Partial Differential Equations

Learning Outcomes

  • Explain the meaning of a partial differential equation and give an example.

In Introduction to Differential Equations, we studied differential equations in which the unknown function had one independent variable. A partial differential equation is an equation that involves an unknown function of more than one independent variable and one or more of its partial derivatives. Examples of partial differential equations are

[latex]\large{u_t=c^{2}\,(u_{xx}+u_{yy})}[/latex]

(heat equation in two dimensions)

[latex]\large{u_{tt}=c^{2}\,(u_{xx}+u_{yy})}[/latex]

(wave equation in two dimensions)

[latex]\large{u_{xx}+u_{yy}=0}[/latex]

(Laplace’s equation in two dimensions)

In the first two equations, the unknown function [latex]u[/latex] has three independent variables[latex]-t,\ x[/latex], and [latex]y-[/latex]and [latex]c[/latex] is an arbitrary constant. The independent variables [latex]x[/latex] and [latex]y[/latex] are considered to be spatial variables, and the variable [latex]t[/latex] represents time. In Laplace’s equation, the unknown function [latex]u[/latex] has two independent variables [latex]x[/latex] and [latex]y[/latex].

Example: A Solution to the Wave Equation

Verify that

[latex]\large{u\,(x,\ y,\ t)=5\sin{(3\pi x)}\sin{(4\pi y)}\cos{(10\pi t)}}[/latex]

is a solution to the wave equation

[latex]\large{u_{tt}=4\,(u_{xx}+u_{yy})}.[/latex]

try it

Verify that [latex]u\,(x,\ y,\ t)=2\,\sin\,(\frac{x}{3})\,\sin\,(\frac{y}{4})\,e^{-25t/16}[/latex] is a solution to the heat equation

[latex]u_{t}=9(u_{xx}+u_{yy})[/latex].

Watch the following video to see the worked solution to the above Try It

Since the solution to the two-dimensional heat equation is a function of three variables, it is not easy to create a visual representation of the solution. We can graph the solution for fixed values of [latex]t[/latex], which amounts to snapshots of the heat distributions at fixed times. These snapshots show how the heat is distributed over a two-dimensional surface as time progresses. The graph of the preceding solution at time [latex]t=0[/latex] appears in the following figure. As time progresses, the extremes level out, approaching zero as [latex]t[/latex] approaches infinity.

A complicated curve in xyz space with many sinusoidally alternating local maxima and minima.

Figure 1.

If we consider the heat equation in one dimension, then it is possible to graph the solution over time. The heat equation in one dimension becomes

[latex]u_{t}=c^{2}u_{xx}[/latex],

where [latex]c^{2}[/latex] represents the thermal diffusivity of the material in question. A solution of this differential equation can be written in the form

[latex]u_{m}\,(x,\ t)=e^{-\pi^{2}m^{2}c^{2}t}\sin{(m\pi{x})}[/latex]

where [latex]m[/latex] is any positive integer. A graph of this solution using [latex]m=1[/latex] appears in Figure 4, where the initial temperature distribution over a wire of length [latex]1[/latex] is given by [latex]u\,(x,\ 0)=\sin{\pi{x}}[/latex]. Notice that as time progresses, the wire cools off. This is seen because, from left to right, the highest temperature (which occurs in the middle of the wire) decreases and changes color from red to blue.

A curve in xtu space with a local maximum at (0.5, 0, 12). From this maximum, the values decrease with increasing t and for any value of x.

Figure 2. Graph of a solution of the heat equation in one dimension over time.

Lord Kelvin and the Age of Earth

This figure consists of two figures marked a and b. Figure a show Lord Kelvin, dressed well and with a beard. Figure b shows an image of the planet Earth taken from space.

Figure 3. (a) William Thomson (Lord Kelvin), 1824-1907, was a British physicist and electrical engineer; (b) Kelvin used the heat diffusion equation to estimate the age of Earth (credit: modification of work by NASA).

During the late 1800s, the scientists of the new field of geology were coming to the conclusion that Earth must be “millions and millions” of years old. At about the same time, Charles Darwin had published his treatise on evolution. Darwin’s view was that evolution needed many millions of years to take place, and he made a bold claim that the Weald chalk fields, where important fossils were found, were the result of [latex]300[/latex] million years of erosion.

At that time, eminent physicist William Thomson (Lord Kelvin) used an important partial differential equation, known as the heat diffusion equation, to estimate the age of Earth by determining how long it would take Earth to cool from molten rock to what we had at that time. His conclusion was a range of [latex]20[/latex] to [latex]400[/latex] million years, but most likely about [latex]50[/latex] million years. For many decades, the proclamations of this irrefutable icon of science did not sit well with geologists or with Darwin.

Interactive

Read Kelvin’s paper on estimating the age of the Earth.

Kelvin made reasonable assumptions based on what was known in his time, but he also made several assumptions that turned out to be wrong. One incorrect assumption was that Earth is solid and that the cooling was therefore via conduction only, hence justifying the use of the diffusion equation. But the most serious error was a forgivable one—omission of the fact that Earth contains radioactive elements that continually supply heat beneath Earth’s mantle. The discovery of radioactivity came near the end of Kelvin’s life and he acknowledged that his calculation would have to be modified.

Kelvin used the simple one-dimensional model applied only to Earth’s outer shell, and derived the age from graphs and the roughly known temperature gradient near Earth’s surface. Let’s take a look at a more appropriate version of the diffusion equation in radial coordinates, which has the form

[latex]\large{\frac{\partial{T}}{\partial t}=K\left[\frac{\partial^{2}T}{\partial^{2}r}+\frac{2}{r}\frac{\partial T}{\partial r}\right]}[/latex].

 

Here, [latex]T\,(r,\ t)[/latex] is temperature as a function of [latex]r[/latex] (measured from the center of Earth) and time [latex]t[/latex]. [latex]K[/latex] is the heat conductivity conductivity—for molten rock, in this case. The standard method of solving such a partial differential equation is by separation of variables, where we express the solution as the product of functions containing each variable separately. In this case, we would write the temperature as

[latex]\large{T\,(r,\ t)=R\,(r)\,f\,(t)}[/latex].

 

  1. Substitute this form into our first equation from this section and, noting that [latex]f(t)[/latex] is constant with respect to distance [latex](r)[/latex] and [latex]R(r)[/latex] is constant with respect to time [latex](t)[/latex], show that
    [latex]\large{\frac{1}{f}\frac{\partial f}{\partial t}=\frac{K}{R}\left[\frac{\partial^{2}R}{\partial r^{2}}+\frac{2}{r}\frac{\partial R}{\partial r}\right]}[/latex].

     

  2. This equation represents the separation of variables we want. The left-hand side is only a function of [latex]t[/latex] and the right-hand side is only a function of [latex]r[/latex], and they must be equal for all values of [latex]r[/latex] and [latex]t[/latex]. Therefore, they both must be equal to a constant. Let’s call that constant [latex]-\lambda^{2}[/latex]. (The convenience of this choice is seen on substitution.) So, we have
    [latex]\large{\frac{1}{f}\frac{\partial f}{\partial t}=-\lambda^{2}}[/latex] and [latex]\large{\frac{K}{R}\left[\frac{\partial^{2}R}{\partial r^{2}}+\frac{2}{r}\frac{\partial R}{\partial r}\right]=-\lambda^{2}}[/latex].

     

    Now, we can verify through direct substitution for each equation that the solutions are [latex]f\,(t)=Ae^{-\lambda^{2}t}[/latex] and [latex]R\,(r)=B\left(\frac{\sin{\alpha r}}{r}\right)+C\left(\frac{\cos{\alpha r}}{r}\right)[/latex], where [latex]\alpha=\lambda /\sqrt{K}[/latex]. Note that [latex]f\,(t)=Ae^{+\lambda n^{2}t}[/latex] is also a valid solution, so we could have chosen [latex]+\lambda^{2}[/latex] for our constant. Can you see why it would not be valid for this case as time increases?

  3. Let’s now apply boundary conditions.
    1. The temperature must be finite at the center of Earth, [latex]r=0[/latex]. Which of the two constants, [latex]B[/latex] or [latex]C[/latex], must therefore be zero to keep [latex]R[/latex] finite at [latex]r=0[/latex]? (Recall that [latex]\sin{(\alpha r)}/r\to a=[/latex] as [latex]r\to 0[/latex], but [latex]\cos{(\alpha r)}/r[/latex] behaves very differently.)
    2. Kelvin argued that when magma reaches Earth’s surface, it cools very rapidly. A person can often touch the surface within weeks of the flow. Therefore, the surface reached a moderate temperature very early and remained nearly constant at a surface temperature [latex]T_s[/latex]. For simplicity, let’s set [latex]T=0[/latex] at [latex]r=R_E[/latex] and find [latex]\alpha[/latex] such that this is the temperature there for all time [latex]t[/latex]. (Kelvin took the value to be [latex]300\text{ K}\approx 80^{\circ}F[/latex]. We can add this [latex]300\text{ K}[/latex] constant to our solution later.) For this to be true, the sine argument must be zero at [latex]r=R_E[/latex]. Note that [latex]\alpha[/latex] represents a valid solution (each with its own value for [latex]A[/latex]). The total or general solution is the sum of all these solutions.
    3. At [latex]t=0[/latex], we assume that all of Earth was at an initial hot temperature [latex]T_0[/latex] (Kelvin took this to be about [latex]7000\text{ K}[/latex].) The application of this boundary condition involves the more advanced application of Fourier coefficients. As noted in part 2. each value of [latex]\alpha_n[/latex] represents a valid solution, and the general solution is a sum of all these solutions. This results in a series solution:
      [latex]\large{T(r,t)=\left(\frac{T_0 R_E}{\pi}\right)\displaystyle\sum_{n}^{}\frac{(-1)^{n-1}}{n}e^{-\lambda n^2t}\frac{\sin{(\alpha_nr)}}{r}}[/latex], where [latex]\large{\alpha_n=n\pi /R_E}[/latex].

       

Note how the values of [latex]\alpha_n[/latex] come from the boundary condition applied in part 2. The term [latex]\frac{-1^{n-1}}{n}[/latex] is the constant [latex]A_n[/latex] for each term in the series, determined from applying the Fourier method. Letting [latex]\beta=\frac{\pi}{R_E}[/latex], examine the first few terms of this solution shown here and note how [latex]\lambda^{2}[/latex] in the exponential causes the higher terms to decrease quickly as time progresses:

[latex]\hspace{4cm}T(r,t)=\frac{T_0 R_E}{\pi r}\Bigg(\begin{alignat}{2} e^{-K\beta^{2}t}(\sin{\beta r})&- \frac{1}{2}e^{-4K\beta^{2}t}(\sin{2\beta r})+\frac{1}{3}e^{-9K\beta^{2}t}(\sin{3\beta r})\\ &-\frac{1}{4}e^{-16K\beta^{2}t}(\sin{4\beta r})+\frac{1}{5}e^{-25K\beta^{2}t}(\sin{5\beta r})\ldots&\quad\\ \end{alignat}\Bigg).[/latex]

Near time [latex]t=0[/latex], many terms of the solution are needed for accuracy. Inserting values for the conductivity [latex]K[/latex] and [latex]\beta=\pi /R_E[/latex] for time approaching merely thousands of years, only the first few terms make a significant contribution. Kelvin only needed to look at the solution near Earth’s surface (Figure 6) and, after a long time, determine what time best yielded the estimated temperature gradient known during his era ([latex]1^{\circ}\text{F}[/latex] increase per [latex]50\text{ ft}[/latex]). He simply chose a range of times with a gradient close to this value. In Figure 6, the solutions are plotted and scaled, with the [latex]300-\text{K}[/latex] surface temperature added. Note that the center of Earth would be relatively cool. At the time, it was thought Earth must be solid.

This figure consists of two figures labeled a and b. Figure a shows three curves labeled 20, 50, and 200 million years on a chart showing fraction of the earth’s radius vs. temperature (K). The highest curve is the 20 million one, then the 50 million one, and then the 200 million one, with all of them starting with a mildly decreasing slope until the slope decreases more steeply around x = 0.2 and then they all intersect at roughly (1, 315). Figure b shows a close up near (1, 315) with the x axis marked 4.0 miles below Earth’s surface; the curves all appear linear in this close up, with the slopes increasing as the value of the curve does.

Figure 4. Temperature versus radial distance from the center of Earth. (a) Kelvin’s results, plotted to scale. (b) A close-up of the results at a depth of [latex]\small{4.0\text{ mi}}[/latex] below Earth’s surface.

Epilogue: On May 20, 1904, physicist Ernest Rutherford spoke at the Royal Institution to announce a revised calculation that included the contribution of radioactivity as a source of Earth’s heat. In Rutherford’s own words:

“I came into the room, which was half-dark, and presently spotted Lord Kelvin in the audience, and realised that I was in for trouble at the last part of my speech dealing with the age of the Earth, where my views conflicted with his. To my relief, Kelvin fell fast asleep, but as I came to the important point, I saw the old bird sit up, open an eye and cock a baleful glance at me.

Then a sudden inspiration came, and I said Lord Kelvin had limited the age of the Earth, provided no new source [of heat] was discovered. That prophetic utterance referred to what we are now considering tonight, radium! Behold! The old boy beamed upon me.”

Rutherford calculated an age for Earth of about [latex]500[/latex] million years. Today’s accepted value of Earth’s age is about [latex]4.6[/latex] billion years.