I am to solve next task.
solve this PDE with boundary conditions.
$\Delta u = \frac{64}{r^5}\sin\varphi, \quad 1<r<2,$
$u'_r|_{r=1} = 2\cos^2\frac{\varphi}{2}, \quad u'_r|_{r=2} = 4\sin^2\frac{\varphi}{2}$
In my solution I am trying to search $u=w+v$, where $w=8\frac{1}{r^3}\sin\varphi$
(because $\Delta_{r,\varphi}w=\frac{64}{r^5}\sin\varphi$, that is my right part of the very first equation)
I will get for function $v(r,\varphi)$ next task with boundary conditions.
$\Delta v = 0$,
$v'_r|_{r=1} = 2\cos^2\frac{\varphi}{2}+24\sin\varphi=1+\cos\varphi+24\sin\varphi,$
$v'_r|_{r=2} = 4\sin^2\frac{\varphi}{2}+3/2\sin\varphi=2-2\cos\varphi+3/2\sin\varphi.$
Let's solve it! we will try to search solution as $v=a\ln r+b+(cr+d/r)\cos\varphi+(er+f/r)\sin\varphi$ and $a,b,c,d,f,e$ are constants.
But take a look at the boundary conditions.
If you apply this to my function $v=a\ln r+...$ you will get $v'_r=a/r+...$
$a/1 = 1$ from the first condition and $a/2 = 2$ from the second... But $4\ne1$
So, does solution exist? Or do I make a mistake?
It upsets me. I also cannot understand conditions for existings of solutions for this PDE and cannot find it.
I don't have even answer for this task. It's just task to solve and show teacher. But it is strange to show him "not existing" or "not correct task" .
By the way, it is not important. I just want to know truth. Help me, please.
Oh no, I was very tired. Of course condition of existing solutions here is not done.
because $2\cdot\int_0^{2\pi} 4\sin^2 (\varphi/2) d\varphi - 1\cdot \int 2\cos^2 (\varphi/2) d\varphi \ne 0.$
Here is incorrect task. I am sorry.