Question: Please Help me to solve my linear partial differential equation.

 

Dear my friends

Hi

I have a linear partial differential equation to solve.

> equ1:=F5*(diff(phi[2](r, theta), r, r)+(diff(phi[2](r, theta), r))/r+(diff(phi[2](r, theta), theta, theta))/r^2)+F3*phi[4](r,theta) = 0;

where F5 and F3 is constant and phi[2](r,theta) is unknown function.

I tried to solve this equation by the following Maple's command:

> pdsolve(equ1,phi[2](r,theta));

The result is obtained by:

phi[2](r,theta) = -Intat(Intat(F3*phi[4](_a,-I*ln(_a)+2*I*ln(_b)-I*ln(r)+theta)/F5*_a,_a = _b)/_b,_b = r)+_F1(I*ln(r)+theta)+_F2(-I*ln(r)+theta)

In the first view, it is obvious that the result of above command is an expression containing imaginary symbol (I). This means that  phi[2](r, theta) is a complex expression; even though, existing complex number in phi[2](r,theta) does not have any physical concept. In fact, this PDE raised from a mechanical problem which dose not have any relationship with imaginary numbers and works in real domain.

It is worth mentioning that phi[4](r,theta) is to obtain by solving the following partial differential equation:

  > equ2:=F2*(diff(phi[4](r, theta), r, r)+(diff(phi[4](r, theta), r))/r+(diff(phi[4](r, theta), theta, theta))/r^2)+F3*phi[4](r, theta) = 0;

by using the following command, phi[4](r,theta) can be obtained:

> pdsolve(equ2,phi[4](r,theta),build);
The result of above is obtained by:

phi[4](r,theta)= _C1*BesselJ((_c[1]/F2)^(1/2),(F3/F2)^(1/2)*r)*_C3*sin(1/F2^(1/2)*_c[1]^(1/2)*theta)+_C1*BesselJ((_c[1]/F2)^(1/2),(F3/F2)^(1/2)*r)*_C4*cos(1/F2^(1/2)*_c[1]^(1/2)*theta)+_C2*BesselY((_c[1]/F2)^(1/2),(F3/F2)^(1/2)*r)*_C3*sin(1/F2^(1/2)*_c[1]^(1/2)*theta)+_C2*BesselY((_c[1]/F2)^(1/2),(F3/F2)^(1/2)*r)*_C4*cos(1/F2^(1/2)*_c[1]^(1/2)*theta);

I used the above relation of phi[4](r,theta) to solve equ1. However, the results is again dependence on complex number which is not acceptable for my work.

Please help me to solve PDE equ1 to determine phi[2](r,theta) in Real domain.

Thanks in advance.

Please Wait...