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Calculus III - 3-Dimensional Space: Equations of Lines

In this section we need to take a look at the equation of a line in  R 3 R 3 . As we saw in the previous section the equation  y = m x + b y = m x + b  does not describe a line in  R 3 R 3 , instead it describes a plane. This doesn’t mean however that we can’t write down an equation for a line in 3-D space. We’re just going to need a new way of writing down the equation of a curve. So, before we get into the equations of lines we first need to briefly look at vector functions. We’re going to take a more in depth look at vector functions later. At this point all that we need to worry about is notational issues and how they can be used to give the equation of a curve. The best way to get an idea of what a vector function is and what its graph looks like is to look at an example. So, consider the following vector function. → r ( t ) = ⟨ t , 1 ⟩ r → ( t ) = ⟨ t , 1 ⟩ A vector function is a function that takes one or more variables, one in this case, and returns a vector. Note as we

Differential Equations - Fourier Series: Eigenvalues and Eigenfunctions - iii


Example 4 Find all the eigenvalues and eigenfunctions for the following BVP.x2y+3xy+λy=0y(1)=0y(2)=0

This is an Euler differential equation and so we know that we’ll need to find the roots of the following quadratic.
r(r1)+3r+λ=r2+2r+λ=0The roots to this quadratic are,
r1,2=2±44λ2=1±1λNow, we are going to again have some cases to work with here, however they won’t be the same as the previous examples. The solution will depend on whether or not the roots are real distinct, double or complex and these cases will depend upon the sign/value of 1λ. So, let’s go through the cases.
1λ<0,λ>1_
In this case the roots will be complex and we’ll need to write them as follows in order to write down the solution.

r1,2=1±1λ=1±(λ1)=1±iλ1By writing the roots in this fashion we know that λ1>0 and so λ1 is now a real number, which we need in order to write the following solution,
y(x)=c1x1cos(ln(x)λ1)+c2x1sin(ln(x)λ1)Applying the first boundary condition gives us,
0=y(1)=c1cos(0)+c2sin(0)=c1c1=0The second boundary condition gives us,
0=y(2)=12c2sin(ln(2)λ1)In order to avoid the trivial solution for this case we’ll require,
sin(ln(2)λ1)=0ln(2)λ1=nπn=1,2,3,This is much more complicated of a condition than we’ve seen to this point, but other than that we do the same thing. So, solving for λ gives us the following set of eigenvalues for this case.
λn=1+(nπln2)2n=1,2,3,Note that we need to start the list of n’s off at one and not zero to make sure that we have λ>1 as we’re assuming for this case.
The eigenfunctions that correspond to these eigenvalues are,
yn(x)=x1sin(nπln2ln(x))n=1,2,3,Now the second case.
1λ=0,λ=1_
In this case we get a double root of r1,2=1 and so the solution is,

y(x)=c1x1+c2x1ln(x)Applying the first boundary condition gives,
0=y(1)=c1The second boundary condition gives,
0=y(2)=12c2ln(2)c2=0We therefore have only the trivial solution for this case and so λ=1 is not an eigenvalue.
Let’s now take care of the third (and final) case.
1λ>0,λ<1_
This case will have two real distinct roots and the solution is,

y(x)=c1x1+1λ+c2x11λApplying the first boundary condition gives,
0=y(1)=c1+c2c2=c1Using this our solution becomes,
y(x)=c1x1+1λc1x11λApplying the second boundary condition gives,
0=y(2)=c121+1λc1211λ=c1(21+1λ211λ)Now, because we know that λ1 for this case the exponents on the two terms in the parenthesis are not the same and so the term in the parenthesis is not the zero. This means that we can only have,
c1=c2=0and so in this case we only have the trivial solution and there are no eigenvalues for which λ<1.
The only eigenvalues for this BVP then come from the first case.
So, we’ve now worked an example using a differential equation other than the “standard” one we’ve been using to this point. As we saw in the work however, the basic process was pretty much the same. We determined that there were a number of cases (three here, but it won’t always be three) that gave different solutions. We examined each case to determine if non-trivial solutions were possible and if so found the eigenvalues and eigenfunctions corresponding to that case.

We need to work one last example in this section before we leave this section for some new topics. The four examples that we’ve worked to this point were all fairly simple (with simple being relative of course…), however we don’t want to leave without acknowledging that many eigenvalue/eigenfunctions problems are so easy.

In many examples it is not even possible to get a complete list of all possible eigenvalues for a BVP. Often the equations that we need to solve to get the eigenvalues are difficult if not impossible to solve exactly. So, let’s take a look at one example like this to see what kinds of things can be done to at least get an idea of what the eigenvalues look like in these kinds of cases.


Example 5 Find all the eigenvalues and eigenfunctions for the following BVP.y+λy=0y(0)=0y(1)+y(1)=0

The boundary conditions for this BVP are fairly different from those that we’ve worked with to this point. However, the basic process is the same. So let’s start off with the first case.
λ>0_
The general solution to the differential equation is identical to the first few examples and so we have,

y(x)=c1cos(λx)+c2sin(λx)Applying the first boundary condition gives us,
0=y(0)=c1c1=0The second boundary condition gives us,
0=y(1)+y(1)=c2sin(λ)+λc2cos(λ)=c2(sin(λ)+λcos(λ))So, if we let c2=0 we’ll get the trivial solution and so in order to satisfy this boundary condition we’ll need to require instead that,
0=sin(λ)+λcos(λ)sin(λ)=λcos(λ)tan(λ)=λNow, this equation has solutions but we’ll need to use some numerical techniques in order to get them. In order to see what’s going on here let’s graph tan(λ) and λ on the same graph. Here is that graph and note that the horizontal axis really is values of λ as that will make things a little easier to see and relate to values that we’re familiar with.
So, eigenvalues for this case will occur where the two curves intersect. We’ve shown the first five on the graph and again what is showing on the graph is really the square root of the actual eigenvalue as we’ve noted.
The interesting thing to note here is that the farther out on the graph the closer the eigenvalues come to the asymptotes of tangent and so we’ll take advantage of that and say that for large enough n we can approximate the eigenvalues with the (very well known) locations of the asymptotes of tangent.
How large the value of n is before we start using the approximation will depend on how much accuracy we want, but since we know the location of the asymptotes and as n increases the accuracy of the approximation will increase so it will be easy enough to check for a given accuracy.
For the purposes of this example we found the first five numerically and then we’ll use the approximation of the remaining eigenvalues. Here are those values/approximations.
λ1=2.0288λ1=4.1160(2.4674)λ2=4.9132λ2=24.1395(22.2066)λ3=7.9787λ3=63.6597(61.6850)λ4=11.0855λ4=122.8883(120.9027)λ5=14.2074λ5=201.8502(199.8595)λn2n12πλn(2n1)24π2n=6,7,8,The number in parenthesis after the first five is the approximate value of the asymptote. As we can see they are a little off, but by the time we get to n=5 the error in the approximation is 0.9862%. So less than 1% error by the time we get to n=5 and it will only get better for larger value of n.
The eigenfunctions for this case are,
yn(x)=sin(λnx)n=1,2,3,where the values of λn are given above.
So, now that all that work is out of the way let’s take a look at the second case.
λ=0_
The general solution is,

y(x)=c1+c2xApplying the first boundary condition gives,
0=y(0)=c1Using this the general solution is then,
y(x)=c2xApplying the second boundary condition to this gives,
0=y(1)+y(1)=c2+c2=2c2c2=0Therefore, for this case we get only the trivial solution and so λ=0 is not an eigenvalue. Note however that had the second boundary condition been y(1)y(1)=0 then λ=0 would have been an eigenvalue (with eigenfunctions y(x)=x) and so again we need to be careful about reading too much into our work here.
Finally let’s take care of the third case.
λ<0_
The general solution here is,

y(x)=c1cosh(λx)+c2sinh(λx)Applying the first boundary condition gives,
0=y(0)=c1cosh(0)+c2sinh(0)=c1c1=0Using this the general solution becomes,
y(x)=c2sinh(λx)Applying the second boundary condition to this gives,
0=y(1)+y(1)=λc2cosh(λ)+c2sinh(λ)=c2(λcosh(λ)+sinh(λ))Now, by assumption we know that λ<0 and so λ>0. This in turn tells us that sinh(λ)>0 and we know that cosh(x)>0 for all x. Therefore,
λcosh(λ)+sinh(λ)0and so we must have c2=0 and once again in this third case we get the trivial solution and so this BVP will have no negative eigenvalues.
In summary the only eigenvalues for this BVP come from assuming that λ>0 and they are given above.
So, we’ve worked several eigenvalue/eigenfunctions examples in this section. Before leaving this section we do need to note once again that there are a vast variety of different problems that we can work here and we’ve really only shown a bare handful of examples and so please do not walk away from this section believing that we’ve shown you everything.


The whole purpose of this section is to prepare us for the types of problems that we’ll be seeing in the next chapter. Also, in the next chapter we will again be restricting ourselves down to some pretty basic and simple problems in order to illustrate one of the more common methods for solving partial differential equations.

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