HW 1, MAE 200B.
Problem 1
by Nasser Abbasi
UCI, Winter 2006.
Question
Solve the following IBVP for the 1D heat conduction problem over the interval
by separation of variables.
for
with the boundary conditions
and the initial conditions specified by the function
Solution
Assume the solution is of the form
where
and
are functions which are independent of each others, i.e.
is a function of
only and
is a function of
only.
Substitute this form in the PDE above we obtain
,
and now divide both sides by
we obtain
.
Since the left hand side is a function which depends on time only and the
right hand side is a function which depends on
only and both are equal, hence they both must equal to a constant, say
.
Hence we obtain the following 2 ODE's to solve:
or
and
or
Start by solving the spatial ODE. Assume the solution to
is of the form
,
and substitute this back in the ODE we obtain the characteristic equation
,
and divide both side by
we get
,
hence the solution can be written as
So one general solution to
is a sum of scaled version of these 2 basic solution, i.e.
case (1):
Apply the spatial B.C. to the above solution to determine
and
,
we get from the B.C. at
the equation
and from the B.C. at
we obtain the equation
.
Hence
and substitute this in the second equation above, we obtain
.
But for a non-trivial solution,
hence this means that
or
and since we assumed
this implies that
for
some positive
but this is not possible unless
,
hence
only gives a trivial solution
or
.
case(2):
When
the
ODE is
which has the solution
,
and when we apply the B.C. we obtain
and
or
,
hence this results also in a trivial solution
case(3)
Let
for some constant
,
hence eq (1) above can be written as
which can be rewritten (using Euler relations
and
)
as
Apply the B.C. at
we obtain
hence
,
and now apply the B.C. at
we obtain
and
to avoid a trivial solution
we must have that
or
or
for
Hence the solution for the spatial ODE is
Where
for
and we assumed the constant
Now we need to find solution to the time ODE. For each
we have a solution.
Since
,
assume solution is of the form
hence the characteristic equation leads to
hence the solution is
but
hence
or
By choosing
Hence the overall solution is
By principle of superposition, the general solution is a scaled sum of these
solutions. where the scaling parameter depends on
which
we call
i.e.
Now apply the initial condition
,
from eq (2) above we obtain that
Compare this an expression where we write a vector in terms of its basis as in
. To find the vector components
we apply the inner product as follows:
hence using similar idea, we apply this to find the components
by considering in this case the vector
as being the function
and it is being decomposed to its basis
as what we do normally in fourier series expansion. Hence we now can find
by writing
and take the inner product over
we obtain
But
But
hence we get that
Now to evaluate
, we rewrite as sum of 2 integrals for each of the ranges of the initial
condition as follows
But
by integration by parts. Apply this to the above we obtain
But
hence above
becomes
Hence
Try for few
values
Hence
Since
for even
,
we only need to sum over odd values of
Hence
When
and
And
the general solution is from eq (2)
:
Now consider the case when
and
hence the solution is
The following is a plot of the initial condition
and the approximation to it as given by eq(4) above by using trying different
values of
.
I tried for
,
we see that as
increases, the summation approximation becomes closer and closer to the
initial condition function
as expected from Fourier series expansion.
To find the number of terms we need so that the approximation is consider
'good', I will find, for each
,
the maximum of the error that occurs between the function
and the function represented by the sum.
Will stop when the maximum error is less than some tolerance, say
or
.
The following is the Matlab code and the plot result and the final
was found to be.
The result shows that if we want maximum error to be less than
then
,
and for maximum error to be less than
then
I will use
since this make the approximation almost the same as
.
Now use this
to find the final solution given by equation (5) above
The following is the plot of
for different
values.
The following is another plot but using smaller time increments of 0.01
seconds, up to