2nd order nonlinear imaginary partial dif eqn

In summary: If you simplify it more and take only the real and imaginary parts, it simplifies to this: u(x,t)=C_1e^{-itx}+C_2e^{itx}where C_1 and C_2 are constants. If you are trying to solve for u(x,t), then you can use the general solution u(x,t)=-C_1e^{-itx}+C_2e^{itx} and plug in the values for x and t to get:u(x,t)=C_1e^{-itx}+C_2e^{itx}=-C_
  • #1
FD2010
3
0
Hi all,

I am having a hard time solving a partial second order differential equation with an imaginary part. I basically took a much bigger function with real and imaginary parts and simplified it down to this. I also know the solution to a similar equation (shown in image). Any help would be appreciated!

Nick
 

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  • #2
Welcome to PF!

Hi FD2010! Welcome to PF! :wink:

Can't you just replace Ω by Ω+iK (and use polar form to find the square-root) ? :smile:
 
  • #3
I am terribly sorry but I wound up writing the original expression wrong. I was so hasty for help I had an error in the equation (I left out a u!). Here is what I am trying to solve (new image).

Any help is sincerely appreciated!

Nick
 

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  • #4
Hi Nick! :wink:

in that case, you need a particular solution to add to the general solution you already have …

try u = a constant :smile:
 
  • #5
FD2010 said:
I am terribly sorry but I wound up writing the original expression wrong. I was so hasty for help I had an error in the equation (I left out a u!). Here is what I am trying to solve (new image).

Any help is sincerely appreciated!

Nick

I don't see how you're getting all that unless the constants are eliminating the complex components of the solution. For starters, it's just one derivative so you could replace it with just an ODE.

[tex]u''-ku=0[/tex]

and keep in mind the i is just a constant and as long as you use correct complex-value arithmetic, it's just like real variables. So the solution to that is:

[tex]u(y)=c_1 e^{\sqrt{k}y}+c_2 e^{-\sqrt{k}y}[/tex]

however, that expression is already (implicitly) taking out the two values of the multi-valued square root. Now in the case of [itex]\sqrt{i}[/itex], it's best to explicitly evaluate the square root of i:

[tex]\sqrt{\frac{i\omega}{\nu}}=\sqrt{\omega/\nu}e^{\pi i/4}{, \sqrt{\omega/\nu}e^{-3\pi i/4}}[/tex]

then the solution with the i would be:

[tex]u(y)=c_1\text{exp}\left(y\sqrt{\omega/\nu}e^{\pi i/4}\right)+c_2\text{exp}\left(y\sqrt{\omega/\nu}e^{-3\pi i/4}\right)[/tex]

And if it was indeed a partial of say u(y,t), then the constants of integration would be functions of t and the solutions would be:

[tex]u(y,t)=c_1(t)\text{exp}\left(y\sqrt{\omega/\nu}e^{\pi i/4}\right)+c_2(t)\text{exp}\left(y \sqrt{\omega/\nu}e^{-3\pi i/4}\right)[/tex]
 
  • #6
no, forget what i said about Ω + iK, that was for the (original) wrong problem …

just put u = A, then ∂2u/dy2 - (iΩ/ν)u + κ/ν = (-iΩA + κ)/ν :wink:
 
  • #7
Sincere thanks for the help. I would go into detail about how I arrived at that very simple equation but it would take awhile.

I wound up having to solve the original equation I was working with using non-dimensionalization. u (a velocity in this case) was unsteady, and depended on a oscillating pressure gradient. Due to this, I couldn't just let du/dt go to 0 because u is not a constant. The result wound up being quite a mess, but I believe I'm good now.

Thanks again!
 
  • #8
FD2010 said:
Hi all,

I am having a hard time solving a partial second order differential equation with an imaginary part. I basically took a much bigger function with real and imaginary parts and simplified it down to this. I also know the solution to a similar equation (shown in image). Any help would be appreciated!

Nick


Hi you can Take (i*omg-K)/v as one whole part(is omg).
 

1. What is a 2nd order nonlinear imaginary partial differential equation (PDE)?

A 2nd order nonlinear imaginary PDE is a mathematical equation that involves multiple variables and their partial derivatives, where the highest order of any derivative is 2 and the equation also contains imaginary numbers. It is a type of PDE commonly used to model nonlinear phenomena in physics, engineering, and other fields.

2. How is a 2nd order nonlinear imaginary PDE different from a regular 2nd order PDE?

A 2nd order nonlinear imaginary PDE differs from a regular 2nd order PDE in that it involves the use of imaginary numbers. This means that the solutions to the equation will involve complex numbers instead of just real numbers, making it a more complex and challenging problem to solve.

3. What are some applications of 2nd order nonlinear imaginary PDEs?

2nd order nonlinear imaginary PDEs have a wide range of applications in various fields such as fluid mechanics, quantum mechanics, and electromagnetism. They are commonly used to model complex phenomena such as turbulence, wave propagation, and quantum tunneling.

4. How do you solve a 2nd order nonlinear imaginary PDE?

The specific method for solving a 2nd order nonlinear imaginary PDE will depend on the specific equation and its boundary conditions. In general, the process involves using techniques such as separation of variables, Fourier transforms, or numerical methods. It may also require the use of software programs or computer simulations.

5. Why are 2nd order nonlinear imaginary PDEs important in scientific research?

2nd order nonlinear imaginary PDEs are important in scientific research because they allow us to model and understand complex phenomena that cannot be described by simpler equations. They provide a powerful tool for studying nonlinear systems and their behavior, which is crucial in many areas of physics, engineering, and other fields.

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