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I was thinking about an integral and got it into the form:

$$\frac{\pi}{2\sqrt{pq}}=\frac{\partial g}{\partial q} + \frac{\partial g}{\partial p}$$

where $g$ is the integral as a function of $q$ and $p$.

I've not really done much on solving PDEs so was wondering if anyone could outline how you'd go about solving this.

I was thinking the symmetry between $q$ and $p$ is important, but really not sure how to approach the problem.

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  • $\begingroup$ Can you provide more information? For instance what kind of function $g$? $\endgroup$ – MathArt Feb 23 at 13:55
  • $\begingroup$ $g$ is the integral $g(p,q)$=$\int_0^\frac{\pi}{2}\ln(p\cos^2 x +q\sin^2 x)$ , then pde comes from the integral $\frac{\pi}{2\sqrt{pq}}=\int_0^\frac{\pi}{2}\frac{1}{p\cos^2x + q\sin^2x}$ $\endgroup$ – Eren Ozturk Feb 23 at 14:16
  • $\begingroup$ The general solution to the PDE can be found with the method of characteristics. See the example given in wiki. $\endgroup$ – user10354138 Feb 23 at 14:47
  • $\begingroup$ Thats great, thanks $\endgroup$ – Eren Ozturk Feb 23 at 19:24
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$$\frac{\partial g(p,q)}{\partial q} + \frac{\partial g(p,q)}{\partial p}=\frac{\pi}{2\sqrt{pq}}$$ The Charpit-Lagrange system of characteristic ODEs is : $$\frac{dp}{1}=\frac{dq}{1}=\frac{dg}{\frac{\pi}{2\sqrt{pq}}}$$ A first characteristic equation comes from solving $\frac{dp}{1}=\frac{dq}{1}$ : $$p-q=c_1$$ A second charactristic equation comes from solving $\frac{dp}{1}=\frac{dg}{\frac{\pi}{2\sqrt{pq}}}$ with $q=p-c_1$ :

$dg=\frac{\pi}{2\sqrt{p(p-c_1)}}dp$

$g=\int \frac{\pi}{2\sqrt{p(p-c_1)}}dp=\frac{1}{2}\ln\left|p-\frac12 c_1+\sqrt{p(p-c_1)}\right|+c_2$ $$g-\frac{1}{2}\ln\left|p-\frac12 c_1+\sqrt{p(p-c_1)}\right|=c_2$$ The general solution of the PDE expressed on implicit form $c_2=F(c_1)$ leads to : $$\boxed{g(p,q)=\frac{1}{2}\ln\left|\frac{p+q}{2}+\sqrt{p\,q}\right|+F\left(p-q\right)}$$ $F$ is an arbitrary function.

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  • $\begingroup$ Great, thanks for this $\endgroup$ – Eren Ozturk Feb 23 at 19:26

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