Science:Math Exam Resources/Courses/MATH257/December 2011/Question 04 (b)

MATH257 December 2011

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  •  December 2011  •  

Question 04 (b)
Consider the following problem for the heat equation with a time-dependent source term, and mixed boundary conditions: $u_{t}=u_{xx}+t,\quad 0<x<1,\ t>0,$ $u_{x}(0,t)=0,\quad u(1,t)=0,\quad u(x,0)=1.$ Find the solution to this problem using the method of eigenfunction expansion.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
How can we formulate this problem in terms of eigenfunctions of a similar problem?

Hint 2
Consider the related problem $u_{t}=u_{xx}$ first.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We must solve this eigenfunction expansion. First, following the hints, we consider the problem without a source term $u_{t}=u_{xx}$ and attempt a separation of variables $u=\Phi (t)\Psi (x)$: $\Psi \Phi _{t}=\Phi \Psi _{xx},\quad {\text{ which can only be true if }}\quad {\frac {\Phi _{t}}{\Phi }}={\frac {\Psi _{xx}}{\Psi }}=-\lambda ,$ for some fixed constant $\lambda$. Let's start with the x-equation. $\Psi _{xx}=-\lambda \Psi ,\qquad \Psi _{x}(0)=\Psi (1)=0.$ From the boundary conditions we require $\lambda$ to be positive. Therefore $\Psi =A\sin({\sqrt {\lambda }}x)+B\cos({\sqrt {\lambda }}x).$ To plug in the boundary conditions we first calculate the derivative of $\Psi$: $\Psi _{x}=A{\sqrt {\lambda }}\cos({\sqrt {\lambda }}x)-B{\sqrt {\lambda }}\sin({\sqrt {\lambda }}x),$ so that $\Psi _{x}(0)=A{\sqrt {\lambda }}$. From the boundary condition $\Psi _{x}(0)=0$, we get A = 0. (The possibility that $\lambda =0$ is ruled out by the other boundary condition). Similarly, from $0=\Psi (1)=B\cos({\sqrt {\lambda }})$, we get ${\sqrt {\lambda }}={\frac {(2n+1)\pi }{2}},\quad n=0,1,2,\dots$ Hence, the eigenfunction for $\Psi _{xx}=-\lambda \Psi ,\quad \Psi _{x}(0)=\Psi (1)=0$ is given (after suppressing the arbitrary constant B) by $\Psi _{n}=\cos \left({\frac {(2n+1)\pi x}{2}}\right).$ At this point we have $u=\Phi (t)\Psi (x)$ where we normally would find $\Phi$ for this problem. This would lead us to a solution of the form: $u=\sum _{n=0}^{\infty }A_{n}\cos \left({\frac {(2n+1)\pi x}{2}}\right)\Phi _{n}(t).$ However, recall that the problem we really want to solve is $u_{t}=u_{xx}+t$, not just $u_{t}=u_{xx}$. We use the same eigenfunctions for x and now allow our coefficients $A_{n}$ to depend on t (which subsumes the $\Phi _{n}(t)$ terms as well). Thus $u(x,t)=\sum _{n=0}^{\infty }A_{n}(t)\Psi _{n}(x)$ so that $u_{t}-u_{xx}=t$ becomes $\sum _{n=0}^{\infty }A_{n}'\Psi _{n}-A_{n}(\Psi _{n})_{xx}=t.$ Recall that $(\Psi _{n})_{xx}=-\lambda _{n}\Psi _{n}$ so we get $\sum _{n=0}^{\infty }(A_{n}'+\lambda _{n}A_{n})\Psi _{n}=t$ Further, recall that our eigenfunctions are orthogonal $\int _{0}^{1}\Psi _{n}\Psi _{m}\,dx={\begin{cases}0,&n\not =m,\\{\frac {1}{2}},&n=m,\end{cases}}$ so we can multiply both sides by $\Psi _{n}$ and then integrate over x to get $(A_{n}'+\lambda _{n}A_{n}){\frac {1}{2}}=t\int _{0}^{1}\Psi _{n}\,dx=\left.{\frac {t}{\sqrt {\lambda }}}\sin \left({\sqrt {\lambda }}x\right)\right|_{0}^{1}={\frac {t}{\sqrt {\lambda }}}\sin({\sqrt {\lambda }}).$ Also recall that ${\sqrt {\lambda _{n}}}={\frac {(2n+1)\pi }{2}}$ so that $\sin {\sqrt {\lambda _{n}}}=(-1)^{n}.$ Therefore, $A_{n}'+\lambda _{n}A_{n}={\frac {2(-1)^{n}}{\sqrt {\lambda }}}t.$ We can solve this by integrating factor, ${\frac {d}{dt}}\left(A_{n}e^{\lambda _{n}t}\right)={\frac {2(-1)^{n}}{\sqrt {\lambda }}}te^{\lambda _{n}t}.$ Recall that $\int te^{\lambda _{n}t}\,dt=(t-1)e^{\lambda _{n}t}/\lambda _{n}+C$ and hence $A_{n}e^{\lambda _{n}t}={\frac {2(-1)^{n}}{\sqrt {\lambda }}}\left(\left({\frac {t-1}{\lambda _{n}}}\right)e^{\lambda _{n}t}+C\right)$ so that $A_{n}(t)={\tilde {C}}e^{-\lambda _{n}t}+{\frac {2(-1)^{n}}{\sqrt {\lambda }}}\left({\frac {t-1}{\lambda _{n}}}\right),$ where we have redefined the constant. To solve for the arbitrary constant recall that $u(x,0)=\sum _{n=0}^{\infty }A_{n}(0)\Psi _{n}(x)=1.$ We use orthogonality once more to obtain $A_{n}(0)=2\int _{0}^{1}\Psi _{n}\,dx={\frac {2\sin {\sqrt {\lambda _{n}}}}{\sqrt {\lambda _{n}}}}={\frac {2(-1)^{n}}{\sqrt {\lambda _{n}}}}.$ Using the above we get $A_{n}(0)={\tilde {C}}-{\frac {2(-1)^{n}}{{\sqrt {\lambda _{n}}}\lambda _{n}}}={\frac {2(-1)^{n}}{\sqrt {\lambda _{n}}}},$ which implies ${\tilde {C}}={\frac {2(-1)^{n}}{\sqrt {\lambda _{n}}}}(1+1/\lambda _{n}).$ Hence, we finally conclude that $u(x,t)=\sum _{n=0}^{\infty }{\frac {2(-1)^{n}}{\sqrt {\lambda _{n}}}}\left[\left(1+{\frac {1}{\lambda _{n}}}\right)e^{-\lambda _{n}t}+{\frac {t-1}{\lambda _{n}}}\right]\cos \left({\sqrt {\lambda _{n}}}x\right),$ for $\lambda _{n}={\frac {(2n+1)\pi }{2}}.$