text: # Partial Differential Equations
  
  $\newcommand\inv{^{-1}}\newcommand\invt{^{-t}} \newcommand\bbP{\mathbb{P}} \newcommand\bbR{\mathbb{R}} \newcommand\defined{ \mathrel{\lower 5pt \hbox{{\equiv\atop\mathrm{\scriptstyle D}}}}}$ 18.1 : Partial derivatives
  18.2 : Poisson or Laplace Equation
  18.3 : Heat Equation
  18.4 : Steady state
  Back to Table of Contents
  
  # 18 Partial Differential Equations
  
  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.
  
  # 18.1 Partial derivatives
  
  Top > 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: $$u_x(x,y) = \lim_{h\rightarrow0}\frac{u(x+h,y)-u(x,y)}h,\quad u_y(x,y) = \lim_{h\rightarrow0}\frac{u(x,y+h)-u(x,y)}h$$
  
  # 18.2 Poisson or Laplace Equation
  
  Top > 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 $$\frac{\delta Q}{\delta t}=c\Delta x\frac{\delta u}{\delta t}=0$$ 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 $$0= \left.\frac{\delta u}{\delta x}\right|_{x+\Delta x}- \left.\frac{\delta u}{\delta x}\right|_{x}$$ 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 .
  
  # 18.3 Heat Equation
  
  Top > 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 $$\frac{\delta Q}{\delta t}=c\Delta x\frac{\delta u}{\delta t}$$ 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 $$\frac{\delta Q}{\delta t}= \left.\frac{\delta u}{\delta x}\right|_{x+\Delta x}- \left.\frac{\delta u}{\delta x}\right|_{x}$$ In the limit of $\Delta x\downarrow0$ this gives $u_t=\alpha u_{xx}$.
  
  # 18.4 Steady state
  
  Top > 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: $$\lim_{t\rightarrow\infty} u(x,t)=u_{\mathrm{steady state}}(x).$$ This solution satisfies a BVP , which can be found by setting $u_t\equiv0$. For instance, for the heat equation $$u_t=u_{xx}+q(x)$$ the steady state solution satisfies $-u_{xx}=q(x)$.
  
  Back to Table of Contents
metadata: {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 7, "x-ck12": 0, "texerror": 0}