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2.Find the particular solution of $u_x+2u_y-4u=e^{x+y}$ satisfying the following side condition $u(x,-x) = x$

I know that under that condition $y = -x$ which is the reflection of the $x$ graph. I have problems with taking the integration with respect to z because my change of variables is causing me to have a wicked exponential.

Anyway, here is my attempt

The slope is $\frac{2}{1}$ which is is 2, so my characteristic lines must be positive. The characteristic lines are in the form of $bx-ay=d$ which is for the pde $au_x+bu_y+cu=f(x,y)$

Using my change of variables, I get the following lines

$ 2x - y = w \rightarrow x=\frac{w+z}{2}$

$z = y \rightarrow y = z$

Now I have to take the partial derivatives and the chain rule.

$W_x = 2, W_y = -1, Z_x = 0, Z_y = 1$

$V_wW_x+V_zZ_x+2(V_wW_y+V_zZ_y) -4(v) = e^{\frac{w+z}{2}+z}$

$2V_w+2(-V_w+V_z) -4(v) = e^{\frac{w+z}{2}+z}$

$2V_w-2V_w+V_z -4(v) = e^{\frac{w+z}{2}+z}$

$V_z -4(v) = e^{\frac{w+z}{2}+z}$

Let $p(a) = -4$, and $q(a) = e^{\frac{w+z}{2}+z}$. Then my integrating factor would be. $e^{\int -4} \rightarrow e^{-4z}$

Multiplying the integrating factor, I get

$e^{-4z}V_z -e^{-4z}4(v) = e^{-4z}e^{\frac{w+z}{2}+z}$

$e^{-4z}V_z -e^{-4z}4(v) = e^{\frac{w+z}{2}+z}e^{-4z}$

$e^{-4z}V_z -e^{-4z}4(v) = e^{\frac{w+z}{2}+z-4z}$

$e^{-4z}V_z -e^{-4z}4(v) = e^{\frac{w+z}{2}-3z}$

$e^{-4z}V_z -e^{-4z}4(v) = e^{\frac{w+z-6z}{2}}$

$e^{-4z}V_z -e^{-4z}4(v) = e^{\frac{w-5z}{2}}$

By reverse product rule

$e^{-4z}(v) = \int e^{\frac{w-5z}{2}}$

Integrating the right hand side, I get

$e^{-4z}(v) = \frac{-2}{5}e^{\frac{w-5z}{2}}+C(w)$

$ v = \frac{\frac{-2}{5}e^{\frac{w-5z}{2}}}{e^{-4z}}$

$ v = \frac{-2}{5}e^{\frac{w-5z+4z}{2}}$

$ v = \frac{-2}{5}e^{\frac{w-z}{2}} + C(w)$

Substituting back I get

$ u = \frac{-2}{5}e^{\frac{2x-y-y}{2}} +C(2x-y)$

$ u = \frac{-2}{5}e^{\frac{2x-2y}{2}}+C(2x-y)$

from side condition of $u(x,-x) = x$, we have

$x = \frac{-2}{5}e^{\frac{2x-2(-x)}{2}}+C(x)$

$x = \frac{-2}{5}e^{\frac{2x+2x}{2}}+C(x)$

$x = \frac{-2}{5}e^{\frac{4x}{2}}+C(x)$

$x = \frac{-2}{5}e^{2x}+C(x)$

The problem is... that's not right because the answer is $u(x,y)=-e^{x+y}+(\frac{2}{3}(x-\frac{1}{2}y+1)e^{\frac{4}{3}(x-\frac{1}{2}y)}e^{2y}$

What happened?

Edit: even if I did rewrite it correctly, the $\frac{-2}{5}$ is still there!!!

$v = \frac{-2}{5}e^{\frac{w-5z}{2}}e^{4z}+C(w)e^{4z}$

Applying change of variables

$u = \frac{-2}{5}e^{\frac{2x-y-5y}{2}}e^{4y}+C(2x-y)e^{4y}$

$u = \frac{-2}{5}e^{\frac{2x-6y}{2}}e^{4y}+C(2x-y)e^{4y}$

$u = \frac{-2}{5}e^{x-3y+4y}+C(2x-y)e^{4y}$

$u = \frac{-2}{5}e^{x+y}+C(2x-y)e^{4y}$

usukidoll
  • 2,074
  • Here is the problem that I've tried to do...I think it has something to do with the integration with respect to z @Thursday – usukidoll Sep 15 '14 at 04:44

2 Answers2

1

I don't know if anyone will read this after 8 years, but I think there is another mistake higher up concerning the following two lines:

$2V_w+2(-V_w+V_z) -4(v) = e^{\frac{w+z}{2}+z}$

$2V_w-2V_w+V_z -4(v) = e^{\frac{w+z}{2}+z}$

In the second line it should instead read:

$2V_w-2V_w+2V_z -4(v) = e^{\frac{w+z}{2}+z}$

You forgot to also multiply the $V_z$ by 2, thus invalidating the rest of the calculation.

Bossel
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I spotted the following mistake, which invalidates all steps after it: from $$e^{-4z}(v) = \frac{-2}{5}e^{\frac{w-5z}{2}}+C(w)$$ you should have gotten $$v = \frac{-2}{5}e^{\frac{w-5z}{2}}e^{4z}+C(w)e^{4z}$$ Instead, you omitted $C$, then multiplied by $e^{4z}$, then brought $C$ back in, un-multiplied.

Why don't you do this in a reasonable way?

  1. Particular solution: try $u=ce^{x+y}$, find $ce^{x+y}+2ce^{x+y}-4ce^{x+y}=e^{x+y}$, so $c=-1$.

  2. Homogeneous equation: $u_x+2u_y-4u=0$. Write $u=Ce^v$, so that $v_x+2v_y-4=0$.

  3. Solve the equation for $v$ as usual: $v(x,y) = 4x+f(2x-y)$ with arbitrary $f$.

  4. Collect and clean up: $u(x,y)=-e^{x+y} + e^{4x}h(2x-y)$ is the general solution of original equation.

  5. Since $u(x,-x)=x$, we have $x=-1+e^{4x}h(3x)$, hence $h(3x)=e^{-4x}(x+1)$. The function $h$ is $h(t) = e^{-4t/3}(t/3+1)$.

  6. Plug known $h$ into $u$ to get the final answer: $$u(x,y)=-e^{x+y} + e^{4x}e^{-4(2x-y)/3}((2x-y)/3+1)$$

If you want to verify the result of 6, simplify it before taking derivatives.

  • so after this line $v = \frac{-2}{5}e^{\frac{w-5z}{2}}e^{4z}+C(w)e^{4z}$ I should substitute back and then apply the $u(x,-x) = x$ condition. By the way, my book is showing this particular method that I am using which I could rarely find a replica online. Every time I search first order PDES I end up with First Order PDEs with Method of Characteristics which wasn't covered in class. Thanks for the help. I'll try using that line, sub back, and apply the condition to see if I really or more like FINALLY get the right answer. – usukidoll Sep 16 '14 at 05:23
  • A word of advice: after substituting back (thus obtaining the general solution), you may want to check it for correctness [by plugging into the PDE] before moving forward to the condition on $u(x,-x)$. Just to spot any errors sooner, before wasting more time on following erroneous computations. –  Sep 16 '14 at 05:26
  • Damn... well right now I'm doing the final step onward to get my equation before the substitution.The substitution I got is $u = \frac{-2}{5}e^{\frac{2x-y-5y}{2}}e^{4y}+C(2x-y)e^{4y} \rightarrow u = \frac{-2}{5}e^{\frac{2x-6y}{2}}e^{4y}+C(2x-y)e^{4y} $ wait why could I just use $e$ rules such that $e^{x+y} \rightarrow e^xe^y$ wouldn't that make it easier? – usukidoll Sep 16 '14 at 05:31
  • Yes, you can use algebra any time you want... –  Sep 16 '14 at 05:36
  • before the side condition, I got $ u = \frac{-2}{5}e^{x+y}+C(2x-y)e^{4y}$ – usukidoll Sep 16 '14 at 05:38
  • Does not look right. Plug $C\equiv 0$; the PDE is not satisfied. My step 4 has the correct general solution [I just checked it myself]. I write $h$ where you write $C$, and the solution need not look exactly the same when undetermined terms are involved. But yours isn't right... –  Sep 16 '14 at 05:42
  • what the heck? I applied the exponential rule you know the one with $\frac{-2}{5}e^{x-3y}e^{4y}+C(2x-y)e^{4y}$ That should be... $ \frac{-2}{5}e^{x-3y+4y}+C(2x-y)e^{4y} \rightarrow\frac{-2}{5}e^{x+y}+C(2x-y)e^{4y} $ >:/ – usukidoll Sep 16 '14 at 05:45
  • it has something to do with the substitution on the right. But by the change of variables $y = z \rightarrow z =y$ and how does the $\frac{-2}{5}$ disappear? ugh – usukidoll Sep 16 '14 at 05:46
  • I could get it to work if the fraction didn't exist in the first place but that would mean that the integration wrt z would be way wrong....unless I don't write the $\frac{-2}{5}$?!?!?! – usukidoll Sep 16 '14 at 06:10
  • I was finally able to get this problem. There were some errors earlier, so I redid it, labeled my 2 with blue and z with red, and then got the right results. I also need some help regarding this question @Thursday http://math.stackexchange.com/questions/936220/pdes-with-variable-coefficents-solve-xu-x-xyu-y-u-0-for-all-x-y I got to the point where I need the integrating factor but I am not sure if it will do more harm than good. maybe the xy's disappear or something like that. – usukidoll Sep 18 '14 at 08:35