yay, figured it out.
letting u(x,t)=e^{-ct}v(x,t), with v(x,t) solving
v_t - \Delta v = f(x,t) e^{ct}
v(x,0) = g(x)
Solves the original equation. I guess I was just thinking too hard.
Be careful when dividing by a variable because x=0 could easily be a solution to this problem. Instead of trying to solve for lambda, try eliminating it. You could try multiplying the first equation by y and the second equation by x. You should get an equation for x and y, and so coupled with...
Homework Statement
So I'm trying to solve Evans - PDE 2.5 # 12...
"Write down an explicit formula for a solution of
u_t - \Delta u + cu=f with (x,t) \in R^n \times (0,\infty)
u(x,0)=g(x)"
Homework Equations
The Attempt at a Solution
I figure if I can a fundamental solution...
So what if P(n, 1) is trivially true for arbitrary n, then surely we can prove P(n, m) by showing P(n, m) => P(n, m+1) without having to do induction on n?
The thing that confuses me is that for example to prove the binomial theorem, which is a statement P(x,y,n), we can simply prove that P(x,y,n) => P(x,y, n+1) and we don't have to prove it for all x and y, as we can just leave them arbitrary. Similarly why can't we prove, P(n,m) by proving...
Suppose we have a statement (say an equation) P(n, m), and we want to prove it works for arbitrary n and m.
Either
(1) It is sufficient to define Q(n) = P(n,m) or R(m) = P(n,m) and prove either Q(n) or R(m) by induction
(2) It is necessary to prove that P(n,m) => P(n+1,m) and P(n, m)...
One of my recommendations actually got misplaced. They (UCLA) were nice enough to write me an e-mail notifying me so. I wasn't expecting on getting into any of the other top schools.
I'm only doing this as a last resort. There's a reason I graduated in 2.5 years, and it's not because I wanted...
They went probably went something like this:
"While µ³ may achieve high grades, he never shows up for class, only turns in about half the assignments, rarely on time. However, he gets just high enough on a final to make an A except for the times he doesn't and gets a C".
I can't really take...
So I'm a big-headed idiot and only applied to top schools. Friday I got my last rejection letter from all the physics programs I applied to (it was UCLA's). I'm guessing it was because one of my weakspots as an applicant showed up on two-three of my recommendations (I'm lazy). I wasn't expecting...
Well, \hat{a}^2 commutes with \hat{a} so any eigenvector of the latter is an eigenvector of the former. So to compute the eigenvalues, just operate \hat{a}^2 on a coherent state.
Well, here's some things that might help:
S(z)S^{\dagger} (z)= I^{\hat}
so S(z) is unitary
e^{A+B}=e^A e^Be^{-1/2[A,B]}=e^B e^A e^{1/2 [A,B]}
provided that
[A,[A,B]]=[B,[A,B]]=0
and given the eigen value equation
\hat{A}|a>=a|a>
Then
e^{\hat{A}}|a> = e^a |a>
I tried it for a while and then I...