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

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.

• Can you provide more information? For instance what kind of function $g$? – MathArt Feb 23 at 13:55
• $g$ 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 $\frac{\pi}{2\sqrt{pq}}=\int_0^\frac{\pi}{2}\frac{1}{p\cos^2x + q\sin^2x}$ – Eren Ozturk Feb 23 at 14:16
• The general solution to the PDE can be found with the method of characteristics. See the example given in wiki. – user10354138 Feb 23 at 14:47
• Thats great, thanks – Eren Ozturk Feb 23 at 19:24

$$\frac{\partial g(p,q)}{\partial q} + \frac{\partial g(p,q)}{\partial p}=\frac{\pi}{2\sqrt{pq}}$$ 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.