Other approach: Implicit Euler
Explicit Euler
- \(u^{k+1}=u^k+h\,\mathcal{F}(u^{\color{red}{k}},t^{\color{red}{k}})\)
- \(u^k\) is known to compute \(u^{k+1}\)
- - In the linear case
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- \(\displaystyle u^{k+1} = (\mathrm{I}+h\,\mathrm{A})\,u^k+h\,b(t^k)\)
Implicit Euler
- \(u^{k+1}=u^k+h\,\mathcal{F}(u^{\color{red}{k+1}},t^{\color{red}{k+1}})\)
- \(u^{k+1}\) appears in RHS
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- \(\Rightarrow\) Need to solve \(\displaystyle u^{k+1}-h\,\mathcal{F}(u^{k+1},t^{k+1}) = u^k\)
- - In the linear case : solve a linear system
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- \(\displaystyle u^{k+1}=u^{k}+h\,\left(\mathrm{A} u^{k+1}+b(t^{k+1})\right)\)
- \(\Rightarrow \displaystyle u^{k+1} = (\mathrm{I}-h\,\mathrm{A})^{-1}\,\left(u^k+h\,b(t^{k+1})\right)\)