Download Solution HW#5

Solution HW#5
We have to solve the non-homogenous heat equation
∂u
∂2u
+ 4e−4t cos3 x
=
∂t
∂x2
with a non-homogenous boundary condition,
u( π2 , t) = 0.
∂u
∂x (0, t)
= 2, and a homogenous one,
We start by finding a simple function r(x) that has the same boundary values. For
example,
r(x) = 2x − π .
Writing u = v + r and substituting it into the heat equation, we get
∂v
∂2v
=
+ 4e−4t cos3 x
∂t
∂x2
with homogenous boundary conditions:
∂v
∂x (0, t)
= 0 and v( π2 , t) = 0.
We expand v(x, t) in terms of the complete set of eigenfuctions to the homogeneous
problem on the given interval
v(x, t) =
∞
X
an (t) cos (1 + 2n)x .
n=0
Because the eigenfunctions satisfy the same boundary conditions as v(x, t), we are allowed to differentiate term by term inside the sum. From the wave equation we obtain
∞ X
dan
n=1
dt
+ λn an cos (1 + 2n)x = 4e−4t cos3 x
where the eigenvalues are given by λn = (1 + 2n)2 .
Using the identity 4 cos3 x = 3 cos x + cos (3x), we can read off the Fourier coefficients of
the source. The wave equation decomposes into an ODE for each term in the Fourier
series,
da0
+ λ0 a0 = 3e−4t ,
dt
da1
+ λ1 a1 = e−4t ,
dt
dan
+ λ n an = 0 ,
dt
1
n ≥ 2.
Multiplying the equations by an integrating factor eλn t , we get
d λ0 t a0 e
= 3e(λ0 −4)t = 3e−3t ,
dt
d λ1 t a1 e
= e(λ1 −4)t = e5t ,
dt
d
an eλn t = 0 ,
n ≥ 2.
dt
The solutions to the above 1st order ODEs are
−λ0 t
a0 (t) = a0 (0) e
+e
−λ0 t
Z
t
3e−3τ dτ ,
0
a1 (t) = a1 (0) e−λ1 t + e−λ1 t
Z
t
e5τ dτ ,
0
an (t) = an (0) e−λn t ,
n ≥ 2,
where the terms containing integrals evaluates to (e−t − e−4t ) and
spectively.
1 −4t
5 (e
− e−9t ), re-
The solution for v(x, t) is then
∞
X
1
v(x, t) = (e−t − e−4t ) cos x + (e−4t − e−9t ) cos (3x) +
an (0) e−λn t cos (1 + 2n)x .
5
n=0
The initial condition v(x, 0) = f (x) − r(x) = f (x) − 2x + π gives the equation,
f (x) − 2x + π =
∞
X
an (0) cos (1 + 2n)x ,
n=0
which fixes the integration constants in the homogenous part of the solution
4
an (0) =
π
Z
π/2
f (x) − 2x + π cos (1 + 2n)x dx .
0
Final answer:
The solution is
∞
−t
−4t
u(x, t) = 2x−π+(e −e
X
1
an (0) e−λn t cos (1+2n)x ,
) cos x+ (e−4t −e−9t ) cos (3x)+
5
n=0
with λn = (1 + 2n)2 , and an (0) given above.
2