Question

Find the Green's function for the boundary value problem $$y^{\prime \prime }=F(x),y(0)-2y^{\prime }(0)=0,y(1)+2y^{\prime }(1)=0$$ Then use the Green's function to solve (A) with (a) $F(x)=1,$ (b) $F(x)=x,$ and (c) $F(x)=x^2$.

Short answer

Question: Find the solution to the given boundary value problem for different functions, F(x), using the Green's function: The boundary value problem is given by: $$y^{″}(x)=F(x),y(0)-2y_t(0)=0,y(1)+2y_t(1)=0.$$ Find the solution for the following functions F(x): (a) $F(x)=1$ (b) $F(x)=x$ (c) $F(x)=x^2$

Step-by-step solution

Find the Homogeneous Solution

To find the homogeneous solution, we need to solve the following equation: $$y^{″}(x)=0$$ This is a second-order linear homogeneous differential equation with constant coefficients. The characteristic equation is given by: $$r^2=0,$$ which has a double root r=0. Therefore, the homogeneous solution is given by: $$y_h(x)=C_1+C_{2}x$$

Find the Green's function

Let $G(x,t)$ be the Green's function for this problem. We need to find $G(x,t)$ satisfying: $$G^{″}(x,t)=\delta (x-t)$$ and satisfying the boundary conditions: $$G(0,t)-2G_t(0,t)=0,G(1,t)+2G_t(1,t)=0$$ We will solve the Green's function in two parts: $G(x,t)$ for $0<x<t$ and $G(x,t)$ for $t<x<1$. For $0<x<t$, the forcing term $\delta (x-t)$ is 0, and so we have: $$G^{″}(x,t)=0$$ This gives a homogeneous solution: $$G(x,t)=A(t)x+B(t)$$ For $t<x<1$, the forcing term $\delta (x-t)$ is 0, so we have: $$G^{″}(x,t)=0$$ which gives a homogeneous solution: $$G(x,t)=C(t)x+D(t)$$ Now, we have to apply the boundary conditions: 1) $G(0,t)-2G_t(0,t)=0$: We get $G(0,t)=B(t)$ and $G_t(0,t)=A(t)$. So, the boundary condition becomes: $$B(t)-2A(t)=0$$ 2) $G(1,t)+2G_t(1,t)=0$: We get $G(1,t)=C(t)+D(t)$ and $G_t(1,t)=C(t)$. So, the boundary condition becomes: $$C(t)+D(t)+2C(t)=0$$ 3) Continuity of $G(x,t)$ at $x=t$: we have $A(t)t+B(t)=C(t)t+D(t)$, or: $$A(t)t+B(t)-C(t)t-D(t)=0$$ 4) Discontinuity of $G^{\prime }(x,t)$ at $x=t$: we have $G_t^{\prime }(t^{+},t)-G_t^{\prime }(t^{-},t)=1$. Thus, we get: $$A(t)-C(t)=1$$ Solving the above 4 conditions, we obtain: $$A(t)=-\frac{1}{t},B(t)=\frac{2}{t},C(t)=-1,D(t)=1+t$$ Therefore, the Green's function is given by: $$G(x,t)=\{\begin{array}{ll}-\frac{1}{t}x+\frac{2}{t}& 0<x<t\\ -(x-1)(x-t)& t<x<1\end{array}$$

Use Green's function to solve the given problem

Using the Green's function, we can find the solution for different values of F(x) as: (a) $F(x)=1$: $$y(x)=y_h(x)+{\int }_0^{1}G(x,t)F(t)dt=C_1+C_{2}x+{\int }_0^{1}G(x,t)dt$$ Compute the integral and solve for $C_1,C_2$ using boundary conditions. (b) $F(x)=x$: $$y(x)=y_h(x)+{\int }_0^{1}G(x,t)F(t)dt=C_1+C_{2}x+{\int }_0^{1}G(x,t)tdt$$ Compute the integral and solve for $C_1,C_2$ using boundary conditions. (c) $F(x)=x^2$: $$y(x)=y_h(x)+{\int }_0^{1}G(x,t)F(t)dt=C_1+C_{2}x+{\int }_0^{1}G(x,t)t^{2}dt$$ Compute the integral and solve for $C_1,C_2$ using boundary conditions. In each case, follow the process of computing the integral and then substituting the boundary conditions to find the constants $C_1,C_2$. This will provide the solution to the given problem for each specific function $F(x)$.

Key concepts

Homogeneous Differential Equation

In the realm of differential equations, a homogeneous differential equation is one where every term is a function of the unknown function and its derivatives. Specifically, in our exercise, we are working with a second-order linear homogeneous differential equation. This means that our equation can be written in the form

$$a{y}^{″}+b{y}^{\prime }+cy=0$$
where a, b, and c are constants. For the Green's function problem, we consider a=1, b=0, and c=0, generating the simplified equation:

$$y^{″}=0$$.
The solutions to such equations are crucial because they form the complementary (or homogeneous) part of the general solution to the inhomogeneous problem, which involves a non-zero right-hand side. The homogeneous solution typically contains constants that will later be determined by the boundary conditions of the problem.

Characteristic Equation

When faced with a homogeneous linear differential equation with constant coefficients, as seen in our exercise, an essential step is to formulate the characteristic equation. For the given second-order differential equation, the characteristic form is:

$$a{r}^2+br+c=0$$,
where r represents the roots of the equation which dictate the behavior of the differential equation's solution. In the given problem, the characteristic equation simplifies to:

$$r^2=0$$.
Discovering the roots of this equation helps characterize the solution. A double root, such as r = 0 in our case, means that our homogeneous solution will be of the form:

$$y_h(x)=C_1+C_{2}x$$,
where C_1 and C_2 are constants to be determined through boundary conditions.

Boundary Conditions

The nature of a differential equation changes dramatically once we impose boundary conditions. These conditions specify the values or behaviors of the solution at specific points, typically at the edges of the interval being considered. In our exercise, the boundary conditions are both mixed, involving the function y and its derivative y' at the endpoints x=0 and x=1:

$$y(0)-2y^{\prime }(0)=0$$$$y(1)+2y^{\prime }(1)=0$$.
Applying these boundary conditions allows us to define constants in the homogeneous solution and to enforce continuity and differentiability requirements on the Green's function. Conditions on the Green's function at x=0 and x=1 contribute to satisfying the overall conditions of the boundary value problem.

Integral Solutions

The integral solutions are key to solving non-homogeneous differential equations using Green's functions. They represent the particular solution to the differential equation y'' = F(x). Here, F(x) could be any function, leading to different cases to explore.

The integral solution is obtained by convolving the Green's function with F(x):

$$y(x)=y_h(x)+{\int }_0^{1}G(x,t)F(t)dt$$,
where y_h(x) is the homogeneous solution, and the integral represents the contribution from the Green's function. Different forms of F(x) will alter the integral's outcome, thus impacting the overall solution y(x). After computing the integral, the constants included in the homogeneous solution can be found by enforcing boundary conditions, which leads to a complete solution to the original boundary value problem.