\[ % mathjax inclusion.
\newcommand\bbP{\mathbb{P}}
\newcommand\bbR{\mathbb{R}}
\newcommand\becomes{\mathop{:=}}
\newcommand\dtdxx{\frac{\alpha\Delta t}{\Delta x^2}}
\newcommand\defined{
\mathrel{\lower 5pt \hbox{${\equiv\atop\mathrm{\scriptstyle D}}$}}}
\newcommand\fp[2]{#1\cdot10^{#2}}
\newcommand\inv{^{-1}}\newcommand\invt{^{-t}}
\newcommand\macro[1]{$\langle$#1$\rangle$}
\newcommand\nobreak{}
\newcommand\Rn{{\cal R}^n}
\newcommand\Rnxn{{\cal R}^{n\times x}}
\newcommand\sublocal{_{\mathrm\scriptstyle local}}
\newcommand\toprule{\hline}
\newcommand\midrule{\hline}
\newcommand\bottomrule{\hline}
\newcommand\multicolumn[3]{#3}
\newcommand\defcolvector[2]{\begin{pmatrix}
#1_0

#1_1

vdots

#1_{#2-1} \end{pmatrix}} % {

left(

begin{array}{c} #1_0

#1_1

vdots

#1_{#2-1}

end{array}

right) } \] 15.1 : Partial derivatives

15.2 : Poisson or Laplace Equation

15.3 : Heat Equation

15.4 : Steady state

Back to Table of Contents# 15 Partial Differential Equations

## 15.1 Partial derivatives

## 15.2 Poisson or Laplace Equation

## 15.3 Heat Equation

## 15.4 Steady state

Back to Table of Contents
#1_1

vdots

#1_{#2-1} \end{pmatrix}} % {

left(

begin{array}{c} #1_0

#1_1

vdots

#1_{#2-1}

end{array}

right) } \] 15.1 : Partial derivatives

15.2 : Poisson or Laplace Equation

15.3 : Heat Equation

15.4 : Steady state

Back to Table of Contents

Partial Differential Equations
are the source of a large fraction of
*HPC*
problems. Here is a
quick derivation of two of the most important ones.

crumb trail: > pde > Partial derivatives

Derivatives of a function $u(x)$ are a measure of the rate of change. Partial derivatives to the same, but for a function $u(x,y)$ of two variables. Notated $u_x$ and $u_y$, these \indexterm{partial derivates} indicate the rate of change if only one variable changes and the other stays constant.

Formally, we define $u_x,u_y$ by: \begin{equation} u_x(x,y) = \lim_{h\rightarrow0}\frac{u(x+h,y)-u(x,y)}h,\quad u_y(x,y) = \lim_{h\rightarrow0}\frac{u(x,y+h)-u(x,y)}h \end{equation}

crumb trail: > pde > Poisson or Laplace Equation

Let $T$ be the temperature of a material, then its heat energy is
proportional to it. A segment of length $\Delta x$ has heat energy
$Q=c\Delta x\cdot u$. If the heat energy in that
segment is constant
\begin{equation}
\frac{\delta Q}{\delta t}=c\Delta x\frac{\delta u}{\delta t}=0
\end{equation}
but it is also the difference between inflow and outflow of the
segment. Since flow is proportional to temperature differences, that
is, to $u_x$, we see that also
\begin{equation}
0=
\left.\frac{\delta u}{\delta x}\right|_{x+\Delta x}-
\left.\frac{\delta u}{\delta x}\right|_{x}
\end{equation}
In the limit of $\Delta x\downarrow0$ this gives $u_{xx}=0$, which is
called the
*Laplace equation*
. If we have a source term, for
instance corresponding to externally applied heat, the equation
becomes $u_{xx}=f$, which is called the
*Poisson equation*
.

crumb trail: > pde > Heat Equation

Let $T$ be the temperature of a material, then its heat energy is proportional to it. A segment of length $\Delta x$ has heat energy $Q=c\Delta x\cdot u$. The rate of change in heat energy in that segment is \begin{equation} \frac{\delta Q}{\delta t}=c\Delta x\frac{\delta u}{\delta t} \end{equation} but it is also the difference between inflow and outflow of the segment. Since flow is proportional to temperature differences, that is, to $u_x$, we see that also \begin{equation} \frac{\delta Q}{\delta t}= \left.\frac{\delta u}{\delta x}\right|_{x+\Delta x}- \left.\frac{\delta u}{\delta x}\right|_{x} \end{equation} In the limit of $\Delta x\downarrow0$ this gives $u_t=\alpha u_{xx}$.

crumb trail: > pde > Steady state

The solution of an
*IBVP*
is a function $u(x,t)$. In cases where
the forcing function and the boundary conditions do not depend on
time, the solution will converge in time, to a function called the
*steady state*
solution:
\begin{equation}
\lim_{t\rightarrow\infty} u(x,t)=u_{\mathrm{steady state}}(x).
\end{equation}
This solution satisfies a
*BVP*
, which can be found by setting
$u_t\equiv\nobreak0$. For instance, for the heat equation
\begin{equation}
u_t=u_{xx}+q(x)
\end{equation}
the steady state solution satisfies $-u_{xx}=q(x)$.