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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...

Differential Equations - Fourier Series: Eigenvalues and Eigenfunctions - ii


Example 2 Find all the eigenvalues and eigenfunctions for the following BVP.y+λy=0y(0)=0y(2π)=0

Here we are going to work with derivative boundary conditions. The work is pretty much identical to the previous example however so we won’t put in quite as much detail here. We’ll need to go through all three cases just as the previous example so let’s get started on that.
λ>0_
The general solution to the differential equation is identical to the previous example and so we have,

y(x)=c1cos(λx)+c2sin(λx)Applying the first boundary condition gives us,
0=y(0)=λc2c2=0Recall that we are assuming that λ>0 here and so this will only be zero if c2=0. Now, the second boundary condition gives us,
0=y(2π)=λc1sin(2πλ)Recall that we don’t want trivial solutions and that λ>0 so we will only get non-trivial solution if we require that,
sin(2πλ)=02πλ=nπn=1,2,3,Solving for λ and we see that we get exactly the same positive eigenvalues for this BVP that we got in the previous example.
λn=(n2)2=n24n=1,2,3,The eigenfunctions that correspond to these eigenvalues however are,
yn(x)=cos(nx2)n=1,2,3,So, for this BVP we get cosines for eigenfunctions corresponding to positive eigenvalues.
Now the second case.
λ=0_
The general solution is,

y(x)=c1+c2xApplying the first boundary condition gives,
0=y(0)=c2Using this the general solution is then,
y(x)=c1and note that this will trivially satisfy the second boundary condition,
0=y(2π)=0Therefore, unlike the first example, λ=0 is an eigenvalue for this BVP and the eigenfunctions corresponding to this eigenvalue is,
y(x)=1Again, note that we dropped the arbitrary constant for the eigenfunctions.
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)=λc1sinh(0)+λc2cosh(0)=λc2c2=0Applying the second boundary condition gives,
0=y(2π)=λc1sinh(2πλ)As with the previous example we again know that 2πλ0 and so sinh(2πλ)0. Therefore, we must have c1=0.
So, for this BVP we again have no negative eigenvalues.
In summary then we will have the following eigenvalues/eigenfunctions for this BVP.
λn=n24yn(x)=cos(nx2)n=1,2,3,λ0=0y0(x)=1Notice as well that we can actually combine these if we allow the list of n’s for the first one to start at zero instead of one. This will often not happen, but when it does we’ll take advantage of it. So the “official” list of eigenvalues/eigenfunctions for this BVP is,
λn=n24yn(x)=cos(nx2)n=0,1,2,3,
So, in the previous two examples we saw that we generally need to consider different cases for λ as different values will often lead to different general solutions. Do not get too locked into the cases we did here. We will mostly be solving this particular differential equation and so it will be tempting to assume that these are always the cases that we’ll be looking at, but there are BVP’s that will require other/different cases.
Also, as we saw in the two examples sometimes one or more of the cases will not yield any eigenvalues. This will often happen, but again we shouldn’t read anything into the fact that we didn’t have negative eigenvalues for either of these two BVP’s. There are BVP’s that will have negative eigenvalues.

Let’s take a look at another example with a very different set of boundary conditions. These are not the traditional boundary conditions that we’ve been looking at to this point, but we’ll see in the next chapter how these can arise from certain physical problems.


Example 3 Find all the eigenvalues and eigenfunctions for the following BVP.y+λy=0y(π)=y(π)y(π)=y(π)

So, in this example we aren’t actually going to specify the solution or its derivative at the boundaries. Instead we’ll simply specify that the solution must be the same at the two boundaries and the derivative of the solution must also be the same at the two boundaries. Also, this type of boundary condition will typically be on an interval of the form [-L,L] instead of [0,L] as we’ve been working on to this point.
As mentioned above these kind of boundary conditions arise very naturally in certain physical problems and we’ll see that in the next chapter.
As with the previous two examples we still have the standard three cases to look at.
λ>0_
The general solution for this case is,

y(x)=c1cos(λx)+c2sin(λx)Applying the first boundary condition and using the fact that cosine is an even function (i.e.cos(x)=cos(x)) and that sine is an odd function (i.e. sin(x)=sin(x)). gives us,
c1cos(πλ)+c2sin(πλ)=c1cos(πλ)+c2sin(πλ)c1cos(πλ)c2sin(πλ)=c1cos(πλ)+c2sin(πλ)c2sin(πλ)=c2sin(πλ)0=2c2sin(πλ)This time, unlike the previous two examples this doesn’t really tell us anything. We could have sin(πλ)=0 but it is also completely possible, at this point in the problem anyway, for us to have c2=0 as well.
So, let’s go ahead and apply the second boundary condition and see if we get anything out of that.
λc1sin(πλ)+λc2cos(πλ)=λc1sin(πλ)+λc2cos(πλ)λc1sin(πλ)+λc2cos(πλ)=λc1sin(πλ)+λc2cos(πλ)λc1sin(πλ)=λc1sin(πλ)2λc1sin(πλ)=0So, we get something very similar to what we got after applying the first boundary condition. Since we are assuming that λ>0 this tells us that either sin(πλ)=0 or c1=0.
Note however that if sin(πλ)0 then we will have to have c1=c2=0 and we’ll get the trivial solution. We therefore need to require that sin(πλ)=0 and so just as we’ve done for the previous two examples we can now get the eigenvalues,
πλ=nπλ=n2n=1,2,3,Recalling that λ>0 and we can see that we do need to start the list of possible n’s at one instead of zero.
So, we now know the eigenvalues for this case, but what about the eigenfunctions. The solution for a given eigenvalue is,
y(x)=c1cos(nx)+c2sin(nx)and we’ve got no reason to believe that either of the two constants are zero or non-zero for that matter. In cases like these we get two sets of eigenfunctions, one corresponding to each constant. The two sets of eigenfunctions for this case are,
yn(x)=cos(nx)yn(x)=sin(nx)n=1,2,3,Now the second case.
λ=0_
The general solution is,

y(x)=c1+c2xApplying the first boundary condition gives,
c1+c2(π)=c1+c2(π)2πc2=0c2=0Using this the general solution is then,
y(x)=c1and note that this will trivially satisfy the second boundary condition just as we saw in the second example above. Therefore, we again have λ=0 as an eigenvalue for this BVP and the eigenfunctions corresponding to this eigenvalue is,
y(x)=1Finally 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 and using the fact that hyperbolic cosine is even and hyperbolic sine is odd gives,
c1cosh(πλ)+c2sinh(πλ)=c1cosh(πλ)+c2sinh(πλ)c2sinh(πλ)=c2sinh(πλ)2c2sinh(πλ)=0Now, in this case we are assuming that λ<0 and so we know that πλ0 which in turn tells us that sinh(πλ)0. We therefore must have c2=0.
Let’s now apply the second boundary condition to get,
λc1sinh(πλ)=λc1sinh(πλ)2λc1sinh(πλ)=0c1=0By our assumption on λ we again have no choice here but to have c1=0.
Therefore, in this case the only solution is the trivial solution and so, for this BVP we again have no negative eigenvalues.
In summary then we will have the following eigenvalues/eigenfunctions for this BVP.
λn=n2yn(x)=sin(nx)n=1,2,3,λn=n2yn(x)=cos(nx)n=1,2,3,λ0=0y0(x)=1Note that we’ve acknowledged that for λ>0 we had two sets of eigenfunctions by listing them each separately. Also, we can again combine the last two into one set of eigenvalues and eigenfunctions. Doing so gives the following set of eigenvalues and eigenfunctions.
λn=n2yn(x)=sin(nx)n=1,2,3,λn=n2yn(x)=cos(nx)n=0,1,2,3,
Once again, we’ve got an example with no negative eigenvalues. We can’t stress enough that this is more a function of the differential equation we’re working with than anything and there will be examples in which we may get negative eigenvalues.
Now, to this point we’ve only worked with one differential equation so let’s work an example with a different differential equation just to make sure that we don’t get too locked into this one differential equation.

Before working this example let’s note that we will still be working the vast majority of our examples with the one differential equation we’ve been using to this point. We’re working with this other differential equation just to make sure that we don’t get too locked into using one single differential equation.

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