Find u such that−u&nbsp;″(x)+b(x)u&nbsp;′(x)+c(x)u(x)=f(x)&nbsp;&nbsp;&nbsp;in&nbsp;&nbsp;(0,1),and conditions&nbsp;u(0)=u(1)=0, whereb(x)=x2,&nbsp;c(x)=1+x,&nbsp;f(x)=−2+13x2+3x3−x4−5x5

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Find u such that−u&amp;nbsp;″(x)+b(x)u&amp;nbsp;′(x)+c(x)u(x)=f(x)&amp;nbsp;&amp;nbsp;&amp;nbsp;in&amp;nbsp;&amp;nbsp;(0,1),and conditions&amp;nbsp;u(0)=u(1)=0, whereb(x)=x2,&amp;nbsp;c(x)=1+x,&amp;nbsp;f(x)=−2+13x2+3x3−x4−5x5

anita1415snck · Accepted Answer

Step 1Looking at the last scheme, the approximations being used are are follows:u″(xi)=1h2(u(xi+1)−2u(xi)+u(xi−1))−h212u(4)(ξi)u′(xi)=12h(u(xi+1)−u(xi−1))+h26u(3)(ζi)Step 2So, when you write the equation at x=xi, and denote Wi=u(xi), you get−dWi+1−2Wi+Wi−1h2+bidWi+1−Wi−12h+ciWi=fi+Eiwhere Ei=−h214u(4)(ξi)−h26⋅bi,u(3)(ζi)Step 3Since the system your will actually solve does not include the Ei terms, the global error can now be estimated by standard perturbation analysis on the rhs of a linear system. Basically, you are solving some system AW=F, instead of A(W+δW)=F+δF, which leads to an (absolute) error estimate |δW|≤|A−1||δF|.

enchantsyseq · Accepted Answer

Explanation:The classical way to get to the truncation error at point i is to expand the terms Ui±1 in a Taylor series around that point i. As a practical advice, you can first just do linear expansion, see if only the desired terms persist (by comparing to the PDE) and then include also higher order terms (quadratic) and so on. At some point there are higher-order derivatives arising which are not vanishing anymore. This is your truncation error, which is accompanied with some power of h, which then defines the order of your local truncation error.

Find u such that PSK \(\displaystyle-{u}^{{''}}{\left({x}\right)}+{b}{\left({x}\right)}{u}^{{'}}{\left({x}\right)}+{c}{\left({x}\right)}{u}{\left({x}\right)}={f{{\left({x}\right)}}}\ \ \text{ in

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