4.2: Partial Differential Equations

In this section we will introduce some generic partial differential equations and see how the discussion of such equations leads naturally to the study of boundary value problems for ordinary differential equations. However, we will not derive the particular equations, leaving that to courses in differential equations, mathematical physics, etc.

For ordinary differential equations, the unknown functions are functions of a single variable, e.g., \(y=y(x)\). Partial differential equations are equations involving an unknown function of several variables, such as \(u=u(x, y), u= u(x, y), u=u(x, y, z, t)\), and its (partial) derivatives. Therefore, the derivatives are partial derivatives. We will use the standard notations \(u_{x}=\dfrac{\partial u}{\partial x}, u_{x x}=\dfrac{\partial^{2} u}{\partial x^{2}}\), etc.

There are a few standard equations that one encounters. These can be studied in one to three dimensions and are all linear differential equations. A list is provided in Table 4.1. Here we have introduced the Laplacian operator, \(\nabla^{2} u=u_{x x}+u_{y y}+u_{z z}\). Depending on the types of boundary conditions imposed and on the geometry of the system (rectangular, cylindrical, spherical, etc.), one encounters many interesting boundary value problems for ordinary differential equations.

Name	2 Vars	3 D
Heat Equation	\(u_{t}=k u_{x x}\)	\(u_{t}=k \nabla^{2} u\)
Wave Equation	\(u_{t t}=c^{2} u_{x x}\)	\(u_{t t}=c^{2} \nabla^{2} u\)
Laplace's Equation	\(u_{x x}+u_{y y}=0\)	\(\nabla^{2} u=0\)
Poisson's Equation	\(u_{x x}+u_{y y}=F(x, y)\)	\(\nabla^{2} u=F(x, y, z)\)
Schrödinger's Equation	\(i u_{t}=u_{x x}+F(x, t) u\)	\(i u_{t}=\nabla^{2} u+F(x, y, z, t) u\)

Table 4.1. List of generic partial differential equations.

Let’s look at the heat equation in one dimension. This could describe the heat conduction in a thin insulated rod of length \(L\). It could also describe the diffusion of pollutant in a long narrow stream, or the flow of traffic down a road. In problems involving diffusion processes, one instead calls this equation the diffusion equation.

A typical initial-boundary value problem for the heat equation would be that initially one has a temperature distribution \(u(x, 0)=f(x)\). Placing the bar in an ice bath and assuming the heat flow is only through the ends of the bar, one has the boundary conditions \(u(0, t)=0\) and \(u(L, t)=0\). Of course, we are dealing with Celsius temperatures and we assume there is plenty of ice to keep that temperature fixed at each end for all time. So, the problem one would need to solve is given as

1D Heat Equation

\[\begin{array}{lcc}
\text { PDE } & u_{t}=k u_{x x} \quad 0<t, \quad 0 \leq x \leq L \\
\text { IC } & u(x, 0)=f(x) & 0<x<L \\
\text { BC } & u(0, t)=0 & t>0 \\
& u(L, t)=0 & t>0
\end{array} \label{4.3} \]

Here, \(k\) is the heat conduction constant and is determined using properties of the bar.

Another problem that will come up in later discussions is that of the vibrating string. A string of length \(L\) is stretched out horizontally with both ends fixed. Think of a violin string or a guitar string. Then the string is plucked, giving the string an initial profile. Let \(u(x, t)\) be the vertical displacement of the string at position \(x\) and time \(t\). The motion of the string is governed by the one dimensional wave equation. The initial-boundary value problem for this problem is given as

1D Wave Equation

\[\begin{array}{lcc}
\mathrm{PDE} & u_{t t}=c^{2} u_{x x} & 0<t, \quad 0 \leq x \leq L \\
\mathrm{IC} & u(x, 0)=f(x) & 0<x<L \\
\mathrm{BC} & u(0, t)=0 & t>0 \\
& u(L, t)=0 & t>0
\end{array} \label{4.4} \]

In this problem \(c\) is the wave speed in the string. It depends on the mass per unit length of the string and the tension placed on the string.

4.2.1 Solving the Heat Equation

We would like to see how the solution of such problems involving partial differential equations will lead naturally to studying boundary value problems for ordinary differential equations. We will see this as we attempt the solution of the heat equation problem 4.3. We will employ a method typically used in studying linear partial differential equations, called the method of separation of variables.

We assume that \(u\) can be written as a product of single variable functions of each independent variable,

\[u(x, t)=X(x) T(t) \nonumber \]

Substituting this guess into the heat equation, we find that

\[X T^{\prime}=k X^{\prime \prime} T. \nonumber \]

Dividing both sides by \(k\) and \(u=X T\), we then get

\[\dfrac{1}{k} \dfrac{T^{\prime}}{T}=\dfrac{X^{\prime \prime}}{X} \nonumber \]

We have separated the functions of time on one side and space on the other side. The only way that a function of \(t\) equals a function of \(x\) is if the functions are constant functions. Therefore, we set each function equal to a constant, \(\lambda:\)

\[\underbrace{\dfrac{1}{k} \dfrac{T^{\prime}}{T}}_{\text {function of } x}=\underbrace{\dfrac{X^{\prime \prime}}{X}}_{\text {constant }} = \underbrace{\lambda}_{\text{constant}} . \nonumber \]

This leads to two equations:

\[T^{\prime}=k \lambda T \label{4.5} \]

\[X^{\prime \prime}=\lambda X \label{4.6} \]

These are ordinary differential equations. The general solutions to these equations are readily found as

\[T(t)=A e^{k \lambda t} \label{4.7} \]

\[X(x)=c_{1} e^{\sqrt{\lambda} x}+c_{2} e^{\sqrt{-\lambda} x} \label{4.8} \]

We need to be a little careful at this point. The aim is to force our product solutions to satisfy both the boundary conditions and initial conditions. Also, we should note that \(\lambda\) is arbitrary and may be positive, zero, or negative. We first look at how the boundary conditions on \(u\) lead to conditions on \(X\).

The first condition is \(u(0, t)=0\). This implies that

\[X(0) T(t)=0 \nonumber \]

for all \(t\). The only way that this is true is if \(X(0)=0\). Similarly, \(u(L, t)=0\) implies that \(X(L)=0\). So, we have to solve the boundary value problem

\[X^{\prime \prime}-\lambda X=0, \quad X(0)=0=X(L). \label{4.9} \]

We are seeking nonzero solutions, as \(X \equiv 0\) is an obvious and uninteresting solution. We call such solutions trivial solutions.

There are three cases to consider, depending on the sign of \(\lambda\).

I. \(\underline{\dfrac{\lambda>0}}\)

In this case we have the exponential solutions

\[X(x)=c_{1} e^{\sqrt{\lambda} x}+c_{2} e^{\sqrt{-\lambda} x} \label{4.10} \]

For \(X(0)=0\), we have

\[0=c_{1}+c_{2}. \nonumber \]

We will take \(c_{2}=-c_{1}\). Then, \(X(x)=c_{1}\left(e^{\sqrt{\lambda} x}-e^{\sqrt{-\lambda} x}\right)=2 c_{1} \sinh \sqrt{\lambda} x\). Applying the second condition, \(X(L)=0\) yields

\[c_{1} \sinh \sqrt{\lambda} L=0 \nonumber \]

This will be true only if \(c_{1}=0\), since \(\lambda>0\). Thus, the only solution in this case is \(X(x)=0\). This leads to a trivial solution, \(u(x, t)=0\).

II. \(\underline{\lambda=0}\)

For this case it is easier to set \(\lambda\) to zero in the differential equation. So, \(X^{\prime \prime}=0\). Integrating twice, one finds

\[X(x)=c_{1} x+c_{2} \nonumber \]

Setting \(x=0\), we have \(c_{2}=0\), leaving \(X(x)=c_{1} x\). Setting \(x=L\), we find \(c_{1} L=0\). So, \(c_{1}=0\) and we are once again left with a trivial solution.

III. \(\underline{\lambda<0}\)

In this case is would be simpler to write \(\lambda=-\mu^{2}\). Then the differential equation is

\[X^{\prime \prime}+\mu^{2} X=0 \nonumber \]

The general solution is

\[X(x)=c_{1} \cos \mu x+c_{2} \sin \mu x. \nonumber \]

At \(x=0\) we get \(0=c_{1}\). This leaves \(X(x)=c_{2} \sin \mu x\). At \(x=L\), we find

\(0=c_{2} \sin \mu L\).

So, either \(c_{2}=0\) or \(\sin \mu L=0. c_{2}=0\) leads to a trivial solution again. But, there are cases when the sine is zero. Namely,

\[\mu L==n \pi, \quad n=1,2, \ldots \nonumber \]

Note that \(n=0\) is not included since this leads to a trivial solution. Also, negative values of \(n\) are redundant, since the sine function is an odd function.

In summary, we can find solutions to the boundary value problem (4.9) for particular values of \(\lambda\). The solutions are

\[X_{n}(x)=\sin \dfrac{n \pi x}{L}, \quad n=1,2,3, \ldots \nonumber \]

for

\[\lambda_{n}=-\mu_{n}^{2}=-\left(\dfrac{n \pi}{L}\right), \quad n=1,2,3, \ldots \nonumber \]

Product solutions of the heat equation (4.3) satisfying the boundary conditions are therefore

\[u_{n}(x, t)=b_{n} e^{k \lambda_{n} t} \sin \dfrac{n \pi x}{L}, \quad n=1,2,3, \ldots, \label{4.11} \]

where \(b_{n}\) is an arbitrary constant. However, these do not necessarily satisfy the initial condition \(u(x, 0)=f(x)\). What we do get is

\[u_{n}(x, 0)=\sin \dfrac{n \pi x}{L}, \quad n=1,2,3, \ldots \nonumber \]

So, if our initial condition is in one of these forms, we can pick out the right \(n\) and we are done.

For other initial conditions, we have to do more work. Note, since the heat equation is linear, we can write a linear combination of our product solutions and obtain the general solution satisfying the given boundary conditions as

\[u(x, t)=\sum_{n=1}^{\infty} b_{n} e^{k \lambda_{n} t} \sin \dfrac{n \pi x}{L} \label{4.12} \]

The only thing to impose is the initial condition:

\[f(x)=u(x, 0)=\sum_{n=1}^{\infty} b_{n} \sin \dfrac{n \pi x}{L} \nonumber \]

So, if we are given \(f(x)\), can we find the constants \(b_{n}\)? If we can, then we will have the solution to the full initial-boundary value problem. This will be the subject of the next chapter. However, first we will look at the general form of our boundary value problem and relate what we have done to the theory of infinite dimensional vector spaces.