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
Example 2 Solve the following heat problem for the given initial conditions.
a)
b)
This is actually easier than it looks like. All we need to do is choose and in the product solution above to get,
and we’ve got the solution we need. This is a product solution for the first example and so satisfies the partial differential equation and boundary conditions and will satisfy the initial condition since plugging in will drop out the exponential.
and we’ve got the solution we need. This is a product solution for the first example and so satisfies the partial differential equation and boundary conditions and will satisfy the initial condition since plugging in will drop out the exponential.
b)
This is almost as simple as the first part. Recall from the Principle of Superposition that if we have two solutions to a linear homogeneous differential equation (which we’ve got here) then their sum is also a solution. So, all we need to do is choose and as we did in the first part to get a solution that satisfies each part of the initial condition and then add them up. Doing this gives,
We’ll leave it to you to verify that this does in fact satisfy the initial condition and the boundary conditions.
We’ll leave it to you to verify that this does in fact satisfy the initial condition and the boundary conditions.
So, we’ve seen that our solution from the first example will satisfy at least a small number of highly specific initial conditions.
Now, let’s extend the idea out that we used in the second part of the previous example a little to see how we can get a solution that will satisfy any sufficiently nice initial condition. The Principle of Superposition is, of course, not restricted to only two solutions. For instance, the following is also a solution to the partial differential equation.
and notice that this solution will not only satisfy the boundary conditions but it will also satisfy the initial condition,
Let’s extend this out even further and take the limit as . Doing this our solution now becomes,
This solution will satisfy any initial condition that can be written in the form,
This may still seem to be very restrictive, but the series on the right should look awful familiar to you after the previous chapter. The series on the left is exactly the Fourier sine series we looked at in that chapter. Also recall that when we can write down the Fourier sine series for any piecewise smooth function on .
So, provided our initial condition is piecewise smooth after applying the initial condition to our solution we can determine the as if we were finding the Fourier sine series of initial condition. So we can either proceed as we did in that section and use the orthogonality of the sines to derive them or we can acknowledge that we’ve already done that work and know that coefficients are given by,
So, we finally can completely solve a partial differential equation.
Example 3 Solve the following BVP.
There isn’t really all that much to do here as we’ve done most of it in the examples and discussion above.
First, the solution is,
The coefficients are given by,
If we plug these in we get the solution,
First, the solution is,
The coefficients are given by,
If we plug these in we get the solution,
That almost seems anti-climactic. This was a very short problem. Of course, some of that came about because we had a really simple constant initial condition and so the integral was very simple. However, don’t forget all the work that we had to put into discussing Fourier sine series, solving boundary value problems, applying separation of variables and then putting all of that together to reach this point.
While the example itself was very simple, it was only simple because of all the work that we had to put into developing the ideas that even allowed us to do this. Because of how “simple” it will often be to actually get these solutions we’re not actually going to do anymore with specific initial conditions. We will instead concentrate on simply developing the formulas that we’d be required to evaluate in order to get an actual solution.
So, having said that let’s move onto the next example. In this case we’re going to again look at the temperature distribution in a bar with perfectly insulated boundaries. We are also no longer going to go in steps. We will do the full solution as a single example and end up with a solution that will satisfy any piecewise smooth initial condition.
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