# How to derive the substituted Partial Differential Equation

I have a second order partial differential equation.

$$\frac{\partial^2 U}{\partial x^2} + \frac{\partial^2 U}{\partial z^2} + \frac{2}{l}\frac{\partial U}{\partial z}=0$$.

I need to introduce a perturbation $$\zeta(x,t)$$ (i.e. introduction of a perturbation) into the above equation in order to shift the coordinates. Boundary conditions $$U(0)=0$$

Here, $$z = \zeta(x,t)=\hat{\zeta}exp(ik\cdot x+\omega t)$$, where, $$k$$ is a two-dimensional wave vector and $$\omega$$ the wave number.

After substitution I get the following equation :

$$-k^2\hat{\zeta}exp(ik\cdot x + \omega t)\frac{\partial^2U}{\partial \zeta^2} + \frac{2}{l}\frac{\partial U}{\partial\zeta}=0$$. As, $$\frac{\partial^2 U}{\partial x^2} = \frac{\partial^2 U}{\partial \zeta^2}\frac{\partial^2 \zeta}{\partial x^2}$$.

But, when I look at the solution for this problem it is given as,

$$U = exp(-2z/l) - 1 + \hat{\zeta}exp(ik\cdot x+\omega t - qz)$$, where $$q$$ is the solution of a quadratic equation.

So I'm stuck at the substituted equation, as it will not result in the given solution.

Can anyone point out where I'm going wrong in the substitution?

• What is z? Where did it go after your substitution? Are x and z related somehow? – Yuriy S Nov 5 '18 at 15:15
• @YuriyS Here, $z = \zeta(x,t)$ – newstudent Nov 5 '18 at 15:30
• Are you absolutely convinced that after your transformation $$\frac{\partial^2 U}{\partial x^2} = \frac{\partial^2 U}{\partial \zeta^2}\frac{\partial^2 \zeta}{\partial x^2}$$? – Kevin Nov 5 '18 at 15:40
• @Kevin Ah..I just applied the chain rule..should it be written as $\frac{\partial^2 U}{\partial x^2} = \frac{\partial}{\partial x} (\frac{\partial U}{\partial\zeta} \frac{\partial \zeta}{\partial x})$ ? – newstudent Nov 5 '18 at 15:45
• You didn't specify the domain, what is your domain and boundary conditions? That's very crucial. – plus1 Nov 5 '18 at 15:51

Actually, this equation is screaming for the substitution $$U(x,z)=e^{-\alpha z}W(x,z)+c.$$ where $$c$$ is an arbitrary constant that will not be essential for the following argument. Then, $$\frac{\partial U}{\partial z}=-\alpha e^{-\alpha z}W(x,z)+e^{-\alpha z}\frac{\partial W(x,z)}{\partial z}$$ and $$\frac{\partial^2 U}{\partial z^2}=\alpha^2e^{-\alpha z}W(x,z)-2\alpha e^{-\alpha z}\frac{\partial W(x,z)}{\partial z}+e^{-\alpha z}\frac{\partial^2 W(x,z)}{\partial z^2}.$$ Putting all this in the original equation one gets $$e^{-\alpha z}\frac{\partial^2 W(x,z)}{\partial x^2}+ \alpha^2e^{-\alpha z}W(x,z)-2\alpha e^{-\alpha z}\frac{\partial W(x,z)}{\partial z}+e^{-\alpha z}\frac{\partial^2 W(x,z)}{\partial z^2} -\frac{2}{l}\alpha e^{-\alpha z}W(x,z)+\frac{2}{l}e^{-\alpha z}\frac{\partial W(x,z)}{\partial z}=0.$$ I can remove the first order derivative by the choice $$\alpha=l^{-1}$$ and I can also remove the exponential. This will yield $$\frac{\partial^2 W(x,z)}{\partial x^2}+\frac{\partial^2 W(x,z)}{\partial z^2} -\frac{1}{l^2}W(x,z)=0$$ where I have substituted the $$\alpha$$ value obtained above. This equation is very well-known and can be solved by variable separation. As already stated in the comments, you will need to specify the boundary conditions. Anyway, to recover your kind of solution, you will need to look for solutions in the form $$W(x,z)=\hat\zeta\exp(ik_xx+ik_zz+i\phi),$$ being $$\phi$$ an arbitrary phase that you have taken to be $$\omega t$$. You will get the dispersion relation $$k_x^2+k_z^2+\frac{1}{l^2}=0.$$ A given domain and the boundary conditions will fix $$k_x$$ and $$k_z$$ properly.
• Thanks for the detailed answer. So the substition I was working on was not correct (was given in an article)? Is it ? As it had to be a function of $z$ too ? – newstudent Nov 7 '18 at 8:50
• Ok. Can you just briefly explain why did you take the perturbation as a function of $z$, was it something to do with first order approximation ? – newstudent Nov 7 '18 at 9:00
• There is no approximation whatsoever. This equation can be solved exactly. It is a homogeneous linear second-order pde reducible to the Helmholtz form. But it can be read in the way you write when the superposition principle is applied, being the equation linear. Then, a given solution can appear to have wave-like perturbations. You can write your solution in the form you gave as $U(x,z)=f(z)+f_1(x,z)$ and now $f_1(x,z)$ is the wave solution presented in my answer. – Jon Nov 7 '18 at 9:13
• Thanks again. So, if the wave is propagating only along $x$, then $W(x,z)=W(x)=\hat{\zeta}exp(ik_x x+ i\phi)$. ? Can I directly take like that ? – newstudent Nov 7 '18 at 17:06