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Integrate the canonical form of a second-order partial differential equation first in $\eta$ and then in $\xi$. The first integration gives $U_{\xi }=\eta +g^{\prime }(\xi )$, and the second givesThusThe two arbitrary twice continuously differentiable functions describe the two characteristic families. Conversely, substituting this expression into $U_{\xi \eta }=1$ verifies every such solution. The formula applies on $x>0$ or $x<0$; independent choices can be made on the two half-planes. Smooth extension across the degenerate line $x=0$ requires extra compatibility and is not supplied by the logarithmic formula alone.

The characteristic curves satisfyIntegrating suggests the characteristic coordinatesTheir Jacobian determinant is $2/x$, so this is an invertible local change of variables whenever $x\ne 0$. Write $u(x,y)=U(\xi ,\eta )$. The chain rule givesSubstitution cancels both unmixed second derivatives and both first derivatives:The canonical form of a second-order partial differential equation is thereforeThe cancellation of the first derivatives is the key feature of these logarithmic characteristic coordinates.

Write the principal part as $e^{2y}{u}_{xx}-u_{yy}$. The characteristic curves satisfyor $dx/dy=\pm e^y$. Integration gives $x+e^y=\text{constant}$ and $x-e^y=\text{constant}$. Thus take $\xi =x+e^y$, $\eta =x-e^y$. Their Jacobian determinant is $-2e^y\ne 0$, and their image is $\xi >\eta$.

For $u(x,y)=U(\xi ,\eta )$, the chain rule givesThe first-derivative terms cancel exactly. The canonical form of a second-order partial differential equation isOn $y=0$, $F(x+1)+G(x-1)=x$. Differentiating this condition in $x$ and using $u_y=1$ givesHence $F^{\prime }=1$ and $G^{\prime }=0$. Substitution into the value condition fixes their combined constant and yieldsDirect substitution verifies the PDE and both initial data. The cancellation illustrates a hyperbolic equation reduced by an exponential characteristic coordinate.

For a second-order linear partial differential equation with principal part $A{u}_{xx}+2B{u}_{xy}+C{u}_{yy}$, the discriminant is $B^2-AC$. Here $A=1$, $B=0$, $C=-x^2$, soThe equation is therefore a hyperbolic partial differential equation throughout its domain. The lower-order term $-u_x/x$ does not affect this classification.

The characteristic curves satisfyIntegrating gives the two families $y\mp x^2/2=\text{constant}$. Choose coordinates constant along these families:Their Jacobian determinant is $8x\ne 0$, so they define an invertible change of variables on $x>0$, with image $\xi +\eta >0$. Write $u(x,y)=U(\xi ,\eta )$. The chain rule givesBoth the pure second partial derivatives and the first partial derivatives cancel in the original equation, leavingThis is the full canonical form of a second-order partial differential equation, including the effect of the lower-order term. Integrating shows $U_{\xi }$ is a function of $\xi$ alone, hencewith arbitrary twice differentiable functions $f,g$. The representation is global on $x>0$: for each fixed $\xi$, the allowed $\eta >-\xi$ form an interval, so $U_{\xi }$ is independent of $\eta$ throughout that interval; after subtracting its antiderivative, the same reasoning in the other coordinate leaves only $g(\eta )$. These are quadratic characteristic coordinates for a variable-speed wave equation.

For the data on $x=1$, set $s=1-2y$, so the other argument is $2-s$. The two initial conditions becomeDifferentiating the first equation with respect to $s$ also gives $f^{\prime }(s)-g^{\prime }(2-s)=0$. Solving these two linear equations yields $f^{\prime }(s)=g^{\prime }(2-s)=1/2$. Since $s$ ranges over all real numbers,ConsequentlyIndeed $u_{xx}=2$, $u_{yy}=0$, $u_x=2x$, so the equation and both initial conditions hold directly.

On $x>0$, the linear partial differential equation $u_{xx}-x^2{u}_{yy}-u_x/x=0$ has characteristic curves $y\pm x^2/2=\text{constant}$. The change of variables $\xi =x^2-2y$, $\eta =x^2+2y$ has Jacobian determinant $8x\ne 0$. The chain rule givesThus its canonical form of a second-order partial differential equation is $U_{\xi \eta }=0$, and its general twice differentiable solution is $u=f(x^2-2y)+g(x^2+2y)$. The first derivative term exactly cancels the derivatives of the coordinate coefficients.

More generally, for positive differentiable $a(x)$, choose $s(x)$ with $s^{\prime }(x)=a(x)$. Then the same cancellation givesThe characteristic coordinates are $s\pm y$. The quadratic example is $a(x)=x$, $s=x^2/2$, with both coordinates multiplied by two.