# 2-u-x-2-v-1-2-u-x-t-v-2-2-2-u-t-2-u-x-0-f-x-u-t-x-0-g-x-

Question Number 776 by 123456 last updated on 12/Mar/15
$$\frac{\partial^{\mathrm{2}} {u}}{\partial{x}^{\mathrm{2}} }={v}_{\mathrm{1}} \frac{\partial^{\mathrm{2}} {u}}{\partial{x}\partial{t}}+{v}_{\mathrm{2}} ^{\mathrm{2}} \frac{\partial^{\mathrm{2}} {u}}{\partial{t}^{\mathrm{2}} } \\$$$${u}\left({x},\mathrm{0}\right)={f}\left({x}\right) \\$$$${u}_{{t}} \left({x},\mathrm{0}\right)={g}\left({x}\right) \\$$
Commented by prakash jain last updated on 12/Mar/15
$${v}_{\mathrm{1}} \:\mathrm{and}\:{v}_{\mathrm{2}} \:\mathrm{are}\:\mathrm{constants}? \\$$$${u}\left({x},{t}\right)={f}\left({x}\right)+{th}\left({x},{t}\right) \\$$$${u}_{{t}} \left({x},{t}\right)={h}\left({x},{t}\right)+{th}_{{t}} \left({x},{t}\right) \\$$$${u}_{{t}} \left({x},\mathrm{0}\right)={h}\left({x},\mathrm{0}\right)+\mathrm{0}={g}\left({x}\right) \\$$$$\mathrm{Special}\:\mathrm{case}\:{h}\left({x},{t}\right)={g}\left({x}\right) \\$$$$\mathrm{General}\:\mathrm{Case}\:{h}\left({x},{t}\right)={g}\left({x}\right)+{ty}\left({x},{t}\right) \\$$$${u}\left({x},{t}\right)={f}\left({x}\right)+{tg}\left({x}\right)+{t}^{\mathrm{2}} {y}\left({x},{t}\right) \\$$
Commented by prakash jain last updated on 12/Mar/15
$$\mathrm{Special}\:\mathrm{case}\:\mathrm{solutin} \\$$$${u}\left({x},{t}\right)={f}\left({x}\right)+{tg}\left({x}\right) \\$$$$\left.{f}''\left({x}\right)+{tg}''\left({x}\right)={v}_{\mathrm{1}} {g}'{x}\right) \\$$$${g}''\left({x}\right)=\mathrm{0}\Rightarrow{g}\left({x}\right)={kx}+{c}_{\mathrm{1}} \\$$$${f}''\left({x}\right)={v}_{\mathrm{1}} {k}\Rightarrow{f}\left({x}\right)={v}_{\mathrm{1}} {k}\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{c}_{\mathrm{2}} {x}+{c}_{\mathrm{3}} \\$$$${u}\left({x},{t}\right)={kv}_{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{c}_{\mathrm{2}} {x}+{c}_{\mathrm{3}} +{t}\left({kx}+{c}_{\mathrm{1}} \right) \\$$
Answered by prakash jain last updated on 12/Mar/15
$${u}\left({x},{t}\right)={f}\left({x}\right)+{tg}\left({x}\right)+{t}^{\mathrm{2}} {y}\left({x},{t}\right) \\$$$$\frac{\partial{u}}{\partial{t}}=\mathrm{2}{ty}+{t}^{\mathrm{2}} \frac{\partial{y}}{\partial{t}}+{g}\left({x}\right) \\$$$$\frac{\partial^{\mathrm{2}} {u}}{\partial{t}^{\mathrm{2}} }=\mathrm{2}{t}\frac{\partial{y}}{\partial{t}}+\mathrm{2}{y}+{t}^{\mathrm{2}} \frac{\partial^{\mathrm{2}} {y}}{\partial{t}^{\mathrm{2}} }+\mathrm{2}{t}\frac{\partial{y}}{\partial{t}}=\mathrm{2}{y}+\mathrm{4}{t}\frac{\partial{y}}{\partial{t}}+{t}^{\mathrm{2}} \frac{\partial^{\mathrm{2}} {y}}{\partial{t}^{\mathrm{2}} } \\$$$${f}''\left({x}\right)+{tg}''\left({x}\right)+{t}^{\mathrm{2}} \frac{\partial^{\mathrm{2}} {y}}{\partial{x}^{\mathrm{2}} } \\$$$$\:\:\:\:\:\:={v}_{\mathrm{1}} \left[\mathrm{2}{t}\frac{\partial{y}}{\partial{x}}+{g}'\left({x}\right)+{t}^{\mathrm{2}} \frac{\partial^{\mathrm{2}} {y}}{\partial{x}\partial{t}}\right]+{v}_{\mathrm{2}} ^{\mathrm{2}} \left[\mathrm{2}{y}+\mathrm{4}{t}\frac{\partial{y}}{\partial{t}}+{t}^{\mathrm{2}} \frac{\partial^{\mathrm{2}} {y}}{\partial{t}^{\mathrm{2}} }\right] \\$$$$\mathrm{This}\:\mathrm{needs}\:\mathrm{to}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{for}\:\mathrm{general}\:\mathrm{case}. \\$$
Commented by prakash jain last updated on 12/Mar/15
$$\mathrm{Try}\:{y}={C} \\$$$${g}\left({x}\right)={kx}+{c}_{\mathrm{1}} \\$$$${f}''\left({x}\right)={kv}_{\mathrm{1}} +\mathrm{2}{Cv}_{\mathrm{2}} ^{\mathrm{2}} \\$$$${f}\left({x}\right)=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\left[{kv}_{\mathrm{1}} +\mathrm{2}{Cv}_{\mathrm{2}} ^{\mathrm{2}} \right]+{c}_{\mathrm{2}} {x}+{c}_{\mathrm{3}} \\$$$${u}\left({x},{t}\right)={f}\left({x}\right)+{tg}\left({x}\right)+{Ct}^{\mathrm{2}} \\$$$$\frac{\partial^{\mathrm{2}} {u}}{\partial{t}^{\mathrm{2}} }=\mathrm{2}{C} \\$$$$\frac{\partial{u}}{\partial{t}}={g}\left({x}\right)+\mathrm{2}{Ct} \\$$$$\frac{\partial^{\mathrm{2}} {u}}{\partial{x}\partial{t}}={k} \\$$$$\frac{\partial^{\mathrm{2}} {u}}{\partial{x}^{\mathrm{2}} }={kv}_{\mathrm{1}} +\mathrm{2}{Cv}_{\mathrm{2}} ^{\mathrm{2}} \\$$$$\mathrm{The}\:\mathrm{solution}\:\mathrm{meets}\:\mathrm{all}\:\mathrm{given}\:\mathrm{conditions}. \\$$$$\mathrm{So}\:\mathrm{given}\:\mathrm{any}\:{y}\left({x},{t}\right)\:\mathrm{a}\:\mathrm{solution}\:\mathrm{for}\:{u}\left({x},{t}\right) \\$$$$\mathrm{can}\:\mathrm{be}\:\mathrm{found}. \\$$