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anonymous

  • 5 years ago

can some the heat the equation for me? Ut=(c^2)Uxx u(t,0)=0 u(t,pi)=(pi)cost u(0,x)=x I want to know the steps

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  1. anonymous
    • 5 years ago
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    it is non homogenous so I think you use solve it as a steady state separation of variables?

  2. anonymous
    • 5 years ago
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    i don't know how to do it with the cost

  3. anonymous
    • 5 years ago
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    I think it can be done that way. So since it's non homogenous, assume u(x,t)= v(x) + w(x,t) Take the partials of your assumption so you can plug back into your pde. So plugging it back in, you get 1. dw/dt=K(v(x)'' + K(d2w/dx2). Now you plug in your boundary conditions from the question into your assumption. So 2. v(0)+w(0,t)=0, v(pi)+w(pi,t)=picost and 3. v(x)+ w(x,0)=x. Now let t-> infinity, so all of w(x,t) and its derivatives of 0. Using this to simplify our newly attained equations, we learn that: 0=v''(x) and v(0)=0, v(pi)=picost. Integrate v'' twice to solve for v(x). v(x)=c1x+c2. Plug in v(0)=0, v(pi)=picost to solve for the constants. C2=0 and C1=cost. Now you have v(x). Now plug in the values for v(x) to simplify the 3 equations i labelled back there. You will get: 1. dw/dt=Kd2w/dx2, 2. w(0,t)=0, w(pi,t)=0 (the cost is gone because v(pi)=picost so subtracting from both sides you get w(pi,t)=0), w(x,0)=x-xcost. With these new BC, the pde is now homongeous. Now you can solve this new pde as a normal separation of variables going through eigenfunctions etc. I think you can use this method, but I'm not 100% sure as I said before. This is the furtherest type of pde I know how to solve so yeah haha

  4. anonymous
    • 5 years ago
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    Sorry, the third equation, 3. w(x,0)=x-xcost is the initial condition on the last part incase you had trouble reading.

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