# Solving a PDE using Characteristics

Let

$$\displaystyle (x,y)\cdot Du = au,\ a>0$$.

I computed the characteristics to be

\begin{align} \dot{\vec{x}}(s) &= \vec{x}(s)\\ \dot{z}(s)&=\vec{x}(s)\cdot\vec{p}(s)=az(s) \end{align}

where I used the notation from Evans - PDE, that is

$$\vec x(s)=(x_1(s),x_2(s))$$ being the initial curve and $$z(s):=u(\vec{x}(s))$$ with $$\vec p(s):=Du(\vec{x}(s))$$, where $$D$$ is the gradient.

With my characteristics I furthermore get,

\displaystyle \begin{align} \frac{\mathrm{d} x_1}{\mathrm{d}s}&=x_1 \Rightarrow x_1(s)=e^{s}c_1 \\ \frac{\mathrm{d} x_2}{\mathrm{d}s}&=x_2 \Rightarrow x_2(s)=e^{s}c_2\\ \frac{\mathrm{d}\, z(s)}{\mathrm{d}s}&=az(s) \Rightarrow z(s)=e^{as}c_3 \end{align}

I am given the condition that along the curve $$\{x,1\},\, x\in\mathbb{R}$$ the initial values $$\phi(x)$$ are given.

From here I am unsure of how to proceed. I somehow have to recover $$u(x,y)$$ from my characteristic equations using the initial values.

Edit: My best attempt:

I read from Evans, that I can construct a (local) solution from

\displaystyle\begin{align} x_1(r,s)&=e^sc_1(r) \\ x_2(r,s)&=e^sc_2(r) \\ z(r,s)&=e^{as}c_3(r) \end{align}

With the initial conditions I get

\displaystyle\begin{align} x_1(r,1)&=r \\ x_2(r,1)&=1 \\ z(r,1)&=\phi(r)\end{align}

yielding

\displaystyle\begin{align} x_1&=e^{s-1}r\\ x_2&=e^{s-1} \\ z&=e^{a(s-1)}\phi(r)\end{align}

$$\displaystyle u=x_2^a\phi(x_1/x_2)$$

We are almost done. We can make $$c_1=1$$, as it represents a simple displacementfor the parameter driving the characteristics, $$s$$. Then, get rid of $$s$$ because we are interested in the relation among $$x,y$$ and $$u$$ for the characteristics.

$$y/x=c_2$$ or $$x/y=c'_2$$

$$z(s)=u(x(s),y(s))=c_3x(s)^a$$ or for easy reading $$u=c_3x^a$$

But the integration constants must be related: $$c_3=f(c'_2)$$

Substituting, $$u=x^af(x/y)$$, being the general solution.

Now $$u(x,1)=\phi(x)$$

$$\phi(x)=x^af(x)\to f(x/y)=(x/y)^{-a}\phi(x/y)$$

leading to $$u(x,y)=x^a(y/x)^a\phi(x/y)$$ or

$$u(x,y)=y^a\phi(x/y)$$

• Since I have never worked with characteristics before, I have no idea what you mean by "driving the characteristics". Also, how did you obtain $u$? Nov 2, 2018 at 14:59
• I've done an edition of my post to correct a mistake (setting $c_1=0$ amounts a displacement of the parameter $s\to s+ c'_1$, not a reescaling). Now, $s$ is that parameter, the one controlling the characteristic curve the equations of which you wrote, so say, "a parameter driving the characteristics". Lastly, for the equations you wrote to have any sense it is necessary identify some variables this way: $x_1=x\,,x_2=y\,,z=u$ Nov 2, 2018 at 16:02
• I did some edition to the post. Nov 2, 2018 at 17:06
• Ok, thanks. I just read a bit further in Evans and tried something myself. But I get something different. I will make an EDIT with my attempt, however! I would appreciate if you could tell me, where my approach went wrong. Nov 2, 2018 at 18:13
• There was another mistake. Your solution and mine are the same. Nov 2, 2018 at 19:37