1-Dimensional Heat Initial Boundary Value Problems 1: Separation of Variables

Let us consider the following assumptions for a heat conduction problem.

  1. The region $\Omega$ is a cylinder of length $L$ centered on the $x$-axis.
  2. The lateral surface is insulated.
  3. The left end ($x=0$) and the right end ($x=L$) have boundary conditions that do not depend on the $y$ and $z$ coordinates.
  4. The initial temperature distribution $\phi$ does not depend on $y$ and $z$ i.e. $\phi=\phi(x)$.

Under these assumptions we want to find temperature distribution $u(\mathbf{r},t)=u(x,y,z,t)$ which satisfies the heat equation
$$\frac{\partial u}{\partial t}=\alpha^2\nabla^2 u+F(\mathbf{r},t)$$
Here $\alpha$ is a constant called the diffusitivity of the material and $F$ is a scalar function called the heat source density. Due to the assumption 2 the  heat equation becomes the 1-dimensional heat equation
$$\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}+F(x,t)$$
The most general form of the boundary conditions is given by the Newton’s law of cooling
$$-\kappa\nabla u\cdot\mathbf{n}=h(u-g)\ \mbox{on}\ \partial\Omega,\ t>0$$
where $\kappa\geq 0, h\geq 0$, $\mathbf{n}$ is unit normal to $\partial\Omega$ and $g$ is the temperature on $\partial\Omega$, $t>0$.

In our 1-dimensional case, the heat flux vector field $\nabla u$ is given by $\nabla u=\left(\frac{\partial u}{\partial x},0,0\right)$. So on the lateral surface $\nabla u\cdot\mathbf{n} =0$ i.e. the assumption that lateral surface is insulated is automatically satisfied. On the left end, $\mathbf{n}=(-1,0,0)$ so
$$-\kappa\nabla u\cdot\mathbf{n}=\kappa\frac{\partial u}{\partial x}.$$
On the right end, $\mathbf{n}=(1,0,0)$ so
$$-\kappa\nabla u\cdot\mathbf{n}=-\kappa\frac{\partial u}{\partial x}.$$
Hence for our 1-dimensional heat problem the general boundary conditions (BCs) are given by
-\kappa_1u_x(0,t)+h_1u(0,t)&=g_1(t),\ t>0\\
\kappa_2u_x(L,t)+h_2u(L,t)&=g_2(t),\ t>0
The BCs are the main ingredients that uniquely determine a specific physical heat conduction phenomenon along with initial condition (IC)
$$u(x,0)=\phi(x),\ 0<x<L$$

The heat equation is said to be homogeneous if $F(x,t)=0$. The BCs are said to be homogeneous if $g_1(t)=g_2(t)=0$.

Separation of Variables Method: The separation of variables method is one of the oldest methods of solving partial differential equations. It reduces a partial differential equation to a number of ordinary differential equations.

Consider the homogeneous 1-dimensional heat equation
$$\frac{\partial u}{\partial t}=\alpha^2\frac{\partial^2 u}{\partial x^2}$$
Assume that $u(x,t)=X(x)T(t)$.
where $’=\frac{\partial}{\partial x}$ and $\dot{}=\frac{\partial}{\partial t}$.
Divide this by $\alpha^2X(x)T(t)$. Then
The LHS depends only on time variable $t$ while the RHS depends on $x$ variable. This is possible when both the LHS and the RHS are the same as a constant, say, $k$. Thereby the 1D heat equation reduces to the ordinary differential equations
The second order linear equation has the following solutions depending on the sign of $k$:

  • If $k=0$, $X(x)=Ax+B$.
  • If $k=\lambda^2>0$, $X(x)=Ae^{\lambda x}+Be^{-\lambda x}$ or  $X(x)=A\cosh\lambda x+B\sinh\lambda x$.
  • If $k=-\lambda^2<0$, $X(x)=A\cos\lambda x+B\sinh\lambda x$.

The first order linear equation has solution
$$T(t)=Ce^{k\alpha^2 t}$$
Here one may assume that $C=1$ without loss of generality. Therefore we have three possible cases of $u(x,t)$:

  • $u(x,t)=Ax+B$
  • $u(x,t)=e^{\lambda^2\alpha^2 t}(Ae^{\lambda x}+Be^{-\lambda x})$ or $u(x,t)=e^{\lambda^2\alpha^2 t}(A\cosh\lambda x+B\sinh\lambda x)$
  • $u(x,t)=e^{-\lambda^2\alpha^2 t}(A\cos\lambda x+B\sinh\lambda x)$


David Betounes, Partial Differential Equations for Computational Science with Maple and Vector Analysis, TELOS, Springer-Verlag

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