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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 - Partial: Laplace's Equation - iii


Example 3 Find a solution to the following partial differential equation.
2u=1rr(rur)+1r22uθ2=0|u(0,θ)|<u(a,θ)=f(θ)u(r,π)=u(r,π)uθ(r,π)=uθ(r,π)

In this case we’ll assume that the solution will be in the form,
u(r,θ)=φ(θ)G(r)Plugging this into the periodic boundary conditions gives,
φ(π)=φ(π)dφdθ(π)=dφdθ(π)|G(0)|<Now let’s plug the product solution into the partial differential equation.
1rr(rr(φ(θ)G(r)))+1r22θ2(φ(θ)G(r))=0φ(θ)1rddr(rdGdr)+G(r)1r2d2φdθ2=0This is definitely more of a mess that we’ve seen to this point when it comes to separating variables. In this case simply dividing by the product solution, while still necessary, will not be sufficient to separate the variables. We are also going to have to multiply by r2 to completely separate variables. So, doing all that, moving each term to one side of the equal sign and introduction a separation constant gives,
rGddr(rdGdr)=1φd2φdθ2=λWe used λ as the separation constant this time to get the differential equation for φ to match up with one we’ve already done.
The ordinary differential equations we get are then,
rddr(rdGdr)λG=0d2φdθ2+λφ=0|G(0)|<φ(π)=φ(π)dφdθ(π)=dφdθ(π)Now, we solved the boundary value problem above in Example 3 of the Eigenvalues and Eigenfunctions section of the previous chapter and so there is no reason to redo it here. The eigenvalues and eigenfunctions for this problem are,
λn=n2φn(θ)=sin(nθ)n=1,2,3,λn=n2φn(θ)=cos(nθ)n=0,1,2,3,Plugging this into the first ordinary differential equation and using the product rule on the derivative we get,
rddr(rdGdr)n2G=0r(rd2Gdr2+dGdr)n2G=0r2d2Gdr2+rdGdrn2G=0This is an Euler differential equation and so we know that solutions will be in the form G(r)=rp provided p is a root of,
p(p1)+pn2=0p2n2=0p=±nn=0,1,2,3,So, because the n=0 case will yield a double root, versus two real distinct roots if n0 we have two cases here. They are,
G(r)=c1+c2lnrn=0G(r)=c¯1rn+c¯2rnn=1,2,3,Now we need to recall the condition that |G(0)|<. Each of the solutions above will have G(r) as r0Therefore in order to meet this boundary condition we must have c2=c¯2=0.
Therefore, the solution reduces to,
G(r)=c1rnn=0,1,2,3,and notice that with the second term gone we can combine the two solutions into a single solution.
So, we have two product solutions for this problem. They are,
un(r,θ)=Anrncos(nθ)n=0,1,2,3,un(r,θ)=Bnrnsin(nθ)n=1,2,3,Our solution is then the sum of all these solutions or,
u(r,θ)=n=0Anrncos(nθ)+n=1Bnrnsin(nθ)Applying our final boundary condition to this gives,
u(a,θ)=f(θ)=n=0Anancos(nθ)+n\(L=π\)=1Bnansin(nθ)This is a full Fourier series for f(θ) on the interval πθπi.e. L=π. Also note that once again the “coefficients” of the Fourier series are a little messier than normal, but not quite as messy as when we were working on a rectangle above. We could once again use the orthogonality of the sines and cosines to derive formulas for the An and Bn or we could just use the formulas from the Fourier series section with to get,
A0=12πππf(θ)dθAnan=1πππf(θ)cos(nθ)dθn=1,2,3,Bnan=1πππf(θ)sin(nθ)dθn=1,2,3,Upon solving for the coefficients we get,
A0=12πππf(θ)dθAn=1πanππf(θ)cos(nθ)dθn=1,2,3,Bn=1πanππf(θ)sin(nθ)dθn=1,2,3,

Prior to this example most of the separation of variable problems tended to look very similar and it is easy to fall in to the trap of expecting everything to look like what we’d seen earlier. With this example we can see that the problems can definitely be different on occasion so don’t get too locked into expecting them to always work in exactly the same way.

Before we leave this section let’s briefly talk about what you’d need to do on a partial disk. The periodic boundary conditions above were only there because we had a whole disk. What if we only had a disk between say αθβ.

When we’ve got a partial disk we now have two new boundaries that we not present in the whole disk and the periodic boundary conditions will no longer make sense. The periodic boundary conditions are only used when we have the two “boundaries” in contact with each other and that clearly won’t be the case with a partial disk.
So, if we stick with prescribed temperature boundary conditions we would then have the following conditions
|u(0,θ)|<u(a,θ)=f(θ)αθβu(r,α)=g1(r)0rau(r,β)=g2(r)0ra


Also note that in order to use separation of variables on these conditions we’d need to have g1(r)=g2(r)=0 to make sure they are homogeneous.
As a final note we could just have easily used flux boundary conditions for the last two if we’d wanted to. The boundary value problem would be different, but outside of that the problem would work in the same manner.
We could also use a flux condition on the r=a boundary but we haven’t really talked yet about how to apply that kind of condition to our solution. Recall that this is the condition that we apply to our solution to determine the coefficients. It’s not difficult to use we just haven’t talked about this kind of condition yet. We’ll be doing that in the next section.

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