If $U ( x , s ) = \mathcal { C } \{ u ( x , t ) \} \text {, then } \mathcal { L } \left\{ \frac { \partial u } { \partial x } \right\}$ is
A) $s U ( x , s ) - u ( x , 0 )$
B) $s U ( x , s ) - u _ { x } ( x , 0 )$
C) $s U ( x , s ) - s u ( x , 0 )$
D) $U _ { x } ( x , s )$
E) $$U _ { s } ( x , s )$$

Question

Quizplus · Accepted Answer

The Laplace transform of a partial derivative with respect to $x$ does not affect the Laplace variable $s$, which is associated with the transformation with respect to time $t$. Therefore, the Laplace transform of $\frac{\partial u}{\partial x}$ is simply the partial derivative of the Laplace transform of $u(x,t)$ with respect to $x$, which is $U_x(x,s)$.

If U(x,s)=C{u(x,t)}, then L{∂u∂x}U ( x , s ) = \mathcal { C } \{ u ( x , t ) \} \text {, then } \mathcal { L } \left\{ \frac { \partial u } { \partial x } \right\}U(x,s)=C{u(x,t)}, then L{∂x∂u​}

If $U ( x , s ) = \mathcal { C } \{ u ( x , t ) \} \text {, then } \mathcal { L } \left\{ \frac { \partial u } { \partial x } \right\}$