# PDE boundary conditions

• January 27th 2008, 03:02 PM
angels_symphony
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!!
• January 27th 2008, 03:53 PM
ThePerfectHacker
Quote:

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

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

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

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

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),
$\frac{\partial^2 w}{\partial x^2} - \frac{\partial w}{\partial t} = \mu (x,y)$.
For a given differenciable function $\mu (x,y)$.
But that is a totally different question.
• January 27th 2008, 08:30 PM
angels_symphony
Quote:

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

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

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

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

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),
$\frac{\partial^2 w}{\partial x^2} - \frac{\partial w}{\partial t} = \mu (x,y)$.
For a given differenciable function $\mu (x,y)$.
But that is a totally different question.

Hi,

I'm not quite understanding this part:

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

Why do the boundaries becomes zero? Sorry I'm pretty new to this and I'm not really understanding my professors notes!
• January 28th 2008, 08:02 AM
ThePerfectHacker
Quote:

Originally Posted by angels_symphony
$\frac{\partial^2 w}{\partial x^2} - \frac{\partial w}{\partial t} = f'(t) + \frac{x}{L} [ g'(t) - f'(t)]$.

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

Quote:

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

Quote:

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