Non homogeneous Cauchy-Neumann Problem -Homogeneous equation

[Camillo]

Solve with the variable separation method the following problem :
\[
\begin{cases}
u_t(x,t) - u_{xx} (x,t) = 0 &\text{, if } 0 < x <\pi \text{ and } t>0\\
u(x,0) = 0 &\text{, if } 0\leq x\leq \pi\\
u_x (0,t) =0 ; u_x(\pi,t) = U &\text{, if } t>0
\end{cases}
\]

If $U ne 0 $, can a stationary solution $u_(oo) =u_(oo)(x )$ exist?

Edit : modified boundary conditions.

[Camillo]

Hint : it is convenient to transform the original problem into one with homogeneous conditions.
Let's put :
$w(x,t)= u(x,t)-v(x) $
where $v_x(0)=0 ; v_x(pi)= U $.

[Luca.Lussardi]

It suffices to set $w(x,t)=w_1(x)w_2(t)$? Hence, when we have $w_1$ and $w_2$ it is easy to construct the Fourier's serie of the solution.

[gugo82]

Camillo, sorry, could you please work out the solution to this old problem when you have some spare time?

[Camillo]

Interesting problem , I will post solution in the near future

[Camillo]

Following my previous posts with the problem and the hint we can choose, for instance :
$v(x)=(U x^2)/(2 pi)$.
Consequently the function $ w $ is solution of the non homogeneous problem (but with homogeneous conditions) :

$w_t(x,t) –w_(x x)(x,t) =U/pi ; 0< x< pi , t >0 $
$ w(x,0) = -(Ux^2)/(2pi) ; 0<=x <=pi $
$w_x(0,t) =0 ; w_x(pi,t) = 0 ; t > 0 $


We first consider solutions of homogeneous equation i.e. :

$w_t(x,t) –w_(x x)(x,t) =0 $.

and look for solutions of the kind :

$w(x,t) = a(x)*c(t) $.

Substituting in the equation we get :
$(c’(t))/(c(t)) =(a’’(x))/(a(x)) = lambda$ , $ lambda in RR $

This is an eigenvalue problem, solving it we find that eigenvalues are $ lambda_k = -k^2 $ and the eigenfunctions are : $a(x)= cos (kx) $. ($k in NN$).
It can be shown that the candidate solution is :

$ w(x,t)= (c_(0)(t))/2 +sum_(k=1)^(oo) c_k(t)*cos(kx) $.

We have to determine the $c_k(t)$ coefficients such that :
· $w_t-w_(x x) = (c_(0)’(t))/2+sum_(k=1)^(oo)[c’_k(t)+k^2c_k(t)]cos (kx)= U/pi.$
· $w(x,0)=(c_0(0))/2+sum_(k=1)^(oo) c_k(0) cos(kx)=-Ux^2/(2pi)$
Writing $ g(x)= (Ux^2)/(2pi) $ in cosine series , we find the expression :
· $(Ux^2)/(2pi)= U/(2pi) [ (pi)^2/3+ 4 sum_(k=1)^(oo) (-1)^k/(k^2)*cos(kx) ]$.
The series is uniformly convergent in $[0, pi] $.

Comparing the last three formulas, we realize that the various $c_k(t) $ must be solutions of the following Cauchy problems (with relevant initial conditions) :

*$c_(0)’(t) =(2U)/(pi) ;c_0(0)=-(Upi)/3 $
* $ c’_k(t) +k^2 c_k(t)= 0 ; c_k(0) =(2U)/(pi) (-1)^(k+1)/(k^2) ; k >=1 $

Solutions for Cauchy problems are :

$c_(0)(t)= (2U)/(pi) t – (Upi)/3$
$c_k(t) = (2U)/(pi)*(-1)^(k+1)/(k^2)*e^(-k^2*t) ; k>=1 $.

The solution of our problem is therefore :
$u(x,t)=w(x,t)+v(x)= (Ut)/pi –(Upi)/6 +(Ux^2)/(2pi)+(2U)/(pi) sum_(k=1)^(oo) (-1)^(k+1)/(k^2) e^(-k^2t)cos(kx)$. (**)

* If $U ne 0 $ a stationary solution $ u_(oo) =u_(oo)(x) $ cannot exist.
In fact it should be solution of the problem ( note that $u_t(x,t)=0 $ ):
$u_(oo)''(x)=0 $
$u_(oo)'(0) =0 $
$u_(oo)' (pi) = U $
and this problem has no solution.

Analysis of the solution (**)
The series is uniformly convergent in $[0,pi] X[0, oo] $ therefore $u $ is there continuous.
The derivatives of any order can be exchanged with the sum sign in $[0, pi]X[t_0,oo] $ for any $t_0 >0$; as consequence $u $ is solution of the diffusion equation in $(0,pi)X(0, oo)$.

[gugo82]

Mmmm... I think there's a problem here: in fact \(u_x(0,t) = Ut/\pi \), hence \(u\) doesn't match the initial condition \(u_x =U\) on the \(t\) axis.


P.S.: Is \(U\) a constant? I guess so, but when I first read the problem I thought it was a given function of \(t\).
Moreover, has \(u_x=U\) to hold only on \(x=0\)? Or also on \(x=\pi\)?

[Camillo]

Sorry, I dont know how happened but I did a mistake in copying the boundary conditions , the correct ones are :

$ u_x(0,t)=0 ; u_x(pi,t) = U ; t >0 $.Now I modify in the first post also.

Yes in the first issue of the problem $U $ was considered as $U(t) $; the correct formulation is that $ U $ is a constant.

[gugo82]

Thanks a lot, Camillo!

[*gs86]

Hi Camillo! Could you solve me this Non homogeneous Cauchy-Neumann Problem?


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