Problem 2

1.1. Let \(Y(u)= y(e^u)\). The chain rule applied twice gives \[ \frac{dY}{du}= e^u y', \qquad \frac{d^2 Y}{du^2}= e^u y'+ e^u y'' e^u= e^u y'+ e^{2u} y'', \] from which we get \[ y'= e^{-u} \frac{dY}{du}, \qquad y''= e^{-2u} \frac{d^2Y}{du^2} - e^{-2u} \frac{dY}{du} \] Now we can write ODE \(x^2y'' + xy' + \lambda y=0\) in terms of \(Y\), \(Y'= dY/du\) and \(Y''= d^2Y/du^2\): \[ e^{2u} (e^{-2u} Y'' - e^{-2u} Y') + e^u (e^{-u}Y') + \lambda Y=0 \] This gives the following Sturm-Liouville problem \[ \boxed{ \begin{cases} Y''+ \lambda Y=0, & \qquad u_0 <u<u_1 \\ Y(u_0)=0, \quad Y(u_1)=0 & \end{cases} } \qquad (2) \]

1.2. The Sturm-Liouville problem (2) can be solved by letting \(\lambda=\kappa^2\) as we know that the eigenvalues are positive. \[ Y(u)= A \cos(\kappa u)+B\sin(\kappa u), \quad u_0 \leq u \leq u_1 \] with \[ \begin{cases} A \cos(\kappa u_0)+B \sin(\kappa u_0) & =0 \\ A \cos(\kappa u_1)+B \sin(\kappa u_1) & =0 \end{cases} \] To obtain non-trivial solutions we must impose \[ \cos(\kappa u_0)\sin(\kappa u_1)-\sin(\kappa u_0)\cos(\kappa u_1)=0 \] or \[ \sin\big(\kappa(u_1-u_0)\big)=0 \] The solutions are given by \[ \boxed{\kappa_n = \frac{n\pi}{u_1-u_0}, \qquad n=1,2,\ldots} \] The eigenfunctions are then \[ Y_n(u)= -B_n \frac{\sin(\kappa_n u_0)\cos(\kappa_n u)}{\cos(\kappa_n u_0)}+ B_n \sin(\kappa_n u) = \frac{B_n}{\cos(\kappa_n u_0)}\Big( \cos(\kappa_n u_0)\sin(\kappa_n u)-\sin(\kappa_n u_0)\cos(\kappa_n u)\Big) \] or \[ Y_n (u)= \frac{B_n}{\cos(\kappa_n u_0)} \sin\big(\kappa_n (u-u_0)\big) \] So, within a multiplicative constant, we can write \[ \boxed{Y_n (u)=\sin\big(\kappa_n (u-u_0)\big)} \] for \(n=1,2,\ldots\). With this expression, we can easily verify that \(Y(u_0)=Y(u_1)=0\).

Remark: It is straightforward (but not necessary) to show that the cases \(\lambda=0\) and \(\lambda=-\kappa^2 <0\) give only trivial solutions.

1.3. We can use the relationship \(y(x)= Y(\ln x)\) to arrive at \[ \boxed{ y_n (x) =\sin\big(\kappa_n (\ln x -\ln x_0)\big)= \sin\Big(\kappa_n \ln\frac{x}{x_0}\Big) } \] for \(n=1,2,\ldots\). The eigenvalues take the same expression: \[ \boxed{ \kappa_n = \frac{n\pi}{\ln\frac{x_1}{x_0}} } \] after using \(u_0=\ln x_0\) and \(u_1=\ln x_1\).

Remark: It is of course possible to solve the original Sturm-Liouville problem \(x^2 y''+xy'+\lambda y=0\) directly by looking for solutions of the type \(y(x)= r^s\). With \(\lambda=\kappa^2\) we find \(s=\pm i\kappa\). However, the complete solution is more problematic.

Problem 2

2.1. The boundary conditions \(u(0,y)=\) and \(u(x, b) = f(x)\) corresponds to fixing the temperature to \(0^\circ\) or to the specific profile \(f(x)\).

The BC \(u_y (x, 0) = 0\) is equivalent to setting the heat flux to \(0\) across the boundary: the boundary is insulated.

Finally the BC \(u_x (a, y) + \beta u (a, y) =0\) correspond to connecting the side \(x=a\) to a large reservoir maintained at constant zero temperature.

2.2. We seek a solution of the form \(u(x,y)= X(x) Y(y)\). Substitution in Laplace equation gives \[ \frac{X''}{X}+ \frac{Y''}{Y}=0 \] which necessarily implies that \[ \frac{X''}{X}= -\lambda, \qquad \frac{Y''}{Y}=\lambda \] where \(\lambda\) is a constant yet to be found.

The boundary conditions \(u(0,y)=0\) and \(u_y(a,y)+\beta u(a,y)=0\) imply the homogeneous BCs \(X(0)=0\) and \(X'(a)+\beta X(a)=0\). So we conclude that the ODE \(X''+ \lambda X=0\) with BCs \(X(0)=0\) and \(X'(a)+\beta X(a)=0\) defines a Sturm-Liouville problem with weight function \(r(x)=1\) on interval \([0,a]\), where \(\lambda\) plays the role of an eigenvalue.

We cannot impose the non-homogeneous BC \(u(x,b)= f(x)\) on \(u(x,y)= X(x)Y(y)\) as it would imply that \(f(x) = X(x) Y(b)\): this cannot provide a BC for \(Y(y)\).

2.3. We need to solve the Sturm-Liouville problem defined in 2.2 by assuming \(\lambda= \kappa^2 >0\): this gives \[ X(x)= A\cos\kappa x+ B \sin\kappa x \] We then impose the BCs to find \[ \begin{cases} & A=0 \\ & B(\kappa\cos a \kappa+\beta \sin a \kappa)=0 \end{cases} \] We must impose \(B\neq 0\): therefore \(\kappa\) is solution of the equation (dividing by \(\cos a \kappa\) which cannot be zero) \[ \boxed{\tan (a\kappa)= -\frac{1}{\beta} \kappa} \] The solution of this equation can be “visualized” by graphing the two functions \(\kappa\to \tan(a\kappa)\) and \(\kappa\to -\frac{1}{\beta} \kappa\). The intersections of the two graphs show the existence of an infinite sequence \(0< \kappa_1 <\kappa_2 < \cdots < \infty\) of solutions.

Each eigenvalue \(\lambda_n =\kappa^2_n\) is associated with the eigenfunction \[ X_n(x) = \sin\kappa_n x \] \(n=1,2,\ldots\).

Now we can solve for the functions \(Y_n (y)\) solution of \(Y''-\kappa_n^2 Y\) with BC \(Y'_n (0)=0\): \[ Y_n (y)= A_n \cosh (\kappa_n y) \]

The final solution is expressed as a linear combination of modes \(X_n(x) Y_n (y)\): \[ \boxed{u(x,y)= \sum_{n=1}^\infty A_n \sin(\kappa_n x)\cosh (\kappa_n y)} \] where the coefficients \(A_n\) are found by imposing the non-homogeneous BC \(u(x,b)=f(x)\): \[ f(x)=\sum_{n=1}^\infty A_n \sin(\kappa_n x)\cosh (\kappa_n b) \quad (0\leq x \leq a) \] We use the orthogonality of the functions \(\{X_n(x)\}_{n\geq 1}\) w.r.t. inner product \(\langle u,v \rangle= \int_0^a u(x)v(x)dx\) to find \[ \boxed{ A_n= \frac{1}{\cosh (\kappa_n b)} \frac{ \int_0^a f(x) \sin(\kappa_n x) dx } { \int_0^a \sin^2 (\kappa_n x) dx } } \]

2.4. The previous analysis is valid for any value of \(\beta\neq 0\). However we did not check for the possibility of \(\lambda= 0\) as an eigenvalue. In this case, we have \(X(x)=A x +B\) with \[ B=0, \qquad A(1+ \beta a)=0 \] which admits a non-trivial solution \(A\) if \(\beta\) satisfies \[ \boxed{ \beta = -\frac{1}{a} } \] The corresponding equation \(Y''=0\) with \(Y'(0)=0\) has solution: \[ Y(x)= Cst. \] In this case, the expansion for \(u\) must take into account this additional eigenfunction: \[ \boxed{ u(x,y)=A_0 x + \sum_{n=1}^\infty A_n \sin(\kappa_n x)\cosh (\kappa_n y) } \] with \(A_0\) given by \[ \boxed{ A_0= \frac{3}{a^3} \int_0^a x f(x) dx } \] This is valid because Sturm-Liouville theory guarantees the orthogonality of the functions \(\{x, \sin\kappa_1 x, \sin\kappa_2 x, \ldots\}\).

Remark: For \(\beta < -1/a\), there is an additional eigenvalue solution of \[ \tanh (a \kappa)= - \frac{1}{\beta} \kappa \] This equation has a single root \(\kappa_0\). The corresponding eigenfunction is \(X_0 (x)=\sinh(\kappa_0 x)\). Now \(u\) takes the form \[ u(x,y)=A_0 \sinh(\kappa_0 x)\cos(\kappa_0 y) + \sum_{n=1}^\infty A_n \sin(\kappa_n x)\cosh (\kappa_n y) \]