# hmk#12_s (1).pdf - MATH 241 - Partial Differential...

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MATH 241-Partial Differential Equations(Homework#12)Fall Semester, 2020M. Carchidi———————————————————————————————————————Problem#1(20 Points) -Time-Dependent Boundary ConditionsSolve completely forux,tin the regionRx,t|0x1, 0tgiven the heat equation,2ux,tx2ux,ttalong with the boundary conditions:u0,t0 andu1,tt2, and the initial condition,ux,0x.———————————————————————————————————————Problem#2(20 Points) -A Non-Homogeneous Heat EquationDetermine the specific solution to the heat equation2ux,tx26xux,ttfor a one-dimensional rod betweenx0andx1 for 0t, assuming the insulatedboundary conditions:ux,txx00andux,txx10for 0t, and the initial conditionux,00 for 0x1. Is your solution unique?Explain.———————————————————————————————————————Problem#3(20 points) -A Strange Boundary ConditionConsider the following boundary-value, initial-value problem in the regionRx,t|0x1, 0t,having partial differential equation2ux,tux,tt,boundary conditions:u0,t0andu1,txu1,tt0,and initial condition:ux,0fx.a.) (10 points) Show that the separation of variables method applied to this problem does notlead to a regular Sturm-Liouville problem.b.) (10 points) Continue with solving this problem using the method of separation ofvariables anyway, and explain why the hard part of this problem is fitting the initialcondition.x2———————————————————————————————————————
———————————————————————————————————————Problem#4(20 points) -A Heat-Like EquationSolve forux,tin the regionRx,t|0x1,0tgiven the partial differential equation,x22ux,tx2xux,tx4ux,tx2ux,tt,the boundary conditions:u0,tfinite,u1,t1and the initial condition,ux,0x21.Note that you need not evaluate the integrals in the dot product in your final solution.———————————————————————————————————————Problem#5(20 points) -A Wave-Like Equation(No Damping)Solve completely forux,tin the regionRx,t|0x1, 0tgiven the equation,x22ux,tx22xux,tx2ux,tx22ux,tt22x3along with the boundary conditions:u0,tx1andu1,t3and the initial conditions:ux,0x3andux,0tx2.Hint: You may use the fact that a general solution to the ODEx2′′x21xx2x2x0is given byxAxsinhxBxcoshx,for0AxBx,for0AxsinxBxcosx,for0forx0.

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