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Consider the following PDE. Where $u(x,t) = X(x)T(t)$

$u_{tt}+u_t = u_{xx}$

$u(0,t)=u(\pi,t)=0$

$u(x,0)=0$

$u_t(x,0)=10$

I am having trouble solving for my $T(t)$, it comes down to an ODE

So just to save some time and work, for my eigenfunctions, I got $X_n(x) = \sin(nx)$ with $\lambda = n$.

$T''+T'+n^2T = 0$

$T(0)=0$

$T'(0)=10$

Now I am having trouble writing my solution as hyperbolic functions. My roots from auxilarily equation is $r=\dfrac{-1 \pm \sqrt{1-4n^2}}{2}$. Any idea on how to get rid of that $\frac{1}{2}$ in front?

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In fact for any Telegraph-type equation, Klein-Gordon-type equation and Helmholtz-type equation, they both have the characteristic that you are impossible to accommodate all of the function parts of the independent variables to get the nice form. –  doraemonpaul Nov 24 '12 at 11:10

1 Answer 1

up vote 1 down vote accepted

Note that $n \geq 1$. Hence, $1-4n^2 < 0$. Let $1-4n^2 = -k_n^2$. This gives us that $r = -\dfrac12 \pm i\dfrac{k_n}2$. Then $T_n(t)$ is of the form $$T_n(t) = \exp(-t/2) \left(A_n \cos(k_nt/2) + B_n \sin(k_n t/2) \right)$$ $T_n(0) = 0 \implies A_n = 0$. Hence, $$T_n(t) = B_n \exp(-t/2) \sin(k_n t/2)$$ $$T_n'(t) = -1/2 B_n \exp(-t/2) \sin(k_n t/2) + B_n k_n/2 \exp(-t/2) \cos(k_n t/2)$$ $$T_n'(0) = B_n k_n/2 = 10 \implies B_n = \dfrac{20}{k_n}$$ Hence, $$T_n(t) = \exp(-t/2) \dfrac{20 \sin(k_n t/2)}{k_n}$$

EDIT

Your final solution is now of the form $$u(x,t) = 20 \exp(-t/2) \left(\sum_{n=1}^{\infty} \dfrac{\sin(k_n t/2) \sin(nx)}{k_n} \right)$$

EDIT

If you get the roots as $r=a,b \in \mathbb{R}$, then the solution is of the form $$c_1 \exp(at) + c_2 \exp(bt)$$ which can be rewritten as $$\exp \left(\dfrac{a+b}2t \right) \left(k_1 \cosh \left(\dfrac{(a-b)t}2 \right) + k_2 \sinh \left(\dfrac{(a-b)t}2 \right) \right)$$

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Oh shoot, that was an oversight...now I have a very stupid question to follow. Suppose $0 < n < 1/4$, how do I write the solution in hyperbolic form? –  sidht Nov 24 '12 at 6:58
    
@sizz Your $n$ is always $\geq 1$ since you need $X(\pi) = 0$. This gives us $\sin(n \pi) = 0 \implies n \in \mathbb{Z}$. $n=0$ gives the solution as $0$. –  user17762 Nov 24 '12 at 7:00
    
No, no, no, I know the restrictions on my 'n's, but now I am just solely interested in an eigenvalue problem where we get something like a distinct root like I mentioned and I want to write it as a hyperbolic sum –  sidht Nov 24 '12 at 7:07
    
Wonderful, thank you –  sidht Nov 24 '12 at 7:25

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