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Heat equation. Fourier method - separation of variables.

Consider the heat equation

\begin{displaymath}
\frac{\partial }{\partial t} u(t,x) = \frac{\partial^2 }{\partial x^2}u(t,x), \;\;\;x\in (0,\;L)
\end{displaymath}

It models the heat propagation in a thin uniform bar or wire of length $L.$ The function $u(t,x)$ describes the temperature at the point $x$ and time $t.$ The heat dynamic depends on the boundary conditions,

\begin{displaymath}
u(t,0)=0,\;\;\;u(t,L)=0
\end{displaymath}

and initial conditions

\begin{displaymath}
u(0,x) = f(x).
\end{displaymath}

To find the solution for the heat equation we use the Fourier method of separation of variables. First, we look for special solutions having the form

\begin{displaymath}
u(t,x)= F(t) G(x).
\end{displaymath}

Substitution of this special type of the solution into the heat equation leads us to

\begin{displaymath}
(\frac{d}{dt}F(t))G(x) = F(t)\frac{d^2}{dx^2}G(x)
\end{displaymath}

and after separation of variables ($x$ and $t$) we obtain

\begin{displaymath}
\frac{\frac{d}{dt}F(t)}{F(t)} = \frac{\frac{d^2}{dx^2}G(x)}{G(x)}
\end{displaymath}

Since $t$ and $x$ are independent variables this equality can hold only when both left and right hand sides are constants. Thus we have two independent equations
\begin{displaymath}
\frac{d}{dt}F(t) = \lambda \cdot F(t)
\end{displaymath} (1.1)

and
\begin{displaymath}
\frac{d^2}{dx^2}G(x) = \lambda \cdot G(x)
\end{displaymath} (1.2)

that are related to each other only through the constant $\lambda.$ Let us first solve the equation (1.2). There are three possibilities to consider:

I.
$\lambda > 0 .$
II.
$\lambda = 0.$
III.
$\lambda < 0 .$



Subsections
next up previous
Next: I. Up: Heat equation. Fourier method Previous: Heat equation. Fourier method
Sergey Nikitin 2004-10-25