I am trying to express the eliptic integral in series expression that depends on $a,b,\alpha$ and without integral

$$L(\alpha)=\int_0^\alpha\sqrt{a^2\sin^2 t+b^2 \cos^2 t}\,dt $$

$$\frac {\partial L(\alpha)}{\partial a}=\int_0^\alpha \frac{a\sin^2 t}{\sqrt{a^2\sin^2 t+b^2 \cos^2 t}}\,dt $$

$$\frac {\partial L(\alpha)}{\partial b}=\int_0^\alpha \frac{b\cos^2 t}{\sqrt{a^2\sin^2 t+b^2 \cos^2 t}}\,dt $$

$$a\frac {\partial L(\alpha)}{\partial a}+b\frac {\partial L(\alpha)}{\partial b}=\int_0^\alpha \frac{a^2\sin^2 t+b^2 \cos^2 t}{\sqrt{a^2\sin^2 t+b^2 \cos^2 t}}\,dt=\int_0^\alpha\sqrt{a^2\sin^2 t+b^2 \cos^2 t}\,dt $$

$$a\frac {\partial L}{\partial a}+b\frac {\partial L}{\partial b}= L \tag 1 $$

Could you please help me how to solve the partial differential equation?

Thanks a lot

Some boundary conditions: $$L(\alpha)|_{a=0}=\int_0^\alpha\sqrt{b^2 \cos^2 t}\,dt=\int_0^\alpha b \cos t\,dt =b \sin \alpha $$

$$L(\alpha)|_{b=0}=\int_0^\alpha\sqrt{a^2 \sin^2 t}\,dt=\int_0^\alpha a \sin t\,dt =-a (\cos \alpha- 1)= a (1- \cos \alpha) $$

$$L(\alpha)|_{b=a}=\int_0^\alpha\sqrt{a^2\sin^2 t+a^2 \cos^2 t}\,dt=\alpha a $$

  • $\begingroup$ Hint: en.wikipedia.org/wiki/Method_of_characteristics $\endgroup$ – Siminore Sep 13 '12 at 14:53
  • $\begingroup$ @Mathlover, although you create the PDE correctly, in fact express $\int_0^\alpha\sqrt{a^2\sin^2t+b^2\cos^2 t}~dt$ in series expression should only be done by other means method and the PDE approach is no helpful. You are an experienced user, why you ask such wrongly-directed question? $\endgroup$ – doraemonpaul Sep 14 '12 at 0:21
  • $\begingroup$ For the PDE aspect, see this thread. But you should not try to reinvent the wheel. There are hundreds of formulas for elliptic integrals, including dozens of series representations. $\endgroup$ – user31373 Sep 14 '12 at 3:06
  • $\begingroup$ @doraemonpaul: I have been self-studying on elliptic integrals. I like to try different approach to a problem. I am not so much experienced about some subjects such as PDEs. Actually I knew binom method but I tried to find a solution via a different way . Maybe I have lost myself in big PDE Ocean. Thanks for answer. $\endgroup$ – Mathlover Sep 14 '12 at 7:20
  • $\begingroup$ @LVK: Thanks for advice. You are right but to rediscover some relations gives me deep understanding. It is just self-learning method. I am trying to learn to catch fish not to wait for a ready one in my stomach. Thank you for your understanding and links. I have been checking them. $\endgroup$ – Mathlover Sep 14 '12 at 7:25

Just to expand on the method of characteristics as Siminore pointed out in the comments: the differential operator $$a \frac{\partial}{\partial a} + b \frac{\partial}{\partial b}$$ is represented by the radial vector field in the $(a,b)$ plane.

This means that all three of your "boundaries", $\{ a = 0\}$, $\{b = 0\}$ and $\{a = b\}$ are tangential the the characteristic curves. This means, in fact, that given just the data you gave us and the partial differential equation, your equation has infinitely many solutions!

Indeed, let $\omega = \arctan b/a$ and $r = \sqrt{a^2 + b^2}$. Then given any $2\pi$ periodic function $F$, the function $(r,\omega)\mapsto r F(\omega)$ solves the differential equation. Your boundary condition specifies $F$ at $\{0,\pi/4, \pi/2, \pi, 5\pi/4, 3\pi/2,\}$ and nowhere else.

  • $\begingroup$ Does your approach really helpful to express $\int_0^\alpha\sqrt{a^2\sin^2t+b^2\cos^2t}~dt$ in series expression? If not, your approach should be a waste to this question. $\endgroup$ – doraemonpaul Sep 19 '12 at 8:55
  • $\begingroup$ (-1): Cannot provide evidence that your approach is really helpful to express $\int_0^\alpha\sqrt{a^2\sin^2 t+b^2\cos^2t}~dt$ in series expression. $\endgroup$ – doraemonpaul Sep 20 '12 at 23:29
  • $\begingroup$ @doraemonpaul: the approach is generally useful for solving hyperbolic differential equations with non-characteristic initial data. In the specific case that the OP has in mind the answer points out that (i) the PDE is not solvable as stated (the data being characteristic) and so (ii) the OP cannot possibly obtain the desired integral by the displayed method of converting to and solving a PDE. $\endgroup$ – Willie Wong Sep 24 '12 at 8:43
  • $\begingroup$ @doraemonpaul: the question actually asked here is on how to solve a PDE. I showed that the PDE is cannot be solved as stated. I find it incredible that you would downvote because I didn't address the motivation behind the question asked. $\endgroup$ – Willie Wong Sep 24 '12 at 8:44

For expressing $\int_0^\alpha\sqrt{a^2\sin^2t+b^2\cos^2t}~dt$ in series expression:





For the binomial series of $\sqrt{1+x}$ , e.g. according to http://en.wikipedia.org/wiki/Square_root#Properties, $\sqrt{1+x}=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^n}{4^n(n!)^2(1-2n)}$




Now for $\int\cos^{2n}t~dt$ , where $n$ is any natural number,

$\int\cos^{2n}t~dt=\dfrac{(2n)!t}{4^n(n!)^2}+\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin t~\cos^{2k-1}t}{4^{n-k+1}(n!)^2(2k-1)!}+C$

This result can be done by successive integration by parts, e.g. as shown as http://hk.knowledge.yahoo.com/question/question?qid=7012022000808


$=\left[|a|t+\sum\limits_{n=1}^\infty\dfrac{(-1)^n((2n)!)^2|a|(b^2-a^2)^nt}{4^{2n}(n!)^4(1-2n)a^{2n}}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n((2n)!)^2((k-1)!)^2|a|(b^2-a^2)^n\sin t~\cos^{2k-1}t}{4^{2n-k+1}(n!)^4(1-2n)(2k-1)!a^{2n}}\right]_0^\alpha$

$=\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^n((2n)!)^2|a|(b^2-a^2)^nt}{4^{2n}(n!)^4(1-2n)a^{2n}}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n((2n)!)^2((k-1)!)^2|a|(b^2-a^2)^n\sin t~\cos^{2k-1}t}{4^{2n-k+1}(n!)^4(1-2n)(2k-1)!a^{2n}}\right]_0^\alpha$



You can also try another approach that





But you should handle $\int\sin^{2n}t~dt$ , where $n$ is any non-negative integer instead


Solve PDE $\;xu_x+yu_y=u$.

  • After change $\;\xi=\log x,\;\eta=\log y\;$ we get PDE with constant coefficients $$u_\xi+u_\eta-u=0.$$
  • Then $$u=e^\eta f(\xi-\eta)\\=yf(\log x-\log y)\\=yf\left(\log \frac xy\right) =yF\left(\frac xy\right)$$
  • Answer: $$u=yF\left(\frac xy\right)$$

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