Solving Second Order Partial Derivative By Changing Variable

  • #1
1. The problem statement, all variables, and given/known data
Given is a second order partial differential equation $$u_{xx} + 2u_{xy} + u_{yy}=0$$ which should be solved with change of variables, namely ##t = x## and ##z = x-y##.

Homework Equations


Chain rule $$\frac{dz}{dx} = \frac{dz}{dy} \cdot \frac{dy}{dx}$$

The Attempt at a Solution


I was able to make the first change of variables $$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial t} \cdot \frac{\partial t}{\partial x} + \frac{\partial u}{\partial z} \cdot \frac{\partial z}{\partial x} = u_t + u_z$$ but I'm stuck at making the second change of variables (for second derivative). If I attempt repeating the same process I end up with $$\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x}(\frac{\partial u}{\partial t} \cdot \frac{\partial t}{\partial x} + \frac{\partial u}{\partial z} \cdot \frac{\partial z}{\partial x}) = \frac{\partial}{\partial x} \cdot \frac{\partial u}{\partial t} + \frac{\partial}{\partial x} \cdot \frac{\partial u}{\partial z}$$I don't think that I can write something like ##u_{tx}##?

I think this task is easy to solve, I'm apparently just missing something, Can somebody please help me? I don't expect anyone to solve homework tasks instead of me, I just need some guidance.

PS: Hope you're having a wonderful Tuesday!
 

Answers and Replies

  • #2
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Under certain conditions, i.e. if the second derivatives of ##u## are continuous, you can according to Schwartz theorem change the order of the mixed derivatives. E.g, $$\frac{\partial^2 u}{\partial x \partial t}=\frac{\partial^2 u}{\partial t \partial x}=\frac{\partial}{\partial t}(\frac{\partial u}{\partial x})$$.
 
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  • #3
Orodruin
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Under certain conditions, i.e. if the second derivatives of ##u## are continuous, you can according to Schwartz theorem change the order of the mixed derivatives. E.g, $$\frac{\partial^2 u}{\partial x \partial t}=\frac{\partial^2 u}{\partial t \partial x}=\frac{\partial}{\partial t}(\frac{\partial u}{\partial x})$$.
This is true when you are taking second derivatives with respect to independent variables - in the case of this problem t and z or x and y. It is not generally true when you make coordinate changes!
 
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  • #4
Orodruin
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1. The problem statement, all variables, and given/known data
Given is a second order partial differential equation $$u_{xx} + 2u_{xy} + u_{yy}=0$$ which should be solved with change of variables, namely ##t = x## and ##z = x-y##.

Homework Equations


Chain rule $$\frac{dz}{dx} = \frac{dz}{dy} \cdot \frac{dy}{dx}$$

The Attempt at a Solution


I was able to make the first change of variables $$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial t} \cdot \frac{\partial t}{\partial x} + \frac{\partial u}{\partial z} \cdot \frac{\partial z}{\partial x} = u_t + u_z$$ but I'm stuck at making the second change of variables (for second derivative). If I attempt repeating the same process I end up with $$\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x}(\frac{\partial u}{\partial t} \cdot \frac{\partial t}{\partial x} + \frac{\partial u}{\partial z} \cdot \frac{\partial z}{\partial x}) = \frac{\partial}{\partial x} \cdot \frac{\partial u}{\partial t} + \frac{\partial}{\partial x} \cdot \frac{\partial u}{\partial z}$$I don't think that I can write something like ##u_{tx}##?

I think this task is easy to solve, I'm apparently just missing something, Can somebody please help me? I don't expect anyone to solve homework tasks instead of me, I just need some guidance.

PS: Hope you're having a wonderful Tuesday!
Your expression for the partial derivative of u with respect to x is true for any function u. In particular, it is true if you replace u by ##u_t##.
 
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  • #5
Ray Vickson
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1. The problem statement, all variables, and given/known data
Given is a second order partial differential equation $$u_{xx} + 2u_{xy} + u_{yy}=0$$ which should be solved with change of variables, namely ##t = x## and ##z = x-y##.

Homework Equations


Chain rule $$\frac{dz}{dx} = \frac{dz}{dy} \cdot \frac{dy}{dx}$$

The Attempt at a Solution


I was able to make the first change of variables $$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial t} \cdot \frac{\partial t}{\partial x} + \frac{\partial u}{\partial z} \cdot \frac{\partial z}{\partial x} = u_t + u_z$$ but I'm stuck at making the second change of variables (for second derivative). If I attempt repeating the same process I end up with $$\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x}(\frac{\partial u}{\partial t} \cdot \frac{\partial t}{\partial x} + \frac{\partial u}{\partial z} \cdot \frac{\partial z}{\partial x}) = \frac{\partial}{\partial x} \cdot \frac{\partial u}{\partial t} + \frac{\partial}{\partial x} \cdot \frac{\partial u}{\partial z}$$I don't think that I can write something like ##u_{tx}##?

I think this task is easy to solve, I'm apparently just missing something, Can somebody please help me? I don't expect anyone to solve homework tasks instead of me, I just need some guidance.

PS: Hope you're having a wonderful Tuesday!
If you introduce the operators ##D_x = \partial / \partial x## and ##D_y = \partial / \partial y## your pde is ##(D_x+D_y)^2 u =0##. Thus, if ##v = (D_x + D_y) u## we have ##(D_x + D_y) v = 0, ## whose general solution is ##v = f(x-y)## for some differentiable univariate function ##f(\cdot)##. The pde becomes ##(D_x+D_y)u(x,y) = f(x-y)##. I think this new pde is fairly straightforward to solve, in terms of the new independent variables ##x-y## and ##x+y##.
 
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  • #6
I apologize for the late response, did read through the post and managed to finally solve the task. I would like to thank everyone here for helping me out on that one, I appreciate your time and effort! Really helped me a lot!

Hope you're all having a fantastic Friday ;)
 

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