Science:Math Exam Resources/Courses/MATH257/December 2011/Question 04 (b)/Solution 1

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 $\lambda _{n}={\frac {(2n+1)\pi }{2}}.$