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I'm working on the following past paper question, and struggling with part (c):

(i) Find the characteristics for the particular equation $$ \frac{\partial z}{\partial x}+z \frac{\partial z}{\partial y}=2 $$ (ii) Find the solution of this equation in parametric form, given that $z=1$ on $x+y=0$ for $x \geqslant 0$.

(iii) Show that the projection into the $(x, y)$-plane of each characteristic is a parabola tangent to the line $$ x+y=-1 $$

This is what I've done so far:

$$ \begin{aligned} & \frac{d x}{d t}=1 \Rightarrow \int 1 d x=\int 1 d t \Rightarrow x=t+A \\ & \frac{d y}{d t}=z \Rightarrow \int 1 d y=\int 2 t+C d t \Rightarrow y=t^2+C t+B \\ & \frac{d z}{d t}=2 \Rightarrow \int 1 d z=\int 2 d t \Rightarrow z=2 t+C \end{aligned} $$ (ii) Data $(s,-s, 1) \quad s \geqslant 0$ $$ \begin{aligned} & x=t+A \quad S=0+A \Rightarrow A=S \\ & y=t^2+C t+B \quad-s=B \Rightarrow B=-S \\ & z=2 t+C \quad 1=C \Rightarrow C=1 \\ & \Rightarrow(x, y, z)=\left(t+s, t^2+t-5,2 t+1\right), \text { parametric form } \end{aligned} $$ (iii) Projection into the $(x, y)$ - plane of each characteristic $$ \left.\begin{array}{l} z=0 \Rightarrow 2 t+1=0 \Rightarrow t=-\frac{1}{2} \\ x=s-\frac{1}{2} \Rightarrow s=x+\frac{1}{2} \\ y=\frac{1}{4}-\frac{1}{2}-s=-\frac{1}{2}-s \end{array}\right\} y=-\frac{1}{2}-x-\frac{1}{2} \Rightarrow x+y=-1 ? $$

I'm not sure what I've done wrong, since I seem to have gotten a linear projection, rather than a parabola like they asked. If anyone can spot my mistake/misunderstanding, then I would appreciate any help, thanks!

yw_2003
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1 Answers1

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Projection and intersection are two different things. By enforcing $z=0$, the latter is obtained. To get the former, apply the projector $(x,y,z)\mapsto (x,y)$. In other words, we get the parabolas $$ (x,y) = (t+s, t^2+t-s) $$ parametrized by $t$, where $(x,y)|_{t=0} = (s,-s)$. The derivative of the parametric curve is $$ (\dot x,\dot y) = (1, 2t+1) . $$ It appears that the characteristic curves intersect the line $x+y=-1$ whenever $t^2+2t=-1$, that is when $t=-1$. At the point of intersection $(-1+s,-s)$, the slope of the parabola is $$ \frac{\text d y}{\text d x}\bigg|_{t=-1} = \frac{\dot y}{\dot x}\bigg|_{t=-1} = -1, $$ which is also the slope of the line $x+y=-1$. Therefore the two curves are tangent at the point of intersection. Below is an illustration obtained for the particular case $s=0$, which leads to the two curves being tangent at the point $(-1, 0)$.

characteristics

EditPiAf
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