Question

6. Determine whether the following PDEs can be reduced to a set of two ODEs by the method
of separation of variables.

i) d^2 u / dx^2 + d^2 u / dy^2 = x

ii) x d^2 u / dx^2 + t du/dt = 0

Answer 1

Answer on Question #58207 – Math – Differential Equations

Question

Determine whether the following PDE can be reduced to a set of two ODEs by the method of separation of variables.

i) $d^2 u / dx^2 + d^2 u / dy^2 = x$

Solution

i) Let

$$u(x, y) = X(x)Y(y).$$

Then,

$$\frac{\partial^2 u}{\partial x^2} = Y \frac{d^2 X}{d x^2} = Y X''; \quad \frac{\partial^2 u}{\partial y^2} = X \frac{d^2 Y}{d y^2} = X \ddot{Y}.$$

Now,

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

can be rewritten in the following form:

$$Y \frac{d^2 X}{d x^2} + X \frac{d^2 Y}{d y^2} = x$$ $$\frac{Y \frac{d^2 X}{d x^2}}{X Y} + \frac{X \frac{d^2 Y}{d y^2}}{X Y} = \frac{x}{X Y}$$ $$\frac{\frac{d^2 X}{d x^2}}{X} + \frac{\frac{d^2 Y}{d y^2}}{Y} = \frac{x}{X Y}$$ $$\frac{X''}{X} + \frac{\ddot{Y}}{Y} = \frac{x}{X Y}$$

We cannot reduce $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = x$ to a set of two ODEs by the method of separation of variables.

Answer: No.

Question

Determine whether the following PDE can be reduced to a set of two ODEs by the method of separation of variables.

ii) $x d^2 u / dx^2 + t du/dt = 0$

Solution

ii) Let $u(x, t) = X(x)T(t)$. Then,

$$\frac {\partial^ {2} u}{\partial x ^ {2}} = T \frac {d ^ {2} X}{d x ^ {2}} = X ^ {\prime \prime} \cdot T; \quad \frac {\partial u}{\partial t} = X \frac {d T}{d t} = X \cdot \dot {T}.$$

Now,

$$x \frac {\partial^ {2} u}{\partial x ^ {2}} + t \frac {\partial u}{\partial t} = 0$$

can be rewritten in the following form:

$$x T \frac {d ^ {2} X}{d x ^ {2}} + t X \frac {d T}{d t} = 0,$$ $$\frac {x T \frac {d ^ {2} X}{d x ^ {2}}}{X T} + \frac {t X \frac {d T}{d t}}{X T} = 0,$$ $$\frac {x \frac {d ^ {2} X}{d x ^ {2}}}{X} + \frac {t \frac {d T}{d t}}{T} = 0,$$ $$\frac {x \frac {d ^ {2} X}{d x ^ {2}}}{X} = - \frac {t \frac {d T}{d t}}{T},$$ $$\frac {x X ^ {\prime \prime}}{X} = - \frac {t \dot {T}}{T}.$$

The left-hand side is the function of $x$ and the right-hand side is the function of $t$ , therefore, now both sides must be constant, so we set

$$\frac {x X ^ {\prime \prime}}{X} = - \frac {t \dot {T}}{T} = - \lambda .$$

From these we get the ordinary differential equations:

$$\frac {x X ^ {\prime \prime}}{X} = - \lambda ,$$ $$\frac {t \dot {T}}{T} = \lambda ,$$

that is,

$$x X ^ {\prime \prime} + \lambda X = 0,$$ $$t \dot {T} - \lambda T = 0.$$

We can reduce $x \frac{\partial^2 u}{\partial x^2} + t \frac{\partial u}{\partial t} = 0$ to a set of two ODEs by the method of separation of variables.

Answer: Yes.

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