Homework Statement
On a long dielectric line a charge with density ##10^{-3}## is applied one half with positive charge and the other half with negative charge. Perpendicular to the first line and 5 cm away from it we have another line with the same charge density and also half of it is...
That is a lot easier, I agree. So the conclusion is that if ##\lambda =0## we can find a function that satisfies boundary conditions. In this case ##y=D## is a constant function, therefore ##\lambda ## can also be ##0##.
b) Is ##L## self-adjoint?
##<f,g>=\int _{-\pi }^{\pi }f(x)g(x)dx## for...
Homework Statement
We have a linear differential operator ##Ly=-y^{''}## working on all ##y## that can be derived at least twice on ##[-\pi ,\pi ]## and also note that ##y(-\pi )=y(\pi )## and ##y^{'}(-\pi )=y^{'}(\pi )##.
a) Is ##0## eigenvalue for ##L##?
b) Is ##L## symmetric? (I think the...
Aaaa, ok, I get it!
Solution is than ##y(z)=z+\sum _{m=1}^{\infty }\frac {(-1)^m z^{3m+1}}{(3m+1)(3m)(3m-2)(3m-3)...(4)(3)}##Just to clear things out: Taking ##y(z)=z^r\sum _{k=0}^{\infty } C_kz^k## in this case where ##z=0## is not a regular singular point is WRONG or does it just complicate...
That seems to be something I should have known... With that in mind, here is version 2.0:
##y=C_0+\sum _{k=1}^{\infty }C_kz^k## and using ##y(0)=0## also ##C_0=0##.
##y^{'}=C_1+\sum _{k=2}^{\infty }kC_kz^{k-1}## and using ##y^{'}(0)=1## also ##C_1=1##.
##y^{''}=\sum _{k=2}^{\infty...
Homework Statement
Solve ##y^{''}+zy=0## where ##y(0)=0## and ##y^{'}(0)=1##Homework Equations
##y(z)=z^r\sum _{k=0}^{\infty } C_kz^k##
The Attempt at a Solution
Well firstly:
##r(r-1)+p_0r+q_0=0## where obviously ##p_0=q_0=0## so ##r_1=0## and ##r_2=1##.
In general ##y(z)=\sum...
No, it is not. ?!
##y>0## also ##a_1\geq 0##. The product therefore ##a_1y \geq 0##. Aha. Ok. I take that back. It is ##b_2 \geq 0##.
That was our goal right? To find such a and b that f will have the desired property. And we found them. Why would it be any different the other way around?
Ok, makes sense.
##a_1y+a_2x+b_2>0## and now we know that ##y>0## also ##a_1\geq 0## therefore ##a_1y \geq 0##.
##b_2>-a_1y-a_2x## where if ##a_2<0## I can find such ##x## that the inequality is violated. The same goes if ##a_2>0##. Does this mean that ##a_2=0##?
I'm so sorry! :D
However, since this everything is an introduction to complex analysis, I think it is safe to assume that ##a,b \mathbb{C}##.
I think there should be##b_2>-a_1y##.
How does this imply anything? ##b_2## can be either positive either negative and the same goes for ##a_1##...
Ooops, I have the right sign in my papers. Sorry for that. Should be ok everywhere else. Anyway, good point.
I will get thousands of conditions...
So for ##x=0## than ##b_2+a_1y>0## now I know nothing about ##b_2## or ##a_1## except the fact that they are both real numbers.
##b_2>-a_1y##...
Here is what I came up with:
##a_1y+a_2x+b_2>0## and therefore ##x>\frac{b_2+a_1y}{a_2}##.
Now we already know that ##y>0##, let's check everything for ##x##.
1.) ##x=0##:
##0>\frac{b_2+a_1y}{a_2}## therefore ##b_2<-a_1y##.
2.) ##x>0##:
##\frac{b_2+a_1y}{a_2}<0## therefore ##b_2<-a_1y##...
##a_1y+a_2x+b_2>0##
The ##x>-\frac{b_2+a_1y}{a_2}##. So ##x## can not be ##x\leq -\frac{b_2+a_1y}{a_2}##.
But, how does this answer:
I could also say that ##a_2 \neq 0## meaning ##a \in \mathbb{C}## for sure!
Hmmm...
Let's say that ##z=x+iy##, ##a=a_1+ia_2## and ##b=b_1+ib_2## for ##x,y,a_i,b_i \in \mathbb{R} ##.
Than ##f(z)=(a_1+ia_2)(x+iy)+b_1+ib_2=a_1x+ia_1y+ia_2x-a_2y+b_1+ib_2=a_1x-a_2y+b_1+i(a_1y+a_2x+b_2)##.
Now ##Im(f(z))=a_1y+a_2x+b_2>0## where we already know that ##Im(z)=y>0##...