I apologize for the outage on the site yesterday and today. Lamar University is in Beaumont Texas and Hurricane Laura came through here and caused a brief power outage at Lamar. Things should be up and running at this point and (hopefully) will stay that way, at least until the next hurricane comes through here which seems to happen about once every 10-15 years. Note that I wouldn't be too suprised if there are brief outages over the next couple of days as they work to get everything back up and running properly. I apologize for the inconvienence.

Paul

August 27, 2020

*i.e.*you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 9-3 : Terminology

We’ve got one more section that we need to take care of before we actually start solving partial differential equations. This will be a fairly short section that will cover some of the basic terminology that we’ll need in the next section as we introduce the method of separation of variables.

Let’s start off with the idea of an operator. An operator is really just a function that takes a function as an argument instead of numbers as we’re used to dealing with in functions. You already know of a couple of operators even if you didn’t know that they were operators. Here are some examples of operators.

\[L = \frac{d}{{dx}}\hspace{0.25in}L = \int{{\,dx}}\hspace{0.25in}L = \int_{a}^{b}{{\,dx}}\hspace{0.25in}L = \frac{\partial }{{\partial t}}\]Or, if we plug in a function, say \(u\left( x \right)\), into each of these we get,

\[L\left( u \right) = \frac{{du}}{{dx}}\hspace{0.25in}L\left( u \right) = \int{{u\left( x \right)\,dx}}\hspace{0.25in}L\left( u \right) = \int_{a}^{b}{{u\left( x \right)\,dx}}\hspace{0.25in}L\left( u \right) = \frac{{\partial u}}{{\partial t}}\]These are all fairly simple examples of operators but the derivative and integral are operators. A more complicated operator would be the **heat operator**. We get the heat operator from a slight rewrite of the heat equation without sources. The heat operator is,

Now, what we really want to define here is not an operator but instead a **linear operator**. A linear operator is any operator that satisfies,

The heat operator is an example of a linear operator and this is easy enough to show using the basic properties of the partial derivative so let’s do that.

\[\begin{align*}L\left( {{c_1}{u_1} + {c_2}{u_2}} \right) & = \frac{\partial }{{\partial t}}\left( {{c_1}{u_1} + {c_2}{u_2}} \right) - k\frac{{{\partial ^2}}}{{\partial {x^2}}}\left( {{c_1}{u_1} + {c_2}{u_2}} \right)\\ & = \frac{\partial }{{\partial t}}\left( {{c_1}{u_1}} \right) + \frac{\partial }{{\partial t}}\left( {{c_2}{u_2}} \right) - k\left[ {\frac{{{\partial ^2}}}{{\partial {x^2}}}\left( {{c_1}{u_1}} \right) + \frac{{{\partial ^2}}}{{\partial {x^2}}}\left( {{c_2}{u_2}} \right)} \right]\\ & = {c_1}\frac{{\partial {u_1}}}{{\partial t}} + {c_2}\frac{{\partial {u_2}\,}}{{\partial t}} - k\left[ {{c_1}\frac{{{\partial ^2}{u_1}}}{{\partial {x^2}}} + {c_2}\frac{{{\partial ^2}{u_2}}}{{\partial {x^2}}}} \right]\\ & = {c_1}\frac{{\partial {u_1}}}{{\partial t}} - k{c_1}\frac{{{\partial ^2}{u_1}}}{{\partial {x^2}}} + {c_2}\frac{{\partial {u_2}\,}}{{\partial t}} - k{c_2}\frac{{{\partial ^2}{u_2}}}{{\partial {x^2}}}\\ & = {c_1}\left[ {\frac{{\partial {u_1}}}{{\partial t}} - k\frac{{{\partial ^2}{u_1}}}{{\partial {x^2}}}} \right] + {c_2}\left[ {\frac{{\partial {u_2}\,}}{{\partial t}} - k\frac{{{\partial ^2}{u_2}}}{{\partial {x^2}}}} \right]\\ & = {c_1}L\left( {{u_1}} \right) + {c_2}L\left( {{u_2}} \right)\end{align*}\]You might want to verify for yourself that the derivative and integral operators we gave above are also linear operators. In fact, in the process of showing that the heat operator is a linear operator we actually showed as well that the first order and second order partial derivative operators are also linear.

The next term we need to define is a **linear equation**. A linear equation is an equation in the form,

where \(L\) is a linear operator and \(f\) is a known function.

Here are some examples of linear partial differential equations.

\[\begin{align*}\frac{{\partial u}}{{\partial t}} & = k\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{Q\left( {x,t} \right)}}{{c\rho }}\\ \frac{{{\partial ^2}u}}{{\partial {t^2}}} & = {c^2}\frac{{{\partial ^2}u}}{{\partial {x^2}}}\\ \frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}} + \frac{{{\partial ^2}u}}{{\partial {z^2}}} & = {\nabla ^2}u = 0\\ \frac{{\partial u}}{{\partial t}} - 4\frac{{{\partial ^2}u}}{{\partial {t^2}}} & = \frac{{{\partial ^3}u}}{{\partial {x^3}}} + 8u - g\left( {x,t} \right)\end{align*}\]The first two from this list are of course the heat equation and the wave equation. The third uses the Laplacian and is usually called **Laplace’s Equation**. We’ll actually be solving the 2-D version of Laplace’s Equation in a few sections. The fourth equation was just made up to give a more complicated example.

Notice as well with the heat equation and the fourth example above that the presence of the \(Q\left( {x,t} \right)\) and \(g\left( {x,t} \right)\) do not prevent these from being linear equations. The main issue that allows these to be linear equations is the fact that the operator in each is linear.

Now just to be complete here are a couple of examples of nonlinear partial differential equations.

\[\begin{align*}\frac{{\partial u}}{{\partial t}} & = k\frac{{{\partial ^2}u}}{{\partial {x^2}}} + {u^2}\\ \frac{{{\partial ^2}u}}{{\partial {t^2}}} - \frac{{\partial u}}{{\partial x}}\frac{{\partial u}}{{\partial t}} & = u + f\left( {x,t} \right)\end{align*}\]We’ll leave it to you to verify that the operators in each of these are not linear however the problem term in the first is the \({u^2}\) while in the second the product of the two derivatives is the problem term.

Now, if we go back to \(\eqref{eq:eq1}\) and suppose that \(f = 0\) then we arrive at,

\[\begin{equation}L\left( u \right) = 0\label{eq:eq2}\end{equation}\]We call this a **linear homogeneous equation** (recall that \(L\) is a linear operator).

Notice that \(u = 0\) will always be a solution to a linear homogeneous equation (go back to what it means to be linear and use \({c_1} = {c_2} = 0\) with any two solutions and this is easy to verify). We call \(u = 0\) the **trivial solution**. In fact, this is also a really nice way of determining if an equation is homogeneous. If \(L\) is a linear operator and we plug in \(u = 0\) into the equation and we get \(L\left( u \right) = 0\) then we will know that the operator is homogeneous.

We can also extend the ideas of linearity and homogeneous to boundary conditions. If we go back to the various boundary conditions we discussed for the heat equation for example we can also classify them as linear and/or homogeneous.

The prescribed temperature boundary conditions,

\[u\left( {0,t} \right) = {g_1}\left( t \right)\hspace{0.25in}u\left( {L,t} \right) = {g_2}\left( t \right)\]are linear and will only be homogenous if \({g_1}\left( t \right) = 0\) and \({g_2}\left( t \right) = 0\).

The prescribed heat flux boundary conditions,

\[ - {K_0}\left( 0 \right)\frac{{\partial u}}{{\partial x}}\left( {0,t} \right) = {\varphi _1}\left( t \right)\hspace{0.25in} - {K_0}\left( L \right)\frac{{\partial u}}{{\partial x}}\left( {L,t} \right) = {\varphi _2}\left( t \right)\]are linear and will again only be homogeneous if \({\varphi _1}\left( t \right) = 0\) and \({\varphi _2}\left( t \right) = 0\).

Next, the boundary conditions from Newton’s law of cooling,

\[ - {K_0}\left( 0 \right)\frac{{\partial u}}{{\partial x}}\left( {0,t} \right) = - H\left[ {u\left( {0,t} \right) - {g_1}\left( t \right)} \right]\hspace{0.25in} - {K_0}\left( L \right)\frac{{\partial u}}{{\partial x}}\left( {L,t} \right) = H\left[ {u\left( {L,t} \right) - {g_2}\left( t \right)} \right]\]are again linear and will only be homogenous if \({g_1}\left( t \right) = 0\) and \({g_2}\left( t \right) = 0\).

The final set of boundary conditions that we looked at were the periodic boundary conditions,

\[u\left( { - L,t} \right) = u\left( {L,t} \right)\hspace{0.25in}\frac{{\partial u}}{{\partial x}}\left( { - L,t} \right) = \frac{{\partial u}}{{\partial x}}\left( {L,t} \right)\]and these are both linear and homogeneous.

The final topic in this section is not really terminology but is a restatement of a fact that we’ve seen several times in these notes already.

#### Principle of Superposition

If \({u_1}\) and \({u_2}\) are solutions to a linear homogeneous equation then so is \({c_1}{u_1} + {c_2}{u_2}\) for any values of \({c_1}\) and \({c_2}\).

Now, as stated earlier we’ve seen this several times this semester but we didn’t really do much with it. However, this is going to be a key idea when we actually get around to solving partial differential equations. Without this fact we would not be able to solve all but the most basic of partial differential equations.