PDE Separation of Variables with Nonzero Boundary Conditions

In summary, the conversation discusses solving the diffusion equation with given boundary and initial conditions using the separation of variables method. The participant suggests making a substitution before separating variables to account for the nonhomogeneous boundary conditions. The conversation ends with the participant successfully solving the problem.
  • #1
jtleafs33
28
0

Homework Statement


Solve the diffusion equation:

[itex]u_{xx}-\alpha^2 u_{t}=0[/itex]

With the boundary and initial conditions:

[itex]u(0,t)=u_{0}[/itex]
[itex]u(L,t)=u_{L}[/itex]
[itex]u(x,0=\phi(x)[/itex]

The Attempt at a Solution



I want to solve using separation of variables...
I start by assuming a solution of the form:

[itex]u(x,t)=X(x)T(t)[/itex]

Differentiating and dropping the notation so f(F) = F :

[itex]u_{x}=X'T[/itex]
[itex]u_{xx}=X''T[/itex]
[itex]u_{t}=XT'[/itex]

Substituting back into the PDE,

[itex]X''T-\alpha^2 XT'=0[/itex]
[itex]X''T=\alpha^2 XT[/itex]

[itex]\frac{X''}{X}=\alpha^2 \frac{T'}{T}[/itex]

Let each side be equal to a constant, [itex]-\lambda^2[/itex]
So now we have two ODE's:

[itex]X''+\lambda^2 X=0[/itex]
[itex]T'+(\frac{\lambda}{\alpha})^2 T=0[/itex]

These ODE's are easy to solve:

[itex]T(t)=Ae^{-(\frac{\lambda}{\alpha})^2 t}[/itex]
[itex]X(x)=Bsin(\lambda x)+Ccos(\lambda x)[/itex]

So now, let AB=A and AC=B so the PDE has the general solution:

[itex]u_{\lambda}(x,t)=e^{-(\frac{\lambda}{\alpha})^2 t}[Asin(\lambda x)+Bcos(\lambda x)][/itex]

So now, I want to match my boundary conditions. All the other PDE's I've solved have zero-valued boundary conditions, so I'm not exactly sure what to do here.

[itex]u(0,t)=u_{0}=e^{-(\frac{\lambda}{\alpha})^2 t}[Asin(0)+Bcos(0)][/itex]
[itex]u(0,t)=u_{0}=Be^{-(\frac{\lambda}{\alpha})^2 t}[/itex]

This is where I'm confused.
To satisfy the B.C., my intuition says [itex]B=u_{0}[/itex] and [itex]e^{-(\frac{\lambda}{\alpha})^2 t}[/itex] must be a multiple of [itex]\pi[/itex].
Even if this is correct, I'm not sure how to express it as a constraint on [itex]\lambda[/itex].
 
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  • #2
jtleafs33 said:

Homework Statement


Solve the diffusion equation:

[itex]u_{xx}-\alpha^2 u_{t}=0[/itex]

With the boundary and initial conditions:

[itex]u(0,t)=u_{0}[/itex]
[itex]u(L,t)=u_{L}[/itex]
[itex]u(x,0=\phi(x)[/itex]

The Attempt at a Solution



I want to solve using separation of variables...

Before you separate variables, make the substitution ##u(x,t) = v(x,t) +\psi(x)##. Substitute that into your problem:$$
v_{xx} + \psi''(x) + v_t = 0$$ $$
v(0,t) + \psi(0) = u_0$$ $$
v(L,t) + \psi(L) = u_L$$ $$
v(x,0) + \psi(x) = \phi(x)$$
Now if you let ##\psi''(x)=0,\ \psi(0)=u_0,\ \psi(L)=u_L##, you can solve for ##\psi(x)## and what is left is a system in ##v(x)## you can separate variables on. Your boundary conditions will be homogeneous and the last equation will be ##v(x,0)=\phi(x) - \psi(x)##. Your final answer will be ##u(x,t) = v(x,t)+\psi(x)##.
 
  • #3
LCKurtz,

I can't find anything like that in my textbook. Did you do that so the solution would satisfy both the nonhomogenous and homogenous parts? I suppose you added the extra function of x because the nonhomogenous boundary conditions only depend on x?
 
  • #4
jtleafs33 said:
LCKurtz,

I can't find anything like that in my textbook. Did you do that so the solution would satisfy both the nonhomogenous and homogenous parts? I suppose you added the extra function of x because the nonhomogenous boundary conditions only depend on x?

I don't know about your text, but it is a standard method. You let the ##\psi(x)## take care of the nonhomogeneous part. So did you try it on your problem?
 
  • #5
I will be trying it soon.
 
  • #6
Got it! It was trivial after I did the substitution you mentioned. Thanks again!
 

Related to PDE Separation of Variables with Nonzero Boundary Conditions

What is PDE Separation of Variables with Nonzero Boundary Conditions?

PDE Separation of Variables with Nonzero Boundary Conditions is a mathematical technique used to solve partial differential equations (PDEs) with non-zero boundary conditions. It involves breaking down the PDE into simpler equations, known as ordinary differential equations (ODEs), and then solving for each variable separately.

How does PDE Separation of Variables with Nonzero Boundary Conditions work?

The first step in PDE Separation of Variables with Nonzero Boundary Conditions is to identify the variables in the PDE and separate them into different equations. These equations can then be solved using standard techniques, such as integration or series expansion. The solutions to each equation are then combined to form a general solution to the PDE.

What types of PDEs can be solved using this method?

PDE Separation of Variables with Nonzero Boundary Conditions can be used to solve a wide range of PDEs, including the heat equation, wave equation, and Laplace's equation. It is most commonly used for PDEs with separable variables, where the PDE can be broken down into simpler ODEs.

What are the advantages of using PDE Separation of Variables with Nonzero Boundary Conditions?

One of the main advantages of this method is that it allows for the solution of complex PDEs by breaking them down into simpler equations. It also provides a systematic approach to solving PDEs, making it easier to follow and understand the solution process. Additionally, it can be used to solve PDEs with non-zero boundary conditions, which makes it more versatile than other solution techniques.

Are there any limitations to using PDE Separation of Variables with Nonzero Boundary Conditions?

PDE Separation of Variables with Nonzero Boundary Conditions is a powerful method, but it does have some limitations. It can only be used for PDEs with separable variables, which means it may not be applicable to all PDEs. Additionally, the boundary conditions must be well-defined and consistent for this method to work effectively. In some cases, the solutions obtained using this method may also be limited to a specific range of the variables involved.

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