Chapter 2
Differential equations

Context: Most of the phenomena we encounter in science and engineering are well described by differential equations. A common example is the discrete network model used in electrical engineering. This mathematical formulation is a linear system of ordinary differential equations, with time as the single independent variable. Another example is the diffusion or heat-transport process. Diffusive transport is best described using a partial differential equations, which have more than one independent variable. In this chapter, we introduce different dimensions of classification, beyond the classification into ordinary and partial differential equations.

2.1 Ordinary differential equations

We begin by recalling the classification of ordinary differential equations (ODEs) and identifying the different types. For any given differential equation, the primary objective is to find the function \(x(t)\) that satisfies a specific initial or boundary value, such as \(x(t=0)=x_0\). This initial value is an integral part of the equation’s definition.

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2.1.1 Linearity

A linear differential equation is, for example \begin {equation} m\ddot {x}(t)+c\dot {x}(t)+kx=f(t) \label {eq:linear} \end {equation} which describes the damped and driven harmonic oscillator, while \begin {equation} \frac {\dif ^2x}{\dif t^2}+\mu (x^2-1)\frac {\dif x}{\dif t}+x= 0 \label {eq:nonlinear} \end {equation} is a non-linear equation of motion for \(x\). It describes the so-called van der Pol oscillator. The non-linearity can be recognized here by the fact that \(x^2\) multiplies the derivative \(\dif x/\dif t\).

Note: The first or higher order derivative is a linear operation, since \begin {equation} \frac {\dif ^n}{\dif x^n} \lambda f(x) = \lambda \frac {\dif ^n}{\dif x^n} f(x) \end {equation} for a constant \(\lambda \) and \begin {equation} \frac {\dif ^n}{\dif x^n} \left [f(x) + g(x)\right ] = \frac {\dif ^n}{\dif x^n} f(x) + \frac {\dif ^n}{\dif x^n} g(x). \end {equation} Time derivatives are shown by a dot, \begin {equation} \dot {x}(t)=\frac {\dif }{\dif t} x(t). \end {equation} For functions of a variable, the derivative is often displayed with a dash, \begin {equation} f'(x)=\frac {\dif }{\dif x} f(x). \end {equation} This is no longer possible for functions of several variables. We will therefore always explicitly use the differential operator here.

2.1.2 Order

The order of a differential equation is given by the highest derivative that appears in the equation. Eq. \eqref{eq:linear} and Eq. \eqref{eq:nonlinear} are examples of second-order differential equations.

2.1.3 Systems

A system of first-order differential equations is formed, for example, by the equations \begin {align} \frac {\dif x}{\dif t} =& x(m - n y), \label {eq:sys1} \\ \frac {\dif y}{\dif t} =& - y(\gamma - \delta x), \label {eq:sys2} \end {align}

the well-known predator-prey or Lotka-Volterra equations. Equations \eqref{eq:sys1} and \eqref{eq:sys2} are still non-linear.

Differential equations of higher order can always be rewritten into a system of 1st order equations. In the example of the damped harmonic oscillator, \begin {equation} m\ddot {x}(t)+c\dot {x}(t)+kx=f(t), \end {equation} we replace \(\dot {x} = y\) and thus obtain two first-order equations instead of the original second-order equation, namely \begin {align} \dot {x} =& y \\ m\dot {y} =& -cy-kx+f(t). \end {align}

2.2 Partial differential equations

Partial differential equations (PDEs) involve derivatives of more than one independent variable. For example, consider a time-dependent heat transport problem in one dimension, represented by the diffusion equation for the local temperature of the system. In this case, the temperature is a function of two independent variables: time \(t\) and spatial position \(x\), denoted as \(T(x, t)\). The time evolution of the temperature is governed by the following equation: \begin {equation} \frac {\partial T(x,t)}{\partial t}=\kappa \frac {\partial ^2 T(x,t)}{\partial x^2}, \label {eq:heateq} \end {equation} where \(\kappa \) denotes the heat conduction coefficient. This equation was developed by Joseph Fourier (*1768, \(\dagger \)1830), whom we will encounter again during this course.

Note: In Eq. \eqref{eq:heateq}, \(\partial /\partial t\) denotes the partial derivative. This is the derivative with respect to one of the arguments (here \(t\)), i.e. the variation of the function if all other arguments are kept constant. ODEs, in contrast to PDEs, are characterized by derivatives with respect to just one variable (usually the time \(t\)), which are then denoted by the differential operator \(\dif /\dif t\).

2.2.1 First order

Quasilinear PDEs of the first order, i.e. equations of the form \begin {equation} P(x,t;u)\frac {\partial u(x,t)}{\partial x}+ Q(x,t;u)\frac {\partial u(x,t)}{\partial t}= R(x,t;u), \label {eq:PDE1Oquasi} \end {equation} for an (unknown) function \(u(x,t)\) and the initial condition \(u(x,t=0)=u_0(x)\) can be systematically traced back to a system of coupled first-order ODEs. We want to investigate this important property.

Note: In Eq. \eqref{eq:PDE1Oquasi}, a representation with two variables \(x\) and \(t\) was chosen for illustration. In general, we can write \begin {equation} \sum \limits _i P_i(\{x_i\};u)\frac {\partial u(\{x_i\})}{\partial x_i}=R(\{x_i\};u) \end {equation} The notation used here is \(u(\{x_i\})=u(x_0, x_1, x_2, \ldots )\), i.e. the curly brackets denote all degrees of freedom \(x_i\).

Equation \eqref{eq:PDE1Oquasi} can be transformed into a system of ODEs. This is called the method of characteristics. We can then apply the formalisms (analytical or numerical) for solving systems of ODEs that we learned about in the lecture “Differential Equations”. The method of characteristics works as follows:

  1. First, we parameterize the independent variables in Eq. \eqref{eq:PDE1Oquasi} with a parameter \(s\) according to \(x(s)\) and \(t(s)\).
  2. We then form the total derivative of \(u(x(s),t(s))\) to \(s\) \begin {equation} \frac {\dif u(x(s),t(s))}{\dif s}= \frac {\partial u(x(s),t(s))}{\partial x}\frac {\dif x(s)}{\dif s}+ \frac {\partial u(x(s),t(s))}{\partial t}\frac {\dif t(s)}{\dif s}. \label {eq:totalderiv} \end {equation}
  3. By comparing the coefficients of the total derivative \eqref{eq:totalderiv} with the PDE \eqref{eq:PDE1Oquasi}, you can see that this DGL is solved exactly when \begin {align} \frac {dx(s)}{ds}&=P(x,t,u),\label {eq:transode1}\\\ \frac {dt(s)}{ds}&=Q(x,t,u)\quad \text {und}\\ \frac {du(s)}{ds} &= R(u(s)).\label {eq:transode3} \end {align}

    is fulfilled. This describes the solution along certain curves in the \((x,t)\)-plane.

We have thus converted the PDE into a set of coupled first-order ODEs, Eq. \eqref{eq:transex1}-\eqref{eq:transex3}.

Example: We want to solve the transport equation \begin {equation} \frac {\partial u(x,t)}{\partial t}+c\frac {\partial u(x,t)}{\partial x}=0 \label {eq:transportexample} \end {equation} with the initial condition \(u(x,t=0)=u_0(x)\). We proceed according to the recipe above:

  1. We parameterize the variables \(x\) and \(t\) with the help of a new variable \(s\), i.e. \(x(s)\) and \(t(s)\). We are now looking for an expression from which we can determine \(x(s)\) and \(t(s)\).
  2. We first ask how the function \(u(x(s),t(s))\) behaves. This function describes the change in an initial value \(u(x(0),t(0))\) with the variable \(s\). The total derivative becomes \begin {equation} \frac {\dif u(x(s),t(s))}{\dif s}=\frac {\partial u}{\partial t}\frac {\dif t(s)}{\dif s}+\frac {\partial u}{\partial x}\frac {\dif x(s)}{\dif s}. \end {equation}
  3. The total derivative is identical to the partial differential equation that we want to solve, if \begin {align} \frac {\dif x(s)}{\dif s} &=c\quad \text {and} \label {eq:transex1}\\ \frac {\dif t(s)}{\dif s} &=1. \end {align}

    In this case, the following applies \begin {equation} \frac {\dif u(s)}{\dif s} = 0.\label {eq:transex3} \end {equation}

  4. The general solution for the three coupled ordinary differential equations \eqref{eq:transex1}-\eqref{eq:transex3} is given by \begin {align} x(s) &= cs + \text {const.},\\ t(s) &= s + \text {const.}\quad \text {and}\\ u(s) &= \text {const.} \end {align}
  5. With the initial conditions \(t(0)=0\), \(x(0)=\xi \) and \(u(x,t=0)=f(\xi )\) you get \(t=s\), \(x=ct+\xi \) and \(u=f(\xi )=f(x-ct)\),

The initial condition \(f(\xi )\) is transported with the speed \(c\) in the positive x-direction. The solution for \(u\) remains constant, as the derivative of \(u\) is zero, so \(u\) retains the value given by the initial condition. The field \(u(x,0)\) is therefore shifted at a constant speed \(c\): \(u(x,t)=u(x-ct,0)\).

2.2.2 Second order

Examples of second-order PDEs are the...

  • ...wave equation: \begin {equation} \frac {\partial ^2 u}{\partial t^2}-\frac {\partial ^2 u}{\partial x^2}=0 \end {equation}
  • ...diffusion equation (which we will look at in more detail in these notes): \begin {equation} \frac {\partial u}{\partial t}-\frac {\partial ^2 u}{\partial x^2}=0 \end {equation}
  • ...Laplace equation (which we will also get to know better): \begin {equation} \frac {\partial ^2 u}{\partial x^2}+\frac {\partial ^2 u}{\partial y^2}=0 \end {equation}

The term “second order” here refers to the second derivative. These examples are formulated for two variables, but these differential equations can also be written down for more degrees of freedom.

For two variables, the general form of second-order linear PDEs is \begin {equation} a(x,y) \frac {\partial ^2 u}{\partial x^2}+ b(x,y)\frac {\partial ^2 u}{\partial x\partial y}+ c(x,y)\frac {\partial ^2 u}{\partial y^2}=F\left (x,y;u,\frac {\partial u}{\partial x},\frac {\partial u}{\partial y}\right ), \end {equation} where \(F\) itself must of course also be linear in the arguments if the entire equation is to be linear. We now classify 2nd order PDEs, but note that this classification is not exhaustive and that it only applies pointwise. The latter means that the PDE can fall into a different categories at different points in space.

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We first assume that \(F=0\) and \(a\), \(b\), \(c\) are constant. Then we get: \begin {equation} a\frac {\partial ^2 u}{\partial x^2}+b\frac {\partial ^2 u}{\partial x\partial y}+ c\frac {\partial ^2 u}{\partial y^2}=0. \label {eq:n2ndoconst} \end {equation} We rewrite this equation as the quadratic form \begin {equation} \begin {pmatrix} \partial /\partial x \\ \partial /\partial y \end {pmatrix} \cdot \begin {pmatrix} a & b/2 \\ b/2 & c \end {pmatrix} \cdot \begin {pmatrix} \partial /\partial x \\ \partial /\partial y \end {pmatrix} u = \nabla \cdot \t {C} \cdot \nabla u =0. \label {eq:quadform} \end {equation} We now diagonalize the coefficient matrix \(\t {C}\). This yields \begin {equation} \t {C} = \t {U} \cdot \begin {pmatrix} \lambda _1 & 0 \\ 0 & \lambda _2 \end {pmatrix}\cdot \t {U}^T, \label {eq:diagquadform} \end {equation} where \(\t {U}\) is unitary (\(\t {U}^T \cdot \t {U}=\t {1}\)) due to the symmetry of \(\t {C}\). The geometric interpretation of the operation \(\t {U}\) is a rotation. We now introduce transformed coordinates \(x'\) and \(y'\) so that \begin {equation} \nabla = \t {U} \cdot \nabla ' \end {equation} with \(\nabla '=(\partial /\partial x', \partial /\partial y')\). In other words, the transformation matrix is given as \begin {equation} \t {U} = \begin {pmatrix} \partial x'/\partial x & \partial y'/\partial x \\ \partial x'/\partial y & \partial y'/\partial y \end {pmatrix}. \end {equation} Equation \eqref{eq:n2ndoconst} becomes \begin {equation} \lambda _1 \frac {\partial ^2 u}{\partial x'^2} + \lambda _2 \frac {\partial ^2 u}{\partial y'^2} = 0. \label {eq:diag2nd} \end {equation} We have diagonalized the coefficients of the differential equation. For any function \(f(z)\) that is twice differentiable, \begin {equation} u(x', y') = f\left (\sqrt {\lambda _2} x' + i\sqrt {\lambda _1} y'\right ) \end {equation} is a solution of Eq. \eqref{eq:diag2nd}.

The analytical expression for the eigenvalues is: \begin {equation} \lambda _{1/2}=\frac {-b\pm \sqrt {b^2-4ac}}{2a} \label {eqnsol4m} \end {equation} We now distinguish three cases:

  • The case \(\det \t {C}=\lambda _1\lambda _2=ac-b^4/4=0\) with \(b\ne 0\) and \(a\ne 0\) leads to a parabolic PDE. This PDE is called parabolic because the quadratic form Eq. \eqref{eq:quadform} or \eqref{eq:diagquadform} describes a parabola. (This is of course an analogy. You have to replace the differential operators with coordinates for this to work). Without restriction of generality, let \(\lambda _2=0\). Then we get \begin {equation} \frac {\partial ^2 u}{\partial x'^2}=0. \label {eqnparab} \end {equation} This is the canonical form of a parabolic PDE.
  • The case \(\det \t {C}=\lambda _1 \lambda _2=ac-b^2/4>0\) leads to an elliptic PDE. This PDE is called elliptic because the quadratic form Eq. \eqref{eq:quadform} or \eqref{eq:diagquadform} describes an ellipse for a constant right-hand side. (For \(\lambda _1=\lambda _2\) it is a circle). We now convert the equation for the elliptical case to a standardized form and introduce the scaled coordinates \(x'=\sqrt {\lambda _1} x''\) and \(y'=\sqrt {\lambda _2} y''\). Eq. \eqref{eq:diag2nd} then becomes the canonical elliptic PDE \begin {equation} \frac {\partial ^2 u}{\partial x''^2}+\frac {\partial ^2 u}{\partial y''^2}=0. \label {eqnelliptic} \end {equation} The canonical elliptic PDE is therefore the Laplace equation, Eq. \eqref{eqnelliptic} (here in two dimensions). Solutions of the Laplace equation are called harmonic functions.
  • The case \(\det \t {C}=\lambda _1\lambda _2=ac-b^2/4<0\) results in the so-called hyperbolic PDE. This PDE is called hyperbolic because the quadratic form Eq. \eqref{eq:quadform} or \eqref{eq:diagquadform} for a constant right-hand side describes a hyperbola. Without restricting the generality, we now require \(\lambda _1>0\) and \(\lambda _2<0\). Then we can again introduce scaled coordinates \(x'=\sqrt {\lambda _1}x''\) and \(y'=\sqrt {-\lambda _2}y''\) so that \begin {equation} \frac {\partial ^2 u}{\partial x''^2} - \frac {\partial ^2 u}{\partial y''^2} = \begin {pmatrix} \partial u/\partial x'' \\ \partial u/\partial y'' \end {pmatrix} \cdot \begin {pmatrix} 1 & 0 \\ 0 & -1 \end {pmatrix} \cdot \begin {pmatrix} \partial u/\partial x'' \\ \partial u/\partial y'' \end {pmatrix} = 0. \label {eq:hyb} \end {equation} We can now use a further coordinate transformation, namely a rotation by \(45^\circ \), to bring the coefficient matrix in Eq. \eqref{eq:hyb} to a form in which the diagonal elements are \(0\) and the secondary diagonal elements are \(1\). This results in the differential equation \begin {equation} \frac {\partial ^2 u}{\partial x''' \partial y'''}=0, \label {eqnd2udxideta0} \end {equation} where \(x''''\) and \(y''''\) are the corresponding rotated coordinates. This equation is the canonical form of a hyperbolic PDE and is equivalent to Eq. \eqref{eq:n2ndoconst} in the new variables \(x'''\) and \(y'''\).

For higher dimensional problems, we need to look at the eigenvalues of the coefficient matrix \(\t {C}\). The PDE is called parabolic if there is an eigenvalue that vanishes, but all other eigenvalues are either greater or less than zero. The PDE is called elliptic if all eigenvalues are either greater than zero or less than zero. The PDE is called hyperbolic if there is exactly one negative eigenvalue and all others are positive or if there is exactly one positive eigenvalue and all others are negative. It is clear that for PDEs with more than two variables, these three classes of PDEs are not exhaustive and there are coefficient matrices that fall outside this classification scheme. For problems with exactly two variables, this classification leads to the conditions on the determinants of the coefficient matrix mentioned above.

Example: These three types of 2nd-order linear PDEs can also be solved analytically for some problems. As an example, we solve the one-dimensional wave equation, \begin {equation} \frac {\partial ^2 u}{\partial x^2}-\frac {1}{c^2}\frac {\partial ^2 u}{\partial t^2}=0, \label {eqn1Dwaveeqn} \end {equation} by separating the variables. We make the ansatz \(u(x,t)=X(x)T(t)\), which leads to \begin {equation} \frac {1}{X}\frac {\partial ^2 X}{\partial x^2}=\frac {1}{c^2}\frac {1}{T}\frac {\partial ^2 T}{\partial t^2}. \label {eqnseparate} \end {equation} In Eq. \eqref{eqnseparate}, the left-hand side depends only on the variable \(x\), while the right-hand side depends only on \(t\). This means that for any \(x\) and \(t\), this equation can only be fulfilled if both sides are equal to a constant and we thus obtain \begin {equation} \frac {1}{X}\frac {\partial ^2 X}{\partial x^2}=-k^2=\frac {1}{c^2}\frac {1}{T}\frac {\partial ^2 T}{\partial t^2}, \end {equation} where \(k\) is our constant. This results in the following two equations \begin {equation} \frac {\partial ^2 X}{\partial x^2}+k^2X=0 \end {equation} with solution \(X(x)=e^{\pm ikx}\) and \begin {equation} \frac {\partial ^2 T}{\partial t^2}+\omega ^2T=0 \end {equation} with solution \(T(t)=e^{\pm i\omega t}\), where we have set \(\omega ^2=c^2k^2\).

Bibliography


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