Differential equations – Partial differential equations – Two-point boundary-value problems

An initial-value problem

\[\left\{\begin{array}{l}y''(t)+p(t)y'(t)+q(t)y(t)=g(t),\quad t\in I\\[2.5mm] y(t_0)=y_0,\quad y'(t_0)=y_0'\quad\text{with}\quad t_0\in I.\end{array}\right.\]

has a unique solution.

For a boundary-value problem

\[\left\{\begin{array}{l}y''(x)+p(x)y'(x)+q(x)y(x)=g(x),\quad x\in(\alpha,\beta)\\[2.5mm] y(\alpha)=y_0,\quad y(\beta)=y_1\end{array}\right.\]

this is completely different: no solutions, a unique solutions and infinitely many solutions are all possible.

Definition: A boundary-value problem is called homogeneous if both the differential equation and the boundary conditions are homogeneous, so: \(g(x)=0\) and \(y_0=0\) and \(y_1=0\).

Examples:

1) \(\left\{\begin{array}{l}y''(x)+2y(x)=0,\quad 0 < x < \pi\\[2.5mm]y(0)=1,\quad y(\pi)=0.\end{array}\right.\)

The general solution of the differential equations is: \(y(x)=c_1\cos(x\sqrt{2})+c_2\sin(x\sqrt{2})\). Using the boundary conditions we now have: \(c_1=1\) and \(c_2=-\displaystyle\frac{\cos(\pi\sqrt{2})}{\sin(\pi\sqrt{2})}\). So there is exactly one solution: \(y(x)=\cos(x\sqrt{2})-\displaystyle\frac{\cos(\pi\sqrt{2})}{\sin(\pi\sqrt{2})}\sin(x\sqrt{2})\).

2) \(\left\{\begin{array}{l}y''(x)+y(x)=0,\quad 0 < x < \pi\\[2.5mm]y(0)=1,\quad y(\pi)=a.\end{array}\right.\)

The general solution of the differential equation is: \(y(x)=c_1\cos(x)+c_2\sin(x)\). Using the boundary conditions we now have: \(c_1=1\) and \(c_1=-a\). So for \(a\neq -1\) there are no solutions. For \(a=-1\) there are infinitely many solutions: \(y(x)=\cos(x)+c_2\sin(x)\) with \(c_2\in\mathbb{R}\) arbitrary.

Homogeneous boundary-value problems (examples):

1) \(\left\{\begin{array}{l}y''(x)+2y(x)=0,\quad 0 < x < \pi\\[2.5mm]y(0)=0,\quad y(\pi)=0.\end{array}\right.\)

The general solution of the differential equation is: \(y(x)=c_1\cos(x\sqrt{2})+c_2\sin(x\sqrt{2})\). Using the boundary conditions we now have: \(c_1=0\) and \(c_2=0\). Hence: \(y(x)=0\), only the trivial solutions.

2) \(\left\{\begin{array}{l}y''(x)+y(x)=0,\quad 0 < x < \pi\\[2.5mm]y(0)=0,\quad y(\pi)=0.\end{array}\right.\)

The general solution of the differential equation is: \(y(x)=c_1\cos(x)+c_2\sin(x)\). Using the boundary conditions we now have: \(c_1=0\) and \(c_2\in\mathbb{R}\). Hence: \(y(x)=c_2\sin(x)\) with \(c_2\in\mathbb{R}\), infintely many solutions.

Homogeneous boundary-value problem (general):

\[\left\{\begin{array}{l}y''(x)+\lambda y(x)=0,\quad 0 < x < \pi\\[2.5mm]y(0)=0,\quad y(\pi)=0.\end{array}\right.\]

1. \(\lambda=0\): \(y''(x)=0\;\;\Longrightarrow\;\;y(x)=a_1x+a_2\). Then we have: \(a_1=a_2=0\;\;\Longrightarrow\;\;y(x)=0\), the trivial solution.


2. \(\lambda=-\mu^2 < 0\): \(y(x)=b_1\cosh(\mu x)+b_2\sinh(\mu x)\). Then we have: \(b_1=b_2=0\;\;\Longrightarrow\;\;y(x)=0\), the trivial solution.


3. \(\lambda=\mu^2 > 0\): \(y(x)=c_1\cos(\mu x)+c_2\sin(\mu x)\). Then we have: \(c_1=0\) and \(c_2\sin(\mu\pi)=0\).

If \(\mu=n\in\{1,2,3,\ldots\}\), then we have: \(\lambda_n=\mu_n^2=n^2\) (eigenvalues) and eigenfunctions \(y_n(x)=\sin(n x)\) with \(n\in\{1,2,3,\ldots\}\).

Yet somewhat more general:

\[\left\{\begin{array}{l}y''(x)+\lambda y(x)=0,\quad 0 < x < L\\[2.5mm] y(0)=0,\quad y(L)=0.\end{array}\right.\]

Eigenvalues: \(\lambda_n=\displaystyle\frac{n^2\pi^2}{L^2}\) with \(n\in\{1,2,3,\ldots\}\). Eigenfunctions: \(y_n(x)=\displaystyle\sin\left(\frac{n\pi x}{L}\right)\) with \(n\in\{1,2,3,\ldots\}\).

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Last modified on August 23, 2021