2.8 A First Look at Partial Diferential Equations

Just evaluate $u_t(x,t)$ and $u_{xx}(x,t)+u(x,t)$ and stare at them for a while.

$g(x) = 1-4x^2$

We start by evaluating $u_t(x,t)$ and $u_{xx}(x,t)+u(x,t)$ when $u(x,t)= e^{-t-x^2}$.

So

For this to equal $g(x)\, u_t(x,t) = -g(x)\, e^{-t-x^2}$, we need $g(x) = 1-4x^2$.

(a), (b) Fix any $y_0$ and set $v(x)=u(x,y_0)$. What is $\diff{v}{x}(x)$?

(a) $u(x,y)=C(y)$ with $C(y)$ being any function of the single variable $y$.

(b) $u(x,y)=F(x)+C(y)$ where $F(x)$ is any function obeying $F'(x)=f(x)$ (i.e. any antiderivative of $f(x)$) and $C(y)$ is any function of the single variable $y$.

(a) Fix any $y_0$ and set $v(x)=u(x,y_0)$. Then

So, for each fixed $y_0$, $v(x)=u(x,y_0)$, which is a function of $x$, has to be a constant. The constant may be different for each different choice of $y_0$. So $u(x,y_0) = C(y_0)$ with $C(y_0)$ depending only on $y_0$, not on $x$. Or, renaming $y_0$ back to $y$, $u(x,y)=C(y)$ with $C(y)$ being any function of the single variable $y$.

(b) Fix any $y_0$ and set $v(x)=u(x,y_0)$. Then

In words, $v(x)$ has to have derivative $f(x)$, i.e. be an antiderivative of $f(x)$. So if $F(x)$ is any function whose derivative is $f(x)$, i.e. if $F(x)$ is any antiderivative of $f(x)$, then, for each fixed $y_0$, $v(x)=u(x,y_0) = F(x) +C$, with $C$ being a constant. The constant may be different for each different choice of $y_0$. So $u(x,y_0) = F(x)+ C(y_0)$ with $C(y_0)$ depending only on $y_0$, not on $x$. Or, renaming $y_0$ back to $y$, $u(x,y)=F(x) + C(y)$ with $F(x)$ being any antiderivative of $f(x)$ and $C(y)$ being any function of the single variable $y$.

(a), (d) and (e) are harmonic. (b) and (c) are not harmonic.

(a) If $u(x,y) = x^3-3xy^2$, then

So $u_{xx}(x,y)+u_{yy}(x,y)=6x-6x=0$ and $x^3-3xy^2$ is harmonic.

(b) If $u(x,y) = x^3-y^3$, then

So $u_{xx}(x,y)+u_{yy}(x,y)=6x-6y$ is not identically zero and $x^3-y^3$ is not harmonic.

(c) If $u(x,y) = \sin(x)\,\cos(y)$, then

So $u_{xx}(x,y)+u_{yy}(x,y)=-2\sin(x)\,\cos(y)$ is not identically zero and $\sin(x)\,\cos(y)$ is not harmonic.

(d) If $u(x,y) = e^{7x}\,\cos(7y)$, then

So $u_{xx}(x,y)+u_{yy}(x,y)=49\,e^{7x}\,\cos(7y)-49\,e^{7x}\,\cos(7y)=0$ and $e^{7x}\,\cos(7y)$ is harmonic.

(e) If $u(x,y) = \ln(x^2+y^2)$, then

So

and $\ln(x^2+y^2)$ is harmonic.

Just substitute the given $u(x,t)$ into the given PDE.

$a=\pm 2$

We evaluate both sides of the given PDE with $u=u(x,t) = e^{t+ax} + e^{t-ax}$. Since

the left hand side of the PDE is

and the right hand side of the PDE is

The left and right hand sides are equal if and only if

Just substitute the given $u(x,y,z)$ into the given PDE.

$a=\pm 5$

We evaluate $u_{xx}+u_{yy}+u_{zz}$ with $u=u(x,t) = e^{3x+4y}\sin(az)$. Since

We have

This is zero (for all $x$, $y$, $z$) if and only if

Just substitute the given $u(x,t)$ into the given PDE.

$a=\pm b$, for any real number $b$.

We evaluate both sides of the given PDE with $u=u(x,t)=\sin(at)\,\cos(bx)$. Since

the left hand side of the PDE is

and the right hand side of the PDE is

The left and right hand sides are equal if and only if

Just substitute the given $z(x,y)$ into the given PDE.

Yes it is. For the justification, see the solution.

We simply evaluate the two terms on the left hand side when $z=z(x,y)=F\big(x^2+y^2\big)$. By the chain rule,

So

and $z(x,y)=F\big(x^2+y^2\big)$ really does solve the PDE $y\pdiff{z}{x} - x\pdiff{z}{y}=0$ for any differentiable function $F$.

Substitute the given $u(x,t)$ into the given PDE. Review Theorem 3.3.2 in the CLP-1 text.

$f(t) = Ce^{-4t}$ with $C$ being an arbitrary constant.

We evaluate both sides of the given PDE with $u=u(x,t)=f(t)\,\cos(2x)$. Since

the left hand side of the PDE is

and the right hand side of the PDE is

The left and right hand sides are equal if and only if

This is the type of ordinary differential equation that we studied in Section 3.3, on exponential growth and decay, in the CLP-1 text. We found in Theorem 3.3.2 there that the general solution to this ODE is $f(t) = Ce^{-4t}$ with $C$ being an arbitrary constant.

See the solution.

Let $u_1(x,t)$ and $u_2(x,t)$ obey $\frac{\partial^2}{\partial t^2}u_1(x,t) =\frac{\partial^2}{\partial x^2}u_1(x,t)$ and $\frac{\partial^2}{\partial t^2}u_2(x,t) =\frac{\partial^2}{\partial x^2}u_2(x,t)$. Then $u(x,t)=a_1u_1(x,t)+a_2u_2(t,x)$ obeys

as desired.

Evaluate $u_{xx}+u_{yy}$ for the given $u(x,y)$.

perpendicular

We evaluate $u_{xx}+u_{yy}$ with $u(x,y)=v(ax+by\,,\,cx+dy)$. Since, by the chain rule,

we have

So if $\llt a,b\rgt$ and $\llt c,d\rgt$ have the same length and are mutally perpendicular, then $u_{xx}+u_{yy}=0$. The missing word is “perpendicular”.

$n=0, -1$

In preparation for substituting into the PDE, we compute $u_{xx}$, $u_{yy}$ and $u_{zz}$.

So

This is zero if and only if

(b) The left hand side is independent of $x$ and the right hand side is independent of $t$.

(c) Review Section 3.3 in the CLP-1 text and Section 2.4 in the CLP-2 text.

(a), (b) See the solutions.

(c) $T(t) = Ce^{\la t}$, $X(x) = K x^\la$, $u(x,t)= D\,e^{\la t}\,x^\la$ with $C$, $D$ and $K$ being arbitrary positive constants.