1. ## PDE boundary conditions

Hi, this is the question

Consider
u_t=u_xx 0<x<l, t>0 for u(x,t)
with boundary conditions u(0,t) = f(t) and u(l,t) = g(t) for t>0
and initial condition u(x,0) = h(x) for 0<x<l.

Show that this problem can be transformed into a new equations problem with a new dependent varaible v(x,t) with a 2nd order partial differential equation, vanishing boundary conditions (v(0,t) = v(l,t) = 0, t>0) and an initial condition.

Assume differentiability when needed, f and g are defined for all t needed.

The main problem I have with this question is I'm not sure how to make the boundary conditions vanish...

Thanks!!

2. Originally Posted by angels_symphony
Hi, this is the question

Consider
u_t=u_xx 0<x<l, t>0 for u(x,t)
with boundary conditions u(0,t) = f(t) and u(l,t) = g(t) for t>0
and initial condition u(x,0) = h(x) for 0<x<l.

Show that this problem can be transformed into a new equations problem with a new dependent varaible v(x,t) with a 2nd order partial differential equation, vanishing boundary conditions (v(0,t) = v(l,t) = 0, t>0) and an initial condition.

Assume differentiability when needed, f and g are defined for all t needed.
Suppose that,
$\displaystyle \frac{\partial u}{\partial t} = \frac{\partial ^2 u}{\partial x^2}$
With boundary value problem $\displaystyle u(0,t) = f(t) \mbox{ and }u(L,t) = g(t) \mbox{ for }t\geq 0$.
And initial value problem $\displaystyle u(x,0) = h(x) \mbox{ for }0\leq x\leq L$.

Write the function $\displaystyle u(x,t)$ as $\displaystyle v(x,t)+w(x,t)$ such that $\displaystyle v(x,t)$ satisfies the boundary value problem. For example, $\displaystyle v(x,t) = f(t) + \frac{x}{L}[g(t) - f(t)]$. And $\displaystyle w(x,t)$ is the modified function that is needed to solve the equation.

Substitute that into the equation,
$\displaystyle \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 w}{\partial x^2} = \frac{\partial v}{\partial t} + \frac{\partial w}{\partial t}$.

Thus,
$\displaystyle \frac{\partial^2 w}{\partial x^2} - \frac{\partial w}{\partial t} = f'(t) + \frac{x}{L} [ g'(t) - f'(t)]$.
With, $\displaystyle w(0,t) = w(L,t) = 0 \mbox{ for }t\geq 0$.
And, $\displaystyle w(x,0) = h(x) - v(x,0) = h(x) - f(0) + \frac{x}{L}[ g(0)-f(0)]$.

Therefore, we have transformed an equation with a non-homogenous boundary value problem into an equation with a homogenous boundary value problem but it has an 'uglier' form. Meaning we have to solve the inhomogenous heat equation.

Now there is a method for solving (with homogenous boundary),
$\displaystyle \frac{\partial^2 w}{\partial x^2} - \frac{\partial w}{\partial t} = \mu (x,y)$.
For a given differenciable function $\displaystyle \mu (x,y)$.
But that is a totally different question.

3. Originally Posted by ThePerfectHacker
Suppose that,
$\displaystyle \frac{\partial u}{\partial t} = \frac{\partial ^2 u}{\partial x^2}$
With boundary value problem $\displaystyle u(0,t) = f(t) \mbox{ and }u(L,t) = g(t) \mbox{ for }t\geq 0$.
And initial value problem $\displaystyle u(x,0) = h(x) \mbox{ for }0\leq x\leq L$.

Write the function $\displaystyle u(x,t)$ as $\displaystyle v(x,t)+w(x,t)$ such that $\displaystyle v(x,t)$ satisfies the boundary value problem. For example, $\displaystyle v(x,t) = f(t) + \frac{x}{L}[g(t) - f(t)]$. And $\displaystyle w(x,t)$ is the modified function that is needed to solve the equation.

Substitute that into the equation,
$\displaystyle \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 w}{\partial x^2} = \frac{\partial v}{\partial t} + \frac{\partial w}{\partial t}$.

Thus,
$\displaystyle \frac{\partial^2 w}{\partial x^2} - \frac{\partial w}{\partial t} = f'(t) + \frac{x}{L} [ g'(t) - f'(t)]$.
With, $\displaystyle w(0,t) = w(L,t) = 0 \mbox{ for }t\geq 0$.
And, $\displaystyle w(x,0) = h(x) - v(x,0) = h(x) - f(0) + \frac{x}{L}[ g(0)-f(0)]$.

Therefore, we have transformed an equation with a non-homogenous boundary value problem into an equation with a homogenous boundary value problem but it has an 'uglier' form. Meaning we have to solve the inhomogenous heat equation.

Now there is a method for solving (with homogenous boundary),
$\displaystyle \frac{\partial^2 w}{\partial x^2} - \frac{\partial w}{\partial t} = \mu (x,y)$.
For a given differenciable function $\displaystyle \mu (x,y)$.
But that is a totally different question.

Hi,

I'm not quite understanding this part:

Thus,
$\displaystyle \frac{\partial^2 w}{\partial x^2} - \frac{\partial w}{\partial t} = f'(t) + \frac{x}{L} [ g'(t) - f'(t)]$.
With, $\displaystyle w(0,t) = w(L,t) = 0 \mbox{ for }t\geq 0$.
And, $\displaystyle w(x,0) = h(x) - v(x,0) = h(x) - f(0) + \frac{x}{L}[ g(0)-f(0)]$.

Why do the boundaries becomes zero? Sorry I'm pretty new to this and I'm not really understanding my professors notes!

4. Originally Posted by angels_symphony
$\displaystyle \frac{\partial^2 w}{\partial x^2} - \frac{\partial w}{\partial t} = f'(t) + \frac{x}{L} [ g'(t) - f'(t)]$.
This comes from $\displaystyle v(x,t) = f(t) + \frac{x}{L}[g(t) - f(t)]$ and thus $\displaystyle \frac{\partial^2 v}{\partial x^2} = 0$ and $\displaystyle \frac{\partial v}{\partial t} = f'(t) + \frac{x}{L}[g'(t) - f'(t)]$.

With, $\displaystyle w(0,t) = w(L,t) = 0 \mbox{ for }t\geq 0$.
Because $\displaystyle v(x,t)+w(x,t) = u(x,t)$ thus $\displaystyle v(x,t)+w(x,t)$ solve this equation. Note $\displaystyle v(0,t) = f(t)$ which means $\displaystyle v(0,t)+w(0,t) = u(0,t) \implies f(t) + w(0,t) = f(t) \implies w(0,t) = 0$ for $\displaystyle t\geq 0$. Similarly $\displaystyle w(L,t) = 0$ for $\displaystyle t\geq 0$.

And, $\displaystyle w(x,0) = h(x) - v(x,0) = h(x) - f(0) + \frac{x}{L}[ g(0)-f(0)]$.
Again $\displaystyle u(x,t) = v(x,t) + w(x,t)$ thus $\displaystyle u(x,0) = v(x,0) + w(x,0) \implies h(x) - v(x,0) = w(x,0)$. Note, $\displaystyle v(x,0) = f(0) + \frac{x}{L}[g(0)-f(0)]$.