Markiyan Hirnyk

## 7163 Reputation

11 years, 162 days

## Corollary...

@ Dr. Venkat Subramanian:  pde2 is still not  solved correctly.

PS.

```solz:-plot3d( x = 0 ..1, t = 0 .. 1,axes=boxed);
```

pdehyperbolic_up_to_1.mw

## Are you serious?...

@ Dr. Venkat Subramanian  Thank you for your opinion. Have you read Preben's comment, in particular,

• The limitation is also shown in a separated version, where w = u-v and z = u+v. The two resulting pdes for w and z can be solved independently. The pde for w gives no problem, but the one for z shows the same problem as the given system.

before posting yours?

## I noticed that...

@Konstantin@ in my comment "Adjustment".

## Sorry for the typo...

@Markiyan Hirnyk I got it on my own. The plot command should be

`plot(U^2+U1^2-2*convert(.9/abs(x-.4)^(1/3)+.1/abs(x-.6)^(1/2), rational)*U, x = 0 .. 1);`

@vv base it.

@vv Thank you for your valuable comment. The approach is right, but the adjusted approach demonstrates the convergence because the singularities at x = 2/5 and x = 3/5 are integrable. Here are details.

```f := convert(.9/abs(t-.4)^(1/3)+.1/abs(t-.6)^(1/2), rational);
G1 := convert(-.9445379894, rational);
a := proc (x) options operator, arrow; int(f*exp(t), t = 0 .. x) end proc;
b := proc (x) options operator, arrow; int(f*exp(-t), t = 0 .. x) end proc;
U1 := -(1/2)*exp(-x)*(a(x)+G1)-(1/2)*exp(x)*(b(x)+G1);
U := -(1/2)*exp(x)*(b(x)+G1)+(1/2)*exp(-x)*(a(x)+G1);

limit((U^(2)+U1^(2)-2*convert(0.9/abs(x-0.4)^(1/3)+0.1/abs(x-0.6)^(1/2),rational)*U)*abs(x-3/(5))^(1/(2)),x=(3)/(5));

(1/10)*exp(3/5)*(int((1/10)*(9*sqrt(abs(t-3/5))+abs(t-2/5)^(1/3))*exp(-t)/(abs(t-2/5)^(1/3)*sqrt(abs(t-3/5))), t = 0 .. 3/5))-(18231/193015)*exp(3/5)-(1/10)*exp(-3/5)*(int((1/10)*(9*sqrt(abs(t-3/5))+abs(t-2/5)^(1/3))*exp(t)/(abs(t-2/5)^(1/3)*sqrt(abs(t-3/5))), t = 0 .. 3/5))+(18231/193015)*exp(-3/5)

evalf(%);
-0.04846374787
limit((U^2+U1^2-2*convert(0.9/abs(x-0.4)^(1/3)+0.1/abs(x-0.6)^(1/2),rational)*U)*abs(x-2/(5))^(1/(2)),x=(2)/(5));
0
plot((U^2+U1^2-2*convert(.9/abs(x-.4)^(1/3)+.1/abs(x-.6)^(1/2), rational))*U, x = 0 .. 1);```

divergence3.mw

PS. If I am not mistaken, the above plot does not confirm    -0.3753314046 .

## Thank you...

I am glad of the workaround by Mariusz. Every big soft includes bugs. In particular, both Maple and Mathematica are not any exception. However, I don't remember a Mathematica's bug which appears in 32-Bit version of OS only.

## If you look in ?list,...

@dellair you will read

• Multiple elements of a list or set S can be extracted. The selection op(i..j,S) selects the sub-sequence (S[i],S[i+1],...,S[j]). If S is a set then the selection S[i..j] selects the subset {op(i..j,S)}. If S is a list then the selection S[i..j] selects the sublist [op(i..j,S)]. Negative subscripts can also be used.  Thus S[2..-2] selects all elements except the first and last.

## Output...

```f := x ->piecewise(1 <= x and x <= 2, c[1]*x+c[2], 2 <= x and x <= 3, c[3]*x+c[4], 3 <= x and x <= 4, c[5]*x+c[6]) :
convert(f(x), pwlist, x);

[0, 1, x*c[1]+c[2], 2, x*c[3]+c[4], 3, x*c[5]+c[6], 4, 0]```

## Indeed,...

@Preben Alsholm today I also obtain the same results. Maybe, that was depending on session result yesterday (BTW up to my experience and mathematica.stackexchange.com, the session depending results never happened with Mma .). Because of this reason, I have changed my mind: I  find Kitonum and you are right.

## Thank you...

@Preben Alsholm for your feedback. The "limitation" in your words demonstrates that the numerical methods used in pdsolve,numeric are weak and unstable. Mma solves it without any problem. The question arises: can we trust Maple in this field?

## Please, look at...

@Axel Vogt the modified example with no assumption.

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