Thomas Dean

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15 years, 134 days

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These are questions asked by Thomas Dean

I have a PDE with boundary conditions, from a NASA paper.  I always seem to have problems expressing the bc.

PDE := diff(u(x, t), t) = (1/4)*exp(2)*exp(-u(x, t))*(diff(diff(u(x, t), x), x))/(x^2+2);

The initial/boundary conditions are

@t=0, u(x, t) = 2-2*ln(-x^2+2)

@x=0, diff(u(x,t),x)=0  ## this is the bc I have problem expressing

@x=1, u(x,t) = 2+ln(1+t)

The exact solution given in the paper:

2 + ln(1+t) - 2*ln(2-x^2)

I tried

ics := u(x, 0) = 2*(1-ln(2-x^2));

bcs := D[1](u(0,t))=0, u(1,t)=2+ln(1+t);

PDEtools[Solve]([PDE,ics,bcs]); ## no solution

How do I do this?

Tom Dean

I have several maple worksheets (from the web) that have discussion blocks mixed within executable blocks.

All the executable blocks are delineated with a single '[' at the left while the discussion blocks do not.

How do I do this?

Tom Dean

I attempted to show that two lines are parallel.  I started with a problem in Geometry for which I do not have the solution.

I tried several ways with Maple to show this to be true.  Most of the time, I ended when maple could not determine if a-b = c-d, etc.

brg_proof.txt contains a statement of the problem and my latest maple code.

Question: How should I approach the proof, by the compass and straight edge method?  Is this possible in maple?

intersection in the geometry package does not seem to recognize assume.

restart: with(geometry):

assume(p[1]<>0, p[2]<>0, p[3]<>0);
assume(q[1]<>0, q[2]<>0, q[3]<>0);
point(T,[p[1],q[1]]);
point(U,[p[2],q[2]]);
point(V,[p[3],q[3]]);
point(Op,[0,0]);

line(OT,[Op,T]);
line(OU,[Op,U]);
line(OV,[Op,V]);

point(B,2*q[2],solve(subs(x=2*q[2],Equation(OU)),y));
coordinates(B);
IsOnLine(B,OU);

PerpendicularLine(AD,B,OT);
ArePerpendicular(AD,OT);
sol:=solve({Equation(AD),Equation(OT)},{x,y});
eval(x,sol);
point(A,eval(x,sol),eval(y,sol));  ## the intersection exists
intersection(xA,AD,OT); ## fails
about(p[1]),about(q[1]);

I have several plots generated with geometry objects:

plt1 := draw(...); plt2 := draw(...)

I want to display all them with a reasonable window.  I have been looping and and using a rather dumb proc to find the window size. I use the view=[...] option with the values calculated to set the plot view.

I wanted to post the proc, but, really messed that up!  I will try the proc later.

 

Some of the plots I am working with have views like

> for itm in [relBrgLinePlot,relTgtPositPlot,relTgtLinePlot] do
    op(-1,itm);
end do;
                     VIEW(0. .. 95., -61.58179461 .. 16.)
                 VIEW(5.48225506 .. 95., -61.58179461 .. 16.)
                 VIEW(5.48225506 .. 95., -61.58179461 .. 16.)

> for itm in [geoSensPositPlot,geoSensLinePlot,\
            geoBrgLinePlot,geoTgtPositPlot,geoTgtLinePLot] do
    op(-1,itm);
end do;
                  VIEW(0. .. 35.35331852, 0. .. 87.13244438)
                  VIEW(0. .. 35.35331852, 0. .. 87.13244438)
                     VIEW(0. .. 95., 0. .. 87.13244438)
                 VIEW(40.83557358 .. 95., 16. .. 25.55064977)
                 VIEW(40.83557358 .. 95., 16. .. 25.55064977)

Is there an easier way to do this?

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