Cauchy-Neumann Problem -Non Homogeneous Equation
Solve with variable separation method the following C-N problem :
$u_t(x,t) -u_(x x)(x,t)= tx $ ; $ 00 $
$u(x,0)= 1 $ ;$ 0<=x<=pi $
$u_x(0,t)=0 ; u_x(pi,t)=0$ ; $ t > 0 $
$u_t(x,t) -u_(x x)(x,t)= tx $ ; $ 0
$u(x,0)= 1 $ ;$ 0<=x<=pi $
$u_x(0,t)=0 ; u_x(pi,t)=0$ ; $ t > 0 $
Risposte
If you have time to compute and present the solution in nice form that would be appreciated.
I enclose a graph ( very rough , only first order terms ) of the solution with $ 0
Uploaded with ImageShack.us
I enclose a graph ( very rough , only first order terms ) of the solution with $ 0
Uploaded with ImageShack.us
Not surprisingly, you're right. What I did was to forget the [tex]${1\over 2}[/tex] in the well-known formula
[tex]$tx=\frac{a_0}{2}+\sum_{n=1}^\infty \left( a_n \cos(nx)+b_n\sin(nx) \right)[/tex];
I wrote
[tex]$tx=a_0+\sum_{n=1}^\infty \left( a_n \cos(nx)+b_n\sin(nx) \right)[/tex]
instead.
I've left the solution in a very raw form and am sure that with more careful computations it can be mellowed. If you think this is necessary for my post to be readable, do let me know: it will be a (boring but) useful exercise for me. I'm always dodging computations and so often forget how to handle them!
[tex]$tx=\frac{a_0}{2}+\sum_{n=1}^\infty \left( a_n \cos(nx)+b_n\sin(nx) \right)[/tex];
I wrote
[tex]$tx=a_0+\sum_{n=1}^\infty \left( a_n \cos(nx)+b_n\sin(nx) \right)[/tex]
instead.
I've left the solution in a very raw form and am sure that with more careful computations it can be mellowed. If you think this is necessary for my post to be readable, do let me know: it will be a (boring but) useful exercise for me. I'm always dodging computations and so often forget how to handle them!
I have a comment on your Fourier series for $tx $ with respect to $x $ (in $0
You write $tx= tpi+ sum_(k=1)^(+oo) (2t)/(pik^2)[(-1)^k-1]cos(kx) $
while I write $tx= tpi/2-(4t)/pi sum_(k=0)^(+oo) cos((2k+1)x)/(2k+1)^2 $
The only difference is $tpi $ against $ tpi/2 $ ( for the other difference I just took advantage of tne fact that for $ k$ even the value is $0$).
For $x=pi/2 rarr tx=tpi/2$ since all terms in $cos $ disappear.
After clarification on this point we can continue in the comparison of your and mine solutions.
You write $tx= tpi+ sum_(k=1)^(+oo) (2t)/(pik^2)[(-1)^k-1]cos(kx) $
while I write $tx= tpi/2-(4t)/pi sum_(k=0)^(+oo) cos((2k+1)x)/(2k+1)^2 $
The only difference is $tpi $ against $ tpi/2 $ ( for the other difference I just took advantage of tne fact that for $ k$ even the value is $0$).
For $x=pi/2 rarr tx=tpi/2$ since all terms in $cos $ disappear.
After clarification on this point we can continue in the comparison of your and mine solutions.
First of all we take into account the homogeneous Neumann problem
[tex]$\begin{cases} u_t-u_{xx}=0 \\ u_x(0, t)=u_x(\pi, t)=0 \end{cases}[/tex] ;
our fist task will be to find a family of solutions via separation of variables. To do this, we put [tex]u(x, t)=X(x)T(t)[/tex] and plug it into the differential equation, getting the relation [tex]X(x)\dot{T}(t)=X''(x)T(t)[/tex]; separating variables we obtain
[tex]\begin{cases} X''(x)+\lambda X(x)=0 \\ \dot{T}(t)+\lambda T(t)=0 \end{cases}[/tex] .
Merging the first equation with the prescribed boundary condition we get
[tex]\begin{cases} X''(x)+\lambda X(x)=0 \\ X'(0)=X'(\pi)=0 \end{cases}[/tex]
this is a well-known example of boundary value problem: its eigenvalues are [tex]\lambda=k^2,\ k\in\mathbb{N}[/tex] and the corresponding eigenfunctions are [tex]X(x)=\cos(kx)[/tex]. ([size=75]*[/size])
Plugging [tex]\lambda=k^2[/tex] into the second equation and integrating we get [tex]T(t)=C e^{-k^2 t}[/tex]. We have thus obtained the family
[tex]$u_k(x, t)=e^{-k^2 t}\cos(kx),\ k\in\mathbb{N}[/tex] ;
every [tex]u_k[/tex] solves the homogeneous Neumann problem.
____________________________
We now turn our attention to the original problem
[tex]$\begin{cases} u_t - u_{xx}=tx \\ u(0, x)=1 \\ u_x(t,0)=u_x(t, \pi)=0 \end{cases}[/tex] .
The idea is to look for a solution in the form
[tex]$u(x, t)=\sum_{k=0}^\infty c_k(t)u_k(x, t)[/tex] .
We have chosen [tex]c_k=c_k(t)[/tex], i.e. as functions of [tex]t[/tex] only, so that the Neumann condition is verified as long as the convergence of the series is nice enough:
[tex]$u_x(0, t)=\sum_{k=0}^\infty c_k(t)\frac{\partial u_k}{\partial x}(0, t)=0[/tex]
[tex]$u_x(\pi, t)=\sum_{k=0}^\infty c_k(t)\frac{\partial u_k}{\partial x}(\pi, t)=0[/tex] .
Plugging the expression for [tex]u[/tex] into the differential equation we get
[tex]$\sum_{k=0}^\infty \dot{c_k}(t)u_k(x, t) + \sum_{k=0}^\infty c_k(t)\frac{\partial u_k}{\partial t}(x, t) - \sum_{k=0}^\infty c_k(t) \frac{\partial^2 u_k}{\partial x^2}(x, t)=tx[/tex] ;
which reduces to
[tex]$\sum_{k=0}^\infty \dot{c_k}(t)u_k(x, t) = tx[/tex].
The preceding equation need hold for [tex]0< x <\pi[/tex]; in this interval we can think of [tex]tx[/tex] as the restriction of the sawtooth function
[asvg]xmin=-2*3.14; xmax=2*3.14; ymin=0; ymax=2*3.14; axes("label"); xmin=-2*3.14; xmax=-1*3.14; plot("x+2*3.14"); xmin=-1*3.14;xmax=1*3.14; plot("abs(x)"); xmin=1*3.14; xmax=2*3.14; plot("-x+2*3.14");[/asvg]
whose Fourier series with respect to [tex]x[/tex] is
[tex]$\frac{t\pi}{2}+\sum_{k=1}^\infty\frac{2t}{\pi k^2}\left( (-1)^k-1 \right) \cos (kx)[/tex] .
Comparing the last two series we get the equations
[tex]$\dot{c}_0=\frac{t \pi}{2},\ \dot{c_k}=\frac{2t}{\pi k^2}\left( (-1)^k-1 \right)t e^{k^2t}[/tex]
integrating we have
[tex]$c_0(t)=\frac{t^2 \pi}{4}+ H_0,\ c_k(t)=\frac{2(k^2t-1)[(-1)^k-1]e^{k^2t}}{k^6 \pi}+H_k[/tex]
where the constants [tex]H_k[/tex] can be determined via the Cauchy condition [tex]u(x, 0)=1[/tex]. Note that for every fixed [tex]t[/tex] we have [tex](c_k(t)e^{-k^2t})_{k\in \mathbb{N}} \in \ell^1[/tex], because [tex]c_k(t)e^{-k^2 t} \sim k^{-4}[/tex].
____________________________
The last step is to show that the function
[tex]$u(x, t)=\sum_{k=0}^\infty c_k(t)e^{-k^2t}\cos(kx)[/tex]
is well defined for all [tex]t>0,\ 0 < x < \pi[/tex] and that the convergence of the series is nice enough.
Fix [tex]T>0[/tex]. From the last note, and the inequality [tex]\lvert \cos(k x) \rvert \le 1[/tex], we see that the series converges uniformly for all [tex]0
Last, we differentiate term by term with respect to [tex]t[/tex], obtaining
[tex]$\sum_{k=0}^\infty \dot{c_k}(t)e^{-k^2 t}\cos(kx)+ \sum_{k=0}^\infty (-k^2) c_k(t)e^{-k^2t}\cos(kx)[/tex];
the two series both converge uniformly for all [tex]0
____________________________
([size=75]*[/size]) I've omitted the computations, I can provide them if needed.
[tex]$\begin{cases} u_t-u_{xx}=0 \\ u_x(0, t)=u_x(\pi, t)=0 \end{cases}[/tex] ;
our fist task will be to find a family of solutions via separation of variables. To do this, we put [tex]u(x, t)=X(x)T(t)[/tex] and plug it into the differential equation, getting the relation [tex]X(x)\dot{T}(t)=X''(x)T(t)[/tex]; separating variables we obtain
[tex]\begin{cases} X''(x)+\lambda X(x)=0 \\ \dot{T}(t)+\lambda T(t)=0 \end{cases}[/tex] .
Merging the first equation with the prescribed boundary condition we get
[tex]\begin{cases} X''(x)+\lambda X(x)=0 \\ X'(0)=X'(\pi)=0 \end{cases}[/tex]
this is a well-known example of boundary value problem: its eigenvalues are [tex]\lambda=k^2,\ k\in\mathbb{N}[/tex] and the corresponding eigenfunctions are [tex]X(x)=\cos(kx)[/tex]. ([size=75]*[/size])
Plugging [tex]\lambda=k^2[/tex] into the second equation and integrating we get [tex]T(t)=C e^{-k^2 t}[/tex]. We have thus obtained the family
[tex]$u_k(x, t)=e^{-k^2 t}\cos(kx),\ k\in\mathbb{N}[/tex] ;
every [tex]u_k[/tex] solves the homogeneous Neumann problem.
____________________________
We now turn our attention to the original problem
[tex]$\begin{cases} u_t - u_{xx}=tx \\ u(0, x)=1 \\ u_x(t,0)=u_x(t, \pi)=0 \end{cases}[/tex] .
The idea is to look for a solution in the form
[tex]$u(x, t)=\sum_{k=0}^\infty c_k(t)u_k(x, t)[/tex] .
We have chosen [tex]c_k=c_k(t)[/tex], i.e. as functions of [tex]t[/tex] only, so that the Neumann condition is verified as long as the convergence of the series is nice enough:
[tex]$u_x(0, t)=\sum_{k=0}^\infty c_k(t)\frac{\partial u_k}{\partial x}(0, t)=0[/tex]
[tex]$u_x(\pi, t)=\sum_{k=0}^\infty c_k(t)\frac{\partial u_k}{\partial x}(\pi, t)=0[/tex] .
Plugging the expression for [tex]u[/tex] into the differential equation we get
[tex]$\sum_{k=0}^\infty \dot{c_k}(t)u_k(x, t) + \sum_{k=0}^\infty c_k(t)\frac{\partial u_k}{\partial t}(x, t) - \sum_{k=0}^\infty c_k(t) \frac{\partial^2 u_k}{\partial x^2}(x, t)=tx[/tex] ;
which reduces to
[tex]$\sum_{k=0}^\infty \dot{c_k}(t)u_k(x, t) = tx[/tex].
The preceding equation need hold for [tex]0< x <\pi[/tex]; in this interval we can think of [tex]tx[/tex] as the restriction of the sawtooth function
[asvg]xmin=-2*3.14; xmax=2*3.14; ymin=0; ymax=2*3.14; axes("label"); xmin=-2*3.14; xmax=-1*3.14; plot("x+2*3.14"); xmin=-1*3.14;xmax=1*3.14; plot("abs(x)"); xmin=1*3.14; xmax=2*3.14; plot("-x+2*3.14");[/asvg]
whose Fourier series with respect to [tex]x[/tex] is
[tex]$\frac{t\pi}{2}+\sum_{k=1}^\infty\frac{2t}{\pi k^2}\left( (-1)^k-1 \right) \cos (kx)[/tex] .
Comparing the last two series we get the equations
[tex]$\dot{c}_0=\frac{t \pi}{2},\ \dot{c_k}=\frac{2t}{\pi k^2}\left( (-1)^k-1 \right)t e^{k^2t}[/tex]
integrating we have
[tex]$c_0(t)=\frac{t^2 \pi}{4}+ H_0,\ c_k(t)=\frac{2(k^2t-1)[(-1)^k-1]e^{k^2t}}{k^6 \pi}+H_k[/tex]
where the constants [tex]H_k[/tex] can be determined via the Cauchy condition [tex]u(x, 0)=1[/tex]. Note that for every fixed [tex]t[/tex] we have [tex](c_k(t)e^{-k^2t})_{k\in \mathbb{N}} \in \ell^1[/tex], because [tex]c_k(t)e^{-k^2 t} \sim k^{-4}[/tex].
____________________________
The last step is to show that the function
[tex]$u(x, t)=\sum_{k=0}^\infty c_k(t)e^{-k^2t}\cos(kx)[/tex]
is well defined for all [tex]t>0,\ 0 < x < \pi[/tex] and that the convergence of the series is nice enough.
Fix [tex]T>0[/tex]. From the last note, and the inequality [tex]\lvert \cos(k x) \rvert \le 1[/tex], we see that the series converges uniformly for all [tex]0
Last, we differentiate term by term with respect to [tex]t[/tex], obtaining
[tex]$\sum_{k=0}^\infty \dot{c_k}(t)e^{-k^2 t}\cos(kx)+ \sum_{k=0}^\infty (-k^2) c_k(t)e^{-k^2t}\cos(kx)[/tex];
the two series both converge uniformly for all [tex]0
____________________________
([size=75]*[/size]) I've omitted the computations, I can provide them if needed.
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