Frobenijusov metod predstavlja jedan od metoda rešavanja diferencijalnih jednačina drugoga reda oblika:
![{\displaystyle z^{2}u''+p(z)zu'+q(z)u=0\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a0835b932fa4e4c68f8df6e78c18d5593c02099)
gde su:
i
u blizini regularnoga singulariteta z=0. Podelimo li sa z2 dobijamo diferencijalnu jednačinu:
![{\displaystyle u''+{p(z) \over z}u'+{q(z) \over z^{2}}u=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/27ea866ba997d4285d0982d6b1ca3d9f2cfb1d11)
Metoda je dobila ime po nemačkom matematičaru Ferdinandu Frobenijusu.
Prema Frobenijusovoj metodi tražimo rešenje u obliku reda:
![{\displaystyle u(z)=\sum _{k=0}^{\infty }A_{k}z^{k+r},\qquad (A_{0}\neq 0)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6f46753d30e83b2272dd6a7be2aae79138d8cd05)
Diferenciranjem dobijamo:
![{\displaystyle u'(z)=\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/24e10ab169a25baf1c39bb544f618ee6500f067f)
![{\displaystyle u''(z)=\sum _{k=0}^{\infty }(k+r-1)(k+r)A_{k}z^{k+r-2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b00a1939f6abf1607bb7ec90122e9635d55eeab7)
Posle toga gore dobiujene redove supstituiramo u diferencijalnu jednačinu i dobijamo:
![{\displaystyle {\begin{aligned}&{}\quad z^{2}\sum _{k=0}^{\infty }(k+r-1)(k+r)A_{k}z^{k+r-2}+zp(z)\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-1}+q(z)\sum _{k=0}^{\infty }A_{k}z^{k+r}\\&=\sum _{k=0}^{\infty }(k+r-1)(k+r)A_{k}z^{k+r}+p(z)\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r}+q(z)\sum _{k=0}^{\infty }A_{k}z^{k+r}\\&=\sum _{k=0}^{\infty }(k+r-1)(k+r)A_{k}z^{k+r}+p(z)(k+r)A_{k}z^{k+r}+q(z)A_{k}z^{k+r}\\&=\sum _{k=0}^{\infty }\left[(k+r-1)(k+r)+p(z)(k+r)+q(z)\right]A_{k}z^{k+r}\\&=\left[r(r-1)+p(z)r+q(z)\right]A_{0}z^{r}+\sum _{k=1}^{\infty }\left[(k+r-1)(k+r)+p(z)(k+r)+q(z)\right]A_{k}z^{k+r}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/59021a42b9f1dec0512b28467b1abfc140730760)
Inicijalni polinom je sledeći izraz:
![{\displaystyle r\left(r-1\right)+p\left(0\right)r+q\left(0\right)=I(r)\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/791b411c8d9868b608dbd2bb984b086919950cf3)
Prema opštoj definiciji inicijalni polinomi su koeficijenti najnižega stepena po z.
Opšti izraz za koeficijente od zk + r je:
![{\displaystyle I(k+r)A_{k}+\sum _{j=0}^{k-1}\left[(j+r)p(k-j)+q(k-j)\right]A_{j}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e7efec331f2e6e77f78455546fc998cbf1e1e21c)
Ti koeficijenti treba da budu jednaki nuli, jer oni treba da predstavljaju rešenja diferencijalne jednačine, pa sledi:
![{\displaystyle I(k+r)A_{k}+\sum _{j=0}^{k-1}\left[(j+r)p(k-j)+q(k-j)\right]A_{j}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a218dde8e50f2d768a034f659ef894b89dc21a40)
![{\displaystyle \sum _{j=0}^{k-1}\left[(j+r)p(k-j)+q(k-j)\right]A_{j}=-I(k+r)A_{k}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6d7682c2b8141a6304aed64358183ee34817b09)
![{\displaystyle {1 \over -I(k+r)}\sum _{j=0}^{k-1}\left[(j+r)p(k-j)+q(k-j)\right]A_{j}=A_{k}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8ff3cd35b251557d0b766f66a2f326b03825140)
Gornje rešenje sa Ak je:
![{\displaystyle U_{r}(z)=\sum _{k=0}^{\infty }A_{k}z^{k+r}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3af0279517f1fdd5143a56a950c9bd09c8d1d09d)
i zadovoljava:
![{\displaystyle z^{2}U_{r}(z)''+p(z)zU_{r}(z)'+q(z)U_{r}(z)=I(r)z^{r}\!\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/13bc17d71d5a75f4c13ed55d5eed3ef8dd30608b)
Odaberemo li jedan od korena inicijalnoga polinoma, tada dobijamo rešenje diferencijalne jednačine.
Pokušamo li da rešimo sledeći diferencijalnu jednačinu:
![{\displaystyle z^{2}f''-zf'+(1-z)f=0\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22ea81ecfbf7c93086254fae1851fdb8c283fa6f)
Podelimo li je sa z2 dobijamo:
![{\displaystyle f''-{1 \over z}f'+{1-z \over z^{2}}f=f''-{1 \over z}f'+\left({1 \over z^{2}}-{1 \over z}\right)f=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3055ca25ecdacccced7e59ac8ca0ec31f6208b94)
Pretpostavljamo rešenja u obliku reda:
![{\displaystyle f=\sum _{k=0}^{\infty }A_{k}z^{k+r}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e26cf5f4cc8761b8b021686f4468a24f2e2caad0)
![{\displaystyle f'=\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0230306c0259afb3b9e6527ef1346ed0324f2151)
![{\displaystyle f''=\sum _{k=0}^{\infty }(k+r)(k+r-1)A_{k}z^{k+r-2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69c619d4ecf4a9770ad3d02346eea1935ea5cfab)
i ta rešenja supstituiramo u gornju jednačinu:
![{\displaystyle {\begin{aligned}&\sum _{k=0}^{\infty }(k+r)(k+r-1)A_{k}z^{k+r-2}-{1 \over z}\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-1}+\left({1 \over z^{2}}-{1 \over z}\right)\sum _{k=0}^{\infty }A_{k}z^{k+r}\\&=\sum _{k=0}^{\infty }(k+r)(k+r-1)A_{k}z^{k+r-2}-{1 \over z}\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-1}+{1 \over z^{2}}\sum _{k=0}^{\infty }A_{k}z^{k+r}-{1 \over z}\sum _{k=0}^{\infty }A_{k}z^{k+r}\\&=\sum _{k=0}^{\infty }(k+r)(k+r-1)A_{k}z^{k+r-2}-\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-2}+\sum _{k=0}^{\infty }A_{k}z^{k+r-2}-\sum _{k=0}^{\infty }A_{k}z^{k+r-1}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3f1f7038b2e0809aa752357863aede597d397b1e)
Pomeramo indekse poslednje sume, tako da se dobija:
![{\displaystyle {\begin{aligned}&=\sum _{k=0}^{\infty }(k+r)(k+r-1)A_{k}z^{k+r-2}-\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-2}+\sum _{k=0}^{\infty }A_{k}z^{k+r-2}-\sum _{k-1=0}^{\infty }A_{k-1}z^{k+r-2}\\&=\sum _{k=0}^{\infty }(k+r)(k+r-1)A_{k}z^{k+r-2}-\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-2}+\sum _{k=0}^{\infty }A_{k}z^{k+r-2}-\sum _{k=1}^{\infty }A_{k-1}z^{k+r-2}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c93b51c1963dd28815e65fc54629984bc169bbc)
Startni indeks za k=0 se posebno piše, pa se dobija:
![{\displaystyle {\begin{aligned}&=((r)(r-1)A_{0}z^{r-2})+\sum _{k=1}^{\infty }(k+r)(k+r-1)A_{k}z^{k+r-2}-((r)A_{0}z^{r-2})-\sum _{k=1}^{\infty }(k+r)A_{k}z^{k+r-2}\\&{}+(A_{0}z^{r-2})+\sum _{k=1}^{\infty }A_{k}z^{k+r-2}-\sum _{k=1}^{\infty }A_{k-1}z^{k+r-2}\\&=(r(r-1)-r+1)A_{0}z^{r-2}+\sum _{k=1}^{\infty }\left(((k+r)(k+r-1)-(k+r)+1)A_{k}-A_{k-1}\right)z^{k+r-2}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6a5ede30f27905452c2a4f1a5e6468bb404c8ab)
Jedno rešenje dobijamo rešavanjem inicijalnoga polinoma r(r − 1) − r + 1 = r2 − 2r + 1 = 0, odnosno dobijamo da je 1 dvostruki koren. Koristeći taj koren koeficijenti od zk + r − 2 treba da budu nula, šta daje rekurziju:
![{\displaystyle ((k+1)(k)-(k+1)+1)A_{k}-A_{k-1}=(k^{2})A_{k}-A_{k-1}=0\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/31d376607ab6f86e907cf5ac17b2719f29f33988)
![{\displaystyle A_{k}={A_{k-1} \over k^{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/db133c628bb06cf19eeed35dce496be4c50a5c28)
Pošto je omer
racionalna funkcija onda se red može napisati kao opšti hipergeometrijski red.