# 2.3: Boundary Value Problems

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You might have only solved initial value problems in your undergraduate differential equations class. For an initial value problem one has to solve a differential equation subject to conditions on the unknown function and its derivatives at one value of the independent variable. For example, for $$x = x(t)$$ we could have the initial value problem

$x''+x=2,\quad x(0)=1,\quad x'(0)=0.\label{eq:1}$

Typically, initial value problems involve time dependent functions and boundary value problems are spatial. So, with an initial value problem one knows how a system evolves in terms of the differential equation and the state of the system at some fixed time. Then one seeks to determine the state of the system at a later time.

## Example $$\PageIndex{1}$$: Initial Value Problem

Solve the initial value problem, $$x'' +4x = \cos t,\quad x(0) = 1,\quad x'(0) = 0$$.

###### Solution

Note that the conditions are provided at one time, $$t = 0$$. Thus, this an initial value problem. Recall from your course on differential equations that we need to find the general solution and then apply the initial conditions. Furthermore, this is a nonhomogeneous differential equation, so the solution is a sum of a solution of the homogeneous equation and a particular solution of the nonhomogeneous equation, $$x(t) = x_h (t) + x_p(t)$$. [See the ordinary differential equations review in the Appendix.]

The solution of $$x'' + 4x = 0$$ is easily found as

$x_h (t) = c_1 \cos 2t + c_2 \sin 2t.\nonumber$

The particular solution is found using the Method of Undetermined Coefficients. We guess a solution of the form

$x_p(t)=A\cos t+B\sin t.\nonumber$

Differentiating twice, we have

$x_p''(t)=-(A\cos t+B\sin t).\nonumber$

So,

$x_p'' + 4x_p = −(A \cos t + B \sin t) + 4(A \cos t + B \sin t).\nonumber$

Comparing the right hand side of this equation with $$\cos t$$ in the original problem, we are led to setting $$B = 0$$ and $$A =\frac{1}{3}\cos t$$. Thus, the general solution is

$x(t)=c_1\cos 2t+c_2\sin 2t+\frac{1}{3}\cos t.\nonumber$

We now apply the initial conditions to find the particular solution. The first condition, $$x(0) = 1$$, gives

$1=c_1+\frac{1}{3}.\nonumber$

Thus, $$c_1 =\frac{2}{3}$$. Using this value for $$c_1$$, the second condition, $$x'(0) = 0$$, gives $$c_2 = 0$$. Therefore,

$x(t)=\frac{1}{3}(2\cos 2t+\cos t).\nonumber$

For boundary values problems, one knows how each point responds to its neighbors, but there are conditions that have to be satisfied at the endpoints. An example would be a horizontal beam supported at the ends, like a bridge. The shape of the beam under the influence of gravity, or other forces, would lead to a differential equation and the boundary conditions at the beam ends would affect the solution of the problem. There are also a variety of other types of boundary conditions. In the case of a beam, one end could be fixed and the other end could be free to move. We will explore the effects of different boundary conditions in our discussions and exercises. But, we will first solve a simple boundary value problem which is a slight modification of the above problem.

## Example $$\PageIndex{2}$$: Boundary Value Problem

Solve the boundary value problem, $$x'' + x = 2,\quad x(0) = 1,\quad x(1) = 0$$.

###### Solution

Note that the conditions at $$t = 0$$ and $$t = 1$$ make this a boundary value problem since the conditions are given at two different points. As with initial value problems, we need to find the general solution and then apply any conditions that we may have. This is a nonhomogeneous differential equation, so the solution is a sum of a solution of the homogeneous equation and a particular solution of the nonhomogeneous equation, $$x(t) = x_h (t) + x_p(t)$$. The solution of $$x'' + x = 0$$ is easily found as

$x_h(t)=c_1\cos t+c_2\sin t.\nonumber$

The particular solution is found using the Method of Undetermined Coefficients,

$x_p(t)=2.\nonumber$

Thus, the general solution is

$x(t)=2+c_1\cos t+c_2\sin t.\nonumber$

We now apply the boundary conditions and see if there are values of $$c_1$$ and $$c_2$$ that yield a solution to this boundary value problem. The first condition, $$x(0) = 0$$, gives

$0=2+c_1.\nonumber$

Thus, $$c_1 = −2$$. Using this value for $$c_1$$, the second condition, $$x(1) = 1$$, gives

$0 = 2 − 2 \cos 1 + c_2 \sin 1.\nonumber$

This yields

$c_2=\frac{2(\cos 1-1)}{\sin 1}.\nonumber$

We have found that there is a solution to the boundary value problem and it is given by

$x(t)=2\left(1-\cos t\frac{(\cos 1-1)}{\sin 1}\sin t\right).\nonumber$

Boundary value problems arise in many physical systems, just as the initial value problems we have seen earlier. We will see in the next sections that boundary value problems for ordinary differential equations often appear in the solutions of partial differential equations. However, there is no guarantee that we will have unique solutions of our boundary value problems as we had found in the example above.

Now that we understand simple boundary value problems for ordinary differential equations, we can turn to initial-boundary value problems for partial differential equations. We will see that a common method for studying these problems is to use the method of separation of variables. In this method the problem of solving partial differential equations is to separate the partial differential equation into several ordinary differential equations of which several are boundary value problems of the sort seen in this section.

This page titled 2.3: Boundary Value Problems is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Russell Herman via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.