Consider the heat problem $k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial u } { \partial t } , 0

Question

Consider the heat problem $k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial u } { \partial t } , 0 < x < \infty , t > 0 , u ( x , 0 ) = 0 , u ( 0 , t ) = u _ { 0 }$ . Apply a Fourier sine transform. The resulting problem for $U ( \alpha , t ) = \mathcal { F } _ { s } \{ u ( x , t ) \}$ is
A) $U _ { t } = - k \alpha U + k \alpha u _ { 0 } , U ( \alpha , 0 ) = 0$
B) $U _ { t } = - k \alpha ^ { 2 } U - k \alpha t _ { 0 } , U ( \alpha , 0 ) = 0$
C) $U _ { t } = - k \alpha ^ { 2 } U + k \alpha t _ { 0 } , U ( \alpha , 0 ) = 0$
D) $U _ { t } = k \alpha ^ { 2 } U + k \alpha u _ { 0 } , U ( \alpha , 0 ) = 0$
E) $U _ { t } = k \alpha ^ { 2 } U - k \alpha u _ { 0 } , U ( \alpha , 0 ) = 0$

Quizplus · Accepted Answer

Applying the Fourier sine transform to the given heat equation, we get:$$\mathcal{F}_s\left[k\frac{\partial^2 u}{\partial x^2}\right] = \mathcal{F}_s\left[\frac{\partial u}{\partial t}\right]$$Using the properties of Fourier sine transform, we get:$$-k\alpha^2 U(\alpha, t) = U_t(\alpha, t)$$With the initial condition, we have:$$U(\alpha, 0) = \mathcal{F}_s[u(x, 0)] = 0$$Therefore, the correct answer is C: $U_t = - k \alpha^2 U + k \alpha u_0, U(\alpha, 0) = 0$.

Consider the Heat Problem Apply a Fourier Sine Transform U(α,t)=Fs{u(x,t)}U ( \alpha , t ) = \mathcal { F } _ { s } \{ u ( x , t ) \}U(α,t)=Fs​{u(x,t)}

Consider the Heat Problem Apply a Fourier Sine Transform $U ( \alpha , t ) = \mathcal { F } _ { s } \{ u ( x , t ) \}$