# 2nd order non-homogenous ODE with delta function & B.C.S

I'm relatively new to the concept of the Dirac Delta function have come across a problem in dealing with ODE with delta

Solve the ODE:

$$A''(y) - λ^2 A(y) = δ(y - ξ)$$

Subject to B.C (Hint: Use Hyperbolic functions) $$A(0) = A(b) = 0,$$

For the solution I have assumed $$δ(ξ)$$ is undefined, So $$A(y) = \begin{cases} Asinh(λy)+BCosh(λy), & y<ξ &(1) \\ Csinh(λy)+DCosh(λy), & y>ξ&(2) \end{cases}$$

Using $$A(0)=0, \ A(y)=Asinh(λy)$$

Using $$A(b)=0, \ A(y)=Csinh(λ(b - y))$$

Such that; $$A(y) = \begin{cases} Asinh(λy), & y<ξ &(1) \\ Csinh(λ(b - y), & y>ξ&(2) \end{cases}$$

What can I do to further simplify my solution, Can I get the constants A , C?

I'm going to assume that $$0 < \xi < b$$. Let $$\epsilon>0$$, then take your differential equation and integrate it like so:

$$\int_{\xi-\epsilon}^{\xi+\epsilon} A''(y) - \lambda^2 A(y) dy = \int_{\xi-\epsilon}^{\xi+\epsilon} \delta(y-\xi)dy$$

$$\implies A'\left(\xi+\epsilon\right) - A'\left(\xi-\epsilon\right) -\lambda^2\int_{\xi-\epsilon}^{\xi+\epsilon} A(y)dy = 1$$

Now take the limit as $$\epsilon\to 0^+$$. By assumption, we want $$A(y)$$ to be continuous, so the integral term goes to $$0$$. What we are left with is the following:

$$A'\left(\xi^+\right)-A'\left(\xi^-\right) = 1$$

This is your new condition to solve for the remaining coefficients. The first derivative is not continuous at $$y=\xi$$, but it has a jump discontinuity that is exactly $$1$$ in size.

• So do I have to solve a system of equations? $$A(y) = \begin{cases} Asinh(λy), & y<ξ &(1) \\ Csinh(λ(b - y), & y>ξ&(2) \end{cases}$$ Since A(y) is continous $\implies$ $Asinh(λy)=Csinh(λ(b - y))$ $$A'(y) = \begin{cases} λAcosh(λy), & y<ξ &(1) \\ -λCcosh(λ(b - y)) +1, & y>ξ&(2) \end{cases}$$ I've added +1 into A'(y) to make it discontinuous (unsure about this) and this $\implies λAcosh(λy)=-λCcosh(λ(b - y)) +1$ and thus I am left with a system of equations $$λAsinh(λy)=Csinh(λ(b - y))\\ λAcosh(λy)=-λCcosh(λ(b - y)) +1$$ – Daniel Chan Nov 15 at 19:28
• @DanielChan that is almost exactly correct, except you should've plugged in $\xi$ for $y$. And the $+1$ is on the wrong side – Ninad Munshi Nov 16 at 2:57