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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 $

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    @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.2012-09-14

3 Answers 3

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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.

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    @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.2012-09-24
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For expressing $\int_0^\alpha\sqrt{a^2\sin^2t+b^2\cos^2t}~dt$ in series expression:

$\int_0^\alpha\sqrt{a^2\sin^2t+b^2\cos^2t}~dt$

$=\int_0^\alpha\sqrt{a^2(1-\cos^2t)+b^2\cos^2t}~dt$

$=\int_0^\alpha\sqrt{a^2+(b^2-a^2)\cos^2t}~dt$

$=\int_0^\alpha|a|\sqrt{1+\dfrac{b^2-a^2}{a^2}\cos^2t}~dt$

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

$\therefore\int_0^\alpha|a|\sqrt{1+\dfrac{b^2-a^2}{a^2}\cos^2t}~dt$

$=\int_0^\alpha\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!|a|(b^2-a^2)^n\cos^{2n}t}{4^n(n!)^2(1-2n)a^{2n}}dt$

$=\int_0^\alpha\left(|a|+\sum\limits_{n=1}^\infty\dfrac{(-1)^n(2n)!|a|(b^2-a^2)^n\cos^{2n}t}{4^n(n!)^2(1-2n)a^{2n}}\right)dt$

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

$\therefore\int_0^\alpha\left(|a|+\sum\limits_{n=1}^\infty\dfrac{(-1)^n(2n)!|a|(b^2-a^2)^n\cos^{2n}t}{4^n(n!)^2(1-2n)a^{2n}}\right)dt$

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

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

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n((2n)!)^2|a|(b^2-a^2)^n\alpha}{4^{2n}(n!)^4(1-2n)a^{2n}}$

$~~~+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n((2n)!)^2((k-1)!)^2|a|(b^2-a^2)^n\sin\alpha~\cos^{2k-1}\alpha}{4^{2n-k+1}(n!)^4(1-2n)(2k-1)!a^{2n}}$

You can also try another approach that

$\int_0^\alpha\sqrt{a^2\sin^2t+b^2\cos^2t}~dt$

$=\int_0^\alpha\sqrt{a^2\sin^2t+b^2(1-\sin^2t)}~dt$

$=\int_0^\alpha\sqrt{b^2+(a^2-b^2)\sin^2t}~dt$

$=\int_0^\alpha|b|\sqrt{1+\dfrac{a^2-b^2}{b^2}\sin^2t}~dt$

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

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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: $u=yF\left(\frac xy\right)$