How to solve Diffusion Equation with time dependant conditions?

In summary: I’ll have to check to see if that works.In summary, U(x,t)=-a(t) at x=0, and U'(x,t)=0, so B(t)=-a(t)/A(0). Using separation of variables and Laplace transform, U(x,t) can be solved for as a weighted sum of e^iωt i cosh(ωx-L) over ##\omega##.
  • #1
Zimbalj
7
1
TL;DR Summary
By which method could i analytically solve Diffusion Equation with this boundary conditions:
U=U(x,t)

Ut=DUxx; 0<=x<=L, t>0

U(x,0)=0 0<x<=L
U(0,t)=a(t); t>0 *a(t) is known function*
(dU/dx)=0 for x=L

I have looked into many ways but not one is usable for diffusion equation with this boundary conditions.
 
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  • #3
Standard method of separation by variables does not match up nicely with boundary conditions. I have seen methods for solving inhomogeneous heat equation using green functions, but that again failed, because it required knowledge of U(L,t) which was not given by any condition
 
  • #4
fresh_42 said:
How far did you come?

You should also check the last attachment in
https://www.physicsforums.com/threads/solution-manuals-for-the-math-challenges.977057/
(problem 14 Nov. 2020)
November 2020 does have diffusion equation problem, but i do not know how to use any of information there to solve my problem. Soluton on picturem from the book is the case that mostly resembles my problem, but again i need boundary condition U(L,t)=b(t), instead i have flux that equals zero at that point U'(L,t)=0, so i can't use that metod
 

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  • #5
The idea behind solving the source-on-the-half-line problem is to transform the PDE for U with nonhomogeneous boundary conditions into another PDE for V with homogeneous boundary conditions, as is done in the attachment that you've posted.
 
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  • #6
I don’t see why separation of variables isn’t a good start? For a separation constant, ##\omega##, the solution with ##u’_\omega(t,L)=0## is,

##u_\omega(t,x) = e^{\omega t}\cosh(\sqrt{\omega}(x-L))##

So, the solution you seek is some weighted sum over ##\omega##. Evaluating at ##x=0## gives,

##u_\omega(t,0)=\cosh(\sqrt{\omega}L)e^{\omega t}##

this will lead to an answer in terms of Laplace transforms.
 
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  • #7
The problem is boundary condition that is time dependant:
U(0,t)=a(t)
With constant boundary condition i get analytical solution relatively easy, but with time dependant boundary condition i always get that some of the integration constants are functions of t which is not right

Lets say:
U(x,t)=A(x)B(t)

U(0,t)=A(0)B(t)=a(t) => B(t)=a(t)/A(0) *A(0) should be some constant


but function a(t) is not in a form that resembles solution for the time part, when the equation is separated

Bt+(Dw2)B=0
 
  • #8
Zimbalj said:
The problem is boundary condition that is time dependant:
U(0,t)=a(t)
With constant boundary condition i get analytical solution relatively easy, but with time dependant boundary condition i always get that some of the integration constants are functions of t which is not right

Lets say:
U(x,t)=A(x)B(t)

U(0,t)=A(0)B(t)=a(t) => B(t)=a(t)/A(0) *A(0) should be some constant


but function a(t) is not in a form that resembles solution for the time part, when the equation is separated

Bt+(Dw2)B=0
Have a look at 16.1. in
https://web.math.ucsb.edu/~grigoryan/124A/lecs/lec16.pdf
And
https://web.math.ucsb.edu/~grigoryan/124A/lecs/
is the list of all lectures of the course.
 
  • #10
Zimbalj said:
With constant boundary condition i get analytical solution relatively easy, but with time dependant boundary condition i always get that some of the integration constants are functions of t which is not right
That is what the Laplace transform is all about. To find the weight in the weighted sum, one must take an inverse Laplace transform. One must also meet the ##t=0## boundary value by selecting appropriate constants.
 
  • #11
Thanks. I tried Laplace transform L[U(x,t)]=V(x,s) and got:

Vxx-s/D*V=0

sorry for my bad writing of formulas, i am new at LaTeX.

With problematic boundary condition getting form

V(0,s)=A(s)
 
  • #12
Zimbalj said:
sorry for my bad writing of formulas, i am new at LaTeX.
You should follow some of the how to docs for this forum.

Zimbalj said:
I tried Laplace transform L[U(x,t)]=V(x,s) and got:
What you tried isn’t what I’m suggesting. I’ve provided explicit solutions to the heat equation that all obey the boundary condition, ##u’(t,L)=0##, for all ##t##. With a slight and obvious modification, we can make these solutions also all obey the boundary condition, ##u(0,x)=0##, for all ##x##. These solution are just linear combinations of the original set,

##u_\omega(t,x)= \cosh(\sqrt{\omega}(x-L)(e^{\omega t}-1)##[1]

notice that ##u_\omega(0,x)=0##. So, I’ve provided a set of solutions to the heat equation indexed by a complex parameter, ##\omega##, all of which obey 2 out of the three boundary conditions. The final boundary condition, ##u(t,0)=a(t)##, may be expressed as an integral equation of the first kind,

##a(t)=\int A(\omega)u_\omega(t,0)d\omega##, (*)

making the solution to the stated problem,

##u(t,x)=\int A(\omega)u_\omega(t,x)d\omega##

Relating Equation (*) to an inverse Laplace transform is left to the reader.

[1] yeah, I see the problem the problem here. ##\cosh(x)## isn’t a solution. Oh well.
 
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  • #13
Even with its flaws the approach in the previous post still has promise, which is why I’ll leave it as is. it is still possible to take,

##u_\omega(t,x)=\cosh(\sqrt{\omega}(x-L))e^{\omega t}##

as the kernel in the previous post. One just has to add the additional condition,

##0 = \int A(\omega)u_\omega(0,x)d\omega##
 
  • #14
The equation is left unsolved, but that is not a big deal. Thanks everyone for insights, it will be helpful in future, because I will try to solve some other similar equations.
 
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FAQ: How to solve Diffusion Equation with time dependant conditions?

How do I determine the initial and boundary conditions for a diffusion equation with time-dependent conditions?

The initial and boundary conditions for a diffusion equation with time-dependent conditions can be determined by considering the physical system being modeled. The initial condition represents the initial distribution of the diffusing substance, while the boundary conditions define the behavior of the substance at the boundaries of the system. These conditions can also be determined experimentally.

Can the diffusion equation with time-dependent conditions be solved analytically?

In most cases, the diffusion equation with time-dependent conditions cannot be solved analytically. However, there are some special cases where an analytical solution can be obtained, such as for simple geometries and boundary conditions. In general, numerical methods are used to solve the diffusion equation with time-dependent conditions.

What are the common numerical methods used to solve the diffusion equation with time-dependent conditions?

The most commonly used numerical methods for solving the diffusion equation with time-dependent conditions are the finite difference method, finite element method, and spectral methods. These methods discretize the diffusion equation into a system of algebraic equations, which can then be solved using computational techniques.

How do I validate the results obtained from solving the diffusion equation with time-dependent conditions?

The results obtained from solving the diffusion equation with time-dependent conditions can be validated by comparing them to experimental data or analytical solutions, if available. Additionally, sensitivity analysis can be performed to assess the impact of different parameters on the results. It is also important to ensure that the numerical method used is stable and converges to the correct solution.

Can the diffusion equation with time-dependent conditions be applied to real-world problems?

Yes, the diffusion equation with time-dependent conditions is widely used in many scientific fields, such as physics, chemistry, biology, and engineering. It can be applied to model various real-world problems, such as heat transfer, mass transport, and chemical reactions. However, it is important to carefully consider the assumptions and limitations of the diffusion equation when applying it to a specific problem.

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