Solve-the-differential-equation-for-the-function-given-by-U-x-t-U-t-2-2-U-x-2-0-lt-x-lt-pi-U-0-t-0-U-pi-t-0-t-gt-0-U-x-0-25x-

Question Number 183761 by MikeH last updated on 29/Dec/22

Solve the differential equation for the function

given by U (x, t) .

{\begin{cases} \frac{\partial U}{\partial t} = 2 \frac{\partial^{2} U}{\partial x^{2}}, 0 < x < π \\ U (0, t) = 0, U (π, t) = 0, t > 0 \end{cases}

U (x, 0) = 25 x

Answered by leodera last updated on 18/May/23

t a k e f o u r i e r s i n e t r a n s f o r m o f b o t h s i d e s

F_{s} {\frac{\partial u}{\partial t}} = 2 F_{s} {\frac{\partial^{2} u}{\partial x^{2}}}

l e t {\bar{u}}_{s} = \int_{0}^{π} u (x, t) \sin (s x) d x

\frac{d}{d t} {\bar{u}}_{s} = 2 [- s^{2} {\bar{u}}_{s} + s {u (0, t) - {(- 1)}^{s} u (π, t)}]

\frac{d}{d t} {\bar{u}}_{s} = - 2 s^{2} {\bar{u}}_{s}

s o l v i n g t h e D . E

{\bar{u}}_{s} = {A e}^{- 2 s^{2} t}

b u t {\bar{u}}_{s} (s, 0) = \int_{0}^{π} 25 x \sin (s x) d x

= - \frac{25 π}{s} \cos (s π)

{\bar{u}}_{s} (s, 0) = {A e}^{- 2 s^{2} \times 0} = - \frac{25 π}{s} \cos (s π)

∴ A = - \frac{25 π}{s} \cos (s π)

∴ {\bar{u}}_{s} (s, t) = - \frac{25 π}{s} \cos (s π) e^{- 2 s^{2} t}

t a k i n g i n v e r s e f o u r i e r t r a n s f o r m

F^{-} {{\bar{u}}_{s} (s, t)} = u (x, t) = \frac{2}{π} \underset{s = 1}{\sum^{\infty}} - \frac{25 π}{s} \cos (s π) e^{- 2 s^{2} t} \sin (s x)

u (x, t) = - 50 \underset{s = 1}{\sum^{\infty}} \frac{{(- 1)}^{s}}{s} e^{- 2 p^{2} t} \sin (s x)

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