HW 3. Math 504. Spring 2008. CSUF by Nasser Abbasi


problem.png

Solution

The PDE equation to solve is MATH

part A

multiply through by $e^{-iyx}$ and integrate w.r.t. x from $-\infty$ to $\infty$ we obtain

MATH

Now do integration by parts on the first and second terms on the RHS above. We start with the first term

MATH

Using the assumption given that MATH as MATH then the first term above will vanish leaving

MATH

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 $y$ and write

MATH

Therefore we see that

MATH

We now use this result to complete the solution. Substitute (2) into (1) we obtain

MATH

And using

MATH

Then (3) becomes

MATH

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

MATH

The term MATH vanishes from the assumption that MATH as MATH, hence the above becomes

MATH

Doing integration by parts again on the above we obtain

MATH

The term MATH vanishes from the assumption that MATH as MATH, hence the above becomes

MATH

Substitute (5) and (4) into (A) we obtain

MATH

Hence

MATH

Simplify

MATH


part B

We need to show that MATH

satisfies the hyperbolic equation shown below in (2), where MATH and MATH MATH MATH

One way to do this is to plug in the expression for MATH given in (1) into the LHS of (2) and see if that gives zero. Hence the LHS of the above pde becomes

MATH

But MATH and MATH and MATH since this is the pde we obtained in part(A), hence putting all these into (3) we obtain

MATH

Which is zero.

Hence MATH satisfies MATH by direct substitution.


Part C

Now we need to solve the first order hyperbolic PDE equation

MATH

In the method of characteristics we convert the PDE to an ODE by looking for parametric path along which solutions for PDE exist. Let MATH and MATH where $s$ is a parameter. So now we can write

MATH

Therefore taking full derivative of $u$ w.r.t. to the parameter $s$ we obtain using the chain rule the following

MATH

Compare (2) to (1) we see that if we set MATH

with the initial condition MATH and if we set MATH

with initial condition MATH, then this would make $\frac{du}{ds}=0,$ which means that the solution is constant along each specific parameter $s$ which is what we want. Let the initial condition MATH Hence solution to $\frac{du}{ds}=0$ is MATH

Now, from (2.1) we have $t=s$ since MATH and from (2.2) we have $y=y_{0}e^{cs}$where $y_{0}$ comes from initial condition of $y\left( s\right) $ as above.

Now, since $t=s\,\ $hence we have

MATH

Hence solve for $y_{0}$ we have MATH

Plug the above into (3) we obtain

MATH

Which is the solution for $t\geq0$ and MATH

Now to show the final part. From part(B) we showed that

MATH

But since MATH then we write (4) as

MATH

But MATH

which is the initial conditions we are given in the problem statement (where I replaced $y$ by $ye^{-ct}$) Hence plug (6) into (5) we obtain

MATH

or

MATH

But MATH

Hence (7) becomes

MATH

Letting MATH, then the above can be rewritten as

MATH


Part (D)

Since now MATH, then MATH where I replaced $y$ by $ye^{-ct}$

Substitute the above into (8)

MATH

Where MATH


part(E)

We need to show that

MATH

where $\digamma $ is the Fourier transform operator.

Since the transform variable is $y$, we can rewrite the above as needing to show the following

MATH

We start by using result from problem (3) part (c) which says the following is true

MATH

Hence, we know (2) is true, and we need to show that (1) is true. Using the hint given, we let MATH and MATH in (2) to arrive at (1). hence starting with (2) we write

MATH

Simplify, we obtain

MATH

Which is the same as (1). QED