Explore BrainMass

Differential equations

Not what you're looking for? Search our solutions OR ask your own Custom question.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

Consider the following problem for u = u(x,t):
,
,

a) Seeking a solution of this problem of the form
,

show that f and g satisfy the coupled system
; where and ;
, ; , .

b) Eliminating g between the differential equations in a), show that f satisfies

c) Assuming that

and noting that

conclude that

d) Applying the boundary conditions for f in a), deduce that
where

e) Now use a) to show that
.

f) Applying the boundary conditions for g in a), finally conclude that
and hence obtain the result.

since

https://brainmass.com/math/calculus-and-analysis/differential-equations-boundary-conditions-177146

SOLUTION This solution is FREE courtesy of BrainMass!

Consider the following problem for u = u(x,t):
,
,

a) Seeking a solution of this problem of the form
,

show that f and g satisfy the coupled system
; where and ;
, ; , .

Now,

Dividing throughout by 

=
Comparing the cos terms, and comparing the sin terms,
For u(0, t) = cos(t), f(0) = 1 and g(0) = 0

b) Eliminating g between the differential equations in a), show that f satisfies

Given that f'' = 2k2g,
Differentiating with respect to x on both sides, f''' = 2k2g'
Differentiating again with respect to x on both sides, f(4) = 2k2g''
But g'' = -2k2f, so f(4) = 2k2(-2k2f) = -4k4f
Therefore,

c) Assuming that

and noting that

conclude that

f'(x) = r erx
f''(x) = r2 erx
f'''(x) = r3 erx
f(4)(x) = r4 erx
Plugging in the equation , we get r4 erx + 4k4 erx = 0
Or (r4 + 4k4) erx = 0, which gives (r4 + 4k4) = 0
But
Therefore,
Taking
Or
Or (r + k)2 = -k2
Or r + k = -ki, where i =
Or r = -k - ki
Therefore, f(x) = e-k-ki
By Euler's formula,
Similarly, by taking , it can be shown that r = k - ki
Or f(x) = ek-ki =
Combining the two,

d) Applying the boundary conditions for f in a), deduce that
where
Given that ,
so
or
or c1 + c3 = 1
Similarly, c3 = c4= 0
Plugging in c3 = 0 in c1 + c3 = 1 gives c1 = 1
Therefore,
=
=
If , we get

e) Now use a) to show that
.
f'' = 2k2 g(x)  g(x) = f"/2k2
But f = e-kx h(x)
Therefore, f'(x) = -ke-kx h(x) + h'(x) e-kx = e-kx(-kh + h') = -kf + e-kx h'(x)
Or f' + kf = e-kx h'(x)
But g(x) = -(f' + kf)/k, therefore,

f) Applying the boundary conditions for g in a), finally conclude that
and hence obtain the result.

since

Given the boundary condition g(0) = 0, 0 = , which means that
h'(0) = 0
But , therefore, h'(x) = -k sin(kx) + c2 cos(kx)
At x = 0, h'(0) = -k*sin(0) + c2*cos(0)
Or 0 = 0 + c2, which gives c2 = 0
Therefore,
Therefore, =
And =
Now, = cos(t) + sin(t) Or .

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!