Separation of Variables

Separation of Variables

The method of separation of variables is effective in solving several types of partial differential equations.

the idea is roughly that we think of a solution, say $u(x,t),$ to a partial differential equation as beinga linear combinatio nof simple component functions $u_n(x,t), n = 0, 1, 2, \dots,$ which also satisfy the equation and certian boundary conditions. This assumption is reasonable provided the partial differential equation and the boundary conditions are linear. To determine a component solution, $u_n(x,t)$, we assume it can be written with its variables separated, that is, as:

\[u_n(x,t) = X_n(X)T_n(t).\]

Substituting this form for a solution into the partial differential equation and using the boundary conditions leads, in many cases, to two ordinary differential equations for the unknown functions $X_n(x)$ and $T_n(t)$.

Solving the Heat Equation

First, see the notes on the heat equation We have the following mathematical model for the heat flow in a uniform wire without internal sources $(P = 0)$ whose ends are kept at the constant temperature $0^\circ c$:

\[\frac{\partial u}{\partial t}(x,t) = \beta \frac{\partial^2 u}{\partial x^2}(x,t), \quad 0 < x < L, \quad t > 0, \tag{1}\] \[u(0, t) = u(L, t) = 0, \quad t>0, \tag{2}\] \[u(x,0) = f(x), \quad 0 < x < L. \tag{3}\]

We wish to find a function $u(x,t)$ that satisfies all three of these conditions.

We being by proposing that the solutions to equation (1) have the form:

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

To determine $X$ and $T$, we first compute the partial derivatives of $u$ to obtain

\[\frac{\partial u}{\partial t} = X(x)T'(t), \quad \frac{\partial^2 u}{\partial x^2} = X''(x)T(t).\]

Substituting these expressions into (1) gives

\[X(x)T'(t) = \beta X''(x)T(t),\]

and separating variables gives

\[\frac{T'(t)}{\beta T(t)} = \frac{X''(x)}{X(x)}. \tag{4}\]

Note that the functions on the left-hand side of (4) depend only on $t$ while those on the right-hand side depend only on $x$. If we fix $t$ and vary $x$, the ratio on the right cannot change; it must be constant. The same applies to fixing $x$ and varying $t$ on the left. Following the convention of this constant being negative, we get:

\[\frac{X''(x)}{X(x)} = - \lambda, \quad \frac{T'(t)}{\beta T(t)} = - \lambda.\]

or

\[X''(x) + \lambda X(x) = 0, \quad T'(t) + \lambda \beta T(t) = 0. \tag{5}\]

which are linear ordinary differential equations, easily solved through a number of means.

Now, turning to the initial conditions in (2), we have, letting $u(x,t) = X(x)T(t):

\[X(0)T(t) = 0, \quad X(L)T(t) = 0, \quad t > 0.\]

Therefore, either $T(t) = 0$ for all $t > 0$, which implies that $u(x,t) \equiv 0$, or

\[X(0) = X(L) = 0. \tag{6}\]

We will ignore the trivial solution $u(x,t) \equiv 0$ and combine the boundary conditions in (6) with the differential equation in (5) to get the boundary value problem

\[X''(x) + \lambda X(x) = 0, \quad X(0) = X(L) = 0, \tag{7}\]

where $\lambda$ can be any constant.

Notice that the function $X(x) \equiv 0$ is a soluton of (7) for every $\lambda$. Depending on the choice of $\lambda$, this may be the only solution to the boundary value problem. Thus, to find a non-trival solution $u(x,t) = X(x)T(t)$, we must first determine those values of $\lambda$ for which the boundary value problem has nontrivial solutions. These solutions are called the eigenfunctions of the problem; the eigenvalues are the special values of $\lambda$.

To solve the equation in (7), we note that it has constant coefficients and follow the method outlined for linear differential equations with constant coefficients, and then solve for the constants using the boundary conditions.

There are three cases:

Case 1: $\lambda < 0$. In this case, the roots of the characteristic equation are $\pm \sqrt{- \lambda}$, so a general solution to (7) is

\[X(x) = c_1 e^{\sqrt{-\lambda}x} + c_2 e^{-\sqrt{-\lambda}x}\]

Applying the boundary conditions and solving for $c_1, c_2$, we find that $c_1 = c_2 = 0$, and hence there are no nontrivial solutions to (7) for $\lambda < 0$.

Case 2: $\lambda = 0$. In this case, $0$ is a repeated root to the characteristic equation, and the general solution to the differential equation is

\[X(x) = C_1 + C_2x.\]

The boundary conditions in (7) again yield $C_1 = C_2 = 0$, so for $\lambda = 0$, there are no nontrivial solutions to (7).

Case 3: $\lambda > 0$. In this case, the roots of the characteristic equation are $\pm i\sqrt{\lambda}$, and the general solution to (7) is

\[X(x) = c_1 \cos{\sqrt{\lambda}x} + c_2 \sin{\sqrt{\lambda}x}. \tag{8}\]

Now, the boundary conditions $X(0) = X(L) = 0$ give the system

\[\begin{align} c_1 = 0 \\ c_1 \cos{\sqrt{\lambda}}L + c_2 \sin{\sqrt{\lambda}}L = 0. \end{align}\]

Because c_1 = 0, the system reduces to solving $c_2 \sin{\sqrt{\lambda}}L = 0.$ Hence, either $\sin{\sqrt{\lambda}L} = $ or $c_2 = 0$.

Now, $\sin{\sqrt{\lambda}L} = 0$ only when $\sqrt{\lambda}L = n \pi$, where $n$ is an integer. Therefore, (7) has a nontrivial solution ($c_2 \neq 0$) when $\sqrt{\lambda}L = n \pi$ or $\lambda = (n \pi / L)^2$, $n = 1,2,3,…$. The nontrivial solutions (eigenfunctions) $X_n$ corresponding to the eigenvalue $\lambda = (n \pi/L)^2$ are given by

\[X_n(x) = a_n \sin{\left ( \frac{n \pi x}{L} \right ) }, \tag{9}\]

where the $a_n$’s are arbitrary nonzero constants.

Now that we’ve determined that $\lambda = (n \pi/L)^2$, for any positive integer $n$, we return our condition to the second equation in (5):

\[T'(t) + \beta \left ( \frac{n \pi}{L} \right )^2 T(t) = 0.\]

For each $n = 1, 2, 3 \dots$, the general solution to this linear first-order equation is

\[T_n(t) = b_n e^{-\beta(n \pi/L)^2}.\]

Combining this with equation (9), we obtain, for each $n = 1,2,3,\dots$, the functions

\[\begin{align} u_n(x_t) = X_n(x)T_n(t) = a_n\sin{(n \pi x / L)} b_n e^{-\beta (n \pi / L)^2 t} \\ = c_n e^{-\beta (n \pi / L)^2 t} \sin{(n \pi x / L)}, \end{align}\]

where $c_n$ is also an arbitrary constant.

Example

Find the soluton to the heat flow problem

\[\frac{\partial u}{\partial t} = 7 \frac{\partial^2 u}{\partial x^2}, \quad 0 < x < \pi, \quad t > 0, \tag{11}\] \[u(0, t) = u(\pi, t) = 0, \quad t>0, \tag{12}\] \[u(x,0) = 3\sin{2x} - 6\sin{5x}, \quad 0 < x < \pi. \tag{13}\]

Comparing (11) with (1), we have that $\beta = 7$ and $L = \pi$. Hence, we need only find a combination of terms like (10) that satisfies the initial condition (13):

\[u(x,0) = \sum c_n e^0 \sin{nx} = 3\sin{2x} - 6\sin{5x}, \quad 0 < x < \pi.\]

The constants we require are $c_2 = 3$ and $c_5 = -6$. The solution to the heat flow problem (11)-(13) is:

\[\begin{align} u(x,t) = c_2e^{- \beta(2 \pi / L)^2t}\sin(2 \pi x/L) + c_5e^{- \beta(5 \pi / L)^2t}\sin(5 \pi x/L) \\ = 3e^{-28t}\sin{2x} - 6e^{-175t} \sin{5x}. \end{align}\]

Complete Solution to Generic Problem

It turns out that almost any function $f(x)$ likely to arise in applications can be expressed as a convergent series of eigenfunctions. For the sines we’ve been using, the Fourier sine series looks like:

\[f(x) = \sum_{n=1}^{\infty} c_n \sin{\left ( \frac{n \pi x}{L} \right )}, \quad 0 < x < L.\]

and the complete solution to the generic problem given by (1)-(3) is

\[u(x,t) = \sum_{n=1}^{\infty} c_n e^{- \beta (n \pi/L)^2 t}\sin{\left ( \frac{n \pi x}{L} \right )}, \tag{2a}\]

as long as this expansion and its first two derivatives converge.

Other homogenous boundary conditions

Ends zero, but with arbitrary f(x)

If our initial condition isn’t just a simple, finite set of $c_n\sin(nx)$ functions, finding our $c_n$’s is just a matter of finding the coefficients for the fourier sine series

\[c_n = \frac{2}{L} \int_0^{L} f(x) \sin{\frac{n \pi x}{L}} dx, \quad n = 1,2, \dots ,\]

and substituting them into (2a) above. We use the sine series because the initial conditions having the ends set to 0 lines up with the behavior of sine.

Ends perfectly insulated, with arbitrary f(x)

If the ends of the wire are perfectly insulated, then no heat flows through them and thus the derivative of them, at each end, with respect to time, is 0:

\[\frac{\partial u}{\partial x}(0,t) = \frac{\partial u }{\partial x}(L, t) = 0, \quad t > 0.\]

Now, we use the fourier cosine series because the derivative of cosine at $x=0$ is 0, which matches our initial conditions. So, our solution is

\[\frac{a_0}{2} + \sum_{n=1}^{\infty} a_n e^{-\beta\frac{n\pi}{L}^2t} \cos{\frac{n \pi x}{L}}\]

where

\[a_n = \frac{2}{L} \int_0^{L} f(x) \cos{\frac{n \pi x}{L}} dx, \quad n = 0, 1, \dots .\]

Nonhomogenous boundary conditions

Ends kept at constant, non-zero temperature

When the ends of the wire are kept a constant, non-zero temperatures:

\[u(0, t) = U_1, \quad U(L,t) = U_2, \quad t > 0,\]

the boundary conditions are called nonhomogenous.

Here, we’ll end up with a solution that’s a combintation of a steady-state solution $v(x)$ that satisfies the nonhomgenous boundary conditions, plus a transient solution $w(x,t)$, and

\[u(x,t) = v(x) + w(x,t),\]

where $w(x,t)$ and its partial derivatives tend to zero as $t \to \inf$. The function $w(x,t)$ will then staisfy homogenous boundary conditions.

For example,

\[\frac{\partial u}{\partial t} = \beta \frac{\partial^2 u}{\partial x^2}, \quad 0 < x < L, \quad t > 0, \tag{11b}\] \[u(0, t) = U_1, ~ u(L, t) = U_2, \quad t>0, \tag{12b}\] \[u(x,0) = f(x), \quad 0 < x < L. \tag{13b}\]

Here,

\[v(x) = U_1 + \frac{(U_2 - U_1)x}{L}, \quad w(x,t) = \sum_{n=1}^{\infty} c_n e^{-\beta(n\pi/L)^2t}\sin{\frac{n \pi x}{L}},\]

with

\[c_n = \frac{2}{L}\int_{0}^{L} \left [ f(x) -U_1 - \frac{(U_2 - U_1)x}{L} \right ] \sin{\frac{n \pi x}{L}}.\]

Vibrating String Problem

The equation of motion for a vibrating string of length $L$ with fixed ends, and a bunch of other simplifying factors, is governed by the following initial-boundary value problem:

\[\frac{\partial^2 u}{\partial t^2} = \alpha^2 \frac{\partial^2 u}{\partial x^2}, \quad 0 < x < L, \quad t > 0 \tag{16}\] \[u(0,t) = u(L,t) = 0, \quad t \geq 0, \tag{17}\] \[u(x,0) = f(x), \quad 0 \leq x \leq L, \tag{18}\] \[\frac{\partial u}{\partial t}(x, 0) = g(x), \quad 0 \leq x \leq L. \tag{19}\]

Applying the method of separation of variables, we end up with

\[u(x,t) = \sum_{n=1}^{\infty} \left [ a_n \cos{\frac{n \pi \alpha}{L}t} + b_n \sin {\frac{n \pi \alpha}{L}t} \right ] \sin{\frac{n \pi x}{L}}.\]

To find $a_n$ and $b_n$, we have

\[f(x) = \sum_{n=1}^{\infty} a_n \sin{\frac{n \pi x}{L}}, \quad g(x) = \sum_{n=1}^{\infty} \frac{n \pi \alpha}{L} b_n \sin{\frac{n \pi x}{L}}.\]

Example

Solve the vibrating string problem with $\alpha = 5$, $L = \pi$, initial functions $f(x) = 2\sin{4x}+7\sin{5x}$ and $g(x) = 11\sin{6x} - 14\sin{13x}$.

Our $a_n$’s are $a_4 = 2,$ $a_5 = 7$, and to find the $b_n$’s:

\[11 = 5 \cdot 6 b_6, ~ b_6 = \frac{11}{30}; \quad -14 = 5 \cdot 13 b_13, ~ b_{13} = \frac{-14}{65}.\]

Then

\[u(x,t) = 2\cos{20t}\sin{4x} + 7\cos{25t}\sin{5x} + \frac{11}{30}\sin{30t}\sin{6x} - \frac{14}{15}\sin{65t}\sin{13x}.\]

Laplace’s Equation

Laplace’s equation:

\[\Delta u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y ^2} = 0\]

Has a lot of applications. You might recall from complex analysis that the real and imaginary parts of an analytic (holomorphic) function must satisfy Laplace’s equation.

Dirichlet and Neumann Boundary Conditions

There are two basic types of boundary conditions that are usually associated with Laplace’s equation. The first is Dirichlet boundary conditions, where the solution $u(x,y)$ to Laplace’s equation in a domain $D$ is required to satisfy:

\[u(x,y) = f(x,y), \quad \text{on} ~ D,\]

with $f(x,y)$ a specified function defined on the boundary $\partial D$ of $D$. The other type of boundary conditions are Neumann boundary coundtiions, where the directional derivative $\partial u/\partial n$ along the outward normal to the boundary is required to satisfy

\[\frac{\partial u}{\partial n}(x,y) = g(x,y), \quad \text{on} ~ D,\]

with $g(x,y)$ a specified function defined on the boundary $\partial D$ of $D.$ We say that a boundary condition is mixed if the solution is required to satisfy $u(x,y) = f(x,y)$ on part of the boundary and $(\partial u / \partial n)(x,y) = g(x,y)$ on the remaining portion of the boundary.

When solving problems like this, we sometime need to pick which type of Fourier series we want to use - sine or cosine.

A Fourier cosine series is typically used for functions that have Neumann (derivative) boundary conditions (especially if the initial value of the derivative on each end is 0) or are even across the domain.

A Fourier sine series is typically used for functions that have Dirichlet boundary conditions (especially if the initial value of the function on each end is 0) or are odd across the domain.