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We must solve this eigenfunction expansion. First, following the hints, we consider the problem without a source term
![{\displaystyle u_{t}=u_{xx}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/b599a36d52d51a6e98a845cf0473c9ca0b3e3d32)
and attempt a separation of variables :
![{\displaystyle \Psi \Phi _{t}=\Phi \Psi _{xx},\quad {\text{ which can only be true if }}\quad {\frac {\Phi _{t}}{\Phi }}={\frac {\Psi _{xx}}{\Psi }}=-\lambda ,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/57e778bc29d16add064d7ac9587ee16ea5b6321c)
for some fixed constant .
Let's start with the x-equation.
![{\displaystyle \Psi _{xx}=-\lambda \Psi ,\qquad \Psi _{x}(0)=\Psi (1)=0.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/8d6d013ec0eaedd607fe34e59d79c084e36dd1a7)
From the boundary conditions we require to be positive. Therefore
![{\displaystyle \Psi =A\sin({\sqrt {\lambda }}x)+B\cos({\sqrt {\lambda }}x).}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/deb14ffcfd1d4ab20b02260332bf3a29f21f1a15)
To plug in the boundary conditions we first calculate the derivative of :
![{\displaystyle \Psi _{x}=A{\sqrt {\lambda }}\cos({\sqrt {\lambda }}x)-B{\sqrt {\lambda }}\sin({\sqrt {\lambda }}x),}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/3b1eb6617b8d66d24d167f9de1b4f9a84b62b0d0)
so that . From the boundary condition , we get A = 0. (The possibility that is ruled out by the other boundary condition).
Similarly, from , we get
![{\displaystyle {\sqrt {\lambda }}={\frac {(2n+1)\pi }{2}},\quad n=0,1,2,\dots }](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/3bbcf794581a445a59069d7b722cd3134191010e)
Hence, the eigenfunction for
![{\displaystyle \Psi _{xx}=-\lambda \Psi ,\quad \Psi _{x}(0)=\Psi (1)=0}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/f0ac8f8a21cd78b2c8eb60d3ee3855dc40d1e07c)
is given (after suppressing the arbitrary constant B) by
![{\displaystyle \Psi _{n}=\cos \left({\frac {(2n+1)\pi x}{2}}\right).}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/f4855602b2d2b3d17d245933e41c8cc78a537b1a)
At this point we have where we normally would find for this problem. This would lead us to a solution of the form:
However, recall that the problem we really want to solve is , not just .
We use the same eigenfunctions for x and now allow our coefficients to depend on t (which subsumes the terms as well). Thus
![{\displaystyle u(x,t)=\sum _{n=0}^{\infty }A_{n}(t)\Psi _{n}(x)}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/1bb643bf1227dfbe7547c2549b53c9e9a5a0708d)
so that becomes
![{\displaystyle \sum _{n=0}^{\infty }A_{n}'\Psi _{n}-A_{n}(\Psi _{n})_{xx}=t.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/ecc63ec88143801ff501133e373fba583178e0a0)
Recall that so we get
![{\displaystyle \sum _{n=0}^{\infty }(A_{n}'+\lambda _{n}A_{n})\Psi _{n}=t}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/5c074ba95d8f5038fcd8f06a64fe539107423619)
Further, recall that our eigenfunctions are orthogonal
![{\displaystyle \int _{0}^{1}\Psi _{n}\Psi _{m}\,dx={\begin{cases}0,&n\not =m,\\{\frac {1}{2}},&n=m,\end{cases}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/e77bf827abcee8b72c6f754ec159a81d52d1ce8c)
so we can multiply both sides by and then integrate over x to get
![{\displaystyle (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 }}).}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/44b047e15653c8df7dea416ed81fe201fc9875b4)
Also recall that so that Therefore,
![{\displaystyle A_{n}'+\lambda _{n}A_{n}={\frac {2(-1)^{n}}{\sqrt {\lambda }}}t.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/c6402175a8bfca24c71e490b64752bd938c2e230)
We can solve this by integrating factor,
![{\displaystyle {\frac {d}{dt}}\left(A_{n}e^{\lambda _{n}t}\right)={\frac {2(-1)^{n}}{\sqrt {\lambda }}}te^{\lambda _{n}t}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/b28480eb807f07a60ce6f309dd28e619772feaa3)
Recall that and hence
![{\displaystyle 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)}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/fbe020801f53bee8f6acd65f7f47139ed0212601)
so that
![{\displaystyle A_{n}(t)={\tilde {C}}e^{-\lambda _{n}t}+{\frac {2(-1)^{n}}{\sqrt {\lambda }}}\left({\frac {t-1}{\lambda _{n}}}\right),}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/e23671f0eb178c33f98aa87a0294d018f279356f)
where we have redefined the constant. To solve for the arbitrary constant recall that
![{\displaystyle u(x,0)=\sum _{n=0}^{\infty }A_{n}(0)\Psi _{n}(x)=1.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/c9a73d1a5aee039814e1aee85ca2b4e8ed98467b)
We use orthogonality once more to obtain Using the above we get
![{\displaystyle A_{n}(0)={\tilde {C}}-{\frac {2(-1)^{n}}{{\sqrt {\lambda _{n}}}\lambda _{n}}}={\frac {2(-1)^{n}}{\sqrt {\lambda _{n}}}},}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/5cc778392a8b23a92ebbe40daa87ec62e30a2a40)
which implies Hence, we finally conclude that
![{\displaystyle 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),}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/cec55b4e17d2d5d32fbb4b157ff02aa19570d2b8)
for
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