Solving Partial Differential Equations Using Separation of Variables

[SqueeSpleen]

I'm having troubles with PDE.

Apply separation of variables, if possible, to found product solutions to the following differential equations.

a)
x\frac{\partial u}{\partial x}=y\frac{\partial u}{\partial y}
I suppose that:
u=X(x) \cdot Y(y)
Then:
xX'Y=yXY'
xX'/X=yY'/Y
So xX'/X=yY'/Y=c because they can't be in function of x or y.

X'/X=c/x=ln(X(x))'
We integrate both sides and then ln(X(x))=c ln(x)
Then X(x)=cx
And Y(y)=cy

b)
\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial xy}+\frac{\partial^2 u}{\partial y^2}=0
Using the same procedure:
u=X(x) \cdot Y(y)

X''Y+X'Y'+XY''=0
?
Edit: I can't solve the exercise b), and I'm not sure if the other ones were solved.

c)
k \cdot \frac{\partial^2 u}{\partial x^2}-u=\frac{\partial u}{\partial t} and k>0
(I changed the t for y because I'm used to)
kX''Y-XY=XY'
k\frac{X''}{X}=\frac{Y'}{Y}+1
Y=e^{c_1 y}
\frac{X''}{X}=\frac{c_1}{k}+\frac{1}{k}
If c_t=\frac{c_1}{k}+\frac{1}{k}>0 then X=c_1 e^{\sqrt{c_t}x} + c_2 e^{-\sqrt{c_t}x}, if it's equal to zero then it's a polynomial of degree 1, and if it's lesser than zero then it's a linear combination of sins and cosines:
X=c_1 cos(\sqrt{c_t}x) +c_2 sin(\sqrt{c_t}x)

d)
\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0

X''Y+XY''=0
X''/X=-Y''/Y=c
X''=cX
X''-cX=0
If c=0 then X=c_1 + c_2x and Y=c_3 + c_4y
If c>0 then X=c_1 \cdot e^{-\sqrt{c}x} +c_2 \cdot e^{\sqrt{c}x}
and Y=c_3 \cdot cos(\sqrt{c}y)+c_4 \cdot sin(\sqrt{c}y)
And c(c_1+c_2)=c
c_1+c_2=1
And c_3+c_4=1
So we could use only 2 constants (I don't know why I got 4 in c=0) ... Can I have up to 4 constants if I have 2 independent variables with their maximum derivative of degree 2?
If c<0 it's the same than the previous case but X and Y swap their places.

e)
\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=u
I guess the solution here is the homogeneous solution of d) plus:
e^{\frac{1}{2}(x+y)}

 

[haruspex]

X&#039;/X=c/x=ln(X(x))&#039;
We integrate both sides and then ln(X(x))=c ln(x)

Don't forget the constant of integration.

Then X(x)=cx

Have another think about that step.

 

[SqueeSpleen]

I'm sometimes play lazy with the constants names, I guess e^c is my new constant is the thing you're speaking of.

 

[haruspex]

I'm sometimes play lazy with the constants names, I guess e^c is my new constant is the thing you're speaking of.

No, there are two separate issues. The constant of integration is not related to c, which is a constant you were given. The step I suggested you have another think about is plain wrong, regardless of any constant of integration.

 

[SqueeSpleen]

I don't really understand. I'll give it another try. Sorry if I was a little rude and didn't thought very much your advice the previous time, I really thought I did right the first exercise.
ln(X(x))=c ln(x) +c_1
If I put a c_1 here, then:
X(x)=c_1 x^{c}
x \frac{X&#039;}{X}=c
And
Y(x)=c_2 y^{c}
This is right?

 

[SqueeSpleen]

I don't have much idea what I'm doing here.
X&#039;&#039;Y+X′Y′+XY′′
\frac{X&#039;&#039;Y}{X&#039;Y&#039;}+1+\frac{XY′′}{X&#039;Y&#039;}
Then
\frac{X&#039;&#039;Y}{X&#039;Y&#039;}+\frac{XY′′}{X&#039;Y&#039;}=-1
Let's try exponential solutions for both sides. Then we have 1/c+c=-1
c^2+c+1=0
k_1=\frac{1 - \sqrt{-3}}{2} k_2=\frac{1 + \sqrt{-3}}{2}
Then if X and Y are both exponentials:
c_1/c_2=c
Now I guess I got
X=e^{c1x}
Y=e^{\frac{k_2}{c1} y}
And I think it works, I got 1/c\frac{X&#039;&#039;Y}{X&#039;Y&#039;}+\frac{XY′′}{X&#039;Y&#039;}=1/k_1+k_1 and I know this is -1. This solution work.
But I'm not sure what can happen if I add it to:
X=e^{c1x}
Y=e^{\frac{k_2}{c1} y}
I guess I'll try to found a solution to this in a book.

 

[haruspex]

x \frac{X&#039;}{X}=c
And
Y(x)=c_2 y^{c}
This is right?

Yes. The question actually asks for product solutions, so maybe put the answer as $c_1(xy)^c$.

For (b), I don't see how to solve it as a product either. It does say "if possible".
(c) looks right.
In (d), for c=0 you didn't really get four constants of integration. When you write out u, there's a redundant one.
Anyway, don't expect to get the same number of arbitrary constants in each case. Remember that the full solution would be a sum of these product solutions, so the actual number of arbitrary constants is unlimited. (In practice, would be determined by boundary conditions.)

Your answer for (e) is wrong. You are to find a single expression of the form XY, so it's not a matter of adding in an inhomogeneous term. Work it exactly as you did in (d).

 

[SqueeSpleen]

\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=u

X&#039;&#039;Y+XY&#039;&#039;=XY
\frac{X&#039;&#039;}{X}+\frac{Y&#039;&#039;}{Y}=1
\frac{X&#039;&#039;}{X}=k_1 --- \frac{Y&#039;&#039;}{Y}=k_2 --- k_1+k_2=1 (How can I put more than one space here? I used the white --- to make it.)
If k_1 &lt;0 then
X=c_1 e^{\sqrt{k_1} x}+c_2 e^{-\sqrt{k_1} x} and Y=c_3 cos(\sqrt{k_2}y)+c_4 sin(\sqrt{k_2}y)
Then the solution is the product of both.
If k_1 &gt;1 then x and y swap their places.
If 0 &lt; k_1 &lt; 1 (*) then:
X=c_1 e^{\sqrt{k_1} x}+c_2 e^{-\sqrt{k_1} x} and
Y=c_3 e^{\sqrt{k_2} y}+c_4 e^{-\sqrt{k_2} y}
If k_1=0 then Y is the same as (*) and X is a polynomial of degree 1.
If k_1=1 then X is the same as (*) and Y is a polynomial of degree 1.In b) the only solution I found is the almost trivial U(x,y)=(x+c_1)(y+c_2)

 

[haruspex]

In b) the only solution I found is the almost trivial U(x,y)=(x+c_1)(y+c_2)

I don't think that is a solution. Do you mean ax+by+c?
You can easily generate others by trying expressions like ax³+bx²y+cxy²+dy³. Some fairly simple constraints on the constants make it a solution. But these cannot be of the form X(x)Y(y).

 

[SqueeSpleen]

Yes, you're right, it's not a solution because \frac{\partial^2}{\partial xy}u=1.
I'll guess if I want to know more I'll have to take a course of PDE (this homework was of a general course of DE with only have 2 weeks to PDEs).