Example 19.1. For given \(a\in \mR \) and \(u_0:\mR \to \mR \), we consider the transport equation

\[ a\partial _xu+\partial _yu=0,\quad u(x,0)=u_0(x). \]

The corresponding characteristic equations are

\[ \dot {x}=a,\qquad \dot {y}=1,\qquad \dot {z}=0. \]

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

\[ x(r,0)=r,\qquad y(r,0)=0,\qquad z(r,0)=u_0(r). \]

The general solutions of the characteristic equations are

\[ x(t)=at+c_1,\qquad y(t)=t+c_2,\qquad z(t)=c_3, \]

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

\[ x(r,t)=at+r,\qquad y(r,t)=t,\qquad z(r,t)=u_0(r). \]

We can solve for \(r\) and \(t\) in terms of \(x\) and \(y\) and obtain \(t=y\), \(r=x-ay\). Substituting in \(z\) we obtain

\[ u(x,y)=u_0(x-ay). \]

We can indeed verify that \(u(x,0)=u_0(x)\) and that since \(\partial _x u=u_0'(x-ay)\) and \(\partial _y u=-au_0'(x-ay)\) we have \(a\partial _x u+\partial _y u=au_0'(x-ay)-au_0'(x-ay)=0\).

Example 19.2. For given \(u_0:\mR \to \mR \), we consider

\[ a\partial _xu+\partial _yu=u^2,\quad u(x,0)=u_0(x). \]

The corresponding characteristic equations are

\[ \dot {x}=a,\qquad \dot {y}=1,\qquad \dot {z}=z^2, \]

with initial conditions

\[ x(r,0)=r,\qquad y(r,0)=0,\qquad z(r,0)=u_0(r). \]

The general solutions of the characteristic equations are

\[ x(t)=at+c_1,\qquad y(t)=t+c_2,\qquad z(t)=\frac {-1}{t+c_3}, \]

for constants (in \(t\)) \(c_1\), \(c_2\) and \(c_3\). Determining the constants from the initial conditions gives

\[ x(r,t)=at+r,\qquad y(r,t)=t,\qquad z(r,t)=\frac {-1}{t-\frac {1}{u_0(r)}}=\frac {u_0(r)}{1-tu_0(r)}. \]

We can solve for \(r\) and \(t\) in terms of \(x\) and \(y\) and obtain \(t=y\) and \(r=x-ay\). Substituting in \(z\) we obtain

\[ u(x,y)=\frac {u_0(x-ay)}{1-yu_0(x-ay)}. \]

Note that the solution will blow up when \(1-yu_0(x-ay)=0\). We note that this blow-up is due to the blow-up in the ODE for \(z\).

Example 19.3. We consider Burgers’ equation

\[ u\partial _xu+\partial _yu=0,\quad u(x,0)=-x. \]

We first more generally consider \(u(x,0)=u_0(x)\). The corresponding characteristic equations are

\[ \dot {x}=z,\qquad \dot {y}=1,\qquad \dot {z}=0. \]

with initial conditions

\[ x(r,0)=r,\qquad y(r,0)=0,\qquad z(r,0)=u_0(r). \]

The general solutions of the characteristic equations are

\[ x(t)=c_1t+c_2,\qquad y(t)=t+c_3,\qquad z(t)=c_1, \]

for constants (in \(t\)) \(c_1\), \(c_2\) and \(c_3\). Determining the constants from the initial conditions gives

\[ x(r,t)=u_0(r)t+r,\qquad y(r,t)=t,\qquad z(r,t)=u_0(r). \]

When \(u_0(r)=-r\), we obtain

\[ x(r,t)=-rt+r,\qquad y(r,t)=t,\qquad z(r,t)=-r. \]

This we can solve for \(r\) and \(t\) in terms of \(x\) and \(y\) and obtain \(t=y\) and \(r=\frac {x}{1-y}\). Substituting in \(z\) we obtain

\[ u(x,y)=\frac {x}{y-1}. \]

The Jacobian determinant equals

\[ \dot {x}\partial _ry-\dot {y}\partial _rx=t-1. \]

This equals zero if \(t=1\), which corresponds to \((x,y)=(0,1)\).

We note that as \(y\uparrow 1\), the solution blows up. We note that in contrast to Example 19.2, in Burgers’ equation, there is no blow-up in the ODEs. The blow-up in the PDE is instead due to the coordinate transformation \((r,t)\mapsto (x,y)\) becoming non-invertible.