# Question:why pdsolve only solves this wave PDE when telling it that speed is positive?

## Question:why pdsolve only solves this wave PDE when telling it that speed is positive?

Maple 2018

This is a Wave PDE inside disk. with ﬁxed edge of disk,  and no theta dependency. initial position and velocity  given.

When using "c^2" for the  wave propagation speed, Maple only gives solution when also using assumptions to tell that c is positive.

```restart;
Physics:-Version()[2];
`2018, November 28, 1:35 hours, version in the MapleCloud: 224, version installed in this computer: 224````

And the PDE is

```pde := diff(u(r, t), t\$2) = c^2*( diff(u(r,t), r\$2)  + (1/r)* diff(u(r,t),r)  ) ;
ic:=u(r,0)=1, eval( diff(u(r,t),t),t=0)=r/3;
bc:=u(1,t)=0;
sol:=pdsolve([pde, ic,bc], u(r, t)) assuming t>0;
```

sol:=()

Now adding assuming c>0, or "c::positive", or "c>=0", it solves it

```restart;
pde := diff(u(r, t), t\$2) = c^2*( diff(u(r,t), r\$2)  + (1/r)* diff(u(r,t),r)  ) ;
ic:=u(r,0)=1, eval( diff(u(r,t),t),t=0)=r/3;
bc:=u(1,t)=0;
sol:=pdsolve([pde, ic,bc], u(r, t)) assuming t>0,c>=0;
```

Any idea why the assumption "c>0"  is needed when the speed is given as "c^2" ? I also tried assumption "c::real", but it still did not solve it.   It seems related to Maple using Laplace transform to solve it.

It seems to me the assumption c>0 should not needed here. But if it is, I'd like to learn why.

Maple 2018.2, windows 10

Update

Thanks to comment below

" so u(r, 0)=1 (for all r), and u(1, t)=0 (for all t). So what is u(1,0)??? "

I should have added "r>0,r<1".  The reason I did not, is that I copied the code to solve this from Mathematica. In Mathematica, it did not need this assumption to solve it. It seems to depend on the method of solution used by Mathematica vs. Maple to cause this difference.

But now I see it helps to have it there. With the above assumption, now, there is no need to do any assumption on "c" at all and it gives solution. adding c>0 or t>0 or c>=0, makes no change now.

```restart;
pde := diff(u(r, t), t\$2) = c^2*( diff(u(r,t), r\$2)  + (1/r)* diff(u(r,t),r)  ) ;
ic:=u(r,0)=1, eval( diff(u(r,t),t),t=0)=r/3;
bc:=u(1,t)=0;
sol:=pdsolve([pde, ic,bc], u(r, t)) assuming t>0,r>0,r<1;
```

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