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Does the equation

$$\frac{d^2y}{dx^2} - \frac{H(x)}{b} y = c H(x)$$

have a solution where $H(x)$ is the Heaviside step function and $b$ and $c$ are constant?


Update: What about the second step function be $H(-x)$: $$\frac{d^2y}{dx^2} - \frac{H(x)}{b} y = c H(-x)$$

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    $\begingroup$ Solve the equation for $x>0$ and $x<0$ seperately and match the solution at $x=0$. $\endgroup$
    – Winther
    Aug 12, 2014 at 19:36
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    $\begingroup$ The answer to your question is yes. For example if $y(0)=y'(0)=0$ and $b=1$ you can easily check that $y(x) = (\cosh(x)-1)H(x)$ is a solution. To find the general solution just follow the steps above. $\endgroup$
    – Winther
    Aug 12, 2014 at 19:50

3 Answers 3

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For $x<0$, $y''=0$, i.e. $$y=C_1x+C_0.$$

For $x\ge0$, $y''-\frac1by=1$.

Assuming $b>0$, the solution of the homogenous equation is $$y=A\exp(\frac x{\sqrt b})+B\exp(-\frac x{\sqrt b}),$$ and a particular solution is $$y=-b.$$ Hence the general solution, $$y=A\exp(\frac x{\sqrt b})+B\exp(-\frac x{\sqrt b})-b.$$

You will ensure continuity of the function by equating the function and first derivative at $x=0$: $$C_0=A+B-b\\C_1=\frac A{\sqrt b}-\frac B{\sqrt b}.$$

In particular, starting from the steady state, $A=B=\frac b2$, and $$y=b\left(\cosh(\frac x{\sqrt b})-1\right)H(x).$$

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  • $\begingroup$ Thanks, but the second derivative of H(x) f(x) wouldn't have continuity and would have Kroeniker delta in it. $\endgroup$
    – Roy
    Aug 12, 2014 at 22:29
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    $\begingroup$ @Hesam: $y''$ must be discontinuous (unless $y(0)=-b$), that's obvious. But there is no $\delta$. $\endgroup$
    – user65203
    Aug 13, 2014 at 6:44
  • $\begingroup$ What if second $H(x)$ be $H(-x)$. I thought maybe it becomes $cosh(-x)$ but it is same as $cosh(x)$ $\endgroup$
    – Roy
    Aug 14, 2014 at 3:50
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I'm going to answer the updated question, which asks for the solution of the problem:

$$ \color{green}{y''(x) - \frac{H(x)}{b} y(x) = H(-x) \, c } \tag{1}$$

which can be rewritten equivalently as follows:

$$ \begin{array}{ll} y_1'' - \frac{1}{b} y_1 = 0 & \quad x > 0 \\ \tag{2} y_2'' = c & \quad x < 0 \end{array}$$

where I have denoted $y_1(x)$ as the solution for $x>0$ and $y_2(x)$ the solution for $x<0$. If we integrate both equations we will arrive at:

$$ \color{blue}{y_1(x) = A e^{r_1 x} + B e^{r_2 x}, \quad y_2(x) = \frac{c x^2}{2} + Dx +E} \tag{3}$$

where $r_i$ are the solutions of the characteristic equation $s^2 - 1/b = 0$ and $A, B, D$ and $E$ are constants of integration. To enforce continuity of both the function and its first derivative, we must provide:

$$y_1(0) = y_2(0), \quad y'_1(0) = y_2'(0), \tag{4}$$

which yields to:

\begin{align} A+B & = E \\ Ar_1 + B r_2 & = D, \tag{5} \end{align} which gives us the solution, provided $r_1 \neq r_2$:

$$\color{blue}{A = \frac{D-r_2 E}{r_1 - r_2} , \quad B = \frac{D - r_1 E}{r_2 - r_1}} \tag{6}$$

The constants $D$ and $E$ remain unkown until any boundary conditions or initial conditions are specified.

Hope this helps!

Cheers.

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  • $\begingroup$ Thanks, Isn't it possible to have $xe^{-x}$ in the solution due to the jump at $x=0$? $\endgroup$
    – Roy
    Aug 14, 2014 at 20:11
  • $\begingroup$ I'm not sure about what you mean with "to have $xe^{ -x}$ in the solution". What are the values of $b$? What is your domain? Is this an IVP or a BVP? As I told you once, it should be extremely helpful for us to help you better that you give us some more detailed description of (the physical meaning of) your problem. $\endgroup$
    – Dmoreno
    Aug 14, 2014 at 20:16
  • $\begingroup$ I mean that with your previous answer about $y_p(x)$, considering $y_1(x)=exp(x)$ I could get $y_p(x) = x e^{-x}$. However, I am wondering if we can have that term too or not. $\endgroup$
    – Roy
    Aug 14, 2014 at 20:18
  • $\begingroup$ I don't remember too much of my questions :)... but notice that the particular solutions of this problems are rather easy to obtain. $\endgroup$
    – Dmoreno
    Aug 14, 2014 at 20:20
  • $\begingroup$ I mean this link: math.stackexchange.com/questions/895420/solution-of-fracd2ydx2-frachx-yb-h-x/897628?noredirect=1#comment1852684_897628 $\endgroup$
    – Roy
    Aug 14, 2014 at 20:20
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I am not sure wether this answer is correct or not, but as far as the step function is only a sign, I brought it out of the integration.

Please note that, for finding the constants, you need the boundary values and compare the value of the function at the point $x=0$.

$$\begin{array}{l}\frac{{{d^2}y}}{{d{x^2}}} - \frac{{H(x)y}}{b} = H(x)\\\frac{{{d^2}y}}{{d{x^2}}} = \frac{{H(x)y}}{b} + H(x)\\\frac{{{d^2}y}}{{d{x^2}}} = \left( {\frac{y}{b} + 1} \right)H(x)\\\frac{{{d^2}y}}{{\left( {\frac{y}{b} + 1} \right)}} = H(x){d^2}x\\\int {\frac{b}{{y + b}}{d^2}y} = H(x)\int {{d^2}x} \\\int {\left\{ {bLn\left( {y + b} \right) + {c_1}} \right\}dy} = H(x)\int {x + {c_3}dx} \\b\left\{ {\left( {y + b} \right)Ln\left( {y + b} \right) - \left( {y + b} \right)} \right\} + {c_1}y + {c_2} = H(x)\left\{ {\frac{1}{2}{x^2} + {c_3}x + {c_4}} \right\}\end{array}$$

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    $\begingroup$ The Heaviside step function is defined as $H(x) = 0$ for $x<0$ and $H(x) = 1$ for $x>1$. Also, you cannot integrate $y''/(y+b)$. Its only for expressions $y'/g(y)$ this works. $\endgroup$
    – Winther
    Aug 12, 2014 at 20:03
  • $\begingroup$ @Winther I know, but it is not type of function that change the integration too much. I mean, we can bring it out of the integration sign. $\endgroup$
    – enthu
    Aug 12, 2014 at 20:05
  • $\begingroup$ Your final solution is also not correct. To get the correct answer you need to solve two equations: one for $x>0$ and one for $x<0$ and then match them. $\endgroup$
    – Winther
    Aug 12, 2014 at 20:06
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    $\begingroup$ I have tried to tell about you the many mistakes you have made and if you don't want to listen and just stick with 'I'm right' then thats OK, I'm done arguing. The downvotes you get should tell you something. $\endgroup$
    – Winther
    Aug 12, 2014 at 20:28
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    $\begingroup$ @Enthusiastic Student: $d(F(y)dy)=f(y)dy^2+F(y)d^2y$. $\endgroup$
    – user65203
    Aug 12, 2014 at 20:40

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