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Simple partial differential equation

by dexter90
Tags: differential, equation, partial, simple
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dexter90
#1
May1-12, 02:52 PM
P: 15
Hello.

I have equation:

[tex]\frac{\partial T}{\partial t}-\frac{1}{2}\cdot \frac{(\partial)^2 T}{\partial x^2}=0[/tex]

I calculated determinant:

[tex]\Delta=(-\frac{1}{2})^2)-4\cdot 1 \cdot 0 \Rightarrow \sqrt{\Delta}=\frac{1}{2} \\ (\frac{dT}{dt})_{1}=-\frac{1}{4} \\ (\frac{dT}{dt})_{2}=\frac{1}{4}[/tex]

next

[tex]T=-\frac{1}{4}t+C_{1} \Rightarrow T+\frac{1}{4}t=C_{1} \\ T=\frac{1}{4}t+C_{2} \Rightarrow T-\frac{1}{4}t=C_{2}[/tex]

I am add a new coefficients [tex]\eta[/tex] and [tex]\xi[/tex], then

[tex]\xi=\frac{1}{4}t+T\\ \eta=-\frac{1}{4}t+T[/tex]

Then I calculated matrix jacobian's =[tex]\frac{1}{2}[/tex]

Good?

I greet

Post edited
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#2
May1-12, 04:11 PM
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Quote Quote by dexter90 View Post
Hello.

I have equation:

[tex]\frac{\partial T}{\partial t}-\frac{1}{2}\cdot \frac{(\partial)^2 T}{\partial x^2}=0[/tex]

I calculated determinant:

[tex]\Delta=(-\frac{1}{2})^2)-4\cdot 1 \cdot 0 \Rightarrow \sqrt{\Delta}=\sqrt{2} \\ (\frac{dT}{dt})_{1}=-\sqrt{2} \\ (\frac{dT}{dt})_{2}=\sqrt{2}[/tex]
I'm not all that clear on what you are doing but that first statement is obviously untrue.
[tex]\Delta= (-\frac{1}{2})^2- 4\cdot 1 \cdot 0= \frac{1}{4}[/tex]
so
[tex]\sqrt{\Delta}= \frac{1}{2}[/tex]
not [itex]\sqrt{2}[itex]

next

[tex]T=-\sqrt{2}t+C_{1} \Rightarrow T+\sqrt{2}t=C_{1} \\ T=\sqrt{2}t+C_{2} \Rightarrow T-\sqrt{2}t=C_{2}[/tex]

I am add a new coefficients [tex]\eta[/tex] and [tex]\xi[/tex], then

[tex]\xi=\sqrt{2}t+T\\ \eta=-\sqrt{2}t+T[/tex]

Then I calculated matrix jacobian's =[tex]2\sqrt{2}[/tex]

Good?

I greet
dexter90
#3
May2-12, 02:02 AM
P: 15
Thanks,

Of course, I made mistake in write. I would like solve partial differential equation but I dont have experience. I edited my post.


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