Solution
The PDE to solve is \[ u_{tt}=ku_{xx}\] With boundary conditions\begin{align} u\left ( 0,t\right ) & =0\tag{1}\\ Ku_{x}\left ( \pi ,t\right ) & =A\nonumber \end{align}
And initial conditions\[ u\left ( x,0\right ) =0 \] The solution to example 2 section 40 is\begin{equation} U\left ( x,t\right ) =\sum _{n=1}^{\infty }B_{2n-1}\exp \left ( \frac{-\left ( 2n-1\right ) ^{2}k}{4}t\right ) \sin \left ( \frac{\left ( 2n-1\right ) x}{2}\right ) \tag{2} \end{equation} With\[ B_{2n-1}=\frac{2}{\pi }\int _{0}^{\pi }f\left ( x\right ) \sin \left ( \frac{\left ( 2n-1\right ) x}{2}\right ) dx \] Now, in this problem, we start by writing \begin{equation} u\left ( x,t\right ) =U\left ( x,t\right ) +\Phi \left ( x\right ) \tag{3} \end{equation} The function \(\Phi \left ( x\right ) \) needs to satisfy the nonhomogeneous B.C. (1). Let \[ \Phi \left ( x\right ) =c_{1}x+c_{2}\] When \(x=0\) this gives \(0=c_{2}\). Hence \(\Phi \left ( x\right ) =c_{1}x\). Taking derivative gives \(\Phi ^{\prime }\left ( x\right ) =c_{1}\). But from (1) \(K\Phi ^{\prime }\left ( \pi \right ) =A\). Hence \(c_{1}=\frac{A}{K}\). Therefore\[ \Phi \left ( x\right ) =\frac{A}{K}x \] Substituting the above back into (3) gives\[ u\left ( x,t\right ) =U\left ( x,t\right ) +\frac{A}{K}x \] But \(U\left ( x,t\right ) \) is given by (2), hence the above becomes\begin{equation} u\left ( x,t\right ) =\frac{A}{K}x+\sum _{n=1}^{\infty }B_{2n-1}\exp \left ( \frac{-\left ( 2n-1\right ) ^{2}k}{4}t\right ) \sin \left ( \frac{\left ( 2n-1\right ) x}{2}\right ) \tag{4} \end{equation} At \(t=0\), the initial conditions is \(0\). Hence the above becomes\[ -\frac{A}{K}x=\sum _{n=1}^{\infty }B_{2n-1}\sin \left ( \frac{\left ( 2n-1\right ) x}{2}\right ) \] Hence\(\ B_{2n-1}\) is the Fourier sine series of\(\ -\frac{A}{K}x\) given by\begin{align*} B_{2n-1} & =\frac{2}{\pi }\int _{0}^{\pi }\left ( -\frac{A}{K}x\right ) \sin \left ( \frac{\left ( 2n-1\right ) x}{2}\right ) dx\\ & =-\frac{2A}{\pi K}\int _{0}^{\pi }x\sin \left ( \frac{\left ( 2n-1\right ) x}{2}\right ) dx \end{align*}
Integration by parts. Let \(u=x,dv=\sin \left ( \frac{\left ( 2n-1\right ) x}{2}\right ) \), hence \(du=1\) and \(v=-\frac{2}{\left ( 2n-1\right ) }\cos \left ( \frac{\left ( 2n-1\right ) x}{2}\right ) \) and the above becomes\begin{align*} B_{2n-1} & =-\frac{2A}{\pi K}\left ( \left [ -\frac{2x}{\left ( 2n-1\right ) }\cos \left ( \frac{\left ( 2n-1\right ) x}{2}\right ) \right ] _{0}^{\pi }+\int _{0}^{\pi }\frac{2}{\left ( 2n-1\right ) }\cos \left ( \frac{\left ( 2n-1\right ) x}{2}\right ) dx\right ) \\ & =-\frac{2A}{\pi K}\left ( -\frac{2}{\left ( 2n-1\right ) }\left [ x\cos \left ( \frac{\left ( 2n-1\right ) x}{2}\right ) \right ] _{0}^{\pi }+\frac{4}{\left ( 2n-1\right ) ^{2}}\left [ \sin \left ( \frac{\left ( 2n-1\right ) x}{2}\right ) \right ] _{0}^{\pi }\right ) \\ & =-\frac{2A}{\pi K}\left ( -\frac{2\pi }{\left ( 2n-1\right ) }\cos \left ( \frac{\left ( 2n-1\right ) \pi }{2}\right ) +\frac{4}{\left ( 2n-1\right ) ^{2}}\sin \left ( \frac{\left ( 2n-1\right ) \pi }{2}\right ) \right ) \end{align*}
Since \(2n-1\) is odd, then the cosine terms above vanish and the above simplifies to\begin{align*} B_{2n-1} & =-\frac{A}{\pi K}\frac{8\left ( -1\right ) ^{n+1}}{\left ( 2n-1\right ) ^{2}}\\ & =\frac{A}{\pi K}\frac{8\left ( -1\right ) ^{n+2}}{\left ( 2n-1\right ) ^{2}}\\ & =\frac{A}{\pi K}\frac{8\left ( -1\right ) ^{n}}{\left ( 2n-1\right ) ^{2}} \end{align*}
Substituting the above in (4) gives\begin{align*} u\left ( x,t\right ) & =\frac{A}{K}x+\sum _{n=1}^{\infty }\frac{A}{\pi K}\frac{8\left ( -1\right ) ^{n}}{\left ( 2n-1\right ) ^{2}}\exp \left ( \frac{-\left ( 2n-1\right ) ^{2}k}{4}t\right ) \sin \left ( \frac{\left ( 2n-1\right ) x}{2}\right ) \\ & =\frac{A}{K}\left \{ x+\frac{8}{\pi }\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n}}{\left ( 2n-1\right ) ^{2}}\exp \left ( \frac{-\left ( 2n-1\right ) ^{2}k}{4}t\right ) \sin \left ( \frac{\left ( 2n-1\right ) x}{2}\right ) \right \} \end{align*}
Which is the result required.
Solution
The PDE is \[ v_{t}=kv_{xx}-bv \] With boundary conditions\begin{align*} v_{x}\left ( 0,t\right ) & =0\\ v_{x}\left ( c,t\right ) & =0 \end{align*}
And initial conditions\[ v\left ( x,0\right ) =f\left ( x\right ) \] Let \(v\left ( x,t\right ) =X\left ( x\right ) T\left ( t\right ) \). Substituting into the PDE gives\[ T^{\prime }X=kX^{\prime \prime }T-bXT \] Dividing by \(XT\neq 0\) gives\begin{align*} \frac{T^{\prime }}{T} & =k\frac{X^{\prime \prime }}{X}-b\\ \frac{T^{\prime }}{T}+b & =k\frac{X^{\prime \prime }}{X}\\ \frac{T^{\prime }}{kT}+\frac{b}{k} & =\frac{X^{\prime \prime }}{X}=-\lambda \end{align*}
Where \(\lambda \) is the separation constant. We obtain the boundary value eigenvalue ODE as\begin{align} X^{\prime \prime }+\lambda X & =0\tag{1}\\ X^{\prime }\left ( 0\right ) & =0\nonumber \\ X^{\prime }\left ( c\right ) & =0\nonumber \end{align}
And the time ODE as\begin{align*} \frac{T^{\prime }}{kT}+\frac{b}{k} & =-\lambda \\ T^{\prime }+\frac{b}{k}kT & =-\lambda kT\\ T^{\prime }+\frac{b}{k}kT+\lambda kT & =0\\ T^{\prime }+T\left ( b+\lambda k\right ) & =0 \end{align*}
Now we solve the space ODE (1) in order to determine the eigenvalues \(\lambda \).
Case \(\lambda <0\)
The solution to (1) becomes\begin{align*} X\left ( x\right ) & =A\cosh \left ( \sqrt{-\lambda }x\right ) +B\sinh \left ( \sqrt{-\lambda }x\right ) \\ X^{\prime } & =A\sqrt{-\lambda }\sinh \left ( \sqrt{-\lambda }x\right ) +B\sqrt{-\lambda }\cosh \left ( \sqrt{-\lambda }x\right ) \end{align*}
Satisfying \(X^{\prime }\left ( 0\right ) =0\) gives\[ 0=B\sqrt{-\lambda }\] Hence \(B=0\) and the solution becomes \(X\left ( x\right ) =A\cosh \left ( \sqrt{-\lambda }x\right ) \). Therefore \(X^{\prime }=A\sqrt{-\lambda }\sinh \left ( \sqrt{-\lambda }x\right ) \). Satisfying \(X^{\prime }\left ( c\right ) =0\) gives\[ 0=A\sqrt{\lambda }\sinh \left ( \sqrt{-\lambda }c\right ) \] But \(\sinh \) is zero only when its argument is zero, which is not the case here since \(\lambda \neq 0\). This implies \(A=0\), leading to trivial solution. Therefore \(\lambda <0\) is not possible.
Case \(\lambda =0\)
The solution to (1) becomes\begin{align*} X\left ( x\right ) & =Ax+B\\ X^{\prime } & =A \end{align*}
Satisfying \(X^{\prime }\left ( 0\right ) =0\) gives\[ 0=A \] And the solution becomes \(X\left ( x\right ) =B\). Therefore \(X^{\prime }=0\). Satisfying \(X^{\prime }\left ( c\right ) =0\) gives\[ 0=0 \] Which is valid for any \(B\). Hence choosing \(B=1\) shows that \(\lambda =0\) is valid eigenvalue with corresponding eigenfunction \(X_{0}\left ( x\right ) =1\).
Case \(\lambda >0\)
The solution to (1) becomes\begin{align*} X\left ( x\right ) & =A\cos \left ( \sqrt{\lambda }x\right ) +B\sin \left ( \sqrt{\lambda }x\right ) \\ X^{\prime } & =-A\sqrt{\lambda }\sin \left ( \sqrt{\lambda }x\right ) +B\sqrt{\lambda }\cos \left ( \sqrt{\lambda }x\right ) \end{align*}
Satisfying \(X^{\prime }\left ( 0\right ) =0\) gives\[ 0=B\sqrt{\lambda }\] Hence \(B=0\) and the solution becomes \(X\left ( x\right ) =A\cos \left ( \sqrt{\lambda }x\right ) \). Therefore \(X^{\prime }=-A\sqrt{\lambda }\sin \left ( \sqrt{\lambda }x\right ) \). Satisfying \(X^{\prime }\left ( c\right ) =0\) gives\[ 0=-A\sqrt{\lambda }\sin \left ( \sqrt{\lambda }c\right ) \] For nontrivial solution we want\begin{align} \sin \left ( \sqrt{\lambda }c\right ) & =0\nonumber \\ \sqrt{\lambda }c & =n\pi \qquad n=1,2,3,\cdots \nonumber \\ \lambda _{n} & =\left ( \frac{n\pi }{c}\right ) ^{2} \tag{2} \end{align}
And the corresponding eigenfunctions\begin{equation} X_{n}\left ( x\right ) =\cos \left ( \sqrt{\lambda _{n}}x\right ) \tag{3} \end{equation} Now that we found \(\lambda _{n}\), we can solve the time ODE \(T^{\prime }+T\left ( b+\lambda k\right ) =0\). The solution is\begin{equation} T_{n}\left ( t\right ) =e^{-\left ( b+\lambda _{n}k\right ) t} \tag{4} \end{equation} Hence the fundamental solution is \begin{align*} v_{n}\left ( x,t\right ) & =X_{n}\left ( x\right ) T_{n}\left ( t\right ) \\ & =\cos \left ( \sqrt{\lambda _{n}}x\right ) e^{-\left ( b+\lambda _{n}k\right ) t} \end{align*}
And the general solution is the superposition of all these solutions\begin{align*} v\left ( x,t\right ) & =A_{0}X_{0}T_{0}+\sum _{n=1}^{\infty }A_{n}X_{n}\left ( x\right ) T_{n}\left ( t\right ) \\ & =A_{0}e^{-bt}+\sum _{n=1}^{\infty }A_{n}\cos \left ( \sqrt{\lambda _{n}}x\right ) e^{-\left ( b+\lambda _{n}k\right ) t} \end{align*}
Which can be written as\[ v\left ( x,t\right ) =u\left ( x,t\right ) e^{-bt}\] Where \(u\left ( x,t\right ) \) is \[ u\left ( x,t\right ) =A_{0}+\sum _{n=1}^{\infty }A_{n}\cos \left ( \sqrt{\lambda _{n}}x\right ) e^{-\lambda _{n}kt}\] Which is the same as given in section 36, page 106. In the above \begin{align*} \lambda _{0} & =0\\ \lambda _{n} & =\left ( \frac{n\pi }{c}\right ) ^{2}\qquad n=1,2,3,\cdots \end{align*}
Solution
The heat PDE in spherical coordinates, assuming no dependency on \(\phi \) nor on \(\theta \) is given by\begin{align} u_{t} & =k\nabla ^{2}u\tag{1}\\ & =k\frac{1}{r}\left ( ru\right ) _{rr}\nonumber \end{align}
Where \(1<r<2\) and \(t>0\). With the boundary conditions\begin{align*} u\left ( 1,t\right ) & =0\\ u\left ( 2,0\right ) & =u_{0} \end{align*}
And initial conditions\[ u\left ( r,0\right ) =0 \]
Let \(v\left ( r,t\right ) =ru\left ( r,t\right ) \). Hence \(v_{t}=ru_{t}\) and \(\frac{1}{r}\left ( ru\right ) _{rr}=\frac{1}{r}v_{rr}\). Substituting these in(1), the PDE simplifies to\begin{equation} v_{t}=kv_{rr} \tag{2} \end{equation} And the boundary conditions \(u\left ( 1,t\right ) =0\) becomes \(v\left ( 1,t\right ) =0\) and \(u\left ( 2,0\right ) =u_{0}\) becomes \(v\left ( 2,t\right ) =2u_{0}\). And initial conditions \(u\left ( r,0\right ) =0\) becomes \(v\left ( r,0\right ) =0\). Hence the new boundary conditions\begin{align*} v\left ( 1,t\right ) & =0\\ v\left ( 2,t\right ) & =2u_{0} \end{align*}
And new initial conditions\[ v\left ( r,0\right ) =0 \] Now let \(s=r-1\). Since \(\frac{\partial r}{\partial s}=1\), then the PDE becomes \(v_{t}=kv_{ss}\). When \(r=1\), then \(s=0\) and the boundary conditions \(v\left ( 1,t\right ) =0\) becomes \(v\left ( 0,t\right ) =0\) and the boundary conditions \(v\left ( 2,t\right ) =2u_{0}\) becomes \(v\left ( 1,t\right ) =2u_{0}\). And initial conditions do not change. Hence the new problem is to solve for \(v\left ( s,t\right ) \) in\begin{align} v_{t} & =kv_{ss}\tag{3}\\ v\left ( 1,t\right ) & =0\nonumber \\ v\left ( 1,t\right ) & =2u_{0}\nonumber \\ v\left ( s,0\right ) & =0\nonumber \end{align}
With \(0<s<1\) and \(t>0\).
The PDE (3) in part(a) is now the same as result of problem 2 section 40. Hence we can use that solution for (3) which gives\[ v\left ( s,t\right ) =2u_{0}\left [ x+\frac{2}{\pi }\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n}}{n}e^{-n^{2}\pi ^{2}kt}\sin \left ( n\pi s\right ) \right ] \] Replacing \(s\) by \(r-1\) in the above gives\[ v\left ( r,t\right ) =2u_{0}\left [ \left ( r-1\right ) +\frac{2}{\pi }\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n}}{n}e^{-n^{2}\pi ^{2}kt}\sin \left ( n\pi \left ( r-1\right ) \right ) \right ] \] But \(v\left ( r,t\right ) =ru\left ( r,t\right ) \), hence \(u\left ( r,t\right ) =\frac{v}{r}\) and therefore\begin{align*} u\left ( r,t\right ) & =2u_{0}\left [ \frac{\left ( r-1\right ) }{r}+\frac{2}{\pi r}\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n}}{n}e^{-n^{2}\pi ^{2}kt}\sin \left ( n\pi \left ( r-1\right ) \right ) \right ] \\ & =2u_{0}\left [ \left ( 1-\frac{1}{r}\right ) +\frac{2}{\pi r}\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n}}{n}e^{-n^{2}\pi ^{2}kt}\sin \left ( n\pi \left ( r-1\right ) \right ) \right ] \end{align*}
Which is the result required.
Solution
Using method of eigenfunction expansion (or method of variation of parameters as the book calls it), we start by assuming the solution to the PDE \(u_{t}=ku_{xx}+q\left ( x,t\right ) \) is given by \begin{equation} u\left ( x,t\right ) =\sum _{n=1}^{\infty }a_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \tag{1} \end{equation} Where \(\Phi _{n}\left ( x\right ) \) are the eigenfunctions associated with the homogeneous PDE \(u_{t}=ku_{xx}\) with the homogeneous boundary conditions \(u\left ( 0,t\right ) =0\) and \(u\left ( c,t\right ) =0\). But we solved this homogeneous PDE before. It has eigenvalues and corresponding eigenfunctions\begin{align*} \lambda _{n} & =\left ( \frac{n\pi }{c}\right ) ^{2}\qquad n=1,2,3,\cdots \\ \Phi _{n}\left ( x\right ) & =\sin \left ( \sqrt{\lambda _{n}}x\right ) \end{align*}
Substituting (1) into the original PDE \(u_{t}=ku_{xx}+q\left ( x,t\right ) \) results in\begin{align*} \frac{\partial }{\partial t}\sum _{n=1}^{\infty }a_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) & =k\frac{\partial ^{2}}{\partial x^{2}}\sum _{n=1}^{\infty }a_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) +q\left ( x,t\right ) \\ \sum _{n=1}^{\infty }a_{n}^{\prime }\left ( t\right ) \Phi _{n}\left ( x\right ) & =k\sum _{n=1}^{\infty }a_{n}\left ( t\right ) \Phi _{n}^{\prime \prime }\left ( x\right ) +q\left ( x,t\right ) \end{align*}
But from the Sturm-Liouville ODE, we know that \(\Phi _{n}^{\prime \prime }\left ( x\right ) +\lambda _{n}\Phi _{n}\left ( x\right ) =0\). Hence \(\Phi _{n}^{\prime \prime }\left ( x\right ) =-\lambda _{n}\Phi _{n}\left ( x\right ) \) and the above reduces to\begin{equation} \sum _{n=1}^{\infty }a_{n}^{\prime }\left ( t\right ) \Phi _{n}\left ( x\right ) =-k\sum _{n=1}^{\infty }a_{n}\left ( t\right ) \lambda _{n}\Phi _{n}\left ( x\right ) +q\left ( x,t\right ) \tag{2} \end{equation} Since the eigenfunctions \(\Phi _{n}\left ( x\right ) \) are complete, we can expand \(q\left ( x,t\right ) \) using them. Therefore \[ q\left ( x,t\right ) =\sum _{n=1}^{\infty }b_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \] Substituting the above back in (2) gives\[ \sum _{n=1}^{\infty }a_{n}^{\prime }\left ( t\right ) \Phi _{n}\left ( x\right ) =-k\sum _{n=1}^{\infty }a_{n}\left ( t\right ) \lambda _{n}\Phi _{n}\left ( x\right ) +\sum _{n=1}^{\infty }b_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \] Since \(\Phi _{n}\left ( x\right ) \) are never zero, we can simplify the above to\begin{align*} a_{n}^{\prime }\left ( t\right ) & =-ka_{n}\left ( t\right ) \lambda _{n}+b_{n}\left ( t\right ) \\ a_{n}^{\prime }\left ( t\right ) +ka_{n}\left ( t\right ) \lambda _{n} & =b_{n}\left ( t\right ) \end{align*}
The above is first order ODE in \(I_{n}\left ( t\right ) \). It is linear ODE. The integrating factor is \(\mu =e^{\int k\lambda _{n}dt}=e^{k\lambda _{n}t}\). Multiplying the above ODE by this integrating factor gives\[ \frac{d}{dt}\left ( a_{n}\left ( t\right ) e^{k\lambda _{n}t}\right ) =b_{n}\left ( t\right ) e^{k\lambda _{n}t}\] Integrating both sides\begin{align*} a_{n}\left ( t\right ) e^{k\lambda _{n}t} & =\int _{0}^{t}b_{n}\left ( \tau \right ) e^{k\lambda _{n}\tau }d\tau \\ a_{n}\left ( t\right ) & =\int _{0}^{t}b_{n}\left ( \tau \right ) e^{-k\lambda _{n}\left ( t-\tau \right ) }d\tau \end{align*}
Now that we found \(a_{n}\left ( t\right ) \), we substitute it back into (1) which gives\begin{equation} u\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( \int _{0}^{t}b_{n}\left ( \tau \right ) e^{-k\lambda _{n}\left ( t-\tau \right ) }d\tau \right ) \Phi _{n}\left ( x\right ) \tag{3} \end{equation} What is left is to find \(b_{n}\left ( t\right ) \). Since \(q\left ( x,t\right ) =\sum _{n=1}^{\infty }b_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \), then by orthogonality we obtain\begin{align*} \int _{0}^{c}q\left ( x,t\right ) \Phi _{m}\left ( x\right ) dx & =\int _{0}^{c}\sum _{n=1}^{\infty }b_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \Phi _{m}\left ( x\right ) dx\\ & =\sum _{n=1}^{\infty }b_{n}\left ( t\right ) \int _{0}^{c}\Phi _{n}\left ( x\right ) \Phi _{m}\left ( x\right ) dx\\ & =b_{m}\left ( t\right ) \int _{0}^{c}\Phi _{m}^{2}\left ( x\right ) dx\\ & =b_{m}\left ( t\right ) \frac{c}{2} \end{align*}
Hence\[ b_{n}\left ( t\right ) =\frac{2}{c}\int _{0}^{c}q\left ( x,t\right ) \Phi _{m}\left ( x\right ) dx \] Substituting this back into (3) gives\begin{align} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }\left ( \int _{0}^{t}e^{-k\lambda _{n}\left ( t-\tau \right ) }\frac{2}{c}\left ( \int _{0}^{c}q\left ( x,\tau \right ) \Phi _{m}\left ( x\right ) dx\right ) d\tau \right ) \Phi _{n}\left ( x\right ) \nonumber \\ & =\frac{2}{c}\sum _{n=1}^{\infty }\left ( \int _{0}^{t}e^{-k\lambda _{n}\left ( t-\tau \right ) }\left ( \int _{0}^{c}q\left ( x,\tau \right ) \Phi _{m}\left ( x\right ) dx\right ) d\tau \right ) \Phi _{n}\left ( x\right ) \tag{4} \end{align}
If we let \[ I_{n}\left ( t\right ) =\int _{0}^{t}e^{-k\lambda _{n}\left ( t-\tau \right ) }\left ( \int _{0}^{c}q\left ( x,\tau \right ) \Phi _{m}\left ( x\right ) dx\right ) d\tau \] Then (4) becomes\[ u\left ( x,t\right ) =\frac{2}{c}\sum _{n=1}^{\infty }I_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \] Since \(\Phi _{n}\left ( x\right ) =\sin \left ( \frac{n\pi }{c}x\right ) \) then the above is\[ u\left ( x,t\right ) =\frac{2}{c}\sum _{n=1}^{\infty }I_{n}\left ( t\right ) \sin \left ( \frac{n\pi }{c}x\right ) \] Which is what required to show.
Solution
The solution in problem \(4\) above us\begin{equation} u\left ( x,t\right ) =\frac{2}{c}\sum _{n=1}^{\infty }I_{n}\left ( t\right ) \sin \left ( \frac{n\pi }{c}x\right ) \tag{1} \end{equation} Where \[ I_{n}\left ( t\right ) =\int _{0}^{t}e^{-k\lambda _{n}\left ( t-\tau \right ) }\left ( \int _{0}^{c}q\left ( x,\tau \right ) \sin \left ( \frac{n\pi }{c}x\right ) dx\right ) d\tau \] And \(\lambda _{n}=\left ( \frac{n\pi }{c}\right ) ^{2}\). Let \(c=1,k=1\) and \(q\left ( x,t\right ) =xp\left ( t\right ) \), then the above becomes\[ I_{n}\left ( t\right ) =\int _{0}^{t}e^{-n^{2}\pi ^{2}\left ( t-\tau \right ) }\left ( \int _{0}^{1}xp\left ( \tau \right ) \sin \left ( n\pi x\right ) dx\right ) d\tau \] Substituting this in (1), using \(c=1\), then (1) becomes\begin{align} u\left ( x,t\right ) & =2\sum _{n=1}^{\infty }\left ( \int _{0}^{t}e^{-n^{2}\pi ^{2}\left ( t-\tau \right ) }\left ( \int _{0}^{1}xp\left ( \tau \right ) \sin \left ( n\pi x\right ) dx\right ) d\tau \right ) \sin \left ( n\pi x\right ) \nonumber \\ & =2\sum _{n=1}^{\infty }\left ( \int _{0}^{t}p\left ( \tau \right ) e^{-n^{2}\pi ^{2}\left ( t-\tau \right ) }\left ( \int _{0}^{1}x\sin \left ( n\pi x\right ) dx\right ) d\tau \right ) \sin \left ( n\pi x\right ) \tag{2} \end{align}
But \(\int _{0}^{1}x\sin \left ( n\pi x\right ) dx\) can now be integrated by parts. Let \(u=x,dv=\sin \left ( n\pi x\right ) \), hence \(du=1,v=-\frac{\cos \left ( n\pi x\right ) }{n\pi }\) and therefore\begin{align*} \int _{0}^{1}x\sin \left ( n\pi x\right ) dx & =-\frac{1}{n\pi }\left [ x\cos \left ( n\pi x\right ) \right ] _{0}^{1}+\frac{1}{n\pi }\int _{0}^{1}\cos \left ( n\pi x\right ) dx\\ & =-\frac{1}{n\pi }\cos \left ( n\pi \right ) +\frac{1}{n\pi }\left [ \frac{\sin \left ( n\pi x\right ) }{n\pi }\right ] _{0}^{1}\\ & =-\frac{1}{n\pi }\left ( -1\right ) ^{n}+\frac{1}{n^{2}\pi ^{2}}\left [ \sin \left ( n\pi \right ) \right ] \\ & =\frac{\left ( -1\right ) ^{n+1}}{n\pi } \end{align*}
Substituting this back in (2) gives\begin{align*} u\left ( x,t\right ) & =2\sum _{n=1}^{\infty }\left ( \int _{0}^{t}p\left ( \tau \right ) e^{-n^{2}\pi ^{2}\left ( t-\tau \right ) }\left ( \frac{\left ( -1\right ) ^{n+1}}{n\pi }\right ) d\tau \right ) \sin \left ( n\pi x\right ) \\ & =\frac{2}{\pi }\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n+1}}{n}\sin \left ( n\pi x\right ) \left ( \int _{0}^{t}p\left ( \tau \right ) e^{-n^{2}\pi ^{2}\left ( t-\tau \right ) }d\tau \right ) \end{align*}
Which is the solution for problem 1.
Solution
The PDE to solve is \[ u_{t}=ku_{xx}+ax^{2}\] With boundary conditions\begin{align*} u_{x}\left ( 0,t\right ) & =0\\ u_{x}\left ( c,t\right ) & =0 \end{align*}
And initial conditions\[ u\left ( x,0\right ) =0 \] Using method of eigenfunction expansion, we start by assuming the solution to the PDE \(u_{t}=ku_{xx}+ax^{2}\) is given by \begin{equation} u\left ( x,t\right ) =\sum _{n=0}^{\infty }a_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \tag{1} \end{equation} Where \(\Phi _{n}\left ( x\right ) \) are the eigenfunctions associated with the homogeneous PDE \(u_{t}=ku_{xx}\) with the homogeneous boundary conditions \(u_{x}\left ( 0,t\right ) =0\) and \(u_{x}\left ( c,t\right ) =0\). But we solved this homogeneous PDE before. It has eigenvalues and corresponding eigenfunctions\begin{align*} \lambda _{0} & =0\\ \Phi _{0}\left ( x\right ) & =1\\ \lambda _{n} & =\frac{n^{2}\pi ^{2}}{c^{2}}\qquad n=1,2,3,\cdots \\ \Phi _{n}\left ( x\right ) & =\cos \left ( \frac{n\pi }{c}x\right ) \end{align*}
Substituting (1) into the original PDE \(u_{t}=ku_{xx}+ax^{2}\) results in\begin{align*} \frac{\partial }{\partial t}\sum _{n=0}^{\infty }a_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) & =k\frac{\partial ^{2}}{\partial x^{2}}\sum _{n=0}^{\infty }a_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) +ax^{2}\\ \sum _{n=0}^{\infty }a_{n}^{\prime }\left ( t\right ) \Phi _{n}\left ( x\right ) & =k\sum _{n=0}^{\infty }a_{n}\left ( t\right ) \Phi _{n}^{\prime \prime }\left ( x\right ) +ax^{2} \end{align*}
But from the Sturm-Liouville ODE, we know that \(\Phi _{n}^{\prime \prime }\left ( x\right ) +\lambda _{n}\Phi _{n}\left ( x\right ) =0\). Hence \(\Phi _{n}^{\prime \prime }\left ( x\right ) =-\lambda _{n}\Phi _{n}\left ( x\right ) \) and the above reduces to\begin{equation} \sum _{n=0}^{\infty }a_{n}^{\prime }\left ( t\right ) \Phi _{n}\left ( x\right ) =-k\sum _{n=0}^{\infty }a_{n}\left ( t\right ) \lambda _{n}\Phi _{n}\left ( x\right ) +ax^{2} \tag{2} \end{equation} Since the eigenfunctions \(\Phi _{n}\left ( x\right ) \) are complete, we can expand \(ax^{2}\) using them. Therefore \[ ax^{2}=\sum _{n=0}^{\infty }b_{n}\left ( x\right ) \Phi _{n}\left ( x\right ) \] Substituting the above back in (2) gives\[ \sum _{n=0}^{\infty }a_{n}^{\prime }\left ( t\right ) \Phi _{n}\left ( x\right ) =-k\sum _{n=0}^{\infty }a_{n}\left ( t\right ) \lambda _{n}\Phi _{n}\left ( x\right ) +\sum _{n=0}^{\infty }b_{n}\left ( x\right ) \Phi _{n}\left ( x\right ) \] Since \(\Phi _{n}\left ( x\right ) \) are never zero, we can simplify the above to\begin{align*} a_{n}^{\prime }\left ( t\right ) & =-ka_{n}\left ( t\right ) \lambda _{n}+b_{n}\left ( x\right ) \\ a_{n}^{\prime }\left ( t\right ) +ka_{n}\left ( t\right ) \lambda _{n} & =b_{n}\left ( x\right ) \end{align*}
The above is first order ODE in \(I_{n}\left ( t\right ) \). It is linear ODE. The integrating factor is \(\mu =e^{\int k\lambda _{n}dt}=e^{k\lambda _{n}t}\). Multiplying the above ODE by this integrating factor gives\[ \frac{d}{dt}\left ( a_{n}\left ( t\right ) e^{k\lambda _{n}t}\right ) =b_{n}\left ( x\right ) e^{k\lambda _{n}t}\] Integrating both sides\begin{align} a_{n}\left ( t\right ) e^{k\lambda _{n}t} & =b_{n}\left ( x\right ) \int _{0}^{t}e^{k\lambda _{n}\tau }d\tau \nonumber \\ a_{n}\left ( t\right ) & =b_{n}\left ( x\right ) \int _{0}^{t}e^{-k\lambda _{n}\left ( t-\tau \right ) }d\tau \tag{3} \end{align}
What is left is to find \(b_{n}\left ( x\right ) \). Since \(ax^{2}=\sum _{n=0}^{\infty }b_{n}\left ( x\right ) \Phi _{n}\left ( x\right ) \), and from example 1 section 8, we found that\begin{align*} b_{0}\left ( x\right ) & =a\frac{c^{2}}{3}\\ b_{n}\left ( x\right ) & =a\frac{4c^{2}}{\pi ^{2}}\frac{\left ( -1\right ) ^{n}}{n^{2}}\qquad n=1,2,3,\cdots \end{align*}
Hence when \(n=0\), then (3) becomes (since \(\lambda _{0}=0\))\begin{align*} a_{0}\left ( t\right ) & =a\frac{c^{2}}{3}\int _{0}^{t}d\tau \\ & =\frac{ac^{2}}{3}t \end{align*}
When \(n>0\) then (3) becomes\begin{align*} a_{n}\left ( t\right ) & =\left ( a\frac{4c^{2}}{\pi ^{2}}\frac{\left ( -1\right ) ^{n}}{n^{2}}\right ) \int _{0}^{t}e^{-k\lambda _{n}\left ( t-\tau \right ) }d\tau \\ & =\frac{\left ( -1\right ) ^{n}}{n^{2}}\frac{4ac^{2}}{\pi ^{2}}\int _{0}^{t}e^{-k\left ( \frac{n\pi }{c}\right ) ^{2}\left ( t-\tau \right ) }d\tau \\ & =\frac{\left ( -1\right ) ^{n}}{n^{2}}\frac{4ac^{2}}{\pi ^{2}}e^{-k\left ( \frac{n\pi }{c}\right ) ^{2}t}\int _{0}^{t}e^{k\left ( \frac{n\pi }{c}\right ) ^{2}\tau }d\tau \\ & =\frac{\left ( -1\right ) ^{n}}{n^{2}}\frac{4ac^{2}}{\pi ^{2}}e^{-k\left ( \frac{n\pi }{c}\right ) ^{2}t}\left [ \frac{e^{k\left ( \frac{n\pi }{c}\right ) ^{2}\tau }}{k\left ( \frac{n\pi }{c}\right ) ^{2}}\right ] _{0}^{t}\\ & =\frac{\left ( -1\right ) ^{n}}{n^{2}}\frac{4ac^{2}}{\pi ^{2}}\frac{e^{-k\left ( \frac{n\pi }{c}\right ) ^{2}t}}{k\left ( \frac{n\pi }{c}\right ) ^{2}}\left [ e^{k\left ( \frac{n\pi }{c}\right ) ^{2}t}-1\right ] \\ & =\frac{\left ( -1\right ) ^{n}}{n^{2}}\frac{4ac^{2}}{\pi ^{2}}\frac{1-e^{-k\left ( \frac{n\pi }{c}\right ) ^{2}t}}{k\frac{n^{2}\pi ^{2}}{c^{2}}}\\ & =\frac{\left ( -1\right ) ^{n}}{n^{4}}\frac{4ac^{4}}{k\pi ^{4}}\left ( 1-e^{-k\left ( \frac{n\pi }{c}\right ) ^{2}t}\right ) \end{align*}
Now that we found \(a_{n}\left ( t\right ) \), we substitute it back into (1) which gives\begin{align*} u\left ( x,t\right ) & =a_{0}\left ( t\right ) +\sum _{n=1}^{\infty }a_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \\ u\left ( x,t\right ) & =\frac{ac^{2}}{3}t+\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n}}{n^{4}}\frac{4ac^{4}}{k\pi ^{4}}\left ( 1-e^{-k\left ( \frac{n\pi }{c}\right ) ^{2}t}\right ) \cos \left ( \frac{n\pi }{c}x\right ) \\ & =\frac{ac^{2}}{3}t+\frac{4ac^{4}}{k\pi ^{4}}\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n}}{n^{4}}\left ( 1-e^{-k\left ( \frac{n\pi }{c}\right ) ^{2}t}\right ) \cos \left ( \frac{n\pi }{c}x\right ) \\ & =ac^{2}\left \{ \frac{t}{3}+\frac{4c^{2}}{k\pi ^{4}}\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n}}{n^{4}}\left ( 1-e^{-k\left ( \frac{n\pi }{c}\right ) ^{2}t}\right ) \cos \left ( \frac{n\pi }{c}x\right ) \right \} \end{align*}
Which is the result required to show.
Solution
Let \(u\left ( x,y\right ) =X\left ( x\right ) Y\left ( y\right ) \). The PDE becomes
\begin{align*} X^{\prime \prime }Y+Y^{\prime \prime }X & =0\\ \frac{X^{\prime \prime }}{X} & =-\frac{Y^{\prime \prime }}{Y}=-\lambda \end{align*}
Hence the eigenvalue problem is
\begin{align} X^{\prime \prime }+\lambda X & =0\tag{1}\\ X^{\prime }\left ( 0\right ) & =0\nonumber \\ X^{\prime }\left ( \pi \right ) & =0\nonumber \end{align}
And the ODE for \(Y\left ( y\right ) \) is\[ Y^{\prime \prime }-\lambda Y=0 \] We start by solving (1) to find the eigenvalues and eigenfunctions.
Case \(\lambda <0\) The solution is\begin{align*} X & =A\cosh \left ( \sqrt{-\lambda }x\right ) +B\sinh \left ( \sqrt{-\lambda }x\right ) \\ X^{\prime } & =A\sqrt{-\lambda }\sinh \left ( \sqrt{-\lambda }x\right ) +B\sqrt{-\lambda }\cosh \left ( \sqrt{-\lambda }x\right ) \end{align*}
At \(x=0\) the above becomes\[ 0=B\sqrt{-\lambda }\] Hence \(B=0\) and the solution becomes\begin{align*} X & =A\cosh \left ( \sqrt{-\lambda }x\right ) \\ X^{\prime } & =A\sqrt{-\lambda }\sinh \left ( \sqrt{-\lambda }x\right ) \end{align*}
At \(x=\pi \) the above gives\[ 0=A\sqrt{-\lambda }\sinh \left ( \sqrt{-\lambda }\pi \right ) \] For nontrivial solution \(\sinh \left ( \sqrt{-\lambda }\pi \right ) =0\) but this is not possible since \(\sinh \) is zero only when its argument is zero and this is not the case here. Hence \(\lambda <0\) is not eigenvalue.
Case \(\lambda =0\) The solution is\begin{align*} X & =Ax+B\\ X^{\prime } & =A \end{align*}
At \(x=0\) the above becomes\[ 0=A \] Hence the solution becomes\begin{align*} X & =B\\ X^{\prime } & =0 \end{align*}
At \(x=\pi \) the above gives\[ 0=0 \] Therefore \(\lambda =0\) is eigenvalue with \(X_{0}\left ( x\right ) =1\).
Case \(\lambda >0\) The solution is\begin{align*} X & =A\cos \left ( \sqrt{\lambda }x\right ) +B\sin \left ( \sqrt{\lambda }x\right ) \\ X^{\prime } & =-A\sqrt{\lambda }\sin \left ( \sqrt{\lambda }x\right ) +B\sqrt{\lambda }\cos \left ( \sqrt{\lambda }x\right ) \end{align*}
At \(x=0\) the above becomes\[ 0=B\sqrt{\lambda }\] Hence \(B=0\) and the solution becomes\begin{align*} X & =A\cos \left ( \sqrt{\lambda }x\right ) \\ X^{\prime } & =-A\sqrt{\lambda }\sin \left ( \sqrt{\lambda }x\right ) \end{align*}
At \(x=\pi \) the above gives\[ 0=-A\sqrt{\lambda }\sin \left ( \sqrt{\lambda }\pi \right ) \] For nontrivial solution \begin{align*} \sin \left ( \sqrt{\lambda }\pi \right ) & =0\\ \sqrt{\lambda }\pi & =n\pi \qquad n=1,2,3,\cdots \\ \lambda _{n} & =n^{2} \end{align*}
And the corresponding eigenfunctions \(X_{n}\left ( x\right ) =\cos \left ( nx\right ) \). Therefore in summary we have
| eigenvalue | eigenfunction |
| \(\lambda _{0}=0\) | \(1\) |
| \(\lambda _{n}=n^{2}\qquad n=1,2,3,\cdots \) | \(\cos \left ( nx\right ) \) |
Hence the \(Y\left ( y\right ) \) ode becomes\begin{align*} Y^{\prime \prime }-\lambda _{n}Y & =0\\ Y^{\prime \prime }-n^{2}Y & =0 \end{align*}
The solution to the above is, when \(n=0\)\[ Y_{0}=A_{0}y+B_{0}\] When \(y=0\) the above gives \(0=B_{0}\). Hence \(Y_{0}=A_{0}y\).
When \(n>0\)\[ Y_{n}\left ( y\right ) =B_{n}\cosh \left ( ny\right ) +A_{n}\sinh \left ( ny\right ) \] When \(y=0\) the above gives \(0=B_{n}\), Hence \[ Y_{n}\left ( y\right ) =A_{n}\sinh \left ( ny\right ) \] Hence the fundamental solution is\[ u\left ( x,y\right ) =X_{n}Y_{n}\] And the general solution is the superposition of these solutions\[ u\left ( x,y\right ) =A_{0}X_{0}Y_{0}+\sum _{n=1}^{\infty }A_{n}Y_{n}X_{n}\] Therefore\begin{equation} u\left ( x,y\right ) =A_{0}y+\sum _{n=1}^{\infty }A_{n}\sinh \left ( ny\right ) \cos \left ( nx\right ) \tag{A} \end{equation} What is left is to determine \(A_{0}\) and \(A_{n}\). At \(y=\pi \) the above gives\[ f\left ( x\right ) =A_{0}\pi +\sum _{n=1}^{\infty }A_{n}\sinh \left ( n\pi \right ) \cos \left ( nx\right ) \] Multiplying both sides by \(\cos \left ( mx\right ) \) and integrating gives\begin{equation} \int _{0}^{\pi }f\left ( x\right ) \cos \left ( mx\right ) dx=\int _{0}^{\pi }A_{0}\pi \cos \left ( mx\right ) dx+\int _{0}^{\pi }\sum _{n=1}^{\infty }A_{n}\sinh \left ( n\pi \right ) \cos \left ( nx\right ) \cos \left ( mx\right ) dx\tag{1} \end{equation} For \(m=0\), (1) becomes\begin{align} \int _{0}^{\pi }f\left ( x\right ) dx & =\int _{0}^{\pi }A_{0}\pi dx\nonumber \\ \int _{0}^{\pi }f\left ( x\right ) dx & =A_{0}\pi ^{2}\nonumber \\ A_{0} & =\frac{1}{\pi ^{2}}\int _{0}^{\pi }f\left ( x\right ) dx\tag{2} \end{align}
For \(m>0\), (1) becomes\begin{align*} \int _{0}^{\pi }f\left ( x\right ) \cos \left ( mx\right ) dx & =\int _{0}^{\pi }\sum _{n=1}^{\infty }A_{n}\sinh \left ( n\pi \right ) \cos \left ( nx\right ) \cos \left ( mx\right ) dx\\ \int _{0}^{\pi }f\left ( x\right ) \cos \left ( mx\right ) dx & =A_{m}\sinh \left ( m\pi \right ) \int _{0}^{\pi }\cos ^{2}\left ( nx\right ) dx\\ & =A_{m}\sinh \left ( m\pi \right ) \frac{\pi }{2} \end{align*}
Hence\begin{equation} A_{n}=\frac{2}{\pi \sinh \left ( n\pi \right ) }\int _{0}^{\pi }f\left ( x\right ) \cos \left ( nx\right ) dx\tag{3} \end{equation} When \(f\left ( x\right ) =u_{0}\) a constant, then (2) becomes\begin{align*} A_{0} & =\frac{1}{\pi ^{2}}\int _{0}^{\pi }u_{0}dx\\ & =\frac{u_{0}}{\pi } \end{align*}
And (3) becomes\begin{align*} A_{n} & =\frac{2}{\pi \sinh \left ( n\pi \right ) }\int _{0}^{\pi }u_{0}\cos \left ( nx\right ) dx\\ & =\frac{2u_{0}}{\pi \sinh \left ( n\pi \right ) }\left [ \frac{\sin \left ( nx\right ) }{n}\right ] _{0}^{\pi }\\ & =0 \end{align*}
Hence the solution (A) becomes\[ \fbox{$u\left ( x,y\right ) =u_0\frac{y}{\pi }$}\] This shows the final solution changes linearly in \(y\). When \(y=0\) then \(u\left ( x,0\right ) =0\) and when \(y=\pi \), then \(u\left ( x,\pi \right ) =u_{0}\).
Solution
The PDE \(\nabla ^{2}u\left ( \rho ,\phi \right ) =0\) in polar coordinates is \[ u_{\rho \rho }+\frac{1}{\rho }u_{\rho }+\frac{1}{\rho ^{2}}u_{\phi \phi }=0 \] For \(0<\rho <a\) and \(0<\phi <\alpha \). With boundary conditions\begin{align*} u\left ( \rho ,0\right ) & =0\\ u\left ( \rho ,\alpha \right ) & =0\\ u\left ( a,\phi \right ) & =f\left ( \phi \right ) \end{align*}
And since \(u\) is bounded, then we have an extra condition \(u\left ( 0,\phi \right ) <\infty \).
Let \(u\left ( \rho ,\phi \right ) =R\left ( \rho \right ) \Phi \left ( \phi \right ) \). Substituting into the above PDE gives\begin{align*} R^{\prime \prime }\Phi +\frac{1}{\rho }R^{\prime }\Phi +\frac{1}{\rho ^{2}}\Phi ^{\prime \prime }R & =0\\ \frac{R^{\prime \prime }}{R}+\frac{1}{\rho }\frac{R^{\prime }}{R}+\frac{1}{\rho ^{2}}\frac{\Phi ^{\prime \prime }}{\Phi } & =0\\ \frac{\Phi ^{\prime \prime }}{\Phi } & =-\left ( \rho ^{2}\frac{R^{\prime \prime }}{R}+\rho \frac{R^{\prime }}{R}\right ) =-\lambda \end{align*}
Where \(\lambda \) is the separation constant. The above gives the boundary values problem to solve for \(\lambda \)\begin{align} \Phi ^{\prime \prime }+\lambda \Phi & =0\tag{1}\\ \Phi \left ( 0\right ) & =0\nonumber \\ \Phi \left ( \alpha \right ) & =0\nonumber \end{align}
And\begin{align} \rho ^{2}\frac{R^{\prime \prime }}{R}+\rho \frac{R^{\prime }}{R} & =\lambda \nonumber \\ \rho ^{2}R^{\prime \prime }+\rho R^{\prime }-\lambda R & =0\tag{2} \end{align}
We start with (1) to find \(\lambda \) then use the result to solve (2). The ODE (1) we solved before, it has the eigenvalues \[ \lambda _{n}=\left ( \frac{n\pi }{\alpha }\right ) ^{2}\qquad n=1,2,3,\cdots \] And corresponding eigenfunctions\begin{equation} \Phi _{n}\left ( \phi \right ) =\sin \left ( \frac{n\pi }{\alpha }\phi \right ) \tag{3} \end{equation} Now (2) can be solved. This is a Euler ODE. Using \(R\left ( \rho \right ) =\rho ^{m}\) and substituting into (2) gives\begin{align*} \rho ^{2}m\left ( m-1\right ) \rho ^{m-2}+\rho m\rho ^{m-1}-\left ( \frac{n\pi }{\alpha }\right ) ^{2}\rho ^{m} & =0\\ m\left ( m-1\right ) \rho ^{m}+m\rho ^{m}-\left ( \frac{n\pi }{\alpha }\right ) ^{2}\rho ^{m} & =0\\ m\left ( m-1\right ) +m-\left ( \frac{n\pi }{\alpha }\right ) ^{2} & =0\\ m^{2} & =\left ( \frac{n\pi }{\alpha }\right ) ^{2} \end{align*}
Hence \[ m=\pm \frac{n\pi }{\alpha }\] Therefore the solution to (2) is\[ R_{n}\left ( \rho \right ) =A_{n}\rho ^{\frac{n\pi }{\alpha }}+B_{n}\rho ^{\frac{-n\pi }{\alpha }}\] We immediately reject the solution \(\rho ^{\frac{-n\pi }{\alpha }}\) since this blows up at origin where \(\rho \rightarrow 0\). Hence the above becomes\begin{equation} R_{n}\left ( \rho \right ) =A_{n}\rho ^{\frac{n\pi }{\alpha }}\tag{4} \end{equation} Now that we found \(\Phi _{n}\left ( \phi \right ) \) and \(R_{n}\left ( \rho \right ) \), then we use superposition to obtain the general solution\begin{align} u\left ( \rho ,\phi \right ) & =\sum _{n=1}^{\infty }R_{n}\left ( \rho \right ) \Phi _{n}\left ( \phi \right ) \nonumber \\ & =\sum _{n=1}^{\infty }A_{n}\rho ^{\frac{n\pi }{\alpha }}\sin \left ( \frac{n\pi }{\alpha }\phi \right ) \tag{5} \end{align}
At \(\rho =a\), \(u\left ( a,\phi \right ) =f\left ( \phi \right ) \), hence the above becomes\[ f\left ( \phi \right ) =\sum _{n=1}^{\infty }A_{n}a^{\frac{n\pi }{\alpha }}\sin \left ( \frac{n\pi }{\alpha }\phi \right ) \] By orthogonality we obtain\begin{align*} \int _{0}^{\alpha }f\left ( \phi \right ) \sin \left ( \frac{m\pi }{\alpha }\phi \right ) d\phi & =\int _{0}^{\alpha }\sum _{n=1}^{\infty }A_{n}a^{\frac{n\pi }{\alpha }}\sin \left ( \frac{n\pi }{\alpha }\phi \right ) \sin \left ( \frac{m\pi }{\alpha }\phi \right ) d\phi \\ & =A_{m}a^{\frac{m\pi }{\alpha }}\int _{0}^{\alpha }\sin ^{2}\left ( \frac{m\pi }{\alpha }\phi \right ) d\phi \\ & =A_{m}a^{\frac{m\pi }{\alpha }}\frac{\alpha }{2} \end{align*}
Solving for \(A_{n}\) from the above gives\[ A_{n}=\frac{2}{\alpha }a^{\frac{-n\pi }{\alpha }}\int _{0}^{\alpha }f\left ( \phi \right ) \sin \left ( \frac{n\pi }{\alpha }\phi \right ) d\phi \] Substituting the above in (5) gives the final solution\begin{align*} u\left ( \rho ,\phi \right ) & =\sum _{n=1}^{\infty }\left ( \frac{2}{\alpha }a^{\frac{-n\pi }{\alpha }}\int _{0}^{\alpha }f\left ( \psi \right ) \sin \left ( \frac{n\pi }{\alpha }\psi \right ) d\psi \right ) \rho ^{\frac{n\pi }{\alpha }}\sin \left ( \frac{n\pi }{\alpha }\phi \right ) \\ & =\frac{2}{\alpha }\sum _{n=1}^{\infty }\left ( \frac{\rho }{a}\right ) ^{\frac{n\pi }{\alpha }}\sin \left ( \frac{n\pi }{\alpha }\phi \right ) \left ( \int _{0}^{\alpha }f\left ( \psi \right ) \sin \left ( \frac{n\pi }{\alpha }\psi \right ) d\psi \right ) \end{align*}
Solution
\[ u_{t}=ku_{xx}\] With \(-\pi <x<\pi ,t>0\) and periodic boundary conditions\begin{align*} u\left ( -\pi ,t\right ) & =u\left ( \pi ,t\right ) \\ u_{x}\left ( -\pi ,t\right ) & =u_{x}\left ( \pi ,t\right ) \end{align*}
And initial conditions\[ u\left ( x,0\right ) =f\left ( x\right ) \] Normal process of separation of variables leads to eigenvalue problem\begin{align} X^{\prime \prime }+\lambda X & =0\tag{1}\\ X\left ( -\pi \right ) & =X\left ( \pi \right ) \nonumber \\ X^{\prime }\left ( -\pi \right ) & =X^{\prime }\left ( \pi \right ) \nonumber \end{align}
And the time ODE\begin{equation} T^{\prime }+k\lambda T=0\tag{2} \end{equation} We start by solving (1) to find the eigenvalues and eigenfunctions.
Case \(\lambda <0\)
Solution is \begin{align*} X\left ( x\right ) & =A\cosh \left ( \sqrt{-\lambda }x\right ) +B\sinh \left ( \sqrt{-\lambda }x\right ) \\ X^{\prime }\left ( x\right ) & =A\sqrt{-\lambda }\sinh \left ( \sqrt{-\lambda }x\right ) +B\sqrt{-\lambda }\cosh \left ( \sqrt{-\lambda }x\right ) \end{align*}
The boundary conditions \(X\left ( -\pi \right ) =X\left ( \pi \right ) \) results in (using the fact that \(\cosh \) is even and \(\sinh \) is odd)\begin{align} A\cosh \left ( \sqrt{-\lambda }\pi \right ) +B\sinh \left ( \sqrt{-\lambda }\pi \right ) & =A\cosh \left ( \sqrt{-\lambda }\pi \right ) -B\sinh \left ( \sqrt{-\lambda }\pi \right ) \nonumber \\ B\sinh \left ( \sqrt{-\lambda }\pi \right ) & =-B\sinh \left ( \sqrt{-\lambda }\pi \right ) \nonumber \\ B\sinh \left ( \sqrt{-\lambda }\pi \right ) & =0\tag{3} \end{align}
The boundary conditions \(X^{\prime }\left ( -\pi \right ) =X^{\prime }\left ( \pi \right ) \) results in (using the fact that \(\cosh \) is even and \(\sinh \) is odd)\begin{align} A\sqrt{-\lambda }\sinh \left ( \sqrt{-\lambda }\pi \right ) +B\sqrt{-\lambda }\cosh \left ( \sqrt{-\lambda }\pi \right ) & =-A\sqrt{-\lambda }\sinh \left ( \sqrt{-\lambda }\pi \right ) +B\sqrt{-\lambda }\cosh \left ( \sqrt{-\lambda }\pi \right ) \nonumber \\ A\sqrt{-\lambda }\sinh \left ( \sqrt{-\lambda }\pi \right ) & =-A\sqrt{-\lambda }\sinh \left ( \sqrt{-\lambda }\pi \right ) \nonumber \\ A\sinh \left ( \sqrt{-\lambda }\pi \right ) & =0\tag{4} \end{align}
So we obtain (3,4) equations, here they are again\begin{align*} B\sinh \left ( \sqrt{-\lambda }\pi \right ) & =0\\ A\sinh \left ( \sqrt{-\lambda }\pi \right ) & =0 \end{align*}
There are two possibility, either \(\sinh \left ( \sqrt{-\lambda }\pi \right ) =0\) or \(\sinh \left ( \sqrt{-\lambda }\pi \right ) \neq 0\). If \(\sinh \left ( \sqrt{-\lambda }\pi \right ) \neq 0\) then this leads to trivial solution, as it implies that both \(A=0\) and \(B=0\). On the other hand, if \(\sinh \left ( \sqrt{-\lambda }\pi \right ) =0\) then this implies that \(\sqrt{-\lambda }\pi =0\) since \(\sinh \) is only zero when its argument is zero which is not the case here. This implies that \(\lambda <0\) is not possible.
Case \(\lambda =0\)
The solution now becomes \(X\left ( x\right ) =Ax+B\). Satisfying the boundary conditions \(X\left ( -\pi \right ) =X\left ( \pi \right ) \) gives\begin{align*} A\pi +B & =-A\pi +B\\ 2A\pi & =0\\ A & =0 \end{align*}
Hence the solution becomes\begin{align*} X\left ( x\right ) & =B\\ X^{\prime } & =0 \end{align*}
Satisfying the boundary conditions \(X^{\prime }\left ( -\pi \right ) =X^{\prime }\left ( \pi \right ) \) gives \(0=0\). Hence \(\lambda =0\) is possible eigenvalue, with corresponding eigenfunction as constant, say \(1\).
Case \(\lambda >0\)
Solution is \begin{align*} X\left ( x\right ) & =A\cos \left ( \sqrt{\lambda }x\right ) +B\sin \left ( \sqrt{\lambda }x\right ) \\ X^{\prime }\left ( x\right ) & =-A\sqrt{\lambda }\sin \left ( \sqrt{\lambda }x\right ) +B\sqrt{\lambda }\cos \left ( \sqrt{\lambda }x\right ) \end{align*}
The boundary conditions \(X\left ( -\pi \right ) =X\left ( \pi \right ) \) results in (using the fact that \(\cos \) is even and \(\sin \) is odd)\begin{align} A\cos \left ( \sqrt{\lambda }\pi \right ) +B\sin \left ( \sqrt{\lambda }\pi \right ) & =A\cos \left ( \sqrt{\lambda }\pi \right ) -B\sin \left ( \sqrt{\lambda }\pi \right ) \nonumber \\ B\sin \left ( \sqrt{\lambda }\pi \right ) & =-B\sin \left ( \sqrt{\lambda }\pi \right ) \nonumber \\ B\sin \left ( \sqrt{\lambda }\pi \right ) & =0\tag{5} \end{align}
The boundary conditions \(X^{\prime }\left ( -\pi \right ) =X^{\prime }\left ( \pi \right ) \) results in (using the fact that \(\cosh \) is even and \(\sinh \) is odd)\begin{align} -A\sqrt{\lambda }\sin \left ( \sqrt{\lambda }\pi \right ) +B\sqrt{\lambda }\cos \left ( \sqrt{\lambda }\pi \right ) & =A\sqrt{\lambda }\sin \left ( \sqrt{\lambda }\pi \right ) +B\sqrt{\lambda }\cos \left ( \sqrt{\lambda }\pi \right ) \nonumber \\ -A\sqrt{\lambda }\sin \left ( \sqrt{\lambda }\pi \right ) & =A\sqrt{\lambda }\sin \left ( \sqrt{\lambda }\pi \right ) \nonumber \\ A\sin \left ( \sqrt{\lambda }\pi \right ) & =0\tag{6} \end{align}
So we obtain (5,6) equations, here they are again\begin{align*} B\sin \left ( \sqrt{\lambda }\pi \right ) & =0\\ A\sin \left ( \sqrt{\lambda }\pi \right ) & =0 \end{align*}
There are two possibility, either \(\sin \left ( \sqrt{\lambda }\pi \right ) =0\) or \(\sin \left ( \sqrt{\lambda }\pi \right ) \neq 0\). If \(\sin \left ( \sqrt{\lambda }\pi \right ) \neq 0\) then this leads to trivial solution, as it implies that both \(A=0\) and \(B=0\). If \(\sin \left ( \sqrt{\lambda }\pi \right ) =0\) then this implies that \(\sqrt{\lambda }\pi =n\pi \) where \(n=1,2,3,\cdots \). Hence \(\lambda >0\) is possible with eigenvalues and corresponding eigenfunctions given by \begin{align*} \lambda _{n} & =n^{2}\qquad n=1,2,3,\cdots \\ X_{n}\left ( x\right ) & =A_{n}\cos \left ( nx\right ) +B_{n}\sin \left ( nx\right ) \end{align*}
Now that we solved the eigenvalue problem (1), we use the eigenvalues found to solve the time ODE (2)\[ T^{\prime }+k\lambda _{n}T=0 \] When \(\lambda =0\), this becomes \(T^{\prime }=0\) or \(T_{0}\left ( t\right ) \) is constant. When \(\lambda >0\) the solution is\begin{align*} T_{n}\left ( t\right ) & =e^{-k\lambda _{n}t}\\ & =e^{-kn^{2}t} \end{align*}
Hence the fundamental solution is\[ u_{n}\left ( x,t\right ) =X_{n}\left ( x\right ) T_{n}\left ( t\right ) \] And by superposition, the general solution is\[ u\left ( x,t\right ) =A_{0}X_{0}\left ( x\right ) T_{0}\left ( t\right ) +\sum _{n=1}^{\infty }\left ( A_{n}\cos \left ( nx\right ) +B_{n}\sin \left ( nx\right ) \right ) e^{-kn^{2}t}\] But \(X_{0}\left ( x\right ) =1\) and \(T_{0}\left ( t\right ) \) is constant. Hence the above simplifies to\[ u\left ( x,t\right ) =A_{0}+\sum _{n=1}^{\infty }\left ( A_{n}\cos \left ( nx\right ) +B_{n}\sin \left ( nx\right ) \right ) e^{-kn^{2}t}\] What is left is to find \(A_{0},A_{n},B_{n}\). At \(t=0\) the above gives\begin{equation} f\left ( x\right ) =A_{0}+\sum _{n=1}^{\infty }A_{n}\cos \left ( nx\right ) +B_{n}\sin \left ( nx\right ) \tag{7} \end{equation} For \(n=0\), by orthogonality we obtain\begin{align*} \int _{-\pi }^{\pi }f\left ( x\right ) dx & =\int _{-\pi }^{\pi }A_{0}dx\\ \int _{-\pi }^{\pi }f\left ( x\right ) dx & =A_{0}\left ( 2\pi \right ) \\ A_{0} & =\frac{1}{2\pi }\int _{-\pi }^{\pi }f\left ( x\right ) dx \end{align*}
For \(n>0\). We start by multiplying both sides of (7) by \(\cos \left ( mx\right ) \) and integrating both sides. This gives\begin{align*} \int _{-\pi }^{\pi }f\left ( x\right ) \cos \left ( mx\right ) dx & =\int _{-\pi }^{\pi }\left ( \sum _{n=1}^{\infty }A_{n}\cos \left ( nx\right ) \cos \left ( mx\right ) +B_{n}\sin \left ( nx\right ) \cos \left ( mx\right ) \right ) dx\\ & =\sum _{n=1}^{\infty }A_{n}\int _{-\pi }^{\pi }\cos \left ( nx\right ) \cos \left ( mx\right ) dx+\sum _{n=1}^{\infty }B_{n}\int _{-\pi }^{\pi }\sin \left ( nx\right ) \cos \left ( mx\right ) dx \end{align*}
But \(\int _{-\pi }^{\pi }\sin \left ( nx\right ) \cos \left ( mx\right ) dx=0\) for all \(n,m\). And \(\int _{-\pi }^{\pi }\cos \left ( nx\right ) \cos \left ( mx\right ) dx=\int _{-\pi }^{\pi }\cos ^{2}\left ( mx\right ) dx\) and zero for all other \(n\neq m\). Hence the above simplifies to\begin{align*} \int _{-\pi }^{\pi }f\left ( x\right ) \cos \left ( mx\right ) dx & =A_{m}\int _{-\pi }^{\pi }\cos ^{2}\left ( mx\right ) dx\\ & =A_{m}\pi \end{align*}
Therefore\[ A_{n}=\frac{1}{\pi }\int _{-\pi }^{\pi }f\left ( x\right ) \cos \left ( nx\right ) dx \] To find \(B_{n}\) we do the same, but now we multiply both sides of (7) by \(\sin \left ( mx\right ) \) and this leads to \begin{align*} \int _{-\pi }^{\pi }f\left ( x\right ) \sin \left ( mx\right ) dx & =\int _{-\pi }^{\pi }\left ( \sum _{n=1}^{\infty }A_{n}\cos \left ( nx\right ) \sin \left ( mx\right ) +B_{n}\sin \left ( nx\right ) \sin \left ( mx\right ) \right ) dx\\ & =\sum _{n=1}^{\infty }A_{n}\int _{-\pi }^{\pi }\cos \left ( nx\right ) \sin \left ( mx\right ) dx+\sum _{n=1}^{\infty }B_{n}\int _{-\pi }^{\pi }\sin \left ( nx\right ) \sin \left ( mx\right ) dx \end{align*}
But \(\int _{-\pi }^{\pi }\cos \left ( nx\right ) \sin \left ( mx\right ) dx=0\) for all \(n,m\). And \(\int _{-\pi }^{\pi }\sin \left ( nx\right ) \sin \left ( mx\right ) dx=\int _{-\pi }^{\pi }\sin ^{2}\left ( mx\right ) dx\) and zero for all other \(n\neq m\). Hence the above simplifies to\begin{align*} \int _{-\pi }^{\pi }f\left ( x\right ) \sin \left ( mx\right ) dx & =B_{m}\int _{-\pi }^{\pi }\sin ^{2}\left ( mx\right ) dx\\ & =B_{m}\pi \end{align*}
Therefore\[ B_{n}=\frac{1}{\pi }\int _{-\pi }^{\pi }f\left ( x\right ) \sin \left ( nx\right ) dx \] This completes the solution. The final solution is \begin{align*} u\left ( x,t\right ) & =A_{0}+\sum _{n=1}^{\infty }\left ( A_{n}\cos \left ( nx\right ) +B_{n}\sin \left ( nx\right ) \right ) e^{-kn^{2}t}\\ & =\frac{1}{2\pi }\int _{-\pi }^{\pi }f\left ( x\right ) dx+\sum _{n=1}^{\infty }e^{-kn^{2}t}\left [ \left ( \frac{1}{\pi }\int _{-\pi }^{\pi }f\left ( x\right ) \cos \left ( nx\right ) dx\right ) \cos \left ( nx\right ) +\left ( \frac{1}{\pi }\int _{-\pi }^{\pi }f\left ( x\right ) \sin \left ( nx\right ) dx\right ) \sin \left ( nx\right ) \right ] \end{align*}