Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For an one-dimensional bar that goes from $x=0$ to $x=1$ and with $t$ belonging to $0$ to $+\infty$.

$u_t=u_{xx}+\sin\pi x+\sin 2\pi x\\ u(0,t)=u(1,t)=0\\ u(x,0)=0$


I recognize that this is a forced heat equation problem, with homogeneous Dirichlet boundary conditions and an initial condition fairly unusual.

I suspect that I have to first solve the homogeneous cousin $u_t=u_{xx}$ with same conditions and later try to find the non-homogeneous solution. But the initial condition puts the fourier coefficients equal zero, meaning trivial solution.

What am I not understanding?

share|improve this question
    
Solve first $u_{xx}=\sin(\pi x)+\sin(2\pi x)$, this will give you the stationary temperature $u_{st}$. Then put $u=u_{st}+v$ and get for $v$ a homogeneous heat equation with nonzero initial conditions. –  Artem Feb 2 '13 at 15:00
    
Ok, let me see if I can get it to work. –  Dan Feb 2 '13 at 15:03
    
@Artem Thanks, I got it. –  Dan Feb 2 '13 at 16:08
    
Are you sure? I made a type in the equation you need to solve. It has to have minus before both sine functions. You can post your own answer if you'd like someone check it. –  Artem Feb 2 '13 at 16:37
    
Well, If i got it straight you want me to define a function $u_{gt}$ so that the equation ends up only $v_t = v_{xx}$ (as the temporal derivative of $u_{gt}$ is zero and the second x derivative cancels the sinuses). I can easily find $w = sin(\pi x)/\pi^2 + sin(2 \pi x)/(4 \pi^2) + c_1x+c_2 $ Then there is a fight for the conditions, the conditions end up: $v(0,t) = -c_2; v(1,t) = - c_1 - c_2; v(x,t) = - w(x)$ And from them on its just a nasty, but ordinary problem with non-homogeneous boundary conditions. –  Dan Feb 2 '13 at 17:01

1 Answer 1

up vote 0 down vote accepted

FIrst solve the equation $$ u''=-\sin \pi x-\sin 2\pi x, $$ subject to the boundary conditions $$ u(0)=u(1)=0. $$ This problem has solution $$ u_{st}=\frac{1}{\pi^2}\sin \pi x+\frac{1}{2^2\pi^2}\sin 2\pi x. $$ Now put $u(x,t)=u_{st}+v(x,t)$. Make sure that you understand that $v(x,t)$ solves the problem $$ v_t=v_{xx},\\ v(0,t)=0,\\ v(1,t)=0,\\ v(x,0)=-u_{st}. $$ Note the initial condition.

Now you can solve it to find $$ v(x,t)=-\frac{1}{\pi^2}e^{-\pi^2 t}\sin \pi x-\frac{1}{2^2\pi^2}e^{-2^2\pi^2t}\sin 2\pi x. $$ Finally write $$ u(x,t)=\underbrace{v(x,t)}_{\mbox{transient part}}+\underbrace{u_{st}}_{\mbox{stationary part}} $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.