This is an example of as advection equation of the form
$$u_t + cu_x +au = f(x,t)$$
$$u(x,0) = u_0,$$
where $c = 1$, $a = -3$, $f(x,t) = 0$, and $u_0 = x^2$. To solve this equation, we rewrite $u(x,t)$ as $U(\xi,\tau)$, where $\xi = x-ct = x-t$ and $\tau = t.$ By applying the chain rule, the PDE becomes
where we abuse $\phi$ to represent $e^{\phi}$ as is done with constants $C$ in calculus. it may seem odd that we can seemingly just transform an exponential function of $\xi$ into a regular one. Recall, however, that we do not yet know what the original $\phi(\xi)$ is, and it could very well be a logarithmic function, canceling out the exponential. We will get the same answer regardless when we plug in our initial conditions.
Hence our formula for U is
$$U(\xi,\tau) = \phi(\xi)e^{3\tau}.$$
Now we rewrite our equation in terms of $x$ and $t$:
$$u(x,t) = \phi(x-t)e^{3t}.$$
Finally, we plug in our initial values to solve for $\phi$:
To solve this question, we first need perform a change of variables with $\chi = x - ct$ and $\tau = t$ giving $U(\chi,\tau)$. This gives $u(x,t)$ ~ $U(\chi,\tau)$ and $x = \chi + ct$.
To perform the partial derivatives, you first take the derivative of $U(\chi,\tau)$ with respect to $\chi$ and then multiple by the derivative of $x - ct$ with respect to $t$. Next repeat with respect to $\tau$, and then repeat both for $\frac{d}{dx}U(x - ct,t)$. This gives:
After the change of variables, $f(x,t) = F(\chi,\tau) = 0$
Plugging the values of $a$ and $F(\chi,\tau)$ into the $U_\tau + aU(x - ct,t) = F(\chi,\tau)$ gives:
$$U_\tau - 3U(x - ct,t) = 0$$
Therefore $U_\tau = 3U(x - ct,t)$ then divide both sides by $U$ and multiply both by $d\tau$ giving:
$$\frac{dU}{U} = 3d\tau$$
Integrating both sides gives $lnU = 3\tau + K(\chi)$, with $K(\chi)$ is a constant function. To remove the natural log, make both sides powers of $e$ giving:
We are able to separate the $e$ exponent because of the product rule. Next treat $K(\chi)$ as you would an unknown constant raised to be a power of $e$:
$$U = Q(\chi)e^{3\tau}$$
Now we move back to terms of $x$ and $t$, with $\tau = t$ and $\chi = x - ct =x - t$ as $c = 1$ giving:
$$u(x,t) = Q(x - t)e^{3t}$$
$$u(x,0) = x^2$$
$$u(x,0) = Q(x - 0)e^{3*0} = x^2$$
Therefore $u(x,0) = Q(x) = x^2$, plugging back into $u(x,t)$ gives a final solution of:
Comments
This is an example of as advection equation of the form
$$u_t + cu_x +au = f(x,t)$$
$$u(x,0) = u_0,$$
where $c = 1$, $a = -3$, $f(x,t) = 0$, and $u_0 = x^2$. To solve this equation, we rewrite $u(x,t)$ as $U(\xi,\tau)$, where $\xi = x-ct = x-t$ and $\tau = t.$ By applying the chain rule, the PDE becomes
$$(U_{\xi}(-1) + U_{\tau}(1)) + U_{\xi}(1) = 3U.$$
SImplifying, we get
$$U_{\tau} = 3U$$
or
$$\frac{\partial U}{\partial \tau} = 3U.$$
We move all $U$ and $\tau$ terms to one side,
$$\frac{dU}{U} = 3d\tau,$$
and integrate:
$$lnU = 3\tau + \phi(\xi),$$
where $\phi$ is an unknown function of $\xi$. We then solve for U by raising $e$ to both sides:
$$U = e^{3\tau + \phi(\xi)}.$$
We simplify the right hand side using a technique familiar to students of calculus and ordinary differential equations:
$$e^{3\tau + \phi(\xi)} = e^{ \phi(\xi)}e^{3\tau} = \phi(\xi)e^{3\tau},$$
where we abuse $\phi$ to represent $e^{\phi}$ as is done with constants $C$ in calculus. it may seem odd that we can seemingly just transform an exponential function of $\xi$ into a regular one. Recall, however, that we do not yet know what the original $\phi(\xi)$ is, and it could very well be a logarithmic function, canceling out the exponential. We will get the same answer regardless when we plug in our initial conditions.
Hence our formula for U is
$$U(\xi,\tau) = \phi(\xi)e^{3\tau}.$$
Now we rewrite our equation in terms of $x$ and $t$:
$$u(x,t) = \phi(x-t)e^{3t}.$$
Finally, we plug in our initial values to solve for $\phi$:
$$u(x,0) = \phi(x) = x^2.$$
Our final solution, then, is
$$u(x,t) = (x-t)^2e^{3t}.$$
@Abs - Most awesome!!!
The initial value problem is
$$u_t + u_x = 3u, u(x,0) = x^2 .$$
General form of 1st order, liner PDE is:
$$u_t + cu_x + au = f(x,t).$$
Therefore adjusting the initial problem gives:
$$u_t + u_x - 3u = 0.$$
With $c = 1, a = -3$ and $f(x,t) = 0$
To solve this question, we first need perform a change of variables with $\chi = x - ct$ and $\tau = t$ giving $U(\chi,\tau)$. This gives $u(x,t)$ ~ $U(\chi,\tau)$ and $x = \chi + ct$.
Rewriting the right half of the general form as:
$$\frac{d}{dt}u(x,t) + c\frac{d}{dx}u(x,t) + a*u(x,t)$$
Which after the change of variables becomes:
$$\frac{d}{dt}U(x - ct,t) + c\frac{d}{dx}U(x - ct,t) +a*U(x - ct,t)$$
To perform the partial derivatives, you first take the derivative of $U(\chi,\tau)$ with respect to $\chi$ and then multiple by the derivative of $x - ct$ with respect to $t$. Next repeat with respect to $\tau$, and then repeat both for $\frac{d}{dx}U(x - ct,t)$. This gives:
$$U_\chi(-c) + U_\tau(1) + cU_\chi(1) + cU_\tau(0) + aU(x - ct,t)$$
$$= U_\tau + aU(x - ct,t)$$
After the change of variables, $f(x,t) = F(\chi,\tau) = 0$
Plugging the values of $a$ and $F(\chi,\tau)$ into the $U_\tau + aU(x - ct,t) = F(\chi,\tau)$ gives:
$$U_\tau - 3U(x - ct,t) = 0$$
Therefore $U_\tau = 3U(x - ct,t)$ then divide both sides by $U$ and multiply both by $d\tau$ giving:
$$\frac{dU}{U} = 3d\tau$$
Integrating both sides gives $lnU = 3\tau + K(\chi)$, with $K(\chi)$ is a constant function. To remove the natural log, make both sides powers of $e$ giving:
$$U = e^{3\tau + K(\chi)} = e^{3\tau}e^{K(\chi)}$$
We are able to separate the $e$ exponent because of the product rule. Next treat $K(\chi)$ as you would an unknown constant raised to be a power of $e$:
$$U = Q(\chi)e^{3\tau}$$
Now we move back to terms of $x$ and $t$, with $\tau = t$ and $\chi = x - ct =x - t$ as $c = 1$ giving:
$$u(x,t) = Q(x - t)e^{3t}$$
$$u(x,0) = x^2$$
$$u(x,0) = Q(x - 0)e^{3*0} = x^2$$
Therefore $u(x,0) = Q(x) = x^2$, plugging back into $u(x,t)$ gives a final solution of:
$$u(x,t) = (x - t)^2 * e^{3t}$$