currently I'm working on a problem in which I need to solve an equation of the following form:

$$\mathcal{L}\left[\frac{\partial f(x,t) }{\partial x} + \frac{\partial f(x,t)}{\partial t}\right] = \frac{1}{y^{3/4}}\mathcal{L}\left[\frac{\partial^{2}f(x,t)}{\partial t^{2}}\right] $$

Where $\mathcal{L}$ stands for the Laplace transformed and $ y $ is corresponding variable to $x$ after performing the Laplace transform.

I have the following doubts:

  1. Although the derivative in the RHS is respect t. Do I have to apply the transform over the two variables $x,\ y$ ?

  2. I've been trying to solve the equation using the following code in mathematical but it seems that it's not working:

A = 10(*Some number*)
RHS = (1/y^(3/4))LaplaceTransform[D[f(x,t),{x,2}],x,s]
LHS = LaplaceTranform[D[f(x,t),x] + D[f(x,t),t],x,s]
ICOND = {f[0,t] == 0, f[1000,t] == 1000 - A, f[x,1] == Max[x-A,0]}

Sol = NSolve[{LHS == RHS,ICOND},c,{x,0.1,1000},{t,0,1}]

What would be a plausible way for solving the first equation I wrote? Are there any books or PDF's which talk about solving this kind of equations? I'm new to Mathematica, I've looked around an I haven't found any that resembles my doubt.

Thank you for reading.

Update 1.

Another doubt. Does someone have any hints on how would this Laplace transform is to be applyed?

$$ \mathcal{L}\left[t\frac{\partial^{2}f(x,t)}{\partial t^{2}}\right]\ =\ ?$$

  • 2
    $\begingroup$ notation are utterly confusing. What is $f$ function of? In Laplace, you should use $s$ as the Laplace variable name, and $t$ as the time domain variable name. Laplace it almost always used for time domain functions. If you know all this, you can do this by hand easily by using the properties of Laplace applied to derivatives. But with what you wrote, I am not able to understand what function depends on what. i.e. does $f$ depend on time? But you wrote $f(x,y)$ in one part and then $f(x,t)$ in another. $\endgroup$
    – Nasser
    Jul 21, 2020 at 22:21
  • 2
    $\begingroup$ fyi, In Mathematica, it is LaplaceTransform and not LaplaceTransformed and it takes 3 arguments, not one like you have. And Laplace takes initial conditions, at one location, typically at $t=0$ $\endgroup$
    – Nasser
    Jul 21, 2020 at 22:22
  • $\begingroup$ Hello, I just checked the equation it had a few typos. As you said f is function of t but also of x. Just x i.e. $f\ =\ f(x,t)$ $\endgroup$ Jul 23, 2020 at 15:57

1 Answer 1


Considering instead

$$\mathcal{L}[\frac{\partial f(x,y,t) }{\partial x} + \frac{\partial f(x,y,t)}{\partial t}] = \frac{1}{y^{3/4}}\mathcal{L}[\frac{\partial^{2}f(x,y,t)}{\partial t^{2}}] $$

and calling

$ g_y(x,t) = f(x,y,t) $

we can write as

$$ \mathcal{L}[\frac{\partial g_y(x,t) }{\partial x} + \frac{\partial g_y(x,t)}{\partial t}] = \frac{1}{y^{3/4}}\mathcal{L}[\frac{\partial^{2}g_y(x,t)}{\partial t^{2}}] $$

with Laplace transform

$$ \frac{\partial}{\partial x}G_y(x,s) + s G_y(x,s)-\frac{s^2}{y^{3/4}}G_y(x,s) = g_y(x,0)-\frac{1}{y^{3/4}}\left(s g_y(x,0)+g_y'(x,0)\right) $$


$$ \frac{\partial}{\partial x}G_y(x,s) + \left(s-\frac{s^2}{y^{3/4}}\right)G_y(x,s) = g_y(x,0)-\frac{1}{y^{3/4}}\left(s g_y(x,0)+g_y'(x,0)\right) $$

now assuming $g_y(x,0) = g'_y(x,0) = 0$ we can proceed with

Gy[x, s] /. DSolve[D[Gy[x, s], x] + (s - s^2/y^(3/4)) Gy[x, s] == 0, Gy, {x, s}]
  • $\begingroup$ Thank you very much. I just fixed the misunderstanding at my equation. f is function of x and t i.e. $f\ =\ f(x,t)$. I assume just as you did I can work on how to write the proper mathematica code $\endgroup$ Jul 23, 2020 at 16:00

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