anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
it is non homogenous so I think you use solve it as a steady state separation of variables?
anonymous
  • anonymous
i don't know how to do it with the cost
anonymous
  • anonymous
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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.