- #1
Mike Jonese
Homework Statement
Boundary conditions are i) V=0 when y=0 ii) V=0 when y=a iii) V=V0(y) when x=0 iv) V=0 when x app infinity.
I understand and follow this problem (separating vars and eliminated constants) until the potential
is found to be V(x,y) = Ce^(-kx)*sin(ky)
Condition ii implies that V(0,a) = Csin(k*a) = 0 so sin(ka) must =0 and k=n*π/a
Question 1) Why can't we just set k=π/a so that V(0,y) = Csin(πy/a) where V(0,a) = 0 and V(0,0) =0? This would not screw up the boundary condition and you wouldn't have to deal with infinite solutions because of n?
Homework Equations
Moving on,
Book give a more general solution that includes all the different solutions due to n
V(x,y) = (n=1---->inf)∑Cn*e^(-nπx/a)*sin(nπy/a)
V(0,y) = (n=1---->inf)∑Cn*sin(nπy/a) = V0(y)
and now we are trying to find the Coefficient C's that make this possible
I'm including the section here to save typing
I didn't realize the forum would kill the resolution
3.Attempt at solution
I tried, I don't understand this in general ( I haven't taken diff eqn yet)
Question 2. How can you solve the integral that leads to (3.32)
Question 3 How does V0(y) pop into the integrand in (3.32)
Question 4 how can the answer be 0 or a/2 based on n and n' (is this some type of delta fxn?)
Question 5 Can someone describe as you would to a 3rd grader why 3.36 is an answer and what it means. Is this just an approximation?
Question 6. What makes something orthogonal in the sense of 3.32
I extremely appreciate and help or insight (or direction to external reference material) anyone might have. Thank you very much! -Mike
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