Theorem (General solutions to a special ODE).
Let \(f\) be a function with continuous first and second derivatives. Consider the differential equation \[ f'' + af' + bf = 0, \] where \(a\) and \(b\) are constants. If \(r_1\) and \(r_2\) are the roots of the characteristic equation \[ r^2 + ar + b = 0, \]
then the general solution to the ODE is
\[ f( x ) = C_1 e^{r_1 x} + C_2 e^{r_2 x}, \]
where \(C_1\) and \(C_2\) are arbitrary constants determined by the initial and boundary conditions.
This ODE is called homogeneous because it has no terms that are functions of \(x\) alone, i.e., the right-hand side is zero.
This ODE is called linear because the differential operator on the left, \[ L(f) = f'' + af' + bf, \] preserves linear combinations, i.e., we have \(L(f+g)= L(f) + L(g)\) and \(L(cf) = cL(f)\), for all functions \(f\) and \(g\) and any scalar \(c\).
This ODE is called second-order because the highest derivative that appears is the second derivative, \(f''\).
This ODE is called constant-coefficient because the coefficients \(a\) and \(b\) are constants, not functions of \(x\).
Convenient expression for the general solutions
The general solution can be written as \[ f(x) = e^{\alpha x} \left( C_1 \sin(\beta x) + C_2 \cos(\beta x) \right), \] where \(C_1\) and \(C_2\) are arbitrary constants.
Recall that the heat equation is \(\displaystyle\frac{\partial u}{\partial t} = k \frac{\partial^2 u}{\partial x^2}\) where \(k\) is a positive (known) constant.
We assume the boundary conditions \[ u(0,t) = 0, \quad u(L,t) = 0, \quad t > 0, \] where \(L\) is some positive (known) constant.
We ignore initial conditions for the moment.
The “separation of variables” strategy
Dividing both sides of the equation by \(k X(x) T(t)\) gives \[ \frac{T'(t)}{k T(t)} = \frac{X''(x)}{X(x)}. \]
Notice that the left-hand side depends only on \(t\) and the right-hand side depends only on \(x\). Therefore, both sides must be equal to a constant, which we denote by \(-\lambda\): \[ \frac{T'(t)}{k T(t)} = \frac{X''(x)}{X(x)} = -\lambda. \]
This leads to a system of two ODEs: \[ T'(t) = - k \lambda T(t) \tag{1} \] and \[ X''(x) + \lambda X(x) = 0, \quad X(0) = 0, \quad X(L) = 0, \tag{2} \]
The general solution to (1) is \[ T(t) = C_0 \exp\left(-k \lambda t\right), \] where \(C_0\) is an arbitrary constant.
Based on our knowledge of ODEs (from the first slide), the general solution to (2) is \[ X(x) = C_1 \sin(\sqrt{\lambda} x) + C_2 \cos(\sqrt{\lambda} x), \] where \(C_1\) and \(C_2\) are arbitrary constants.
But now the boundary condition \(X(0)=0\) forces \(C_2 = 0\), so \[ X(x) = C_1 \sin(\sqrt{\lambda} x). \]
The boundary condition \(X(L)=0\) then gives \[ C_1 \sin(\sqrt{\lambda} L) = 0. \]
Since \(C_1 \neq 0\) (otherwise the solution is trivial), we must have \[ \sin(\sqrt{\lambda} L) = 0, \] which implies \[ \sqrt{\lambda} L = n \pi, \quad n = 0, \pm 1, \pm 2, \pm3, \dots \]
Thus, \[ \lambda = \left( \frac{n \pi}{L} \right)^2. \]
Therefore, for each \(n=1,2,3,\dots\), we have a solution \[ X_n(x) = \sin\left( \frac{n \pi x}{L} \right) \] to (2). (We ignore the negative values of \(n\)—why?)
Basic solutions to the heat equation (w/zero temperature at the boundaries)
Consider the heat equation \[ \frac{\partial u}{\partial t} = k \frac{\partial^2 u}{\partial x^2} \] with boundary conditions \(u(0,t) = 0\) and \(u(L,t) = 0\) for all \(t>0\). Then for each \(n=1,2,3,\dots\), the function \[ u_n(x,t) = \exp\left(\frac{-n^2 \pi^2 kt}{L^2} \right) \sin\left( \frac{n \pi x}{L} \right), \quad 0 \leq x \leq L, \ t\geq 0 \] is a solution to the boundary-value problem above.
Theorem (Superposition principle for the heat equation).
If \(u_1(x,t), u_2(x,t), \dots, u_n(x,t)\) are solutions to the heat equation, then any linear combination \[ u(x,t) = a_1 u_1(x,t) + a_2 u_2(x,t) + \dots + a_n u_n(x,t) \] is also a solution, where \(a_1, a_2, \dots, a_n\) are constants.
Thought: If linear combinations of finitely many solutions are also solutions, what about infinite sums?
If \(u_1(x,t), u_2(x,t), \dots\) are solutions to the heat equation, then perhaps the infinite sum \[ u(x,t) = \sum_{n=1}^{\infty} a_n u_n(x,t) \] is also a solution, where \(a_n\) are constants.
In particular, if we take the basic solutions \[ u_n(x,t) = \exp\left(\frac{-n^2 \pi^2 kt}{L^2} \right) \sin\left( \frac{n \pi x}{L} \right), \] to the heat equation, then is the infinite sum \[ u(x,t) = \sum_{n=1}^{\infty} a_n \exp\left(\frac{-n^2 \pi^2 kt}{L^2} \right) \sin\left( \frac{n \pi x}{L} \right) \] also a solution? (Even more pressing—what about convergence of the series?)
If we now bring in an initial condition specifying that \[ u(x,0) = f(x), \quad 0 \leq x \leq L, \] for some initial temperature distribution \(f(x)\), then our “infinite superposition” solution must satisfy \[ f(x) = \sum_{n=1}^{\infty} a_n \sin\left( \frac{n \pi x}{L} \right), \quad 0 \leq x \leq L. \]
But for what functions \(f(x)\) does this series converge? What are the \(a_n\)’s?
Quetions raised by separation of variables
Assuming it converges, is an infinite linear combination of solutions to the heat equation also a solution?
For what types of functions \(f(x)\) can we write \[ f(x) = \sum_{n=1}^{\infty} a_n \sin\left( nx \right), \quad (0 \leq x \leq \pi)? \] (This is just a “rescaled” version of the series on the previous slide.)