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Question Number 183761 by MikeH last updated on 29/Dec/22
$Solve the differential equation for the function given by U(x,t). { (((∂U/∂t) = 2(∂^2 U/∂x^2 ) , 0 < x < π)),((U(0,t) = 0, U(π,t) = 0, t > 0)) :} U(x,0) = 25x$
$Solve the differential equation for the function$ $given by U (x, t) .$ ${\begin{cases} \frac{\partial U}{\partial t} = 2 \frac{\partial^{2} U}{\partial x^{2}}, 0 < x < π \\ U (0, t) = 0, U (π, t) = 0, t > 0 \end{cases}$ $U (x, 0) = 25 x$
	Answered by leodera last updated on 18/May/23
	$take fourier sine transform of both sides F_s {(∂u/∂t)} = 2F_s {(∂^2 u/∂x^2 )} let u_s ^− = ∫_0 ^π u(x,t)sin (sx)dx (d/dt)u_s ^− = 2[−s^2 u_s ^− + s{u(0,t) − (−1)^s u(π,t)}] (d/dt)u_s ^− = −2s^2 u_s ^− solving the D.E u_s ^− = Ae^(−2s^2 t) but u_s ^− (s,0) = ∫_0 ^π 25xsin (sx)dx = −((25π)/s)cos (sπ) u_s ^− (s,0) = Ae^(−2s^2 ×0) = −((25π)/s)cos (sπ) ∴ A = −((25π)/s)cos (sπ) ∴ u_s ^− (s,t) = −((25π)/s)cos (sπ)e^(−2s^2 t) taking inverse fourier transform F^− {u_s ^− (s,t)} = u(x,t) = (2/π)Σ_(s=1) ^∞ −((25π)/s)cos (sπ)e^(−2s^2 t) sin (sx) u(x,t) = −50Σ_(s=1) ^∞ (((−1)^s )/s)e^(−2p^2 t) sin (sx)$
	$t a k e f o u r i e r s i n e t r a n s f o r m o f b o t h s i d e s$ $F_{s} {\frac{\partial u}{\partial t}} = 2 F_{s} {\frac{\partial^{2} u}{\partial x^{2}}}$ $l e t {\bar{u}}_{s} = \int_{0}^{π} u (x, t) \sin (s x) d x$ $\frac{d}{d t} {\bar{u}}_{s} = 2 [- s^{2} {\bar{u}}_{s} + s {u (0, t) - {(- 1)}^{s} u (π, t)}]$ $\frac{d}{d t} {\bar{u}}_{s} = - 2 s^{2} {\bar{u}}_{s}$ $s o l v i n g t h e D . E$ ${\bar{u}}_{s} = {A e}^{- 2 s^{2} t}$ $b u t {\bar{u}}_{s} (s, 0) = \int_{0}^{π} 25 x \sin (s x) d x$ $= - \frac{25 π}{s} \cos (s π)$ ${\bar{u}}_{s} (s, 0) = {A e}^{- 2 s^{2} \times 0} = - \frac{25 π}{s} \cos (s π)$ $∴ A = - \frac{25 π}{s} \cos (s π)$ $∴ {\bar{u}}_{s} (s, t) = - \frac{25 π}{s} \cos (s π) e^{- 2 s^{2} t}$ $t a k i n g i n v e r s e f o u r i e r t r a n s f o r m$ $F^{-} {{\bar{u}}_{s} (s, t)} = u (x, t) = \frac{2}{π} \underset{s = 1}{\sum^{\infty}} - \frac{25 π}{s} \cos (s π) e^{- 2 s^{2} t} \sin (s x)$ $u (x, t) = - 50 \underset{s = 1}{\sum^{\infty}} \frac{{(- 1)}^{s}}{s} e^{- 2 p^{2} t} \sin (s x)$
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