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
This will be the final partial differential equation that we’ll be solving in this chapter. In this section we’ll be solving the 1-D wave equation to determine the displacement of a vibrating string. There really isn’t much in the way of introduction to do here so let’s just jump straight into the example.
Example 1 Find a solution to the following partial differential equation.
One of the main differences here that we’re going to have to deal with is the fact that we’ve now got two initial conditions. That is not something we’ve seen to this point but will not be all that difficult to deal with when the time rolls around.
We’ve already done the separation of variables for this problem, but let’s go ahead and redo it here so we can say we’ve got another problem almost completely worked out.
So, let’s start off with the product solution.
Plugging this into the two boundary conditions gives,
Plugging the product solution into the differential equation, separating and introducing a separation constant gives,
We moved the to the left side for convenience and chose for the separation constant so the differential equation for would match a known (and solved) case.
The two ordinary differential equations we get from separation of variables are then,
We solved the boundary value problem above in Example 1 of the Solving the Heat Equation section of this chapter and so the eigenvalues and eigenfunctions for this problem are,
The first ordinary differential equation is now,
and because the coefficient of the is clearly positive the solution to this is,
Because there is no reason to think that either of the coefficients above are zero we then get two product solutions,
The solution is then,
Now, in order to apply the second initial condition we’ll need to differentiate this with respect to so,
If we now apply the initial conditions we get,
Both of these are Fourier sine series. The first is for on while the second is for on with a slightly messy coefficient. As in the last few sections we’re faced with the choice of either using the orthogonality of the sines to derive formulas for and or we could reuse formula from previous work.
It’s easier to reuse formulas so using the formulas form the Fourier sine series section we get,
Upon solving the second one we get,
We’ve already done the separation of variables for this problem, but let’s go ahead and redo it here so we can say we’ve got another problem almost completely worked out.
So, let’s start off with the product solution.
Plugging this into the two boundary conditions gives,
Plugging the product solution into the differential equation, separating and introducing a separation constant gives,
We moved the to the left side for convenience and chose for the separation constant so the differential equation for would match a known (and solved) case.
The two ordinary differential equations we get from separation of variables are then,
We solved the boundary value problem above in Example 1 of the Solving the Heat Equation section of this chapter and so the eigenvalues and eigenfunctions for this problem are,
The first ordinary differential equation is now,
and because the coefficient of the is clearly positive the solution to this is,
Because there is no reason to think that either of the coefficients above are zero we then get two product solutions,
The solution is then,
Now, in order to apply the second initial condition we’ll need to differentiate this with respect to so,
If we now apply the initial conditions we get,
Both of these are Fourier sine series. The first is for on while the second is for on with a slightly messy coefficient. As in the last few sections we’re faced with the choice of either using the orthogonality of the sines to derive formulas for and or we could reuse formula from previous work.
It’s easier to reuse formulas so using the formulas form the Fourier sine series section we get,
Upon solving the second one we get,
So, there is the solution to the 1-D wave equation and with that we’ve solved the final partial differential equation in this chapter.
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