Consider the problem $c ^ { 2 } \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } , u ( 0 , t ) = 0 , u ( 1 , t ) = 0 , u ( x , 0 ) = \sin ( \pi x ) , u _ { t } ( x , 0 ) = g ( x )$ . Replace $\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } }$ with a central difference approximation with $h = 1 / 4$ and $\frac { \partial ^ { 2 } u } { \partial t ^ { 2 } }$ with a central difference approximation with $k = 1 / 3$ The resulting equation is
A) $c ^ { 2 } [ u ( x + h , t ) - 2 u ( x , t ) + u ( x - h , t ) ] / h ^ { 2 } = ( u ( x , t + k ) - u ( x , t ) + u ( x , t - k ) ) / k ^ { 2 }$
B) $c ^ { 2 } [ u ( x + h , t ) - 2 u ( x , t ) + u ( x - h , t ) ] / h ^ { 2 } = ( u ( x , t + k ) - 2 u ( x , t ) + u ( x , t - k ) ) / k ^ { 2 }$
C) $c ^ { 2 } [ u ( x + h , t ) - 2 u ( x , t ) + u ( x - h , t ) ] / h ^ { 2 } = ( u ( x , t + k ) - u ( x , t ) ) / k ^ { 2 }$
D) $c ^ { 2 } [ u ( x + h , t ) - 2 u ( x , t ) + u ( x - h , t ) ] / h ^ { 2 } = ( u ( x , t + k ) + u ( x , t ) ) / k$
E) $c ^ { 2 } [ u ( x + h , t ) - 2 u ( x , t ) + u ( x - h , t ) ] / h ^ { 2 } = ( u ( x , t + k ) - u ( x , t ) ) / k$

Question

Quizplus · Accepted Answer

The correct central difference approximations for the second derivatives are $\frac{u(x+h,t) - 2u(x,t) + u(x-h,t)}{h^2}$ for space and $\frac{u(x,t+k) - 2u(x,t) + u(x,t-k)}{k^2}$ for time, which matches option B.

Consider the Problem Replace ∂2u∂x2\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } }∂x2∂2u​

Consider the Problem Replace $\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } }$