$$\frac{\partial z}{\partial x}+2\frac{\partial z}{\partial y}+3z=0$$

Find the characteristics associated with this PDE and find an explicit solution $z(x,y)$ which satisfies the initial condition $z=e^{x}$ on the line segment $\{(x,y):0 \leq x \leq 1, y=0 \}$.

State the region on the $(x,y)$-plane in which $z(x,y)$ is uniquely determined by the initial condition.


To find the characteristics:

$$\frac{dx}{ds}=1, \quad \frac{dy}{ds}= 2, \quad \frac{dz}{ds}=-3z$$ which yields that $$x=s+A, \quad y=2s+B, \quad z=Ce^{-3s}$$

Are we now OK to parametrise the line segment as $L=(t,0,e^t)$ using our initial condition, giving rise to explicit parametric solutions

$$\begin{cases} x=s+t \\ y=2s \\ z=e^{t-3s} \end{cases}$$ for $0 \leq t \leq 1$.

I'm not convinced that having $z=e^x$ allows us to parametrise the line segment in the way. Also, to establish where the solution is uniquely determined:

At $t=0$,

$$x=s, \quad y=2s, \quad \text{so } y=2x$$ and at $t=1$,

$$ x=s+1, \quad, y=2s, \quad \text{so } y=2x-2$$

Not sure how much guidance this offers.

Any help with the various parts of this would be much appreciated. Not sure what's right/wrong really. Regards as always, MM.


The solution using the method of characteristic is a way to reduce a partial differential equation to a set of ordinary differential equations. Originally, you wanted to find a function $z(x,y)$ which satisfies $$\frac{\partial z}{\partial x}+2\frac{\partial z}{\partial y}+3z=0\qquad\qquad(1)$$ as well as the boundary condition.

We make the ansatz, that $x=x(s)$ and $y=y(s)$ where $s$ is a parameter and $(x,y)$ is a curve in $\mathbb{R}^2$. Furthermore, let $z(s)=z(x(s),y(s))$. Then $$z'(s) = \frac{\partial z}{\partial x} x'(s) + \frac{\partial z}{\partial y} x'(s).$$ Note that this equation is equivalent to (1) if $$x'(s)=1, \qquad y'(s)=2, \qquad z'(s)=-3z.$$ (this is the same set of equations as you have in your post)

The solutions to the equations for $x(s)$ and $y(s)$ give us a family of curves $$ x(s)= x_0 +s \qquad y(s) = 2s,$$ parameterized by $x_0$ (note that the additive constant in $y$ can be absorbed in a reparameterization of the curve). These curves go through the point $(x_0,0)$ at $s=0$ for which we have the boundary value $z= e^{x_0}$ specified.

Along each of the curves (characteristics), the partial differential equation (1) becomes an ordinary differential equation $$z'(s)=z$$ as noted above. The solution is given by $z(s)=Ce^{-3s}$. The boundary value read $z(0)=C =e^{x_0}$. Thus we have $$z(s)= e^{x_0-3s}$$ as a solution along the characteristic.

It you want to obtain the (original) function $z(x,y)$, you have to find $x_0(x,y)$ and $s(x,y)$. In this case this is rather trivial and we obtain $x_0=x-y/2$ and $s=y/2$ such that $$ z(x,y) = e^{x-2y}.$$

  • $\begingroup$ Many thanks, that's a great explanation. How can I determine the region of the $(x,y)$-plane in which $z(x,y)$ is uniquely determined? $\endgroup$ – Mathmo Jan 13 '12 at 10:39
  • $\begingroup$ @MiamiMaths: All $\mathbb{R}^2$ (as both $x(s)$ and $y(s)$ are invertible. $\endgroup$ – Fabian Jan 13 '12 at 16:33

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