### Math 421 diary, fall 2005 Partial Differential Equations and Boundary Value Problems

 Confession

### Monday, December 12

I "reviewed" the two-dimensional heat equation on the Pi-by-Pu square. We discussed the Steady State and Transient solutions, both qualitatively (flow lines, isothermals, etc. and -->0 for steady state) and quantitatively (the formulas involving trig/hyperbolic series for the Dirichlet problem, and involving double sine series for the transient solution).

Waving
A thin plate made of homogeneous material vibrating with no friction (no dissipation) would move according to the two-dimensional wave equation:
utt=uxx+uyy. A simple boundary value problem for this PDE on the Pi-by-Pi unit square is:
(PDE) utt=uxx+uyy. (BC) u is zero on all of the boundary all of the time: that is, u(x,0,t)=0 and u(x,Pi,t)=0 for all x in [0,Pi] and u(0,y,t)=0 and u(Pi,y,t)=0 for all y in [0,Pi].
(IC) Here both an initial position (displacement) and velocity will be specified: u(x,y,0)=J(x,y) and ut(x,y,0)=K(x,y).
The boundary condition of course corresponds to the edges of the square plate somehow fixed to the equilibrium position. This is a rather simple boundary condition. Boundary conditions for the wave equation can be much more complicated. In particular, the BC here is not what is called a clamped plate which is important in practice. That problem is much more complicated.
The additional information in the initial conditions satisfies the physical "intuiion" that the motion of an object should depend on two initial conditions (position and velocity). Also, from the mathematical point of view, the PDE is second order in time, t, and initial value problems for such equations need two chunks of information. When we try initial data like sin(3x)sin(8y)FUNC(t), which satisfies all the (BC)'s, then FUNC´´(t)=-[32+82]FUNC(t) if we want the wave equation to be true. Thus FUNC(t) should be a linear combination of sin(sqrt(32+82)t) and cos(sqrt(32+82)t).

Just J
Let's look for u(x,y,t) satisfying
(PDE) utt=uxx+uyy.
(BC) u is zero on all of the boundary all of the time: that is, u(x,0,t)=0 and u(x,Pi,t)=0 for all x in [0,Pi] and u(0,y,t)=0 and u(Pi,y,t)=0 for all y in [0,Pi].
(IC) u(x,y,0)=J(x,y)=sin(3x)sin(8y) and ut(x,y,0)=K(x,y)=0.
FUNC is a function chosen to be 1 at t=0 and to have derivative 0 at t=0. Also its second derivative should almost be itself. We hesitated for a while and then, in a dramatic display of delayed competence, we asserted:
u(x,y,t)=sin(3x)sin(8y)cos(sqrt(32+82)t)
You can (you should!) check this. By now, though, you should appreciate that everything is chosen to work out correctly. Two derivatives of cos(sqrt(32+82)t) will "spit out" two copies of sqrt(32+82) and a minus sign, and this is exactly what uxx+uyy does to u(x,y,t). Wonderful!
Additionally, the t part shows that the plate vibrates up and down, just as we could imagine an ideal plate would behave.

and now K alone
Let's look for u(x,y,t) satisfying
(PDE) utt=uxx+uyy.
(BC) u is zero on all of the boundary all of the time: that is, u(x,0,t)=0 and u(x,Pi,t)=0 for all x in [0,Pi] and u(0,y,t)=0 and u(Pi,y,t)=0 for all y in [0,Pi].
(IC) u(x,y,0)=J(x,y)=0 and ut(x,y,0)=K(x,y)=sin(44x)sin(83y).
u(x,y,t)=sin(44x)sin(83y)cos(sqrt(442+832)t)
Again, I don't think that this is too surprising after all of our experience. You could go back and consider the one-dimensional formulas again, and check that they are gotten using the same ideas.

Together? Does linearity ever stop?
So now (PDE) utt=uxx+uyy.
(BC) u is zero on all of the boundary all of the time: that is, u(x,0,t)=0 and u(x,Pi,t)=0 for all x in [0,Pi] and u(0,y,t)=0 and u(Pi,y,t)=0 for all y in [0,Pi].
(IC) u(x,y,0)=J(x,y)=135sin(3x)sin(8y) and ut(x,y,0)=K(x,y)=-403sin(44x)sin(83y).
The solution is
u(x,y,t)=135sin(3x)sin(8y)cos(sqrt(32+82)t)-403sin(44x)sin(83y)cos(sqrt(442+832)t)/sqrt(442+832)
and I might write in class on the board the word, Clearly! and this is almost true if you have followed the discussion.

I wrote an answer out in disgusting detail in termms of the double Fourier sine series of what is called here J(x,y) and K(x,y).

Conservation of energy with this wave equation

### Diary entry in progress! More to come.

One more detail
But it isn't a tiny detail. We've just seen how to solve what are called initial value problems with zero boundary data provided that the initial conditions are given as linear combinations of sin(n x)sin(m y) with n and m positive integers. But what if there aren't any natural expressions for the initial data? Look at the following:
Suppose J(x,y) is xy when x is between 0 and Pi/2 and when y is between 0 and Pi/3. Otherwise J(x,y) is 0.
To use the methods we discussed above, we need to represent J(x,y) as a sum of constants multiplying sin(n x)sin(m y) with n and m positive integers. How can we do this?
Well I just happen to know that if we take the function J1(x), defined by J1(x)=x for x in [0,Pi/2] and 0 otherwise, then J1(x)=SUMn=1infinityansin(n x) where an=(2/Pi)0PiJ1(x)sin(n x)dx=(2/Pi)0Pi/2x·sin(n x)dx (hey, I could compute that if I wanted to, but I've already done 205 integrations by parts in this course).
y stuff
Well I just happen to know that if we take the function J2(y), defined by J2(y)=y for y in [0,Pi/3] and 0 otherwise, then J2(y)=SUMm=1infinitybmsin(m y) where bm=(2/Pi)0PiJ2(y)sin(m y)dy=(2/Pi)0Pi/3y·sin(m y)dy (and I'm not going to do this computation either).
So we have the Fourier sine series for J1 and J2. If we multiply the two series and notice (!) that J(x,y)=J1(x)J2(y), then
J(x,y)=SUMn,m=1infinityanbmsin(n x)sin(m y)
This is a double Fourier sine series for J(x,y) and if we had to (!) we could now solve (or approximate solutions of) the PDE's we just looked at.

Double sine series
Suppose that J(x,y) is defined on the square with x and y between 0 and Pi. Compute
cnm=(4/Pi20Pi0PiJ(x,y)sin(n x)sin(m y)dxdy
Then J(x,y)"="SUMn,m=1infinitycnmsin(n x)sin(m y)
This is called the double Fourier sine series for J(x,y). The (4/Pi2) comes from squaring the one-dimensional normalization constant 2/Pi. The reason for the quotes around the equal sign is that this Fourier expansion has the same defects and virtues as the one-dimensional example. For J(x,y) you are likely to see, the expansion will converge to J(x,y) except maybe on a very small collection of points in the square, and you would have to be very unlikely (or be in a math course [maybe that's the same!]) to even observe the difference.

 One last example Here is a slightly more complicated example than the xy function I started with. I began this example in class. I cheated with the xy function because I got the double sine series as a product of two one-dimensional sine series. But what if I wanted a double sine series for, say, a function like J(x,y) which equals x+y when x+y is between 0 and Pi and which is 0 otherwise. Let me try to use Maple on this function. The function is a tilted plane over a triangular region in the xy-plane, and is 0 otherwise. As you can see, Maple does not handle the two-dimensional discontinuities of the function nicely. I don't know how to fix that. I wrote Maple commands to compute the two-dimensional Fourier sine coefficients for this function, and then assemble these coefficients and the appropriate sine functions into a partial sum of the two-dimensional Fourier sine series. I had Maple graph a partial sum. Plotting this function took a lot of time. The 25th partial sum has 252=625 Fourier coefficients, and then I asked that the plot be done on a 50-by-50 grid (2500 evaluations). I did cheat a little bit. If you examine the plot instruction closely (the instruction is written below), you will see that I had the partial sum plotted on the square whose sides both run from .01 to Pi, and not from 0 to Pi. That's because there is horrible misbehavior of the graph near the boundary, since this is a graph of sines, and the boundary values all have to be 0. Additionally, there is Gibbs bouncing around on these borders. You can still see some of the Gibbs stuff in the graph here, though. I tried the graph from 0 to Pi on each side, and the shape of the graph away from the edges was difficult to see.

Here are the Maple commands I used. The plot3d function is part of the package plots and one way of using it is to type with(plots): earlier in the Maple session.

```>f:=(x,y)->evalf(piecewise((x+y<Pi),(x+y),0));
>a:=(n,m)->evalf((4/Pi^2)*int(int(f(x,y)*sin(n*x)*sin(m*y),x=0..Pi),y=0..Pi));
>plot3d(T(25,x,y),x=.1..Pi,y=.1..Pi,axes=normal,style=hidden,color=N,grid=[50,50],orientation=[-56,74]);
```
HOMEWORK
Come to review sessions if you can:
Wednesday, December 14 at 4-6 PM in Hill 425 Laplace transforms & linear algebra.
Thursday, December 15 at 4-6 PM in Hill 425 Fourier series, the wave/heat equations, & associated boundary value problems.
and please come to the final exam:
Math 421:01, Friday, December 16, 8 AM-11 AM, SEC 216

 Confession

### Thursday, December 8

What every child should know about trig and hyperbolic functions ...
TRIG
onometric
cosinecos(0)=1
cos´(0)=0
y´´=-ycos(t)=([eit+e-it]/2) It wiggles between -1 and 1 always
sinesin(0)=0
sin´(0)=1
y´´=-ysin(t)=([eit-e-it]/2i) It wiggles between -1 and 1 always
HYPER
bolic
coshcosh(0)=1
cosh´(0)=0
y´´=ycosh(t)=([et+e-t]/2) It gets big both ways
both sides are +
sinhsinh(0)=0
sinh´(0)=1
y´´=ysinh(t)=([et-e-t]/2) It gets big both ways
+ on the right; - on the left

Let's try to do some two dimensional problems. I'll try the heat equation first. Consider a thin homogeneous object, lying over a region R in the plane. The same analysis as we went through for heat in a bar will work here: Newton's law of cooling, heat proportional to temperature, etc. If u(x,y,t) is the temperature at a point (x,y) in the region R at time t, then (setting all the physical constants equal to 1, again, which should irritate those who live in the real world!), we have ut=uxx+uyy. This is the two-dimensional heat equation. Again just setting up a good problem to study takes some preparation. There will be the (PDE) and certain (BC)'s, boundary conditions, and, of course, an initial condition, (IC). One arrangement which has been studied and might be useful in the "real world" is the following:
(PDE) u satisfies the two-dimensional heat equation, ut=uxx+uyy.
(BC) We prescribe the boundary temperature for all time for points, (x,y), on the boundary of R.
(IC) We give an initial temperature distribution, u(x,y,0), for all points (x,y) inside R.

This works. It's turns out to be what's called a "well-posed boundary value problem". There's exactly one solution, and it has properties which are expected from physical models. And, again, the solution turns out to be the sum of a transient solution depending more or less on the (IC) and a steady-state solution, which really depends on the (BC)'s. The steady state solution has no dependence on time. In the one-dimensional case, with ut=uxx, solutions which don't depend on time just solve uxx=0. So these functions are Ax+B for constants A and B. We can understand these well. Geometrically their graphs are just straight lines. For example, they are simply and totally determined by the correct combination of temperatures at the end of the rod or by insulating conditions.

More complicated things occur in two dimensions. I will try to study just one steady-state solution to give an example. In two dimensions, with ut=uxx+uyy, if there is no t in the u function, then the equation
uxx+uyy=0
must be satisfied. This is called Laplace's equation. I remarked that 6x2-5y+44xy-12t was a solution to Laplace's equation, and we checked this by differentiating. I then said, "So what?" By that I meant that I wanted to find physically meaningful solutions of Laplace's equation. For this I need (BC)'s. I don't need initial conditions, since there is no t in this problem.

I'll look at a very simple choice of region R in the plane. In the plane, this usually means one of two domains: either the inside of a circle (a disc) or a rectangular region. Studying the disc introducing the complications of Bessel's equation. You should know about Bessel's equation (see chapter 15), but I won't tell you. I will instead choose the somewhat simpler rectangular region.

Here is the boundary value problem I want to study:
(PDE) u satisfies the two-dimensional heat equation, uxx+uyy=0 for x and y between 0 and Pi.
(BC) The boundary temperature is 0 when y=0 and y=Pi and x=0, and it is 1 when x=1. A picture is really good here.
This should be enough information to determine a unique steady-state solution. What should this steady-state solution "look like"? I hope that class discussion will suggest:

• The temperature range on the inside will always be between 0 and 1.
• Close to the edge with temperature 1, the temperatures should be close to 1. Close to the other edges, the temperatures should be close to 0.
• There should be no interior maximums or minimums to the temperature (this is steady-state, and otherwise there would be heat flow and there is no reason for spontaneous lumps of heat to form inside the plate).
• The temperature graph should be really smooth: no jumps, no creases or shocks. Just smooth, locally averaging again.
Additionally, we might think of two ants, crawling say from left to right and from down to up.
From Phillipsburg to Bayonne, I would expect the temperature to increase from 0 to 1. I don't know much more than that.
From Cape May to Alpine, I would expect the temperature to increase from 0 to a maximum and then back down to 0. The max would of course be something less than 1. And I would expect the graph to be symmetric about 1/2.

This is all nice and qualitative, but an engineer might also want numbers: what the heck is the temperature at, say, (1,2)? So I'd like to at least roughly check the qualitative predictions, and get some guess about u(1,2).

Consider uxx+uyy=0 and separate variables. Now we will guess that u(x,y) is X(x)Y(y), and Laplace's equation becomes X´´(x)Y(y)+X(x)Y´´(y)=0 or Y´´(y)/Y(y)=-X´´(x)/X(x). Again, this must be a constant. I'll call the constant, C.

What is Y(y)?
So we've got Y´´(y)=CY(y). The boundary conditions at y=0 and y=Pi are 0.
What if C=0?
Now the solutions of Y´´(y)=0 are Ay+B. If Y(0)=0 then B=0. If Y(Pi)=0 then A=0. So this solution is only the identically 0 solution, which is not of much help in finding a non-zero product solution.
What if C>0?
The solutions of Y´´(y)=CY(y) are the linear combinations of sinh(sqrt(C)y) and cosh(sqrt(C)y). Thus Y(y)=Asinh(sqrt(C)y)+Bcosh(sqrt(C)y). Since Y(0)=0, we know that B=0 (since sinh(0)=0 and cosh(0)=1). The Pi boundary condition gives Bcosh(sqrt(C)Pi)=0 so B must be 0.
We're left with C<0.
I guess I should rename C: it will be -2. The solutions of Y´´(y)=-2Y(y) are then Asin(y)+Bcos(y). The boundary condition Y(0)=0 means that B must be 0. The boundary condition Y(Pi)=0 means that sin(Pi)=0. We know (repetition is the foundation of learning?) that is a a positive integer, which we will call n.
So now -2 is -n2 and Y(y)=sin(n y).

What is X(x)?
We know Y´´(y)/Y(y)=-X´´(x)/X(x) and our analysis of the separation constant for Y(y) shows that -X´´(x)/X(x)=-n2, so that we need to consider X´´(x)=n2X(x): a positive number. Be careful! Things are more complicated now because the separation constant has suddenly turned positive. Therefore X(x)=Asinh(n x)+Bcosh(n x). X(0)=0 forces B to be 0. So the function must be sinh(n x).

Building up from the product solutions by superposition
Now we know that sinh(n x)sin(n y) is a solution to Laplace's equation for each positive integer n. You can check this as we did in class by finding two x derivatives and two y derivatives. n2 comes out in each case because the chain rule acts on sinh and on sin, and the yy derivative has a minus sign, while the xx derivative has a plus sign.

We take linear combinations, since Laplace's equation is linear. A solution u(x,y) is, therefore, SUMn=1infinityvnsinh(n x)sin(n y) and we don't know what the numbers vn are. But, wait, we have not used the last boundary condition: u(Pi,y)=1. The sum is then constrained. Plug in x=Pi and get:
SUMn=1infinityvnsinh(n Pi)sin(n y)=1.
This means, as a function of y, the Fourier sine series of 1 should be equal to a sine series whose coefficients are vnsinh(n Pi). But the coefficients of the Fourier sine series of 1 are [2/Pi]0Pi1·sin(nx) dx. So vn must be [2/{Pi sinh(n Pi)}]0Pi1·sin(nx) dx.

These are lots of numbers. We can compute the integrals, and either by hand or with Maple guess at what they are. It turns out that for even n's the integrals are 0 (all the sine bumps cancel) and for odd n's the integrals are 4/(nPi). So the solution as explicitly as I can conveniently write it seems to be:
SUMn=1 infinity(4/[{2n+1} Pi sinh({2n+1}Pi)])sinh({2n+1} x)sin({2n+1} y)
ODD INTEGERS ONLY because the even sine Fourier coefficients of 1 are 0.

What about the sinh numbers? 1/sinh(n Pi) is about 2e-n Pi and for n large this is very small. So the series converges fast. I did some experiments with the first 50 or so terms.

The graphs exhibited below are from the class handout. There's a graph of the surface z=u(x,y), and also showed the contour lines. The graph and the contour lines near the corners where the boundary values 0 and 1 meet are sort of a mess and hard to understand, and even hard to picture and compute. The contour lines in this case are traditionally called isothermals, lines of constant temperature. We also saw that the trip from Phillipsburg to Bayonne did go from 0 to 1. Maybe (?) it is clear to some people why this graph is concave up. The trip from Cape May to Alpine is as predicted qualitatively. I even evaluated u(1,2) and got the approximation .1176183537. Because of the rapid growth of sinh, I'm sure this is quite accurate.

The surface is a graph of u(x,y) The contour lines (isothermals) of u(x,y)
From Cape May to Alpine From Phillipsburg to Bayonne
And in general?
It isn't too hard to see how everything would work if we replaced 1 on the boundary by some function of y. We would just compute the Fourier sine coefficients, multiply by the appropriate [2/{Pi sinh(n Pi)}], add, etc. Here we go:
(PDE) Laplace's equation: uxx+uyy=0.
(BC)u(x,0)=0 and u(x,Pi)=0 and u(0,y)=0 and u(Pi,y)=Q(y), where Q(y) is some weird function of y.
I think the solution will look like u(x,y)=SUMn=1infinityvnsinh(n x)sin(n y) where the coefficients are obtained as slight modifications of the Fourier sine coefficients of Q(y): vn=[2/{Pi sinh(n Pi)}]0PiQ(y)·sin(nx) dx.

The sin(n y) comes from the zero boundary conditions on the top and bottom of the square. The zero boundary condition on the left, together with the + sign in Laplace's equation give a positive separation constant, and therefore sinh(n x). Finally, the formula for vn gives me the righthand boundary condition.

Now we sat and thought a while. I suggested that we could equally well solve similar boundary value problems where three edges were 0 and some sort of function was specified along the other edge. A picture of the situation is shown below.

The Dirichlet problem
If we can solve these separate problems, then we could solve the Dirichlet problem, which is probably the single most important and most basic boundary value problem in all of partial differential equations (Google has more than 45,000 references for "Dirichlet problem"):
(PDE) uxx+uyy=0 (Laplace's equation)
(BC) Boundary values for u are specified on all of the boundary.
There's no initial conditions here, because time doesn't come into it. The method outlined above would allow you to solve this problem in the square, or, at least, find a good approximation to the solution. If accuracy is important, using software that's been previously used and validated is probably a good idea! But if you wanted to do it yourself, you could divide up the problem into 4 subproblems, and solve each of them as we did the first, and then use linearity and add up the solutions. Since the boundary data is 0 in lots of places, summing the solutions won't harm the boundary values you already have. So we can get steady-state heat flows, and by taking partial sums of the series, we can even approximate these solutions quite well.

And now for transient solutions
Suppose we want to solve the heat equation on the Pi-by-Pi square, but now we specify initial data. Here is the problem I would like to solve:
(PDE) ut=uxx+uyy. Here u is a function of x and y and t, u(x,y,t). The PDE is called the two-dimensional heat equation (the dimension count refers to the number of space dimensions, which confuses me).
(BC) u is zero on all of the boundary all of the time: that is, u(x,0,t)=0 and u(x,Pi,t)=0 for all x in [0,Pi] and u(0,y,t)=0 and u(Pi,y,t)=0 for all y in [0,Pi].
(IC) There is an initial heat distribution, J(x,y), defined for (x,y) in the square (x and y are both in the interval [0,Pi]), and we want u(x,y,0)=J(x,y).

By the way, what are our "physical" expectations (?) for this solution, a transient temperature distribution? I think that if we hold the temperature at 0 on all of the boundary, we should expect that rather rapidly the temperature distribution inside the plate should --> 0 as t --> infinity. So we should check that our mathematical model does this.

Simple, always simple ...
I tried to find some simple solutions to the boundary value problem with an initial condition. We wanted a function J(x,y) which satisfied the boundary conditions. So our initial J(x,y) was something like sin(4x)sin(17y). This is not too wild a guess, since sine at integer multiples of Pi is 0, and we need boundary values equal to 0 at 0 and Pi. Now what should u(x,y,t) be? The simplest guess is that u(x,y,t) should be sin(4x)sin(17y) multiplied by FUNC(t), some function of t. So I would like to try u(x,y,t)= sin(4x)sin(17y)FUNC(t). What should be true about FUNC(t)? If we want (IC) to be true, then certainly FUNC(0)=1. We can easily compute uxx+uyy. The result will be -(42+172)sin(4x)sin(17y)FUNC(t). And what about ut? It should be sin(4x)sin(17y)FUNC'(t). For things to agree, and for the heat equation to have a solution, we need FUNC'(t)=-(42+172)FUNC(t) with FUNC(0)=1. You should recognize this ordinary differential equation.

But not too simple ...
Hey. I now know the solution: u(x,y,t)=sin(4x)sin(17y)e-(42+172)t. The pieces all work together. It is not an accidental mess, but carefully "guessed" or assembled or something. Now that we have an idea, we shnould exploit it further. For example, we could try to solve the BVP (boundary value problem)
(PDE) ut=uxx+uyy
(BC) u is zero on all of the boundary all of the time: that is, u(x,0,t)=0 and u(x,Pi,t)=0 for all x in [0,Pi] and u(0,y,t)=0 and u(Pi,y,t)=0 for all y in [0,Pi].
(IC) u(x,y,0)=J(x,y)=-3sin(6x)sin(8y)+9sin(11x)sin(42y)
Then the solution is u(x,y,t)=-3sin(6x)sin(8y)e-(62+82)t+9sin(11x)sin(42y)e-(112+422)t.
I am using linearity (the principle of superposition) together with our wonderful idea about creating a companion exponential which when multiplied together with the sine/sine initial conditions will solve the heat equation. Also you should notice that the negative signs in the arguments of the exponentials drive down the amplitudes of the initial heat distribution. More mathematically, as t-->infinity, u(x,y,t)-->0 as we guessed earlier.

I then tried to indicate (not describe precisely) how we could solve the whole mess:
(PDE) ut=uxx+uyy
(BC) u is specified on all of the boundary all of the time: that is, u(x,0,t)=S(x) and u(x,Pi,t)=P(x) for all x in [0,Pi] and u(0,y,t)=R(y) and u(Pi,y,t)=Q(y) for all y in [0,Pi].
(IC) There is an initial heat distribution, J(x,y), defined for (x,y) in the square (x and y are both in the interval [0,Pi]), and we want u(x,y,0)=J(x,y).
First I would solve the Dirichlet problem for uxx+uyy=0 with the indicated boundary conditions. I would get some sort of steady-state solution, which I'll call SS(x,y). Then I would solve the heat equation with zero boundary conditions and with initial conditions equal to J(x,y)-SS(x,y). I would add the solution of that problem to SS(x,y). Because of various seros in boundary conditions, etc. (linearity again) the function of x and y and t would actually solve the whole problem stated above. Of course, I would hope that this could all be implemented by a properly tested computer program. There are lots of details which an unassisted human being could (would?) screw up.

For the initial condition, I would assemble a
Double sine series
Suppose that J(x,y) is defined on the square with x and y between 0 and Pi. Compute
cnm=(4/Pi20Pi0PiJ(x,y)sin(n x)sin(m y)dxdy
Then J(x,y)"="SUMn,m=1infinitycnmsin(n x)sin(m y)
This is called the double Fourier sine series for J(x,y). The (4/Pi2) comes from squaring the one-dimensional normalization constant 2/Pi. The reason for the quotes around the equal sign is that this Fourier expansion has the same defects and virtues as the one-dimensional example. For J(x,y) you are likely to see, the expansion will converge to J(x,y) except maybe on a very small collection of points in the square, and you would have to be very unlikely (or be in a math course [maybe that's the same!]) to even observe the difference.

HOMEWORK
1. Please give me on Monday, the next (and last) class a sheet of paper with all of your homework grades and your name. Some of the homework grades have been misplaced and I want to give people all the credit they deserve.
2. I handed out a sheet of homework problems on two dimensional PDE's. You may try these yourself. Answers to these questions and to some questions from the textbook are given here.
3. Our final exam is on Friday, December 16, from 8 to 11 AM in SEC 216, our usual classroom. Review sessions are scheduled as follows:
Wednesday, December 14 at 4-6 PM in Hill 425
Laplace transforms & linear algebra.
Thursday, December 15 at 4-6 PM in Hill 425
Fourier series, the wave/heat equations, & associated boundary value problems.
4. Please also see here for more review material.

 Confession

### Monday, December 5

The instructor made a number of silly mistakes. No one can equal his ineptitude when he gets going!

Here is the solution to the the wave equation:
(PDE) The wave equation: uxx=utt
(BC) u(0,t)=0 and u(Pi,t)=0 for all t.
(IC) u(x,0)=f(x) for x between 0 and Pi (initial position) and ut(x,0)=g(x) for x between 0 and Pi (initial velocity)
has solution u(x,t)=SUMn=1infinitycnsin(nx)cos(nt)+SUMn=1infinitydnsin(nx)sin(nt). where cn=(2/Pi)0Pif(x)sin(nx) dx and dn=[2/(nPi)]0Pig(x)sin(nx) dx.

We need the n's in the dn formula because that part deals with a derivative initial condition, and the n's make things work out correctly. As I remarked in class, this is wonderful. Well, maybe it is. If you are given f(x) and g(x) and you need to compute u(.2,.7), then you certainly can get a good approximation using the solution above. But it turns out that there are other ways of looking at the solution which are very useful. These other ways are already in the pictures, if you look closely. I tried to motivate the other ways algebraically by looking at, say, sin(nx)cos(nt), which is a piece of one of the sums.

Look at the Fourier solution again ...
Trig identities tell me that sin(nx)cos(nt)=(1/2)sin(nx+nt)+(1/2)sin(nx-nt). Now this is (1/2)sin(n(x+t))+(1/2)sin(n(x-t). A similar result is true for sin(nx)sin(nt). You could imagine that we then reassemble the infinite sums and separate them, with one chunk involving x-t and one chunk involving x+t. Huh. So let's try "something completely different" (quoted from Monty Python).

Make a guess
Suppose Func and Otherfunc are two functions of one variable. Then consider the function
u(x,t)=Func(x-t)+Otherfunc(x+t)
. The chain rule tells me that
ux(x,t)=Func´(x-t)(1)+Otherfunc´(x+t)(1)
and I can't write more than that since I haven't told you much about Func and Otherfunc. But I will differentiate again with respect to x:
uxx(x,t)=Func´´(x-t)(12)+Otherfunc´´(x+t)(12)
These computations will get somewhere, soon. Because we now try things with respect to t:
ut(x,t)=Func´(x-t)(-1)+Otherfunc´(x+t)(1)
and most important, notice the -1 which happens when the chain rule, differentiating with respect to t, meets x-t. But now the second t derivative:
utt(x,t)=Func´´(x-t)(-1)2+Otherfunc´´(x+t)(12)
Since -1 squared is 1, well, golly, we have found solutions of uxx=utt.

Waves moving
What's going on? Well, I then tried to draw some silly pictures. Suppose I give you only graphical information about Func. Its graph is as shown to the right. So Func is 0 when its input is less than 1 or greater than 5, and in between it has the strange profile shown.

What will Func(x-t) look like for various times? Well, we discussed this and below is the result of sketching Func(x-t) when t=0 and t=1 and t=2. Please look carefully at the numbers on the horizontal axes.

Apparently Func(x-t) is really a wave which is moving to the right. A similar analysis shows that Otherfunc(x+t) is a wave moving left. Maybe I should relabel these pieces, and we see that this is a solution of the wave equation:
u(x,t)=Right(x-t)+Left(x+t)
This is the D'Alembert form of the solution to the wave equation. I think I now introduced another classic parameter of the wave equation, a positive number called c, which will be the speed of propagation of the waves (in context, this could be, for example, the speed of light or the speed of sound or ... lots of things). Then consider the function
u(x,t)=Right(x-ct)+Left(x+ct)
It turns out that this function satisfies the wave equation with a slight change:
c2uxx=utt
Problems 13 and 14 of section 13.4 of the text discuss the D'Alembert solution and even give the following relationship between the initial position/velocity solutions:
If u(x,0)=f(x) and ut(x,0)=g(x), then
u(x,t)=(1/2)[f(x+ct)+f(x-ct)]+[1/(2c)]x-ctx+ctg(s)ds.

I don't believe that you need to memorize many of these formulas. You should know that they are around, though. If you look at the pictures which were distributed and look at the moving images, I think you should be able to guess at the left and right traveling waves.

Qualitative comparison of solutions to the wave and heat equations
Behavior for large time Speed of propagation Rough vs. smooth Reversibility
Solutions of the heat equation The solutions are sums of transient and steady-state. Here (BC)'s give different steady-state solutions. (IC)'s give transient solutions which decay rapidly with time so that long term solutions are very close to steady-state. In this model, speed seems infinite, but effects are exponentially small for small time as you move away from where the (IC)'s are not 0. Even if (IC)'s are very rough, the heat equation averages, and so for any positive time, even very small positive time, the temperature is very smooth. Time can't be reversed: if u(x,t) is a solution to the heat equation, u(x,-t) is rarely also a solution. Things don't undiffuse: entropy increases.
Solutions of the wave equation This is idealized motion, and there is conservation of energy: no dissipation. The solutions are all periodic, and there is no equilibrium solution except for constants. There is a fixed propagation speed (in homogeneous media). Waves moving right and left can be very rough: there can be shocks, and these shocks may persist. If u(x,t) solves the wave equation, so does u(x,-t) (because (-1)2=1). Therefore a film run in reverse of a solution of the ideal wave equation would also show a picture of a solution to the wave equation. Solutions are reversible in time.

Comments on colors should be directed to the Math 421 webmaster, Hieronymus Bosch.

The instructor discussed how to solve a homework problem. He didn't do the computational parts.

Light cones
Light cones weigh much less than heavy cones. Sorry. Back to work: imagine a very long string, which is somehow held tight along the x-axis. Suppose I tweak the string a little bit around a number x (I change the position and/or the velocity). What could possibly happen at later time? If you really "buy into" the D'Alembert form of the solution, then you see that the information about the tweaking can't be detected until the waves (moving left or right) get there.
The possible influence doesn't get felt outside of the interval [x-ct,x+ct]. This is called the light cone. If you believe, really believe, in special relativity, and so you think that the speed of light is an absolute limit, then you can't signal faster than the speed of light, so a person standing at x at time 0 can only influence events at time t in the interval [x-ct,x+ct].

The speed limit: 299,792,458 meters/second. It's not just a good idea, it's the law!

Well, the idea is important. So now you engineers have lots to remember about the wave equation. There are Fourier series solutions, D'Alembert's traveling wave solutions, the pictures, etc.: lots of stuff.

I began the study of the two dimensional heat equation, which I will continue next time.

HOMEWORK
Our final exam is on Friday, December 16, from 8 to 11 AM in SEC 216, our usual classroom. More information about the exam will appear shortly.
Please do think about when a review session can be held, and when office hours might be convenient.

 Confession

### Thursday, December 1

The notes for the class follow, but also see here, please.

The basic assumptions for the Wave Equation are described on pp. 693-4 of the textbook. I have been sworn to and at by the Mech Engineering faculty who assert that they will teach the derivation of the wave equation in their courses and I should stay away. By the way, there are many nice descriptions of this derivation on the web. In any case, the wave equation models the motion of an idealized string. We'll assume the string is stretched between two points on the x-axis. I'll call the points x=0 and x=Pi (not a big surprise!). We will assume that the motion of the string is perpendicular to the x-axis, and stays in the xy-plane. The function u(x,t) is supposed to model the height (displacement?) of the portion of the string which is over the point x at time t. Then it turns out that uxx=utt: this is the one-dimensional wave equation. I have forced the physical constants to be 1. This turns out to be a rather serious (and sometimes misleading!) assumption, but I'll work with it for today since it will make some of the algebra easier.

I should emphasize that this vibrating string is supposed to be ideal. So there is no friction, and everything "works". As I said in class, it is probably a good idea to review a much simpler model, the vibrating sPring (not sTring, very easy to mispronounce for me!). Here is a discussion I wrote of the standard model (Hooke's Law, F=ma, etc.) of the spring. It is useful to note that a picture of the spring of any time doesn't have complete information: both the position (in relation to equilibrium) and the velocity of the spring must be considered. Also in this simple model, these is also no friction, and so no dissapation of the energy: energy is conserved, and this has very important consequences for the spring. It never stops vibrating!

O.k., back to the string. The simplest boundary value problem for the string has the ends fastened: u(0,t)=0 and u(Pi,t)=0 for all t. Then we would expect to have the string vibrate, depending what the initial condition(s) are.

So I will study the following boundary value problem:
(PDE) The wave equation: uxx=utt
(BC) u(0,t)=0 and u(Pi,t)=0 for all t.
(IC) Well, let me skip this for a second.

Now we separate variables: if u(x,t)=X(x)T(t), then uxx=utt implies that X´´(x)T(t)=X(x)T´´(t) so that X´´(x)/X(x)=T´´(t)/T(t). The standard logical dance (is it a function of t -- left-hand side says no     is it a function of x -- right-hand side says no) shows that this is a constant, called the separation constant.

It is not true that such constants always turn out to be negative if they arise in physical problems: we'll see this next week. Last time we analyzed this constant using "math": now let's look at some physical reasoning. Hey: if T´´(t)=(a positive constant)T(t) then T(t) would have exponential growth or exponential decay. This does not fit our physical intuition, so a positive separation constant should be rejected. If, by the way, this offends your sense of mathematical propriety (!), then you can eliminate such constants using strictly math reasoning. If the separation constant is 0, then, say, X´´(x)=0, so X(x) is Ax+B and (because of the boundary conditions) A=0 and B=0: the string doesn't move at all. The situation is not very interesting.

Word of the day: fret
One definition:
n.[Mus] each of a sequence of bars or ridges on the finger-board of some stringed musical instruments (esp. the guitar) fixing the positions of the fingers to produce the desired notes.
Another definition:
v. a. be greatly and visibly worried or distressed.
b. be irritated or resentful.

So the separation constant is negative, and, like the textbook, we will call it -2. Therefore the PDE separates into these two ODE's: X´´(x)=-2X(x) and T´´(t)=-2T(t). I'll analyze X(x) first.

Since X´´(x)=-2X(x) we know that any solution X(x) must be a linear combination of cos(x) and sin(x). The boundary conditions now help us. When x=0 we should get 0, so there can be no term with cos(x) (because this is 1 at 0 while the sine term is 0 at 0). When x=Pi, we also get 0, so sin(x)=0. Draw a picture of sine in your head, and that should convince you that must be an integer. Since sin(-v)=-sin(v), we can push a negative sign onto the scalar multiplier. Therefore we learn that
For this boundary-value problem, the eigenvalues are all positive integers, n, and the eigenfunctions are sin(nx).

What are the possible partners of sin(nx) in u(x,t)=X(x)T(t)? Well, T´´(t)=-n2T(t) so that T(t) must be a linear combination of sin(nt) and cos(nt). Notice that both sin(nx)cos(nt) and sin(nx)sin(nt) satisfy uxx=utt. You can check this by direct differentiation! The principal of superposition (linearity!) then applies. The solution u(x,t) must be a sum of two kinds of terms (as just mentioned). So:
u(x,t)=SUMn=1infinitycnsin(nx)cos(nt)+SUMn=1infinitydnsin(nx)sin(nt).

Now we do need to discuss initial conditions. I put this off because the initial conditions are more complex than with the heat equation. Here an initial position is not enough. Go back and think about the vibrating spring model again: a snapshot of where the spring is does not have full information. We also need to know something about the motion of the spring at the time to predict the future behavior. Here we can specify both an initial position and an initial velocity for the string.

We can specify these:
(IC) u(x,0)=f(x) for x between 0 and Pi (initial position) and ut(x,0)=g(x) for x between 0 and Pi (initial velocity)

For simplicity, I will first study the case where g(x) is the zero function (no velocity) and f(x) is specified (the initial position). What happens now? If u(x,t)=SUMn=1infinitycnsin(nx)cos(nt)+SUMn=1infinitydnsin(nx)sin(nt), then u(x,0)=b>SUMn=1infinitycnsin(nx) since cosine is 1 at 0 and sine is 0 at 0. Thus SUMn=1infinitycnsin(nx)=f(x) and we should recognize this equation. The series is simply the Fourier sine series for f(x), and the coefficients cn are given by cn=(2/Pi)0Pif(x)sin(nx) dx. The (2/Pi) is a normalization constant. But we also can get information about the dn's. Since u(x,t)=SUMn=1infinitycnsin(nx)cos(nt)+SUMn=1infinitydnsin(nx)sin(nt), ut=SUMn=1infinitycnsin(nx)(-n) sin(nt)+SUMn=1infinitydnsin(nx)n cos(nt). Since ut(x,0)=0 in our simple model, we get SUMn=1infinitydnsin(nx)n=0. But the functions sin(nx) are all linearly independent, and therefore the dn coefficients are all 0. Hey!

We have solved (?) the boundary value problem:
(PDE) The wave equation: uxx=utt
(BC) u(0,t)=0 and u(Pi,t)=0 for all t.
(IC) u(x,0)=f(x) and ut(x,0)=0 for x between 0 and Pi.
The solution is u(x,t)=SUMn=1infinitycnsin(nx)cos(nt) where cn=(2/Pi)0Pif(x)sin(nx) dx.

So let's look at the velocity case:
(IC) u(x,0)=0 for x between 0 and Pi (initial position) and ut(x,0)=g(x) for x between 0 and Pi (initial velocity).
Now we have u(x,t)=SUMn=1infinitycnsin(nx)cos(nt)+SUMn=1infinitydnsin(nx)sin(nt). When t=0, this becomes SUMn=1infinitycnsin(nx) and if this sum is 0, then (linear independence again) all of the cn's are 0. But ut(x,t) is SUMn=1infinitycnsin(nx)[-sin(nt)]n+SUMn=1infinitydnsin(nx)cos(nt)n. Plug in t=0 and get SUMn=1infinityndnsin(nx). So we want g(x)=SUMn=1infinityndnsin(nx). Well, this is correct if ndn is the nth Fourier sine coefficient of g(x). That is, if dn=[2/(nPi)]0Pig(x)sin(nx) dx.

Does this solution give you all the intuition you would want? I don't find it too comforting. Certainly, if you give me an initial profile for the string, f(x), and an initial velocity, g(x), I could (maybe!) compute the Fourier sine series coefficients. Then the wonderful formulas above would allow me to predict the future behavior of the string. Of course, in the real world, I would have to use partial sums, get numerical approximations, etc.

Then we looked at some examples.

The first example
Here f(x) was a function which was piecewise linear, 0 on [0,Pi/3] and [Pi/2,Pi], and had slope 1 on [Pi/3,Pi/3+Pi/12] and had slope -1 on [Pi/3+Pi/12,Pi/2]: a sort of little triangular tent, and g(x) was the 0 function We discussed what how this initial data would evolve with the wave equation. Here are some things we could see:

• The initial peak was actually a sum of two waves, one moving right and one moving left.
• For later time, the waves kept their shape and their non-differentiability ("shocks").
• When the waves hit the ends, the waves were reflected back upside down.
• After an advance of 2Pi, the initial picture returned.
I tried to lead a discussion and convince people that these observations predicted by the model actually could be seen experimentally. Sigh. I even brought some string to class. The qualitative aspects are very different from what happens in the heat equation. It might help to think of the string as sort of a mattress of springs tied together at the tops, but each of them sort of behaving mostly like a Hooke's Law spring. This is not a very good analogy. Oh well. The picture shown is supposed to have "tiny" springs (in blue, if you can see the colors) tied together by a thick brown string. The profile of the top of this "mattress" is supposed to be like the f(x) which is in the handout. Oh well. The middle of the bump moves down faster because the spring is stretched more there. The springs at the edge pull up the nearby mattress points. Oh well. Not a very good analogy, maybe.

Wow! Now I tried to look with students at the second page of the handout. Here maybe we could imagine that the string's initial position is 0 for all x, but that I have given an impulse up to the portion of the string between Pi/3 and Pi/2. I think this is slightly hard to imagine, but we discussed the results anyway.

The first picture (for t=.1) shows that part of the string is moving up. The tilted part of the profile, at the left and the right, shows that the string is tied together (remember the mattress analogy: the string is sort of made up of springs, but there's a tension between adjoining "springs"). So there is resistence at the edge of the impulse up. Then the effect begins to spread. At t=1.2, what I think is a peculiar thing happens on the left. There are two tilted line segments of slightly different slope. What is happening? Well, the initial "up" hit the end of the string. We learned that the profile is reflected down and fed back. The result in this case is that the feedback of the wave front is subtracted from what is still there, and we get the piecewise linear behavior. Something similar happens on the right in the t=1.7 picture. The final picture, at t=3, shows something very much like the initial behavior. But that is correct, since when we increase t by 2Pi in the Fourier series, all of the functions sin(nt) in the Fourier series have the same value: sin(n(t+2Pi)=sin(nt+2nPi)=sin(nt). So u(x,t+2Pi)=u(x,t): this is perfect, ideal wave behavior. There is no friction, no gravity, and no energy dissipation

I find all this very difficult to understand. I asked if some moving gifs would help, and was told, "Yes," so here they are.

HOMEWORK
I'll discuss the free boundary problem for the wave equation Monday, as well as the D'Alembert solution for the wave equation (see problems 13 and 14 in section 13.4).
Please hand in (the last problems to hand in!) these problems:
13.3: 4 13.4: 3, 5 AND problem 9 from 13.4. I thank Mr. Clark for pointing out that I had already assigned problem 1 in 13.3. I apologize.

 Confession

### Monday, November 28

I want to try another "physical" problem and see if this mathematical model predicts what our intuition suggests.

Fixed (possibly non-zero) end temperatures
Now, again, let's have a bar overlaying the interval [0,Pi] whose lateral ends are kept at specific temperatures. So the model looks like this:
u(x,t) is the temperature at position x and at time t.
PDE The heat equation: ut=uxx
IC Initial condition: u(x,0)=f(x) for 0=<x<=Pi.
BC Boundary condition(s): u(0,t)=u0 and u(Pi,t)=uPi.

Guessing long-term behavior
So what is the anticipated long-time behavior of these solutions? Locally, the heat equation tries desperately to average temperature as time progresses: if I am at position x, I will "feel" positions x+(delta)x and x-(delta)x and try to update (?) my temperature based on what's happening at those places compared to my own. This means that the heat equation would urge (??) the long-term temperature distribution to interpolated between the end temperatures which are fixed. That is actually what the model will predict.

Let me assume some fixed values for u0 and uPi. This will decrease the amount of algebra I'll need to do. So I will assume that u(0)=2 and u(Pi)=5. There are no particular vices or virtues about these numbers. They're just convenient small integers.

Suppose I seek functions, u(x,t), which satisfy the heat equation, ut=uxx but which don't depend on time. Such solutions would be called steady-state. If there is no time dependence, then uxx=0. I know all the solutions of this equation: Ax+B. These are straight lines. If I now want to have the solution satisfy u(0,t)=2 and u(Pi,t)=5, then I compute (actually, students computed!) that the steady-state solution should be 2+(5/Pi)x. The simplicity of this formula maybe is not something one could have guessed from a verbal description of the mathematical model. I might readily have guessed that the solution would increase from to 2 to 5, but the precise shape of the curve (and details about the formula) could well have been different.

Now here's a wonderful little trick to go from the "2 and 5" boundary conditions to "0 and 0" boundary conditions. Suppose that u(x,t) satisfies
PDE The heat equation: ut=uxx
IC Initial condition: u(x,0)=f(x) for 0=<x<=L.
BC Boundary condition(s): u(0,t)=2 and u(Pi,t)=5.
Let's call the function 2+(5/Pi)x, SS(x). "SS" means "steady-state" here. Then consider the function U(x,t)=u(x,t)-SS(x). What can we say about U(x,t)? Since SS(x) and u(x,t) both satisfy the heat equation, and the heat equation is linear, the function U(x,t), which is a linear combination of u(x,t) and SS(x), also satisfies the heat equation. What about the boundary conditions? Well, U(x,t) is assumed to satisfy the "2 and 5" boundary conditions, and we know that SS(x) satisfies the "2 and 5" boundary conditions, and these conditions are linear. Therefore the difference satisfies the "0 and 0" boundary conditions. Hey!

The U(x,t) function is a solution to the old problem. And we know that the solutions to the old "0 and 0" problem all -->0 as t-->infinity. This means that U(x,t)=u(x,t)-SS(x)-->0 so that u(x,t)-->SS(x) as t-->infinity. This is exactly what we guessed should happen. So again this model predicts the expected behavior.

Flux=0
The last situation I'd like to look was mentioned previously:
The setup Think of a bar with a certain distribution of heat initially, and that somehow both ends of the bar are insulated: no flow of heat is possible in or out of the ends. Maybe a diffusion setup is simpler to understand here. We sprinkle (?) sugar into a tube of water, and no further additions or subtractions to the tube are made. Then we try to study how the concentration of the sugar evolves. Presumably there is Brownian motion (caused by "random" ambient heat) which moves things around.
The expected result Over a long time (a long, long time ...) I would expect the sugar to be close to uniformly distributed in concentration. There's no reason for lots of suger (or too little sugar) to be in any one chunk of the tube: this would be remarkably unlikely. So the concentration of sugar, u(x,t), would be close to constant as t-->infinity.

I call this "Flux=0" because flux is another word for flow.

The mathematical model
Here is what we will work with:
PDE The heat equation: ut=uxx
IC Initial condition: u(x,0)=f(x) for 0=<x<=L.
BC Boundary condition(s): ux(0,t)=0 and ux(Pi,t)=0.
We first will separate variables. This phrase is the name for assuming that u(x,t) is a product function: X(x)T(t). The result of using the PDE is then X´´(x)T(t)=X(x)T´(t). And then we get X´´(x)/X(x)=T´(t)/T(t). This is now a constant, depending on neither x nor t. What can happen?

The separation constant
There are several possibilities.
Irving The separation constant is positive.
Jessica The separation constant is zero.
Thrag The separation constant is negative.
What happens if we assume Irving? Well, then T´(t)=(positive #)T(t), so that T(t)=(const)e(positive #)t. I think that this is physically unlikely: the temperature can't just increase and increase (in this model heat is neither created nor destroyed inside the rod -- other, more complicated models deal with such situations). I will throw out the possibility of Irving.
What happens if we assume Jessica? Well, then T´(t)=0. And the temperature doesn't change with time at all. This means that u(x,t) is a steady-state temperature distribution, and is Ax+B. But the flux, ux(x,t) is A, so A must be 0 for this no-flux example. So if Jessica occurs, the temperature is totally constant.
We are left with Thrag (a very common name, I think, on the planet Zorkle). This is why the book essentially assumes that the separation constant is -2. In this case I tried to argue using physical considerations that this separation constant must be negative (expect in the simple case of constant temperature). In the previous analysis I tried to show how the mathematics alone led to the separation constant being positive: you can think about either or both methods. The model will be more useful to you if you know more ways to play with it, though.

Thrag, or the analysis of the negative separation constant
So we assume that X´´(x)=-2X(x) and T´(t)=-2T(t). The more interesting equation is the first, whose solutions are the linear combinations of cos(x) and sin(x): X(x)=Acos(x)+Bsin(x). Now we use the boundary conditions:
ux(0,t)=0 and ux(Pi,t)=0.
X´(x)=-Asin(x)+Bcos(x). X´(0)=0 means B=0. The case =0 will be covered by Jessica, so we know that B=0. Therefore X(x)=Acos(x) with X´(x)=-Asin(x). Now X´(Pi)=0 means Asin(Pi)=0. Again I'll assume A and are not 0 (dear Jessica!) so we must have sin(PI)=0. For which is this going to be true? must be an integer. Since sine is an odd function, negative integers are the "same" as positive ones in this use (change the sign of A). Therefore the acceptable values of are 0 and positive integers. These are the eigenvalues of this boundary value problem

Thus we have the eigenfunctions cos(n x) where n is a non-negative integer. Each of these functions has a partner which helps it solve the heat equation. That partner is obtained from T´(t)=-n2T(t). Thus u(x,t)=cos(n x)e-n2t is a solution of the heat equation satisfying these boundary conditions. Now the principle of superposition (an old-fashioned way of declaring linearity again) says that u(x,t)=(1/2)a0e-02tcos(0x)+SUMn=1infinityane-n2tcos(nx) should also be a solution, where, since u(x,0)=f(x) and e0=1, we know that (t=0)
f(x)=(1/2)a0+SUMn=1infinityane-n2tcos(nx) is the sum of the Fourier cosine series for f(x).

What do we know? Well, the an's are given by an=(2/Pi)0Pif(x)cos(n x) dx. The reason for the (1/2) in front of a0 in the previous formula is because the normalization constant for 12 on [0,Pi] is just Pi, while for (cos(n x))2 (n>0) the constant is Pi/2. What about the various pieces of the formula as t-->infinity? Well look at
ane-n2tcos(nx)
If n>0, then the exponential decreases rapidly, and the maximum amplitude of the cosine goes to 0. So the n>0 terms are not likely to contribute much to the limiting behavior. But what about n=0? If n=0 the term is (1/2)a0e-02tcos(0x)=(1/2)a0
but a0=(2/Pi)0Pif(x)cos(0 x) dx=(2/Pi)0Pif(x) dx
so as t-->infinity, we suspect that the solution to this "no flux" boundary value problem approaches (1/Pi)0Pif(x) dx. This is the average value of f(x) over the interval [0,Pi]. So, the total amount of heat (or sugar in solution) stays the same (the total area) but the temperature (concentration) approaches a constant. Notice how the 2's cancel. Wow! I'll show some pictures below.

One last example
I tried to think about a bar whose left end was kept at temperature 0 and whose right end was insulated. This leads to the following boundary value problem:
u(x,t) is the temperature at position x and at time t.
PDE The heat equation: ut=uxx
IC Initial condition: u(x,0)=f(x) for 0=<x<=L.
BC Boundary condition(s): u(0,t)=0 and ux(Pi,t)=0.
I went through this rather rapidly. Here we go:

Step 1: separate variables and use boundary conditions
If u(x,t)=X(x)T(t), then X´´(x)/X(x)=T´(t)/T(t). This is a constant, and, as before (using either mathematical or physical considerations), the constant will be called -2.

Step 2: use the boundary conditions to get eigen{values|functions}
X´´(x)=-2, so X(x) is in the span of cos(x) and sin(x). Since X(0)=0, we drop the cos(x) term. Now we consider sin(x), whose derivative is cos(x). The second (flux) boundary condition gives cos(Pi)=0. Now =0 gives Ax+B but this has B=0 (first BC) and has A=0 (second BC). So we need cos(Pi)=0. Wow. The 's satisfying this are (1/2)(odd positive integer). Look at the graph of cosine to see this, please! So the eigenvalues are =(1/2)(odd positive integer)=(1/2)(2n+1) for n at least 0 and the associated eigenfunctions are sin((1/2)(2n+1)x).

Section 12.5 of the text, entitled Sturm-Liouville Problem, describes the general situation. We have studied three examples. Generally, if you have a second-order ODE and give two specifications of boundary behavior at two distinct points (the function, the derivative, or some linear combination) there will be a sequence of eigenvalues (tending to +infinity) and a sequence of associated eigenfunctions. I'll list these below. There is also an associated Fourier-like series in each case.
You should know this result and be able to compute the eigenvalues and eigenfunctions in specific situations.
There are many computational situations where such problems arise. That's why there are so many sections in the book devoted to Legendre functions and Bessel functions and ...

Step 3: use the eigenfunctions to write a solution of the heat equation and also use them to satisfy the initial condition
The eigenfunction sin((1/2)(2n+1)x) on [0,Pi] has a normalization constant: 0Pi(sin((1/2)(2n+1)x))2 dx. Again, the cosine and sine squares have the same shape on the interval [0,Pi], so the normalization constant is 2/Pi. It isn't always true that 2/Pi will be the result, but it is here. So in fact, if the initial temperature distribution is f(x), we will have
SUMn=1infinitycne-n2tsin((1/2)(2n+1)x)
with cn=(2/Pi)0Pif(x)sin((1/2)(2n+1)x) dx. Then the solution to this heat equation boundary value problem is u(x,t)=SUMn=0infinitycne- ((1/2)(2n+1))2tsin((1/2)(2n+1)x).

Too many pictures?
Wow. You may think this is all too difficult to comprehend. I claim you can work with this, really. If you need to consider such a problem in an engineering application, the theory will lead you through all the details, and something like Maple can be relied upon to compute good numerical approximations and/or to graph the results. It isn't hard to modify the Maple commands I wrote to handle other situations.
Warning! Lots of pictures in the following link, so lots of bandwidth is needed!
Here is a link to pictures showing the time evolution for the three boundary value problems we analyzed when the initial temperature distribution, is a unit height function on the interval [Pi/3,Pi/2].

 Confession

### Monday, November 21

Separation of variables

Example

Applied to the heat equation
Here we looked at the boundary value problem
The {Heat|Diffusion} equation ut=uxx
Boundary conditions u(0,t)=0 and u(L,t)=0 for all t>=0.
Initial condition u(x,0)=f(x) for 0=<x<=L.

This is almost the simplest problem we could consider which has some physical meaning. So we have an initial heat distribution (or some initial concentration, if we see this as a diffusion problem) and we keep the ends of the bar at 0 temperature (in diffusion, the ends of the bar have access to large amount of "stuff" with 0 contentration). The physical intuition is that for T "large", u(x,T) should be very close to 0.

The first three equations in the boundary value problem are homogeneous linear equations:
ut-uxx=0.
u(0,t)=0.
u(L,t)=0.
This means that:
If u1 and u2 satisfy all three equations, then u1+u2 does also.
If u1 satisfies all three equations, and c is a constant, then cu1 also satisfies all three equations.

This is nice. But we still need to find physically meaningful solutions which we can understand easily and compute efficiently.

The heat equation with ends at temperature 0
Look for product solutions. Here I will intentionally be somewhat ...obtuse ("dull-witted; slow to understand.") and try to go slowly. We will use similar approaches to solve several other boundary value problems, though. So assume that u(x,t) is a product of P(x)T(t), a product of a solution involving position and a solution involving time. This breaks the linearity framework which we have been exploiting all semester, but it turns out to be extremely successful.

If u(x,t)=P(x)T(t) then ut=uxx becomes P(x)T´(t)=P´´(x)T(t). Now put all the x-stuff on one side and all the t-stuff on the other side. Then T´(t)/T(t)=P´´(x)/P(x). But (more cleverness!) what's on the left is a function only of t. When we change x, the left-hand side doesn't change. When we change t, look at the right: it doesn't change. Therefore the value of the two sides of T´(t)/T(t)=P´´(x)/P(x) must be a constant which I will call (brilliantly!), CONST.

So we know that T´(t)/T(t)=CONST and P´´(x)/P(x)= CONST. We have separated variables and somehow changed considering a partial differential equation with 2 variables to a pair of ordinary differential equations each involving 1 variable.
This idea is called separating variables. The CONST is called the separation constant.

The t equation
T´(t)/T(t)=CONST is T´(t)=CONSTT(t). You should recognize that solutions of this equation are all multiples of eCONSTt.

The x equation
P´´(x)/P(x)= CONST is P´´(x)= CONSTP(x). This should be almost as familiar, but the solutions (when we ignore complex numbers!) look different depending on the sign of CONST.
If CONST<0
then solutions are all linear combinations of sin(sqrt(-CONST)x) and cos(sqrt(-CONST)x). It might help to consider a specific example, say P´´(x)=-78P(x). Then solutions are sums of multiples (linear combinations) of sine and cosine of ... sqrt(78)x. Notice that we have CONST=-78, so that sqrt(-CONST) is sqrt(-(-(78))=sqrt(78). The signs work out, since the second derivative of sines and cosines emits (!?) a minus sign compared to the original function.
If CONST>0
then solutions are all linear combinations of certain exponentials, esqrt(CONST)x and e-sqrt(CONST)x. Many people would prefer to use other functions (sinh and cosh) as a basis of the solution space to this ODE. But "you folks" don't seem to like the hyperbolic functions very much. This may be a sign of inexperience, because sinh and cosh are wonderful functions.
If CONST=0
Then P(x)=A+Bx.

Now the boundary conditions
So far we've only used the heat equation itself to find out about P(x)T(t). Since T(t) is an exponential and (except when always 0) is therefore never 0, the boundary condition u(0,t)=0 means P(0)=0 and u(L,t)=0 means P(L)=0. The boundary condition u(0,t)=0 means P(0)=0 and u(L,t)=0 means P(L)=0. First let's consider
If CONST>0
Then P(x)=Aesqrt(CONST)t+Be-sqrt(CONST)t. Now P(0)=0 gives A+B=0 (since e0=1) and P(L)=0 gives Aesqrt(CONST)L+Be-sqrt(CONST)L=0. These constraints on A and B are a system of two linear equations in two unknowns. Since this is a homogeneous system, a solution is certainly A=0 and B=0. This doesn't give a very interesting solution of the original heat equation (the temperature is always 0!) but I wonder if there are any non-trivial solutions. Well, there's only the trivial solution if the determinant of the coefficient matrix is non-zero. So compute the det of

```(     1            1      )
( esqrt(CONST)L e-sqrt(CONST)L)```
and this determinant is e-sqrt(CONST)L)-esqrt(CONST)L. Could this be 0? Well, if the inputs to the exponentials are the same this happens.So could -sqrt(CONST)L be equal to -sqrt(CONST)L? L isn't 0 (a rod of zero length is not physically interesting) so this means -sqrt(CONST)=-sqrt(CONST), but CONST>0, and this equation is impossible. So there are no solutions when CONST<0.

What happens if CONST=0? If x=0 then we see that A=0, and the other boundary condition shows that B=0. So there are no non-trivial solutions for this alternative.

What happens if CONST<0? If x=0 the linear combination Asin(sqrt(-CONST)x)+Bcos(sqrt(-CONST)x) just becomes B. So the condition u(0,t)=0 means that B must be 0. Now look at sin(sqrt(-CONST)x) and check when this is 0 at L.

Making it a bit easier ...
Some of the algebra is easier if we set L=Pi. So when is sin(sqrt(-CONST)Pi)=0? Sine is 0 exactly when the argument is an integer multiple of Pi. Therefore (sqrt(-CONST)Pi=nPi, so that CONST=-n2. The function P(x) must be sin(nx) and its companion function, T(t), must be e-n2t. Notice that we may as well just use positive integers, n, because sin(nx) for n negative is equal to -sin(-nx).

Surely (!?) we now see that e-n2tsin(nx) is a solution of the heat equation,ut=uxx. Well, you can check this: two x deriviatives "spit out" (the chain rule and the double derivative of sine) -(-n)2 and one t derivative multiplies the whole formula by -n2. In fact, things do work.

Therefore 5e-222tsin(22x)-307e-72tsin(7x) is a solution to the heat equation, using linearity. And it is a solution which satisfies the boundary conditions. The only thing to do is to fuss and try to get a solution satisfying the heat equation and the boundary conditions and the initial conditions. Well, if u(x,t)=SUMn=1infinitybne-n2tsin(nx) then the heat equation and boundary conditions are correct, and, when t=0, all of the exponentials are equal to 1. Therefore, the initial condition will be satisfied if
f(x)=SUMn=1infinitybnsin(nx)
The initial condition will be satisfied if we use the Fourier sine expansion of f(x)! The important thing to note is that
sin(nx)'s "companion" for this boundary value problem involving the heat equation is e-n2t.

An example which maybe is almost a real example
I looked at an initial "heat" distribution which was 1 if x is between Pi/3 and Pi/2 and was 0 elsewhere in [0,Pi]. I actually computed the Fourier sine coefficients "by hand" but gave it all up and handed out some work done by Maple.
Then we discussed the solution. The partial sum for the initial data displays the standard phenomena (Gibbs, wiggling, etc.). But notice that for t positive, even fairly small t, the curves shown (page 2) are all rather smooth, and you can almost see the heat "ooze" away. It does ooze away rather quickly (hey, e-n2t even for moderately large n and t small positive is rather small). The initial data is not symmetric, and so the later solution is not symmetric, but it does rather rapidly approach a symmetric solution, which goes down to 0. Here are more pictures.

What's wrong with this model? A subtlety
In fact, the model is good. There are many situation where this PDE/BVP and its solutions are close to measurements which can be observed. There is one uncomfortable feature which can be seen even in the simple example done by Maple. Look at u(x,t). If t is small and positive, and if x is small and positive, then u(x,t)>0. This is true for any small x and t. This may violates some simple "thought experiments".

1. Put a drop of red ink in an enormous swimming pool full of water. A billionth of a second later, every part of the pool will be tinted, even far away, some light pink color (of varying intensity). This seems unrealistic to me.
2. Relativity: imagine a metal bar 1 light year long. Apply heat to the middle of it. A billionth of a second later, just near the edge of the bar, there will be warmth (not much, but warmth). Hey, this seems to violate some simple relativistic signaling ideas, such as the speed of light being a universal limit.
In spite of this, the heat equation is a wonderful model and very useful. We will do a bit more with it.

Another physical model and its flaws
Here is another problem from calc 1 with somewhat similar flaws, but they don't seem to really bother anyone. In fact, many people seem irritated when they are asked to think about it. But to understand the usefulness of mathematical models, you really should "push" them and see where they might break down. So here's one model.
 A 13 foot long ladder is placed on level ground and leans against a vertical wall. The bottom of the ladder is pulled steadily away from the wall at 2 ft/sec. When the bottom of the ladder is 5 ft from the wall, how fast is the height of the top of the ladder dropping?
Such a problem is almost always in the Related Rates chapter of calculus books, maybe about problem #7 or 8 of 20 problems: not considered really difficult.
If H=the height of the top of the ladder, and if x=the distance of the foot of the ladder to the base of the wall, then H2+x2=132. When x=5, then H=12 (wow, a textbook problem with a 5-12-13 right triangle). Then d/dt the equation. The result is 2HH´+2xx´=0, so H´=-5/H. So the answer to the problem posed is -5/12 ft per sec. The alert student will report that the minus sign means that the ladder's top is moving down.

This is all nice, but I can make it weird. It becomes a workshop problem (remember those) if I ask in addtion: what's the height of the ladder when the velocity of the top of the ladder breaks the speed of light? So let's see: 5/H should be (miles/sec·ft/mile) 186,000·5,280 etc. The height is about 5 times 10-7 feet (uhhhh ... 1500 angstroms? infrared light??). Ask your local calculus teacher about this situation.

We will do more with the heat equation.

HOMEWORK
Read sections 13.1, 13.2, and 13.3 of the text.
Please hand in on Monday, November 28, the following problems:
13.1: 3, 15; 13.2: 1; 13.3: 1

 Confession

### Monday, November 14

Textbook problem presentations
People presented problems from section 12.3. I reminded students of the review session Wednesday evening.

I view this part of the course as the payoff. I hope you will feel that it is "worth the trip". Here I will show you certain well-known and simple models of physical situations. The assumptions needed to understand the models are not elaborate, and, most amazingly, the predicted consequences seem intuitively correct and can even, in many situations, be experimentally verified, with very good accuracy.

The 1-dimensional heat equation
This is probably the simplest model so it is an easy choice as the first to be discussed. The heat equation is applicable to a very wide variety of situations. The one-dimensional heat equation describes the flow of heat through a homogeneous bar of constant cross-section, with the "lateral" (long) sides insulated. It also can be used to describe diffusion (put a sugar into water and see what happens), the migration of moose in a narrow valley, or even the spread of gossip. What I did was reproduce what is in the text on pages 692-693. Some students had already seen such a discussion, but it may be new to others. I think seeing the derivation is good, because the assumptions of the physical model are fundamental to applying the whole theory. Even the names of the various physical constants can be useful to know. There are many similar sources available on the web. See, for example, this web page. Another derivation is given on this page. and yet another is here.

If you are someone who finds pictures very helpful (I certainly do), you may find the various animations of solutions of the initial value problem for the heat equation on this web page useful.

Suppose we have the heat equation
So after I went through the discussion, and tried to ask people if steps of the derivation seemed sensible (for example, why a minus sign relating the time derivative of heat to the space derivative of temperature?) the usefulness of the equation may not be apparent. We know ut=uxx where the subscripts represent derivatives. Here u(x,t) is supposed to be the temperature at time t at position x in a thin homogeneous bar whose lateral side are kept insulated (no heat flow through the long side!). The bar is supposed to overlay the interval [0,L]. Since this is a math course, I have set all of the darn physical parameters equal to 1. The function u(x,t)=3x2+5x+6t is a solution of the heat equation (how to check this: two x derivatives give just 6, and one t derivative gives 6). But what the heck does this mean? Is this physically useful? Who could care?

Some physical settings for the heat equation
Here are versions of the first problems usually considered with this physical model.

1. The setup Think of a bar with a certain distribution of heat initially, and that somehow one end of the bar (at x=0) is kept at temperature u0 and the other end (at x=L) is kept at temperature uL. For example, we want a house heated inside to some nice warm temperature, while, in winter, the outside is rather chillier. A thin tube of the barrier between the inside and the outside could be modeled by our equation. Notice that the heating system of the house wants to keep u(0,t)=u0 for all time, t, and the winter forces u(L,t)=uL for all time, t.
The expected result We would guess that the initial temperature distribution would likely not matter very much. The heat in the bar will "evolve" as t-->infinity and tend towards a simple linear interpolation between u0 and uL.
The math problem There is an initial temperature distribution, u(x,0)=f(x) given for x between 0 and L (an initial condition). Also required are u(0,t)=u0 and u(L,t)=uL for all time. These are boundary conditions. The phrase boundary value problem is applied to all of these complicated specifications.

2. The setup Think of a bar with a certain distribution of heat initially, and that somehow both ends of the bar are insulated: no flow of heat is possible in or out of the ends. Maybe a diffusion setup is simpler to understand here. We sprinkle (?) sugar into a tube of water, and no further additions or subtractions to the tube are made. Then we try to study how the concentration of the sugar evolves. Presumably there is Brownian motion (caused by "random" ambient heat) which moves things around.
The expected result Over a long time (a long, long time ...) I would expect the sugar to be close to uniformly distributed in concentration. There's no reason for lots of suger (or too little sugar) to be in any one chunk of the tube: this would be remarkably unlikely. So the concentration of sugar, u(x,t), would be close to constant as t-->infinity.
The math problem There is an initial {temperature|sugar concentration} distribution, u(x,0)=f(x) given for x between 0 and L (an initial condition). Also required are that the {heat|sugar} doesn't flow out. What math condition matches this? ux(0,t)=0 and ux(L,t)=0 for all time. These are conditions on the flux (a neat word to use instead of flow so that people won't understand). This is another boundary value problem.
What we want is to solve these boundary value problems for the heat equation and check that the solutions are as predicted. It is truly remarkable that things will work out well: we'll see that the model does predict this behavior. Wait 'til next time!

Phlogiston
Here's a link to an old theory of heat. You may find it particularly interesting if you have a good chemistry background. Enjoy!

HOMEWORK
Study for the exam.