Kansas State University

Discussion

At first glance, there appears to be little different about such problems. You just find the general solution, then plug in the two conditions and solve to find the values of the constants. This is indeed how you go about solving the problem. But there turn out to be various interesting difficulties that arise in boundary value problems that don't arise in initial value problems.

EXAMPLE: Solve the boundary value problem $y''+y=0$, $y(0)=1$, $y(1)=2$.

FIRST: Find the general solution. This is a homogeneous equation so it is easy to find that the general solution is $y(x)=c_1\sin(x)+c_2\cos(x)$.

SECOND: Plug in the boundary values and solve for the constants. $$ \eqalign { y(0)&=c_1\sin(0)+c_2\cos(0)=c_2{\buildrel \text{set}\over =}1 \cr y(1)&=c_1\sin(1)+c_2\cos(1){\buildrel \text{set}\over =}2 \cr } $$ Solving the equations we get $$ \eqalign { c_1&=\frac{2-\cos(1)}{\sin(1)} \cr c_2&=1 \cr } $$ so the solution is $y(x)=\frac{2-\cos(1)}{\sin(1)}\sin(x)+\cos(x).$

That didn't seem very exciting. But consider the next example.

FIRST: As before, we find the general solution is $y(x)=c_1\sin(x)+c_2\cos(x)$.

SECOND: This time we plug in and get $$ \eqalign { y(0)=c_1\sin(0)+c_2\cos(0)=c_2&{\buildrel \text{set}\over =}1 \cr y(\pi)=c_1\sin(\pi)+c_2\cos(\pi)=-c_2&{\buildrel \text{set}\over =}2 \cr } $$ This can't happen since $c_2$ can't equal both 1 and 2. In this case there is no solution to the boundary value problem. That never happened with initial value problems, and there is a theorem that it can't happen for any reasonable initial value problem. But boundary value problems are a whole new ball game. Another weird possibility is in the following example.

FIRST: As always, the general solution is $y(x)=c_1\sin(x)+c_2\cos(x)$.

SECOND: We plug in and get $$ \eqalign { y(0)=c_1\sin(0)+c_2\cos(0)=c_2&{\buildrel \text{set}\over =}0 \cr y(\pi)=c_1\sin(\pi)+c_2\cos(\pi)=-c_2&{\buildrel \text{set}\over =}0 \cr } $$

So we see that $c_2$ emphatically must be 0. But we don't know anything about $c_1$. Any function of the form $y(x)=c_1\sin(x)$ satisfies the given boundary value problem. Again, with initial value problems we don't have this difficulty. Any reasonable initial value problem has exactly one solution. With boundary value problems, we might have just one solution, as in the first example, or no solutions, as in the second example, or infinitely many solutions, as in the third example.

Boundary value problems often arise in connection with partial differential equations. In this case, it is usually most interesting to know when you get infinitely many solutions. This isn't quite so standardized a problem that I can give you a paradigm to mimic, but I can work one to give you the general idea.

QUESTION: For what values of $k$ does $y''+k^2y=0$, $y(0)=0$, $y(\pi)=0$, have infinitely many solutions?

Observe that $y(x)=0$ is always a solution to the given problem, but we are interested in when there are other solutions as well (once there is a second solution you are guaranteed that there are infinitely many solutions).

The way to begin such a problem is to write out the general solution and plug in the boundary values. The general solution to $y''+k^2y=0$ is $y(x)=c_1\sin(kx)+c_2\cos(kx)$. We plug in the boundary values to get $$ \eqalign { y(0)=c_1\sin(0)+c_2\cos(0)=c_2&{\buildrel \text{set}\over =}0 \cr y(\pi)=c_1\sin(k\pi)+c_2\cos(k\pi)&{\buildrel \text{set}\over =}0 \cr } $$ From the first equation we know $c_2$ must be 0. Plugging this into the second equation we get $c_1\sin(k\pi)=0$, so either $c_1=0$ or $\sin(k\pi)=0$. If $c_1=0$, we just get the solution $y(x)=0$, so if we want infinitely many solutions we need $\sin(k\pi)=0$. This happens exactly when $k$ is an integer.