Chapter 18 The method of characteristics: first order PDEs I
A first order partial differential equation is an equation of the form
\[ F(x,y,u,\partial _xu,\partial _yu)=0, \]
where \(F:\mR ^5\to \mR \) is given and \(u:\mR ^2\to \mR \) is to be found. This equation is called quasi-linear if
\[ F(x,y,u,\partial _xu,\partial _yu)=a(x,y,u)\partial _xu+b(x,y,u)\partial _yu-c(x,y,u). \]
To the quasi-linear PDE
\(\seteqnumber{0}{18.}{0}\)\begin{equation} \label {eq:quasi} a(x,y,u)\partial _xu+b(x,y,u)\partial _yu=c(x,y,u), \end{equation}
we associate the following system of ODEs
\[ \dot {x}(t)=a(x(t),y(t),z(t)),\qquad \dot {y}(t)=b(x(t),y(t),z(t)),\qquad \dot {z}(t)=c(x(t),y(t),z(t)). \]
These are called the characteristic equations of (18.1) and the parametric curves \(t\mapsto (x(t),y(t),z(t))\) in \(\mR ^3\) obtained from them are called characteristic curves.
We assume that we have an initial-boundary condition for our PDE in the sense that a curve \(\Gamma \subset \mR ^2\) (called the data curve) and \(u|_{\Gamma }\) are given. Assume that our data curve is given by a parametrization:
\[ \gamma (r)=(\gamma _1(r),\gamma _2(r)), \]
so that the initial-boundary condition is given by
\[ u(\gamma (r))=u_0(r), \]
for a given function \(u_0:\mR \to \mR \). We then add the following initial conditions to the characteristic equations:
\[ x(0)=\gamma _1(r),\qquad y(0)=\gamma _2(r),\qquad z(0)=u_0(r). \]
Note that \(x\), \(y\) and \(z\) then depend on both \(t\) and \(r\). So we really have (here the overdot represents differentiation with respect to \(t\))
\(\seteqnumber{0}{18.}{1}\)\begin{gather*} \dot {x}(r,t)=a(x(r,t),y(r,t),z(r,t)),\qquad x(r,0)=\gamma _1(r), \\ \dot {y}(r,t)=b(x(r,t),y(r,t),z(r,t)),\qquad y(r,0)=\gamma _2(r), \\ \dot {z}(r,t)=c(x(r,t),y(r,t),z(r,t)),\qquad z(r,0)=u_0(r). \end{gather*} Once we have solved these equations, we can invert the map \((r,t)\mapsto (x,y)\) to obtain \(r\) and \(t\) as functions of \(x\) and \(y\). We can then substitute this in the equation for \(z\) to obtain \(z\) as a function of \(x\) and \(y\). This is the solution \(u\) of the PDE with the given initial-boundary condition
\[ a(x,y,u)\partial _xu+b(x,y,u)\partial _yu=c(x,y,u),\qquad u(\gamma _1(r),\gamma _2(r))=u_0(r). \]
This solution method is called the method of characteristics.
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Remark 18.1. That \(u(x,y)=z(r,t)\) is a solution of
\[ a(x,y,u)\partial _xu+b(x,y,u)\partial _yu=c(x,y,u), \]
can be seen by using the chain rule: we have
\[ \dot {z}=\frac {d}{dt}u(x,y)=\dot {x}\partial _x u+\dot {y}\partial _yu, \]
which by the characteristic equations equals
\[ c=a\partial _xu+b\partial _yu, \]
as desired.
A crucial assumption is that we can invert the map \((r,t)\mapsto (x,y)\). A condition for being able to do this is given by the Inverse Function Theorem (see MA40254 Differential and Geometric Analysis) and is the invertibility of the Jacobian matrix
\[ \bbm {\dot {x}&\partial _r x\\\dot {y}&\partial _r y}. \]
This invertibility condition is equivalent to the determinant being nonzero:
\[ \dot {x}\partial _r y-\dot {y}\partial _r x\neq 0. \]
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\[ y\partial _xu-x\partial _yu=0,\quad u(x,0)=x^2. \]
The corresponding characteristic equations are
\[ \dot {x}=y,\qquad \dot {y}=-x,\qquad \dot {z}=0. \]
The data curve \(\mR \times \{0\}\) is parametrized by \(\gamma (r)=(r,0)\) with \(r\in \mR \), so that the initial condition gives rise to
\[ x(r,0)=r,\qquad y(r,0)=0,\qquad z(r,0)=r^2. \]
The general solutions of the characteristic equations are
\[ x(t)=c_1\sin (t)+c_2\cos (t),\qquad y(t)=c_1\cos (t)-c_2\sin (t),\qquad z(t)=c_3, \]
for constants (in \(t\)) \(c_1\), \(c_2\) and \(c_3\); we emphasize that these “constants” may depend on \(r\). Determining the constants from the initial conditions gives
\[ x(r,t)=r\cos (t),\qquad y(r,t)=-r\sin (t),\qquad z(r,t)=r^2. \]
We can solve for \(r\) in terms of \(x\) and \(y\) and obtain \(r^2=x^2+y^2\). Substituting in \(z\) we obtain
\[ u(x,y)=x^2+y^2. \]
We can directly verify that this satisfies \(u(x,0)=x^2\) and since \(\partial _xu=2x\) and \(\partial _yu=2y\), we have \(y\partial _xu=2xy=x\partial _yu\).
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Example 18.3. Consider the same PDE as in Example 18.2 but with now initial condition
\[ u(x,0)=x. \]
The only change in the characteristic equations and initial conditions is that now
\[ z(r,0)=r. \]
We then obtain the same \(x\) and \(y\) and
\[ z(r,t)=r. \]
We then obtain
\[ u(x,y)=\pm \sqrt {x^2+y^2}. \]
This gives \(u(x,0)=\pm \sqrt {x^2}=\pm |x|\). We therefore obtain the solution
\[ u(x,y)= \begin {cases} \sqrt {x^2+y^2}& x>0\\ -\sqrt {x^2+y^2}& x<0. \end {cases} \]
Note that the Jacobian determinant equals
\[ r\sin ^2(t)+r\cos ^2(t)=r. \]
Therefore there is a problem with the invertibility of the map \((r,t)\mapsto (x,y)\) when \(r=0\), i.e. when \((x,y)=(0,0)\). Note that this point lies on the data curve and this is what causes an issue. In Example 18.2 this issue didn’t arise because of the special nature of the initial-boundary condition.