question archive Case 3: λ = γ2 (In your answers below use gamma instead of lambda) X(x) = Plugging in the boundary values into this formula gives 0 = X(0) = 0 = X(10) = Which leads us to the eigenvalues γn = and eigenfunctions Xn(x) = (Notation: Eigenfunctions should not include any constants a or b
Case 3: λ = γ2 (In your answers below use gamma instead of lambda) X(x) = Plugging in the boundary values into this formula gives 0 = X(0) = 0 = X(10) = Which leads us to the eigenvalues γn = and eigenfunctions Xn(x) = (Notation: Eigenfunctions should not include any constants a or b
Subject:Mechanical EngineeringPrice: Bought3
Case 3: λ = γ2
(In your answers below use gamma
instead of lambda)
X(x) =
Plugging in the boundary values into this formula
gives
0 = X(0) =
0 = X(10) =
Which leads us to the eigenvalues γn =
and eigenfunctions Xn(x) =
(Notation: Eigenfunctions should not include any
constants a or b.)
Solve for T(t).
Plug the eigenvalues λn = γ2
n
from Case 3 into the differential
equation for T(t) and solve:
Tn(t) =
(Notation: use c for the unknown constant.)
Combining all of the Xn and Tn we get that
u(x,t) =
∞
∑
n=1
Bn
where Bn are unknown constants.
Fourier Coefficients.
We compute Bn by plugging t = 0 into the formula for u(x,t)
and setting equal to the initial heat distribution given in the
problem.
u(x,0) =
∞
∑
n=1
Bn
= (0
,
0
<
x
<
5
3
,
5
≤
x
<
10
So the Bn are Fourier coefficients.
Bn =
2
10
Z5
10
dx
=
Remember that cos(nπ) = (−1)
