# non-constant coefficient Differential Equation with Dirac delta (unsure how to properly write the solution)

$$\zeta$$ is a constant, $$g(M) < M$$, $$g^{-1}(\zeta) > 0$$, and the equation is: $$a'(M) - a(M)\frac{g'(M)}{M-g(M)} + \frac{\delta_{g^{-1}(\zeta)}(M)}{M-g(M)} = 0$$

Things can be left in terms of $$g(M)$$ and $$g'(M)$$ and the goal is to calculate $$a(M)$$ (by integrating from $$0$$ to $$M$$)

The reason for solving this equation is to get a solution for $$a(M)$$ in the following equation $$\partial_M{\Large[}a(M)(x-g(M)) + 1_{M \geq g^{-1}(\zeta)}(M){\Large]}_{x=M} = 0.$$ So I expanded the derivative and used the identity
$$\partial_M1_{M \geq g^{-1}(\zeta)}(M) = \delta_{g^{-1}(\zeta)}(M)$$ in order to get to the above differential equation.

The correct answer is:

$$a(M) = a(0)*e^{\int_0^M\frac{g'(y)}{y-g(y)}dy} -\frac{e^{\int_{g^{-1}(\zeta)}^M\frac{g'(r)}{r-g(r)}dr}}{g^{-1}(\zeta)-\zeta}*1_{M \geq g^{-1}(\zeta)}$$

This is my work:

The homogeneous equation $$a'(M) - a(M)\frac{g'(M)}{M-g(M)} = 0$$ has solution: $$a(0)*e^{\int_0^M\frac{g'(y)}{y-g(y)}dy}$$ which I can easily calculate using the integrating factor.

Now for when I am solving it with the dirac delta included, I am trying to solve it as an equation in the form: $$y' + p(t)y = q(t)$$ (so it is a linear first order ODE as described here)

The form of the general solution is then: $$a(M) = a(0)*e^{\int_0^M\frac{g'(y)}{y-g(y)}dy} -e^{\int_0^M\frac{g'(y)}{y-g(y)}dy} *\int_0^M\delta_{g^{-1}(\zeta)}(y) * e^{-\int_0^y\frac{g'(r)}{r-g(r)}dr}*\frac{1}{y-g(y)}dy$$ which I simplified by just bringing inside the exponential: $$a(M) = a(0)*e^{\int_0^M\frac{g'(y)}{y-g(y)}dy} -\int_0^M\delta_{g^{-1}(\zeta)}(y) * e^{\int_y^M\frac{g'(r)}{r-g(r)}dr}*\frac{1}{y-g(y)}dy$$

So now, using the formal relation $$\int_X f(y)\delta_x(y)\mathrm{d}y= f(x)$$ from the Wikipedia entry on Dirac measure, I calculated the integral to be $$-\int_0^M\delta_{g^{-1}(\zeta)}(y) * e^{\int_y^M\frac{g'(r)}{r-g(r)}dr}*\frac{1}{y-g(y)}dy = -e^{\int_{g^{-1}(\zeta)}^M\frac{g'(r)}{r-g(r)}dr}*\frac{1}{g^{-1}(\zeta)-\zeta}$$

So my final answer is:

$$a(M) = a(0)*e^{\int_0^M\frac{g'(y)}{y-g(y)}dy} -\frac{e^{\int_{g^{-1}(\zeta)}^M\frac{g'(r)}{r-g(r)}dr}}{g^{-1}(\zeta)-\zeta}$$

which is nearly the actual answer, but it is missing the indicator function in the second term.

Anyone can point out where I went wrong? I think it might have to do with the initial equation that is being solved as well, so maybe that is why the indicator function reappears.

Thanks!

• When you use the formula from wikipedia, this only works for $g^{-1}(\zeta) \in (0,M)$, if $g^{-1}(\zeta)$ lies outside the interval $[0,M]$ this integral is just zero. I think that should give you the missing indicator function. – quarague Feb 5 '19 at 15:49
• Hi @Slade, I apologize for the mess. I tried to improve your already nice question. – Daniele Tampieri Feb 5 '19 at 15:57
• Thanks! Dw. It looks good! – Slade Feb 5 '19 at 15:57
• @quarague, it looks to me like that is it! Thanks a lot! – Slade Feb 5 '19 at 16:09

Thanks to @quarague for helping get to this answer!

So the expressions: $$-\int_0^M\delta_{g^{-1}(\zeta)}(y) * \frac{e^{\int_y^M\frac{g'(r)}{r-g(r)}dr}}{y-g(y)}dy = \frac{-e^{\int_{g^{-1}(\zeta)}^M\frac{g'(r)}{r-g(r)}dr}}{g^{-1}(\zeta)-\zeta} =$$

are only valid if the dirac measure's fixed element ($$g^{-1}(\zeta)$$ in our case) is in the set the integral is done over, so in this case, $$[0,M]$$, and otherwise the integral will be $$0$$. So we need for $$g^{-1}(\zeta) \in [0,M]$$

So basically the expression can be written:

$$-\int_0^M\delta_{g^{-1}(\zeta)}(y) * \frac{e^{\int_y^M\frac{g'(r)}{r-g(r)}dr}}{y-g(y)}dy = 1_{0 \leq g^{-1}(\zeta) \leq M}(M)\frac{-e^{\int_{g^{-1}(\zeta)}^M\frac{g'(r)}{r-g(r)}dr}}{g^{-1}(\zeta)-\zeta}$$ and since we use the condition $$g^{-1}(\zeta) > 0$$, we can rewrite $$1_{0 \leq g^{-1}(\zeta) \leq M}(M)$$ as $$1_{g^{-1}(\zeta) \leq M}(M)$$, and therefore we have as final answer:

$$a(M) = a(0)*e^{\int_0^M\frac{g'(y)}{y-g(y)}dy} -\frac{e^{\int_{g^{-1}(\zeta)}^M\frac{g'(r)}{r-g(r)}dr}}{g^{-1}(\zeta)-\zeta}*1_{M \geq g^{-1}(\zeta)}(M)$$

• Note that you have to be careful with the integral bounds here. If $g^{-1}(\zeta)=M$ the integral is not well defined. As long as you are integrating functions it doesn't matter whether you have include the bounds or not. With distributions this is no longer the case. – quarague Feb 6 '19 at 7:41
• So how would this change the answer I wrote? The 'correct solution' I wrote in the post needs to match and this way makes the answers match, but I am unsure what rationale to use. I was hoping whatever you mentioned would lead to the correct answer – Slade Feb 7 '19 at 6:33
• It wouldn't change the answer. If you wanted to be pendatic you would have to exclude the case $g^{-1}(\zeta)=M$ from the computation (because your equations are not well defined for this case) and in the end consider what happens in the limit $M \rightarrow g^{-1}(\zeta)$ but this limit exists so you are good to go. – quarague Feb 7 '19 at 9:17
• Okay that makes sense. My teacher says that the following is true: $(1_{g(M) \geq\zeta})' f(M) = (1_{g(M)\geq\zeta})' f(g^{-1}(\zeta))$ for any function $f$. Do you know where this comes from? I looked up a lot of pages on the dirac measure/mass, but can't find something that matches this identity. I am mostly having trouble understanding why the dirac mass/derivative of the indicator function still appears in the right hand side of the equation. – Slade Feb 9 '19 at 19:33
• I think I understand the equality but now I see two possible interpretations of $-\int_0^M\delta_{g^{-1}(\zeta)}(y) * \frac{e^{\int_y^M\frac{g'(r)}{r-g(r)}dr}}{y-g(y)}dy$. As above,$1_{0 \leq g^{-1}(\zeta) \leq M}(M)\frac{-e^{\int_{g^{-1}(\zeta)}^M\frac{g'(r)}{r-g(r)}dr}}{g^{-1}(\zeta)-\zeta}$ vs $-\frac{-e^{\int_{g^{-1}(\zeta)}^M\frac{g'(r)}{r-g(r)}dr}}{g^{-1}(\zeta)-\zeta}*\int_0^M\delta_{g^{-1}(\zeta)}(y)dy$, where $\delta_{g^{-1}(\zeta)}(y) = 1_{y \geq g^{-1}(\zeta)}(y)'$, and so $\int_0^M\delta_{g^{-1}(\zeta)}(y)dy = \int_0^M1_{y \geq g^{-1}(\zeta)}(y)'$. I'm not sure what that equals – Slade Feb 10 '19 at 1:10