7.3: Singular Points and the Method of Frobenius

We seek a Frobenius-type solution of the form y=∞∑k=0akxk+r. · The obtained series must be zero. · If the indicial equation has two real roots r ... Skiptomaincontent ExamplesExample\(\PageIndex{1}\)Example\(\PageIndex{2}\)MethodofFrobeniusExample\(\PageIndex{3}\):ExpansionaroundaregularsingularpointTheorem\(\PageIndex{1}\)BesselFunctionsExercise\(\PageIndex{1}\)Exercise\(\PageIndex{2}\)Example\(\PageIndex{4}\):UsingBesselfunctionstoSolveaODEFootnotes Examples WhilebehaviorofODEsatsingularpointsismorecomplicated,certainsingularpointsarenotespeciallydifficulttosolve.Letuslookatsomeexamplesbeforegivingageneralmethod.Wemaybeluckyandobtainapowerseriessolutionusingthemethodoftheprevioussection,butingeneralwemayhavetotryotherthings. Example\(\PageIndex{1}\) Letusfirstlookatasimplefirstorderequation \[2xy'-y=0.\label{ex1eq1}\] Notethat\(x=0\)isasingularpoint.Ifweonlytrytoplugin \[y=\sum_{k=0}^\inftya_kx^k,\label{ex1eq2}\] weobtain \[\begin{align}\begin{aligned}0=2xy'-y&=2x\,\left(\sum_{k=1}^\inftyka_kx^{k-1}\right)-\left(\sum_{k=0}^\inftya_kx^k\right)\\&=a_0+\sum_{k=1}^\infty(2ka_k-a_k)\,x^{k}.\end{aligned}\end{align}\label{ex1eq3}\] First,\(a_0=0\).Next,theonlywaytosolve\(0=2ka_k-a_k=(2k-1)\,a_k\)for\(k=1,2,3,\dots\)isfor\(a_k=0\)forall\(k\).Thereforeweonlygetthetrivialsolution\(y=0\).Weneedanonzerosolutiontogetthegeneralsolution. Letustry\(y=x^r\)forsomerealnumber\(r\).Consequentlyoursolution---ifwecanfindone---mayonlymakesenseforpositive\(x\).Then\(y'=rx^{r-1}\).So \[0=2xy'-y=2xrx^{r-1}-x^r=(2r-1)x^r.\nonumber\] Therefore\(r=\dfrac{1}{2}\),orinotherwords\(y=x^{1/2}\).Multiplyingbyaconstant,thegeneralsolutionforpositive\(x\)is \[y=Cx^{1/2}.\nonumber\] If\(C\not=0\)thenthederivativeofthesolution"blowsup"at\(x=0\)(thesingularpoint).Thereisonlyonesolutionthatisdifferentiableat\(x=0\)andthat'sthetrivialsolution\(y=0\). Noteveryproblemwithasingularpointhasasolutionoftheform\(y=x^r\),ofcourse.Butperhapswecancombinethemethods.Whatwewilldoistotryasolutionoftheform \[y=x^rf(x)\nonumber\] where\(f(x)\)isananalyticfunction. Example\(\PageIndex{2}\) Supposethatwehavetheequation \[4x^2y''-4x^2y'+(1-2x)y=0,\label{ex2eq1}\] andagainnotethat\(x=0\)isasingularpoint.Letustry \[y=x^r\sum_{k=0}^\inftya_kx^k=\sum_{k=0}^\inftya_kx^{k+r},\label{ex2eq2}\] where\(r\)isarealnumber,notnecessarilyaninteger.Againifsuchasolutionexists,itmayonlyexistforpositive\(x\).Firstletusfindthederivatives \[\begin{align}y'&=\sum_{k=0}^\infty(k+r)\,a_kx^{k+r-1},\nonumber\\y''&=\sum_{k=0}^\infty(k+r)\,(k+r-1)\,a_kx^{k+r-2}.\label{ex2eq3}\end{align}\] PluggingEquations\ref{ex2eq2}-\ref{ex2eq3}intoouroriginaldifferentialequation(Equation\ref{ex2eq1})weobtain \[\begin{align}\begin{aligned}0&=4x^2y''-4x^2y'+(1-2x)y\\&=4x^2\,\left(\sum_{k=0}^\infty(k+r)\,(k+r-1)\,a_kx^{k+r-2}\right)-4x^2\,\left(\sum_{k=0}^\infty(k+r)\,a_kx^{k+r-1}\right)+(1-2x)\left(\sum_{k=0}^\inftya_kx^{k+r}\right)\\&=\left(\sum_{k=0}^\infty4(k+r)\,(k+r-1)\,a_kx^{k+r}\right)-\left(\sum_{k=0}^\infty4(k+r)\,a_kx^{k+r+1}\right)+\left(\sum_{k=0}^\inftya_kx^{k+r}\right)-\left(\sum_{k=0}^\infty2a_kx^{k+r+1}\right)\\&=\left(\sum_{k=0}^\infty4(k+r)\,(k+r-1)\,a_kx^{k+r}\right)-\left(\sum_{k=1}^\infty4(k+r-1)\,a_{k-1}x^{k+r}\right)+\left(\sum_{k=0}^\inftya_kx^{k+r}\right)-\left(\sum_{k=1}^\infty2a_{k-1}x^{k+r}\right)\\&=4r(r-1)\,a_0x^r+a_0x^r+\sum_{k=1}^\infty\left(4(k+r)\,(k+r-1)\,a_k-4(k+r-1)\,a_{k-1}+a_k-2a_{k-1}\right)\,x^{k+r}\\&=\left(4r(r-1)+1\right)\,a_0x^r+\sum_{k=1}^\infty\left(\left(4(k+r)\,(k+r-1)+1\right)\,a_k-\left(4(k+r-1)+2\right)\,a_{k-1}\right)\,x^{k+r}.\end{aligned}\end{align}\nonumber\] Tohaveasolutionwemustfirsthave\(\left(4r(r-1)+1\right)\,a_0=0\).Supposingthat\(a_0\not=0\)weobtain \[4r(r-1)+1=0.\nonumber\] Thisequationiscalledtheindicialequation.Thisparticularindicialequationhasadoublerootat\(r=\dfrac{1}{2}\). OK,soweknowwhat\(r\)hastobe.Thatknowledgeweobtainedsimplybylookingatthecoefficientof\(x^r\).Allothercoefficientsof\(x^{k+r}\)alsohavetobezeroso \[\left(4(k+r)\,(k+r-1)+1\right)\,a_k-\left(4(k+r-1)+2\right)\,a_{k-1}=0.\nonumber\] Ifweplugin\(r=\dfrac{1}{2}\)andsolvefor\(a_k\)weget \[a_k=\dfrac{4(k+\dfrac{1}{2}-1)+2}{4(k+\dfrac{1}{2})\,(k+\dfrac{1}{2}-1)+1}\,a_{k-1}=\dfrac{1}{k}\,a_{k-1}.\nonumber\] Letusset\(a_0=1\).Then \[a_1=\dfrac{1}{1}a_0=1,\qquada_2=\dfrac{1}{2}a_1=\dfrac{1}{2},\qquada_3=\dfrac{1}{3}a_2=\dfrac{1}{3\cdot2},\qquada_4=\dfrac{1}{4}a_3=\dfrac{1}{4\cdot3\cdot2},\qquad\dots\nonumber\] Extrapolating,wenoticethat \[a_k=\dfrac{1}{k(k-1)(k-2)\cdots3\cdot2}=\dfrac{1}{k!}.\nonumber\] Inotherwords, \[y=\sum_{k=0}^\inftya_kx^{k+r}=\sum_{k=0}^\infty\dfrac{1}{k!}x^{k+1/2}=x^{1/2}\sum_{k=0}^\infty\dfrac{1}{k!}x^{k}=x^{1/2}e^x.\nonumber\] Thatwaslucky!Ingeneral,wewillnotbeabletowritetheseriesintermsofelementaryfunctions.Wehaveonesolution,letuscallit\(y_1=x^{1/2}e^x\).Butwhataboutasecondsolution?Ifwewantageneralsolution,weneedtwolinearlyindependentsolutions.Picking\(a_0\)tobeadifferentconstantonlygetsusaconstantmultipleof\(y_1\),andwedonothaveanyother\(r\)totry;weonlyhaveonesolutiontotheindicialequation.Well,therearepowersof\(x\)floatingaroundandwearetakingderivatives,perhapsthelogarithm(theantiderivativeof\(x^{-1}\))isaroundaswell.Itturnsoutwewanttotryforanothersolutionoftheform \[y_2=\sum_{k=0}^\inftyb_kx^{k+r}+(\lnx)y_1,\nonumber\] whichinourcaseis \[y_2=\sum_{k=0}^\inftyb_kx^{k+1/2}+(\lnx)x^{1/2}e^x.\nonumber\] Wenowdifferentiatethisequation,substituteintothedifferentialequationandsolvefor\(b_k\).Alongcomputationensuesandweobtainsomerecursionrelationfor\(b_k\).Thereadercan(andshould)trythistoobtainforexamplethefirstthreeterms \[b_1=b_0-1,\qquadb_2=\dfrac{2b_1-1}{4},\qquadb_3=\dfrac{6b_2-1}{18},\qquad\ldots\nonumber\] Wethenfix\(b_0\)andobtainasolution\(y_2\).Thenwewritethegeneralsolutionas\(y=Ay_1+By_2\). MethodofFrobenius Beforegivingthegeneralmethod,letusclarifywhenthemethodapplies.Let \[p(x)y''+q(x)y'+r(x)y=0\nonumber\] beanODE.Asbefore,if\(p(x_0)=0\),then\(x_0\)isasingularpoint.If,furthermore,thelimits \[\lim_{x\tox_0}~(x-x_0)\dfrac{q(x)}{p(x)}\qquad\text{and}\qquad\lim_{x\tox_0}~(x-x_0)^2\dfrac{r(x)}{p(x)}\nonumber\] bothexistandarefinite,thenwesaythat\(x_0\)isaregularsingularpoint. Example\(\PageIndex{3}\):Expansionaroundaregularsingularpoint Often,andfortherestofthissection,\(x_0=0\).Consider \[x^2y''+x(1+x)y'+(\pi+x^2)y=0.\nonumber\] Write \[\begin{align}\begin{aligned}\lim_{x\to0}~x\dfrac{q(x)}{p(x)}&=\lim_{x\to0}~x\dfrac{x(1+x)}{x^2}=\lim_{x\to0}~(1+x)=1,\\\lim_{x\to0}~x^2\dfrac{r(x)}{p(x)}&=\lim_{x\to0}~x^2frac{(\pi+x^2)}{x^2}=\lim_{x\to0}~(\pi+x^2)=\pi\end{aligned}\end{align}.\nonumber\] So\(x=0\)isaregularsingularpoint. Ontheotherhandifwemaketheslightchange \[x^2y''+(1+x)y'+(\pi+x^2)y=0,\nonumber\] then \[\lim_{x\to0}~x\dfrac{q(x)}{p(x)}=\lim_{x\to0}~x\dfrac{(1+x)}{x^2}=\lim_{x\to0}~\dfrac{1+x}{x}=\text{DNE}.\nonumber\] HereDNEstandsfordoesnotexist.Thepoint\(0\)isasingularpoint,butnotaregularsingularpoint. LetusnowdiscussthegeneralMethodofFrobenius\(^{1}\).Letusonlyconsiderthemethodatthepoint\(x=0\)forsimplicity.Themainideaisthefollowingtheorem. Theorem\(\PageIndex{1}\) MethodofFrobenius Supposethat \[\label{eq:26}p(x)y''+q(x)y'+r(x)y=0\] hasaregularsingularpointat\(x=0\),thenthereexistsatleastonesolutionoftheform \[y=x^r\sum_{k=0}^\inftya_kx^k.\nonumber\] AsolutionofthisformiscalledaFrobenius-typesolution. Themethodusuallybreaksdownlikethis. WeseekaFrobenius-typesolutionoftheform\[y=\sum_{k=0}^\inftya_kx^{k+r}.\nonumber\]Weplugthis\(y\)intoequation\(\eqref{eq:26}\).Wecollecttermsandwriteeverythingasasingleseries. Theobtainedseriesmustbezero.Settingthefirstcoefficient(usuallythecoefficientof\(x^r\))intheseriestozeroweobtaintheindicialequation,whichisaquadraticpolynomialin\(r\). Iftheindicialequationhastworealroots\(r_1\)and\(r_2\)suchthat\(r_1-r_2\)isnotaninteger,thenwehavetwolinearlyindependentFrobenius-typesolutions.Usingthefirstroot,weplugin\[y_1=x^{r_1}\sum_{k=0}^\inftya_kx^{k},\nonumber\]andwesolveforall\(a_k\)toobtainthefirstsolution.Thenusingthesecondroot,weplugin\[y_2=x^{r_2}\sum_{k=0}^\inftyb_kx^{k},\nonumber\]andsolveforall\(b_k\)toobtainthesecondsolution. Iftheindicialequationhasadoubledroot\(r\),thentherewefindonesolution\[y_1=x^{r}\sum_{k=0}^\inftya_kx^{k},\nonumber\]andthenweobtainanewsolutionbyplugging\[y_2=x^{r}\sum_{k=0}^\inftyb_kx^{k}+(\lnx)y_1,\nonumber\]intoEquation\(\eqref{eq:26}\)andsolvingfortheconstants\(b_k\). Iftheindicialequationhastworealrootssuchthat\(r_1-r_2\)isaninteger,thenonesolutionis\[y_1=x^{r_1}\sum_{k=0}^\inftya_kx^{k},\nonumber\]andthesecondlinearlyindependentsolutionisoftheform\[y_2=x^{r_2}\sum_{k=0}^\inftyb_kx^{k}+C(\lnx)y_1,\nonumber\]whereweplug\(y_2\)into\(\eqref{eq:26}\)andsolvefortheconstants\(b_k\)and\(C\). Finally,iftheindicialequationhascomplexroots,thensolvingfor\(a_k\)inthesolution\[y=x^{r_1}\sum_{k=0}^\inftya_kx^{k}\nonumber\]resultsinacomplex-valuedfunction---allthe\(a_k\)arecomplexnumbers.Weobtainourtwolinearlyindependentsolutions\(^{2}\)bytakingtherealandimaginarypartsof\(y\). ThemainideaistofindatleastoneFrobenius-typesolution.Ifweareluckyandfindtwo,wearedone.Ifweonlygetone,weeitherusetheideasaboveorevenadifferentmethodsuchasreductionoforder(Exercise2.1.8)toobtainasecondsolution. BesselFunctions AnimportantclassoffunctionsthatarisescommonlyinphysicsaretheBesselfunctions\(^{3}\).Forexample,thesefunctionsappearwhensolvingthewaveequationintwoandthreedimensions.FirstwehaveBessel'sequationoforder\(p\): \[x^2y''+xy'+\left(x^2-p^2\right)y=0.\nonumber\] Weallow\(p\)tobeanynumber,notjustaninteger,althoughintegersandmultiplesof\(\dfrac{1}{2}\)aremostimportantinapplications.Whenweplug \[y=\sum_{k=0}^\inftya_kx^{k+r}\nonumber\] intoBessel'sequationoforder\(p\)weobtaintheindicialequation \[r(r-1)+r-p^2=(r-p)(r+p)=0.\nonumber\] Thereforeweobtaintworoots\(r_1=p\)and\(r_2=-p\).If\(p\)isnotanintegerfollowingthemethodofFrobeniusandsetting\(a_0=1\),weobtainlinearlyindependentsolutionsoftheform \[\begin{align}\begin{aligned}y_1&=x^p\sum_{k=0}^{\infty}\dfrac{(-1)^kx^{2k}}{2^{2k}k!(k+p)(k-1+p)\cdots(2+p)(1+p)},\\y_2&=x^{-p}\sum_{k=0}^{\infty}\dfrac{(-1)^kx^{2k}}{2^{2k}k!(k-p)(k-1-p)\cdots(2-p)(1-p)}.\end{aligned}\end{align}\nonumber\] Exercise\(\PageIndex{1}\) VerifythattheindicialequationofBessel'sequationoforder\(p\)is\((r-p)(r+p)=0\). Supposethat\(p\)isnotaninteger.Carryoutthecomputationtoobtainthesolutions\(y_1\)and\(y_2\)above. Besselfunctionswillbeconvenientconstantmultiplesof\(y_1\)and\(y_2\).Firstwemustdefinethegammafunction \[\Gamma(x)=\int_0^\inftyt^{x-1}e^{-t}\,dt.\nonumber\] Noticethat\(\Gamma(1)=1\).Thegammafunctionalsohasawonderfulproperty \[\Gamma(x+1)=x\Gamma(x).\nonumber\] Fromthisproperty,onecanshowthat\(\Gamma(n)=(n-1)!\)when\(n\)isaninteger,sothegammafunctionisacontinuousversionofthefactorial.Wecompute: \[\begin{align}\begin{aligned}\Gamma(k+p+1)=(k+p)(k-1+p)\cdots(2+p)(1+p)\Gamma(1+p),\\\Gamma(k-p+1)=(k-p)(k-1-p)\cdots(2-p)(1-p)\Gamma(1-p).\end{aligned}\end{align}\nonumber\] Exercise\(\PageIndex{2}\) Verifytheaboveidentitiesusing\(\Gamma(x+1)=x\Gamma(x)\). WedefinetheBesselfunctionsofthefirstkindoforder\(p\)and\(-p\)as \[\begin{align}\begin{aligned}J_p(x)&=\dfrac{1}{2^p\Gamma(1+p)}y_1=\sum_{k=0}^\infty\dfrac{{(-1)}^k}{k!\Gamma(k+p+1)}{\left(\dfrac{x}{2}\right)}^{2k+p},\\J_{-p}(x)&=\dfrac{1}{2^{-}\Gamma(1-p)}y_2=\sum_{k=0}^\infty\dfrac{{(-1)}^k}{k!\Gamma(k-p+1)}{\left(\dfrac{x}{2}\right)}^{2k-p}.\end{aligned}\end{align}\nonumber\] Astheseareconstantmultiplesofthesolutionswefoundabove,thesearebothsolutionstoBessel'sequationoforder\(p\).Theconstantsarepickedforconvenience. When\(p\)isnotaninteger,\(J_p\)and\(J_{-p}\)arelinearlyindependent.When\(n\)isanintegerweobtain \[J_n(x)=\sum_{k=0}^\infty\dfrac{{(-1)}^k}{k!(k+n)!}{\left(\dfrac{x}{2}\right)}^{2k+n}.\nonumber\] Inthiscaseitturnsoutthat \[J_n(x)={(-1)}^nJ_{-n}(x),\nonumber\] andsowedonotobtainasecondlinearlyindependentsolution.Theothersolutionistheso-calledBesselfunctionofsecondkind.Thesemakesenseonlyforintegerorders\(n\)andaredefinedaslimitsoflinearcombinationsof\(J_p(x)\)and\(J_{-p}(x)\)as\(p\)approaches\(n\)inthefollowingway: \[Y_n(x)=\lim_{p\ton}\dfrac{\cos(p\pi)J_p(x)-J_{-p}(x)}{\sin(p\pi)}.\nonumber\] Aseachlinearcombinationof\(J_p(x)\)and\(J_{-p}(x)\)isasolutiontoBessel'sequationoforder\(p\),thenaswetakethelimitas\(p\)goesto\(n\),\(Y_n(x)\)isasolutiontoBessel'sequationoforder\(n\).Italsoturnsoutthat\(Y_n(x)\)and\(J_n(x)\)arelinearlyindependent.Thereforewhen\(n\)isaninteger,wehavethegeneralsolutiontoBessel'sequationoforder\(n\) \[y=AJ_n(x)+BY_n(x),\nonumber\] forarbitraryconstants\(A\)and\(B\).Notethat\(Y_n(x)\)goestonegativeinfinityat\(x=0\).Manymathematicalsoftwarepackageshavethesefunctions\(J_n(x)\)and\(Y_n(x)\)defined,sotheycanbeusedjustlikesay\(\sin(x)\)and\(\cos(x)\).Infact,theyhavesomesimilarproperties.Forexample,\(-J_1(x)\)isaderivativeof\(J_0(x)\),andingeneralthederivativeof\(J_n(x)\)canbewrittenasalinearcombinationof\(J_{n-1}(x)\)and\(J_{n+1}(x)\).Furthermore,thesefunctionsoscillate,althoughtheyarenotperiodic.SeeFigure\(\PageIndex{1}\)forgraphsofBesselfunctions. Figure\(\PageIndex{1}\):Plotofthe\(J_{0}(x)\)and\(J_{1}(x)\)inthefirstgraphand\(Y_{0}(x)\)and\(Y_{1}(x)\)inthesecondgraph. Example\(\PageIndex{4}\):UsingBesselfunctionstoSolveaODE OtherequationscansometimesbesolvedintermsoftheBesselfunctions.Forexample,givenapositiveconstant\(\lambda\), \[xy''+y'+\lambda^2xy=0,\nonumber\] canbechangedto\(x^2y''+xy'+\lambda^2x^2y=0\).Thenchangingvariables\(t=\lambdax\)weobtainviachainruletheequationin\(y\)and\(t\): \[t^2y''+ty'+t^2y=0,\nonumber\] whichcanberecognizedasBessel'sequationoforder0.Thereforethegeneralsolutionis\(y(t)=AJ_0(t)+BY_0(t)\),orintermsof\(x\): \[y=AJ_0(\lambdax)+BY_0(\lambdax).\nonumber\] Thisequationcomesupforexamplewhenfindingfundamentalmodesofvibrationofacirculardrum,butwedigress. Footnotes [1]NamedaftertheGermanmathematicianFerdinandGeorgFrobenius(1849–1917). [2]SeeJosephL.Neuringera,TheFrobeniusmethodforcomplexrootsoftheindicialequation,InternationalJournalofMathematicalEducationinScienceandTechnology,Volume9,Issue1,1978,71–77. [3]NamedaftertheGermanastronomerandmathematicianFriedrichWilhelmBessel(1784–1846).