- Question: Where is the exam?
The exam will be in 309 BH.

- Question: In problem 2, we are given the initial
condition u(x,0)=sin x, when x>0. Should this be the partial with
respect
to x? I factored the wave equation into two first order PDEs involving
w
and v, where w=Ut - Ux and v=Ut + Ux. To determine w and v I need Ux.
Or should I assume U is differentiable at t=0 to get Ux(x,0)=cos x?
I believe that the problem is stated correctly. You only need to use the factorization of the wave equation to conclude that the solution can be written as u(x,t)=F(x-ct)+G(x+ct). You should use this form as your starting point. You can differentiate u(x,0)=sin(x) with respect to x, if needed.

- Question: Is there a misprint for problem 1?
Shouldn't the boundary
condition Wx=h(x)
on x=0 be replaced with Wx=h(y) on x=0?
Yes there is a misprint in question 1. The boundary condition should be h(y) since it is on the x-axis.

- Question: I have two doubts about the final questions:
In problem no.1, shouldn't the boundary condition h be a function of y
(h(y))?
Also, problem no.6 part c is a bit vague for me. Shouldn't we construct
G
using eigenfunction expansion method first and then generate the
solution u
utilizing Lagrange identity? I guess it's not correct if we use
e-function
expansion method to solve for u directly since it's not homogeneous in
BC's
and therefore we can't differentiate Fourier series w.r.t. x.
Thanks for your considerations,
The typo in problem 1 is addressed above. In question 6, I wanted you to construct the Green's function in part (a) by solving the homogeneous ODE for x not equal to x_0 using the general solution. This leads to solution defined in two parts. In part (c), I wanted you to construct the solution using eigenfunction expansion and hence get a series solution. Part (b) requires the Lagrange identity.

- In problem 4, we first use Laplace transform to solve the problem,
than we
need to use
Green's function. Would you please talk a little bit about how to use
green's
function to
solve that heat eqution? Does the result look the same as the one by
using
Laplace
transform?
The second part of question 4 asks you to construct the solution using the Green's function. The Green's function solves the problem with homogeneous bc and ic, and 1 replaced by the product of delta functions. The Green's function for the semi-infinite interval problem can be constructed by the method of images. The formula is given in (11.3.34). With the Green's function, you then construct the solution u using the Lagrange-type identity for the heat equation (11.3.21), which is summarized for the one dimensional case on the top of page 530 (boundary terms only). Showing the two solutions are the same is a bit difficult and I am not expecting you to be able to that on the exam. Constructing either form of the solution is fair game.

- In question 4 you ask us to use Laplace transforms to solve the
diffusion
equation. However, in the book and in the notes that you did in class
you
only did it with Utt=Uxx and Ut=Ux. Is the way if solving them the
exact same
way or is it a typo? Also with the +1 at the end of the equation do you
just
make the transform and then us it to solve?
I believe I solved a heat equation problem 13.4.4 in class before the wine cellar problem. You apply the Laplace tranform in t to the PDE to get an ODE in x with a boundary condition at x=0 and boundedness at x=infinity. You use L[1]=1/s in the PDE.

- For number 3, I got the resulting ODE's using the method of
characteristics
are a. x'=-tx and b. w'=-w. I solved for a. using the integrating
factor
(1/2)t^2 and I solved for be by assuming the solutions was Ce^rt. and
got
r=-1. I used x(0)= n to solve for the constants but I don't know what
to do
with the boundary condition. I solved for w without it and now I don't
know
if my solution is correct. Any suggestions?
You need to solve the characteristic equations a. and b. but the initial conditions for a. and b. depend on whether the curves are starting on the initial curve t=0 or x=0. You must parameterize these curves differently, leading to different forms for the characteristics. I worked an example in class similar to 12.2.4.

- On problem 6 part c), I'm a little confused. Do you want us to
construct u
directly via eigenfunction expansion or construct the Green's function
from
eigenfunction expansion and then construct u?
I guess that I wanted you to construct the Green's function first and then construct u. By constructing u directly you should see the Green's function.

- Question: Just another question... This time number 4. After you
apply the Laplace
transform and end up with the ODE Uxx-sU= 1/s, the only way I know to
solve
this is solving the homog. problem and then using variation of
parameters.
Unfortunately, the Wronskian is ghastly. Did I make some horrible
mistake, or
is this what you wanted us to do?
The independent variable in the transformed ODE is x and the Laplace transform variable s is treated as a constant with respect to x. So you can just use undetermined coefficients. In other words, treat s as a constant and it has an easy solution.

- Question: I am still having trouble with number 3. Unlike 12.2.4
these aren't linear
characteristics. I don't know how to find the characteristics.
Normally, we
looked at the solution for x. In this case, I got x=ne^(-t^2/2). This
is on
the t=0 line. However, I have no idea what the characteristics look
like.I
got the corresponding w=xe^(t^2/2)e^-t. This is true if x> than what?
Looking
on the x=1 line, I got x(1)=r=ce^(-1/2), then I got x=re^1/2e^(-t^2/2)
But
how do I solve for w(1,t). w still equals ke^-t. If I continue to solve
using
the fact that w(1,t)= w(1,r)=r= ke^-r, I can't get rid of the
exponentials and
I get e raised to xe raised to a function of t. and r is also
xe^((1-t^2)/2).
The characteristics are the curves in the x-t plane and the general solution is x=ne^(t^2/2), use separation of variables. The characteristics from t=0 have an initial condition like x(0)=xi while the characteristics starting on x=1 have the initial condition x(eta)=1. The critical characteristic that separates the two types starts at t=0 and x=1. You need to develop the similar types of initial conditions for the w equation. The w solution depends on the characteristic variable xi or eta which is a function of x and t thru the x solution. Does this help. To plot the characteristics, you can graph x=ce^(t^2/2) in the t-x plane and then flip and rotate the graph to get the x-t plane.

- I am signing off for the night. I will check again in the morning.

©2004 Charles Tier