HW 3. Math 504. Spring 2008. CSUF by Nasser Abbasi
The PDE equation to solve is
part A
multiply through by and integrate w.r.t. x from to we obtain
Now do integration by parts on the first and second terms on the RHS above. We start with the first term
Using the assumption given that as then the first term above will vanish leaving
Now we need to solve the RHS of the above. To do that, we take the derivative of the Fourier transform itself with respect to its variable and write
Therefore we see that
We now use this result to complete the solution. Substitute (2) into (1) we obtain
And using
Then (3) becomes
Now going back to equation (A) above, we do integration by parts twice on the second term on the RHS of that equation and obtain
The term vanishes from the assumption that as , hence the above becomes
Doing integration by parts again on the above we obtain
The term vanishes from the assumption that as , hence the above becomes
Substitute (5) and (4) into (A) we obtain
Hence
Simplify
part B
We need to show that
satisfies the hyperbolic equation shown below in (2), where and
One way to do this is to plug in the expression for given in (1) into the LHS of (2) and see if that gives zero. Hence the LHS of the above pde becomes
But and and since this is the pde we obtained in part(A), hence putting all these into (3) we obtain
Which is zero.
Hence satisfies by direct substitution.
Part C
Now we need to solve the first order hyperbolic PDE equation
In the method of characteristics we convert the PDE to an ODE by looking for parametric path along which solutions for PDE exist. Let and where is a parameter. So now we can write
Therefore taking full derivative of w.r.t. to the parameter we obtain using the chain rule the following
Compare (2) to (1) we see that if we set
with the initial condition and if we set
with initial condition , then this would make which means that the solution is constant along each specific parameter which is what we want. Let the initial condition Hence solution to is
Now, from (2.1) we have since and from (2.2) we have where comes from initial condition of as above.
Now, since hence we have
Hence solve for we have
Plug the above into (3) we obtain
Which is the solution for and
Now to show the final part. From part(B) we showed that
But since then we write (4) as
But
which is the initial conditions we are given in the problem statement (where I replaced by ) Hence plug (6) into (5) we obtain
or
But
Hence (7) becomes
Letting , then the above can be rewritten as
Part (D)
Since now , then where I replaced by
Substitute the above into (8)
Where
part(E)
We need to show that
where is the Fourier transform operator.
Since the transform variable is , we can rewrite the above as needing to show the following
We start by using result from problem (3) part (c) which says the following is true
Hence, we know (2) is true, and we need to show that (1) is true. Using the hint given, we let and in (2) to arrive at (1). hence starting with (2) we write
Simplify, we obtain
Which is the same as (1). QED