Answer to Question #128946 in Differential Equations for Sehrish

Question #128946
Solve the boundary value problem:du/dt=βd^2u/dx^2,0<x<l,t>0
U(0,t)=0,u(l,t)=0,t>0
U(x,0)=f(x)0<x<l
1
Expert's answer
2020-08-12T18:02:33-0400

"u = u(x,t)=a(x)b(t)\\neq0.\\\\\na(x)b'(t)=\\beta a''(x)b(t)\\Rightarrow\n\\\\ \\frac{b'(t)}{\\beta b(t)}=\\frac{a''(x)}{a(x)}=-(\\frac{\\pi n}{l})^2 \\Rightarrow\\\\\nb(t)=e^{-\\beta \\cdot (\\frac{\\pi n}{l})^2 \\cdot t},\\quad a(x)=c_1sin(\\frac{\\pi n}{l}x)+c_2cos(\\frac{\\pi n}{l}x)\\\\\nu(0,t)=u(l,t)=0 \\Rightarrow c_2=0 \\Rightarrow u_n(x,t)=e^{-\\beta \\cdot (\\frac{\\pi n}{l})^2 \\cdot t}\\cdot sin(\\frac{\\pi n}{l}x).\\\\\nU(x,t)=\\sum\\limits_{n=0}^\\infty a_nu_n(x,t)=\\sum\\limits_{n=0}^\\infty a_n\\cdot e^{-\\beta \\cdot (\\frac{\\pi n}{l})^2 \\cdot t}\\cdot sin(\\frac{\\pi n}{l}x).\\\\\nU(x,0)=f(x)=\\sum\\limits_{n=0}^\\infty a_n\\cdot sin(\\frac{\\pi n}{l}x) \\Rightarrow\na_n=\\frac{2}{l}\\int\\limits_0^lf(t)sin(\\frac{\\pi n}{l}t)dt\\Rightarrow\\\\\nU(x,t)=\\sum\\limits_{n=0}^\\infty a_nu_n(x,t)=\\sum\\limits_{n=0}^\\infty e^{-\\beta \\cdot (\\frac{\\pi n}{l})^2 \\cdot t}\\cdot sin(\\frac{\\pi n}{l}x) \\cdot \\frac{2}{l}\\int\\limits_0^lf(t)sin(\\frac{\\pi n}{l}t)dt."


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