Show that the Diffusion Eqn after substitution gives the Helmholtz Eqn

In summary: algebraic manipulation means for example that you ll transfer the denominator ##\psi(r)## on the right hand side, then you also may transfer ##\kappa## on the right hand side and then transfer all the right hand side...
  • #1
jkthejetplane
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4
Homework Statement
Question is shown below. Basically i am wondering if i am over thinking it or am i just plugging it in? I do not see where psi or theta are possible given to do any further work. Same would go for part b.
Relevant Equations
Helmholtz as defined in my text given below as well
Question:
1607446953575.png

Helmholtz as defined in text:
1607447047932.png

My attempt so far
1607447396059.png
 
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  • #2
So you conclude that each side of your last equation must be equal to some constant [itex]C[/itex], which means that [itex]\psi[/itex] satisfies...

And then you need to justify why [itex]C \leq 0[/itex].

(The constant [itex]k[/itex] which appears in the Helmholtz equation is not necessarily equal to the diffusivity [itex]\kappa[/itex] which appears in the diffusion equation. You need to clearly distinguish between these in your answer.)
 
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  • #3
pasmith said:
So you conclude that each side of your last equation must be equal to some constant [itex]C[/itex], which means that [itex]\psi[/itex] satisfies...

And then you need to justify why [itex]C \leq 0[/itex].

(The constant [itex]k[/itex] which appears in the Helmholtz equation is not necessarily equal to the diffusivity [itex]\kappa[/itex] which appears in the diffusion equation. You need to clearly distinguish between these in your answer.)
Ok so that makes sense as that is how i have started similar problems in the pas that i had boundary conditions for.
I got Θ(t) = Θ(0)eCt and for the other ψ(r) = sin(ψC/[itex]\kappa[/itex]) + cos(ψC/[itex]\kappa[/itex])
Then i do the same for Helmholtz eq? OR
Do i set the C/[itex]\kappa[/itex] = [itex]k[/itex]? Then plug back in my general solution of ψ(r) and the second derivative of it to the Helmholtz eq hoping to yield 0? Showing that it is only true when [itex]C \leq 0[/itex] ?
 
  • #4
No , i believe its not the easiest path to solve for ##\psi(\mathbf{r})## and then show that this ##\psi## satisfies Helmholtz. It is also wrong that you decide to replace the operator ##\nabla^2## with ##\frac{\partial^2}{\partial r^2}## cause we might be in 2 or 3 dimensions where the ##\nabla^2## operator has different expression. You have shown that $$\kappa\frac{\nabla^2\psi(\mathbf{r})}{\psi(\mathbf{r})}=C$$

From the above equation , how can you manipulate it algebraically to show that it is equivalent to Helmholtz equation ,I repeat again, without solving that equation for ##\psi##, just some algebraic manipulation is all that is needed and replacement of ##\frac{C}{\kappa}## with a proper constant so that we have the Helmholtz equation at the end of the algebraic manipulation.
 
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  • #5
Delta2 said:
No , i believe its not the easiest path to solve for ##\psi(\mathbf{r})## and then show that this ##\psi## satisfies Helmholtz. It is also wrong that you decide to replace the operator ##\nabla^2## with ##\frac{\partial^2}{\partial r^2}## cause we might be in 2 or 3 dimensions where the ##\nabla^2## operator has different expression. You have shown that $$\kappa\frac{\nabla^2\psi(\mathbf{r})}{\psi(\mathbf{r})}=C$$

From the above equation , how can you manipulate it algebraically to show that it is equivalent to Helmholtz equation ,I repeat again, without solving that equation for ##\psi##, just some algebraic manipulation is all that is needed and replacement of ##\frac{C}{\kappa}## with a proper constant so that we have the Helmholtz equation at the end of the algebraic manipulation.
So I am just a bit confused still. I need to solve the diff equation for ψ, then just manipulate it to look like Helmholtz?
 
  • #6
jkthejetplane said:
So I am just a bit confused still. I need to solve the diff equation for ψ, then just manipulate it to look like Helmholtz?
Seems like your question doesn't ask you to solve the equation.

If you really want to solve a differential equation, you can manipulate it to look like Helmholtz, because it is easy to solve Helmholtz equation (Your teacher told you the formal solution, didn't he?). This also applies to other diff equations that are easy to solve. We always do this when solving Schroedinger equation.
 
  • #7
pai535 said:
Seems like your question doesn't ask you to solve the equation.

If you really want to solve a differential equation, you can manipulate it to look like Helmholtz, because it is easy to solve Helmholtz equation (Your teacher told you the formal solution, didn't he?). This also applies to other diff equations that are easy to solve. We always do this when solving Schroedinger equation.
No he hasn't. This is an upper level class but i had taken a 5 year break from school so i think he assumes we remember everything.
 
  • #8
jkthejetplane said:
No he hasn't. This is an upper level class but i had taken a 5 year break from school so i think he assumes we remember everything.

For this question, you only need to give the equation (and maybe don't need to solve it).

For Helmholtz equation, you can solve it by separation of variables.
 
  • #9
I am not really getting anywhere. Do you have to be super vague on here? I feel like no one gives me direct answers haha. I just want to understand the problem
 
  • #10
No you don't have to solve any equation for ##\psi## (I repeat myself for 3rd time and others have said this as well).
Just algebraic manipulation of this equation $$\kappa\frac{\nabla^2\psi(\mathbf{r})}{\psi(\mathbf{r})}=C$$ and you ll have Helmholtz equation.

Algebraic manipulation means for example that you ll transfer the denominator ##\psi(r)## on the right hand side, then you also may transfer ##\kappa## on the right hand side and then transfer all the right hand side to the left hand side so you ll have equation of "something"=0. That something will look like Helmholtz equation.
 
  • #11
Delta2 said:
$$\kappa\frac{\nabla^2\psi(\mathbf{r})}{\psi(\mathbf{r})}=C$$

From the above equation , how can you manipulate it algebraically to show that it is equivalent to Helmholtz equation ,I repeat again, without solving that equation for ##\psi##, just some algebraic manipulation is all that is needed and replacement of ##\frac{C}{\kappa}## with a proper constant so that we have the Helmholtz equation at the end of the algebraic manipulation.

If you want to know how to get Helmholtz from diffusion equation, this is what you want. You only need to figure out how you can transform the above equation into the form of Helmholtz.

If you want to solve the Helmholtz equation, I will give a hint in Cartesian coordinates.

Do separation of variables: ##\psi(\vec{r}) = X(x)Y(y)Z(z)##
Then: $$YZ\frac{d^2 X}{dx^2} + ZX\frac{d^2 Y}{dy^2} + XY\frac{d^2 Z}{dz^2} + k^2XYZ = 0.$$
Divide ##XYZ## on both sides: $$\frac{X''}{X}+\frac{Y''}{Y}+\frac{Z''}{Z} + k^2 = 0.$$
You can try to solve it now.
 
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  • #12
You can't solve the [itex]\psi[/itex] equation; you don't know what the boundary is or what condition you have to apply there.

You can solve the [itex]\theta[/itex] equation, which will tell you what values of [itex]C[/itex] are physically realistic (because you can rule out values of [itex]C[/itex] which lead to solutions which grow exponentially with time...).
 
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  • #13
I think this discussion overcomplicates things. You just make an separation ansatz
$$T(t,\vec{x})=\Theta(t) \psi(\vec{r})$$
as in the text shown in the scan in #1. Plugging this into the diffusion equation then leads to the Helmholtz equation for ##\psi(\vec{r})## by just defining ##k## in the right way.
 
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Related to Show that the Diffusion Eqn after substitution gives the Helmholtz Eqn

1. What is the Diffusion Equation?

The Diffusion Equation is a mathematical equation that describes the process of diffusion, which is the movement of particles from an area of high concentration to an area of low concentration. It is often used in physics, chemistry, and other scientific fields to model the spread of substances through a medium.

2. What is the Helmholtz Equation?

The Helmholtz Equation is a mathematical equation that describes the behavior of waves in a medium. It is used in many areas of science and engineering, such as acoustics, electromagnetics, and fluid dynamics.

3. How are the Diffusion and Helmholtz Equations related?

The Diffusion and Helmholtz Equations are related through substitution. By substituting the diffusion coefficient and the diffusion flux into the Helmholtz Equation, we can show that the resulting equation is equivalent to the Diffusion Equation.

4. Why is it important to show that the Diffusion Equation after substitution gives the Helmholtz Equation?

Showing that the Diffusion Equation after substitution gives the Helmholtz Equation is important because it helps us understand the underlying principles of diffusion and wave behavior. It also allows us to apply the same mathematical techniques and solutions to both equations, making it easier to solve complex problems in various fields of science and engineering.

5. Can you provide an example of how the Diffusion Equation and Helmholtz Equation are used in real-world applications?

One example is in the study of heat transfer. The Diffusion Equation is used to model the diffusion of heat through a material, while the Helmholtz Equation is used to study the propagation of thermal waves. By understanding these equations and their relationship, we can better predict and control heat transfer in various systems, such as in building insulation or electronic devices.

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