebookDirectory[]] (*Axis symmetric*) (*definitions*) ClearAll[a,c,n,m,r,theta,f,g,u] A0[n_,a_,lam0_]:= Module[{num,den,theta,r,f}, f=r; num=N[(BesselJ[1,lam0] (2 lam0-Pi StruveH[0,lam0])+ Pi BesselJ[0,lam0] StruveH[1,lam0])/(2 lam0^2)]; den=0.5 (BesselJ[0,lam0]^2+BesselJ[1,lam0]^2); num/den ]; u[r_,theta_,t_]:=Sum[A0tbl[[n]]*Cos[c lamtbl[[n]] t] *BesselJ[0,lamtbl[[n]]*r],{n,1,maxN}]; maxN=6; a=1; c=.2; lam[n_,a_]:=Module[{x}, x=BesselJZero[0,n]; N[(x/a)] ]; lamtbl=Table[lam[n,a],{n,1,maxN}]; A0tbl=Table[A0[n,a,lamtbl[[n]]],{n,1,maxN}]; t=.1 ParametricPlot3D[{r Cos[theta],r Sin[theta],Evaluate[u[r,theta,t]]},{r,0,1}, {theta,0,2 Pi},AxesLabel->{"r","theta","u"},ImageMargins->5, PerformanceGoal->"Quality",BoxRatios->{1,1,1}, PlotRange->{Automatic,Automatic,{-5,5}},Mesh->10,MaxRecursion->1] Animate[ParametricPlot3D[{r Cos[theta],r Sin[theta], Evaluate[u[r,theta,t]]},{r,0,1},{theta,0,2 Pi}, AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, PerformanceGoal->"Speed",BoxRatios->{1,1,1},PlotRange->{Automatic,Automatic,{-5,5}}, Mesh->10,MaxRecursion->1],{t,0,50,.01}] r=Table[ Labeled[ParametricPlot3D[{r Cos[theta],r Sin[theta], Evaluate[u[r,theta,t]]},{r,0,1},{theta,0,2 Pi},AxesLabel->{"r","theta","u"}, BaseStyle->15,ImageMargins->5,PerformanceGoal->"Speed",BoxRatios->{1,1,1}, PlotRange->{Automatic,Automatic,{-5,5}},Mesh->10,MaxRecursion->1], Row[{"time (sec)",Round@t}]],{t,0,40,.1}]; Export["anim_axis.gif",r,"DisplayDurations"->Table[.05,{Length[r]}]]
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Added Oct 6, 2019.
Taken from https://www.mapleprimes.com/posts/211274-Integral-Transforms-revamped-And-PDE
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 ✗
ClearAll["Global`*"]; pde = D[u[r, t], {r, 2}] + D[u[r, t], r]/r + D[u[r, t], {t, 2}] == -Q0*q[r]; ic = u[r, 0] == 0; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic}, u[r, t], {r, t}], 60*10]];
Failed
Maple ✓
restart; pde := diff(u(r, t), r$2) + diff(u(r, t), r)/r + diff(u(r, t), t$2) = -Q__0*q(r); iv := u(r, 0) = 0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,iv],u(r,t))),output='realtime')); sol:=convert(sol,Int,only = hankel);
\[u \left ( r,t \right ) =Q_{0}\, \left ( -\int _{0}^{\infty }\!{\frac {{{\rm e}^{-st}}\int _{0}^{\infty }\!q \left ( r \right ) r\BesselJ \left ( 0,sr \right ) \,{\rm d}r\BesselJ \left ( 0,sr \right ) }{s}}\,{\rm d}s+\int _{0}^{\infty }\!{\frac {\int _{0}^{\infty }\!q \left ( r \right ) r\BesselJ \left ( 0,sr \right ) \,{\rm d}r\BesselJ \left ( 0,sr \right ) }{s}}\,{\rm d}s \right ) \]
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Added Oct 6, 2019.
Taken from https://www.mapleprimes.com/posts/211274-Integral-Transforms-revamped-And-PDE
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 ✗
ClearAll["Global`*"]; pde = c^2*(D[u[r, t], {r, 2}] + D[u[r, t], r]/r) == D[u[r, t], {t, 2}]; ic = {u[r, 0] == A*a*(a^2 + r^2)^(-1/2), Derivative[0, 1][u][r, 0] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic}, u[r, t], {r, t}], 60*10]];
Failed
Maple ✓
restart; pde := c^2*(diff(u(r, t), r, r) + diff(u(r, t), r)/r) = diff(u(r, t), t, t); iv := u(r, 0) = A*a*(a^2 + r^2)^(-1/2), D[2](u)(r, 0) = 0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,iv],u(r,t),method = Hankel) assuming (0 < r, 0 < t, 0 < a) ),output='realtime'));
\[u \left ( r,t \right ) =1/2\,{\frac {Aa \left ( \sqrt {-{c}^{2}{t}^{2}+2\,iact+{a}^{2}+{r}^{2}}+\sqrt {-{c}^{2}{t}^{2}-2\,iact+{a}^{2}+{r}^{2}} \right ) }{\sqrt {-{c}^{2}{t}^{2}-2\,iact+{a}^{2}+{r}^{2}}\sqrt {-{c}^{2}{t}^{2}+2\,iact+{a}^{2}+{r}^{2}}}}\]
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Added January 11, 2020
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 ✓
ClearAll["Global`*"]; pde = D[u[r, theta, t], {t, 2}] == c^2*Laplacian[u[r,theta,t],{r,theta},"Polar"]; ic = {u[r, theta, 0] == f[r, theta], Derivative[0, 0, 1][u][r, theta, 0] == g[r,theta]}; bc = {u[a, theta, t] == 0, u[r, -Pi, t] == u[r, Pi, t], Derivative[0, 1, 0][u][r, -Pi, t] == Derivative[0, 1, 0][u][r, Pi, t]}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[r, theta, t], {r, theta, t}, Assumptions -> {0 < r < a, a > 0, t > 0, -Pi < theta < Pi}], 60*10]];
\[\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 ✗
restart; pde := diff(u(r, theta, t), t$2) = c^2*VectorCalculus:-Laplacian(u(r,theta,t),'polar'[r,theta]); ic := u(r, theta, 0) = f(r, theta) , (D[3](u))(r, theta, 0) = g(r,theta); bc := u(a, theta, t) = 0, u(r, -Pi, t) = u(r, Pi, t), (D[2](u))(r, -Pi, t) = (D[2](u))(r, Pi, t); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic,bc], u(r, theta ,t),HINT = boundedseries(r=0))),output='realtime'));
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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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 ✓
ClearAll["Global`*"]; pde = D[u[r, theta, t], {t, 2}] == c^2*Laplacian[u[r,theta,t],{r,theta},"Polar"]; ic = {u[r, theta, 0] == f[r, theta], Derivative[0, 0, 1][u][r, theta, 0] == 0}; bc = {u[a, theta, t] == 0, u[r, -Pi, t] == u[r, Pi, t], Derivative[0, 1, 0][u][r, -Pi, t] == Derivative[0, 1, 0][u][r, Pi, t]}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[r, theta, t], {r, theta, t}, Assumptions -> {0 < r < a, a > 0, t > 0, -Pi < theta < Pi}], 60*10]];
\[\left \{\left \{u(r,\theta ,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[3]=1}{\overset {\infty }{\sum }}\frac {2 J_0\left (\frac {r j_{0,K[3]}}{a}\right ) \cos \left (\frac {\sqrt {c^2} t j_{0,K[3]}}{a}\right ) \int _0^a\int _{-\pi }^{\pi }r J_0\left (\frac {r j_{0,K[3]}}{a}\right ) f(r,\theta )d\theta dr}{a^2 \pi J_1\left (j_{0,K[3]}\right ){}^2}+\underset {K[3]=1}{\overset {\infty }{\sum }}\left (\underset {K[1]=1}{\overset {\infty }{\sum }}\left (\frac {2 J_{K[1]}\left (\frac {r j_{K[1],K[3]}}{a}\right ) \cos \left (\frac {\sqrt {c^2} t j_{K[1],K[3]}}{a}\right ) \cos (\theta K[1]) \int _0^a\int _{-\pi }^{\pi }r J_{K[1]}\left (\frac {r j_{K[1],K[3]}}{a}\right ) \cos (\theta K[1]) f(r,\theta )d\theta dr}{a^2 \pi J_{K[1]-1}\left (j_{K[1],K[3]}\right ){}^2}+\frac {2 J_{K[1]}\left (\frac {r j_{K[1],K[3]}}{a}\right ) \cos \left (\frac {\sqrt {c^2} t j_{K[1],K[3]}}{a}\right ) \left (\int _0^a\int _{-\pi }^{\pi }r J_{K[1]}\left (\frac {r j_{K[1],K[3]}}{a}\right ) f(r,\theta ) \sin (\theta K[1])d\theta dr\right ) \sin (\theta K[1])}{a^2 \pi J_{K[1]-1}\left (j_{K[1],K[3]}\right ){}^2}\right )\right ) & (K[1]|K[3])\in \mathbb {Z}\land K[1]\geq 1\land K[3]\geq 1\land c^2 \left (j_{K[1],K[3]}\right ){}^2>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
Maple ✗
restart; pde := diff(u(r, theta, t), t$2) =c^2*VectorCalculus:-Laplacian(u(r,theta,t),'polar'[r,theta]); ic := u(r, theta, 0) = f(r, theta) , (D[3](u))(r, theta, 0) = 0; bc := u(a, theta, t) = 0, u(r, -Pi, t) = u(r, Pi, t), (D[2](u))(r, -Pi, t) = (D[2](u))(r, Pi, t); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic,bc], u(r, theta ,t),HINT = boundedseries(r=0))),output='realtime'));
sol=()
Hand solution
The basic solution for this type of PDE was already given in problem 5.2.2.6 on page 1557 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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Added January 11 2020.
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 ✓
ClearAll["Global`*"]; f[r_,theta_]:=r*theta; c=2/10; a=1; pde = D[u[r, theta, t], {t, 2}] == c^2*Laplacian[u[r,theta,t],{r,theta},"Polar"]; ic = {u[r, theta, 0] == f[r, theta], Derivative[0, 0, 1][u][r, theta, 0] == 0}; bc = {u[a, theta, t] == 0, u[r, -Pi, t] == u[r, Pi, t], Derivative[0, 1, 0][u][r, -Pi, t] == Derivative[0, 1, 0][u][r, Pi, t]}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[r, theta, t], {r, theta, t}, Assumptions -> {0 < r < a,t > 0, -Pi < theta < Pi}], 60*10]];
\[\left \{\left \{u(r,\theta ,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[3]=1}{\overset {\infty }{\sum }}\left (\underset {K[1]=1}{\overset {\infty }{\sum }}\frac {(-1)^{K[1]+1} J_{K[1]}\left (r j_{K[1],K[3]}\right ) j_{K[1],K[3]} \cos \left (\frac {1}{5} t j_{K[1],K[3]}\right ) \text {Gamma}\left (\frac {1}{2} (K[1]+3)\right ) \, _1\tilde {F}_2\left (\frac {1}{2} (K[1]+3);\frac {1}{2} (K[1]+5),K[1]+1;-\frac {1}{4} \left (j_{K[1],K[3]}\right ){}^2\right ) \sin (\theta K[1])}{J_{K[1]-1}\left (j_{K[1],K[3]}\right ) \, _0\tilde {F}_1\left (;K[1];-\frac {1}{4} \left (j_{K[1],K[3]}\right ){}^2\right ) K[1]}\right ) & (K[1]|K[3])\in \mathbb {Z}\land K[1]\geq 1\land K[3]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
Maple ✗
restart; f:=(r,theta)->r*theta; c:=2/10; a:=1; pde := diff(u(r, theta, t), t$2) = c^2*VectorCalculus:-Laplacian(u(r,theta,t),'polar'[r,theta]); ic := u(r, theta, 0) = f(r, theta) , (D[3](u))(r, theta, 0) = 0; bc := u(a, theta, t) = 0, u(r, -Pi, t) = u(r, Pi, t), (D[2](u))(r, -Pi, t) = (D[2](u))(r, Pi, t); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic,bc], u(r, theta ,t),HINT = boundedseries(r=0))),output='realtime'));
sol=()
Hand solution
The basic solution for this type of PDE was already given in problem 5.2.2.6 on page 1557 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)
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Source code for all the above animations
(*By Nasser M. Abbasi*) SetDirectory[NotebookDirectory[]] X:\data\public_html\my_notes\PDE_animations\problems\4 (*definitions*) ClearAll[a,c,n,m,r,theta,f,g,u,maxM,maxN,t] maxN=4; maxM=4; a=1; c=.2; minZ=-8; maxZ=10; A0[n_,m_]:= Module[{num,den,r,theta,f}, f=r*theta; num=Integrate[f* BesselJ[n,lamtbl[[n+1,m]] r] Cos[n theta] r,{r,0,a},{theta,0,2Pi}]; den=Integrate[(BesselJ[n,lamtbl[[n+1,m]] r])^2 (Cos[n theta])^2 r,{r,0,a},{theta,0,2Pi}]; num/den ]; C0[n_,m_]:= Module[{num,den,f,r,theta}, f=r*theta; num=Integrate[f* BesselJ[n,lamtbl[[n+1,m]] r] Sin[n theta] r,{r,0,a},{theta,0,2Pi}]; den=Integrate[(BesselJ[n,lamtbl[[n+1,m]] r])^2 (Sin[n theta])^2 r,{r,0,a},{theta,0,2Pi}]; num/den ]; lam[n_,m_]:=Module[{x}, x=BesselJZero[n,m]; N[(x/a)] ]; u[r_,theta_,t_,maxN_,maxM_]:=Module[{tmp,n,m}, tmp=Sum[A0tbl[[n+1,m]]*Cos[c lamtbl[[n+1,m]] t] *BesselJ[n,lamtbl[[n+1,m]]*r]* Cos[n*theta],{n,0,maxN},{m,1,maxM}]; tmp=tmp+Sum[C0tbl[[n+1,m]]*Cos[c lamtbl[[n+1,m]] t] *BesselJ[n,lamtbl[[n+1,m]]*r]* Sin[n*theta],{n,1,maxN},{m,1,maxM}] ]; lamtbl=Table[lam[n,m],{n,0,maxN},{m,1,maxN}]; A0tbl=Table[A0[n,m],{n,0,maxN},{m,1,maxM}] C0tbl=Table[C0[n,m],{n,1,maxN},{m,1,maxM}] (*n=0,m=1*) ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0],Evaluate[u[r0,theta0,12,0,1]]}, {r0,0,1},{theta0,0,2 Pi},AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1}, PlotRange->{Automatic,Automatic,{minZ,maxZ}}] r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], Evaluate[u[r0,theta0,t,0,1]]},{r0,0,1},{theta0,0,2 Pi},AxesLabel->{"r","theta","u"}, BaseStyle->15,ImageMargins->5,PerformanceGoal->"Quality",Mesh->10, MaxRecursion->1, BoxRatios->{1,1,1},PlotRange->{Automatic,Automatic,{minZ,maxZ}}], Row[{"time (sec)",Round@t,", N = ", 0, ", M = ",1}]],{t,0,80,.5}]; Export["anim_n_0_m_1.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] (*n=0,m=2*) ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0],Evaluate[u[r0,theta0,0,0,2]]},{r0,0,1}, {theta0,0,2 Pi},AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1}, PlotRange->{Automatic,Automatic,{minZ,maxZ}}] (*to speed it up, make Z in {t,0,100,Z} larger, and then make Z in Table[Z,{Length[r]}] smaller.*) r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], Evaluate[u[r0,theta0,t,0,2]]},{r0,0,1},{theta0,0,2 Pi},AxesLabel->{"r","theta","u"}, BaseStyle->15,ImageMargins->5,PerformanceGoal->"Quality",Mesh->10, MaxRecursion->1, BoxRatios->{1,1,1},PlotRange->{Automatic,Automatic,{minZ,maxZ}}], Row[{"time (sec)",Round@t,", N = ", 0, ", M = ",2}]],{t,0,80,.5}]; Export["anim_n_0_m_2.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] (*n=0,m=3*) r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], Evaluate[u[r0,theta0,t,0,3]]},{r0,0,1},{theta0,0,2 Pi}, AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1}, PlotRange->{Automatic,Automatic,{minZ,maxZ}}], Row[{"time (sec)",Round@t,", N = ", 0, ", M = ",3}]],{t,0,80,.5}]; Export["anim_n_0_m_3.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] (*n=0,m=4*) t=1; ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0],Evaluate[u[r0,theta0,t,0,4]]}, {r0,0,1},{theta0,0,2 Pi},AxesLabel->{"r","theta","u"},BaseStyle->15, ImageMargins->5,PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1},PlotRange->{Automatic,Automatic,{minZ,maxZ}}] r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], Evaluate[u[r0,theta0,t,0,4]]},{r0,0,1},{theta0,0,2 Pi},AxesLabel->{"r","theta","u"}, BaseStyle->15,ImageMargins->5,PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1},PlotRange->{Automatic,Automatic,{minZ,maxZ}}], Row[{"time (sec)",Round@t,", N = ", 0, ", M = ",4}]],{t,0,80,.5}]; Export["anim_n_0_m_4.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] (*n=1,m=1*) t=1; ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], Evaluate[u[r0,theta0,t,1,1]]},{r0,0,1},{theta0,0,2 Pi}, AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, PerformanceGoal->"Speed",Mesh->10, BoxRatios->{1,1,1}, PlotRange->{Automatic,Automatic,{minZ,maxZ}}] r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], Evaluate[u[r0,theta0,t,1,1]]},{r0,0,1},{theta0,0,2 Pi}, AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1},PlotRange->{Automatic,Automatic,{minZ,maxZ}}], Row[{"time (sec)",Round@t,", N = ", 1, ", M = ",1}]],{t,0,80,.5}]; Export["anim_n_1_m_1.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] (*n=1,m=2*) r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], Evaluate[u[r0,theta0,t,1,2]]},{r0,0,1},{theta0,0,2 Pi}, AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1},PlotRange->{Automatic,Automatic,{minZ,maxZ}}], Row[{"time (sec)",Round@t,", N = ", 1, ", M = ",2}]],{t,0,80,.5}]; Export["anim_n_1_m_2.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] (*n=1,m=3*) r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], Evaluate[u[r0,theta0,t,1,3]]},{r0,0,1},{theta0,0,2 Pi}, AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1}, PlotRange->{Automatic,Automatic,{minZ,maxZ}}], Row[{"time (sec)",Round@t,", N = ", 1, ", M = ",3}]],{t,0,80,.5}]; Export["anim_n_1_m_3.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] (*n=1,m=4*) r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], Evaluate[u[r0,theta0,t,1,4]]},{r0,0,1},{theta0,0,2 Pi}, AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1},PlotRange->{Automatic,Automatic,{minZ,maxZ}}], Row[{"time (sec)",Round@t,", N = ", 1, ", M = ",4}]],{t,0,80,.5}]; Export["anim_n_1_m_4.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] (*n=2,m=1*) r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], Evaluate[u[r0,theta0,t,2,1]]},{r0,0,1},{theta0,0,2 Pi}, AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1}, PlotRange->{Automatic,Automatic,{minZ,maxZ}}], Row[{"time (sec)",Round@t,", N = ", 2 ", M = ",1}]] ,{t,0,80,.5}]; Export["anim_n_2_m_1.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] (*n=2,m=2*) r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], Evaluate[u[r0,theta0,t,2,2]]},{r0,0,1},{theta0,0,2 Pi}, AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1}, PlotRange->{Automatic,Automatic,{minZ,maxZ}}], Row[{"time (sec)",Round@t,", N = ", 2 ", M = ",2}]],{t,0,80,.5}]; Export["anim_n_2_m_2.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] (*n=2,m=3*) r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], Evaluate[u[r0,theta0,t,2,3]]},{r0,0,1},{theta0,0,2 Pi},AxesLabel->{"r","theta","u"}, BaseStyle->15,ImageMargins->5,PerformanceGoal->"Quality",Mesh->10, MaxRecursion->1, BoxRatios->{1,1,1},PlotRange->{Automatic,Automatic,{minZ,maxZ}}], Row[{"time (sec)",Round@t,", N = ", 2 ", M = ",3}]],{t,0,80,.5}]; Export["anim_n_2_m_3.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] (*n=2,m=4*) r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], Evaluate[u[r0,theta0,t,2,4]]},{r0,0,1},{theta0,0,2 Pi}, AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1}, PlotRange->{Automatic,Automatic,{minZ,maxZ}}], Row[{"time (sec)",Round@t,", N = ", 2 ", M = ",4}]],{t,0,80,.5}]; Export["anim_n_2_m_4.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] (*n=3,m=1*) r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], Evaluate[u[r0,theta0,t,3,1]]},{r0,0,1},{theta0,0,2 Pi}, AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1}, PlotRange->{Automatic,Automatic,{minZ,maxZ}}], Row[{"time (sec)",Round@t,", N = ", 3 ", M = ",1}]],{t,0,80,.5}]; Export["anim_n_3_m_1.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] (*n=3,m=2*) r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], Evaluate[u[r0,theta0,t,3,2]]},{r0,0,1},{theta0,0,2 Pi}, AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1},PlotRange->{Automatic,Automatic,{minZ,maxZ}}], Row[{"time (sec)",Round@t,", N = ", 3 ", M = ",2}]],{t,0,80,.5}]; Export["anim_n_3_m_2.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] (*n=3,m=3*) r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], Evaluate[u[r0,theta0,t,3,3]]},{r0,0,1},{theta0,0,2 Pi}, AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1}, PlotRange->{Automatic,Automatic,{minZ,maxZ}}], Row[{"time (sec)",Round@t,", N = ", 3 ", M = ",3}]],{t,0,80,.5}]; Export["anim_n_3_m_3.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] (*n=3,m=4*) r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], Evaluate[u[r0,theta0,t,3,4]]},{r0,0,1},{theta0,0,2 Pi}, AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1}, PlotRange->{Automatic,Automatic,{minZ,maxZ}}], Row[{"time (sec)",Round@t,", N = ", 3 ", M = ",4}]],{t,0,80,.5}]; Export["anim_n_3_m_4.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]]
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Added January 11, 2020
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 ✓
ClearAll["Global`*"]; a=1; c=1; epsilon=1/2; f[r_,theta_]:=0; g[r_,theta_]:=Piecewise[{{1/(Pi*epsilon^2),r<epsilon},{0,True}}]; c=1; a=1; pde = D[u[r, theta, t], {t, 2}] == c^2*Laplacian[u[r,theta,t],{r,theta},"Polar"]; ic = {u[r, theta, 0] == f[r, theta], Derivative[0, 0, 1][u][r, theta, 0] == g[r,theta]}; bc = {u[a, theta, t] == 0, u[r, -Pi, t] == u[r, Pi, t], Derivative[0, 1, 0][u][r, -Pi, t] == Derivative[0, 1, 0][u][r, Pi, t]}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[r, theta, t], {r, theta, t}, Assumptions -> {0 < r < a,t > 0, -Pi < theta < Pi}], 60*10]];
\[\left \{\left \{u(r,\theta ,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[3]=1}{\overset {\infty }{\sum }}\frac {2 J_0\left (r j_{0,K[3]}\right ) \, _0\tilde {F}_1\left (;2;-\frac {1}{16} \left (j_{0,K[3]}\right ){}^2\right ) \sin \left (t j_{0,K[3]}\right )}{\pi J_1\left (j_{0,K[3]}\right ){}^2 j_{0,K[3]}} & (K[1]|K[3])\in \mathbb {Z}\land K[1]\geq 1\land K[3]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
Maple ✗
restart; c:=1; a:=1; epsilon:=1/2; f:=(r,theta)->r*theta; g:=(r,theta)->piecewise(r<epsilon,1/(Pi*epsilon^2),true,0); pde := diff(u(r, theta, t), t$2) = c^2*VectorCalculus:-Laplacian(u(r,theta,t),'polar'[r,theta]); ic := u(r, theta, 0) = f(r, theta) , (D[3](u))(r, theta, 0) = g(r,theta); bc := u(a, theta, t) = 0, u(r, -Pi, t) = u(r, Pi, t), (D[2](u))(r, -Pi, t) = (D[2](u))(r, Pi, t); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic,bc], u(r, theta ,t),HINT = boundedseries(r=0))),output='realtime'));
sol=()
Hand solution
The basic solution for this type of PDE was already given in problem 5.2.2.6 on page 1557 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
(*By Nasser M. Abbasi. Animation of problem 4 solution*) ClearAll[t,r,m]; padIt2[v_,f_List]:=AccountingForm[v,f,NumberSigns->{"",""}, NumberPadding->{"0","0"},SignPadding->True]; nTerms=40; lam = Table[ BesselJZero[0,m],{m,1,nTerms}]//N; c = Table[1/lam[[m]]^2 BesselJ[1,lam[[m]]/2]/BesselJ[1,lam[[m]] ]^2,{m,1,nTerms}]; mySol[r_,t_]:=4/Pi Sum[c[[m]]BesselJ[0,lam[[m]] r] Sin[lam[[m]] t],{m,1,nTerms}]; frames=Table[ Print["t=",t]; Grid[{ {Row[{"time = ",padIt2[t,{3,2}]}]}, {ParametricPlot3D[{r Cos[theta],r Sin[theta],mySol[r,t]},{r,0,1},{theta,0,2 Pi}, PlotRange->{Automatic,Automatic,{-0.6,0.6}}, PerformanceGoal->"Speed",Boxed->True, Axes->True,Mesh->20, ViewPoint->{2.17,-2.4,1}, ImageSize->400, BoxRatios->{1, 1, 1}] }}], {t,0,5,0.05} ]; Manipulate[ frames[[i]], {{i,1,"time"},1,Length@frames,1,Appearance->"Labeled"} ] Export["anim.gif",frames,"DisplayDurations"->Table[.2,{Length@frames}]]
Here is the same animation made in Maple 2018
Maple source code for all the above animations
#by Nasser M. Abbasi, May 23,2018 restart; currentdir("X:/data/public_html/my_notes/PDE_animations/problems/wave_disk_exam_problem_4"); nTerms := 20: lam := evalf([BesselJZeros(0,1..nTerms)]): c := seq(1/lam[n]^2*BesselJ(1,lam[n]/2)/BesselJ(1,lam[n])^2,n=1..nTerms): mySol := proc(r,t) local n; 4/Pi*sum(c[n]*BesselJ(0,lam[n]*r)*sin(lam[n]*t),n=1..nTerms); end proc: maxTime := 5: (*seconds*) delay := 0.05: nFrames := round(maxTime/delay): frames := [seq( plot3d([ r, theta, mySol(r,(i*delay)) ], r = 0..1, theta = 0..2*Pi, coords = cylindrical, axes = none, title = sprintf("%s %3.2f %s","time ",(i*delay),"seconds") ), i=0..nFrames-1) ]: plots:-display(convert(frames,list),insequence=true);
Here is the same animation made in Matlab 2016a
Matlab source code for all the above animations
function nma_HW4_math_322 %By Nasser M. Abbasi, May 23, 2018 close all; GENERATE_GIF=true; %turn to false to not generate animated gif lam = zeros(80,1); %eigenvalues for i = 1:80 lam(i) = fzero(@(x)besselj(0,x),i); end; lam = uniquetol(lam); %must use uniquetol nTerms = 20; c = zeros(nTerms,1); for i = 1:nTerms c(i) = 1/lam(i)^2*besselj(1,lam(i)/2)/besselj(1,lam(i))^2; end %------------- inner function -------- function tot = mySol(r,t) tot = 0; for ii =1:nTerms tot = tot + (c(ii)*besselj(0,lam(ii).*r).*sin(lam(ii)*t)); end; tot = 4/pi*tot; end %------------------- maxTime = 5; %seconds delay = 0.05; nFrames = round(maxTime/delay); r = 0:.05:1; phi = 0:pi/20:2*pi; [R,PHI] = meshgrid(r,phi); fig_handle = figure(); set(fig_handle,'Name',.... 'Math 322, Final exam problem 4 animations by Nasser M. Abbasi'); for i=1:nFrames Z = mySol(R,((i-1)*delay)); surf(R.*cos(PHI), R.*sin(PHI), Z); set(gca,'nextplot','replacechildren','visible','on'); colormap cool ; title(sprintf('time = %3.2f', (i-1)*delay)); zlim([-0.6 0.6]); drawnow; pause(.01); if GENERATE_GIF frame = getframe(gcf); im = frame2im(frame); [imind,cm] = rgb2ind(im,256); if i == 1 imwrite(imind,cm,'matlab_animations.gif','gif', ... 'DelayTime',0.1,'LoopCount',0); else imwrite(imind,cm,'matlab_animations.gif','gif',... 'WriteMode','append','DelayTime',0.1,'LoopCount',0); end end end end
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Added January 15, 2020
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 ✗
ClearAll["Global`*"]; pde = D[u[r, theta, t], {t, 2}] == c^2*Laplacian[u[r,theta,t],{r,theta},"Polar"]+Q[r,theta,t]; ic = {u[r, theta, 0] == f[r, theta], Derivative[0, 0, 1][u][r, theta, 0] == 0}; bc = {u[a, theta, t] == 0, u[r, -Pi, t] == u[r, Pi, t], Derivative[0, 1, 0][u][r, -Pi, t] == Derivative[0, 1, 0][u][r, Pi, t]}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[r, theta, t], {r, theta, t}, Assumptions -> {0 < r < a,t > 0, -Pi < theta < Pi}], 60*10]];
Failed
Maple ✗
restart; pde := diff(u(r, theta, t), t$2) = c^2*VectorCalculus:-Laplacian(u(r,theta,t),'polar'[r,theta])+Q(r,theta,t); ic := u(r, theta, 0) = f(r, theta) , (D[3](u))(r, theta, 0) = 0; bc := u(a, theta, t) = 0, u(r, -Pi, t) = u(r, Pi, t), (D[2](u))(r, -Pi, t) = (D[2](u))(r, Pi, t); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic,bc], u(r, theta ,t),HINT = boundedseries(r=0)) assuming r>0,r<a),output='realtime'));
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.