Question

Solve the IBVP using Fourier series method utt −c2uxx = 100, 0 < x < l, t > 0 u(x, 0) = x(1 −x), 0 < x < l, ∂u ∂t (x, 0) = x, u(0, t) = u(l, t) = 0, t > 0. Show all steps even integration

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Hermann

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73 Answers

To solve this IBVP using the method of Fourier series, we first express the solution as an infinite series:

u(x, t) = \sum_{n=1}^{\infty}(A_n\cos(\omega_nt) + B_n\sin(\omega_nt))\sin(\frac{n\pi x}{l})

where

\omega_n = \frac{nc\pi}{l}

and the coefficientsA_n, B_n are to be determined.

Step 1:

Substitute the given initial conditions into the Fourier series solution:

u(x, 0) = \sum_{n=1}^{\infty} A_n \sin(\frac{n\pi x}{l}) = x(1-x)

\frac{\partial u}{\partial t}(x, 0) = \sum_{n=1}^{\infty} B_n\omega_n \sin(\frac{n\pi x}{l}) = x

Step 2:

Solve forA_n, B_n by using the orthogonality property of sine functions, we multiply both sides by \sin(\frac{m\pi x}{l}) and integrate over [0, l] :

ForA_n :

\int_{0}^{l} x(1-x) \sin(\frac{m\pi x}{l}) dx = \int_{0}^{l} (\sum_{n=1}^{\infty} A_n \sin(\frac{n\pi x}{l})) \sin(\frac{m\pi x}{l}) dx

A_m = \frac{2}{l} \int_{0}^{l} x(1-x)\sin(\frac{m\pi x}{l}) dx

ForB_n :

\int_{0}^{l} x \sin(\frac{m\pi x}{l}) dx = \int_{0}^{l} (\sum_{n=1}^{\infty} B_n \omega_n \sin(\frac{n\pi x}{l})) \sin(\frac{m\pi x}{l}) dx

B_m\omega_m = \frac{2}{l} \int_{0}^{l} x \sin(\frac{m\pi x}{l}) dx

Step 3:

Solve forA_n and B_n using the integrals defined above.

Step 4:

Substitute the obtained coefficients back into the Fourier series solution and calculate the solution foru(x, t) .

Answer: u(x, t) = \ldots

where

and the coefficients

Step 1:

Substitute the given initial conditions into the Fourier series solution:

Step 2:

Solve for

For

For

Step 3:

Solve for

Step 4:

Substitute the obtained coefficients back into the Fourier series solution and calculate the solution for

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