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MathGroup Archive 2006

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Re: Why doesn't DSolve find all the solutions of this pde

  • To: mathgroup at smc.vnet.net
  • Subject: [mg72468] Re: [mg72459] Why doesn't DSolve find all the solutions of this pde
  • From: Devendra Kapadia <dkapadia at wolfram.com>
  • Date: Sat, 30 Dec 2006 05:31:42 -0500 (EST)
  • References: <200612291011.FAA29956@smc.vnet.net>

On Fri, 29 Dec 2006, Valeri Astanoff wrote:

> Good day,
>
> I wonder why DSolve doesn't find all the solutions
> of this pde :
>
> In[1]:=eq = r D[f[r,th],r] - Tan[2th] D[f[r,th],th] == 0;
>
> In[2]:=DSolve[eq ,f[r,th],{r,th}]
>
>> From In[2]:=
> Solve::ifun Inverse functions are being used by Solve,
> so some solutions may not be found; use Reduce
> for complete solution information.
>
> Out[2]={{f[r, th] -> C[1][Log[(-r)*Sqrt[Sin[2*th]]]]},
> {f[r, th] -> C[1][Log[r*Sqrt[Sin[2*th]]]]}}
>
> In[3]:=g[r_,th_] = r^2*Sin[2th];
>
> In[4]:=eq/.f -> g
> Out[4]=True
>
> In[5]:=$Version
> Out[5]=5.1 for Microsoft Windows (January 28, 2005)
>
>
> With my early thanks for any advice
>
>
> V.Astanoff
>
Hello Valeri,

For linear first-order partial differential equations such as the
one considered by you above, DSolve returns a general solution
containing an arbitrary function C[1]. Particular solutions
of the equation can be obtained by specifying the expression
for C[1].

Thus, the solution r^2*Sin[2th] requested by you can be
found by setting C[1][x_]:=E^(2x) as shown below (see In[4]).

============================

In[1]:= $Version

Out[1]= 5.2 for Microsoft Windows (June 20, 2005)

In[2]:= eq = r D[f[r, th], r] - Tan[2 th] D[f[r, th], th] == 0;

In[3]:= (sol = DSolve[eq, f, {r, th}]) // InputForm

Solve::ifun: Inverse functions are being used by Solve, so some solutions 
may not be found;
      use Reduce for complete solution information.

Solve::ifun: Inverse functions are being used by Solve, so some solutions 
may not be found; use Reduce for complete solution information.

Out[3]//InputForm=
{{f -> Function[{r, th}, C[1][Log[-(r*Sqrt[Sin[2*th]])]]]},
  {f -> Function[{r, th}, C[1][Log[r*Sqrt[Sin[2*th]]]]]}}

In[4]:= (f[r, th] /. sol[[1]]) /. {C[1][x_] :> E^(2 x)}

          2
Out[4]= r  Sin[2 th]

============================

I hope that this helps in answering your question.

Sincerely,

Devendra Kapadia,
Wolfram Research, Inc.


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