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

Question 11

Multiple Choice

Consider the problem c2∂2u∂x2=∂2u∂t2,u(0,t) =0,u(1,t) =0,u(x,0) =sin⁡(πx) ,ut(x,0) =g(x) 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 ∂2u∂x2\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } with a central difference approximation with h=1/4h = 1 / 4 and ∂2u∂t2\frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } with a central difference approximation with k=1/3k = 1 / 3 The resulting equation is


A) c2[u(x+h,t) −2u(x,t) +u(x−h,t) ]/h2=(u(x,t+k) −u(x,t) +u(x,t−k) ) /k2c ^ { 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) c2[u(x+h,t) −2u(x,t) +u(x−h,t) ]/h2=(u(x,t+k) −2u(x,t) +u(x,t−k) ) /k2c ^ { 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) c2[u(x+h,t) −2u(x,t) +u(x−h,t) ]/h2=(u(x,t+k) −u(x,t) ) /k2c ^ { 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) c2[u(x+h,t) −2u(x,t) +u(x−h,t) ]/h2=(u(x,t+k) +u(x,t) ) /kc ^ { 2 } [ u ( x + h , t ) - 2 u ( x , t ) + u ( x - h , t ) ] / h ^ { 2 } = ( u ( x , t + k ) + u ( x , t ) ) / k
E) c2[u(x+h,t) −2u(x,t) +u(x−h,t) ]/h2=(u(x,t+k) −u(x,t) ) /kc ^ { 2 } [ u ( x + h , t ) - 2 u ( x , t ) + u ( x - h , t ) ] / h ^ { 2 } = ( u ( x , t + k ) - u ( x , t ) ) / k

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