The aim is to plot the solution of a PDE with initial value along with its characteristics and initial value.

$x u_x - y u_y = 0$ on the line $x=y=u=t$, i.e. $u(t,t)=t$ $(t > 0)$.

Then the general solution is given by $u(x,y) = \Phi(xy)$, on the line by $u_p(x,y) = \sqrt{xy}$.

First I define the solution of the PDE by

In[1]:= sol7212a = DSolve[{x*D[u[x, y], x] - y*D[u[x, y], y] == 0}, u, {x, y}]

Out[1]= {{u -> Function[{x, y}, C[1][x y]]}}

and then the plot.

In[2]:= Show[

Evaluate[u[x, y] /. sol7212a /. {C[1][t_] -> Sqrt[t]}[[1]]], {x, -4, 
4}, {y, -4, 4}, Mesh -> True, 
ColorFunction -> (ColorData["LakeColors"][#3] &)], 
 ParametricPlot3D[{t, t, t}, {t, -4, 4}, PlotStyle -> Cyan], 
Table[ContourPlot3D[{z == Sqrt[x*y], x* y == C}, {x, -4, 4}, {y, -4, 
4}, {z, -4, 4}, Contours -> {0}, ContourStyle -> Opacity[0], Mesh -> None, 
BoundaryStyle -> {1 -> None, 2 -> None, {1, 2} -> {{Green, Tube[.03]}}}, Boxed -> False], {C, 1, 10, 1}] ], LabelStyle -> (FontFamily -> "Museo Sans"), Boxed -> False]

This produces the desired result.


Thus, the first Evaluate[] draws the solution surface of the PDE and ParametricPlot3D[] the line $(t,t,t)$. The second Evaluate[] calculates the characteristics for constants 1,2,...,10.

But the ContourPlot3D command is very slow. I am not even able to resize or turn the produced image. Is there a better way to implement the characteristics?

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The problem seems to be the ContourStyle option. After changing

ContourStyle -> Opacity[0]


ContourStyle -> None

the plot is generated normally and in a fraction of time. The result can be manipulated and exported normally as well.

Thanks, wueb.


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