How to determine the integration constants in solving the Klein Gordon equation?

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SUMMARY

The discussion focuses on solving the Klein-Gordon equation and determining the integration constants involved. The solution presented involves integrating the equation to obtain the form ##\phi(t,x) = x t + c_1 t + c_2##, where ##c_2## is determined to be 0 using the initial condition ##\phi(0,x)=0##. The participants emphasize that the integration constants are not merely constants but can also be functions of other variables, necessitating boundary and initial conditions to fully define them. The final solution for ##\phi(t,x)## is expressed as a product of exponential functions, with specific constants derived from boundary and initial conditions.

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Safinaz
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Homework Statement
How to solve the following wave equation for the scalar ##\phi(t,x)## :
Relevant Equations
##\partial_i \dot{\phi}=0 ##

Where ##\partial_i= \partial/\partial x##. And (.) is the derivative with respect for time ## \partial/\partial t##
I solved by

##
\int d \dot{\phi} = \int d x \to
\dot{\phi} = x+ c_1 \to \int d \phi = \int d t ( x+c_1)
\to \phi = x t + c_1 t + c_2
##

Is this way correct? To determine ##c_2## use initial condition: ##\phi(0,x)=0## that yields ##c_2=0##, but how to get ##c_1## ?
 
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The kernel of the operator \partial_x consists of all functions f for which \partial_x f = 0; this includes constants, but also includes functions which depend only on t. Therefore the most general element of this kernel is A(t). Hence \int \partial_x \dot \phi\,dx = \int 0\,dx \Rightarrow \dot\phi = A(t). Now integrate with respect to t. This time, the most general element of the kernel of \partial/\partial t is a function of x alone.
 
Safinaz said:
Homework Statement: How to solve the following wave equation for the scalar $\phi(t,x)$ :
Relevant Equations: ##\partial_i \dot{\phi}=0 ##

Where ##\partial_i= \partial/\partial x##. And (.) is the derivative with respect for time ## \partial/\partial t##

I solved by

##
\int d \dot{\phi} = \int d x \to
\dot{\phi} = x+ c_1 \to \int d \phi = \int d t ( x+c_1)
\to \phi = x t + c_1 t + c_2
##

Is this way correct? To determine ##c_2## use initial condition: ##\phi(0,x)=0## that yields ##c_2=0##, but how to get ##c_1## ?
Your function provably does not satisfy ##\partial_i\partial_t \phi = 0##. Just try to differentiate it!

(apart from what was already said)
 
pasmith said:
The kernel of the operator \partial_x consists of all functions f for which \partial_x f = 0; this includes constants, but also includes functions which depend only on t. Therefore the most general element of this kernel is A(t). Hence \int \partial_x \dot \phi\,dx = \int 0\,dx \Rightarrow \dot\phi = A(t). Now integrate with respect to t. This time, the most general element of the kernel of \partial/\partial t is a function of x alone.
You mean

##\dot{\phi} =A(t) \to \int \partial_t \phi = \int A(t) dt \to \phi = A(t) t + c ? ##
but what is A(t) ?
 
Safinaz said:
You mean

##\dot{\phi} =A(t) \to \int \partial_t \phi = \int A(t) dt \to \phi = A(t) t + c ? ##
but what is A(t) ?
First of all, you cannot integrate a general function A(t) with respect to t and obtain A(t) t. Not even in single variable calculus.

Second, you are still missing what was said. If ##\partial_t \phi = A(t)##, then ##\phi = a(t) + f(x)##, where ##a’(t) = A(t)##. The “integration constant” when integrating a partial derivative is generally a function of all of the other variables.
 
Orodruin said:
First of all, you cannot integrate a general function A(t) with respect to t and obtain A(t) t. Not even in single variable calculus.

Second, you are still missing what was said. If ##\partial_t \phi = A(t)##, then ##\phi = a(t) + f(x)##, where ##a’(t) = A(t)##. The “integration constant” when integrating a partial derivative is generally a function of all of the other variables.
Okay. But now how to get the definition of ##f(x)## and ##a(t)## ?
 
Safinaz said:
Okay. But now how to get the definition of ##f(x)## and ##a(t)## ?
Just as you need boundary or initial conditions to fix integration constants for ODEs, you will need boundary/initial conditions to fix those functions.
 
Orodruin said:
Just as you need boundary or initial conditions to fix integration constants for ODEs, you will need boundary/initial conditions to fix those functions.
Hello. Thanks so much for your answer. I was trying to find proper IC and BC to find ## \phi(t,x)## . Assuming:

##
bc={\phi[t,0]==1,(D[\phi[t,x],x]/.x->Pi)==0}
##
##
ic={\phi[0,x]==0,(D[\phi[t,x],t]/.t->0)==1}
##

Also ## \phi(t,x)## obays the Klein Gordon’s equation :
## \left( \frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial x^2} \right) \phi = 0 ##

in which solution:
##
\phi(t,x) = ( c_1 e^{kt} + c_2 e^{-kt} ) ( c_3 e^{kx} + c_4 e^{-kx} ) …………(1)
##

To find the constants in Eq. (1) , BC leads to ##c_3 = c_4= 1/2 ## and the IC leads to to ##c_1=- c_2= 1/2 ## is that correct? But how to know ## k ## ?
 
Last edited:

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