Spoony
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Homework Statement
I have the damped wave equation;
u_{tt} = 4 u_{xx} -2 u_{t}
which is to be solved on region 0 < x < 2
with boundary conditions;
u(0,t) = 2, u(2,t) = 1.
i must;
1) find steady state solution u_{steady}(x) and apply boundary conditions.
2) find \theta(x,t) = u(x,t) - u_{steady} and find PDE and boundary conditions obeyed by theta.
3) use theta = f(t)g(x) to obtain 2 separate ODE's with separation constant
4) then i must solve the homogeneous problem for theta and then recover the general solution for u
5) finally i must find the solution for u using intial conditions
u(x,0) = 2 - x/2 , u_{t} (x,0) = x^{2}(2-x)
and I am given
(this is between 2 and 0) \int x^{2} (2-x) sin(\frac{n\pi}{2} x) dx = \frac{32(1+2(-1)^{n}}{n^{3}\pi^{3}}
The Attempt at a Solution
1) steady state solution u_{steady} implies u_{t} = 0 which reduces the PDE to 4 u_{xx} = 0 \rightarrow u_{xx} = 0 after intergration, u_{steady} = ax + b applying boundary conditions leads to u_{steady} = - \frac{1}{2}x + 2
2) As
\theta (x,t) = u(x,t) - u_{st}
and u_{st} is a function of x only then \theta _{t} = \theta _{tt} = 0 . Using the fact that u_{st} = - \frac{1}{2}x + 2 leads to \frac{\partial^{2} u_{st}}{\partial x^{2}} = 0 and therefore theta_{xx} = 0
which doesn't affect the PDE for u, only changes u to theta.
Oh and both boundary conditions become 0.
3) Inserting
\theta (x,t) = f(t)g(x) into the PDE leads to;
f''(t)g(x) = 4f(t)g''(x) - 2g(x)f'(t)
reranging leads to;
\frac{f''(t) + 2f(t)}{4f(t)} = \frac{g''(x)}{g(x)}
And each side is a function of its argument only and are equal to each other, so each side equals the same constant, which for the sake of this question must be -\Lambda to create a trig solution for g(x) (need trig solution as boundary conditions require that g must equal 0 in 2 places g(0) = g(2) = 0) .
4) this is where i start to struggle, the ODE for g(x) is;
(note; i have used \Lambda = \lambda ^{2}
g''(x) + \Lambda g(x) = 0 which leads to g(x) = Acos(\lambda x) + Bsin(\lambda x) with boundary conditions g(0) = g(2) = 0.
Using those boundary conditions, A = 0 and B = 0 OR sin(2 \lambda) = 0
B=0 is trivial case, so it is ignored, if sin(2 \lambda) = 0 then; 2 \lambda = n \pi which leads to \lambda = \frac{n \pi}{2}
this is all fine and dandy, but then when i look at the ODE for f(t)
f''(t) + 2f'(t) + 4 \Lambda f(t) = 0 and applying the value for \Lambda = \lambda ^{2} leads to
f''(t) + 2f'(t) + 4 \frac{n^{2} \pi ^{2}}{4} f(t) = 0
Once i cancel the 4 I am left with an ODE, i try using A e^{mx} as a solution but it leaves me with;
m^{2} + 2m = -n^{2} \pi ^{2} which has no solutions for m unless m = 0, so I am stuck there, i need to find the separation constant such that i can get solutions to both of these ODE's :(
Help is really appreciated, i think its the fact that there is an extra partial u by t in there, only examples we have really are when both partials are second order.
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