The problem under consideration is a linear convection PDE.<\/p>\n
\\(\\) From Farlow’s “Partial Differential Equations for Scientists and Engineers”, chapter 15 problem 3 is $$u_t = -2u_x \\quad -\\infty < x < \\infty \\quad 0<t<\\infty $$$$ u(x,0)=e^{-x^2}$$<\/p>\n
Intuition<\/strong><\/p>\n We see that the governing equation is a linear convection problem. The characteristic velocity is 2. We expect the solution to be a shifted initial condition $$u(x,t)=e^{-(x-2t)^{2}}$$<\/p>\n <\/p>\n Laplace<\/strong><\/p>\n The Laplace transform of the governing equation with respect to the temporal coordinate t\u00a0 is $$L[u_{t}]=L[-2u_{x}]$$ The Fourier transform of the governing equation with respect to spatial coordinate x is $$F(u_{t})=F(-2u_{x})$$ The solution is $$u(x,t)\u00a0\u00a0 \u00a0=\u00a0\u00a0 \u00a0e^{-(x-a)^{2}} The problem under consideration is a linear convection PDE. From Farlow’s “Partial Differential Equations for Scientists and Engineers”, chapter 15 problem 3 is $$u_t = -2u_x \\quad -\\infty < x < \\infty \\quad 0<t<\\infty $$$$ u(x,0)=e^{-x^2}$$ Intuition We see that the governing equation is a linear convection problem. The characteristic velocity is 2. We expect […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[150,11,151,149],"_links":{"self":[{"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/posts\/142"}],"collection":[{"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/comments?post=142"}],"version-history":[{"count":13,"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/posts\/142\/revisions"}],"predecessor-version":[{"id":165,"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/posts\/142\/revisions\/165"}],"wp:attachment":[{"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/media?parent=142"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/categories?post=142"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/tags?post=142"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nExpanding with definitions gives $$U(x,s)-u(x,0)=-2\\frac{dU(x,s)}{dx}$$
\nThe transform of the IC is not found in Farlow’s tables $$L[u(x,0)]\u00a0\u00a0 \u00a0=\u00a0\u00a0 \u00a0L[e^{-x^{2}}]\u00a0\u00a0 =\u00a0\u00a0 \u00a0?$$
\nThe Laplace ODE looks like the non-homogeneous equation $$\\frac{dU(x,s)}{dx}+\\frac{1}{2}sU(x,s)=\\frac{1}{2}u(x,0)$$
\nTwo difficult steps appear before we even see the inverse Laplace transform.
\nAt this point, you should bail out and try another approach.
\nFourier<\/strong><\/p>\n
\nExpanding with identities gives $$\\frac{dU(t)}{dt}=-2i\\zeta U(t)$$
\nThe solution to U(t) , being a 1st order ODE, is $$U(t)\u00a0\u00a0 \u00a0=\u00a0\u00a0 \u00a0U(0)e^{-2i\\zeta t}$$
\nConvolution $$u(x,t)\u00a0\u00a0 \u00a0=\u00a0\u00a0 \u00a0F^{-1}(U(0)e^{-2i\\zeta t})
\n=\u00a0\u00a0 \u00a0e^{-x^{2}}\\star F^{-1}(e^{-2i\\zeta t})$$
\nFinding $$e^{-2i\\zeta t}$$
\nin Farlow’s table A gives $$f(x-a)$$
\nwhen $$a=2t$$<\/p>\n
\n=\u00a0\u00a0 \u00a0e^{-(x-2t)^{2}}$$<\/p>\n","protected":false},"excerpt":{"rendered":"