- #1

- 141

- 5

I'm having troubles with PDE.

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

a)

[itex]x\frac{\partial u}{\partial x}=y\frac{\partial u}{\partial y}[/itex]

I suppose that:

[itex]u=X(x) \cdot Y(y)[/itex]

Then:

[itex]xX'Y=yXY'[/itex]

[itex]xX'/X=yY'/Y[/itex]

So [itex]xX'/X=yY'/Y=c[/itex] because they can't be in function of x or y.

[itex]X'/X=c/x=ln(X(x))'[/itex]

We integrate both sides and then ln(X(x))=c ln(x)

Then [itex]X(x)=cx[/itex]

And [itex]Y(y)=cy[/itex]

b)

[itex]\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial xy}+\frac{\partial^2 u}{\partial y^2}=0[/itex]

Using the same procedure:

[itex]u=X(x) \cdot Y(y)[/itex]

[itex]X''Y+X'Y'+XY''=0[/itex]

????

Edit: I can't solve the exercise b), and I'm not sure if the other ones were solved.

c)

[itex]k \cdot \frac{\partial^2 u}{\partial x^2}-u=\frac{\partial u}{\partial t}[/itex] and [itex]k>0[/itex]

(I changed the t for y because I'm used to)

[itex]kX''Y-XY=XY'[/itex]

[itex]k\frac{X''}{X}=\frac{Y'}{Y}+1[/itex]

[itex]Y=e^{c_1 y}[/itex]

[itex]\frac{X''}{X}=\frac{c_1}{k}+\frac{1}{k}[/itex]

If [itex]c_t=\frac{c_1}{k}+\frac{1}{k}>0[/itex] then [itex]X=c_1 e^{\sqrt{c_t}x} + c_2 e^{-\sqrt{c_t}x}[/itex], 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:

[itex]X=c_1 cos(\sqrt{c_t}x) +c_2 sin(\sqrt{c_t}x)[/itex]

d)

[itex]\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0[/itex]

[itex]X''Y+XY''=0[/itex]

[itex]X''/X=-Y''/Y=c[/itex]

[itex]X''=cX[/itex]

[itex]X''-cX=0[/itex]

If c=0 then [itex]X=c_1 + c_2x[/itex] and [itex]Y=c_3 + c_4y[/itex]

If c>0 then [itex]X=c_1 \cdot e^{-\sqrt{c}x} +c_2 \cdot e^{\sqrt{c}x}[/itex]

and [itex]Y=c_3 \cdot cos(\sqrt{c}y)+c_4 \cdot sin(\sqrt{c}y)[/itex]

And [itex]c(c_1+c_2)=c[/itex]

[itex]c_1+c_2=1[/itex]

And [itex]c_3+c_4=1[/itex]

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)

[itex]\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=u[/itex]

I guess the solution here is the homogeneous solution of d) plus:

[itex]e^{\frac{1}{2}(x+y)}[/itex]

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

a)

[itex]x\frac{\partial u}{\partial x}=y\frac{\partial u}{\partial y}[/itex]

I suppose that:

[itex]u=X(x) \cdot Y(y)[/itex]

Then:

[itex]xX'Y=yXY'[/itex]

[itex]xX'/X=yY'/Y[/itex]

So [itex]xX'/X=yY'/Y=c[/itex] because they can't be in function of x or y.

[itex]X'/X=c/x=ln(X(x))'[/itex]

We integrate both sides and then ln(X(x))=c ln(x)

Then [itex]X(x)=cx[/itex]

And [itex]Y(y)=cy[/itex]

b)

[itex]\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial xy}+\frac{\partial^2 u}{\partial y^2}=0[/itex]

Using the same procedure:

[itex]u=X(x) \cdot Y(y)[/itex]

[itex]X''Y+X'Y'+XY''=0[/itex]

????

Edit: I can't solve the exercise b), and I'm not sure if the other ones were solved.

c)

[itex]k \cdot \frac{\partial^2 u}{\partial x^2}-u=\frac{\partial u}{\partial t}[/itex] and [itex]k>0[/itex]

(I changed the t for y because I'm used to)

[itex]kX''Y-XY=XY'[/itex]

[itex]k\frac{X''}{X}=\frac{Y'}{Y}+1[/itex]

[itex]Y=e^{c_1 y}[/itex]

[itex]\frac{X''}{X}=\frac{c_1}{k}+\frac{1}{k}[/itex]

If [itex]c_t=\frac{c_1}{k}+\frac{1}{k}>0[/itex] then [itex]X=c_1 e^{\sqrt{c_t}x} + c_2 e^{-\sqrt{c_t}x}[/itex], 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:

[itex]X=c_1 cos(\sqrt{c_t}x) +c_2 sin(\sqrt{c_t}x)[/itex]

d)

[itex]\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0[/itex]

[itex]X''Y+XY''=0[/itex]

[itex]X''/X=-Y''/Y=c[/itex]

[itex]X''=cX[/itex]

[itex]X''-cX=0[/itex]

If c=0 then [itex]X=c_1 + c_2x[/itex] and [itex]Y=c_3 + c_4y[/itex]

If c>0 then [itex]X=c_1 \cdot e^{-\sqrt{c}x} +c_2 \cdot e^{\sqrt{c}x}[/itex]

and [itex]Y=c_3 \cdot cos(\sqrt{c}y)+c_4 \cdot sin(\sqrt{c}y)[/itex]

And [itex]c(c_1+c_2)=c[/itex]

[itex]c_1+c_2=1[/itex]

And [itex]c_3+c_4=1[/itex]

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)

[itex]\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=u[/itex]

I guess the solution here is the homogeneous solution of d) plus:

[itex]e^{\frac{1}{2}(x+y)}[/itex]

Last edited: