As an example of a hyperbolic equation we study the wave equation. One of the systems it can describe is a transmission line for high frequency signals, 40m long.
Separate variables,
|
V (x,t) = X(x)T(t).
| (5.19) |
We find
|
{X''\over
X} = LC{T''\over
T} = −λ.
| (5.20) |
Which in turn shows that
We can also separate most of the initial and boundary conditions; we find
|
X'(0) = X'(40) = 0,\kern 2.77695pt \kern 2.77695pt T'(0) = 0.
| (5.22) |
Once again distinguish the three cases λ > 0,
λ = 0, and
λ < 0:
λ > 0 (almost identical to
previous problem) {λ}_{n} = {α}_{n}^{2},
{α}_{n} = {nπ\over
40} ,
{X}_{n} =\mathop{ cos}\nolimits ({α}_{n}x). We find
that
|
{T}_{n}(t) = {D}_{n}\mathop{ cos}\nolimits \left ( {nπt\over
40\sqrt{LC}}\right ) + {E}_{n}\mathop{ sin}\nolimits \left ( {nπt\over
40\sqrt{LC}}\right ).
| (5.23) |
T'(0) = 0 implies {E}_{n} = 0, and taking both together we find (for n ≥ 1)
|
{V }_{n}(x,t) =\mathop{ cos}\nolimits \left ( {nπt\over
40\sqrt{LC}}\right )\mathop{cos}\nolimits \left ({nπx\over
40} \right ).
| (5.24) |
λ = 0 X(x) = A + Bx. B = 0 due to the boundary conditions. We find that T(t) = Dt + E, and D is 0 due to initial condition. We conclude that
|
{V }_{0}(x,t) = 1.
| (5.25) |
λ < 0 No solution.
Taking everything together we find that
|
V (x,t) = {{a}_{0}\over
2} +{ \mathop{∑
}}_{n=1}^{∞}{a}_{
n}\mathop{ cos}\nolimits \left ( {nπt\over
40\sqrt{LC}}\right )\mathop{cos}\nolimits \left ({nπx\over
40} \right ).
| (5.26) |
The one remaining initial condition gives
|
V (x,0) = f(x) = {{a}_{0}\over
2} +{ \mathop{∑
}}_{n=1}^{∞}{a}_{
n}\mathop{ cos}\nolimits \left ({nπx\over
40} \right ).
| (5.27) |
Use the Fourier cosine series (even continuation of f) to find