A one-dimensional domain is discretized into N sub-domains of width Dx with node numbers i = 0, 1, 2, 3…………, N. If the time scale is discretized in steps of Dt, the forward-time and centered-space finite difference approximation at ith node and nth time step, for the partial differential equation $\frac{{\partial v}}{{\partial t}} = \beta \frac{{{\partial ^2}v}}{{\partial {x^2}}}$ is

Question

A one-dimensional domain is discretized into N sub-domains of width Dx with node numbers i = 0, 1, 2, 3…………, N. If the time scale is discretized in steps of Dt, the forward-time and centered-space finite difference approximation at ith node and nth time step, for the partial differential equation $\frac{{\partial v}}{{\partial t}} = \beta \frac{{{\partial ^2}v}}{{\partial {x^2}}}$ is

TuteeHUB · Accepted Answer

$\frac{{V_i^{\left( n \right)} - V_i^{\left( n \right)}}}{{{\rm{\Delta }}t}} = \beta \left[ {\frac{{V_{i + 1}^{\left( n \right)} - 2V_i^{\left( n \right)} + V_{i - 1}^{\left( n \right)}}}{{{{\left( {{\rm{\Delta }}x} \right)}^2}}}} \right]$

TuteeHUB · Answer

$\frac{{V_{i + 1}^{\left( {n + 1} \right)} - V_i^{\left( N \right)}}}{{{\rm{\Delta }}t}} = \beta \left[ {\frac{{V_{i + 1}^{\left( n \right)} - 2V_i^{\left( n \right)} + V_{i - 1}^{\left( n \right)}}}{{2{\rm{\Delta }}x}}} \right]$

TuteeHUB · Answer

$\frac{{V_i^{\left( n \right)} - V_i^{\left( {n - 1} \right)}}}{{2{\rm{\Delta }}t}} = \beta \left[ {\frac{{V_{i + 1}^{\left( n \right)} - 2V_i^{\left( n \right)} + V_{i - 1}^{\left( n \right)}}}{{2{\rm{\Delta }}x}}} \right]$

TuteeHUB · Answer

$\frac{{V_i^{\left( {n + 1} \right)} - V_i^{\left( n \right)}}}{{{\rm{\Delta }}t}} = \beta \left[ {\frac{{V_{i + 1}^{\left( n \right)} - 2V_i^{\left( n \right)} + V_{i - 1}^{\left( n \right)}}}{{{{\left( {{\rm{\Delta }}x} \right)}^2}}}} \right]$

1.	A one-dimensional domain is discretized into N sub-domains of width Dx with node numbers i = 0, 1, 2, 3…………, N. If the time scale is discretized in steps of Dt, the forward-time and centered-space finite difference approximation at ith node and nth time step, for the partial differential equation \(\frac{{\partial v}}{{\partial t}} = \beta \frac{{{\partial ^2}v}}{{\partial {x^2}}}\) is
A.	\(\frac{{V_{i + 1}^{\left( {n + 1} \right)} - V_i^{\left( N \right)}}}{{{\rm{\Delta }}t}} = \beta \left[ {\frac{{V_{i + 1}^{\left( n \right)} - 2V_i^{\left( n \right)} + V_{i - 1}^{\left( n \right)}}}{{2{\rm{\Delta }}x}}} \right]\)
B.	\(\frac{{V_i^{\left( n \right)} - V_i^{\left( {n - 1} \right)}}}{{2{\rm{\Delta }}t}} = \beta \left[ {\frac{{V_{i + 1}^{\left( n \right)} - 2V_i^{\left( n \right)} + V_{i - 1}^{\left( n \right)}}}{{2{\rm{\Delta }}x}}} \right]\)
C.	\(\frac{{V_i^{\left( {n + 1} \right)} - V_i^{\left( n \right)}}}{{{\rm{\Delta }}t}} = \beta \left[ {\frac{{V_{i + 1}^{\left( n \right)} - 2V_i^{\left( n \right)} + V_{i - 1}^{\left( n \right)}}}{{{{\left( {{\rm{\Delta }}x} \right)}^2}}}} \right]\)
D.	\(\frac{{V_i^{\left( n \right)} - V_i^{\left( n \right)}}}{{{\rm{\Delta }}t}} = \beta \left[ {\frac{{V_{i + 1}^{\left( n \right)} - 2V_i^{\left( n \right)} + V_{i - 1}^{\left( n \right)}}}{{{{\left( {{\rm{\Delta }}x} \right)}^2}}}} \right]\)
Answer» D. \(\frac{{V_i^{\left( n \right)} - V_i^{\left( n \right)}}}{{{\rm{\Delta }}t}} = \beta \left[ {\frac{{V_{i + 1}^{\left( n \right)} - 2V_i^{\left( n \right)} + V_{i - 1}^{\left( n \right)}}}{{{{\left( {{\rm{\Delta }}x} \right)}^2}}}} \right]\)

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