Kronecker Delta and Things (Term 2) Flashcards
What is the integral from 0 to L of sin(nπx/L)*sin(mπx/L) dx equal to?
L/2 * δnm, where δnm is the kronecker delta, which equals 1 if n=m and 0 otherwise
What can we do with the kronecker delta form of the integral?
Substitute it into the integral equation from before (orthogonality thing) and get rid of the sum (change n for m
What is the equation for Cn?
Cn = 2/L * integral between 0 and L of f(x)*sin(mπx/L) dx
How do you work out the final equation for u(x,t)?
-Split the integral into 2 part (0 to L/2 and rest) and substitute in f(x) for these domains -Do the integrals to find the final value of Cn -Substitute Cn into general solution from before
What does u(x,t) equal finally?
u(x,t) = sum of 2/L * (4*E*L*sin(nπ/2))/(π^2*n^2) * sin(nπx/L)*cos(ncπt/L) (for n odd)
If you strike a string with hammer with head of width w, what is du(x,0)/dt
-0 for x < x0-w/2 -v for x0-w/2 < x < x0+w/2 -0 for x > x0 + w/2
What do we do to the Cn constant in the hammer problem?
Set them to 0, otherwise the displacement at t=0 will not vanish, so only have Dn constants this time.
What do we do with the initial velocity condition du(x,0)/dt = v0*δ(x-x0)?
-Take time derivative of u(x,t) with only Dn in and then substitute t=o in -Make this answer equal to the initial condition: v0*δ(x-x0)
What do we do with the initial condition after we have subbed it in?
-Multiply both sides by sin(mπx/L) and integrate again, and rearrange to get the orthogonality relationship.
What is the orthogonality relationship and what does it equal?
integral from 0 to L of sin(nπx/L)*sin(mπx/L) = L/2*δnm
What do we do after using the orthogonality relationship in the hammer problem?
Find Dm by taking the sum out and the kronecker delta out (n=m), and rearrange for Dm.
How do you do separation of variables for the diffusion equation?
u(x,t) = X(x)T(t), and sub this in. Then move around until same variables on same side, and set equal to lambda, a constant.
What happens when lambda = 0 for the diffusion equation?
-d^2X/dx^2 = 0, so X(x) = A0x+ -dT/dt = 0, so T(t) = G0 -Therefore u(x,t) = (A0x+B0)*G0 = A0’x+B0’
For the diffusion solution when lambda = 0, when are the equilibrium and steady-state solutions?
-Equilibrium when u(x,t) = B0 -Steady-state when u(x,t) = A0x+B0 (A0 not equal to zero)
How can you maintain the steady state? What is the flux equal to?
Feed particles in at the left and pull them out on the right with exactly the right rate. Flux j=-D du(x,t)/dx
How do you work out the equation for λ?
- Relabel m to n in the equation for Dm, and sub this new Dn into u(x,t)
- Set this new equation equal to v0*sin(πx/L)
- Only D1 is not equal to zero, so rearrange this and find u(x,t) = v0L/cπ * sin(πx/L)*sin(cπt/L)
- Only have one pure frequency ω = cπ/L, so λ = 2L
What happens when lambda > 0 (lambda = k^2) for the diffusion equation?
-dT/dt = D*k^2*T(t) -Solution is T(t) = G*e^(D*k^2*t) -Not physical as t goes to 0, so exclude this choice
What happens when lambda < 0 (lambda = -k^2)?
-Both sides equal -k^2 -T(t) = G*e^(D*k^2*t) again -X(x) = Ak*cos(kx)+Bk*sin(kx)
What is the final solution for the diffusion equation?
u(x,t) = sum over k of (Ak*cos(kx)+Bk*sin(kx))*exp(-D*k^2*t) + A0+B0x
What do we do if we specify boundary conditions of u(x=0,t)=u(x=L,t)=0 for the diffusion equation general solution? (lasers which bleach dye molecules at each end)
-Substitute in x=0 and x=L and find the values of Ak and k from this. -Substitute thee new values in
What if there are no lasers? What do we do to work out the solution?
-No flux at ends (j=0) -Take x-derivative of solution and sub in x=0 and x=L -These must both equal zero so find Bk and k values from this -Sub back in
What is A0 equal to in the general solution?
A0 = N/L (equilibrium concentration)
